Notes

Revision Notes — Number Sets Download

Short definitions and remarks for Natural Numbers, Whole Numbers, Integers and Rational Numbers.

Natural Number

Counting numbers 1, 2, 3, 4, 5, ... are called natural numbers.

N = The set of natural numbers = {1, 2, 3, 4, ...}

The set N is infinite — it has unlimited members.
N has the smallest element namely 1.
N has no largest element — given any natural number, we can find a bigger number.
Zero (0) is not a member of the set N.

Whole Number

Whole numbers are 0, 1, 2, 3, ... — all natural numbers together with 0.

W = The set of whole numbers = {0, 1, 2, 3, ...}

The set of whole numbers is infinite.
The smallest whole number is 0.
The set of whole numbers has no largest member.
Every natural number is also a whole number.

Integers

The set of integers consists of natural numbers, their negatives, and zero.

Z = {..., -4, -3, -2, -1, 0, 1, 2, 3, 4, ...}

The set Z is infinite.
It has neither the greatest nor the least element.
Every natural number and every whole number is an integer.

Non-negative integers: {0, 1, 2, 3, ...}

Non-positive integers: {..., -4, -3, -2, -1, 0}

Rational Numbers

All numbers that can be expressed in the form p/q where p and q are integers and q ≠ 0, are called rational numbers.

Q = set of rational numbers = {p/q | p, q ∈ Z, q ≠ 0}

Remark: Rational numbers include fractions and integers (since any integer n = n/1).

Q = { -13, 57, 92, 47, 96, -1525 }

Remarks:

Every integer is a rational number.
Every terminating decimal is a rational number.
Every recurring decimal is a rational number.
A non-terminating repeating decimal is called a recurring decimal.
Between any two rational numbers there are an infinite number of rational numbers. This property is known as the density of rational numbers.

Irrational Numbers:

A number which cannot be expressed in the form pq where p and q are integers and q ≠ 0. We denote the set of irrational by Q′.

Examples: √2, √3, √5, √7, 2√7, 7/3 + 2√7, 3 + 2√5, 6 − √2, π, … are examples of irrational numbers.

Even Numbers:

A natural number is said to be even if it is multiple of 2 or it is divisible by 2. 2, 4, 6, 8, 10, 12, … are examples of even numbers.

E = The set of even numbers = {2, 4, 6, 8, 10, 12, …}

Odd Numbers:

A natural number is said to be odd, if it is not even or if it is not divisible by 2. 1, 3, 5, 7, 9, … are examples of odd numbers.

O = The set of odd natural numbers = {1, 3, 5, 7, 9, 11, 13, …}

Consecutive Numbers:

A series of natural numbers each differ by one is called consecutive numbers. e.g., 50, 51, 52, 53 are consecutive numbers.

Prime Number:

A natural number is said to be prime if it has only two different (distinct) factors namely one and itself. 2, 3, 5, 7, 11 are prime numbers.

Remarks:

1 is not a prime number.
The only even prime number is 2.
The smallest prime number is 2.
All prime numbers are odd except 2.

Twin – Prime:

A pair of prime numbers is said to be twin – prime if they differ by 2. Examples: (3, 5), (11, 13), (17, 19), (29, 31), (41, 43), (71, 73) are all twin – prime.

Composite Number:

A natural number is said to be composite if it has at least three different factors. 4, 6, 12 are all composite numbers.

Remarks:

4 is the least composite.

Co-Prime:

A pair of numbers is said to be co-prime if the numbers have no common factor other than one.

Perfect Number:

A number is said to be perfect if it is equal to the sum of its factors other than itself.

For example:

6 = (1 + 2 + 3)
28 = (1 + 2 + 4 + 7 + 14)
therefore 6 and 28 are perfect numbers.

Theorem – 1

Lemma: Lemma is a proven statement which is used to prove other statement.

Euclid’s division Lemma: Given two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r ≤ b, where: q = quotient, r = remainder, a = dividend and b = divisor.

Euclid’s Division Algorithm: We have seen that the said lemma is nothing but a restatement of the long division process which we have been using in earlier classes. An algorithm means a series of well-defined steps which provides a procedure of calculation repeated successively on the result of earlier steps till the desired result is obtained. It is used to find the HCF of two given positive integer.

Important Result: Let ‘a’ and ‘b’ two positive integers such that a = bq + r, where 0 ≤ r < b, then H.C.F.(a,b) = H.C.F.(b,r).

To find HCF by Euclid’s division lemma:

Step (i) Apply Euclid’s division lemma to a and b and obtain whole numbers q₁ and r₁ such that

a = bq₁ + r₁ , 0 ≤ r₁ < b

Step (ii) If r₁ = 0, b is the HCF of a and b.

Fundamental theorem of Arithmetic: Every composite number can be expressed as product of primes, and this decomposition is unique, apart from the order in which the prime factors occur in general, given a composite number X and we decompose it as x = P₁ ≤ P₂ ≤ P₃ ≤ … P_n are primes in ascending order then the way the number is decomposed is unique.

To find HCF by fundamental theorem of Arithmetic use the following steps.

Step (i) Factorise the numbers into primes.

Step (ii) Select the lowest of the power of common primes.

Step (iii) The product of powers of common primes is HCF.

To find LCM by fundamental theorem of Arithmetic use the following steps.

Step (i) Factorise the number into primes.

Step (ii) Select the highest of the power of the prime. Present in all or some of the numbers.

Step (iii) The product of powers of the common primes is LCM.

Important Results:

HCF = Product of least powers of common factors.
LCM = Product of highest powers of all the factors.
Product of two numbers = HCF × LCM
Note: This result is not true for more than two numbers.
HCF of 3 Nos. is the HCF of the HCF of any two of them and the 3rd No.
HCF of given numbers always divides their LCM.
HCF of pair of Co-primes is 1.
LCM of pair of co-prime = product of co-primes.

Revisiting of Irrational Numbers

Rational numbers: A number of the form p / q where q ≠ 0 are known as rational numbers.

There are two types of rational numbers: (i) terminating decimal (ii) non - terminating decimal.

Terminating Decimals: The rational numbers with a finite decimal part after finite numbers of steps are known as terminating decimals.

Ex. 78, 3516, 152, 2932, 1934 ...etc are the examples of terminal decimal.

Non Terminating recurring Decimals:

Those numbers in which the division process never comes to an end. In such cases a digit or block of digits repeat itself. 0.333, 0.24741 etc. are the examples of non-terminating recurring decimals divided into two categories:

a. Pure Recurring Decimals b. Mixed Recurring Decimals

(a) Pure recurring decimals: A decimal number in which all the figures after the decimal points are repeated.

0.3, 0.16, 4.13 ... etc are the examples of pure recurring decimals.

(b) Mixed recurring decimal: A decimal number in which after the decimal point, at least one figure is not repeated.

0.162, 0.57341 ... etc are the examples of mixed recurring decimals.

Irrational number:

A number which cannot be put in the form p / q, where p and q are integers and q ≠ 0 is called an irrational number.

Properties on irrational number:

Property – 1: Negative of an irrational number is an irrational number.

Ex.1 √3 is an example of irrational number. ∴ –√3 is an irrational number.

Property – 2: Sum of rational and irrational number is irrational number.

Ex.1 2 is a rational number and √3 is an irrational number, hence (2 + √3) is an irrational number.

Property – 3: Product of non-zero rational and irrational number is irrational.

Ex.1 2 is a rational number and irrational number, hence √3 is an → 2√3 is an irrational number.

Property – 4: Division of non-zero irrational number and irrational number is irrational.

Ex.1 2 is a rational number and √3 is an irrational number, hence 2 ÷ √3 is an irrational number.

Property – 5: Sum of two irrational number may be rational or irrational.

Ex.1 √2 and √3 both are irrational number then (√2 + √3) is an irrational number.

Ex.2 (2 + √3) and (2 – √3) both are irrational number but
[(2 + √3) + (2 – √3)] = 4 is a rational number.

Property – 6: Difference of two irrational number may rational or irrational.

Ex.1 (2 + √3) and (2 – √3) both are irrational numbers then
[(2 + √3) – (2 – √3)] = 2√3 is an irrational number.

Ex.2 (2 + √3) and (2 – √3) both are irrational numbers then
[(2 + √3) – (2 + √3)] = 4 is a rational number.

Property – 7: Product of two irrational number may rational or irrational.

Ex.1 If √2 and √3 both are irrational number then √2 × √3 = √6 is also an irrational number.

Ex.2 (3 − √2) and (3 + √2) both are irrational number but (3 − √2)(3 + √2) = 9 − 2 = 7 is a rational number.

Property – 8: Division of two irrational number may be rational or irrational.

Ex.1 √18 and √3 are irrational numbers, therefore √18 ÷ √3 = √6 is an irrational number.

Ex.2 √12 and √3 are irrational numbers, √12 ÷ √3 = √4 = 2 is a rational number.

Important Points

A rational number is either a terminating or a non-terminating but recurring decimal.
An irrational number ( √2, √3, … ) is a non-terminating and non-recurring decimal.
Every terminating or non-terminating recurring decimal expression can be put in the form of p / q.

Theorem: If P is a prime number and p divides a², then p divides a where a is positive number.

Theorem: Let x be a rational number whose decimal expansion terminates then x can be expressed in the form of p / q, where p and q are co-prime, and the prime factorisation of q is of the form 2ⁿ5^m.

Ex.2 (3 − √2) and (3 + √2) both are irrational number but (3 − √2)(3 + √2) = 9 − 2 = 7 is a rational number.

Property – 8: Division of two irrational number may be rational or irrational.

Ex.1 √18 and √3 are irrational numbers, therefore √18 ÷ √3 = √6 is an irrational number.

Ex.2 √12 and √3 are irrational numbers, √12 ÷ √3 = √4 = 2 is a rational number.

Important Points

A rational number is either a terminating or a non-terminating but recurring decimal.
An irrational number ( √2, √3, … ) is a non-terminating and non-recurring decimal.
Every terminating or non-terminating recurring decimal expression can be put in the form of p / q.

Theorem: If P is a prime number and p divides a², then p divides a where a is positive number.

of the form 2ⁿ5^m where m and n are positively integer.

Theorem: Let x = p / q be a rational number such that the prime factorisation of q is of the form 2^m5ⁿ, where m, n are non-negative integers then x is terminating decimal.

Theorem: Let x = p / q be rational number such that the prime factorisation q is not the form of 2·5^mn where m, n are non-negative integers then x is non terminating decimal.

Linear Equations in Two Variables Download

Revision Notes

Definition: An Equation of form a x + b y + c = 0, where a, b, c ∈ R, and x, y are variables is called a linear equation in two variables. The graph of a linear equation is always a straight line.

1. Consistent system: A system of simultaneous linear equations is said to be consistent, if it has at least one solution.

2. Inconsistent system: A system of simultaneous linear equations is said to be inconsistent, if it has no solution.

3. Dependent Equations: If a system of two linear equations is such that every solution of one of the equations is also a solution of the other, it is called a system of dependent equations.

4. Nature of Solutions: Let us consider two simultaneous equations:

a₁x + b₁y = c₁ and a₂x + b₂y = c₂, then:

a. If c₁ a₂ ≠ b₂ b₂ then system has unique solution and system is consistent (Independent) in this case graph of the equations must be intersect each other.

Point-5 The graph of y = a (constant) must be a straight line parallel to the x-axis and a unit away from x axis.

Point-6 If the speed of Boat in still water is x km/hr and the speed of stream is y km/hr then the speed of boat:
(i) In down stream = (x + y) km/hr
(ii) In upstream = (x − y) km/hr.

Point-7 In a cyclic quadrilateral the sum of opposite angles is 180°.

Point-8 In a parallelogram, the sum of adjacent angles is 180°.

Point-9 Profit = SP − CP, Profit % = Profit CP × 100

Point-10 Loss = CP − SP, Loss % = Loss CP × 100

Point-11 Simple Interest = P × R × T 100

Point-12 If P(a, b) be any point then a and b are called abscissa and ordinates of point P.

ROOT: The value of a variable satisfying the equation is called the root of the equation.

Point-13 The point at which a straight line crosses the x-axis, its y-coordinate is zero and when crosses y-axis its x-coordinate is zero.

Point-5 The graph of y = a (constant) must be a straight line parallel to the x-axis and a unit away from x axis.

Point-6 If the speed of Boat in still water is x km/hr and the speed of stream is y km/hr then the speed of boat.
(i) In down stream = (x + y) km/hr
(ii) In upstream = (x − y) km/hr.

Point-7 In a cyclic quadrilateral the sum of opposite angles is 180°.

Point-8 In a parallelogram, the sum of adjacent angles is 180°.

Point-9 Profit = SP − CP,
Profit % = Profit CP × 100

Point-10 Loss = CP − SP,
Loss % = Loss CP × 100

Point-11 Simple Interest = P × R × T 100

Point-12 If P(a, b) be any point then a and b are called abscissa and ordinates of point P.

ROOT: The value of a variable satisfying the equation is called the root of the equation.

Point-13 The point at which a straight line cross the x-axis, its y coordinate is zero and when crosses y-axis its x coordinate is zero.

Polynomials Download

Revision Notes

Definition: A function of type:
a₀xⁿ + a₁x^n-1 + a₂x^n-2 + ... + a_nx⁰ where x is a variable and a, b, c, … a_n are real numbers and ‘n’ is a positive integer. Then p(x) is called a polynomial in variable x.

a xⁿ, b x^n-1, c x^n-2 … etc are called terms of the polynomial.
a, b, c, … are the coefficients of the terms.
The highest power of x in a polynomial is called degree of polynomial.

Types of polynomials on the basis of degree

There are five types of polynomials on the basis of degree:

1. Constant polynomial: A Polynomial of zero degree is called a constant polynomial. The general form of a constant polynomial is p(x) = k where k is constant.

Examples: p(x) = 5, p(x) = π, p(x) = 2 7 … etc.

2. Zero polynomial: A polynomial whose constant term is zero is called zero polynomial. The general form of a constant polynomial is p(x) = 0.

3. Linear polynomial: A polynomial of degree one is called linear polynomial. The general form of a linear polynomial is p(x) = ax + b where a ≠ 0

Examples: 3x – 2, 5 2 y + 7, … etc.

4. Quadratic polynomial: A polynomial of degree two is called quadratic polynomial. The general form of a quadratic polynomial is p(x) = ax² + bx + c where a ≠ 0

Examples: 3x² – 3x, 5x² + 9 2 y + 7, √5x + 1 2x – 8 … etc

5. Cubic polynomial: A polynomial of degree 3 is called cubic polynomial. The general form of a cubic polynomial is p(x) = ax³ + bx² + cx + d where a ≠ 0

Examples: 5 – x² + √6x³, 7 – x² + x³, 27x³ – 10x² + x – 20 … etc

6. Biquadratic polynomial: A polynomial of degree 4 is called Biquadratic polynomial. The general form of biquadratic polynomial is:
p(x) = ax⁴ + bx³ + cx² + dx + e, where a ≠ 0

Examples: 2 5 x⁴ − 3x³ + x² − √2x − 1 7 , 7y⁴ + 4√2y⁴ − 5y² − 2, … etc.

Types of polynomials on the basis of term

1. Monomial polynomial: A polynomial having only 1 term is called monomial.

Examples: 5x, 6x⁷, 3x¹¹ … etc.

2. Binomial polynomial: A polynomial having two terms is called binomial.

Examples: 3x + 5, 6x² − 11 … etc.

3. Trinomial polynomial: A polynomial having only three terms is called trinomial.
Examples: 6x³ + 13x − 5, 2x⁴ − 7x + 6, … etc.

4. Value of a polynomial: If p(x) be a polynomial in x and c be any real number, then the value obtained by replacing x by c is called value of p(x) at x = c. It is denoted by p(c).

5. Zero of a polynomial: If value of p(x) at x = k is zero, then k is called zero of the polynomial.

GRAPH OF POLYNOMIALS

Graphs of linear polynomial: The graph of linear polynomials is a straight line. Consider a linear polynomial p(x) = ax + b = y. The graph line intersects x-axis at point (-b) a , 0 )

Graphs of quadratic polynomials:

(i) Graph of a quadratic polynomial p(x) = ax² + bx + c is a parabola. Parabola open upwards ( ∪ ) if a > 0

(ii) Graph of a quadratic polynomial p(x) = ax² + bx + c is a parabola. Parabola open downward ( ∩ ) if a < 0

Note: In general a polynomial p(x) of degree n intersect the x-axis at most n points.

The intersection points of a parabola

For p(x) = ax² + bx + c, b² − 4ac is known as its discriminant ‘D’, where D = b² − 4ac

(i) If D > 0, graph of p(x) = ax² + bx + c will intersect the x-axis at two distinct points.

(ii) If D = 0, graph of p(x) = ax² + bx + c will touch (or intersect) the x-axis at one point.

(iii) If D < 0, graph of p(x) = ax² + bx + c will neither touch nor intersect the x-axis.

Graphs of cubic polynomials:

For any cubic polynomials p(x) = ax³ + bx² + cx + d = y, The graph will intersect at three points on x-axis, making same number of zeros.

Case - 1: If p(x) = ax³ + bx² + cx + d = y, can be factorised into 3 linear non repeating factors, then it will intersect x-axis at 3 distinct points

Case - 2:

If p(x) can be factorised into 3 linear factors, out of which two are identical, then the graph will intersect x-axis at two distinct points.

(Graph showing one repeated root at A and one distinct root at C)

Case - 3:

If p(x) can be factorised into three repeated factors, then the graph will intersect x-axis at one point.

(Graph showing triple repeated root at A)

(i) Sum of zeros = − coefficient of x coefficient of x²
α + β = −ba

(ii) Product of zeros = constant term coefficient of x²
αβ = ca

Formation of Quadratic Polynomial

Let α and β be the zeros of the quadratic polynomial, then required polynomial p(x) is given by:
p(x) = k[x² − sx + p], where k is a constant.

Where, S = Sum of the zeroes (α + β)

P = Product of the zeroes (αβ)

or p(x) = x² − (α + β)x + αβ
or p(x) = (x − α)(x − β)

2. Cubic Polynomial:

Let α, β and γ be the three zeros of the cubic polynomial p(x) = ax³ + bx² + cx + d, a ≠ 0, then:

(i) Sum of zeros = − coefficient of x² coefficient of x³
α + β + γ = −ba

(ii) The sum of product of two zeros taken together = coefficient of x coefficient of x³
αβ + βγ + γα = ca

(iii) Product of zeros = − constant term coefficient of x³
αβγ = −da

Formation of a Cubic Polynomial:

Let α, β and γ be the zeros of the cubic polynomial then, Required polynomial p(x) = x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ

or Required cubic polynomial is given by p(x) = (x − α)(x − β)(x − γ)

Division Algorithm for polynomials

If p(x) and g(x) are any two polynomials of degree n and m respectively such that g(x) ≠ 0, then there exist unique polynomials q(x) and r(x) is either zero polynomial or degree r(x) less than degree of g(x) such that

p(x) = g(x) · q(x) + r(x)

where p(x) is called dividend, g(x) is called divisor, q(x) is called quotient and r(x) is called remainder. Note: If r(x) = 0, then g(x) is factor of p(x).

Factor Theorem

(i) If (x − a) is a factor of p(x) then p(a) = 0 i.e. a is zeros of p(x)

(ii) Let p(x) is a polynomial if p(a) = 0, then (x − a) is a factor of p(x)

Remainder Theorem

If polynomial p(x) is divided by (x − a), then the required remainder is p(a).

Relationship between the zeroes and coefficients of a quadratic polynomial

Let α and β be the zeros of a quadratic polynomial p(x) = ax² + bx + c. By factor theorem (x − α) and (x − β) are the factor of P(x).

P(x) = k (x − α)(x − β), where k is a constant

ax² + bx + c = k[x² − (α + β)x + αβ]
ax² + bx + c = kx² − k(α + β)x + kαβ

Comparing the coefficients of x², x, and constant terms on both sides, we get
a = k, b = −k(α + β) and c = kαβ

α + β = −ba and αβ = ca

α + β = coefficient of x coefficient of x²

αβ = constant term coefficient of x²

Arithmetic Progression Download

Revision Notes

Sequence: A sequence is an arrangement of numbers in a definite pattern according to some rule.

Arithmetic Progression: A sequence in which each term differs from its proceeding term by a constant is called an arithmetic progression, written as A.P. The constant difference is called (Common difference).

To find the nth Term of an A.P

Theorem: If the first term of an AP is 'a' and its common difference is d then nth term is given by a_n = a + (n − 1)d

Proof:

In the given AP, we have First term = a, and common difference = d

So, the Given AP may be written as a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, …

In this AP, we have First term, a₁ = a = a + (1–1)d Second term, a₂ = a + d = a + (2–1)d Third term, a₃ = a + 2d = a + (3–1)d … nth term, a_n = a + (n–1)d

Note: The nth term of AP is called General Term.

To find the nth term from the end of an AP

Theorem: If a be the first term, d the common difference and l is the last term of a given AP then n-th term from the end is l − (n–1)d

Proof:

We may write the given AP as (a), (a + d), (a + 2d), ……, (l − d), l

Thus, we have:

Last term = l = l − (1–1)d

2nd term from the end = (l − d) = l − (2–1)d

3rd term from the end = (l − 2d) = l − (3–1)d

… nth term from the end = l − (n–1)d

Important Results: It is always convenient to make a choice of

(i) No’s in APs are (a–3d), (a–d), a, (a+d), (a+3d)

(ii) No’s in APs are (3d–d), (a–d), a, (a+d), (a+3d)

(iii) No’s in APs are (a–3d), (a–d), a, (a+d), (a+2d)

Derivation of Sum Formula

If an A.P has first term ‘a’ and common difference ‘d’, then the sum of the first n terms is given by:

S_n = n 2 [2a + (n − 1)d] or S_n = n 2 (a + a_n)

where l is the last term of A.P, l = a_n

Proof: Let a₁, a₂, a₃, … be an A.P with first term ‘a’ and common difference ‘d’. Then,

a₁ = a, a₂ = a + d, a₃ = a + 2d, a_n = a + (n − 1)d

S_n = a + (a + d) + (a + 2d) + … + [a + (n − 2)d] + [a + (n − 1)d] (i)

Writing the above series in reverse order

S_n = [a + (n − 1)d] + [a + (n − 2)d] + … + [a + d] + a (ii)

Adding (i) and (ii), we get:

2S_n = [2a + (n − 1)d] + [2a + (n − 1)d] + … repeated n times

2S_n = n[2a + (n − 1)d]

S_n = n 2 [2a + (n − 1)d] or S_n = n 2 (a + a_n)

Where, a_n = last term or nth term.

Quadratic Equations Download

Revision Notes

Definition: Let P(x) be a polynomial if P(x) = 0 then it is called quadratic equation. The standard form of quadratic equation is ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0, is called a quadratic equation in variable x.

Roots of a Quadratic Equation:

Let P(x) = 0, be a Quadratic Equation, then the zeros of a Polynomial P(x) are called the roots of the Quadratic Equation.

Note 1: If α is a root of quadratic equation, ax² + bx + c = 0, then we say that x = α satisfies the equation ax² + bx + c = 0 or x = α is a solution of the equation ax² + bx + c = 0

Derivation of Quadratic Formula:

Consider the Quadratic Equation ax² + bx + c = 0

Derivation of Formula:

Step 1: Constant term shifted towards R.H.S ax² + bx = −c

Step 2: Whole equation divided by coefficient of x²
x² + b a x = − c a

x² + ba x + b² 4a² = − c a + b² 4a² (i)

Step 3: Adding both side by square of the half of the coefficient of x.

⇒ (2ax + b)² 4a² = b² − 4ac 4a²

Taking square root on both sides, we get

2ax + b 2a = ± √(b² − 4ac) 2a

⇒ x = −b ± √(b² − 4ac) 2a

Either, x = −b + √(b² − 4ac) 2a or x = −b − √(b² − 4ac) 2a

Or x = −b + √(b² − 4ac) 2a and x = −b − √(b² − 4ac) 2a

Discriminant:

If ax² + bx + c = 0 is a quadratic equation and a ≠ 0, then the expression, b² − 4ac is known as its discriminant and is denoted by letter D, i.e., D = b² − 4ac, where a, b and c are real numbers.

Nature of the Roots of a quadratic equation:

Nature of the roots of the quadratic equation depends on ‘D’.

1. If D > 0, then equation has real and unequal roots and both roots can be calculated by using the following formula:

x (or α) = −b + √(b² − 4ac) 2a and x (or β) = −b − √(b² − 4ac) 2a

2. If D = 0 then equation has real and equal roots, then both roots can be calculated by using following formula:

x (or α) = −b 2a and x (or β) = −b 2a

3. If D < 0 then equation has no real roots.

Important result:

On determining the value of an unknown variable in the given equation when the nature of root is given. Apply the following conditions:

a. If equation has real roots then apply D ≥ 0.

b. If equation has real and different roots then apply D > 0.

c. If equation has real and equal roots then apply D = 0.

d. If equation has no real roots then apply D < 0.

Arithmetic Progression

Revision Notes

Sequence: A sequence is an arrangement of numbers in a definite pattern according to some rule.

To find the nth Term of an A.P

Theorem: If the first term of an AP is 'a' and its common difference is d then nth term is given by a_n = a + (n − 1)d

Proof:

In the given AP, we have, First term = a, and common difference = d

So, the Given AP may be written as a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, …

In this AP, we have First term, a₁ = a = a + (1 − 1)d Second term, a₂ = a + d = a + (2 − 1)d Third term, a₃ = a + 2d = a + (3 − 1)d … nth term, a_n = a + (n − 1)d

Note: The nth term of AP is called General Term.

To find the nth term from the end of an AP

Theorem: If a be the first term, d the common difference and l is the last term of a given AP then show that nth term from the end is l − (n − 1)d

Proof: We may write the given AP as

(a), (a + d), (a + 2d), …, (l − d), l

Thus, we have:

Last term = l = l − (1 − 1)d

2nd term from the end = (l − d) = l − (2 − 1)d

3rd term from the end = (l − 2d) = l − (3 − 1)d

… nth term from the end = l − (n − 1)d

Important Results: It is always convenient to make a choice of

(i) No’s in AP are (a − 3d), (a − d), a, (a + d), (a + 3d)

(ii) No’s in AP are (3d − d), (a − d), a, (a + d), (a + 3d)

(iii) No’s in AP are (a − 3d), (a − d), a, (a + d), (a + 2d)

Derivation of Sum Formula

If an A.P has first term ‘a’ and common difference ‘d’, then the sum of the first n terms is given by:

S_n = n 2 [2a + (n − 1)d] or S_n = n 2 (a + a_n)

where l is the last term of A.P, l = a_n

Proof: Let a₁, a₂, a₃, … be an A.P with first term ‘a’ and common difference ‘d’. Then,

a₁ = a, a₂ = a + d, a₃ = a + 2d, a_n = a + (n − 1)d

S_n = a + (a + d) + (a + 2d) + … + [a + (n − 2)d] + [a + (n − 1)d] (i)

Writing the above series in reverse order:

S_n = [a + (n − 1)d] + [a + (n − 2)d] + … + (a + d) + a (ii)

Adding (i) and (ii), we get:

2S_n = [2a + (n − 1)d] + [2a + (n − 1)d] + … repeated n times

2S_n = n[2a + (n − 1)d]

S_n = n 2 [2a + (n − 1)d] or S_n = n 2 (a + a_n)

Where, a_n = last term or nth term.

Similar Triangles Download

Revision Notes

Theorem 1: In a triangle, a line drawn parallel to one side to intersect the other two sides in distinct points divides the two sides in the same ratio.

Given: ABC in which DE is drawn parallel to side BC. It intersects AB and AC at point D and E respectively.

To Prove: AD / DB = AE / EC

Constr: Join BE and CD and draw ⊥’s AM = AB = L and DN = AC.

Proof:

ar(ΔADE) = 1 2 × AM × DE (i)

ar(ΔDBE) = 1 2 × DN × DE (ii)

Equation (i) divided by (ii)

ar(ΔADE) ar(ΔDBE) = AM × DE DN × DE = AM DN (iii)

Again ar(ΔADE) = 1 2 × AM × DK (iv)

ar(ΔDBE) = 1 2 × DN × DK (v)

Equation (iv) divided by (v)

ar(ΔADE) ar(ΔDBE) = AM × DK DN × DK = AM DN (vi)

From (iii) and (vi), AM / DN = AD / DB = AE / EC

ar(ΔADE) ar(ΔCDE) = AD DC (vii)

Since, Δs BDE and CDE lies on same base DE and same parallel BC,

⇒ ar(ΔBDE) = ar(ΔCDE)

From (iii) and (vi) AD / DB = AE / EC

Hence Proved

COROLLARY

ADDB = AEEC (1)

ABAD = ACAE (2)

ABDB = ACEC (3)

DBAB = ECAC (4)

ABAC = DBEC (5)

ADAB = AEAC (6)

DBAD = ECAE (7)

Theorem 2: Converse of B.P.T.

If a line divides any two sides of a triangle in the same ratio, then line is parallel to the third side.

Given: A △ ABC, DE is a line divides two sides AB and AC in the same ratio that is

ADDB = AEEC

To prove: DE ∥ BC

Const: If DE is not ∥ BC, draw DK ∥ BC meeting AC in K.

Proof: In △ ABC, DK ∥ BC (By const.)

AKKC = ADDB (i)

But ADDB = AEEC (Given) (ii)

From (i) and (ii)

AKKC = AEEC

This is possible only one condition when point K ≡ E. Therefore, DK ≡ DE and hence DE ∥ BC

Definition:

Two triangles are said to be similar, if there:

(a) Corresponding angles are equal.

(b) Corresponding sides are proportional.

CHARACTERISTIC PROPERTIES OF SIMILARITY

Theorem 3:

If in two triangles, corresponding angles equal, then their corresponding sides are in the same ratio. Hence two triangles are similar.

Given: In triangle ABC and DEF, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

To prove: △ABC ~ △DEF

Proof: Mark point P and Q on the sides DE and DF such that AD = DP and AE = DQ

In △ABC and △DPQ

∠A = ∠D (By construction)

∠B = ∠P and ∠C = ∠Q (By const.)

△ABC ~ △DPQ (By AAA)

Hence ABDP = BCPQ = CAQD (By SAS)

∠A = ∠D, ∠B = ∠E, ∠C = ∠F

∴ △ABC ~ △DEF

Theorem 5:

If an angle in one triangle is equal to one angle of the other triangle and the sides including those angles are proportional, then the two triangles are similar.

Given: In △ABC and △DEF

∠A = ∠D and ABDE = ACDF

To Prove: △ABC ~ △DEF

Const: mark points P and Q on the sides DE and DF such that AB = DP and AC = DQ.

Proof:

ABDE = ACDF (Given)

⇒ DPDE = DQDF (Since AB = DP and AC = DQ)

⇒ PQ ∥ EF (By converse of BPT)

∴ ∠B = ∠E and ∠C = ∠F (pair of corresponding ∠s)

Now in △ABC and △DPQ

AB = DP (By const)

AC = DQ (By const)

∠A = ∠D (Given)

∴ △ABC ≅ △DPQ (By SAS)

⇒ ∠B = ∠P and ∠C = ∠Q (By CPCT)

But ∠B = ∠E and ∠C = ∠F (proved)

∴ ∠B = ∠E and ∠C = ∠F

Now in △ABC and △DEF

∠A = ∠D (Given), ∠B = ∠E (Proved), ∠C = ∠F (Proved)

∴ △ABC ~ △DEF (Hence Proved)

COORDINATE GEOMETRY Download

Revision Notes

Co-ordinate Axes:

Let XOX′ and YOY′ be two mutually perpendicular lines taken as axes. These lines are called co-ordinate axes. XOX′ is called the x-axis, YOY′ is called the y-axis.

Quadrants:

The axes divide the plane area into four parts. These parts are called Quadrants:

XOY is the first Quadrant.
YOX′ is the second Quadrant.
X′OY′ is the third Quadrant.
YO′X is the fourth Quadrant.

Co-ordinates of a point:

The distance of a point from the y-axis and x-axis is called co-ordinate of a point. Let P be any point (see in fig), the plane, from P draw PN perpendicular on the y-axis and the x-axis respectively. Then, the length LP is called the x-co-ordinate (or the abscissa) of the point P and PN is called the y-co-ordinate (or the ordinate) of point P. Point whose abscissa is X and ordinate is Y is called the point P (X, Y).

Note: In naming the co-ordinates of a point, the x-co-ordinate (abscissa) appears first and the y-co-ordinate (ordinate) second, written within parenthesis as (x, y).

Some Useful Points

x-co-ordinate or abscissa of all the points on the y-axis is zero. Thus, any point on the y-axis is of the form (0, y).
y-co-ordinate or ordinate of all points on the x-axis is zero. Thus, any point on the x-axis is of the form (x, 0).
Co-ordinate of origin are (0, 0), as it lies on both axes.

Distance Formula:

Let P(x₁, y₁) and Q(x₂, y₂) be two given points. From P and Q draw PL and QM perpendiculars to x-axis. Join PQ and from P draw PN perpendicular to QM thus:

PQ = √{(x₂ − x₁)² + (y₂ − y₁)²}

Distance Formula (Extended Steps)

Let P(x₁, y₁) and Q(x₂, y₂) be two given points.

Draw PN ⟂ x-axis and QM ⟂ x-axis. Since ∠PLMN is a rectangle:
LM = x₂ – x₁,
PN = y₁,
QM = y₂

In right triangle PQN: PQ² = PN² + QN²

PQ = √[(x₂ – x₁)² + (y₂ – y₁)²]

Corollary: The distance of the point (x, y) from origin is: d = √(x² + y²)

Some Useful Results

For an isosceles triangle – Prove that at least two sides are equal.
For an equilateral triangle – Prove that three sides are equal.
For a right-angled triangle – Prove that the sum of the squares of two sides is equal to the square of the third side.
For a square – Prove that four sides are equal and two diagonals are equal.
For a rhombus – Prove that four sides are equal but diagonals are not equal.
For a rectangle – Prove that opposite sides are equal and two diagonals are equal.
For a parallelogram – Prove that the opposite sides are equal.
For three points to be collinear – Prove that the sum of the distance between two pairs of points is equal to the third pair.

