# NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2

Question 1.
If you subtract $$\frac{1}{2}$$ from a number and multiply the result by $$\frac{1}{2}$$, you get $$\frac{1}{8}$$. What is the number.
Solution:
Let the number be ‘x’
According to the given condition, we get
$$\left(x-\frac{1}{2}\right) \frac{1}{2}=\frac{1}{8}$$
Multiplying both sides by 2
$$\mathrm{x}-\frac{1}{2}=\frac{1}{8} \times 2=\frac{1}{4}$$
Transposing $$\left(-\frac{1}{2}\right)$$ to R.H.S. we get
x = $$\frac{1}{4}+\frac{1}{2}$$
x = $$\frac{1+2}{4}=\frac{3}{4}$$
∴ The required number is $$\frac{3}{4}$$. Question 2.
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution:
Let the breadth of the pool be x m
∴ Length of the pool = (2 + 2x) m = (2x + 2) m
Perimeter of the pool = 154 m
2(2x + 2 + x) = 154 m
[∵ Perimeter of the rectangle = 2(l + b)]
2(3x + 2) = 154
6x + 4 = 154
Transposing 4 to the R.H.S.
6x = 154 – 4
6x = 150
Dividing both sides by 6,
$$\frac{6 x}{6}=\frac{150}{6}$$
∴ x = 25
∴ Breadth of the pool = 25m
Length of the pool = 2(25) + 2 = 50 + 2 = 52 m

Question 3.
The base of an isosceles triangle is $$\frac{4}{3}$$ cm and the perimeter of the triangle is 4$$\frac{2}{15}$$ cm. What is the length of the remaining equal sides?
Solution:
Let the length of the equal sides of a triangle be ‘x’
Base of an isosceles triangle = $$\frac{4}{3}$$ cm
Perimeter of the triangle = 4$$\frac{2}{15}$$ cm ∴ Length of the equal sides is 1$$\frac{2}{5}$$ cm. Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let the smaller number be x
then the greater number = x + 15
Sum of two numbers = 95
x + (x + 15) = 95
2x + 15 = 95
Transposing 15 to R.H.S, we get
2x = 95 – 15 = 80
Dividing both sides by 2
x = $$\frac{80}{2}$$ = 40
∴ The smaller number = 40
The greater number = (40 + 15) = 55

Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Solution:
Let the two number be 5x and 3x
Difference of two numbers = 18
5x – 3x = 18
2x = 18
Dividing both sides by 2, we get
$$\frac{2 \mathrm{x}}{2}=\frac{18}{2}$$
∴ x = 9
The two numbers are (5 × 9) and (3 × 9) i.e. 45 and 27.
Hence, the required numbers are 45 and 27. Question 6.
Three consecutive integers add up to 51. What are these integers?
Solution:
Let the three consecutive integers be x, x + 1 and x + 2
Sum of three integers = 51
x + x + 1 + x + 2 = 51
3x + 3 = 51
Transposing 3 to RHS, we get
3x = 51 – 3 = 48
Dividing both sides by 3, we get
$$\frac{3 \mathrm{x}}{3}=\frac{48}{3}$$
x = 16
Now x = 16,
x + 1 = 16 + 1 = 17
x + 2 = 16 + 2 = 18
∴ The required three consecutive integers are 16, 17 and 18.

Question 7.
The sum of three consecutive multiple of 8 is 888. Find the multiples.
Solution:
Let the three consecutive multiples of 8 be x, x + 8, x + 16
Sum of three consecutive multiples = 888
x + x + 8 + x + 16 = 888
3x + 24 = 888
Transposing 24 to R.H.S., we get
3x = 888 – 24 = 864
Dividing both sides by 3, we get
$$\frac{3 x}{3}=\frac{864}{3}$$
x = 288
x + 8 = 288 + 8 = 296
x + 16 = 288 + 16 = 304
∴ The required multiples of 8 are 288, 296 and 304. Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let the three consecutive integers be x, x + 1 and x + 2
According to the given condition
2(x) + 3(x + 1) + 4(x + 2) = 74
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
Transposing 11 to R.H.S., we get
9x = 74 – 11
9x = 63
Dividing both sides by 9, we get
$$\frac{9 x}{9}=\frac{63}{9}$$
x = 7
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9
∴ The consecutive integers are 7, 8 and 9.

