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.