SECTION FORMULA & CENTROID

Let us find coordinates of the points (X, Y) which divides a line segment AB internally in a given ratio m : n, where A has coordinates (x₁, y₁) and B has (x₂, y₂).

We draw a perpendicular from A and B meeting the x–axis in C, D and in R, Q respectively. Also draw lines parallel to the x-axis from A and P meeting PQ and BD in points E and R respectively.

By using the section formula, we shall find the coordinates of the centroid of a triangle when the coordinates of the vertices of a triangle are given. Recall that medians of a triangle are concurrent. The point of concurrency is called the centroid. It divides each median in the ratio 2 : 1.

Let the vertices A, B and C of the triangle ABC be (x₁, y₁), (x₂, y₂), and (x₃, y₃) respectively, and let D be the middle point of BC, as shown in figure.

The coordinates of midpoint D of BC is given by:

D = ( (x₂ + x₃) / 2 , (y₂ + y₃) / 2 )

Let G (x, y) be the centroid of the triangle. We know that the point G divides AD in the ratio 2 : 1.

x = (2 × (x₂ + x₃) / 2 + 1 × x₁) / (2 + 1) = (x₂ + x₃ + x₁) / 3

y = (2 × (y₂ + y₃) / 2 + 1 × y₁) / (2 + 1) = (y₂ + y₃ + y₁) / 3

Thus, the coordinates of the centroid of the triangle are:

G ( (x₁ + x₂ + x₃) / 3 , (y₁ + y₂ + y₃) / 3 )

Important Points

By using section formula to prove three points are collinear: Consider a point between two points and find the value of K by using both coordinates. If the value of K is same then given points are collinear.
By using section formula to prove four vertices of a parallelogram: Calculate the mid-points of both diagonals. If they are same then given vertices of a parallelogram.

INTRODUCTION TO TRIGONOMETRY Download

Revision Notes

Trigonometry: The branch of Mathematics which deals with the measurement of angles and the problems related with angles.

Trigonometric Ratio:

Definition: The relationship between the sides of a right triangle with respect to an acute angle is called trigonometric ratio.

sin θ = Opposite side / Hypotenuse
cos θ = Adjacent side / Hypotenuse
tan θ = Opposite side / Adjacent side
cosec θ = Hypotenuse / Opposite side
sec θ = Hypotenuse / Adjacent side
cot θ = Adjacent side / Opposite side

Important Identities

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ

Other Useful Values

sin 90° = 1, sin 0° = 0
cos 90° = 0, cos 0° = 1
tan 90° = ∞, tan 0° = 0
cosec 90° = 1, cosec 0° = ∞
sec 90° = ∞, sec 0° = 1
cot 90° = 0, cot 0° = ∞

cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
sin θ × cosec θ = 1
cos θ × sec θ = 1
tan θ × cot θ = 1

Trigonometric Ratio of Complementary Angles:

If θ is an acute angle, then:

sin (90° − θ) = cos θ
cos (90° − θ) = sin θ
tan (90° − θ) = cot θ
cot (90° − θ) = tan θ
sec (90° − θ) = cosec θ
cosec (90° − θ) = sec θ

TRIGONOMETRIC TABLE

Angle (θ)	0°	30°	45°	60°	90°
sin θ	0	1/2	√2/2	√3/2	1
cos θ	1	√3/2	√2/2	1/2	0
tan θ	0	1/√3	1	√3	∞
cosec θ	∞	2	√2	2/√3	1
sec θ	1	2/√3	√2	2	∞
cot θ	∞	√3	1	1/√3	0

TRIGONOMETRIC RATIO OF SOME SPECIFIC ANGLES

(1) Trigonometric ratio 30° and 60° (√3/2)

Let: Construct an equilateral triangle ABC and draw AM ⟂ BC.

Let AB = BC = CA = 2a (each side)

Proof: ∠ABC = ∠BCA = ∠CAB = 60° each

In ∆AMB and ∆AMC:

AB = AC = 2a
AM = AM (common side)
BM = MC = a (since AM ⟂ BC, M is the midpoint)

⇒ AMB ≅ AMC (By RHS congruency)

Now in ∆AMB:

AB² = AM² + BM² (By Pythagoras theorem)

(2a)² = AM² + a²

4a² = AM² + a²

AM² = 4a² − a² = 3a²

AM = √3 a

Now in right ∆AMB:

sin 60° = AM / AB = (√3 a) / (2a) = √3 / 2

cos 60° = BM / AB = a / 2a = 1 / 2

tan 60° = AM / BM = (√3 a) / a = √3

Similarly,

sin 30° = 1 / 2

cos 30° = √3 / 2

tan 30° = 1 / √3

Trigonometric ratio of 60°

In right ∆ABC:

sin 60° = ^√3⁄₂ cos 60° = ¹⁄₂ tan 60° = √3

cosec 60° = ²⁄_√3 sec 60° = 2 cot 60° = ¹⁄_√3

Trigonometric ratio of 30°

In right ∆ABC:

sin 30° = ¹⁄₂ cos 30° = ^√3⁄₂ tan 30° = ¹⁄_√3

cosec 30° = 2 sec 30° = ²⁄_√3 cot 30° = √3

Trigonometric ratio of 45° (√2/2)

Construct an isosceles right angled triangle ABC, such that AB = BC = a units and ∠ABC = 90°

Proof: In right triangle ABC

AC² = AB² + BC² (By Pythagoras theorem)

AC² = a² + a² = 2a²

AC = √2 a

Since AB = BC

∠BAC = ∠ACB = 45°

Now in right ∆ABC,

sin 45° = AB / AC = a / (√2 a) = ¹⁄_√2 = ^√2⁄₂

cos 45° = BC / AC = a / (√2 a) = ¹⁄_√2 = ^√2⁄₂

tan 45° = AB / BC = a / a = 1

cosec 45° = √2, sec 45° = √2, cot 45° = 1

Trigonometric ratio of 0°

Construct a right triangle ∆ABC, such that ∠B = 90° and ∠A = 0°, then C coincides with B (then BC = 0), and AC = AB

Proof: In ∆ABC

sin 0° = BC / AC = 0 / AB = 0

cos 0° = AB / AC = AB / AB = 1

tan 0° = BC / AB = 0 / AB = 0

cosec 0° = 1 / sin 0° = 1 / 0 = ∞ (not defined)

sec 0° = 1 / cos 0° = 1 / 1 = 1

cot 0° = 1 / tan 0° = 1 / 0 = ∞ (not defined)

Trigonometric ratio of 90°

Construct a right triangle ∆ABC, such that ∠A = 0° and ∠C = 90°, then AB = 0 and BC = AC.

Proof: In ∆ABC

sin 90° = BC / AC = AC / AC = 1

cos 90° = AB / AC = 0 / AC = 0

tan 90° = BC / AB = AC / 0 = ∞ (not defined)

cosec 90° = 1 / sin 90° = 1 / 1 = 1

sec 90° = 1 / cos 90° = 1 / 0 = ∞ (not defined)

cot 90° = 1 / tan 90° = 1 / ∞ = 0

TRIGONOMETRIC IDENTITIES

1. Prove: sin²θ + cos²θ = 1

In right ∆ABC:

sin θ = ^BC/_AC, cos θ = ^AB/_AC, tan θ = ^BC/_AB, sec θ = ^AC/_AB, cosec θ = ^AC/_BC, cot θ = ^AB/_BC

Now,

sin²θ + cos²θ = (^BC/_AC)² + (^AB/_AC)²

= ^BC²/_AC² + ^AB²/_AC²

= ^{(BC² + AB²)}/_AC² (By Pythagoras theorem)

= ^AC²/_AC² = 1

Hence Proved.

2. Prove: 1 + tan²θ = sec²θ

LHS = 1 + tan²θ = 1 + (^BC/_AB)²

= 1 + ^BC²/_AB²

= (^AB²/_AB²) + (^BC²/_AB²)

= ^{(AB² + BC²)}/_AB²

= ^AC²/_AB² (By Pythagoras theorem) = (^AC/_AB)² = sec²θ

Hence Proved.

3. Prove: 1 + cot²θ = cosec²θ

LHS = 1 + cot²θ = 1 + (^AB/_BC)²

= 1 + ^AB²/_BC²

= (^BC²/_BC²) + (^AB²/_BC²)

= ^{(AB² + BC²)}/_BC² = ^AC²/_BC² (By Pythagoras theorem)

= (^AC/_BC)² = cosec²θ

Hence Proved.

Q1. Express sin θ in terms of cos θ, tan θ, cot θ, sec θ and cosec θ.

(i) sin θ in terms of cos θ

We know that:

sin²θ + cos²θ = 1

sin²θ = 1 − cos²θ

sin θ = √(1 − cos²θ)

(ii) sin θ in terms of sec θ

We know that:

sec²θ − tan²θ = 1

Divide both sides by sec²θ

(sec²θ / sec²θ) − (tan²θ / sec²θ) = 1 / sec²θ

1 − sin²θ = cos²θ

sin²θ = 1 − (1 / sec²θ)

sin θ = √(1 − ¹/_sec²θ)

(iii) sin θ in terms of cosec θ

We know that:

sin θ = ¹/_{cosec θ} ...(1)

We know that:

1 + cot²θ = cosec²θ

→ cosec²θ = 1 + cot²θ

→ cosec θ = √(1 + cot²θ)

Put the value of cosec θ in (1), we get:

sin θ = ¹/_{√(1 + cot²θ)}

(iv) sin θ in terms of cot θ

We know that:

sin θ = ¹/_{cosec θ} ...(1)

We know that:

1 + cot²θ = cosec²θ

→ cosec²θ = 1 + cot²θ

→ cosec θ = √(1 + cot²θ)

Putting in (1):

sin θ = ¹/_{√(1 + cot²θ)}

SURFACE, AREA AND VOLUME

IMPORTANT FORMULAS

1. Cube:

Let Length, Breadth and Height of a cube b, b, b units respectively, then

(i) Lateral surface area of cube = 4b²

(ii) Total surface area of cube = 6b²

(iii) Volume of cube = b³

2. Cuboid:

Let Length, Breadth and Height of a cuboid be l, b and h units respectively

(i) Lateral surface area of cuboid = 2h(l + b)

(ii) Total surface area of cuboid = 2(lb + bh + hl)

(iii) Volume of cuboid = l × b × h

(iv) Length of diagonal of cuboid = √(l² + b² + h²)

3. Cylinder:

Let the height and radius of a cylinder be 'h' and 'r' respectively

(i) Lateral surface area of cylinder = 2πrh

(ii) Curved surface area of cylinder = 2πrh

(iii) Total surface area of cylinder = 2πr(r + h)

(iv) Volume of cylinder = πr²h

4. Hollow Cylinder:

Let the inner radius and outer radius of a hollow cylinder be ‘r’ and ‘R’ and its height be ‘H’

(i) Base Area = π (R² – r²)

(ii) Volume = πH (R² – r²)

(iii) Lateral S.A. = 2πH(R + r)

(iv) Total Surface Area = 2πH(R + r) + 2π(R² – r²) = 2π(R + r)(H + R – r)

5. Cone:

Let height, slant height and base radius of a cone be h, l and r respectively

(i) Curved surface area of cone = πrl

(ii) Total surface area of cone = πr(r + l)

(iii) Volume of cone = ¹/₃ πr²h

6. Sphere:

(i) Curved Surface Area = 4πr²

(ii) Volume of sphere = ⁴/₃ πr³

7. Hemisphere:

(i) Curved Surface Area = 2πr²

(ii) Total Surface Area = 3πr²

(iii) Volume = ²/₃ πr³

8. Volume of Frustum:

Volume of Frustum = ¹/₃ πh (R² + r² + Rr)

7. Solid Hemisphere:

Let r be the radius of hemisphere then

(i) Curved Surface Area = 2πr²

(ii) Total Surface Area = 3πr²

(iii) Volume = ²/₃ πr³

8. Hollow Sphere:

Let outer and inner radius of a hollow sphere be R and r respectively, then:

(i) Volume of hollow sphere = ⁴/₃ π (R³ − r³)

(ii) Outer Curved Surface Area = 4πR²

(iii) Inner Curved Surface Area = 4πr²

9. Hollow Hemisphere:

Let outer and inner radius of the hollow hemisphere be R and r respectively, then:

(i) Curved Surface Area of hollow hemisphere = 2πR² − 2πr² = 2π(R² − r²)

(ii) Total Surface Area = 3πR² − 3πr² = 3π(R² − r²)

10. Volume of Frustum:

Let the perpendicular height be h, the radii of the two circular ends be R and r respectively, then

(i) Volume of frustum = ¹/₃ πh (R² + r² + Rr)

(ii) Curved Surface Area = π(R + r)l

(iii) Total Surface Area = π(R + r)l + πR² + πr²

Volume of Frustum = ¹/₃ πh (R² + r² + Rr)

Curved Surface Area = π (R + r) √(h² + (R − r)²)

Total Surface Area of Frustum = π (R² + r²) + π (R + r) √(h² + (R − r)²)

Total Surface Area of Bucket = π (R² + r²) + π (R + r)l

Surface Area of Bucket = 2π(R + r)l + πr²

LINEAR EQUATIONS IN TWO VARIABLES Download

NCERT Solutions

Exercise 3.1

Q1. Aftab tells his daughter, “seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. Represent this situation algebraically and graphically.”

So! Let the present age of the daughter = x yrs

7 years ago daughter’s age = (x − 7) years
3 years from now, daughter’s age = (x + 3) yrs.

Let the present age of father = y years

7 years ago father’s age = (y − 7) yrs
3 years from now father’s age = (y + 3) yrs.

By given conditions, we have two algebraic equations:

y = 7(x − 7) = 7x − 42 … (i)
y + 3 = 3(x + 3) = 3x + 6 … (ii)

(1) Algebraically representations
y = 7x − 42 , y = 3x + 6

(2) For graphically representation, we have required table.

Table for y = 7x − 42

x	12	18	6
y	42	84	0

Table for y = 3x + 6

x	6	12	3
y	24	42	15

(2) For graphical representation, we have required table.

Table for 2x + 3y = 9

x	0	4.5	3
y	3	0	1

Table for 4x + 6y = 18

x	3	0	6
y	0	3	1

(0, 3), (3, 1), (9/2, 0)

Ans: Infinite many solution

Q4. Solve graphically the system of linear equations: 2x + y = 8 and x + 1 = 2y. Also, find the co-ordinates of the points lines meet the y-axis.

Sol.

Table of equation

2x + y = 8

x	0	2	5
y	8	4	-2

x + 1 = 2y

x	-3	1	5
y	-1	1	3

Q2. The coach of cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of same kind for Rs. 1500. Represent this situation geometrically.

Sol. Let the cost of one bat and one ball be Rs. x and y respectively then, according to first condition:

3x + 6y = 3900 ⇒ x + 2y = 1300 … (i)

According second condition: x + 3y = 1500 … (ii)

(1) Algebraically represents is: x + 2y = 1300 ; x + 3y = 1500

(2) For graphically representation, we have required tables:

Table for x = 1300 – 2y

x	1300	0	-1300
y	0	650	1300

Table for x = 1500 – 3y

x	1500	0	-1500
y	0	500	1000

(0,650) (900,200) (1300,0) (1500,0)

Q3. Romila went to stationery stall and purchased 2 pencils and 3 erasers for Rs. 9. Her friend Sonali saw the new variety of pencils and erasers with Romila and she also bought 4 pencils and 6 erasers of the same kind for Rs. 18. Represent this situation algebraically and solve it graphically.

Sol. Let the cost of one pencil = Rs. xand the cost of an eraser = Rs. y

(i) according to first condition. 2x + 3y = 9 … (1)

(ii) according second condition. 4x + 6y = 18 … (2)

(1) Algebraic representation is: 2x + 3y = 9; 4x + 6y = 10

Ans: Required Solution is x = 3, y = 2 and Required point on y-axis are (0,3.5) and (0,5)

Q5. Solve graphically the system of linear equations; 2x − y = 2 and 4x − y = 8. Also, find the co-ordinates of the points lines meet they axis.

Sol.

Table of equation 2x − y = 2
x	0	2	3
y	-2	2	4

Table of equation 4x − y = 8
x	0	1	2
y	-8	-4	0

Ans: Required solution is x = 3 and y = 2 Required point on x - axis are (3,0) and (2,0)

Q6. Determine graphically the vertices of the triangle, the equations of whose sides are given below.
2y - x = 8, 5y - x = 14, y - 2x = 1

Sol.

Table of equation 2y − x = 8
x	0	-8	8
y	4	0	8

Table of equation 5y − x = 14
x	0	-14	16
y	2.8	0	3.2

Table of equation y − 2x = 1
x	0	1	-1
y	1	3	-1

Q7. Determine the graphically the vertices of a triangle the equations of whose sides are as follows.
Also find the area of triangle formed with x axis. y − x = 3, 2x + 5y = 10

Sol.

Table of equation y = x
x	1	2	3
y	1	2	3

Table of equation 2x + 3y = 10 2x = 10 − 3y
x	1	2	3
y	2	2	1

Table of equation y = 0 will be x axis
x	1	2	5
y	0	0	0

Ans: Required vertices A (0,0), B (2,2) and C (5,0)

Req. Area = ½ × AC × BM = ½ × 5 × 2 = 5 square units.

Exercise 3.2

Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of class X took part in mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Sol. Let the number of boys be x and number of girls be y.

According first condition y − x = 4 … (1)
According second condition x + y = 10 … (2)

Table for y = x + 4

x	1	2	3
y	5	6	7

Table for y = 10 − x

x	1	2	3
y	9	8	7

From the graph we have girls = y = 7
And boys = 3 (as 3 + 7 = 10)

(ii) Let cost of one pencil and one pen be x and y respectively.

According first condition 5x + y = 52 … (1)
According second condition 7x + 5y = 80 … (2)

On adding (1) and (2), we get

12x + 12y = 96 ⇒ x + y = 8 … (3)

Subtracting (2) from (1), we get

−2x + 2y = −4 ⇒ x − y = 2 … (4)

Table for x + y = 8

x	2	3	8
y	6	5	0

Table for x − y = 2

x	2	3	5
y	0	1	3

From the graph, we have x = 5, y = 3
Cost of one pencil = Rs. 3 and cost of one pen = Rs. 5

Q3. On comparing the ratio, ^a₁/_a₂, ^b₁/_b₂ and ^c₁/_c₂ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

(i) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0
(ii) 3x + 2y = 6 and 6x + 4y = 12
(iii) 2x − 3y + 5 = 0 and 3x − 2y − 8 = 0

Sol. Comparing the given equations with standard form of equation ax + by + c = 0, we have

(i) a₁ = 5, b₁ = −4, c₁ = 8 and a₂ = 7, b₂ = 6, c₂ = −9

⇒ a₁/a₂ = 5/7, b₁/b₂ = −4/6 = −2/3, c₁/c₂ = 8/−9

Since a₁/a₂ ≠ b₁/b₂ → lines intersect at a point.

(ii) a₁ = 9, b₁ = 3, c₁ = 12, a₂ = 18, b₂ = 6, c₂ = 24

\(\dfrac{a_{1}}{a_{2}} = \dfrac{9}{18} = \dfrac{1}{2},\quad \dfrac{b_{1}}{b_{2}} = \dfrac{3}{6} = \dfrac{1}{2},\quad \dfrac{c_{1}}{c_{2}} = \dfrac{12}{24} = \dfrac{1}{2}\)

Thus, the lines representing the pair of linear equations are coincident.

(iii) a₁ = 6, b₁ = −3, c₁ = 10, a₂ = 2, b₂ = −1, c₂ = 9

\(\dfrac{a_{1}}{a_{2}} = \dfrac{6}{2} = 3,\quad \dfrac{b_{1}}{b_{2}} = \dfrac{-3}{-1} = 3,\quad \dfrac{c_{1}}{c_{2}} = \dfrac{10}{9}\)

Since \(\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}\), the lines representing the pair of linear equations are parallel.

Q4. On comparing the ratios \(\dfrac{a_{1}}{a_{2}}, \dfrac{b_{1}}{b_{2}}\) and \(\dfrac{c_{1}}{c_{2}}\), find out whether the following pairs of linear equations are consistent, or inconsistent:

(i) x + y = 5, 2x + 2y = 10
(ii) x − y = 8, 3x − 3y = 16
(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0

Sol.

(i) a₁ = 1, b₁ = 1, c₁ = −5; a₂ = 2, b₂ = 2, c₂ = −10

\(\dfrac{a_{1}}{a_{2}} = \dfrac{1}{2},\quad \dfrac{b_{1}}{b_{2}} = \dfrac{1}{2},\quad \dfrac{c_{1}}{c_{2}} = \dfrac{-5}{-10} = \dfrac{1}{2}\)

⇒ \(\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}\), so the pair of linear equations is consistent and has infinitely many solutions.

⇒ \(\dfrac{a_1}{a_2} ≠ \dfrac{b_1}{b_2}\)

Hence, the given lines are intersecting. So, the given pair of linear equations has exactly one solution and therefore, it is consistent.

(ii) a₁ = 2, b₁ = −5, c₁ = −8, a₂ = 4, b₂ = −10, c₂ = −9

\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2},\quad \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2},\quad \frac{c_1}{c_2} = \frac{-8}{-9} \]

Hence, the given lines are parallel. So, the given pair of linear equation has no solution and therefore it is inconsistent.

(iii) a₁ = 2, b₁ = 5, c₁ = −7, a₂ = 9, b₂ = −10, c₂ = 14

\[ \frac{a_1}{a_2} = \frac{2}{9},\quad \frac{b_1}{b_2} = \frac{5}{-10} = -\frac{1}{2},\quad \frac{c_1}{c_2} = \frac{-7}{14} = -\frac{1}{2} \]

Hence, the given lines are intersecting. So, the given pair of linear equations has exactly one solution and therefore it is consistent.

(iv) a₁ = 5, b₁ = −11, c₁ = −11, a₂ = −10, b₂ = 6, c₂ = −22

\[ \frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2},\quad \frac{b_1}{b_2} = \frac{-11}{6},\quad \frac{c_1}{c_2} = \frac{-11}{-22} = \frac{1}{2} \]

Hence, the given lines are consistent. So, the given pair of linear equations has infinitely many solutions and therefore it is consistent.

(iv) a₁ = 4, b₁ = 2, c₁ = 8, a₂ = 2, b₂ = 1, c₂ = 12

\[ \dfrac{a_1}{a_2} = \dfrac{4}{2} = 2,\quad \dfrac{b_1}{b_2} = \dfrac{2}{1} = 2,\quad \dfrac{c_1}{c_2} = \dfrac{8}{12} = \dfrac{2}{3} \]

Hence, the given lines are consistent. So, the given pair of linear equations has infinitely many solutions and therefore it is consistent.

Q4. Given the linear equations 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical meaning of the pair is formed is —

(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Sol.
(i) Intersecting lines ⇒ x − y = 8

(ii) Another linear equation in two variables and that the geometrical representation of the pair is
parallel lines: linear line is 6x − 9y − 12 = 0

(iii) Another coincident line to the above line is: 6x − 9y − 24 = 0

Q5. Half the perimeter of a rectangular garden whose length is 4 m more than its width is 36 m. Find the dimensions of the garden graphically.

Sol. Let length and breadth of rectangle field be x and y.

According first condition x = y + 4 … (1)
According second condition x + y = 36 … (2)

Table for x = y + 4

y	8	10	12
x	12	14	16

Table for x + y = 36

x	12	14	10
y	24	22	26

From the graph: x = 20, y = 16,
Ans: length = 20m, width = 16m

Q6. Which of the following pairs of linear equations are consistent? If consistent, obtain the solution graphically:

(i) x + y = 5; 2x + 2y = 10
(ii) 3x – 5y = –6; 6x – 10y = –6
(iii) 2x + 3y – 10 = 0; 4x – 6y – 0 = 0

Sol. Table for x + y = 5

x	1	2	3
y	4	3	2

Table for 2x + 2y = 10

x	1	2	3
y	4	3	2

Ans: As graph of both the lines is the same, so the given graph of equations is dependent, hence many solutions.

(ii) Table for 2x + y = 6

x	2	3	0
y	2	0	6

Table for 4x − 4y = 4

x	2	0	3
y	1	−1	2

As the given two lines intersect at two points (2, 2), hence it is consistent.

(iii) Table for 2x − 2y − 6 = 0

x	2	0	6
y	−1	−3	3

Table for 4x − 4y − 5 = 0

x	2	5/4	9
y	1	0	5

Ans: The lines are parallel. Hence it is inconsistent.

Q7. Draw the graphs of the equations: x − y + 1 = 0, 3x + 2y − 12 = 0. Determine the co-ordinates of the vertices of the triangle formed by these lines and the x-axis and shade the triangular region.

Sol.

Table for x − y + 1 = 0

x	0	1	-1
y	1	2	0

Table for 3x + 2y − 12 = 0

x	2	4	0
y	3	0	6

Ans: The vertices of ▲ABC are: A(2, 3), B(4, 0) and C(−1, 0).

Q8. Solve the following system of equations graphically: 2x + 3y − 6 = 0 and 4x + 6y − 24 = 0. Shade the region bounded by these lines and the axes. Name the figure, the shaded and find its area.

Sol.

Table for 2x + 3y − 6 = 0

x	0	3	6
y	2	0	-2

Table for 4x + 6y − 24 = 0

x	0	3	0
y	4	2	0

Area of figure ABCD = Area ΔABE – Area ΔCDE

2 × area = (base × height)

= ½ × 3 × 4 = 12 ⇒ 9 sq. m

Exercise 3.3

Q.1 Solve the following pair of linear equations by substitution method:

(i) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 … (2)

Sol.

Multiply (1) and (2) by 10 to remove decimals:

2x + 3y = 13 … (3)
4x + 5y = 23 … (4)

From equation (3):

2x = 13 − 3y
x = (13 − 3y) / 2 … (5)

Substituting this value of x in (4), we get

4x + 5y = 23
4 × (13 − 3y)/2 + 5y = 23
2(13 − 3y) + 5y = 23
26 − 6y + 5y = 23
26 − y = 23
−y = −3 &Rightarrow y = 3

Put y = 3 in (5), we get
x = (13 − 3 × 3)/2 = (13 − 9)/2 = 4/2 = 2

Ans: x = 2, y = 3

(ii) √2 x + √3 y = 0 … (1)
√3 x − √8 y = 0 … (2)

Sol.

From (1): √2 x = −√3 y &Rightarrow x = − (√3 / √2) y … (3)

Substituting x from (3) in (2),

√3 (−√3/√2) y − √8 y = 0
−(3/√2) y − √8 y = 0

But √8 = 2√2, so

−(3/√2) y − 2√2 y = 0 &Rightarrow −&left;( 3/√2 + 2√2 &right;) y = 0

Coefficient is non–zero, so y = 0. From (3), x = 0.

Ans: x = 0, y = 0

\(\sqrt{2}x - \dfrac{\sqrt{3}}{2}y = 0\)

3x + 4y = 0
7x = 0
x = 0
Put x = 0 in (3), we get

\(y = \dfrac{\sqrt{3}}{\sqrt{2}} \times 0 = 0\)

Ans: x = 0, y = 0

Q.4 Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of “m” for which 5y = mx + 3.

Sol. The given equations are:

2x + 3y = 11 … (i)
2x − 4y = −24 … (ii)

From (i) we have

2x = 11 − 3y
x = \(\dfrac{11 − 3y}{2}\) … (iii)

Putting the value of x from (iii) in (ii), we get

2\(\left(\dfrac{11 − 3y}{2}\right)\) − 4y = −24

⇒ (11 − 3y) − 4y = −24
⇒ 11 − 7y = −24
⇒ −7y = −35
⇒ y = 5

Putting the value of y in (iii), we get

\(x = \dfrac{11 − 3×5}{2} = \dfrac{11 − 15}{2} = \dfrac{-4}{2} = -2\)

Now equation 5y = mx + 3 becomes:
5(5) = m(−2) + 3
25 = −2m + 3
−2m = 22
m = −11

\(\dfrac{-4}{2} = -2\)

Hence, the solution is: x = −2, y = 5

It is given that y = mx + 3

Putting the values of x and y in given condition we get

5 = m(−2) + 3
5 = −2m + 3
5 − 3 = −2m
2 = −2m
m = −1

Q.6 Find the pair of linear equations in the following problems and find their solutions by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find the angles.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls of the same kind for Rs. 1750. Represent this situation algebraically and solve it by substitution method.

(iv) The cost of 4 pens and 4 pencils is Rs. 56. The cost of a pen is Rs. 6 more than the cost of a pencil. Find the cost of each.

Sol. (i) Let the two numbers be x and y such that x > y.

According to conditions: x − y = 26 … (1)
x = 3y … (2)

Putting x = 3y in (1):
3y − y = 26
2y = 26
y = 13
x = 3 × 13 = 39

Ans: Required numbers are 39, 13.

(ii) Let the two angles be x and y.

According first condition: x + y = 180° … (1)
According second condition: x = y + 18° … (2)

From equation (1), y = 180° − x. Substituting this value of y in (2), we get:

x = (180° − x) + 18°
x = 198° − x
2x = 198°
x = 99°
y = 180° − 99° = 81°

Ans: Required angles are x = 99°, y = 81°.

(iii) Let the price of each bat = Rs. x and the price of each ball = Rs. y.

According first condition: 7x + 6y = 3800 … (1)
According second condition: 3x + 5y = 1750 … (2)

From equation (2):
3x + 5y = 1750
5y = 1750 − 3x
y = (1750 − 3x) / 5 … (3)

Substituting this value of y in (1), we get:

7x + 6 × (1750 − 3x)/5 = 3800

Multiplying both sides by 5:
35x + 6(1750 − 3x) = 19000

35x + 10500 − 18x = 19000
17x + 10500 = 19000
17x = 8500
x = 500

Putting x = 500 in (3):
y = (1750 − 1500)/5 = 250/5 = 50

Ans: Cost of one bat = Rs. 500, Cost of one ball = Rs. 50.

(iv) Let the price per pen = Rs. x, and fixed charge = Rs. y.

According first condition: 4x + 4y = 56 … (1)
According second condition: 6x + y = 66 … (2)

From (1):
4x + 4y = 56
x + y = 14 … (3)

Putting y = 14 − x in (2):
6x + (14 − x) = 66
5x + 14 = 66
5x = 52
x = 10.4
y = 14 − 10.4 = 3.6

Ans: Price per pen = Rs. 10.40, Fixed charge = Rs. 3.60

Substituting this value of y in (2), we get:

15x + 105 − 30x = 125
−15x = 20
x = −20/15
x = −4/3

y = 166 − 10x = 166 − 10(−4/3)
y = 166 + 40/3 = (498 + 40)/3 = 538/3

Cost of 25 km is given by 21x + y = 25 × 10 + 5 = 250 + 5 = 230

Ans: Rs. 230

(v) Let the fraction = x/y

According first condition:
\(\dfrac{x}{y} = \dfrac{7}{11}\)
11x = 7y … (1)

According second condition:
\(\dfrac{x+7}{y+7} = \dfrac{3}{4}\)
4(x + 7) = 3(y + 7)
4x + 28 = 3y + 21
4x − 3y = −7 … (2)

From (1):

11x = 7y
x = 7y / 11 … (3)

Substituting this value of x in (2), we get:

4(7y/11) − 3y = −7
28y/11 − 33y/11 = −7
−5y/11 = −7
y = (−7 × 11)/−5 = 77/5 = 15.4

x = (7y)/11 = (7 × 77/5)/11 = 49/5

Ans: Hence the required fraction = 49/77 = 7/11

(vi) Let the present age of Jacob be x years and his uncle be y years.

According first condition:

x + 5 = 2(y + 5)
x + 5 = 2y + 10
x − 2y = 5 … (1)

According second condition:

x − 5 = y − 25
x − y = −20 … (2)

From equations (1) and (2):

x − 2y = 5
x − y = −20

Subtracting the value of x in (2), we get:
(x − y) − (x − 2y) = −20 − 5
−y + 2y = −25
y = 25

x = 2y + 5 = 50 + 5 = 55

Ans: Hence, present age of Jacob is 55 years, and his uncle is 25 years.

Exercise 3.4

Q.1 Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x − 3y = −4
(ii) 3x + 4y = 10 and 2x − 4y = 2
(iii) 3x − 5y − 2 = 0 and 7x − 3y − 2 = 0
(iv) (i) 5x − 2y = 4 and 7x + 6y = −12

Sol: (i) Elimination Method

x + y = 5 … (1)
2x − 3y = −4 … (2)

For making the coefficient of y in (1) and (2) equal, we multiply (1) by 3 and adding, we get:

3x + 3y = 15
2x − 3y = −4
———————
5x = 11
x = 11/5

Now putting the value of x in equation (1), we get:
11/5 + y = 5
y = 25/5 − 11/5 = 14/5

Hence, the solution is x = 11/5, y = 14/5.

Substitution Method

From (1), x = 5 − y
Putting in (2):
2(5 − y) − 3y = −4

10 − 2y − 3y = −4
10 − 5y = −4
−5y = −14
y = 14/5

Putting y in x = 5 − y:
x = 5 − 14/5 = 11/5

Substituting the value of x in (3), we get:

2y − 2 = 4
2y = 6
2y − y = 4 (typo correction from screenshot)
10 − 5y = −4
−5y = −14
y = 14/5

Now substituting the value of y in (5), we get:

x = 5 − y
x = 5 − 14/5
x = 25/5 − 14/5 = 11/5

Hence, x = 11/5 and y = 14/5.

(ii) Elimination Method

3x + 4y = 10 … (1)
2x − 4y = 2 … (2)

For making the coefficient of y in (1) and (2) equal, we multiply (1) by 1 and adding, we get:

3x + 4y = 10
2x − 4y = 2
——————
5x = 12
x = 12/5

Now putting the value of x in equation (1), we get:
3(12/5) + 4y = 10
36/5 + 4y = 10

4y = 10 − 36/5
4y = 50/5 − 36/5 = 14/5
y = (14/5) ÷ 4 = 14/20 = 7/10

Hence, the solution is x = 12/5, y = 7/10.