Question 9.
The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later, the sum of their ages will be 56 years. What are their present ages?
Solution:
Let the present age of Rahul be ‘5x’
and the present age of Harron be 7x;
4 years later
Rahul’s age = (5x + 4) years
Harron’s age = (7x + 4) years
Sum of their ages = 56 years
5x + 4 + 7x + 4 = 56
12x + 8 = 56
Transposing 8 to R.H.S. we get
12x = 56 – 8 = 48
Dividing both sides by 12
$$\frac{12 \mathrm{x}}{12}=\frac{48}{12}$$
x = 4
Present age of Rahul = 5 × 4 = 20 years
Present age of Haroon = 7 × 4 = 28 years Question 10.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength.
Solution:
Let the number of boys in the class be 7x
and the number of girls in the class be 5x
As number of boys is 8 more than the number of girls.
7x = 5x + 8
Transposing 5x to L.H.S.
7x – 5x = 8
2x = 8
Dividing both sides by 2
$$\frac{2 \mathrm{x}}{2}=\frac{8}{2}$$
x = 4
∴ Number of boys = 7 × 4 = 28
Number of girls = 5 × 4 = 20
Total class strength = 28 + 20 = 48 students.

Question 11.
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Solution:
Let Baichung’s present age be x years
Baichung father’s age = (x + 29) years
Baichung grandfather’s age = (x + 29 +26) years = (x + 55) years
Sum of the ages of all the three =135 years
x + x + 29 + x + 55 = 135
3x + 84 = 135
Transposing 84 to RHS, we have
3x = 135 – 84 = 51
Dividing both sides by 3, we get
$$\frac{3 x}{3}=\frac{51}{3}$$
x = 17
Baichung’s age =17 years
Baichung fathers age = 17 + 29 = 46 years
Baichung grandfather’s age = 17 + 55 = 72 years. Question 12.
Fifteen years from now, Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let Ravi’s present age be ‘x’ years
4 times his present age = 4x years
15 years from now, Ravi’s age = (x + 15) years
According to the given condition
x + 15 = 4x
Transposing 15 and 4x, we get
-4x + x = -15
-3x = -15
Dividing both sides by (-3) we get
$$\frac{-3 x}{-3}=\frac{-15}{-3}$$
x = 5
∴ Ravi’s present age = 5 years.

Question 13.
A rational number is such that when you multiply it by $$\frac{5}{2}$$ and add $$\frac{2}{3}$$ to the product, you get $$\frac{7}{12}$$. What is the number?
Solution:
Let the required rational number be ‘x’
According to the given condition x = $$\frac{-1}{2}$$
∴ The rational number is $$\frac{-1}{2}$$. Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10 respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00, 000. How many notes of each denomination does she have?
Solution:
Let the number of ₹ 100 notes be 2x
the number of ₹ 50 notes be 3x
and the number of ₹ 10 notes be 5x
Value of ₹ 100 notes = 2x × 100 = ₹ 200x
Value of ₹ 50 notes = 3x × 50 = ₹ 150x
Value of ₹ 10 notes = 5x × 10 = ₹ 50x
According to the given condition,
₹ 200x + ₹ 150x + ₹ 50x = ₹ 4,00000
400x = 4,00,000
Dividing both sides by 400, we get
$$\frac{400 \mathrm{x}}{400}=\frac{4,00,000}{400}$$
x = 1000
∴ Number of ₹ 100 notes = 2 × 1000 = 2000
Number of ₹ 50 notes = 3 × 1000 = 3000
Numbers of ₹ 10 notes = 5 × 1000 = 5000

Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of ₹ 5 coins be x
then number of ₹ 2 coins = 3x
and the number of coins of ₹ 1 = 160 – (x + 3x) = 160 – 4x
Now Value of ₹ 5 coins = ₹ 5 × x = 5x
Value of ₹ 2 coins = ₹ 2 × 3x = 6x
Value of ₹ 1 coins = ₹ 1 × (160 – 4x) = (160 – 4x)
According to the given condition,
160 – 4x + 5x + 6x = 300
160 + 7x = 300
Transposing 160 to the R.H.S.
7x = 300 – 160
7x = 140
Dividing both sides by 7
$$\frac{7 \mathrm{x}}{7}=\frac{140}{7}$$
x = 20
∴ Number of ₹ 5 coins = 20
Number of ₹ 2 coins = (3 × 20) = 60
Number of ₹ 1 coin = 160 – (4 × 20) = 80 Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners be ‘x’
The total number of participants = 63
The number of non-winners = 63 – x
Prize money given to winners = ₹ 100 × x
Prize money given to non-winner participants = ₹ 25(63 – x)
= ₹ 25 × 63 – ₹ 25x
= ₹ 1575 – ₹ 25x
According to the condition given in the question,
100x + 1575 – 25x = 3000
75x + 1575 = 3000
Transposing 1575 to R.H.S. we get
75x = 3000 – 1575
75x = 1425
Dividing both sides by 75
$$\frac{75 \mathrm{x}}{75}=\frac{1425}{75}$$
x = 19
∴ The number of winners = 19.

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