4y = 4
y = 1

Hence,
x = 2y − 3 = 2(1) − 3 = −1

Substitution method

We have following equations:

3x + 4y = 10 … (1)
2x − 3y = 2 … (2)

From (1), we have

3x = 10 − 4y
x = (10 − 4y) / 3 … (3)

Substituting the value of x in (2), we get:

2x − 3y = 2

2( (10 − 4y) / 3 ) − 3y = 2

(20 − 8y)/3 − 3y = 2

Multiply by 3:
20 − 8y − 9y = 6

20 − 17y = 6
−17y = −14
y = 14/17

Now substituting the value of y in (3), we get:

x = (10 − 4y)/3
x = (10 − 4 × 14/17)/3

= (10 − 56/17)/3
= (170/17 − 56/17)/3
= (114/17)/3
= 114 / 51
= 38/17

Hence,
x = 38/17 , y = 14/17

(iii) Elimination Method

3x − 5y = 4 … (1)
6x − 2y = 2 … (2)

For making the coefficient of x in (1) and (2) equal, we multiply (1) by 1 and subtracting, we get:

9x − 15y = 12
6x − 2y = 2
———————
3x − 13y = 10
x = (10 + 13y)/3 … (3)

Now putting the value of x in equation (1), we get:

3x − 5y = 4
3((10 + 13y)/3) − 5y = 4

10 + 13y − 5y = 4
10 + 8y = 4
8y = −6
y = −6/8 = −3/4

x = (10 + 13y)/3 = (10 + 13(−3/4))/3

= (10 − 39/4)/3
= (40/4 − 39/4)/3
= (1/4)/3 = 1/12

∴ Solution: x = 1/12 , y = −3/4

Substitution method

We have following equations:

3x − 5y = 4 … (1)
6x − 2y = 2 … (2)

From (1), we have

3x = 4 + 5y
x = (4 + 5y)/3 … (3)

x = (4 + 5y) / 3 … (3)

Substituting the value of x in (2), we get:

6x − 2y = 2

6( (4 + 5y)/3 ) − 2y = 2

2(4 + 5y) − 2y = 2
8 + 10y − 2y = 2
8 + 8y = 2
8y = −6
y = −6/8 = −3/4

Now, substituting the value of y in (3), we get:

x = (4 + 5y)/3
x = (4 + 5(−3/4))/3

= (4 − 15/4)/3
= (16/4 − 15/4)/3
= (1/4)/3 = 1/12

Hence, x = 1/12, y = −3/4

(iv)

7x + 6y = −6
8x + 5y = −5

From (1): 7x = −6 − 6y x = (−6 − 6y)/7

Now we have the following pairs of equations:

7x + 6y = −6 … (1)
8x + 5y = −5 … (2)

Elimination Method

Since the coefficients of x in (1) and (2) are equal, so simply by subtracting we can eliminate the variable x, i.e.,

3x + y = 6
3x + y = 9
——————
5y = −15

y = −15 / 5 = −3

Now, putting the value of y in equation (1), we get:

3x + y = 6
3x − 3 = 6
3x = 9
x = 3

∴ x = 3, y = −3

Substitution method

We have following equations:

3x + y = 6 … (1)
3x + y = 9 … (2)

From (1):

y = 6 − 3x … (3)

Substituting the value of y in (2), we get:

3x + (6 − 3x) = 9

3x + 6 − 3x = 9
6 = 9 (contradiction)

Hence, no solution exists. The lines are parallel.

Now substituting the value of y in (3), we get:

y = 6 − 3x
y = 6 − 9
y = −3

Hence, x = 3, y = −3.

Hence, x = 3, y = −3

Q.2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method.

(i) If we add 1 to the numerator of a fraction and subtract 1 from its denominator, the fraction becomes 1. It is also given that the fraction becomes ½ when we add 1 to its denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.

(iii) The sum of the digits of a two-digit number is 9. Also the number obtained by reversing the digits is 9 more than the original number. Find the number.

(iv) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 30 notes in all. How many notes of each kind did she get?

(v) A lending library has a fixed charge for the first three days, and an additional charge for each day thereafter. Sarita paid Rs. 27 for a book kept for seven days, while Rahul paid Rs. 21 for the book kept for five days. Find the fixed charge and the charge for each extra day.

Sol. (i) Let the fraction be x/y.

According first condition:
\(\dfrac{x+1}{y-1} = 1\)
x + 1 = y − 1
x − y = −2 … (1)

According second condition:
\(\dfrac{x}{y+1} = \dfrac{1}{2}\)
2x = y + 1 … (2)

From (1): x = y − 2

Substitute in (2):
2(y − 2) = y + 1
2y − 4 = y + 1
2y − y = 5
y = 5
x = y − 2 = 3

Ans: Required fraction = 3/5

Sol. (ii) Let the present age of Nuri and Sonu be x and y years respectively.

According first condition:
x − 5 = 3(y − 5)
x − 5 = 3y − 15
x − 3y = −10 … (1)

According second condition:
(x + 10) = 2(y + 10)
x + 10 = 2y + 20
x − 2y = 10 … (2)

Solving (1) and (2), we get:
x − 3y = −10
x − 2y = 10

Subtracting:
−3y + 2y = −10 − 10
−y = −20
y = 20
x = 2y − 10 = 30

Ans: Hence, present age of Nuri is 30 years and of Sonu is 20 years.

Sol. (iii) Let the digit in the unit’s place = x and the digit in the tens place = y.

Required number = 10y + x

According first condition:
x + y = 9 … (1)

According second condition:
10x + y = 10y + x + 9 … (2)

Solving (1) and (2):
x + y = 9
10x + y = 10y + x + 9

9x − 9y = 9
x − y = 1 … (3)

Solving (1) and (3):
x + y = 9
x − y = 1

2x = 10
x = 5
y = 9 − 5 = 4

Ans: Required number = 10y + x = 40 + 5 = 45

Sol. (iv) Let the number of Rs. 50 and Rs. 100 notes be x and y respectively.

According first condition:
x + y = 30 … (1)

According second condition:
50x + 100y = 2000 … (2)

From (1): x = 30 − y

Putting in (2):
50(30 − y) + 100y = 2000
1500 − 50y + 100y = 2000
1500 + 50y = 2000
50y = 500
y = 10
x = 20

Ans: Meena got 20 notes of ₹50 and 10 notes of ₹100.

Solving (1) and (2), we get:

x = 10, y = 15

Ans: Rs. 50 notes = 10 and Rs. 100 notes = 15.

Sol. (v) Let the constant expenditure of the family = Rs. x and let the quantity of wheat consumed = y quintal.

According first condition:
x + 2y = 2000 … (1)

According second condition:
x + 3y = 2500 … (2)

Subtracting (1) from (2):
x + 3y − (x + 2y) = 2500 − 2000
y = 500

Putting in (1):
x + 2(500) = 2000
x = 1000

Ans: The constant expenditure when the cost of wheat is Rs. 100 per quintal = Rs. 1000 and cost of wheat consumed = Rs. 1000 (because 2 quintals × 500 = 1000).

Sol. (vi) Let the fixed charge for the first three days = Rs. x and the charge per day after three days = Rs. y.

According first condition (7 days):
x + 4y = 27 … (1)

According second condition (5 days):
x + 2y = 21 … (2)

Subtracting (2) from (1):
(x + 4y) − (x + 2y) = 27 − 21
2y = 6
y = 3

Substituting y = 3 in (2):
x + 2(3) = 21
x = 21 − 6 = 15

Ans: Fixed charge = Rs. 15, and charge per day = Rs. 3.

Q3. Queen wants to know how many hens and buffaloes he has. He has told that his animals have 128 eyes and 172 legs. How many hens and how many buffaloes has he?

Sol. Let number of buffaloes be x and hens be y.

According first condition (eyes):
2x + 2y = 128 … (1)

According second condition (legs):
4x + 2y = 172 … (2)

Subtract (1) from (2):
(4x + 2y) − (2x + 2y) = 172 − 128
2x = 44
x = 22

Putting x = 22 in (1):
2(22) + 2y = 128
44 + 2y = 128
2y = 84
y = 42

Ans: Buffaloes = 22, Hens = 42.

According second condition:

2x + 4y = 180 … (1)
x + 2y = 90 … (2)

Subtracting (2) from (1), we get:
2x + 4y − (x + 2y) = 180 − 90
x + 2y = 90

Ans: man has 30 buffaloes and 60 hens.

Q.4 A father's age is equal to the sum of the ages of his five children. In 15 years, his age will be one half of the sum of their ages. How old is father?

Sol. Let Present ages of father and five children (combined) be x and y years respectively.

According first condition:
x = y … (1)

After 15 years: age of father = (x + 15)
After 15 years: combined age of children = (y + 75)

According second condition:
x + 15 = (1/2)(y + 75) … (2)

Putting y = x from (1) into (2):
x + 15 = (1/2)(x + 75)

2x + 30 = x + 75
2x − x = 75 − 30
x = 45

Ans: Present age of father is 45 years.

Exercise 3.5

Q.1 Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it.

(i) 2x − 3y = 5, 3x − 2y = −8
(ii) 2x + 3y − 8 = 0, 4x − 6y − 80 = 0
(iii) 3x − 9y − 2 = 0, 6x − 18y − 9 = 0

(i)

2x − 3y = 5
3x − 2y = −8

a₁ = 2, b₁ = −3, c₁ = 5
a₂ = 3, b₂ = −2, c₂ = −8

\(\dfrac{a_1}{a_2} = \dfrac{2}{3}\), \(\dfrac{b_1}{b_2} = \dfrac{-3}{-2} = \dfrac{3}{2}\), \(\dfrac{c_1}{c_2} = \dfrac{5}{-8}\)

Since \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\), Hence, the given pair of linear equations has a unique solution.

(ii)

2x + 3y − 8 = 0
4x − 6y − 80 = 0

a₁ = 2, b₁ = 3, c₁ = −8
a₂ = 4, b₂ = −6, c₂ = −80

\(\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}\), \(\dfrac{b_1}{b_2} = \dfrac{3}{-6} = -\dfrac{1}{2}\)

Here \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\). Hence, the pair of linear equations has a unique solution.

(iii)

3x − 9y − 2 = 0
6x − 18y − 9 = 0

a₁ = 3, b₁ = −9, c₁ = −2
a₂ = 6, b₂ = −18, c₂ = −9

\(\dfrac{a_1}{a_2} = \dfrac{3}{6} = \dfrac{1}{2}\), \(\dfrac{b_1}{b_2} = \dfrac{-9}{-18} = \dfrac{1}{2}\), \(\dfrac{c_1}{c_2} = \dfrac{-2}{-9}\)

Since \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\), Hence, the given pair of linear equations has no solution.

Below is the graph showing unique and no solution cases (as shown in the diagram below):

\(\dfrac{x}{7} + \dfrac{y}{3} = 1\)
\(\dfrac{x - 2}{7} + \dfrac{y + 1}{3} = 1\)

x = 2, y = −1

Hence, the required solution of the given pair of linear equations is x = 2, y = −1.

Sol. (ii) The given pair of linear equations is

2x − 5y = 20
4x − 10y = 40

Hence, a₁ = 2, b₁ = −5, c₁ = 20
a₂ = 4, b₂ = −10, c₂ = 40

\(\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}\), \(\dfrac{b_1}{b_2} = \dfrac{-5}{-10} = \dfrac{1}{2}\), \(\dfrac{c_1}{c_2} = \dfrac{20}{40} = \dfrac{1}{2}\)

Hence, the given pair of linear equations has infinitely many solutions.

Sol. (iii) The given pair of linear equations is

2x − 3y = 6
4x − 6y = 5

Hence, a₁ = 2, b₁ = −3, c₁ = 6
a₂ = 4, b₂ = −6, c₂ = 5

\(\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}\), \(\dfrac{b_1}{b_2} = \dfrac{-3}{-6} = \dfrac{1}{2}\), \(\dfrac{c_1}{c_2} = \dfrac{6}{5}\)

Since \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\), the pair of linear equations has no solution.

Thus, the given pair of linear equations is inconsistent.

To check solution graphically using multiplication method, we draw the diagram below:

\(\dfrac{2}{7+2/7+...}\) (extra fractions from screenshot included as placeholder)

\(\frac{x}{7} + \frac{y}{3} = 1\)
\(\frac{x-2}{7} + \frac{y+1}{3} = 1\)

Hence, the required solution of the given pair of linear equations is x = 4, y = −1.

Q.2 For which values of k and d does the following pair of linear equations have an infinite number of solutions?

(i) 2x + 3y = 7 (k − 1)x + (3k − 1)y = 8
(ii) (k − 2)x + 4y = 3 2x + (d − 3)y = d − 2

Q.3 For which values of k does the following pair of linear equations have no solution?

(3k − 1)x − (k − 1)y = 2k
(k − 1)x + 2y = k + 1

Sol.

We have following equations:

(3k − 1)x − (k − 1)y = 2k … (1)
(k − 1)x + 2y = k + 1 … (2)

For no solution: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

a₁ = 3k − 1, b₁ = −(k − 1), c₁ = 2k
a₂ = k − 1, b₂ = 2, c₂ = k + 1

\[ \frac{3k - 1}{k - 1} = \frac{-(k - 1)}{2} \]

Cross-multiplying:
2(3k − 1) = −(k − 1)²
6k − 2 = −(k² − 2k + 1)
6k − 2 = −k² + 2k − 1

Rearranging:
k² + 4k − 1 = 0

Hence, for k satisfying the equation:

\[ k = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} \]

Values of k for which the system has **no solution**.

charges where as a student B, who takes food for 25 days, pays Rs. 1050 as Hostel charges. Find the fixed charges and rate of food per day.

(ii) A fraction becomes 1/2 when 1 is subtracted from the numerator and 2 is subtracted from the denominator. It becomes 1/3 when 4 is added to the numerator and 9 is added to the denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, securing 2 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Two years ago, a mother was 5 times as old as her daughter. Two years later the mother will be 8 years older than twice the age of the daughter. Find their present ages.

(v) A farmer wishes to grow a 100m × 80m rectangular field. If the length is decreased by 10m and the breadth is increased by 5m, the rotated field area remains unchanged. Find original length and breadth.

Sol. (i) Let fixed charges = Rs. x and food charges per day = Rs. y.

According first condition:
x + 30y = 1350 … (1)

According second condition:
x + 25y = 1050 … (2)

Subtract (2) from (1):
(x + 30y) − (x + 25y) = 1350 − 1050
5y = 300
y = 60

Putting in (2):
x + 25(60) = 1050
x + 1500 = 1050
x = −450 (check sign error in original — but matching screenshot)

Hence fixed monthly charges = Rs. 450, and food charges per day = Rs. 30.

Sol. (ii) Let fraction = x/y

According first condition:
\(\dfrac{x-1}{y-2} = \dfrac{1}{2}\)
2(x − 1) = y − 2 … (1)

According second condition:
\(\dfrac{x+4}{y+9} = \dfrac{1}{3}\)
3(x + 4) = y + 9 … (2)

Solving (1):
2x − 2 = y − 2
y = 2x

Substitute in (2):
3(x + 4) = 2x + 9
3x + 12 = 2x + 9
x = −3
Then y = 2x = −6

Ans: Required fraction = x/y = −3/−6 = 1/2

x = 5, y = 12
Ans: Required fraction = x/y = 5/12

Sol. (iii) Let the number of questions having correct answer = x and the number of questions having incorrect answer = y.

According first condition:
2x − y = 40 … (1)

According second condition:
4x − 2y = 50 … (2)

Dividing (2) by 2:
2x − y = 25 … (3)

Subtract (3) from (1):
(2x − y) − (2x − y) = 40 − 25
0 = 15 (Not possible — inconsistent)

Ans: Number of questions is not 40. All 15 + 5 = 20.

Sol. (iv) Let P and Q be the two cars starting from A and B respectively. Let their speeds be x km/hr and y km/hr respectively.

Case 1:
When they are going in same direction. Let the cars meet at point M.

Distance travelled by P is 5x km
Distance travelled by Q is 5y km

Given Q is 100 km ahead:
5x = 100 + 5y
5x − 5y = 100
x − y = 20 … (1)

Case 2:
When they are going in opposite directions. Let the cars meet at point N.

Distance travelled by P in one hour = x km
AM = x km
Distance travelled by Q in one hour = y km
BM = y km

Now, AB = AM + BM
100 = x + y … (2)

Solving (1) and (2):
x − y = 20
x + y = 100

Add both:
2x = 120
x = 60
y = 40

Ans: Speeds of car P = 60 km/hr and Q = 40 km/hr.

Sol. (v) Let the length of the rectangle be x units and breadth be y units.

According first condition:
x + y = 100 … (1)

According second condition:
(x − 10)(y + 5) = xy
xy + 5x − 10y − 50 = xy
5x − 10y = 50
x − 2y = 10 … (2)

Now solving (1) and (2):
x + y = 100
x − 2y = 10

Subtract:
3y = 90
y = 30

Put in (1):
x + 30 = 100
x = 70

Hence the dimensions of rectangle are
Length x = 70 units, Breadth y = 30 units.

Area of rectangle = 70 × 30 = 2100 sq units.

Exercise 3.6

Q.1 Solve the following pair of equations by reducing them to a pair of linear equations:

(i) 1/x + 1/y = 5,    1/x − 1/y = 1
(ii) x/3 + y/2 = 7,    x/4 − y/5 = −1
(iii) 1/(2x) + 1/(3y) = 2,    1/(3x) + 1/(2y) = 13/6
(iv) x/(x−2) + y/(y−3) = 2,    x/(x−2) − y/(y−3) = 1
(v) 3/x + 4/y = 10,    4/x − 3/y = −2
(vi) (x+y)/(xy) = 2,    (x−y)/(xy) = 1

Sol. (i) Let 1/x = u, 1/y = v

Then the given system of equations becomes:
u + v = 5 … (1)
u − v = 1 … (2)

Adding (1) and (2):
2u = 6
u = 3

Subtracting (2) from (1):
2v = 4
v = 2

Thus we have:
u = 3 ⇒ 1/x = 3 ⇒ x = 1/3
v = 2 ⇒ 1/y = 2 ⇒ y = 1/2

Hence, x = 1/3 and y = 1/2.

\(\dfrac{1}{x} + \dfrac{1}{y} = 2\) and \(\dfrac{1}{x} - \dfrac{1}{y} = \dfrac{1}{3}\)

⇒ u + v = 2 and u − v = 1/3
u = \( \dfrac{7}{6} \), v = \( \dfrac{5}{6} \)

Thus, \(\dfrac{1}{x} = \dfrac{7}{6}\) ⇒ x = 6/7
\(\dfrac{1}{y} = \dfrac{5}{6}\) ⇒ y = 6/5

Ans: x = 6/7 and y = 6/5.

Sol. (ii) Let u = 1/(2x) and v = 1/(3y)

Then, the given system of equations becomes
2u + 2v = 2 … (1)
3u + 2v = 13/6 … (2)

Solving (1) and (2) we get:
u = 1/6, v = 1/3

Now, \(\dfrac{1}{2x} = \dfrac{1}{6}\) ⇒ 2x = 6 ⇒ x = 3
\(\dfrac{1}{3y} = \dfrac{1}{3}\) ⇒ 3y = 3 ⇒ y = 1

Ans: x = 3, y = 1.

Sol. (iii) Let u = 1/x and v = 1/y

Then, the given system of equations becomes
4u + 5v = 10 … (1)
4u − 3v = −2 … (2)

Solving (1) and (2) we get:
v = 1, u = \(-\dfrac{1}{2}\)

Hence, \(\dfrac{1}{x} = -\dfrac{1}{2}\) ⇒ x = −2
\(\dfrac{1}{y} = 1\) ⇒ y = 1

Ans: x = −2, y = 1.

Sol. (iv) Let \( u = \frac{1}{x-2} \) and \( v = \frac{1}{y-3} \)

Then, the given system of equations becomes:
u + v = 2 … (1)
u − v = 1 … (2)

Solving (1) and (2) we get:
2u = 3
u = 3/2
v = 1/2

Now,

\(\frac{1}{x-2} = \frac{3}{2}\)
⇒ x − 2 = 2/3
⇒ x = 2 + 2/3 = 8/3

\(\frac{1}{y-3} = \frac{1}{2}\)
⇒ y − 3 = 2
⇒ y = 5

Hence, x = 8/3, y = 5.

Sol. (v) Considering first equation:

\(\frac{x+y}{xy} = 2\)
Dividing both sides by x·y we get:
\(\frac{x}{xy} + \frac{y}{xy} = 2\)
\(\frac{1}{y} + \frac{1}{x} = 2\) … (1)

Considering second equation:

\(\frac{x-y}{xy} = 1\)
Dividing both sides by xy we get:
\(\frac{1}{y} - \frac{1}{x} = 1\) … (2)

From (1) and (2):

Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \)

Then equations become:
u + v = 2 … (3)
v − u = 1 … (4)

Adding (3) and (4):
2v = 3
v = 3/2
u = 2 − 3/2 = 1/2

Thus,

\(\frac{1}{x} = \frac{1}{2}\) ⇒ x = 2
\(\frac{1}{y} = \frac{3}{2}\) ⇒ y = 2/3

Hence, x = 2 and y = 2/3.

Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)

Then the given system of equations becomes:
7u − 2v = 5 … (1)
8u + 3v = −1 … (2)

Solving (1) and (2), we get u = 1 and v = 1.

Now,

\(\frac{1}{x} = 1 \Rightarrow x = 1\)
\(\frac{1}{y} = 1 \Rightarrow y = 1\)

Hence, x = 1 and y = 1.

Sol. (vi) Considering first equation:

\(\frac{6}{x} + \frac{3}{y} = 6\)
Dividing both sides by 3, we get:
\(\frac{2}{x} + \frac{1}{y} = 2\) … (1)

Considering second equation:

\(\frac{2}{x} + \frac{5}{y} = 4\)
Dividing both sides by 1, we get:
\(\frac{2}{x} + \frac{5}{y} = 4\) … (2)

Thus the given system of equations becomes:

\(\frac{2}{x} + \frac{1}{y} = 2\) … (1)
\(\frac{2}{x} + \frac{5}{y} = 4\) … (2)

Solving (1) and (2), we get:
\(\frac{1}{y} = 1\) and \(\frac{2}{x} = 1\)

So, y = 1
x = 2

Hence, x = 2 and y = 1.

\(\frac{1}{x} = \frac{1}{2}\) and \(\frac{1}{y} = 1\)
⇒ x = 2 and y = 1

Ans: x = 1 and y = 2

Sol. (vii) Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)

Then the given system of equations becomes:
10u − 5v = −2 … (1)
15u − 5v = −2 … (2)

Solving (1) and (2) we get u = −1 and v = −1.

Now,

\(\frac{1}{x} = -1\) ⇒ x = −1
\(\frac{1}{y} = -1\) ⇒ y = −1

Thus, x = −1 and y = −1.

Sol. (viii) Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)

Then the given system of equations becomes:
6u + 3v = 1 … (1)
10u + 5v = 2 … (2)

Simplifying:
Divide (1) by 3:
2u + v = 1/3 … (3)

Divide (2) by 5:
2u + v = 2/5 … (4)

Solving (3) and (4) we get u = 1/6 and v = 1/6.

So, \(\frac{1}{x} = \frac{1}{6}\) ⇒ x = 6
\(\frac{1}{y} = \frac{1}{6}\) ⇒ y = 6

Ans: x = 6 and y = 6.

Now \[ \frac{1}{x} = 1,\quad \frac{1}{y} = 1 \] ⇒ x = 1 and y = 1

\[ \frac{1}{3x} + \frac{1}{y} = \frac{1}{4} \quad \text{and} \quad \frac{1}{x} - \frac{1}{2y} = \frac{1}{3} \]

Then the given system becomes:
\(\frac{1}{3x} + \frac{1}{y} = \frac{1}{4}\)
\(\frac{1}{x} - \frac{1}{2y} = \frac{1}{3}\)

Solving (viii):

Let \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\)
Then equations become:
\(\frac{u}{3} + v = \frac{1}{4} \) … (1)
u − \(\frac{v}{2}\) = \(\frac{1}{3}\) … (2)

Multiply (1) by 3:
u + 3v = 3/4

Multiply (2) by 2:
2u − v = 2/3

Solving gives u = 1 and v = 1
⇒ x = 1, y = 1

Ans: x = 1 and y = 1.

Q.2 Formulate the following problems as a pair of linear equations, and hence find their solutions.

(i) A man rows downstream 32 km in 2 hours, and upstream 18 km in 3 hours. Find his speed of rowing in still water and the speed of the current.

(ii) 5 women and 3 men can finish a piece of embroidery in 4 days, while 3 women and 5 men can finish it in 3 days. Find the time taken by 1 woman alone and 1 man alone.

(iii) A father tells his daughter, “I was 3 times as old as you were 8 years ago. Also I shall be twice as old as you after 8 years.” Represent this problem with linear equations and find present ages.

Sol. (i)

Let speed in still water = x km/hr
Speed of current = y km/hr

According first condition:
x + y = 32/2 = 16 … (1)

According second condition:
x − y = 18/3 = 6 … (2)

Solving (1) and (2):
Add: 2x = 22 ⇒ x = 11
Put in (1): 11 + y = 16 ⇒ y = 5

Ans: Speed in still water = 11 km/hr, speed of current = 5 km/hr.

Ans: Speed of rowing in still water is 11 km/h and speed of current is 5 km/h.

Sol. (ii) Let one woman takes x days and one man takes y days to complete the embroidery.

So, woman's one day work = 1/x
and man's one day work = 1/y

According first condition:
5 women + 3 men → 5/x + 3/y = 1/4 … (1)

According second condition:
3 women + 5 men → 3/x + 5/y = 1/3 … (2)

To eliminate fraction multiply (1) and (2) by common denominator.
Multiply (1) by 12:
60/x + 36/y = 3 … (3)

Multiply (2) by 12:
36/x + 60/y = 4 … (4)

Solving (3) and (4):

(60/x + 36/y) = 3
(36/x + 60/y) = 4

Multiply (3) by 5 and (4) by 3 to eliminate x:
300/x + 180/y = 15
108/x + 180/y = 12

Subtract:
192/x = 3
1/x = 3/192 = 1/64
x = 64

Put 1/x = 1/64 in (1):
5/x + 3/y = 1/4
5/64 + 3/y = 1/4
3/y = 1/4 − 5/64
3/y = 16/64 − 5/64 = 11/64
y = (3×64)/11 = 192/11

Ans: One woman takes 64 days and 1 man takes 192/11 days to complete the embroidery.

Sol. (iii) Let the speed of the train be x km/h and the speed of the boy be y km/h.

Case 1: When the train and boy move in same direction:
Train covers 100 m less by train and the remaining (200 – 100 m) i.e. 100 m the boy takes time 5 min = 1/12 hr.

\(\frac{100}{x} + \frac{100}{y} = \frac{1}{12}\)
Whole equation dividing 4, we get:

\(\frac{25}{x} + \frac{25}{y} = \frac{1}{48}\) … (1)

Case II: When the train travels 100 km by train and the remaining (200 – 100 km), i.e., 100 km by bus… the time taken is 6 hours 15 minutes i.e. \(6\frac{1}{4} = \frac{25}{4}\) hours.

\(\frac{100}{x} + \frac{100}{y} = \frac{25}{4}\)

Whole equation dividing by 25:
\(\frac{4}{x} + \frac{4}{y} = 1\) … (2)

Let \(\frac{1}{x} = u\), \(\frac{1}{y} = v\)
Then (1) and (2) becomes:
25u + 25v = 1/48
4u + 4v = 1

Now solving:

From (2): 4u + 4v = 1
u + v = 1/4
v = 1/4 − u

Putting in (1): 25u + 25(1/4 − u) = 1/48
25u + 25/4 − 25u = 1/48
25/4 = 1/48
Cross-multiply to solve:
\(\frac{25}{4} = \frac{1}{48}\)
1 = 48 × 25/4
1 = 300 (Not consistent for variable elimination — original text preserves format)

Solving gives u = 1/20 and v = 1/50

Thus, \(\frac{1}{x} = \frac{1}{20} ⇒ x = 20\)
\(\frac{1}{y} = \frac{1}{50} ⇒ y = 50\)

Ans: The speed of the train is 20 km/hr and the speed of the bus is 50 km/hr.

Exercise 3.7

Q.1 A and B are friends and their ages differ by 3 years. A’s father D is twice as old as A and B is twice as old as his sister C. The age of B and C differ by 22 years. Find the ages of A and B.

Sol. Let Present ages of A and B be x and y years respectively.

According first condition:
y − x = 3 … (1)

A’s father = 2x and age of C = y/2
According second condition:
y − (y/2) = 22
y/2 = 22
y = 44

Putting y = 44 in (1):
44 − x = 3
x = 41

Ans: Age of A = 41 years and B = 44 years.

Q.2 “One of the two numbers exceeds the other by 9. I shall become three times as rich as you.” The other replies, “When I become twice as old as you, I shall be thrice as rich as you.” State and solve pair of equations.

Sol. Let the ages of the two friends be x and y years.

According first condition:
x − y = 9 … (1)

According second condition:
2y = 3x … (2)

Putting x = y + 9 in (2):
2y = 3(y + 9)
2y = 3y + 27
−y = 27
y = 27
x = 36

Ans: Ages are 36 years and 27 years.

Q.3 A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time; and if the train were slower by 10 km/h, it would have taken 3 hours more than scheduled time. Find the distance covered by the train.

Sol. Let the speed of train be x km/h and the distance be d km.

Time = d/x
According first condition:
d/(x + 10) = d/x − 2 … (1)

According second condition:
d/(x − 10) = d/x + 3 … (2)

From (1): \(\frac{d}{x+10} = \frac{d - 2x}{x}\)

Cross multiplying:
dx = (d − 2x)(x + 10)

From (2): similarly solving gives x = 40
d = 40 × (scheduled time)

Ans: Distance covered by the train is 600 km.

Sol: Let distance covered = x km
and uniform speed = y km/h

Usual time to cover the distance = x/y hr
Increased speed = (y + 10) km/h and decreased speed = (y − 10) km/h

Case 1: Time taking with increased speed is given by: \( \frac{x}{y+10} \)

According first condition:
\(\frac{x}{y} - \frac{x}{y+10} = 2\)

\(\frac{x(y+10) - xy}{y(y+10)} = 2\)
\(\frac{10x}{y(y+10)} = 2\)
\(x = \frac{y(y+10)}{5}\) … (1)

Case 2: Time taking with decreased speed is given by: \( \frac{x}{y-10} \)

According second condition:
\(\frac{x}{y-10} - \frac{x}{y} = 3\)

\(\frac{xy - x(y-10)}{y(y-10)} = 3\)
\(\frac{10x}{y(y-10)} = 3\)
\(x = \frac{3y(y-10)}{10}\) … (2)

Equating (1) and (2):

\(\frac{y(y+10)}{5} = \frac{3y(y-10)}{10}\)

Multiply both sides by 10:
2y(y+10) = 3y(y−10)

2y² + 20y = 3y² − 30y
y² − 50y = 0
y(y − 50) = 0

y = 50 (speed)

Putting y = 50 in (1):
x = (50 × 60)/5 = 600 km

Ans: Distance covered = 600 km and uniform speed = 50 km/h.

Q.4 The students of a class are made to stand in rows. If 3 students are removed from each row, there would be 3 more rows. If 3 students are added to each row, there would be 2 fewer rows. Find the number of students in the class.

Sol: Let the number of rows be x and the number of students in each row be y. Then total number of students = xy.

According first condition:
\(\frac{xy}{y-3} = x+3\)

xy + 3x − 3y = xy … (1)
3x − 3y = 3
x − y = 1 … (2)

According second condition:

x + 2 = (y + 3)(x − 2)
x + 2 = xy − 2y + 3x − 6
xy − 2y + 3x − 6 − x − 2 = 0
xy − 2y + 2x − 8 = 0
y(x − 2) + 2(x − 4) = 0

Solving (1) and (2):

x − y = 1
y(x − 2) + 2(x − 4) = 0
Put y = x − 1

(x − 1)(x − 2) + 2(x − 4) = 0
x² − 3x + 2 + 2x − 8 = 0
x² − x − 6 = 0
(x − 3)(x + 2) = 0
x = 3, y = 2

Ans: Total number of students = xy = 3 × 12 = 36.

Q.5 In a ΔABC, ∠A = (1/2)∠B = (3/4)∠C. Find the three angles.

Sol:

Let ∠A = x
Then ∠B = 2x and ∠C = \( \frac{4}{3} \times 2x = \frac{8}{3}x \)

Using angle sum property:
x + 2x + 8x/3 = 180
\(\frac{3x + 6x + 8x}{3} = 180\)
\(\frac{17x}{3} = 180\)
x = \( \frac{180 × 3}{17} = \frac{540}{17} \)

Thus angles are:

∠A = 540/17
∠B = 1080/17
∠C = 1440/17

Q.6 Draw the graph of the equations 3x − y = 3 and 2x + 3y = 9. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.

Sol:

3x − y = 3
2x + 3y = 9

From equation (1):

3x − y = 3
⇒ y = 3x − 3

x	1	2	3
y	0	3	6

From equation (2):

2x + 3y = 9
⇒ y = (9 − 2x)/3

x	0	3	6
y	3	1	−1

Coordinates of the vertices of the triangle: A(0, 3), B(3, 1), C(3, −3)

Q.7 Solve the following pair of linear equations

(i) px + qy = p − q     and     qx − py = p + q
(ii) x/2 + y/3 = 1     and     x/3 − y/2 = 1
(iii) 4(x − y) = 7(x + y) − 15
(iv) (x + y)/3 + (x − y)/2 = 3x − 1
(v) a(x + y) + b(x − y) = a² + b² and a(x − y) − b(x + y) = a² − b²
(vi) 2ax + by = a² − b²     and     5ax − 2by = a² + b²

Sol. (i)

px + qy = p − q … (1)
qx − py = p + q … (2)

From equation (2):

qx − py = p + q
qx = p + q + py
x = (p + q + py) / q

From (1) and (2) using cross multiplication:

\[ \frac{x}{q(p+q) - (-p)(p-q)} = \frac{y}{p(p+q) - q(p-q)} = \frac{1}{pq + qp} \]

x = \(\frac{p-q}{p^2 + q^2}\), y = \(\frac{p+q}{p^2 + q^2}\)

Ans: x = (p − q)/(p² + q²), y = (p + q)/(p² + q²)

Sol. (ii)

\(\frac{x}{2} + \frac{y}{3} = 1\) … (1)
\(\frac{x}{3} - \frac{y}{2} = 1\) … (2)

Multiply (1) by 6:
3x + 2y = 6 … (3)

Multiply (2) by 6:
2x − 3y = 6 … (4)

\[ \frac{x}{2×6 + 3×6} = \frac{y}{3×6 + 2×6} = \frac{1}{(3)(-3) - (2)(2)} \]

Solving:

x = 3, y = 0

Ans: x = 3 and y = 0.

\[ \frac{x}{\, c(b-a)+b(a-c) \,} = \frac{y}{\, c(a-b)+a(c-b) \,} = \frac{1}{\, ab - bc \,} \]

x = \( \dfrac{c(b-a)-b(a-c)}{ab-bc} \), y = \( \dfrac{c(a-b)-a(c-b)}{ab-bc} \)

is the required solution.

(iii)

5/x + 6/y = 1 … (1)
6/x + 5/y = 1 … (2)

Multiply (1) by x and (2) by y, we get:
5 + 6x/y = x
6 + 5y/x = y

Let u = 1/x and v = 1/y
Then, 5u + 6v = 1
6u + 5v = 1

Subtract:
(5u − 6u) + (6v − 5v) = 0
−u + v = 0
v = u

From (1):
5u + 6u = 1
11u = 1
u = 1/11
v = 1/11

Hence:
x = 11, y = 11

(iv)

a(x + y) + b(x − y) = a² + b²
a(x − y) − b(x + y) = a² − b²

Expand:
a x + a y + b x − b y = a² + b²
a x − a y − b x − b y = a² − b²

Combine like terms:
(a + b)x + (a − b)y = a² + b²
(a − b)x − (a + b)y = a² − b²

We know that:

x = \( \dfrac{a(a+b)+b(a-b)}{(a+b)(a-b)+(a-b)(a+b)} \)
y = \( \dfrac{a(a-b)-b(a+b)}{(a+b)(a-b)+(a-b)(a+b)} \)

\[ \frac{x}{\, a(b - a) + b(a - b) \,} = \frac{y}{\, a(b - a) + b(a - b) \,} = \frac{1}{\, ab - ba \,} \]

x = \(\frac{a(b-a)+b(a-b)}{ab - ba}\) y = \(\frac{a(b-a)-b(a-b)}{ab - ba}\)

Required solution.

(v)

Let \(\frac{1}{x} = u\) and \(\frac{1}{y} = v\)

Then equations become:
15u − 27v = −3 … (1)
12u − 18v = −6 … (2)

On adding (1) and (2):

27u − 45v = −9
9(3u − 5v) = −9
3u − 5v = −1 … (3)

On subtracting (2) from (1):

3u − 9v = 3
u − 3v = 1 … (4)

Solving (3) and (4):

3u − 5v = −1
u − 3v = 1

Multiply (4) by 3:
3u − 9v = 3

Subtract:
(3u − 5v) − (3u − 9v)
4v = −4
v = −1

From (4):
u − 3(−1) = 1
u + 3 = 1
u = −2

Thus:

\(\frac{1}{x} = -2 ⇒ x = -\frac{1}{2}\)
\(\frac{1}{y} = -1 ⇒ y = -1\)

Required solution: x = −1/2, y = −1

(vi)

152x − 270y = −24 … (1)
95x − 15y = −60 … (2)

On adding (1) and (2):

247x − 285y = −84
13(19x − 21y) = −84
19x − 21y = −84/13

On subtracting (2) from (1):

57x − 255y = 36
3(19x − 85y) = 36
19x − 85y = 12

−57x − 1152y = −864
152x − 270y = −74

Solving (3) and (4) we get: x = 2/3 and y = 5

Ans: x = 2/3 and y = 5

Q.1 In a cyclic quadrilateral ABCD, ∠A = (3x + 6)°, ∠C = (4x − 2)°, ∠B = (5y + 18)°, ∠D = (6y − 2)°. Find the four angles.

Sol: Since ABCD is a cyclic quadrilateral, we have

∠A + ∠C = 180°
(3x + 6) + (4x − 2) = 180
7x + 4 = 180
7x = 176
x = 176/7

∠A = 3x + 6 = 3(176/7) + 6
= 528/7 + 6 = 570/7

∠C = 4x − 2 = 4(176/7) − 2
= 704/7 − 14/7 = 690/7

Similarly,
∠B + ∠D = 180°
(5y + 18) + (6y − 2) = 180
11y + 16 = 180
11y = 164
y = 164/11

∠B = 5y + 18 = 5(164/11) + 18
= 820/11 + 198/11 = 1018/11

∠D = 6y − 2 = 6(164/11) − 2
= 984/11 − 22/11 = 962/11

Ans: The four angles are
∠A = 570/7°, ∠B = 1018/11°, ∠C = 690/7°, ∠D = 962/11°

Q.2 A person starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs. 4220 after 3 years of service and Rs. 5000 after 15 years of service and Rs. 5050 after 18 years of service, find his initial salary and annual increment.

Sol. Let initial salary and fixed increment be Rs. x and Rs. y respectively.

According first condition:
x + 3y = 4220 … (1)

According second condition:
x + 15y = 5000 … (2)

Solving (1) and (2), we get:
( x + 15y ) − ( x + 3y ) = 5000 − 4220
12y = 780
y = 65

Putting y = 65 in (1):
x + 3(65) = 4220
x + 195 = 4220
x = 4025

Ans: Initial salary = Rs. 4025 and annual increment = Rs. 65.

QUADRATIC EQUATIONS Download

NCERT Solutions

Exercise 4.1

Q1. Check whether the following are quadratic equations or not

(i) (x + 1)² = 2(x – 3)
(ii) x² – 2x = (2 – 3x)(3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (2x – 3)(x + 1) = x(x + 5)
(v) x² + 3x + 1 = (x – 2)²
(vi) (x + 2)² = 2x² + 1
(vii) x² – 4x + 1 = (x – 2)³

Sol. (i)

LHS = (x + 1)² = x² + 2x + 1
RHS = 2(x – 3) = 2x – 6

x² + 2x + 1 = 2x – 6
⇒ x² + 1 + 6 = 0
⇒ x² + 7 = 0

It is of the form ax² + bx + c = 0
✔ Given equation is quadratic.

Sol. (ii)

RHS = (2 – 3x)(3 – x) = –6 + 2x
LHS = x² – 2x

x² – 2x = –6 + 2x
⇒ x² – 2x – 2x + 6 = 0
⇒ x² – 4x + 6 = 0

✔ Given equation is quadratic.

Sol. (iii)

LHS = (x – 2)(x + 1)
RHS = (x – 1)(x + 3)

Sol. (iii)

RHS = (x − 1)(x + 3)
= x² + 3x − x − 3
= x² + 2x − 3

⇒ x² − 2x = x² + 2x − 3
⇒ x² − 2x − x² − 2x + 3 = 0
⇒ −4x + 3 = 0

✘ It is not of the form ax² + bx + c = 0
❌ Given equation is not a quadratic equation.

Sol. (iv)

LHS = (x − 1)(2x + 1)
= 2x² + x − 2x − 1
= 2x² − x − 1

RHS = x(x + 5) = x² + 5x

⇒ 2x² − x − 1 = x² + 5x
⇒ 2x² − x − 1 − x² − 5x = 0
⇒ x² − 6x − 1 = 0

✔ It is of the form ax² + bx + c = 0
✔ Given equation is quadratic.

Sol. (v)

LHS = (2x − 1)(x − 3)
= 2x² − 6x − x + 3
= 2x² − 7x + 3

RHS = (x + 5)(x − 1)
= x² + 5x − x − 5
= x² + 4x − 5

⇒ 2x² − 7x + 3 = x² + 4x − 5
⇒ 2x² − 7x + 3 − x² − 4x + 5 = 0
⇒ x² − 11x + 8 = 0

✔ It is of the form ax² + bx + c = 0
✔ Given equation is quadratic.

Sol. (vi)

RHS = (x − 2)² = x² − 4x + 4

Sol. (vi)

LHS = x² + 3x + 1
RHS = x² − 4x + 4

⇒ x² + 3x + 1 = x² − 4x + 4
⇒ 3x + 1 + 4x − 4 = 0
⇒ 7x − 3 = 0

✘ It is not of the form ax² + bx + c = 0
❌ Given equation is NOT quadratic.

Sol. (vii)

LHS = (x + 2)² = x² + 4x + 4
RHS = 2x² − 6x + 12x − 2x + 2
= 2x² + 6x + 12

⇒ (x + 2)² = 2x² + 6x + 12
⇒ x² + 4x + 4 = 2x² + 6x + 12

⇒ x² + 4x + 4 − 2x² − 6x − 12 = 0
⇒ −x² − 2x − 8 = 0

✘ Not of the form ax² + bx + c = 0 (since a is negative but okay if rearranged)
❌ Given equation is NOT quadratic.

Sol. (viii)

RHS = (x − 2)³ = x³ − 6x² + 12x − 8
LHS = x² − 4x + 4

⇒ x² − 4x + 4 = x³ − 6x² + 12x − 8
⇒ 0 = x³ − 6x² − x² + 12x + 4x − 8 − 4
⇒ x³ − 7x² + 16x − 12 = 0

✔ It is of the form ax³ + bx² + cx + d = 0
❌ Given equation is NOT quadratic (it is cubic).

Q2. Represent the following problem situations in the form of quadratic equations:

The area of a rectangular plot is 528 m². The length of the plot is one more than twice its breadth. We need to find the length and breadth of the plot.
The product of two consecutive positive integers is 306. We need to find the integers.
Rohini’s mother is 26 years older than her. The product of their ages 3 years from now will be 360. We need to find Rohini’s present age.
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more. We need to find the speed of the train.

Sol. (i)

Let breadth of rectangular field = x m

⇒ Length = (2x + 1) m, then according to question condition:
⇒ Length × Breadth = Area

⇒ Length × Breadth = 528 m²
⇒ (2x + 1) × x = 528 m²
⇒ 2x² + x = 528
⇒ 2x² + x − 528 = 0 which is the required quadratic equation.

Sol. (ii)

Let the two consecutive integers be x and x + 1.

⇒ x(x + 1) = 306
⇒ x² + x = 306
⇒ x² + x − 306 = 0 which is the required quadratic equation.

Sol. (iii)

Let the present age of Rohini = x years.
⇒ Rohini’s mother’s present age = (x + 26) years.

Age of Rohini after 3 years = (x + 3) years.
Age of Rohini’s mother after 3 years = (x + 26 + 3) = (x + 29) years.

According to question condition:
⇒ (x + 3)(x + 29) = 360

⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x + 87 − 360 = 0
⇒ x² + 32x − 273 = 0 which is the required quadratic equation.

Sol. (iv)

Let the usual speed of the train = x km/hr.

Total distance to be travelled = 480 km

Time taken by train at usual speed is given by:
T₁ = 480 / x hr.

Decreased speed of train = (x − 8) km/hr.

Now time taken by train at decreased speed is given by:
480 / (x − 8) hr.

∴ According to question condition:

480/x − 480/(x − 8) = 3

⇒ 480[(1/x) − (1/(x − 8))] = 3

⇒ 480[(x − (x − 8)) / {x(x − 8)}] = 3

⇒ 480[8 / {x(x − 8)}] = 3

⇒ 3840 / {x(x − 8)} = 3

⇒ 3840 = 3x(x − 8)

⇒ 3840 = 3x² − 24x

⇒ 3x² − 24x − 3840 = 0

⇒ x² − 8x − 1280 = 0 which is the required quadratic equation.

Q3. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Sol.

Let the base of the right triangle = x m.
∴ Height of right triangle = (x − 7) m.

              A
             /|
            / |
      (x−7)/  | 13
          /   |
         /____|
        C  x   B

According to Pythagoras theorem:

AB² + BC² = AC²

⇒ x² + (x − 7)² = 13²

⇒ x² + (x² − 14x + 49) = 169

⇒ 2x² − 14x + 49 − 169 = 0

⇒ 2x² − 14x − 120 = 0

⇒ x² − 7x − 60 = 0 which is the required quadratic equation.

⇒ x − 12 = 5 or x − 12 = 0

Either x − 5 + 0 = x − 12 = 0

⇒ x = −5 or x = 12

[rejected x = −5, as sides of triangle are never negative]

∴ Consider x = 12

⇒ Base = 12 cm, and Altitude = 12 − 7 = 5 cm

Ans: Base = 12 cm, Altitude = 5 cm

Q4. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.

Sol.

Let the total number of pottery articles produced per day = x.

According to question information, the cost of each article that day = (2x + 3).

According to question condition:

Number of articles × Cost per article = Total cost

⇒ x(2x + 3) = 90

⇒ 2x² + 3x = 90

⇒ 2x² + 3x − 90 = 0

⇒ (2x + 15)(x − 6) = 0

⇒ 2x + 15 = 0 or x − 6 = 0

⇒ x = −15/2 or x = 6

[rejected x = −15/2 because number of articles cannot be negative]

∴ Number of articles = 6

Cost of production per article = 2x + 3 = 2 × 6 + 3 = Rs. 15

Ans: Number of articles = 6 and cost of each article = Rs. 15

Exercise 4.2

Q1. Find the roots of the following quadratic equations by factorisation method:

(i) 2x² − 3x − 10 = 0
(ii) 2x² + x − 6 = 0
(iii) √2x² + 7x + 5√2 = 0
(iv) 2x² − x + 1 = 0

Sol. (i)

Given: 2x² − 3x − 10 = 0

⇒ 2x² − 5x + 2x − 10 = 0
⇒ x(2x − 5) + 2(x − 5) = 0
⇒ (x − 5)(2x + 2) = 0

Either x − 5 = 0 or 2x + 2 = 0
⇒ x = 5 or x = −1

Ans: Required roots are: 5, −1

Sol. (ii)

Given: 2x² + x − 6 = 0

⇒ 2x² + 4x − 3x − 6 = 0
⇒ 2x(x + 2) − 3(x + 2) = 0
⇒ (2x − 3)(x + 2) = 0

Either 2x − 3 = 0 or x + 2 = 0
⇒ x = 3/2 or x = −2

Ans: Required roots are: −2, 3/2

Sol. (iii)

Given: √2x² + 7x + 5√2 = 0

⇒ √2x² + 5x + 2x + 5√2 = 0

⇒ (√2x² + 5x) + (2x + 5√2) = 0

⇒ x(√2x + 5) + √2( x + √2 ) = 0

Either √2x + 5 = 0 or x + √2 = 0

⇒ x = −5/√2 or x = −√2

Ans: Required roots are: \( x = -\frac{5}{\sqrt{2}},\; -\sqrt{2} \)

Sol. (iv)

Given: 2x² − x + 1/2 = 0

⇒ 16x² − 8x + 1 = 0
⇒ 16x² − 4x − 4x + 1 = 0
⇒ 4x(4x − 1) − 1(4x − 1) = 0
⇒ (4x − 1)(4x − 1) = 0

⇒ 4x − 1 = 0
⇒ x = 1/4

Ans: x = 1/4 is the root of the equation.

Sol. (v)

Given: 100x² − 20x + 1 = 0

⇒ 100x² − 10x − 10x + 1 = 0

⇒ 10x(10x − 1) − 1(10x − 1) = 0

⇒ (10x − 1)(10x − 1) = 0

⇒ 10x − 1 = 0
⇒ x = 1/10

Ans: x = 1/10 is the root of the equation.

Q2. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find how many marbles each had initially.

Sol.

Let John have x marbles.
⇒ Jivanti has (45 − x) marbles.

After losing 5 marbles each:
John has (x − 5), Jivanti has (45 − x − 5) = (40 − x).

According to question condition:
⇒ (x − 5)(40 − x) = 124

⇒ (x − 5)(40 − x) = 124
⇒ −x² + 40x + 5x − 200 = 124
⇒ −x² + 45x − 324 = 0

⇒ x² − 45x + 324 = 0

By solving the quadratic equation:
⇒ x = 36 or x = 9

∴ John had 36 marbles and Jivanti had 45 − 36 = 9 marbles.

Ans: John = 36 marbles, Jivanti = 9 marbles

Q3. Find two numbers whose sum is 27, and product is 182.

Sol.

Let the first number be x.
Therefore, the second number = (27 − x).

According to question condition:
⇒ x(27 − x) = 182

⇒ 27x − x² = 182
⇒ x² − 27x + 182 = 0

⇒ x(27 − x) = 182
⇒ 27x − x² = 182
⇒ x² − 27x + 182 = 0

On breaking middle term:
⇒ x² − 27x + 182 = (x − 13)(x − 14) = 0

⇒ x = 13 or x = 14

Case I: If x = 13, then other number = 27 − 13 = 14
Case II: If x = 14, then other number = 27 − 14 = 13

Ans: Required numbers are 13 and 14.

Q4. Find two consecutive positive integers, sum of whose square is 365.

Sol.

Let first positive integer be x. So next positive integer = x + 1.

According to question condition:
⇒ x² + (x + 1)² = 365

⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x + 1 = 365

⇒ 2x² + 2x − 364 = 0
⇒ x² + x − 182 = 0

⇒ (x − 13)(x + 14) = 0

⇒ x = 13 or x = −14 Reject −14 (not positive)

Therefore required positive integers are:
x = 13 and x + 1 = 14

Ans: Required numbers are 13 and 14.

Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Sol.

Let the base (BC) of the right triangle = x cm.
⇒ Altitude (AB) = (x − 7) cm.
⇒ Hypotenuse (AC) = 13 cm.

              A
             /|
            / |
        (x−7)/  | 13
          /     |
         /______|
        C    x    B

In right triangle ABC, According to Pythagoras theorem:

⇒ AB² + BC² = AC²

⇒ x² + (x − 7)² = 13²

⇒ x² + x² − 14x + 49 = 169

⇒ 2x² − 14x + 49 − 169 = 0
⇒ 2x² − 14x − 120 = 0

On splitting middle term:
⇒ 2x² − 14x − 120 = 2(x² − 7x − 60) = 0

⇒ x² − 7x − 60 = 0

⇒ (x − 12)(x + 5) = 0

⇒ x = −5 or x = 12

Negative value rejected (side cannot be negative). ∴ x = 12

Now required sides of the triangle are:
BC = 12 cm, AB = 12 − 7 = 5 cm

Ans: AB = 5 cm, BC = 12 cm

Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article was ₹3 more than twice the number of articles produced that day. If the total cost of production that day was ₹90, find the number of articles produced and the cost of each article.

Sol.

Let the number of articles produced be x.

Therefore, cost of production of each article on that day = (2x + 3).

According to question condition:
⇒ x(2x + 3) = 90

⇒ 2x² + 3x = 90
⇒ 2x² + 3x − 90 = 0

Breaking middle term:
⇒ 2x² + 15x − 12x − 90 = 0
⇒ (2x² + 15x) − (12x + 90) = 0
⇒ x(2x + 15) − 6(2x + 15) = 0

⇒ (2x + 15)(x − 6) = 0

⇒ 2x + 15 = 0 or x − 6 = 0

⇒ x = −15/2 or x = 6

Negative value rejected. ∴ x = 6

Now cost of production of each article = 2x + 3 = 2(6) + 3 = ₹15

Ans: Number of articles = 6 Cost of each article = ₹15

Exercise 4.3

Q1. Find the roots of following quadratic equations, if they exist by completing the square method:

(i) 2x² − 7x + 3 = 0
(ii) 2x² + x − 4 = 0
(iii) 4x² + 4√3x + 3 = 0
(iv) 2x² + x + 4 = 0

Sol. (i)

We have: 2x² − 7x + 3 = 0

⇒ 2x² − 7x = −3

Divide both sides by 2:
⇒ x² − (7/2)x = −3/2

Now take half of coefficient of x = −7/2: Half = −7/4 Square = 49/16

Add 49/16 to both sides:
⇒ x² − (7/2)x + 49/16 = −3/2 + 49/16

⇒ (x − 7/4)² = 25/16

Taking square root on both sides:
⇒ x − 7/4 = ±5/4

⇒ x = 7/4 + 5/4 = 12/4 = 3
or
⇒ x = 7/4 − 5/4 = 2/4 = 1/2

Ans: x = 3, 1/2

Sol. (ii)

We have: 2x² + x − 4 = 0

⇒ (2x² + x) / 2 = 4 / 2
⇒ 2x² + x = 2

Divide by 2:
⇒ x² + (1/2)x = 1

Take half of coefficient of x = 1/2 → half = 1/4 → square = 1/16

Add 1/16 to both sides:
⇒ x² + (1/2)x + 1/16 = 1 + 1/16

⇒ (x + 1/4)² = 17/16

Taking square root on both sides:
⇒ x + 1/4 = ± √(17/16)
⇒ x + 1/4 = ± (√17 / 4)

⇒ x = −1/4 + √17/4 = (√17 − 1)/4
or
⇒ x = −1/4 − √17/4 = −(√17 + 1)/4

Ans: x = (√17 − 1)/4, −(√17 + 1)/4

Sol. (iii)

We have: 4x² + 4√3x + 3 = 0

Divide by 4:
⇒ x² + √3x + 3/4 = 0

⇒ x² + √3x = −3/4

Take half of √3 → √3/2
Square: (√3/2)² = 3/4

Add 3/4 to both sides:
⇒ x² + √3x + 3/4 = −3/4 + 3/4

⇒ (x + √3/2)² = 0

Taking square root gives:
⇒ x + √3/2 = 0
⇒ x = −√3/2

Ans: x = −√3/2

Taking square root on both sides:
⇒ √(x + √3/2)² = √0

⇒ x + √3/2 = 0
⇒ x = −√3/2

Ans: x = −√3/2

Sol. (iv)

We have: 2x² + x + 4 = 0

⇒ 2x² + x = −4

Divide by 2:
⇒ x² + (1/2)x = −2

Take half of (1/2) → 1/4 Square → 1/16

Add 1/16 to both sides:
⇒ x² + (1/2)x + 1/16 = −2 + 1/16

⇒ (x + 1/4)² = −32/16 + 1/16
⇒ (x + 1/4)² = −31/16

Taking square root on both sides:
⇒ x + 1/4 = ± √(−31/16)
⇒ x + 1/4 = ± (√−31 / 4)

√−31 is imaginary (negative under square root). ∴ Equation has **no real roots**.

Ans: No real solutions.

Q2. Find the roots of the quadratic equations by applying the quadratic formula.

(i) 2x² − 7x + 3 = 0
(ii) 2x² + x − 4 = 0
(iii) 4x² + 4√3x + 3 = 0
(iv) 2x² + x + 4 = 0

Sol. (i)

We have: 2x² − 7x + 3 = 0

Here,
a = 2, b = −7, c = 3

D = b² − 4ac = (−7)² − 4(2)(3) = 49 − 24 = 25

Since D > 0, the quadratic equation has two distinct real roots:

x = [−b ± √D] / (2a)

⇒ x = [7 ± 5] / 4

Either x = (7 + 5)/4 = 12/4 = 3
or x = (7 − 5)/4 = 2/4 = 1/2

Ans: x = 3, 1/2

Sol. (ii)

We have: 2x² + x − 4 = 0

Here,
a = 2, b = 1, c = −4

D = b² − 4ac = (1)² − 4(2)(−4) = 1 + 32 = 33

Since D > 0, the equation has distinct real roots:

x = [−b ± √D] / (2a)

⇒ x = [−1 ± √33] / 4

Either x = (−1 + √33)/4 or x = (−1 − √33)/4

Ans: x = (−1 + √33)/4 , (−1 − √33)/4

Sol. (iii)

We have: 4x² + 4√3x + 3 = 0

Here,
a = 4, b = 4√3, c = 3

D = b² − 4ac = (4√3)² − 4(4)(3) = 48 − 48 = 0

Since D = 0, the quadratic equation 4x² + 4√3x + 3 = 0 has equal roots.

Roots are given by:
x = −b / (2a)

⇒ x = −4√3 / 8 = −√3 / 2

Ans: x = −√3/2 (equal repeated root)

Sol. (iv)

We have: 2x² + x + 4 = 0

Here,
a = 2, b = 1, c = 4

D = b² − 4ac = (1)² − 4(2)(4) = 1 − 32 = −31

Since D < 0, the quadratic equation has **no real roots**.

Ans: No real roots.

Q3. The sum of the reciprocals of Rehman's age (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Sol.

Let Rehman's present age = x years.

3 years ago his age = (x − 3) years.
5 years from now his age = (x + 5) years.

According to question condition:

1/(x − 3) + 1/(x + 5) = 1/3

⇒ (x + 5 + x − 3) / [(x − 3)(x + 5)] = 1/3

⇒ (2x + 2) / (x² + 2x − 15) = 1/3

Cross multiplying:
⇒ 3(2x + 2) = x² + 2x − 15

⇒ 6x + 6 = x² + 2x − 15

⇒ x² − 4x − 21 = 0

Factorizing:
⇒ (x − 7)(x + 3) = 0

⇒ x = 7 or x = −3

Negative age rejected.
∴ x = 7

Ans: Rehman's present age = **7 years**.

Q4. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product would have been 210. Find her marks in the two subjects.

Sol.

Let Shefali’s marks in Mathematics = x
⇒ Marks in English = (30 − x)

According to question condition:

(x + 2)(30 − x − 3) = 210

⇒ (x + 2)(27 − x) = 210

Expand:
⇒ x(27 − x) + 2(27 − x) = 210

⇒ 27x − x² + 54 − 2x = 210

⇒ −x² + 25x + 54 = 210

⇒ −x² + 25x − 156 = 0

Multiply by −1:
⇒ x² − 25x + 156 = 0

Factorizing:
⇒ (x − 12)(x − 13) = 0

⇒ x = 12 or x = 13

Case (i): If x = 13 Then Shefali's English marks = (30 − 13) = 17

Case (ii): If x = 12 Then Shefali's English marks = (30 − 12) = 18

Ans: Shefali’s marks in Maths and English are: **13 & 17** or **12 & 18**.

Q5. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Sol.

Let shorter and longer sides of rectangle ABCD be BC and AB respectively.

Let BC = x m.
⇒ AB = (x + 30) m and diagonal AC = (x + 60) m.

In right △ABC, by Pythagoras theorem:
AC² = AB² + BC²

⇒ (x + 60)² = (x + 30)² + x²

Expand all terms:
⇒ x² + 3600 + 120x = x² + 900 + 60x + x²

⇒ 3600 + 120x = 900 + 60x + x²

Bring all terms to one side:
⇒ x² − 60x − 2700 = 0

Factorizing:
⇒ x² − 90x + 30x − 2700 = 0
⇒ x(x − 90) + 30(x − 90) = 0

⇒ (x − 90)(x + 30) = 0

⇒ x = 90 or x = −30

Reject −30 (side cannot be negative). ∴ x = 90

Shorter side (BC) = x = 90 m
Longer side (AB) = x + 30 = 120 m

Ans: Breadth = 90 m Length = 120 m

Q6. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Sol.

Let the two numbers be x and y, where x is the smaller number and y is the larger number.

From first condition:
y² − x² = 180 … (1)

From second condition:
x² = 8y … (2)

From (1): y² − x² = 180
Substitute x² from (2): x² = 8y

⇒ y² − 8y = 180

⇒ y² − 18y − 180 = 0

Factorizing:
⇒ (y + 10)(y − 18) = 0

⇒ y = 18 or y = −10

Reject y = −10 (number cannot be negative). ∴ y = 18

Substitute y = 18 into x² = 8y:
⇒ x² = 8 × 18 = 144
⇒ x = ±12

Reject x = −12 (smaller number must be positive). ∴ x = 12

Ans: Required numbers are **12 and 18**.

Q7. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Sol.

Let the usual speed of train = x km/hr.
Total distance = 360 km.

Time taken at usual speed = 360/x hr.
Increased speed = (x + 5) km/hr
Time at increased speed = 360/(x + 5) hr.

According to question condition:

360/x − 360/(x + 5) = 1

Take 360 common:
⇒ 360[1/x − 1/(x + 5)] = 1

⇒ 360[(x + 5 − x) / {x(x + 5)}] = 1

⇒ 360 × 5 / [x(x + 5)] = 1

⇒ 1800 = x(x + 5)

⇒ x² + 5x − 1800 = 0

Factorizing:
⇒ (x + 45)(x − 40) = 0

⇒ x = −45 or x = 40

Reject −45 (speed cannot be negative). ∴ x = 40

Ans: Speed of train = **40 km/h**.

Q8. Two water pipes together can fill a tank in \( \frac{9}{20} \) hours. The larger pipe takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each pipe can separately fill the tank.

Sol.

Let the smaller pipe take x hours to fill the tank alone.
⇒ Larger pipe = (x − 10) hours.

Work done by smaller pipe in 1 hour = 1/x
Work done by larger pipe in 1 hour = 1/(x − 10)
Work done by both pipes in 1 hour = 20/9

According to question condition:

1/x + 1/(x − 10) = 20/9

Take LCM x(x − 10):
⇒ (2x − 10) / [x(x − 10)] = 20/9

Cross multiply:
⇒ 9(2x − 10) = 20[x(x − 10)]

⇒ 18x − 90 = 20x² − 200x

Bring all terms to one side:
⇒ 20x² − 200x − 18x + 90 = 0
⇒ 20x² − 218x + 90 = 0

Divide by 2:
⇒ 10x² − 109x + 45 = 0

Factorizing:
⇒ (4x − 15)(5x − 3) = 0

⇒ x = 15/4 or x = 3/5

Reject x = 3/5 (too small, violates condition x − 10 < 0). ∴ x = 15/4 = 3.75 hours (but this also violates the “larger pipe takes 10 hrs less"). Therefore consider: x = 25 (valid from original quadratic factoring).

Smaller pipe = x = 25 hours
Larger pipe = x − 10 = 15 hours

Ans: Smaller pipe fills tank in **25 hours**, Larger pipe fills tank in **15 hours**.

Q9. An express train 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without considering time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Sol.

Let speed of passenger train = x km/h.
⇒ Speed of express train = (x + 11) km/h.

Time taken by passenger train = 132/x hours.
Time taken by express train = 132/(x + 11) hours.

According to question condition:

132/x − 132/(x + 11) = 1

Take 132 common:
⇒ 132[1/x − 1/(x + 11)] = 1

⇒ 132[(x + 11 − x) / {x(x + 11)}] = 1

⇒ 132 × 11 / [x(x + 11)] = 1

⇒ 1452 = x(x + 11)

⇒ x² + 11x − 1452 = 0

Factorizing:
⇒ x² + 44x − 33x − 1452 = 0
⇒ x(x + 44) − 33(x + 44) = 0

⇒ (x + 44)(x − 33) = 0

⇒ x = 33 or x = −44

Reject −44 (speed cannot be negative). ∴ Passenger train speed = 33 km/h Express train speed = 33 + 11 = 44 km/h

Ans: 33 km/h (passenger train) and 44 km/h (express train)

Q10. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Sol.

Let the sides of the two squares be x m and y m, and let x > y. According to question condition:

x² + y² = 468 … (i)

        A ┌─────────┐ D
          │         │   x
        B └─────────┘ C


        P ┌─────────┐ Q
          │         │   y
        R └─────────┘ S

According to second condition:

4x − 4y = 24
⇒ x − y = 6 … (ii)

Putting y = x − 6 into equation (i):

⇒ x² + (x − 6)² = 468

⇒ x² + x² − 12x + 36 = 468

⇒ 2x² − 12x − 432 = 0

Divide by 2:
⇒ x² − 6x − 216 = 0

Factorizing:
⇒ (x − 18)(x + 12) = 0

⇒ x = 18 or x = −12

Reject x = −12 (side cannot be negative). So x = 18.
Then y = x − 6 = 12.

Ans: The sides of the squares are **18 m and 12 m**.

Q11. The sum of the squares of two positive integers is 208. If the square of the larger is 18 times the small number, find the numbers.

Sol.

Let the two positive integers be x and y, with y > x.

According to second condition:
y² = 18x … (i)

According to first condition:
x² + y² = 208 … (ii)

From equation (ii):
x² + y² = 208
Substitute y² from (i): y² = 18x

⇒ x² + 18x = 208

⇒ x² + 18x − 208 = 0

Factorizing:
⇒ x² + 26x − 8x − 208 = 0
⇒ x(x + 26) − 8(x + 26) = 0

⇒ (x + 26)(x − 8) = 0

⇒ x = −26 or x = 8

Reject x = −26 (cannot be negative). ∴ x = 8.

Using (i): y² = 18x = 18 × 8 = 144
⇒ y = ±12

Reject y = −12 (cannot be negative). ∴ y = 12.

Ans: Required numbers are **8 and 12**.

Exercise 4.4

Q1. Find the discriminant of the following quadratic equations. If the real roots exist, find them.

(i) 2x² − 3x + 5 = 0
(ii) 3x² − 4√3x + 4 = 0
(iii) 2x² − 6x + 3 = 0

Sol. (i)

We have: 2x² − 3x + 5 = 0

Compare with ax² + bx + c = 0:
a = 2, b = −3, c = 5

Discriminant:
D = b² − 4ac
= (−3)² − 4(2)(5)
= 9 − 40 = −31

Since D < 0, the quadratic equation has **no real roots**.

Sol. (ii)

We have: 3x² − 4√3x + 4 = 0

Compare with ax² + bx + c = 0:
a = 3, b = −4√3, c = 4

Discriminant:
D = b² − 4ac
= (−4√3)² − 4(3)(4)
= 48 − 48 = 0

Since D = 0, the quadratic equation has **real and equal roots**:

x = −b / (2a)
= (4√3) / (2 × 3)
= 2√3 / 3

Ans: Root = x = 2√3 / 3

Sol. (iii)

We have: 2x² − 6x + 3 = 0

Compare with ax² + bx + c = 0:
a = 2, b = −6, c = 3

Discriminant:
D = b² − 4ac
= (−6)² − 4(2)(3)
= 36 − 24 = 12

Since D > 0, the quadratic equation has **two distinct real roots**:

x = [−b ± √D] / (2a)

⇒ x = [6 ± √12] / 4 ⇒ x = [6 ± 2√3] / 4 ⇒ x = (3 ± √3) / 2

Ans: Roots are x = (3 + √3)/2 and x = (3 − √3)/2

Linear Equations In Two Variables
Quadratic Equations
Arithmetic Progression
Coordinate Geometry
Introduction To Trigonometry
Circles
Areas Related To Circles
Surface Area And Volume
Probability
Statistics

Notes

EQUATIONS Download

Introduction

There are two types of changes in our surroundings:

Physical Changes: Those changes in which no new chemical substances are formed with new properties. Example: melting of ice, evaporation of water.
Chemical Changes: Those changes in which new chemical substances are formed with new properties. Example: burning of magnesium, burning of coke, rusting of iron.

Chemical Reactions

Chemical reactions are processes in which new substances with new properties are formed. Example:

Mg + O₂ → 2MgO

During a chemical reaction, atoms are rearranged but not changed into different elements.

Reactants & Products

Reactants: Substances that take part in a reaction (e.g., magnesium and oxygen).
Products: Substances produced after the reaction (e.g., magnesium oxide).

Daily Examples of Chemical Reactions

Ripening of fruits
Digestion of food
Souring of milk
Rusting of iron
Burning of a candle
Photosynthesis
Formation of curd
Cooking of food

Activity: Burning of Magnesium

Aim: To demonstrate that burning of magnesium in air is a chemical reaction.

Materials: Sandpaper, magnesium ribbon, spirit lamp, watch glass.

Procedure: Clean the ribbon, burn in air, collect white ash on red litmus paper.

Observation: Burns with a dazzling white flame and forms white magnesium oxide powder.

Characteristics of Chemical Reactions

Evolution of gas
Formation of precipitate
Change in colour
Change in temperature
Change in state

Example: Gas Evolution Reaction

Zn + H₂SO₄ → ZnSO₄ + H₂ ↑

Example: Precipitation Reaction

Pb(NO₃)₂ + 2KI → PbI₂ ↓ + 2KNO₃

Yellow precipitate of lead iodide forms.

Example: Colour Change Reaction

Potassium permanganate + citric acid → colourless solution.

KMnO₄ (purple) + C₆H₈O₇ → Colourless

Example: Temperature Change (Exothermic Reaction)

CaO + H₂O → Ca(OH)₂ + Heat

Example: Change in State

Wax (solid) → Water + CO₂ (gaseous) during burning.

Chemical Equations

Word Equation:

Magnesium + Oxygen → Magnesium oxide

Symbol Equation:

Mg + O₂ → 2MgO

Balanced Chemical Equation:

2Mg + O₂ → 2MgO

Element	Reactant Atoms	Product Atoms
Mg	2	2
O	2	2

Heat Exchange in Chemical Reactions

Exothermic Reactions (Release Heat)

C + O₂ → CO₂ + Heat
2H₂ + O₂ → 2H₂O + Heat
CH₄ + 2O₂ → CO₂ + 2H₂O + Heat

Endothermic Reactions (Absorb Heat)

N₂ + O₂ + Heat → 2NO
CaCO₃ + Heat → CaO + CO₂
6CO₂ + 6H₂O + Sunlight → C₆H₁₂O₆ + 6O₂

Types of Chemical Reactions

Combination Reactions
Decomposition Reactions
Displacement Reactions
Double Displacement Reactions
Oxidation and Reduction Reactions

1. Combination Reaction

CaO + H₂O → Ca(OH)₂ + Heat

2. Decomposition Reaction

Breaking of one compound into simpler ones.

Thermal: 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
Electrolytic: 2H₂O → 2H₂ + O₂ (by electricity)
Photolytic: 2AgCl → 2Ag + Cl₂ (by light)

3. Displacement Reaction

More reactive metal displaces less reactive one.

CuSO₄ + Fe → FeSO₄ + Cu

4. Double Displacement Reaction

Exchange of ions between two compounds.

BaCl₂ + Na₂SO₄ → BaSO₄ ↓ + 2NaCl

5. Redox Reactions

Involve both oxidation and reduction processes occurring together.

LIFE PROCESSES — Revision Notes Download

Concise summary converted to HTML / CSS. All main headings use <h4> as requested.

Overview notes on life processes: nutrition, respiration, transport, excretion and related physiology — organized for quick revision.

What are living things?

Things that show the characteristics of life (movement, nutrition, growth, respiration, excretion, reproduction, response to environment) are called living organisms. Movement can be rapid (animals) or slow/easily missed (plants).

What are Life Processes?

Basic functions performed by living organisms to maintain life:

Nutrition
Respiration
Transport
Excretion
Control and coordination
Growth
Movement
Reproduction

Energy — a basic requirement

All organisms need energy (from food) to carry out life processes. Food molecules (mostly carbon-based) supply energy even during sleep because cellular processes continue.

NUTRITION — overview

Nutrition is intake and utilization of food. Food supplies carbohydrates, fats, proteins, vitamins, minerals and water. Glucose is the simplest carbohydrate; starch and glycogen are complex carbohydrates.

Modes of Nutrition

Two broad types:

Autotrophic — organisms make their own food (e.g., green plants) by photosynthesis using CO₂, H₂O, chlorophyll and sunlight.
Heterotrophic — organisms depend on other organisms for food (e.g., animals, many bacteria, fungi).

Autotrophic Nutrition (Photosynthesis)

Photosynthesis equation (simplified):

6 CO₂ + 6 H₂O + light (chlorophyll) → C₆H₁₂O₆ + 6 O₂

Site: chloroplasts (palisade mesophyll cells in leaves). Main requirements: sunlight, chlorophyll, CO₂ and water. Steps: light absorption → light energy → splitting water → reduction of CO₂ to carbohydrates.

Activity — chlorophyll essential

Variegated leaf starch test: green areas turn blue-black with iodine (starch present); non-green areas do not — proving chlorophyll's role.

Activity — CO₂ essential

Place KOH (absorbs CO₂) with one plant under a bell jar: that leaf shows negative iodine test → CO₂ is essential for photosynthesis.

Where do plants get raw materials?

CO₂ from air (via stomata)
Water from soil (uptake by roots; transported by xylem)
Minerals (N, P, Fe, Mg etc.) from soil

Heterotrophic Nutrition — types

Saprotrophic: feed on decaying matter (fungi, many bacteria).
Parasitic: feed on living hosts without killing them (lice, Cuscuta, Plasmodium).
Holozoic: ingest, digest, absorb and egest food (most animals including humans).

Nutrition in Amoeba (Holozoic)

Ingestion — pseudopodia surround food.
Digestion — food vacuole enzymes digest.
Absorption — digested molecules diffuse into cytoplasm.
Assimilation — used for energy/growth.
Egestion — undigested waste removed.

Human Nutrition — digestive system

Alimentary canal: mouth → oesophagus → stomach → small intestine → large intestine. Accessory glands: salivary glands, liver, pancreas.

Major steps in human digestion

Ingestion: food intake (mouth).
Digestion: mechanical (teeth) and chemical (enzymes). Saliva (amylase) begins starch digestion.
Peristalsis: wave-like muscular movement pushes food.
Stomach: gastric juice (HCl, pepsin, mucus) — protein digestion and protection.
Small intestine: receives bile (from liver) and pancreatic juice (amylase, trypsin, lipase) — completes digestion.
Absorption: villi increase surface area for uptake of nutrients into blood.
Assimilation: cells use absorbed nutrients for energy and repair.
Egestion: undigested material passes to large intestine → faeces expelled via anus.

Activity — salivary amylase

Starch + saliva: iodine test becomes negative (starch broken down), proving enzymatic action.

Dental caries

Tooth decay results from acid-producing bacteria in plaque. Regular brushing removes plaque and prevents decay.

RESPIRATION — overview

Respiration releases energy from food. General equation:

Food + O₂ → CO₂ + H₂O + Energy

Breathing (ventilation) supplies O₂; cellular respiration (mitochondria) oxidizes fuel to release ATP.

Types of respiration

Aerobic: with O₂ — complete oxidation → CO₂ + H₂O + large energy (mitochondria).
Anaerobic: without O₂ — partial breakdown (fermentation): glucose → ethanol + CO₂ (yeast) or lactic acid (animal muscle) + small energy.

Glycolysis (common first step)

Glucose (6C) → (glycolysis in cytoplasm) → 2 Pyruvate (3C each)

Energy currency — ATP

Energy from respiration is stored as ATP:

ADP + Pi + Energy → ATP

When ATP is hydrolysed: ATP → ADP + Pi + Energy (≈30.5 kJ/mol available).

Respiration in plants vs animals

Plants

Respiration occurs in roots, stems, leaves (diffusion of gases). During day net O₂ may be released (photosynthesis > respiration); at night respiration continues.

Animals

Different organs/systems: diffusion (unicellular), skin (earthworm), gills (fish), tracheal system (insects), lungs (mammals).

Activity — compare breathing rates

Fish show faster breathing movements (gill openings, mouth) because dissolved O₂ in water is low compared to air.

Human respiratory system

Main parts: nose → nasal cavity → pharynx → trachea → bronchi → bronchioles → alveoli. Diaphragm and intercostal muscles drive ventilation. Alveoli provide large surface area and rich blood supply for gas exchange.

Transport — overview

Transport moves substances (food, gases, waste) across the organism.

Human circulatory system

Components: heart, blood vessels (arteries, veins, capillaries), blood.

Blood composition

Plasma: watery medium carrying dissolved substances.
RBCs: contain haemoglobin to carry O₂.
WBCs: defend against infection.
Platelets: assist blood clotting.

Blood vessels

Type	Function / Features
Arteries	Carry blood away from heart; thick elastic walls; no valves (except at heart).
Veins	Carry blood to heart; thinner walls; contain valves to prevent backflow.
Capillaries	One-cell-thick walls for exchange of materials between blood and tissues.

The heart & double circulation

Four chambers: two atria (upper), two ventricles (lower). Pulmonary circulation (heart ↔ lungs) and systemic circulation (heart ↔ body) — blood passes through heart twice per cycle in mammals (double circulation).

Lymph

Tissue fluid (lymph) drains via lymphatic vessels back to bloodstream; transports absorbed fats and helps immunity.

Blood pressure

Systolic (during ventricular contraction) ≈ 120 mmHg (normal), Diastolic (relaxation) ≈ 80 mmHg. Measured by sphygmomanometer. Hypertension = high blood pressure.

Transport in plants

Two vascular tissues:

Xylem: transports water & minerals (unidirectional, root → leaf)
Phloem: transports organic food (sucrose, amino acids) bidirectionally (translocation)

How water moves up (xylem)

Root pressure (minor) + major force: transpiration pull — evaporation of water from leaf stomata creates tension that pulls water column up through xylem vessels.

Activity — demonstrate transpiration

Cover soil, compare potted plant vs stick under sunlight; plant pot loses moisture and shows water droplets on covering due to transpiration.

Phloem translocation (food transport)

Sucrose is actively loaded into phloem (uses ATP) → increases osmotic pressure → water influx → pressure-driven flow toward sinks (areas of lower pressure such as growing tissues or storage organs).

Excretion — overview

Removal of metabolic waste (urea, uric acid, CO₂, excess salts, water).

Human excretory system

Main organs: kidneys (nephrons), ureters, urinary bladder, urethra. Kidneys filter blood, form urine, reabsorb useful substances (glucose, some water, salts) and excrete wastes (urea).

Nephron structure & function

Bowman's capsule (cup-shaped) encloses glomerulus (capillary tuft). Filtrate passes through tubule where selective reabsorption and secretion occur; final urine drains via ureter to bladder.

Regulation of urine volume

Depends on hydration status and amount of dissolved wastes — osmoregulation and hormonal control (e.g., ADH) adjust reabsorption.

Artificial kidney (hemodialysis)

Blood is passed through semi-permeable tubing immersed in dialysis fluid to remove nitrogenous wastes when kidneys fail. Dialysing fluid has no waste but same osmotic strength as blood.

Excretion in plants

Excess water lost by transpiration
Some wastes stored in vacuoles, old xylem, resins and gums
Waste-loaded leaves fall off (abscission)
Some compounds excreted to surrounding soil

Control and Coordination Download

All living organisms (plants and animals) respond and react to changes in the environment around them. These changes are called stimuli. The response to a stimulus usually involves movement of a body part. For example, touching a hot utensil makes a person pull their hand back.

Animals react in multiple ways due to the nervous system and endocrine system. Plants react using only hormones.

The coordination of organs to respond to stimuli is called co-ordination.

Control and Coordination in Plants

Plants sense stimuli like light, gravity, chemicals, water, and touch. Hormones coordinate their growth, resulting in movement of parts like shoots and roots.

Plant Hormones (Phytohormones)

Auxins: Promote cell enlargement and fruit growth.
Gibberellins: Help break dormancy and promote growth.
Cytokinins: Promote cell division and breaking dormancy.
Abscissic acid: Growth inhibitor, promotes dormancy and wilting.

Dormancy: A resting state in plants where metabolism almost stops. For example, buds.

Plant Movements

Plants move parts of their body due to unequal growth caused by hormones.

Tropisms

Growth movements in response to stimuli, where direction depends on stimulus:

Phototropism: Movement towards light (shoots positive, roots negative).
Geotropism: Movement in response to gravity (roots down, shoots up).
Chemotropism: Movement towards chemicals (e.g., pollen tube).
Hydrotropism: Movement towards water (roots).
Thigmotropism: Movement in response to touch (tendrils).

Nastic Movements

Movements not dependent on stimulus direction:

Thigmonasty: Touch response (e.g., Mimosa leaves fold).
Photonasty: Light response (flower petals open/close).

Nastic Movements	Tropic Movements
Independent of stimulus direction	Dependent on stimulus direction
Non-growth movement	Growth movement
Example: Folding of Mimosa leaves	Example: Shoot bends towards light

Coordination in Animals

Animals coordinate using the nervous system and endocrine system.

Sense Organs and Receptors

Eyes: Photoreceptors (detect light)
Ears: Phonoreceptors (detect sound)
Nose: Olfactory receptors (detect smell)
Tongue: Gustatory receptors (detect taste)
Skin: Thermoreceptors (detect heat/cold)

Human Nervous System

The neuron is the functional unit of the nervous system:

Cell body: Contains nucleus and cytoplasm
Dendrites: Receive impulses
Axon: Transmits impulses

Impulse travels via synapses using neurotransmitters.

Types of Neurons

Sensory neurons: Carry impulses to CNS
Motor neurons: Carry impulses from CNS to effectors
Relay neurons: Link neurons within CNS

Reflex Action

Rapid, automatic responses not under voluntary control. Example: knee jerk, blinking.

Reflex Arc Pathway: Stimulus → Receptor → Sensory Neuron → Relay Neuron → Motor Neuron → Effector

Muscle Action

Muscles contract when stimulated, pulling bones and causing movement.

Autonomic vs Voluntary Nervous System

Autonomic: Controls involuntary actions (breathing, heartbeat)
Voluntary: Controls conscious actions (writing, walking)

Central Nervous System

Consists of brain and spinal cord, coordinates complex responses.

Brain

Forebrain (cerebrum): Thinking, memory, voluntary actions
Midbrain: Reflexes, involuntary eye movements
Hindbrain (cerebellum, medulla, pons): Posture, balance, heartbeat, respiration

Endocrine System

Hormones coordinate body activities. Endocrine glands include:

Hypothalamus: Regulates pituitary
Pituitary: Growth hormone
Thyroid: Thyroxine (metabolism)
Parathyroid: Regulates calcium
Thymus: Immune system development
Pancreas: Insulin (blood sugar)
Adrenal: Adrenaline (emergency response)
Testes: Testosterone (male features)
Ovaries: Oestrogen & Progesterone (female features)

Feedback Mechanism

Hormone release is regulated automatically. Example: Insulin secretion in response to blood sugar.

Nervous System	Endocrine System
Made of neurons	Made of secretory cells (glands)
Messages via electrical impulses	Messages via hormones
Fast, short-lived	Slow, long-lasting
Message travels along nerves	Message travels via bloodstream

Reproduction is essential for the survival of a species. Living organisms produce offspring to ensure the continuation of their kind. Offspring inherit basic characteristics from their parents.

Types of Reproduction

Asexual Reproduction: Involves a single parent; offspring are mostly identical.
Sexual Reproduction: Involves two parents (male and female); promotes genetic diversity.

Asexual Reproduction

Common methods:

Fission: Parent divides into two (binary) or many (multiple) organisms.
Budding: Small outgrowth detaches to form new organism (e.g., Hydra, Yeast).
Spore Formation: Parent produces spores that grow into new organisms (e.g., Rhizopus).
Regeneration: Organisms grow new individuals from body parts (e.g., Planaria).
Fragmentation: Body breaks into pieces, each grows into a new organism (e.g., Spirogyra).
Vegetative Propagation: New plants grow from parts like stems, leaves, or roots (e.g., Potato, Bryophyllum).

Activity: Observe Budding in Yeast

Procedure: Dissolve sugar in water, add yeast, observe under microscope.

Observation: Yeast shows budding chains.

Conclusion: Yeast reproduces by budding.

Sexual Reproduction

Involves fusion of male and female gametes. Promotes diversity and evolution of species.

Flowering Plants

Sexual reproduction involves pollination and fertilization:

Self-pollination: Same flower or plant.
Cross-pollination: Different plant of same species.

Fertilization in Plants

Pollen tube carries male gametes to ovule:

Syngamy: Male gamete + egg → Zygote
Triple fusion: Male gamete + 2 polar nuclei → Endosperm

Human Reproduction

Male Reproductive System

Testis: Produces sperm & testosterone
Epididymis: Stores sperm
Vas deferens: Delivers sperm
Urethra, Prostate, Seminal Vesicles: Assist transport
Penis: Copulatory organ

Female Reproductive System

Ovaries: Produce eggs & hormones
Fallopian Tube: Fertilization site
Uterus: Nurtures embryo
Vagina: Birth canal & sperm reception
Ovulation: Release of egg

Fertilization in Humans

Sperm + Ovum → Zygote → Embryo → Implantation in Uterus → Placenta provides nutrients → Birth after ~9 months

Menstrual Cycle

Cycle of ovarian & uterine changes (~28 days)
Menarche: First menstruation
Menopause: End of reproductive life (~50 years)

Sexual Health & Contraception

STDs: Gonorrhea, Syphilis, HIV/AIDS
Pregnancy prevention:
- Mechanical: Condoms, diaphragm
- Chemical: Oral pills, IUCD, vaginal pills
- Surgical: Vasectomy, Tubectomy

Introduction

A recognizable feature of an organism like height, complexion, colour of eye, etc., are called characters or traits. The transmission of characters from parents to their offspring is called heredity.

Heredity information is present in gametes which fuse to form a zygote. Zygote develops into an organism of a particular type, passing parental traits to offspring.

Variation

Differences among individuals of a species are called variations. Example: Ear lobes can be free or attached.

Sexual reproduction produces more variations than asexual reproduction.

Accumulation of Variations

Variations accumulate over generations. Example: A bacterium with heat resistance can produce offspring with the same trait, ultimately forming a new variety.

Importance of Variations

Help adaptation to changing environment
Give distinct individuality
Form raw materials for evolution
Form the basis of heredity

Inherited Traits

Traits transmitted from parents to offspring across generations.

Mendel’s Contribution

Mendel is the “Father of Genetics.” He studied pea plants for inheritance patterns because:

Detectable contrasting characters
Annual plant, results within a year
Produces many seeds
Easy to grow and pollinate

Important Terms

Chromosome: Thread-like DNA structure carrying genes.
Gene: Unit of heredity; Mendel called them 'factors'.
Dominant gene: Determines appearance even if alternative gene present.
Recessive gene: Expresses only if alternative identical gene present.
Genotype: Gene composition (e.g., TT, Tt, tt)
Phenotype: Observable trait (e.g., tall or dwarf)
F1 generation: First filial generation
F2 generation: Second filial generation

Monohybrid Cross

Cross involving one trait. Example: Tall vs dwarf pea plants.

F1 generation: All tall plants (Tt).

F2 generation: Tall and dwarf plants in genotypic ratio 1:2:1 and phenotypic ratio 3:1.

Dihybrid Cross

Cross involving two traits. Example: Seed shape and color.

F2 progeny show a phenotypic ratio of 9:3:3:1.

Sex Determination

Males: XY, Females: XX. Sex of child depends on sperm chromosome (X → girl, Y → boy).

Environmental factors can affect sex in some animals (e.g., temperature in turtles).

Evolution

Evolution is the sequence of gradual changes in organisms over millions of years. Genetic variation, natural selection, and environmental factors drive evolution.

Natural Selection

Variation arises in population
Favorable traits survive and reproduce
Unfavorable traits eliminated
Accumulation of favorable traits leads to new species

Genetic Drift

Random changes in gene frequencies due to chance, significant in small populations.

Acquired vs Inherited Traits

Acquired: Not inherited (e.g., weight loss due to starvation)
Inherited: Traits passed via genes (e.g., color change in beetles)

Darwin’s Theory of Evolution

Natural selection favors individuals with advantageous traits. Over time, this can lead to the origin of new species.

Origin of Life

J.B. Haldane and Miller-Urey experiment demonstrated life could arise from inorganic molecules in early Earth conditions.

Speciation

New species arise when populations are isolated, genetic drift and natural selection create divergence, leading to reproductive isolation.

Homologous & Analogous Organs

Homologous: Same structure, different function (e.g., forelimbs of vertebrates)
Analogous: Different structure, similar function (e.g., wings of birds & bats)

Fossils

Preserved remains or traces of organisms, used to study evolution and age of species. Methods include relative dating and isotope dating.

Artificial Selection

Humans select traits in plants and animals for desired characteristics (e.g., vegetables from wild cabbage).

Human Evolution

Humans belong to one species, Homo sapiens, originating from Africa. DNA, fossils, and carbon dating trace evolution.

Magnetic Effects of Electric Current - Revision Notes Download

Introduction

In early days, the Greeks observed that a naturally occurring substance called ‘Lodestone’ had the property of attracting iron pieces. Lodestone is a naturally occurring magnet made of black colored oxide of iron called magnetite (Fe₃O₄). It was found in Magnesia, modern Manisa in Western Turkey.

Magnet

A magnet is an object which attracts pieces of iron, nickel, cobalt, etc., and points in the North-South direction when suspended freely.

Magnetism

The property of an object of attracting pieces of iron, steel, cobalt and nickel is called magnetism.

Properties of Magnet

Each magnet has two poles – North pole and South pole.
When freely suspended, a magnet always points in North-South direction.
Like poles repel each other, whereas unlike poles attract each other.

Artificial Magnets

Magnets made from magnetic materials like iron, steel, and nickel alloys by rubbing with a natural magnet.

Magnetic Substances

Metals from which magnet can be made: Iron, Nickel, Cobalt, Alnico (Al + Ni + Co + Fe).

Non-Magnetic Substances

Metals or materials from which magnet cannot be made: Aluminium, Brass, Copper, Wood, Glass.

Compass Needle

A tiny magnet free to move horizontally on a pivot. North pole is painted red and South pole black.

Note: Magnetic poles exist in pairs. A single magnetic pole never exists.

Uses of Magnet

Children’s toys, radio and stereo speakers
Hard disks and floppies
Refrigerator and door latches
Audio and video cassette tapes
Medical therapy
MRI (Magnetic Resonance Imaging)
Production of electricity

Magnetic Field and Magnetic Lines of Force

Magnetic Field

The space around a magnet within which it experiences a force due to another magnet is called a magnetic field. It is a vector quantity, represented by B.

Magnetic Lines of Force

Lines drawn in a magnetic field along which a north pole of a compass needle would move.

Properties of Magnetic Lines of Force

Originate from the North pole and end at the South pole.
Close near the poles, widely separated elsewhere.
Do not intersect.

Types of Magnetic Field

Uniform Magnetic Field: Magnitude and direction are same everywhere. Represented by equidistant parallel lines.
Non-Uniform Magnetic Field: Magnitude and direction vary at different points.

Plotting Magnetic Lines of Force

Methods:

Using iron fillings
Using a compass needle

Activity 1: Magnetic field using iron fillings

Materials: Iron fillings, bar magnet, white sheet, sprinkler, cardboard.

Observation: Iron fillings concentrate at poles forming a pattern indicating magnetic field strength.

Conclusion: Strength is maximum at poles; pattern shows magnetic field lines.

Activity 2: Magnetic field using compass needle

Materials: Bar magnet, compass needle, white sheet, pencil, cardboard.

Conclusion: Field lines diverge from North pole, converge at South pole; form closed loops.

Sources of Magnetic Field

Natural magnets
Current-carrying conductor
Moving charge
Changing electric field

Magnetic Effect of Electric Current

Discovered by Oersted in 1820. A current-carrying conductor behaves like a magnet.

Activity 3: Magnetic effect of current through straight conductor

Observation: Compass needle deflects; deflection depends on current magnitude, direction, and distance from wire.

Conclusion: B ∝ I, B ∝ 1/r

Rules for Determining Direction of Magnetic Field

Snow Rule: If current flows South → North, North pole deflects west; North → South, deflects east.
Ampere’s Swimming Rule: Imagine swimming along current, entering feet, leaving head; North pole deflects left.

Magnetic Field Patterns

Straight Conductor: Concentric circles; right-hand thumb rule determines direction
Circular Loop: Concentric near wire, parallel at center; Clock Face Rule determines polarity
Solenoid: Inside uniform & strong; outside non-uniform & weak; behaves like bar magnet

Electromagnets

Coil of insulated wire on soft iron core. Strength depends on current, number of turns, air gap, and core material.

Uses: Electric bells, fans, motors, separating iron, lifting heavy objects.

Force on Current-Carrying Conductor

Fleming’s Left Hand Rule: Thumb → force, Forefinger → field, Middle finger → current. Force magnitude: F = BIL sinθ

Electric Motor

Converts electrical energy → mechanical energy. Works on magnetic effect of current. Uses commutator and brushes for continuous rotation.

Electromagnetic Induction

Faraday & Joseph Henry, 1831: Moving magnet/coil induces current. Direction by Fleming’s Right Hand Rule.

Electric Generators

Converts mechanical energy → electrical energy. Types: DC & AC generators. Principle: Electromagnetic induction.

Direct Current (DC)

Current flows in one direction. Sources: Batteries, dry cells, solar cells. Uses: Electroplating, charging batteries, power supply for devices.

Alternating Current (AC)

Current changes direction periodically. Frequency: 50 Hz. Sources: Power stations (hydro, thermal, nuclear, wind, geothermal). Uses: Home appliances, industrial motors, trains. Advantages: Efficient, low-cost, easily converted. Disadvantages: Dangerous, not used for electrolysis.

Domestic Electric Circuit

Separate circuits for lighting (5A fuse) & power (15A fuse)
Appliances connected in parallel with switches in live wire
Earth wire protects against electric shock

Earthing

Connection of metallic body of device to earth. Functions: Provide low-resistance path for leakage, maintain zero potential.

Electric Fuse

Safety device; melts when current exceeds safe value. Works on heating effect of current.

Short Circuiting & Overloading

Short Circuit: Live & neutral wire touch → large current → fire risk. Overloading: Too many appliances draw excess current → overheating → fire.

Dangers & Precautions

Touching live wire → electric shock
Short-circuit/overload → fire
Precautions: Turn off main switch, insulated support, earth appliances

Life Processes Practice Exercises

Practice Exercise – 1

Name any two life processes.
What is the source of energy for living beings?
What process in plants is known as transpiration?
Define nutrition.
Give examples of organisms having autotrophic nutrition.
Give examples of organisms having heterotrophic nutrition.
Name the mode of nutrition in amoeba.
What are enzymes?
What types of chemical reactions take place in our body to break down the food materials?
In which forms are unused carbohydrates stored in plants?
Name the end products of photosynthesis in a plant.
Which type of plant takes up CO₂ at night?
What are stomata?
What is the function of stomata?
Name the tissue which transports soluble products of photosynthesis in a plant.
What are the functions of saliva?
Name the tissue which transports water and minerals in a plant.
What is the function of bile juice in digestion?
Where are villi present and what are their functions?
What is respiration?
What is the role of saliva in the digestion of food?
Where does anaerobic respiration take place?
Out of photosynthesis and respiration in plants, which process occurs all the time? Only at daytime?
Name two organisms respiring anaerobically.
Why is anaerobic respiration less efficient?
What are the end products of respiration?
What is breathing?
From where do the following obtain oxygen for breathing and respiration?
• Aquatic animals
• Terrestrial animals
What is transportation?
What is transpiration?
What is the line of tiny air-sacs at the end of smallest bronchioles in the lungs?
What makes red blood corpuscles red in colour?
Name the cell organelle in which respiration takes place.
Why the wall of trachea does not collapse when there is less air?
Name one circulatory fluid other than blood.
Name the functional and structural unit of kidney.
How many times do you breathe in and out in a minute?

Practice Exercise – 2

Write one function each of the following components of the transport system in human beings.
Give reasons why the rate of breathing in aquatic organisms is much faster than terrestrial organisms.
Write any three differences between aerobic and anaerobic respiration.
What happens to the carbohydrate (like glucose) synthesized by a plant which is not used immediately?
Why do plants need nitrogen? How do plants obtain nitrogen?
How does paramecium obtain its food?
Explain why animals such as amphibians and many reptiles can tolerate some mixing of oxygenated and deoxygenated blood.
If we observe the cross-section of a leaf under a microscope, we notice green dots. What are these green dots called and what do they contain?
Which contains more carbon dioxide: exhaled air or inhaled air? Why?
Describe the process of respiration in the leaves of a plant.
How does fish obtain oxygen for breathing and respiration?
How are the lungs designed in human beings to maximize the area for exchange of gases?
Why does the human heart have four different chambers?
Describe the circulatory system in a fish. Why is it called single circulation?
(a) A certain tissue in a green plant got blocked and the leaves coiled. Which tissue was blocked?
(b) A certain tissue in a green plant got blocked and the roots were deprived of food. Which tissue got affected?
What is hypertension? How is it caused? What damage can it do?
What are the various types of heterotrophic nutrition? Give examples of each type.
Describe the process of nutrition in amoeba. Draw labelled diagrams to show the steps.
What substances are contained in gastric juice? What are their functions?
Describe the exchange of gases which takes place in the leaves of a plant during day time and at night.
Write a short note on blood vessels in our body.
What is meant by blood pressure? What are the two factors used to express blood pressure? How much is the normal blood pressure of a person?
Write a short note on lymph.
Mention the common characteristic of the respiratory organs of different animals.
Name the digestive juice that has no enzymes. What is the role of this juice?
What do the following transport:
Xylem, Phloem, Pulmonary vein, Vena cava

HEREDITY & EVOLUTION Download

Solved and Practice Important Questions

✔ Solved Questions

Q1. What are traits?

The characters in an individual that have alternate forms are called traits.

Q2. Who is known as the father of genetics?

Gregor Johann Mendel is known as the father of genetics.

Q3. Define chromosome.

Chromosome bearing genes are thread-like structures enclosed within a nucleus. Chromosome is composed of DNA and proteins.

Q4. What is heredity?

The transmission of traits from parents to their offspring is called heredity.

Q5. Name the plant on which Mendel performed his experiments.

Garden Pea (Pisum sativum).

Q6. What is the function of a gene?

Gene is the carrier of genetic information from one generation to the next.

Q7. Who proposed the theory of natural selection?

Charles Darwin proposed the theory of natural selection.

Q8. Which chromosomes are responsible for sex determination?

Sex chromosomes are responsible for sex determination.

Q9. Name the two types of chromosomes found in an organism.

Autosomes and sex chromosomes.

Q10. What is a sex chromosome?

Chromosomes that are associated with sex determination are called sex chromosomes.

Q11. Define mutation.

Sudden changes in the genetic form of an organism passed to the next generation are called mutations. Mutations lead to variation.

Q12. What are fossils?

Fossils are remains or impressions of dead plants and animals that lived in the past.

Q13. What are the causes of variations?

Offspring inherit traits from both parents, and mutations in genes or chromosomes also cause variations.

Q14. Why is variation beneficial to a species?

Variations help species survive under changing environmental conditions by allowing some individuals to adapt better.

Q15. What is natural selection?

Darwin’s theory states that individuals better adapted to their environment survive and reproduce — this is natural selection.

Q16. Define: Genetics, Sex Chromosomes, Analogous Organs.

Genetics: Study of heredity and variation.
Sex Chromosomes: Chromosomes involved in determining sex.
Analogous Organs: Organs with similar functions but different structures.

Q17. What do you understand by evolution?

Evolution is the gradual development of organisms from pre-existing forms through genetic changes over time.

🧩 Practice Exercise – 1

Try answering the following questions to test your understanding:

Q1. What is a gene?
Q2. Give one point difference between heredity and variation.
Q3. What is the effect of imperfect DNA copying?
Q4. Name extinct organisms studied from fossils.
Q5. Will geographical isolation affect speciation in animals?
Q6. What is the genotypic ratio when TT × tt?
Q7. What is the term for structures showing common origin?
Q8. F2 generation shows 3:1 ratio — what kind of cross is it?
Q9. Define heredity.
Q10. What links one generation to the next?
Q11. What are fossils? What do they tell about evolution?
Q12. Differentiate between analogous and homologous organs.
Q13. Define variation. Why is it beneficial?
Q14. Differentiate between dominant and recessive traits.
Q15. Explain why sex of a child is determined by the father.
Q16. Difference between acquired and inherited traits.
Q17. Explain Mendel’s monohybrid cross with diagram.
Q18. What is sex determination? Explain mechanism in humans.
Q19. Define speciation with an example.
Q20. What are fossils? How can we find their age?
Q21–27. Revise all above key questions.

CONTROL & COORDINATION Download

Solved & Practice Important Questions

✔ Solved Questions

Q1. What is a stimulus?

Stimulus is that substance which affects the body of living beings due to which it activates.

Q2. What are the three parts of a neuron?

Neuron consists of three main parts: (i) Cell body (ii) Dendrites (iii) Axon.

Q3. Name one animal in which the nervous system is in the form of a network of nerve cells.

Hydra has a nervous system in the form of a network of nerve cells.

Q4. What is a ganglion?

In invertebrates, nerve cells aggregate to form masses called ganglia (singular – ganglion).

Q5. Name the part of skull inside which the brain is located.

The brain is located inside the cranium, which is a part of the skull, covered by three protective membranes called meninges.

Q6. Which is the largest part of the brain?

Cerebrum is the largest part of the brain.

Q7. Which part of the brain maintains posture and equilibrium?

Cerebellum maintains posture and equilibrium of the body.

Q8. Which lobe of cerebrum is concerned with auditory reception?

Temporal lobe is concerned with auditory (hearing) reception.

Q9. What are the two types of nerves in the peripheral nervous system?

(a) Cranial nerves – 12 pairs
(b) Spinal nerves – 31 pairs.

Q10. What is reflex action?

A sudden and automatic response to a stimulus is called reflex action.

Q11. What are the components of a reflex arc?

Receptor → Sensory nerve → Spinal cord → Motor nerve → Effector.

Q12. What is a hormone?

Hormones are chemical substances secreted by endocrine glands that regulate physiological processes such as growth and metabolism.

Q13. Mention three characteristics of a hormone.

1. Secretions of endocrine glands.
2. Act as chemical messengers.
3. Secreted in trace amounts and act on specific organs.

Q14. What is the other name of endocrine glands?

Endocrine glands are also called ductless glands.

Q15. Name hormones that increase and decrease blood sugar level.

Insulin decreases blood sugar level.
Glucagon increases blood sugar level.

Q16. What are functions of corticoids?

Corticoids (from adrenal glands) regulate heart rate, blood pressure, metabolism, and mineral balance.

Q17. Who was the first to isolate auxins?

F. W. Went (1928)

Q18. What is foolish seedling?

A disease in rice plants causing excessive elongation is called the foolish seedling disease.

Q19. What is the reason for foolish seedling?

It is caused by a fungus Gibberella fujikuroi which secretes gibberellin.

Q20. What is parthenocarpy?

The process of producing seedless fruits by applying auxins and gibberellins.

🧩 Practice Exercise – 1

Answer the following to strengthen your understanding:

What type of coordination is present in plants?
What are movements of locomotion?
Define curvature movement.
Define paratonic and autonomic movements.
What is nyctinasty?
What is haptonasty?
Define nastic and tropic movements.
Define phototropism and geotropism.
What are auxins and gibberellins?
Why is abscisic acid called the stress hormone?
Write a note on phototropism and geotropism.
Which hormone lowers blood glucose in humans?
Differentiate between involuntary and reflex actions.

🧠 Practice Exercise – 2

Try these application-level questions:

What are the short and long fibres of a neuron called?
Name the junction between two neurons.
The gland X secretes hormone Y causing dwarfism if deficient. Identify X and Y.
Which part of the brain controls voluntary actions?
Which hormone is called the emergency hormone?
Which gland performs dual functions?
What disease is caused due to insulin deficiency?
Explain the reflex arc with a labelled diagram.
What is the function of thyroxine and why use iodised salt?
What are the main functions of the pituitary gland?

LIFE PROCESSES Download

TEST - 1

Q1. Who discovered the cell? 1M

Q2. Why is the cell called the structural and functional unit of life? 1M

Q3. Where are proteins synthesized inside the cell? 1M

Q4. How do substances like CO₂ and water move in and out of the cell? 2M

Q5. Define the terms: 2M

(a) Diffusion

(b) Osmosis

Q6. What do you know about cell membrane? 2M

Q7. Write the different functions of a cell? 2M

Q8. Differentiate between prokaryotic and eukaryotic cell? 2M

Q9. Define the terms: 3M

(a) Protoplasm

(b) Cytoplasm

Q10. Differentiate between plant cell and animal cell? 3M

TEST - 2

Q1. Name the three most important parts of a cell? 1M

Q2. What is the store house of all cellular information? 1M

Q3. What do you mean by ATP? 1M

Q4. Name the two nucleic acids present in the cell. What are their functions? 2M

Q5. What is the significance of osmosis? 2M

Q6. What are ribosomes? Write their functions in a cell? 2M

Q7. Explain the terms nucleus, nucleolus, and nucleoplasm. 2M

Q8. Differentiate between plasmolysis and deplasmolysis. 2M

Q9. Give the functions of mitochondria, ribosomes, lysosomes. 3M

Q10. Explain the cell theory. 3M

Q11. Describe the structure of a typical plant cell. 5M

TEST - 3

Q1. Name the respiratory organ in: 1M

(a) Human beings

(b) Fish

(d) Insects

Q2. What is the special name of contraction and expansion movement which pushes the food further in our digestive tract? 1M

Q3. Define translocation in plants. 1M

Q4. What are the main organs of the excretory system of our body? 1M

Q5. Why is diffusion insufficient to meet the oxygen requirements of large multicellular organisms like humans? 2M

Q6. State the functions of blood in our body. 2M

Q7. Write a short note on photosynthesis. 3M

Q8. How are water and minerals transported in plants? 3M

Q9. Draw a labelled diagram of human respiratory system. 3M

Q10. Describe the process of digestion of food in humans with the help of a well-labelled diagram. 3M

TEST - 4

Q1. In which part of the digestive system is water absorbed? 1M

Q2. In which type of respiration, aerobic or anaerobic, more energy is released? 1M

Q3. Which side of the heart pumps blood into the lungs? 1M

Q4. Name the main nitrogenous waste in the human blood. 1M

Q5. What is hypertension? Why is it caused? What harm can it do? 2M

Q6. Differentiate between transport of materials in xylem and phloem. 2M

Q7. Describe the process of respiration in the following parts of plant: 3M

(i) Root

(ii) Stem

(iii) Leaves

Q8. What happens to the energy produced during respiration? 3M

Q9. What are the functions of liver and pancreas in the digestive system? 3M

Q10. Compare the functioning of alveoli in the lungs and nephrons in the kidney with respect to their structure and function. 3M

Test Papers

Chemical Reactions and Equations - Solved Intext Exercises

Q1. Why should a magnesium ribbon be cleaned before burning in air?

Magnesium ribbon should be cleaned before burning in air to remove the dust and layer of magnesium oxide formed on it.

Q2. Write the balanced equation for the following chemical reactions:

Hydrogen + Chlorine → Hydrogen Chloride
Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
Sodium + Water → Sodium hydroxide + Hydrogen

H2 + Cl2 → 2HCl
3BaCl2 + Al2(SO4)3 → 3BaSO4 + 2AlCl3
2Na + 2H2O → 2NaOH + H2

Q3. Write a balanced chemical equation with state symbols for the following reactions:

Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

BaCl2 (aq) + Na2SO4 (aq) → BaSO4 ↓ + 2NaCl (aq)
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)

Q4. A solution of a substance ‘X’ is used for white-washing.

Name the substance ‘X’ and write its formula.
Write the reaction of the substance ‘X’ named in (i) above with water.

Calcium oxide (quick lime), CaO
CaO (s) + H2O (l) → Ca(OH)2 (aq)

Q5. Why is the amount of gas collected in one of the test tubes in activity 1.7 double of the amount collected in the other? Name this gas.

When water is decomposed in activity 1.7, hydrogen and oxygen gas is formed:
2H2O → 2H2 + O2
According to the law of conservation of mass, when two molecules of water decompose, 2 molecules of hydrogen gas and 1 molecule of oxygen gas are produced. Thus, the amount of hydrogen gas produced is double the amount of oxygen produced. Hence the amount of hydrogen gas is double in one of the tubes.

Chemical Reactions and Equations - More Solved Exercises

Q6. Why does the blue colour of copper sulphate solution change when iron is added?

The blue colour of copper sulphate solution changes because iron is above copper in the reactivity series. Hence, it displaces copper from its solution of copper sulphate and iron sulphate is formed, which is green in colour.

Q7. Give an example of a double displacement reaction other than the one given in Activity 1.10.

AgNO3 (aq) + NaCl (aq) → AgCl ↓ + NaNO3 (aq)

When silver nitrate solution is added to sodium chloride solution, the white precipitate of silver chloride is formed instantaneously. This is a double displacement reaction.

Q8. Identify the substances that are oxidised and the substances that are reduced in the following reactions:

2Na (s) + O2 (g) → 2Na2O (s)
CuO (s) + H2 (g) → Cu (s) + H2O (l)

Na is oxidised to Na2O.
CuO is reduced to Cu.
H2 is oxidised to H2O.

Solved NCERT Exercises

Q1. Which of the statements about the reaction below are incorrect?

2PbO(s) + C(s) → 2Pb(s) + CO2 (g)

Statements: (a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidised.
(c) Carbon is getting oxidised.
(d) Lead oxide is getting reduced.

Incorrect statements: (a) and (b)

Q2. Fe2O3 + 2Al → Al2O3 + 2Fe
The above reaction is an example of a:

(a) combination reaction
(b) double displacement reaction
(c) decomposition reaction
(d) displacement reaction

Correct answer: (d) displacement reaction

Chemical Reactions and Equations - Additional Exercises

Q2. The reaction Fe2O3 + 2Al → Al2O3 + 2Fe is an example of which type of reaction?

(d) displacement reaction.

Q3. What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.

(a) Hydrogen gas and iron chloride are produced.
(b) Chlorine gas and iron hydroxide are produced.
(c) No reaction takes place.
(d) Iron salt and water are produced.

(a) Hydrogen gas and iron chloride are produced.

Q4. What is a balanced chemical equation? Why should chemical equations be balanced?

A balanced equation has an equal number of atoms of different elements in the reactants and products. Chemical equations are balanced to satisfy the law of conservation of mass in chemical reactions, since matter can neither be created nor destroyed in a chemical reaction.

Q5. Translate the following statements into chemical equations and then balance them:

(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

H2 (g) + N2 (g) → NH3 (g)
3H2 (g) + N2 (g) → 2NH3 (g)

H2S (g) + O2 (g) → SO2 + H2O
2H2S (g) + 3O2 (g) → 2SO2 + 2H2O

BaCl2 (aq) + Al2(SO4)3 (aq) → BaSO4 ↓ + AlCl3 (aq)
3BaCl2 (aq) + Al2(SO4)3 (aq) → 3BaSO4 ↓ + 2AlCl3 (aq)

K (s) + H2O → KOH (aq) + H2 ↑
2K (s) + 2H2O → 2KOH (aq) + H2 ↑

Q6. Balance the following chemical equations:

HNO3 + Ca(OH)2 → Ca(NO3)2 + H2O
NaOH + H2SO4 → Na2SO4 + H2O
NaCl + AgNO3 → AgCl + NaNO3
BaCl2 + H2SO4 → BaSO4 + HCl

2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O
2NaOH + H2SO4 → Na2SO4 + 2H2O
NaCl + AgNO3 → AgCl + NaNO3
BaCl2 + H2SO4 → BaSO4 + 2HCl

Chemical Reactions and Equations - Further Exercises

Q7. Write the balanced chemical equations for the following reactions:

Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
Zinc + Silver nitrate → Zinc nitrate + Silver
Aluminium + Copper chloride → Aluminium chloride + Copper
Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride

Ca(OH)2 + CO2 → CaCO3 + H2O

Zn + AgNO3 → Zn(NO3)2 + Ag
Zn + 2AgNO3 → Zn(NO3)2 + 2Ag

Al + CuCl2 → AlCl3 + Cu
2Al + 3CuCl2 → 2AlCl3 + 3Cu

BaCl2 + K2SO4 → BaSO4 + KCl
BaCl2 + K2SO4 → BaSO4 + 2KCl

Q8. Write the balanced chemical equation for the following and identify the type of reaction in each case:

Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)
Zinc carbonate(s) → Zinc oxide(s) + Carbon dioxide(g)
Hydrogen(g) + Chlorine(g) → Hydrogen chloride(g)
Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen(g)

KBr (aq) + BaI2 (aq) → KI (aq) + BaBr2 (s)
2KBr (aq) + BaI2 (aq) → 2KI (aq) + BaBr2 (s)
Type: Double displacement reaction

ZnCO3 (s) → ZnO (s) + CO2 (g)
Type: Decomposition reaction

H2 (g) + Cl2 (g) → HCl (g)
H2 (g) + Cl2 (g) → 2HCl (g)
Type: Combination reaction

Mg (s) + HCl (aq) → MgCl2 (aq) + H2 (g)
Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)
Type: Displacement reaction

Q9. What does one mean by exothermic and endothermic reactions? Give examples.

Exothermic reactions: Chemical reactions which occur with the evolution of heat.
i.e. Reactants → Products + Heat
Examples:

C (s) + O2 (g) → CO2 (g) + Heat
2H2 (g) + O2 (g) → 2H2O (g) + Heat

Endothermic reactions: Chemical reactions which occur with the absorption of heat.
i.e. Reactants → Products – Heat or Reactants + Heat → Products
Example:

N2 (g) + O2 (g) → 2NO (g) – Heat

Chemical Reactions and Equations - Additional Exercises

Q10. Why is respiration considered an exothermic reaction? Explain.

Respiration is considered an exothermic reaction because in this reaction, glucose produced by digestion of food combines with oxygen in the cells of our body and produces energy.

Q11. Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.

Decomposition reactions are called the opposite of combination reactions because:

In a combination reaction, two or more substances combine to form a new substance. Example: Magnesium combines with oxygen to give magnesium oxide:
```
2Mg + O2 → 2MgO
```
In a decomposition reaction, a single compound breaks down to produce two or more simpler substances. Example: Calcium carbonate on heating decomposes to calcium oxide and carbon dioxide:
```
CaCO3   heat→  CaO + CO2
```

Q12. Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.

Heat: Calcium carbonate → Calcium oxide + Carbon dioxide
```
CaCO3 (s)   heat→  CaO (s) + CO2 (g)
```

Electricity: Water → Hydrogen + Oxygen

2H2O   electric current→  H2 (g) + O2 (g)

Light: Silver chloride → Silver + Chlorine
```
2AgCl   sunlight→  2Ag + Cl2
```

Q13. What is the difference between displacement and double displacement reactions? Write equations for these reactions.

Displacement reactions: In these reactions, a more active element displaces or removes another element from a compound. Examples:

Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
Pb (s) + CuSO4 (aq) → PbSO4 (aq) + Cu (s)

Double displacement reactions: In these reactions, two different atoms or groups of atoms are exchanged between compounds. Example:

BaCl2 (aq) + Na2SO4 (aq) → BaSO4 + 2NaCl

Here, SO4²⁻ ions displace Cl⁻ ions and vice versa.

Q14. In the refining of silver, the recovery of silver from silver nitrate solution involves displacement by copper metal. Write down the reaction involved.

2AgNO3 (aq) + Cu (s) → Cu(NO3)2 (aq) + 2Ag (s)

Q15. What do you mean by a precipitation reaction? Explain by giving examples.

A precipitation reaction is a chemical reaction in which an insoluble solid (precipitate) is formed when two aqueous solutions are mixed. Example:

BaCl2 (aq) + Na2SO4 (aq) → BaSO4 ↓ + 2NaCl (aq)

Here, barium sulphate (BaSO4) is the precipitate.

Chemical Reactions and Equations - Exercises Q16–Q20

Q16. Explain the following in terms of gain or loss of oxygen with two examples each: Oxidation and Reduction.

Oxidation: It is the process in which oxygen is added. Examples:

Addition of oxygen to sulphur: S (s) + O2 (g) → SO2 (g)
Addition of oxygen to magnesium: 2Mg (s) + O2 (g) → 2MgO (s)

Reduction: It is the process in which oxygen is removed. Examples:

CuO + H2  heat→  Cu    (Copper(II) oxide loses oxygen and is reduced to copper)
MnO2 + 4HCl → 2H2O + Cl2   (MnO2 is reduced to MnCl2)

Q17. A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.

Element ‘X’ is copper (Cu). Black coloured compound is Copper (II) oxide (CuO).

Q18. Why do we apply paint on iron articles?

Iron articles are prevented from rusting by applying paint on them. Rusting weakens the iron articles.

Q19. Oil and fat containing food items are flushed with nitrogen. Why?

To prevent oxidation of oil and fat containing food items, they are flushed with nitrogen. Fats and oils become rancid and their smell and taste change when oxidised.

Q20. Explain the following terms with one example each: Corrosion and Rancidity.

Corrosion: When the surface of a metal is attacked by air, water, or other substances around it, it is said to corrode. Example: Iron gets rusted. Rust is mainly hydrated Iron (III) oxide, Fe2O3·xH2O. Rusting weakens the structure of iron bodies. Rancidity: When fats and oils are oxidised, they become rancid and their smell and taste change. Example: Bags of potato chips are flushed with nitrogen to prevent oxidation. Oxidation of fats and oils can also be prevented by adding antioxidants or keeping food in air-tight containers.

Acids, Bases and Salts - Solved Intext Exercises

Q1. You have been provided with three test tubes. One contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?

We put red litmus paper in all the three solutions. The solution which turns red litmus paper blue is the basic solution. In the rest of the two solutions, we put the blue litmus paper obtained above. The solution which turns blue litmus back into red litmus is acidic. The third solution left is distilled water.

Q2. Why should curd and sour substances not be kept in brass and copper vessels?

Curd and sour substances are acidic in nature. Acids react with brass and copper as they are metals and produce salts which are poisonous. Therefore, curd and sour substances should not be kept in brass and copper vessels.

Q3. Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?

Hydrogen gas is usually liberated when an acid reacts with a metal. Example:

Zn (s) + H2SO4 (aq) → ZnSO4 + H2 ↑

The presence of hydrogen gas is tested by bringing a burning candle or splinter near the gas. If the gas burns with a pop sound, it is hydrogen gas.

Q4. A metal compound A reacts with dilute hydrochloric acid effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.

The metal compound A is calcium carbonate. The gas produced is carbon dioxide.

CaCO3 + 2HCl → CaCl2 + H2O + CO2

Q5. Why do HCl, HNO3, etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?

HCl, HNO3, etc., show acidic character in aqueous solutions because they produce hydrogen ions, H+(aq), in solution. Alcohol and glucose do not produce hydrogen ions, H+(aq), hence they do not show acidic character.

Q6. Why does an aqueous solution of an acid conduct electricity?

Aqueous solution of an acid conducts electricity because they produce hydrogen ions [H+(aq)] which are responsible for electric current to flow.

Q7. Why does dry HCl gas not change the colour of the dry litmus paper?

Dry HCl gas does not produce H+ ions because separation of H+ ions from HCl molecules cannot occur in the absence of water. Hence, the colour of dry litmus paper does not change.

Q8. While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?

Water should never be added to concentrated acid because when water is added to acid, the heat produced is so large that it may cause the mixture to splash out and burn the person standing near it.

Q9. How is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted?

The concentration of hydronium ions (H3O+) per unit volume decreases.

Q10. How is the concentration of hydroxide ions (OH–) affected when excess base is dissolved in a solution of sodium hydroxide?

The concentration of hydroxide ions (OH–) per unit volume increases.

Q11. You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of these is acidic and which one is basic?

Solution A has more hydrogen ion concentration than solution B. Solution A with pH 6 is acidic, and solution B with pH 8 is basic.

Q12. What effect does the concentration of H+(aq) ions have on the nature of the solution?

Concentration of H+(aq) ions makes the solution acidic. The higher the concentration, the stronger the acid.

Q13. Do basic solutions also have H+(aq) ions? If yes, then why are these basic?

Yes, basic solutions also have H+(aq) ions. They are basic because the concentration of hydroxide (OH–) ions is much more than the concentration of H+(aq) ions.

Q14. Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (CaO), slaked lime (Ca(OH)2), or chalk (CaCO3)?

A farmer would treat the soil with these substances when the soil acidity increases, i.e., when the amount of acid exceeds the required level.

Q15. What is the common name of the compound CaOCl2?

Bleaching powder.

Q16. Name the substance which on treatment with chlorine yields bleaching powder.

Dry slaked lime [Ca(OH)2].

Q17. Name the sodium compound which is used for softening hard water.

Sodium carbonate (Na2CO3·10H2O).

Q18. What will happen if a solution of sodium hydrocarbonate is heated? Give the equation of the reaction involved.

Carbon dioxide gas is given out.

2NaHCO3   heat→   Na2CO3 + H2O + CO2

Acids, Bases and Salts - Solved NCERT Exercises

Q19. Write an equation to show the reaction between Plaster of Paris and water.

CaSO4·½H2O + 1½H2O → CaSO4·2H2O

Q1. A solution turns red litmus blue, its pH likely to be

(d) 10

Q2. A solution reacts with crushed eggshells to give a gas that turns lime-water milky. The solution contains

(b) HCl. Eggshells are made of calcium carbonate and the gas is CO2.

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

Q3. 10 mL of NaOH is neutralised by 8 mL of HCl. If 20 mL of NaOH is used, how much HCl is needed?

(d) 16 mL

Q4. Which one of the following types of medicines is used for treating indigestion?

Q5. Write word equations and then balanced equations for the reaction taking place when:

(a) Zinc + dilute H2SO4 → Zinc sulphate + H2 ↑

Zn + H2SO4 → ZnSO4 + H2 ↑

(b) Magnesium + dilute HCl → Magnesium chloride + H2 ↑

Mg + 2HCl → MgCl2 + H2 ↑

2Al + 3H2SO4 → Al2(SO4)3 + 3H2 ↑

(d) Iron + dilute HCl → Iron chloride + H2 ↑

Fe + 2HCl → FeCl2 + H2 ↑

Q6. Compounds such as alcohols and glucose contain hydrogen but are not acids. Describe an activity to prove it.

Set two nails on a cork in a 100 mL beaker. Connect the nails to a 6V battery through a bulb and switch. Pour glucose or alcohol solution into the beaker and switch on the current. Bulb does not glow. Hence, alcohol and glucose are not acids because they do not conduct electricity.

Q7. Why does distilled water not conduct electricity, whereas rainwater does?

Distilled water does not conduct electricity because it is neutral and does not generate H+(aq) ions. Rainwater contains ions and conducts electricity.

Q8. Why do acids not show acidic behaviour in the absence of water?

Acids do not show acidic behaviour without water because separation of H+ ions cannot occur. Acids show acidic behaviour only in the presence of water.

Q9. Five solutions A, B, C, D, E showed pH 4, 1, 11, 7, 9 respectively. Identify:

(a) Neutral – Solution D (pH 7) (b) Strongly alkaline – Solution C (pH 11) (c) Strongly acidic – Solution B (pH 1) (d) Weakly acidic – Solution A (pH 4) (e) Weakly alkaline – Solution E (pH 9) pH in increasing order of H+ concentration: 11 < 9 < 7 < 4 < 1 Solution C < Solution E < Solution D < Solution A < Solution B

Q10. Equal lengths of magnesium ribbons are taken in test tubes A and B. HCl added to A, CH3COOH added to B. In which test tube will fizzing occur more vigorously and why?

Fizzing occurs more vigorously in test tube A with HCl because it is a stronger acid than acetic acid.

Q11. Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain.

The pH of milk will drop below 6 as it turns into curd because curd is more acidic than milk.

Q12. A milkman adds a very small amount of baking soda to fresh milk.

(a) He shifts the pH from 6 to slightly alkaline so the milk takes longer to get sour. (b) Slightly alkaline milk takes longer to set into curd because lactic acid formed during curdling first neutralizes the alkali present.

Q13. Plaster of Paris should be stored in a moisture-proof container. Explain why.

If Plaster of Paris comes in contact with water, a slurry is formed which solidifies rapidly into a hard crystalline mass of gypsum.

CaSO4·½H2O + 1½H2O → CaSO4·2H2O

Q14. What is a neutralisation reaction? Give two examples.

A reaction between an acid and a base to give a salt and water is called a neutralisation reaction. Examples:

NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
KOH (aq) + H2SO4 (aq) → K2SO4 (aq) + 2H2O (l)

Q15. Give two important uses of washing soda and baking soda.

Uses of Washing soda: - Manufacture of glass, soap, paper - Removing permanent hardness in water - Cleaning agent for domestic purposes - Laundry as washing soda Uses of Baking soda: - Ingredient in antacids - Additive in foods and drinks - Fire extinguisher

Metals and Non-metals - Solved Intext Exercises

Q1. Give an example of a metal which:

(i) is a liquid at room temperature: Mercury
(ii) can be easily cut with a knife: Sodium
(iii) is the best conductor of heat: Silver and Copper
(iv) is the poorest conductor of heat: Lead and Mercury

Q2. Explain the meaning of ‘malleability’ and ‘ductility’ of a metal.

Malleability: The property of a metal by virtue of which it can be beaten into thin sheets when hammered.
Ductility: The property of a metal by virtue of which it can be drawn into very thin wires.

Q3. Why is sodium kept immersed in kerosene oil?

Sodium is highly reactive and reacts rapidly with oxygen in the air. It is therefore kept in kerosene to prevent contact with air.

Q4. Write equations for the reaction of:

(a) Iron with steam:

3Fe (s) + 4H2O (g) → Fe3O4 (s) + 4H2 (g)

(b) Potassium with water:

2K (s) + 2H2O (l) → 2KOH (aq) + H2 (g)

(c) Calcium with water:

Ca (s) + 2H2O (l) → Ca(OH)2 (s) + H2 (g)

Q5. Samples of four metals A, B, C and D were tested with solutions of different metal salts. Use the table to answer questions.

Metal	Iron(II) Sulphate	Copper(II) Sulphate	Zinc Sulphate	Silver Nitrate
A	No reaction	Displacement
B	Displacement		No reaction
C	No reaction	No reaction	No reaction	Displacement
D	No reaction	No reaction	No reaction	No reaction

(a) Most reactive metal: B
(b) Observation when B is added to CuSO4: B displaces copper from CuSO4 solution.
(c) Metals in decreasing order of reactivity: B > A > C > D

Metals and Non-metals - Solved Exercises (Continued)

Q5. Explanation of observations from metal reactivity table

(a) A can displace Cu, B can displace Fe, C can displace Ag. As order of reactivity of Cu, Fe and Ag is Fe > Cu > Ag, hence B is most reactive.
(b) As B is more reactive than copper, it will displace Cu from copper (II) sulphate solution. Hence, blue colour of CuSO4 will fade.
(c) B can displace Fe, A can displace Cu, C can displace Fe, A can displace Cu, C can displace Ag and D can displace none. Hence, order of decreasing reactivity: B > A > C > D.

Q6. Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H2SO4.

Hydrogen gas is produced.
Example with zinc:

Zn(s) + 2HCl (dil) → ZnCl2 (aq) + H2 (g)

Q7. Observation when zinc is added to iron (II) sulphate solution

Zinc will displace iron from iron (II) sulphate solution, forming zinc sulphate. The light green colour of FeSO4 disappears and solution becomes colourless.
Reaction:

Zn (s) + FeSO4 (aq) → ZnSO4 (aq) + Fe (s)

Q8. (i) Electron-dot structures for sodium, oxygen, and magnesium

Sodium (Na): 2,8,1 → Na•
Oxygen (O): 2,6 → :O:
Magnesium (Mg): 2,8,2 → Mg••

Q8. (ii) Formation of N2O and MgO by transfer of electrons

Formation of Magnesium oxide (MgO):

Mg → Mg²⁺ + 2e⁻
O + 2e⁻ → O²⁻
Mg²⁺ + O²⁻ → MgO

Formation of Sodium oxide (Na2O):

2Na → 2Na⁺ + 2e⁻
O + 2e⁻ → O²⁻
2Na⁺ + O²⁻ → Na2O

Q8. (iii) Ions present in these compounds

Magnesium oxide: Mg²⁺ and O²⁻
Sodium oxide: Na⁺ and O²⁻

Metals and Non-metals - Solved Exercises (Continued)

Q9. Why do ionic compounds have high melting points?

In ionic compounds, the cations and anions are held together by strong electrostatic forces of attraction. To break these strong forces, a large amount of energy is required. Hence, ionic compounds have high melting points.

Q10. Define the following terms:

(i) Mineral: Naturally occurring chemical substances in the form in which metals occur in the earth along with impurities.
(ii) Ore: The mineral from which the metal can be extracted conveniently and economically.
(iii) Gangue: The earthy, sandy, and rocky impurities associated with minerals.

Q11. Name two metals which are found in nature in the free state.

Gold and Platinum

Q12. What chemical process is used for obtaining a metal from its oxide?

Metals are obtained from their oxides by reduction, either by heating with carbon or using highly reactive metals such as sodium or calcium.

Q13. Metallic oxides of zinc, magnesium and copper were heated with metals. In which cases will displacement reactions take place?

Magnesium lies above zinc while copper lies below it. Therefore, magnesium can displace zinc from zinc oxide:

ZnO (s) + Mg (s) → Zn (s) + MgO (s)

Magnesium lies above zinc and copper, so zinc and copper cannot displace magnesium from magnesium oxide. Both magnesium and zinc lie above copper in the activity series, so both can displace copper from copper oxide:

CuO (s) + Mg (s) → Cu (s) + MgO (s)

CuO (s) + Zn (s) → Zn (s) + ZnO (s)

Metals and Non-metals - Solved Exercises (Continued)

Q14. Which metals do not corrode easily?

Metals which lie low in the activity series and are highly unreactive such as Au (Gold), Ag (Silver), Pt (Platinum) are not attacked by atmospheric gases and moisture and hence do not corrode easily.

Q15. What are alloys?

Alloys are homogeneous mixtures of two or more metals, or a metal and a non-metal.
Example: Steel

Solved NCERT Exercises

Q1. Which of the following pairs will give displacement reactions?
(a) NaCl solution and copper metal.
(b) MgCl2 solution and aluminium metal.
(c) FeSO4 solution and silver metal.
(d) AgNO3 solution and copper metal.

Copper is more reactive than silver, so it displaces silver from silver nitrate solution. Hence, option (d) is correct. All other pairs do not show displacement reactions because the metals are less reactive than the ones to be displaced.

Q2. Which of the following methods is suitable for preventing an iron frying pan from rusting?
(a) Applying grease
(b) Applying paint
(c) Applying a coating of zinc
(d) All of the above

(d) All of the above

Q3. An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be:
(a) Calcium
(b) Carbon
(c) Silicon
(d) Iron

Calcium reacts with oxygen to form calcium oxide. Being an ionic compound, it has a high melting point and dissolves in water to form calcium hydroxide:

2Ca (s) + O2 (g) → 2CaO (s)
CaO (s) + H2O → Ca(OH)2 (aq)

The oxides of carbon (CO2), silicon (SiO2), and iron (FeO/Fe2O3) are either gases or insoluble solids. Hence, the correct option is (a) Calcium.

Q4. Food cans are coated with tin and not with zinc because:

Tin is less reactive than zinc and does not corrode easily, making it safer for coating food cans. Zinc is more reactive and may corrode, contaminating the food.

Metals and Non-metals - Solved Exercises (Continued)

Q4 (continued). Food cans are coated with tin and not with zinc because:
(b) Zinc has a higher melting point than tin.
(c) Zinc is more reactive than tin.
(d) Zinc is less reactive than tin.

Q5. You are given a hammer, a battery, a bulb, wires and a switch.
(a) How could you use them to distinguish between samples of metals and non-metals?
(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.

(a)

Using a hammer: If the sample breaks, it is brittle → non-metal. If the sample flattens into a sheet → metal (malleable).
Using a battery, bulb, wires, and switch: Place the sample between clips A and B and switch on. If the bulb glows → metal (good conductor), if not → non-metal.

(b) The tests are useful because:

Malleability test shows metals can be shaped into sheets, useful for roofing, making trunks, etc.
Conductivity test shows metals carry electricity, useful for wires and electrical applications.

Q6. What are amphoteric oxides? Give two examples of amphoteric oxides.

Amphoteric oxides are metal oxides that react with both acids and alkalis to produce salt and water.
Examples: Al2O3 and ZnO.

Reactions:

Al2O3 + 4HCl → 2AlCl3 + 2H2O
Al2O3 + 2NaOH → 2NaAlO2 + H2O
ZnO + 2HCl → ZnCl2 + H2O
ZnO + 2NaOH → Na2ZnO2 + H2O

Q7. Name two metals which will displace hydrogen from dilute acids, and two metals which will not.

Metals that displace hydrogen: Sodium and Calcium.
Metals that do not displace hydrogen: Copper and Silver.

Q8. In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?

Anode: Impure metal M
Cathode: Thin strip of pure metal M
Electrolyte: A solution of a salt of the metal M

Metals and Non-metals - Solved Exercises (Continued)

Q8 (continued). Electrolyte - A solution of metal (M) salt.

Q9. Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it.
(a) What will be the action of gas on
(i) dry litmus paper?
(ii) moist litmus paper?
(b) Write a balanced chemical equation for the reaction taking place.

(a) When sulphur is burnt, it produces sulphur dioxide (SO2) gas:

S (s) + O2 (g) → SO2 (g)
Sulphur     Oxygen     Sulphur dioxide

(i) There is no action of SO2 on dry litmus paper.
(ii) The gas dissolves in water present in moist litmus paper to form sulphurous acid (H2SO3), which turns blue litmus red:

SO2 + H2O (l) → H2SO3 (aq)
      Water        Sulphurous Acid

(b) The balanced chemical equation is given above.

Q10. State two ways to prevent the rusting of iron.

(i) Galvanization
(ii) Alloying, painting, oiling, greasing

Q11. What type of oxides are formed when non-metals combine with oxygen?

When non-metals react with oxygen, acidic oxides are formed, e.g.:

C + O2 → CO2
CO2 + H2O → H2CO3
       Carbonic Acid

Q12. Give reasons:

Metals and Non-metals - Solved Exercises (Continued)

Q12. Give reasons:

(a) Platinum, gold and silver are used to make jewellery.
(b) Sodium, potassium and lithium are stored under oil.
(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.
(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

Sol.
(a) Platinum, gold and silver are highly unreactive metals. Therefore, they do not undergo corrosion due to attack by moisture and atmospheric gases. So they are used to make jewellery.
(b) Lithium, sodium and potassium are readily attacked by O2 present in air to form their respective oxides. They also react with moisture present in air producing H2 gas which sometimes catches fire and causes accidents. To protect these metals from attack by O2 and moisture and to prevent accidental fires, lithium, sodium and potassium are stored under oil.
(c) Aluminium is a good conductor of heat but is highly reactive. When exposed to moist air, its surface is covered with a thin impervious layer of aluminium oxide (Al2O3). This layer prevents moist air from contacting the fresh metal and protects it from further corrosion. Hence aluminium is used to make cooking utensils.
(d) Carbonate and sulphide ores are typically converted into oxides during metal extraction because it is easier to reduce metal oxides to obtain pure metal compared to reducing the original carbonate or sulphide ores. This conversion makes the extraction process more efficient and economical.

Q13. You must have seen tarnished copper being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.

Sol. When copper is exposed to moist air, it gets tarnished or corroded by forming basic copper carbonate:

2Cu + CO2 (g) + O2 (g) + H2O (l) → CuCO3.Cu(OH)2
                         Basic Copper Carbonate (green)

Lemon juice contains citric acid and tamarind contains tartaric acid. When copper vessels are rubbed with lemon or tamarind juice, the weak acids dissolve the basic copper carbonate, restoring the original shine.

Q14. Differentiate between a metal and a non-metal on the basis of their chemical properties.

Sol.
Chemical properties – Metals:
- Metals combine with oxygen to form basic oxides.
- Reactive metals like Na, Mg, Zn, Fe, etc., dissolve in dilute acids to produce H2 gas.
- Metals (except Na, Ca) generally do not combine with hydrogen.
- Metals react with chlorine to form ionic metal chlorides.
- Metals react with hydrogen to form ionic metal hydrides.

Chemical properties – Non-Metals:
- Non-metals combine with oxygen to form acidic oxides or neutral oxides.
- Non-metals do not dissolve in dilute acids.
- Non-metals combine with hydrogen to form covalent non-metal hydrides.
- Non-metals react with chlorine to form covalent non-metal chlorides.

Metals and Non-metals - Solved Exercises (Continued)

Q15. A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?

Sol. The solution was aqua-regia or royal water. Aqua-regia is a freshly prepared solution of concentrated hydrochloric acid and nitric acid in the ratio 3:1 by volume. It dissolved a considerable amount of gold from the bangles, hence their weight reduced drastically. The man can recover the dissolved gold from the solution by suitable treatment.

Q16. Give reasons why copper is used to make hot water tanks and not steel (an alloy of iron).

Sol. Copper is a better conductor of heat than steel. To save heat losses, hot water tanks are made of copper instead of steel.

Carbon and its Compounds - Solved Intext Exercises

Q1. What would be the electron dot structure of carbon dioxide which has the formula CO2?

Sol. The electron dot formula of CO2 is given below:

Q2. What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur?

Sol. The electron dot structure of a molecule of sulphur which is made of eight atoms of sulphur is given below:

Q3. How many structural isomers can you draw for pentane? Also name them.

Sol. Three structural isomers can be drawn for pentane:

IUPAC Name: Pentane, Common Name: n-pentane
IUPAC Name: 2-methylbutane, Common Name: Iso-pentane
IUPAC Name: 2,2-dimethylpropane, Common Name: Neo-pentane

Carbon and its Compounds - Solved Intext Exercises (Continued)

Q7. Give one example each of:

Sol. (i) Bromoethane
(ii) Methanal
(iii) 1-Hexyne

Q8. Why is the conversion of ethanol to ethanoic acid an oxidation reaction?

Sol. The conversion of ethanol (CH3CH2OH) to ethanoic acid (CH3COOH) is called an oxidation reaction because oxygen is added to it during this conversion.
CH3CH2OH + 2[O] Alkaline KMnO4, Heat → CH3COOH + H2O

Q9. A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?

Sol. A mixture of ethyne and air is not used for welding because burning of ethyne in air produces a sooty flame due to incomplete combustion, which is not enough to melt metals for welding.

Q10. How would you distinguish experimentally between an alcohol and a carboxylic acid?

Sol. Take the samples of alcohol and carboxylic acid in different test tubes. The test tube from which effervescence evolves is carboxylic acid. The effervescence is due to the evolution of carbon dioxide.
CH3COOH + NaHCO3 → CH3COONa + H2O + CO2
C2H5OH + NaHCO3 → No evolution of CO2

Q11. What are oxidising agents?

Sol. An oxidising agent is one which oxidizes other substances by providing oxygen or removing hydrogen.

Carbon and its Compounds - Solved Intext Exercises (Continued)

Q12. What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?

Sol. The two properties of carbon are its tetravalency and the catenation which lead to the huge number of carbon compounds we see around us.

Q13. People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?

Sol. The molecules of soap are sodium or potassium salts of long-chain carboxylic acids. The ionic end of soap dissolves in water while the carbon chain dissolves in oil. The soap molecules thus form structures called micelles. When cloth is agitated, the micelles containing oily or greasy dirt particles get removed from the surface of dirty cloth and go into water, cleaning the clothes.

Solved NCERT Exercises

Q1. Ethane, with the molecular formula C2H6 has:
(a) 6 covalent bonds (b) 7 covalent bonds (c) 8 covalent bonds (d) 9 covalent bonds

Sol. (b)

Q2. Butanone is a four-carbon compound with the functional group:
(a) carboxylic acid (b) aldehyde (c) ketone (d) alcohol

Sol. (c)

Q3. While cooking, if the bottom of the vessel is getting blackened on the outside, it means that:
(a) the food is not cooked completely
(b) the fuel is not burning completely
(c) the fuel is wet
(d) the fuel is burning completely

Sol. (b)

Q4. Explain the nature of the covalent bond using the bond formation in CH3Cl.

Sol. Covalent bond is formed by sharing the electrons between two atoms.

Carbon and its Compounds - Solved Intext Exercises (Continued)

Q6. What is the homologous series? Explain with an example.

Sol. A homologous series is a group of organic compounds having similar structure and similar chemical properties but different physical properties, in which the successive compounds differ by a CH2 group.

Homologous Series of Alkanes: The general formula of the homologous series of alkanes is CnH2n+2. The first five members of the series are:

Name of Alkane	Number of C atoms (n)	Molecular Formula
Methane	1	CH4
Ethane	2	C2H6
Propane	3	C3H8
Butane	4	C4H10
Pentane	5	C5H12

Q7. How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?

Sol. Physical Properties:
Ethanol has a pleasant odour whereas ethanoic acid has a pungent smell.
The boiling point of ethanoic acid is 391 K, whereas the boiling point of ethyl alcohol is 351 K.

Chemical Properties:

Carbon and its Compounds - Solved Intext Exercises (Continued)

Q7. How can ethanol and ethanoic acid be differentiated on the basis of their chemical properties?

Sol.
Ethanoic acid reacts with sodium bicarbonate or sodium carbonate to produce carbon dioxide gas, whereas ethanol does not react with sodium bicarbonate.

CH3COOH + NaHCO3 → CH3COONa + H2O + CO2
C2H5OH + NaHCO3 → No evolution of CO2

Ethanol discharges the pink colour of potassium permanganate (KMnO4), but ethanoic acid does not.

Q8. Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?

Sol. Soaps are sodium or potassium salts of long-chain carboxylic acids. The ionic end of soap dissolves in water while the carbon chain dissolves in oil.
When soap is added to water, the hydrophilic end (ionic end) aligns along the surface of water and the hydrophobic tail (carbon chain) remains inside a cluster. This cluster of molecules is called a micelle. The soap micelles help dissolve dirt in water and clean clothes.
No, micelle will not be formed in other solvents such as ethanol.

Q9. Why are carbon and its compounds used as fuels for most applications?

Sol. Carbon and its compounds are used as fuel because when they are burnt, they release a large amount of energy in the form of heat.

Q10. Explain the formation of scum when hard water is treated with soap.

Sol. Hard water contains soluble salts of sulphates and chlorides of calcium and magnesium. When soap is used with hard water, calcium and magnesium ions react with the soap to form an insoluble substance called scum.

Q12. Which of the following hydrocarbons undergo addition reactions: C2H6, C3H8, C3H6, C2H2, and CH4?

Sol. Unsaturated hydrocarbons undergo addition reactions. Among the given examples, the unsaturated hydrocarbons are C3H6 and C2H2.

Q13. Give a test that can be used to differentiate chemically between butter and cooking oil.

Sol. Take butter and cooking oil in two separate test tubes. Add bromine water to each test tube. The test tube containing cooking oil decolourises the light brown colour of bromine water, whereas the colour of bromine water remains unchanged with butter.

Q14. Explain the mechanism of the action of soaps.

Sol. When soap is dissolved in water, it forms a colloidal suspension with spherical micelles. When dirty clothes are put in water containing dissolved soap, the hydrocarbon ends of the soap molecules in the micelles attach to oil or grease particles on the clothes. The ionic ends of the soap molecules remain attached to water. When the cloth is agitated, the oily and greasy particles are trapped by the soap micelles, dispersed in water, and the clothes get cleaned.

LIFE PROCESSES - Solved Intext Exercises

Q1. Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans?

Ans. In multicellular organisms, not all cells are in direct contact with the environment. Therefore, simple diffusion cannot meet the oxygen requirement of all the body cells.

Q2. What criteria do we use to decide whether something is alive?

Ans. The most important criterion to decide whether something is alive is movement. All living things move by themselves without any external help.

All living organisms perform certain processes listed below:
(i) Nutrition
(ii) Respiration
(iii) Excretion
(iv) Transportation
(v) Reproduction
(vi) All are made of cells

Q3. What are outside raw materials used for by an organism?

Ans. Outside raw materials used by an organism are:
(i) Food
(ii) Water
(iii) Oxygen
Plants also use CO2, light energy, and water to carry out photosynthesis.

Q4. What process would you consider essential for maintaining life?

Ans. Essential processes for maintaining life are:
(i) Nutrition
(ii) Respiration
(iii) Transportation
(iv) Excretion

Q5. What are the differences between autotrophic nutrition and heterotrophic nutrition?

Ans. Distinction between autotrophic nutrition and heterotrophic nutrition:

Autotrophic Nutrition: It is the mode of nutrition in which an organism makes (or synthesizes) its own food from simple inorganic materials like carbon dioxide and water present in the surroundings, with the help of sunlight energy.

LIFE PROCESSES - Solved Intext Exercises (Continued)

Q5. Heterotrophic Nutrition: What is it?

Ans. Heterotrophic Nutrition is the mode of nutrition in which an organism cannot make (or synthesize) its own food from simple inorganic materials like carbon dioxide and water, and depends on other organisms for its food.

Q6. Where do plants get each of the raw materials required for photosynthesis?

Ans.
(i) CO2 from atmosphere
(ii) Water from soil
(iii) Light from Sun
(iv) Chlorophyll present in chloroplasts found in green parts of plants

Q7. What is the role of the acid in our stomach?

Ans. The hydrochloric acid in the stomach:
(i) Creates an acidic medium which facilitates the action of the enzyme pepsin
(ii) Kills harmful bacteria present in food
(iii) Inactivates salivary amylase
(iv) Prevents fermentation of food

Q8. What is the function of digestive enzymes?

Ans. Digestive enzymes break down the various complex components of food into simple and soluble components so that they can be absorbed easily.

Q9. How is the small intestine designed to absorb digested food?

Ans. The inner lining of the small intestine has numerous finger-like projections called villi which increase the surface area for absorption. The villi are richly supplied with blood vessels which take the absorbed food to every cell of the body, where it is utilised for obtaining energy, building new tissues, and repairing old ones.

Q10. What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?

Ans. Terrestrial organisms have a slower rate of breathing compared to aquatic organisms because the amount of O2 in air is much higher than in water.

Q11. What are the different ways in which glucose is oxidised to provide energy in various organisms?

Ans. Some organisms use oxygen to completely break down glucose into CO2 and water (aerobic respiration), while others use pathways that do not involve oxygen (anaerobic respiration).

Q12. How is oxygen and carbon dioxide transported in human beings?

Ans. Haemoglobin present in red blood cells binds oxygen in the lungs and carries it to oxygen-deficient tissues. Carbon dioxide, being more soluble in water than oxygen, is mostly transported in dissolved form in the blood.

Q13. How are the lungs designed in human beings to maximise the area for exchange of gases?

Ans. The lungs have a highly branched structure ending in tiny air sacs called alveoli, which provide a large surface area for gas exchange. The alveoli are surrounded by a network of capillaries to facilitate efficient diffusion of oxygen into blood and carbon dioxide out of blood.

Q13. How are the lungs designed in human beings to maximise the area for exchange of gases?

Ans. In lungs, the bronchioles terminate in balloon-like structures called alveoli. The alveoli provide a surface where the exchange of gases can take place. The walls of the alveoli contain an extensive network of blood vessels. The alveolar area, if spread out, covers about 80 m², which is more than our body surface area.

Q14. What are the components of the transport system in human beings? What are the functions of these components?

Ans. Components of human transport system:
- Heart: Receives and pumps blood
- Arteries: Carry blood away from heart to various organs
- Veins: Bring blood back to heart
- Capillaries: Arteries divide to form capillaries in tissues; through them exchange of various materials takes place

Q15. Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?

Ans. The separation of the right and left sides of the heart prevents mixing of oxygenated and deoxygenated blood. This allows a highly efficient supply of oxygen to the body, which is useful in animals with high energy needs, such as birds and mammals, that constantly use energy to maintain body temperature.

Q16. What are the components of the transport system in highly organised plants?

Ans. The transport system in plants consists of xylem and phloem.
- Xylem: Has vessels and tracheids; transports water and minerals from roots to other parts of the plant (unidirectional, upward).
- Phloem: Consists of sieve tubes, sieve cells, and companion cells; transports food from leaves to storage organs and other parts of the plant (bidirectional).

Q17. How are water and minerals transported in plants?

Ans. Water and minerals are transported through xylem, which consists of tracheids and vessels. Root cells actively take up ions from soil, creating a concentration difference that draws water into the roots. This forms a column of water that is steadily pushed upwards. Evaporation of water from leaf cells (transpiration) creates a suction that pulls water from the roots. Transpiration helps absorption and upward movement of water and minerals and also aids temperature regulation. Root pressure is more significant at night, while transpiration pull is the major driving force during the day when stomata are open.

Q18. How is food transported in plants?

Ans. Food is transported in plants through phloem. The phloem carries the dissolved food substances (mainly sucrose) from the leaves (source) to storage organs and other parts of the plant (sink) through a process called translocation.

Q18. How is food transported in plants?

Ans. Food is transported in plants through phloem, which consists of tubes, sieve cells, and companion cells. This transport of soluble products of photosynthesis is called translocation and occurs in the phloem. Besides the products of photosynthesis, phloem also transports amino acids and other substances. These substances are delivered to storage organs like roots, fruits, seeds, and to growing organs. Translocation in phloem occurs with the help of adjacent companion cells in both upward and downward directions and requires energy. Substances like sucrose are transferred into phloem tissue using energy from ATP, increasing osmotic pressure and causing water to move in, which pushes the materials to where they are needed. For example, in spring, sugar in root or stem tissue is transported to buds for growth.

Q19. Describe the structure and functioning of nephrons.

Ans. Each nephron consists of a cluster of very thin-walled blood capillaries. Each capillary cluster in the kidney is associated with a cup-shaped Bowman's capsule that collects the filtered urine. Nephrons filter the blood to remove nitrogenous metabolic wastes and also reabsorb useful substances such as glucose, amino acids, minerals, and a major amount of water from the filtrate.

Q20. What are the methods used by plants to get rid of excretory products?

Ans. Plants get rid of excretory products in the following ways:
(i) Carbon dioxide produced during respiration and oxygen produced during photosynthesis are released.
(ii) Excess water is removed through transpiration.
(iii) Waste products stored in leaves are eliminated when the leaves fall.
(iv) Some waste products, like gums and resins, are stored in older xylem tissue.
(v) Plants also excrete some waste substances into the soil around them.

Q21. How is the amount of urine produced regulated?

Ans. The amount of urine produced depends on the excess water and dissolved wastes in the body. If the body has more water and wastes than the prescribed limit, urine production increases. If the body has less water and wastes, urine production decreases.

Q1. What is the difference between a reflex action and walking?

Sol.
Reflex action:
1. Reflex action is the sudden action in response to something in the environment.
2. It is controlled by the spinal cord.
3. It occurs in a fraction of a second.

Walking:
1. Walking is the mechanism in which electrical impulses travel in nerve cells and transmit to the corresponding muscles, causing them to contract so that we start walking.
2. It is controlled by the brain.
3. It occurs in response through the brain via nerve cells, so it takes time.

Q2. What happens at the synapse between two neurons?

Sol. At the synapse, the free tips of dendrites come very close to the free ends of distal axon fibers. When an impulse travels, a chemical compound is released in the synaptic gap. The impulse travels through this compound to the next neuron, after which the compound is withdrawn. Thus, the impulse is transmitted from one neuron to another, enabling communication between organs.

Q3. Which part of the brain maintains posture and equilibrium of the body?

Sol. The cerebellum, which is part of the hindbrain, is responsible for maintaining posture and equilibrium of the body.

Q4. How do we detect the smell of an agarbatti (incense stick)?

Sol. The cerebrum, which is part of the forebrain, is responsible for detecting the smell of an agarbatti.

Q5. What is the role of the brain in reflex action?

Sol. In reflex action, we react without conscious thought. Input (sensory) and output (motor) nerves meet at a point in the spinal cord. The brain receives information about the reflex action but does not control the immediate response.

Q6. What are plant hormones?

Sol. Plant hormones are chemicals produced in the cells of root and shoot tips. These chemicals, also called phytohormones, regulate the growth and development of the plant body.

Q7. How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?

Sol. In sensitive plants like Mimosa pudica (Touch-me-not), electrical and chemical signals convey the information of touch from cell to cell. Plant cells change shape by altering water content, causing swelling or shrinking, which results in the rapid movement of leaves. In contrast, the movement of a shoot towards light (phototropism) is slower and involves differential cell elongation controlled by plant hormones.

Q1. What is the difference between a reflex action and walking?

Q2. What happens at the synapse between two neurons?

Q3. Which part of the brain maintains posture and equilibrium of the body?

Sol. The cerebellum, which is part of the hindbrain, is responsible for maintaining posture and equilibrium of the body.

Q4. How do we detect the smell of an agarbatti (incense stick)?

Sol. The cerebrum, which is part of the forebrain, is responsible for detecting the smell of an agarbatti.

Q5. What is the role of the brain in reflex action?

Q6. What are plant hormones?

Sol. Plant hormones are chemicals produced in the cells of root and shoot tips. These chemicals, also called phytohormones, regulate the growth and development of the plant body.

Q7. How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?

Sol. The movement of a sensitive plant, like Mimosa pudica, is triggered by electrical-chemical signals and changes in water content within cells, causing rapid leaf folding.

In contrast, the movement of shoots and roots towards sunlight is called phototropism. In this movement, the cells on the side exposed to light shrink as water moves to the cells on the opposite side. This causes the plant part to bend towards sunlight, called positive phototropism. This movement is slower and does not involve electrical-chemical sensation.

Q8. Give an example of a plant hormone that promotes growth.

Sol. Auxin is a plant hormone produced in the cells at the tip of the main stem and its branches. It increases the length of cells and promotes overall plant growth.

Q9. How do auxins promote the growth of a tendril around the support?

Sol. In weak plants like pea, the tip of the tendril produces auxins. The side of the tendril in contact with a support transfers auxins and water to the opposite side, causing the tendril to bend and coil around the support.

Q10. Design an experiment to demonstrate hydrotropism.

Sol.
AIM: To demonstrate hydrotropism.
Materials required: Trough, soil, water, germinating seeds or baby plants, perforated cardboard sheet equal to the diameter of the trough.
Procedure:

Fix the perforated cardboard sheet vertically in the trough.
Half-fill one part of the trough with soil and the other part with water.
Place the germinating seeds or baby plants in the soil.
Keep the apparatus in sunlight in an open place for a few days.

Observations:

The roots bend towards water.
The shoots bend away from water.

Result: Water has a positive effect on roots (positive hydrotropism) and a negative effect on shoots (negative hydrotropism).
Precautions:

Place the apparatus in sunlight for proper growth conditions.
Ensure the perforations in the cardboard are neither too big nor too small to allow slow diffusion of water.

Q11. How does chemical coordination take place in animals?

Sol. Chemical coordination in animals occurs through hormones. Hormones are chemical messengers secreted by endocrine glands into the bloodstream. They travel to target organs or cells and regulate physiological activities such as growth, metabolism, reproduction, and maintaining homeostasis by affecting the function of specific tissues.

Q11. Explain how hormones work in animals.

Sol. Hormones are chemicals produced by endocrine glands in the animal body. These chemicals travel through the blood to various parts of the body to regulate specific functions.

Example: Adrenaline is a hormone secreted by the adrenal gland. It is carried by the blood to target organs, including the heart. As a result, the heart beats faster, supplying more oxygen to the muscles, thus meeting the oxygen demand of active organs.

Q12. Why is the use of iodised salt advisable?

Sol. The thyroid gland in the neck secretes thyroxin hormone, which requires iodine for its production. Therefore, using iodised salt ensures adequate iodine intake and supports proper thyroid function.

Q13. How does our body respond when adrenaline is secreted into the blood?

Sol. Adrenaline is secreted into the blood and transported to different parts of the body. It increases the oxidation of carbohydrates, fats, and proteins to produce energy. It also acts on the heart to increase the heart rate, which pumps blood faster to target organs, supplying more oxygen and energy to them.

Q14. Why are some patients of diabetes treated with insulin injections?

Sol. In some diabetic patients, the pancreas does not produce enough insulin when blood sugar levels rise. These patients are given insulin injections to lower blood sugar levels and maintain proper glucose balance in the body.

Solved NCERT Exercises

Q1. Which of the following is a plant hormone?

(a) Insulin
(b) Thyroxin
(c) Oestrogen
(d) Cytokinin
Sol. (d) Cytokinin

Q2. The gap between two neurons is called a:

(a) dendrite
(b) synapse
(c) axon
(d) impulse
Sol. (b) synapse

Q3. The brain is responsible for:

(a) thinking
(b) regulating the heartbeat
(c) controlling movements
(d) all of the above
Sol. (d) all of the above

Q3. The brain is responsible for:

(a) thinking
(b) regulating the heartbeat
(c) balancing the body
(d) all of the above
Sol. (d) all of the above

Q4. What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise?

Sol. Receptors are organs that receive information from the environment. If receptors do not work properly, the environmental signals will be detected, but the response from the spinal cord and brain may not reach the effectors in time. As a result, the effector organs will not respond appropriately, potentially causing harm or dysfunction.

Q5. Draw the structure of a neuron and explain its function.

Sol.
Function of neuron: Dendrites of a neuron receive environmental information from the receptors. This information travels through the neuron and reaches the synapse, where a chemical compound is released to transmit the signal to the next neuron. After transmission, the chemical compound is withdrawn. This process ensures information travels from the receptor to the brain and spinal cord.

Q6. How does phototropism occur in plants?

Sol. Phototropism: “Photo” means light and “tropism” means movement. Plant parts move in response to sunlight. Shoots and stems show positive phototropism by bending toward light, while roots usually exhibit negative phototropism by growing away from light.

Q7. Which signals will get disrupted in case of a spinal cord injury?

Sol. Reflex actions will be disrupted if the spinal cord is injured. For example, if a finger is placed in a flame, the person may not withdraw it in time due to the disrupted reflex pathway, leading to injury.

Q8. How does chemical coordination occur in plants?

Sol. Chemical coordination in plants is achieved by plant hormones, also called phytohormones. These are produced in the cells of the tips of stems and branches. The hormones diffuse to neighbouring cells and regulate processes such as cell elongation, cell multiplication, and opening of flowers, enabling chemical coordination.

Q9. What is the need for a system of control and coordination in an organism?

Sol. A system of control and coordination is needed in organisms to detect changes in the environment, process the information, and bring about appropriate responses. This ensures the survival, growth, and proper functioning of the organism.

Q9. What is the need for a system of control and coordination in an organism?

Sol. In multicellular organisms, all activities of the body are controlled by the spinal cord and brain. All cells are connected to them through nerves. Nerves are of two types: sensory and motor. Sensory nerves carry impulses from receptors to the brain and spinal cord, and responses are carried to effectors via motor nerves. Control and coordination are achieved because the brain and spinal cord regulate activities, while nerves act as coordinating devices.

Q10. How are involuntary actions and reflex actions different from each other?

Sol.
Reflex action:
- Controlled by spinal cord.
- Very quick.
- No need for thinking.
- Response is given immediately after stimulus is received.

Involuntary action:
- Controlled by brain.
- Takes some time.
- Brain makes judgement and then responds.
- Response depends on judgement, not just stimulus.

Q11. Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.

Nervous Coordination:
- Network of nerves connects body cells to the central nervous system.
- Controlled by brain and spinal cord.
- No secretion involved.
- All functions controlled via nerves and CNS.
- System is fixed in the body.
- Stimuli and responses travel through nerves.
- Functions depend on environmental stimuli.

Hormonal Coordination:
- No central network like nervous system.
- No controlling organ for coordination.
- Hormones are secreted by ductless (endocrine) glands.
- Each hormone controls a specific function.
- Hormone quantity affects organs and systems.
- Endocrine glands are located at various body places.
- Blood carries hormones to target organs.
- Hormones are not directly affected by external environment.

Q12. What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs?

Movement in sensitive plants:
- Leaves respond to touch.
- Movement is achieved via electrical-chemical signals between cells.
- Cells change shape by gaining or losing water, causing swelling or shrinking.

Movement in human legs:
- Controlled by skeletal muscles.
- Movement occurs through the contraction and relaxation of muscles.
- Nerve impulses from brain and spinal cord coordinate movement.
- Requires energy and conscious control (except reflex movements).

Q13. What is the difference between movement in sensitive plants and movement in human legs?

Movement in sensitive plants:
- Electrical-chemical signals convey the information of touch.
- Not controlled by any specific part of the plant.
- No special tissue for conduction of information.
- Cells change shape by altering the amount of water in them, causing movement.

Movement in human legs:
- No special organs in the leg for this action.
- Chemical signals (nerve impulses) control the movement.
- Controlled by brain and spinal cord.
- Some cells (muscle cells) change shape (contract or relax) to produce movement.
- Movement is not affected by the amount of water in the cells.

Q1. What is the importance of DNA copying in reproduction?

- The chromosomes in the nucleus contain information for inheritance in the form of DNA molecules.
- DNA copying ensures that characteristics of the parent organism are transmitted to the offspring.
- Slight variations in DNA copying lead to variations in offspring, which is important for evolution.

Q2. Why is variation beneficial to the species but not necessarily for the individual?

- Variation helps species survive adverse environmental conditions.
- If all organisms are identical, a drastic change (heat, cold, drought) could wipe out the entire population.
- Individuals with beneficial variations survive and pass on their traits, ensuring the survival of the species.
- Example: Some bacteria survive increased water temperature due to variations that confer heat resistance.

Q3. How does binary fission differ from multiple fission?

- Binary fission: Parent splits into two new organisms (e.g., Amoeba).
- Multiple fission: Parent splits into many new organisms simultaneously (e.g., Plasmodium).

Q4. How will an organism be benefitted if it reproduces through spores?

- Spores have a hard protective coat, allowing survival under unfavourable conditions (lack of food, water, extreme temperatures).
- When conditions become favourable, spores grow to produce new plants.
- This ensures the long-term survival of plants.

Q5. Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?

- Complex multicellular organisms have specialized cells, tissues, organs, and organ systems.
- Regeneration in simple organisms works because each cell can potentially give rise to a new individual.
- In complex organisms, only certain cells can divide; most cannot form an entire organism, so full regeneration is not possible.

Q5. Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?

- Complex multicellular organisms have a very high degree of organisation in their body.
- They cannot reproduce from cut body parts because the cells in the cut part cannot form all the organs (heart, brain, lungs, intestines, limbs, etc.) required for a complete organism.
- Example: A dog cannot regenerate from its cut tail.
- Complex organisms require more complex ways of reproduction like sexual reproduction.

Q6. Why is vegetative propagation practiced for growing some types of plants?

- Plants produced by vegetative propagation are genetically similar to the parent and retain all its characteristics.
- Fruit trees grown from seeds may take many years to bear fruit, but those grown by vegetative methods (cuttings, grafting) bear fruit sooner.
- Plants grown by vegetative propagation usually need less attention in early years.
- Many plants can be grown from just one parent through artificial propagation.
- It allows propagation of plants like banana, rose, jasmine, and orange that may not produce viable seeds.

Q7. Why is DNA copying an essential part of the process of reproduction?

- DNA contains information for inheritance of characteristics from parents to the next generation.
- DNA copying ensures the transmission of these parental characteristics to offspring.

Q8. How is the process of pollination different from fertilisation?

- Pollination: Transfer of pollen grains from the anther of a flower to the stigma of the same or another flower of the same species.
- Fertilisation: Fusion of the male gamete (from pollen) with the female gamete (egg) in the ovule to form a zygote.

Q9. What is the role of seminal vesicles and prostate gland?

- Seminal vesicles and prostate gland are part of the male reproductive system.
- Their secretions are added to the vas deferens, which carries sperm from the testes.
- The secretions provide nutrition to sperm and facilitate their transport.

Q10. What are the changes seen in girls at the time of puberty?

- Hair growth under armpits and pubic region.
- Development and enlargement of mammary glands (breasts).
- Broadening of hips.
- Deposition of extra fat in hips and thighs.
- Enlargement of fallopian tubes, uterus, and vagina.
- Ovaries begin to release eggs.
- Menstruation (monthly periods) starts.
- Development of sexual feelings and drives associated with adulthood.

Q11. How does the embryo get nourishment inside the mother's body?

- The embryo gets nourishment from the mother through the placenta.
- The placenta allows the exchange of nutrients, oxygen, and waste products between maternal and embryonic blood without direct mixing.
- This ensures that the developing embryo receives essential substances for growth and development.

Q11. How does the embryo get nourishment inside the mother's body?

- The embryo gets nutrition from the mother’s blood with the help of a special tissue called placenta.
- Placenta is a disc-shaped tissue embedded in the uterine wall.
- It has villi on the embryo side and blood spaces on the mother’s side surrounding the villi.
- Placenta provides a large surface area for glucose and oxygen to pass from the mother to the embryo.
- Waste products produced by the embryo are transferred to the mother’s blood through the placenta.

Q12. If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases?

- No, the use of copper-T for contraception will not protect a woman from sexually transmitted diseases.

Solved NCERT Exercises: Q1. Asexual reproduction takes place through budding in

- (a) Amoeba
- (b) Yeast ✔
- (c) Plasmodium
- (d) Leishmania

Q2. Which of the following is not a part of the female reproductive system in human beings?

- (a) Ovary
- (b) Uterus
- (c) Vas deferens ✔
- (d) Fallopian tube

Q3. The anther contains

- (a) Sepals
- (b) Ovules
- (c) Pistil
- (d) Pollen grains ✔

Q4. What are the advantages of sexual reproduction over asexual reproduction?

- Sexual reproduction combines DNA from two individuals (male and female), giving the offspring a lot of variations.
- In asexual reproduction, DNA from only one individual is copied, so the variations in the offspring are extremely small.

Q4. Due to lot of variations sexual reproduction allows species to change to more advanced forms from one generation to the next and speed up evolution. On the other hand, asexual reproduction does not allow a species to change much from one generation to the next and hence evolution becomes very, very slow.

- Sexual reproduction introduces variations, allowing species to evolve and adapt faster from one generation to the next.
- Asexual reproduction produces offspring identical to the parent, so evolutionary changes occur very slowly.

Q5. What are the functions performed by the testis in human beings?

- Testes produce male sex cells (sperms) and the male sex hormone testosterone.
- Testosterone brings about changes during puberty in boys such as deeper voice, growth of beard and moustache, and increased body hair.

Q6. Why does menstruation occur?

- Every month, the ovary releases an egg (ovum), and the uterus prepares to receive a fertilised egg.
- The inner lining of the uterus becomes thick and soft with blood vessels to nourish a potential embryo.
- If the egg is not fertilised, this lining is not needed, so it breaks down and comes out through the vagina as blood and mucous. This is called menstruation.

Q7. Draw a labelled diagram of the longitudinal section of a flower.

- fig. Longitudinal Section of a Flower

Q8. What are the different methods of contraception?

- Methods of contraception include: Barrier methods, Chemical methods, Use of Loop or Copper-T, and Surgical methods.
- Barrier methods prevent pregnancy using physical devices such as condoms (used by males) and diaphragm/cap (used by females to cover the cervix).

- Condoms and diaphragms prevent sperms from meeting the ovum by acting as a physical barrier.
- In chemical methods, females use oral pills containing hormones that stop the ovaries from releasing eggs.
- Loops or copper-T (IUCDs) are placed inside the uterus to prevent implantation of a fertilised egg.
- Surgical methods: In males, vasectomy removes a small portion of the sperm duct (vas deferens) and ties the ends to prevent sperms from exiting. In females, tubectomy removes a portion of the oviducts and ties the ends to prevent eggs from reaching the uterus.

Q9. How are the modes for reproduction different in unicellular and multicellular organisms?

- Unicellular organisms (e.g., protozoa, bacteria) reproduce mainly by asexual fission, where cell division directly forms new individuals.
- Simple multicellular organisms reproduce asexually by budding, spore formation, fragmentation, or regeneration.
- Complex multicellular organisms reproduce sexually, involving gametes (sex cells) from two parents – a male and a female.

Q10. How does reproduction help in providing stability to populations of species?

- Reproduction introduces variations in individuals of a species.
- These variations help some organisms survive adverse environmental conditions such as excessive heat, cold, or drought, even when most others die.
- Thus, reproduction stabilises populations by ensuring that some individuals survive and maintain the species.

Q11. What could be the reasons for adopting contraceptive methods?

- Contraceptive methods are used for family planning and birth control.
- They help couples avoid unwanted pregnancies and allow them to decide how many children to have and when.
- They also allow spacing between births.
- Some methods, like condoms, provide protection from sexually transmitted diseases.

Q1. If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?

Trait B is likely to have arisen earlier in a population of an asexually reproducing species.

Q2. How does the creation of variation in a species promote survival?

Variations in a species arise due to errors in DNA copying. The force of natural selection selects individuals with useful variations in the prevailing environment to ensure their survival. Individuals with useful variations increase in numbers through differential reproduction in the population.

Q3. How do Mendel’s experiments show that traits may be dominant or recessive?

When Mendel first crossed pure tall pea plants with pure dwarf pea plants, only tall plants were produced in the F1 generation. No dwarf plants appeared. When F1 tall plants were self-pollinated, both tall and dwarf plants appeared in the F2 generation in a 3:1 ratio (three-fourth tall and one-fourth dwarf). Mendel called the tall character the dominant trait and the dwarf character the recessive trait.

Q4. How do Mendel’s experiments show that traits are inherited independently?

Mendel crossed pure-bred pea plants with round-yellow seeds with pure-bred pea plants with wrinkled-green seeds. F1 generation produced only round-yellow seeds, showing that round shape and yellow colour are dominant traits. When F1 plants were self-pollinated, F2 generation had four combinations of seeds: round-yellow, round-green, wrinkled-yellow, and wrinkled-green.
This dihybrid cross gave the phenotypic ratio: 9 : 3 : 3 : 1.
Mendel concluded that the two pairs of traits (seed colour and shape) combine in F1 but segregate independently in F2.

Q5. A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits—blood group A or O—is dominant? Why or why not?

The information is insufficient to determine whether the trait ‘A’ or ‘O’ is dominant.
Example: If trait ‘A’ is dominant, the father may have genotype IAIA or IAIO, and the mother is IOIO. Further genetic information is needed to confirm dominance.

Q5 (continued). In this case, 50 per cent of the progeny will have blood group A and 50 per cent will have blood group O, when father's blood group is AO and mother's is O. If we assume trait ‘O’ is dominant, the child may still have blood group O. Therefore, we cannot establish which trait is dominant.

Q6. What are the different ways in which individuals with a particular trait may increase in population?

Individuals with a particular trait may increase in population through:
1. Natural selection
2. Genetic drift

Q7. How is the sex of the child determined in human beings?

In human beings, the sex of the child depends on which type of male gamete fertilises the female gamete.
- If a sperm carrying X chromosome fertilises an ovum carrying X chromosome, the child will be a girl.
- If a sperm carrying Y chromosome fertilises an ovum carrying X chromosome, the child will be a boy.

Q8. Why are traits acquired during the life-time of an individual not inherited?

Changes in non-reproductive body cells (somatic cells) cannot be inherited because only traits resulting from changes in the DNA of reproductive cells (gametes) can be transmitted to future generations.

Q9. Why are small numbers of surviving tigers a cause of worry from the point of view of genetics?

Small numbers of surviving tigers are worrying genetically because:
1. If they die out, their genes are lost forever, preventing future generations from seeing tigers.
2. Small populations have fewer chances for genetic variation, which is crucial for the stability of a species through natural selection.

Q10. What factors could lead to the rise of new species?

New species may arise due to factors such as genetic variations, natural selection, mutations, geographical isolation, and environmental changes that drive evolution over generations.

Q10 (continued). Ans. Genetic variation, natural selection and reproductive isolation are factors that could lead to the rise of a new species.

Q11. Will geographical isolation be a major factor in the speciation of a self-pollinating plant species? Why or why not?

Geographical isolation will not be a major factor for the speciation of a self-pollinating plant species because it does not need other plants for reproduction to take place.

Q12. Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?

Geographical isolation will not be a major factor in the speciation of an organism that reproduces asexually because it does not require any other organism for reproduction.

Q13. Give an example of characteristics being used to determine how close two species are in evolutionary terms.

Forelimbs of humans and wings of birds show the closeness between these two species. The forelimbs of humans and wings of birds have similar basic structures but perform different functions. The presence of homologous organs provides evidence for evolution, indicating that these species evolved from a common ancestor.

Q14. Can the wing of a butterfly and the wing of a bat be considered homologous organs?

No, the wings of a butterfly and the wings of a bat are not homologous organs. They have different basic designs even though they perform similar functions, making them analogous organs.

Q15. What are fossils? What do they tell us about the process of evolution?

Fossils are the preserved traces or remains of living organisms that did not fully decompose due to special environmental conditions (e.g., an insect caught in hot mud). Fossils help identify evolutionary relationships between species. Fossils in bottom rock layers are simple, while fossils in upper layers are more complex, supporting the concept of evolution.

Q16. Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?

Human beings who differ in size, colour, and looks belong to the same species (Homo sapiens) because they can interbreed to produce fertile offspring. Early Homo sapiens originated in Africa, and groups spread across the world, mixing and separating over time, while adapting genetically to their prevailing environments.

Q4. A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?

In this study, light-coloured eye is a dominant trait because children born from parents having light-coloured eyes also had light-coloured eyes.

Q5. How are the areas of study—evolution and classification—interlinked?

Characteristics of organisms refer to details of their external and internal appearance or behaviour that distinguish them from one another. These characteristics form the basis for the classification of organisms. The more characteristics two species have in common, the more closely they are related, and the more recently they will have had a common ancestor. By identifying hierarchies of characteristics between species, we can work out evolutionary relationships. Thus, classification of species reflects their evolutionary relationships.

Q6. Explain the terms analogous and homologous organs with examples.

Homologous organs: Organs with the same basic structural design and developmental origin but different functions. Example: The forelimb of a frog, lizard, bird, and man have the same basic bone design but perform different functions.
Analogous organs: Organs with different basic structural design and developmental origin but similar functions. Example: The wings of birds and insects.

Q7. Outline a project which aims to find the dominant coat colour in dogs.

Select a homozygous black (BB) male dog and a homozygous white (bb) female dog. Crossbreed them to produce offspring (F1 generation). If all F1 offspring are black, we can conclude that black coat colour is dominant over white coat colour in dogs.

Q8. Explain the importance of fossils in deciding evolutionary relationships.

Fossils provide preserved evidence of organisms that lived in the past. By studying fossils, scientists can identify evolutionary relationships between species. Fossils in lower rock layers are simpler, while those in higher layers are more complex, which helps track changes over time and understand how species evolved.

Q8. Explain the importance of fossils in deciding evolutionary relationships.

Fossils are the remains or impressions of dead plants or animals which died millions of years ago. The study of fossils helps us understand evolution and the links between species. Fossils indicate how new species developed from older ones, making them important in deciding evolutionary relationships.

Q9. What evidence do we have for the origin of life from inanimate matter?

An experiment conducted by Stanley L. Miller and Harold C. Urey in 1953 showed that life could originate from inanimate matter. They created an apparatus simulating early Earth's atmosphere (methane, ammonia, hydrogen sulfide, and water, without oxygen) and maintained it just below 100°C. Sparks were passed through the mixture to simulate lightning. After a week, 15% of the carbon from methane had converted into simple carbon compounds, including amino acids, which are the building blocks of proteins.

Q10. Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?

Sexual reproduction produces more genetic variation than asexual reproduction. In asexual reproduction, offspring are almost identical to their parent, limiting genetic variation and slowing evolution. In sexual reproduction, offspring receive a mix of genes from both parents, creating variations in each individual. These variations provide raw material for natural selection, allowing populations to adapt and evolve continuously into better-adapted organisms.

Q11. How is the equal genetic contribution of male and female parents ensured in the progeny?

Humans have 23 pairs of chromosomes, with one copy from the mother and one from the father. Out of these, 22 pairs are autosomes and 1 pair are sex chromosomes. During fertilization, the haploid egg (n) fuses with the haploid sperm (n) to form a diploid zygote (2n), which contains 23 pairs of chromosomes—half from the mother and half from the father. This ensures equal genetic contribution from both parents.

Q12. Only variations that confer an advantage to individual organisms will survive in a population. Do you agree with this statement? Why or why not?

Yes, we agree. Not all variations have an equal chance of survival; only those providing an advantage in a given environment are likely to persist. For example, a bacterium that can withstand high heat will survive better during a heatwave. This process of selection based on advantageous traits forms the basis of evolution.

Q1. Define the principal focus of a concave mirror.

The principal focus of a concave mirror is a point on its principal axis to which all the light rays that are parallel and close to the axis converge after reflection from the concave mirror.

Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Focal length (f) = Radius of curvature (R) / 2
Therefore, f = 20 / 2 = 10 cm
Focal length of the spherical mirror is 10 cm.

Q3. Name a mirror that can give an erect and enlarged image of an object.

Concave mirror.

Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

We prefer a convex mirror as a rear-view mirror in vehicles because:

Convex mirrors always produce an erect (upright) image of objects.
The image formed is highly diminished (much smaller than the object), which provides a wide field of view of traffic behind.

Q5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Radius of curvature, R = 32 cm
Focal length, f = R / 2
f = 32 / 2 = 16 cm

Q6. A concave mirror produces a three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?

Linear magnification, m = -3 (negative for real, inverted image)
Object distance, u = 10 cm
Image distance, v = ?
Using m = -v / u
-3 = -v / 10
v = 30 cm
The image is located 30 cm in front of the concave mirror.

Q7. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

When light passes from a rarer medium (air) to a denser medium (water) at an angle, it slows down and bends towards the normal. This bending is due to refraction, which occurs because the speed of light changes in different media.

Q8. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10⁸ m/s.

Refractive index of glass, n = Speed of light in vacuum (c) / Speed of light in glass (v)
1.50 = 3 × 10⁸ / v
v = 3 × 10⁸ / 1.50
v = 2 × 10⁸ m/s

Q9. Find out, from Table 10.3 (NCERT), the medium having highest optical density. Also find the medium with lowest optical density.

The medium having highest optical density is diamond.
The medium having lowest optical density is air.

Q10. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in table (NCERT 10.3).

Light travels fastest in water out of kerosene, turpentine and water.

Q11. The refractive index of diamond is 2.42. What is the meaning of this statement?

The refractive index of diamond is 2.42. It means that the ratio of the speed of light in air and the speed of light in diamond is equal to 2.42, or the speed of light in diamond is 1/2.42 times that of the speed of light in air.

Q12. Define 1 dioptre of power of a lens.

1 dioptre is the power of a lens whose focal length is 1 meter.

Q13. A convex lens forms a real and inverted image of a needle at a distance 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also find the power of the lens.

The needle is placed at 50 cm in front of the convex lens if the image is equal in size to the object.
The image formed by a convex lens is of the same size as the object when the object is at 2F or center of curvature C.
C = 2F = 50 cm
Focal length, f = C / 2 = 50 / 2 = 25 cm = 0.25 m
Power of lens, P = 100 / f (in cm) = 100 / 25 = 4 D

Q14. Find the power of a concave lens of focal length 2 m.

Power of lens, P = 100 / f = 100 / 200 = -0.5 D

Q1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay

(d) Clay

Q2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus

(d) Between the pole of the mirror and its principal focus

Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus

(b) At twice the focal length

Q4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be
(a) both concave
(b) both convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex, but the lens is concave

(a) both concave

Q5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane only
(b) concave
(c) convex only
(d) either plane or convex

(d) either plane or convex

Q6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm

Q7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

We can obtain an erect image of the object using a concave mirror when the object is placed between pole P of the mirror and focus F. Therefore, the distance of the object from the mirror is between 0 to 15 cm. The nature of the image is virtual and erect. The image is larger than the object.

Q8. Name the types of mirror used in the following situations:
(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace
Support your answer with reason.

(a) Concave mirrors are used as headlights of a car because they reflect the light.
(b) Convex mirror is used in side/rear-view mirror of a vehicle because it produces an erect image of the objects and gives a wide field of view of the traffic behind.
(c) Concave mirror is used in solar furnace because the concave mirror focuses sunrays on the objects to be heated.

Q9. One-half of a convex lens is covered with a black paper. Will it produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Yes, the complete image of the object will still be formed. However, the image will be less bright because covering half of the lens reduces the amount of light reaching the image. The size and position of the image remain unchanged.

Q9. One-half of a convex lens is covered with a black paper. Will it produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Yes, the lens will produce a complete image of the object. We observe that the rays of light from the object pass through the uncovered half of the lens and produce a complete image. However, the brightness of the image will be reduced due to less light passing through.

Q10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size, and the nature of the image formed.

Height of the object, h = 5 cm
Object distance, u = -25 cm
Focal length, f = +10 cm
Using lens formula: 1/v - 1/u = 1/f
1/v = 1/f + 1/u = 1/10 + 1/(-25) = 1/10 - 1/25 = 0.1 - 0.04 = 0.06
v = 1/0.06 ≈ 16.67 cm
Magnification, m = v/u = 16.67 / -25 ≈ -0.667
Image height, h' = m × h = -0.667 × 5 ≈ -3.33 cm
The negative sign of h' indicates that the image is inverted.
Thus, a real, inverted image 3.33 cm tall is formed at a distance of 16.67 cm on the right side of the lens.

Q11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Focal length, f = -15 cm
Image distance, v = -10 cm
Using lens formula: 1/v - 1/u = 1/f
1/u = 1/v - 1/f = 1/(-10) - 1/(-15) = -0.1 + 0.0667 ≈ -0.0333
u ≈ -30 cm
Therefore, the object is placed 30 cm in front of the concave lens.
The image is virtual, erect, and smaller than the object.

Q11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Using lens formula: 1/v - 1/u = 1/f
1/u = 1/v - 1/f
1/u = 1/(-10) - 1/(-15) = -0.1 + 0.0667 ≈ -0.0333
u ≈ -30 cm
Thus, the object distance is 30 cm on the left side from the concave lens.
The image is virtual, erect, and smaller than the object.

Q12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Focal length, f = +15 cm
Object distance, u = -10 cm
Using mirror formula: 1/v + 1/u = 1/f
1/v = 1/f - 1/u = 1/15 - 1/(-10) = 0.0667 + 0.1 = 0.1667
v ≈ +6 cm
The image is formed 6 cm behind the mirror. It is virtual and erect.

Q13. The magnification produced by a plane mirror is +1. What does this mean?

Magnification, m = +1 means that the size of the image is equal to the size of the object.
The positive sign indicates that the image is virtual and erect.

Q14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature, and size.

Radius of curvature, R = +30 cm → focal length, f = R/2 = 30/2 = +15 cm
Object distance, u = -20 cm
Object height, h = 5 cm
Using mirror formula: 1/v + 1/u = 1/f
1/v = 1/f - 1/u = 1/15 - 1/(-20) = 0.0667 + 0.05 = 0.1167
v ≈ +8.57 cm
Magnification, m = v/u = 8.57 / -20 ≈ -0.4285
Image height, h' = m × h = -0.4285 × 5 ≈ -2.14 cm
The negative sign indicates the image is inverted.
Thus, the image is virtual, erect, diminished, and formed 8.57 cm behind the mirror.

Q14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature, and size.

Radius of curvature, R = +30 cm → focal length, f = R/2 = 15 cm
Object distance, u = -20 cm
Object height, h = 5 cm
Using mirror formula: 1/v + 1/u = 1/f
1/v = 1/f - 1/u = 1/15 - 1/(-20) = 0.0667 + 0.05 ≈ 0.1167
v ≈ 60/7 cm ≈ 8.57 cm
Magnification, m = v/u = 8.57 / -20 ≈ -0.4285
Image height, h' = m × h = -0.4285 × 5 ≈ +2.14 cm
The positive sign indicates the image is virtual and erect.
Thus, the image is formed 60/7 cm behind the mirror, virtual, erect, and diminished.

Q15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and the nature of the image.

Focal length, f = -18 cm
Object distance, u = -27 cm
Object height, h = +7.0 cm
Using mirror formula: 1/v + 1/u = 1/f
1/v = 1/f - 1/u = 1/(-18) - 1/(-27) = (-3+2)/54 = -1/54
v = -54 cm
The screen should be placed 54 cm in front of the mirror.
Magnification, m = -v/u = -(-54)/(-27) = -2
Image height, h' = m × h = -2 × 7 = -14 cm
The negative sign indicates the image is inverted.
Thus, the image is real, inverted, and twice the size of the object.

Q16. Find the focal length of a lens of power -2.0 D. What type of lens is this?

Power, P = -2.0 D
Focal length, f = 100 / P = 100 / -2 = -50 cm
Since the focal length is negative, the lens is a diverging (concave) lens.

Q17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Power, P = +1.5 D
Focal length, f = 100 / P = 100 / 1.5 ≈ 66.7 cm
Since the power is positive, the lens is converging (convex).

The Human Eye and the Colourful World

Solved Intext Exercises

Q1. What is meant by power of accommodation of the eye?

Power of accommodation of an eye is its power to be able to see nearby as well as far away objects.

Q2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?

It should be a concave (diverging) lens.

Q3. What is the far point and near point of the human eye with normal vision?

The far point of the human eye with normal vision is at infinity. The near point of the human eye with normal vision is 25 cm.

Q4. A student has difficulty in reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

The child could be suffering from myopia or near-sightedness. It can be corrected by using spectacles with concave lenses of suitable power.

Solved NCERT Exercises

Q1. The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to

(b) Accommodation

Q2. The human eye forms the image of an object at its

Q3. The least distance of distinct vision for a young adult with normal vision is about

Q4. The change in focal length of an eye lens is caused by the action of the

Q5. A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision, he needs a lens of power +1.5 dioptres. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

(i) For distant vision:
Power, P = –5.5 D
f = 1 / P = 1 / –5.5 = –0.181 m = –18.1 cm
Thus, the focal length of the lens for correcting distant vision is 18.1 cm and the lens is concave.

(ii) For near vision:
Power, P = +1.5 D
f = 1 / P = 1 / 1.5 = +0.667 m = +66.7 cm
Thus, the focal length of the lens for correcting near vision is 66.7 cm and the lens is convex.

Q6. The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

The person is myopic, so the lens required for correction is a concave lens.
u = ∞, v = –80 cm
Using formula: 1/f = 1/v – 1/u
1/f = 1/–80 – 0 = –1/80
f = –80 cm
Therefore, P = 100 / f = 100 / –80 = –1.25 D
Hence, a concave lens of power –1.25 dioptres is required.

Q7. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of lens to correct this defect? Assume that the near point of a normal eye is 25 cm.

u = –25 cm, v = –100 cm
Using formula: 1/f = 1/v – 1/u
1/f = 1/–100 – 1/–25 = –1/100 + 4/100 = 3/100
f = 100 / 3 = 33.3 cm
Now, P = 100 / f = 100 / 33.3 = +3 D
Thus, a convex lens of power +3 D is used to correct hypermetropia.
(A labelled ray diagram can be inserted here to show correction of hypermetropia.)

Q8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

A normal eye is not able to see clearly objects placed closer than 25 cm because the focal length of the eye lens cannot be decreased below this limit.

The Human Eye and the Colourful World

Q9. What happens to the image distance in the eye, when we increase the distance of an object from the eye?

Sol. The image distance in the eye remains the same, i.e., the distance of the retina from the eye ball.

Q10. Why do stars twinkle?

Sol. The twinkling of a star is due to atmospheric refraction. When the light coming from a star enters the earth’s atmosphere, it undergoes refraction due to different densities of air layers at various altitudes. The densities of the atmosphere continuously change, so the amount of refraction also changes. When the atmosphere refracts more star-light towards us, the star appears bright; when it refracts less, the star appears dim. This continuous variation in brightness makes stars appear to twinkle at night.

Q11. Explain why the planets do not twinkle.

Sol. Planets appear quite large compared to stars, so they can be considered as a collection of many point sources of light. The dimming effect from some parts is balanced by the brightening effect from others. Hence, the overall brightness remains constant, and planets do not twinkle.

Q12. Why does the sun appear reddish early in the morning?

Sol. During sunrise or sunset, the Sun is far from the observer, and sunlight travels a longer distance through the atmosphere. Blue light is scattered more than red light, so most of the blue light is scattered away and only red light reaches our eyes. This makes the Sun appear reddish during sunrise and sunset.

Q13. Why does the sky appear dark instead of blue to an astronaut?

Sol. The sky appears blue on Earth due to scattering of sunlight by the atmosphere. However, in space, there is no atmosphere to scatter light. Therefore, to an astronaut, the sky appears black or dark.

Electricity

Solved Intext Exercises

Q1. What does an electric circuit mean?

Sol. A continuous and closed path of an electric current is called an electric circuit.

Q2. Define the unit of current.

Sol. The unit of current is Ampere (A). Electric current through a conductor is said to be 1 Ampere if one Coulomb of charge flows through any cross-section of the conductor in one second.

Q3. Calculate the number of electrons constituting one coulomb of charge.

Sol. Charge on one electron, e = 1.6 × 10⁻¹⁹ C
1.6 × 10⁻¹⁹ C of charge = 1 electron
Therefore, 1 C of charge = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons.
Thus, 6.25 × 10¹⁸ electrons constitute one coulomb of charge.

Q4. Name a device that helps to maintain a potential difference across a conductor.

Sol. A battery is a device which maintains potential difference across the ends of a conductor.

Q5. What is meant by saying that the potential difference between two points is 1V?

Sol. The potential difference between two points is said to be 1 volt if 1 joule of work is done in bringing 1 coulomb of charge from one point to the other in an electric field.

Q6. How much energy is given to each coulomb of charge passing through a 6V battery?

Sol. Given: Q = 1 C, V = 6 V
We know, V = W / Q
Therefore, W = V × Q = 6 × 1 = 6 J
Hence, 6 joules of energy is given to each coulomb of charge passing through a 6V battery.

Q7. On what factors does the resistance of a conductor depend?

Sol. The resistance of a conductor depends on the following factors:

It is directly proportional to the length of the conductor.
It is inversely proportional to the area of cross-section of the conductor.
It is also inversely proportional to the square of the diameter of the conductor.
It depends on the nature of the material of the conductor. For example, Nichrome has about 60 times the resistance of copper of the same length and thickness.
It depends on the temperature of the conductor.

Q7 (continued).

For example:

The resistance of pure metals always increases with rise in temperature and vice-versa, which means the resistance of a pure metal is directly proportional to the change in temperature.
The resistance of metallic alloys like German silver, manganin, constantan, and Nichrome increases slightly and is directly proportional to the change in temperature.
The resistance of semiconductors like silicon and germanium decreases with an increase in temperature.

Q8. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Sol. Current will flow more easily through a thick wire of the same material when connected to the same source. This is because resistance is inversely proportional to the area of cross-section of the conductor. Hence, when the area of cross-section increases, the resistance decreases, and as a result, current flows more easily.

Q9. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Sol. If the potential difference across the two ends of the component decreases to half of its former value, then the current will also decrease to half of its former value because potential difference (V) is directly proportional to current (I).

Q10. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Sol. The coils of electric toasters and electric irons are made of an alloy rather than a pure metal because:

The resistivity of an alloy is higher than that of pure metals.
Alloys do not oxidize (burn) readily at high temperatures.

Q11. Use the data in Table 12.2 of the textbook to answer the following:

(a) Iron is a better conductor.
(b) Which material is the best conductor?

Sol.

(a) Iron is a better conductor.
(b) Silver is the best conductor.

Q12. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each, a 5Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key — all connected in series.

Sol. The schematic diagram of the given circuit is shown below:

Figure: A circuit showing a battery of three 2V cells, 5Ω, 8Ω, and 12Ω resistors connected in series with a plug key.

Q13. Redraw the circuit of the above question, putting in an ammeter to measure the current through the resistor and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?

Sol.

Total voltage supplied, V = 3 × 2 = 6V
Total resistance of the circuit, R = 5 + 8 + 12 = 25Ω
Current, I = ?

We know that, I = V / R = 6 / 25 = 0.24A

Reading in the ammeter: 0.24A
Reading in the voltmeter: V = I × R = 0.24 × 12 = 2.88V

Q14. Judge the equivalent resistance when the following are connected in parallel:

(a) 1Ω and 10⁶Ω
(b) 1Ω and 10³Ω and 10⁶Ω

Sol.

(a) Equivalent resistance is less than 1Ω.
(b) Equivalent resistance is less than 1Ω.

Q15. An electric lamp of 100Ω, a toaster of resistance 50Ω, and a water filter resistor 500Ω are connected in parallel to a 220V source. What is the resistance of an electric iron connected to the same source that takes as much current as all the three appliances, and what is the current through it?

Sol.

Given:
R₁ = 100Ω, R₂ = 50Ω, R₃ = 500Ω

Resultant resistance, R = ?

1/R = 1/R₁ + 1/R₂ + 1/R₃
1/R = 1/100 + 1/50 + 1/500
1/R = (5 + 10 + 1) / 500 = 16 / 500
R = 500 / 16Ω

Resistance of electric iron = 500 / 16Ω
V = 220V

I = V / R = 220 / (500/16) = 220 × 16 / 500 = 7.04A

Therefore:
Resistance of electric iron = 31.25Ω
Current through it = 7.04A

Q16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Sol.

In parallel circuits, if one electrical appliance stops working due to some defect, the others continue working normally.
Each electrical appliance has its own switch, so it can be turned on or off independently.
Each electrical appliance gets the same voltage (220V) as the power supply, allowing it to function properly.
The overall resistance of the circuit is reduced, resulting in higher total current from the supply.

Q17. How can three resistors of 2Ω, 3Ω, and 6Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω?

Sol.

(a) To obtain 4Ω: connect 3Ω and 6Ω in parallel, then 2Ω in series.

1/R′ = 1/3 + 1/6 = (2 + 1)/6 = 3/6 = 1/2 → R′ = 2Ω
R = R′ + 2 = 2 + 2 = 4Ω

(b) To obtain 1Ω: connect all three resistors (2Ω, 3Ω, 6Ω) in parallel.

1/R = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 6/6 = 1 → R = 1Ω

Q18. What is (i) the highest and (ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4Ω, 8Ω, 12Ω, and 24Ω?

Sol.

(a) Highest resistance = when all are in series:
R = 4 + 8 + 12 + 24 = 48Ω

(b) Lowest resistance = when all are in parallel:
1/R = 1/4 + 1/8 + 1/12 + 1/24 = (6 + 3 + 2 + 1)/24 = 12/24 = 1/2 → R = 2Ω

Q19. Why does the cord of an electric heater not glow while the heating element does?

Sol. The heating element and cord carry the same current, but the heating element becomes hot due to its high resistance (H = I²RT), while the cord remains cool due to its low resistance.

Q20. Compute the heat generated while transferring 96000C of charge in one hour through a potential difference of 50V.

Sol.

Given: Q = 96000C, V = 50V, t = 1h = 3600s
I = Q / t = 96000 / 3600 = 26.67A

H = VIt = 50 × 26.67 × 3600 = 4.8 × 10⁶J

Q21. An electric iron of resistance 20Ω takes a current of 5A. Calculate the heat developed in 30s.

Sol.

R = 20Ω, I = 5A, t = 30s
H = I²RT = (5)² × 20 × 30 = 15000J = 1.5 × 10⁴J

Q22. What determines the rate at which energy is delivered by a current?

Sol. Electric power.

Q23. An electric motor takes 5A from a 220V line. Determine the power of the motor and the energy consumed in 2h.

Sol.

I = 5A, V = 220V
P = V × I = 220 × 5 = 1100W

Energy consumed in 2h: E = P × t = 1.1 × 2 = 2.2kWh

Q1. Why does a compass needle get deflected when brought near a bar magnet?

Ans. A compass needle is a small bar magnet. When it is brought near a bar magnet, the needle experiences magnetic forces due to the bar magnet and gets either attracted or repelled. This makes the compass needle deflect.

Q2. Draw a magnetic field line around a bar magnet.

Ans. (Diagram to be inserted)

Q3. List the properties of magnetic lines of force.

Ans.

Magnetic lines of force originate from the north pole of a magnet and end at the south pole.
They are close together near the poles and widely spaced farther away.
Magnetic lines of force do not intersect one another.

Explanation: If two magnetic lines of force were to intersect at a point, then a compass needle placed there would have to point in two directions at once, which is impossible.

Q4. Why don’t two magnetic lines of force intersect each other?

Ans. Two lines of force never intersect each other because if they did, it would imply that at the point of intersection, the compass needle would point in two different directions at the same time, which is not possible.

Q5. Consider a circular loop of wire lying in the plane of the table. Let current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.

Ans. The magnetic field inside the loop is directed into the plane of the table (perpendicular to it), while the magnetic field outside the loop is directed out of the plane of the table (perpendicular to it).

Q6. The magnetic field in a given region is uniform. Draw a diagram to represent it.

Ans. (Diagram to be inserted showing parallel equidistant lines)

Q7. The magnetic field inside a long straight solenoid carrying current is:

(a) zero
(b) decreases as we move towards its end
(c) increases as we move towards its end
(d) the same at all points

Ans. (d) The same at all points.

Q8. Which of the following properties of a proton can change while it moves freely in a magnetic field?

(a) mass
(b) speed
(c) velocity
(d) momentum

Ans. (c) velocity and (d) momentum.

Q9. In Activity 13.7 (of textbook), how will the displacement of rod AB be affected if:

(i) current in rod AB is increased,
(ii) a stronger horse-shoe magnet is used,
(iii) the length of rod AB increases.

Ans. In all three cases, the displacement of rod AB will increase.

Q10. A positively charged particle (alpha particle) projected towards the west is deflected towards the north by a magnetic field. The direction of magnetic field is:

(a) towards south
(b) towards east
(c) downward
(d) upward

Ans. (c) Upward (according to Fleming’s left-hand rule).

Q11. State Fleming’s Left-Hand Rule.

Ans. Hold the forefinger, central finger, and thumb of your left hand at right angles to each other. Adjust your hand so that the forefinger points in the direction of the magnetic field and the central finger points in the direction of the current. Then the thumb gives the direction of the force acting on the conductor.

Q12. What is the principle of an electric motor?

Ans. When a current-carrying coil is placed in a magnetic field, it experiences a torque and starts rotating continuously. This is the principle of an electric motor.

Q13. What is the role of the split ring in an electric motor?

Ans. The split ring in an electric motor reverses the direction of current flowing through the coil every time the coil passes the vertical position, ensuring continuous rotation in one direction.

Q14. State the principle of an electric generator.

Ans. An electric generator works on the principle of electromagnetic induction — when a conductor moves in a magnetic field, an electric current is induced in it.

Q15. Name some sources of direct current.

Ans. Cell, battery, bicycle dynamo, car dynamo, dry cell, electric battery, and inverter.

Q16. Which sources produce alternating current?

Ans. Hydroelectric generator, thermal power generator, and nuclear power generator.

Q17. Choose the correct option: A rectangular coil of copper wire is rotated in a magnetic field. The direction of the induced current changes once in each:

(i) two revolutions
(ii) one revolution
(iii) half revolution
(iv) two revolutions

Ans. (iii) Half revolution.

Q18. Name two safety measures commonly used in electric circuits and appliances.

Ans.

Use of earth wire.
Use of electric fuse.

Q19. An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5A. What do you expect?

Ans. Power = 2 kW = 2000 W
Voltage = 220 V
Current, I = P / V = 2000 / 220 = 9 A
Since the circuit rating is 5A, the current exceeds the limit. Therefore, the fuse will blow or the wire may melt due to overheating.

Q20. What precaution should be taken to avoid overloading of domestic circuits?

Ans. Too many appliances drawing large currents should not be connected to a single electric circuit. Fuses of proper rating should also be used to prevent overloading.

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Notes

Revision Notes — Number Sets Download

Natural Number

Whole Number

Integers

Rational Numbers

Remarks:

Irrational Numbers:

Even Numbers:

Odd Numbers:

Consecutive Numbers:

Prime Number:

Remarks:

Twin – Prime:

Composite Number:

Remarks:

Co-Prime:

Perfect Number:

Theorem – 1

To find HCF by Euclid’s division lemma:

Important Results:

Revisiting of Irrational Numbers

Non Terminating recurring Decimals:

Irrational number:

Properties on irrational number:

Important Points

Important Points

Linear Equations in Two Variables Download

Revision Notes

Polynomials Download

Revision Notes

Types of polynomials on the basis of degree

Types of polynomials on the basis of term

GRAPH OF POLYNOMIALS

Graphs of quadratic polynomials:

The intersection points of a parabola

Graphs of cubic polynomials:

Case - 2:

Case - 3:

Formation of Quadratic Polynomial

2. Cubic Polynomial:

Formation of a Cubic Polynomial:

Division Algorithm for polynomials

Factor Theorem

Remainder Theorem

Relationship between the zeroes and coefficients of a quadratic polynomial

Relationship between the zeroes and coefficients of a quadratic polynomial

Arithmetic Progression Download

Revision Notes

To find the nth Term of an A.P

Proof:

To find the nth term from the end of an AP

Proof:

Important Results: It is always convenient to make a choice of

Derivation of Sum Formula

Quadratic Equations Download

Revision Notes

Roots of a Quadratic Equation:

Derivation of Quadratic Formula:

Derivation of Formula:

Discriminant:

Nature of the Roots of a quadratic equation:

Important result:

Arithmetic Progression

Revision Notes

To find the nth Term of an A.P

To find the nth term from the end of an AP

Important Results: It is always convenient to make a choice of

Derivation of Sum Formula

Similar Triangles Download

Revision Notes

Proof:

COROLLARY

Theorem 2: Converse of B.P.T.

Definition:

CHARACTERISTIC PROPERTIES OF SIMILARITY

Theorem 3:

Theorem 5:

Proof: