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

• Last Updated : 02 Dec, 2020

### Question 1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8 what is the number?

Solution:

Let the number be ‘a’.

According to the question,

(a â€“ 1/2) Ã— 1/2 = 1/8

a/2 â€“ 1/4 = 1/8

a/2 = 1/8 + 1/4

a/2 = 1/8 + 2/8

a/2 = (1 + 2)/8

a/2 = 3/8

a = (3/8) Ã— 2

So,

a = 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 breadth of the pool?

Solution:

Given that,

Perimeter of rectangular swimming pool = 154 m

Let the breadth of rectangle be ‘a’

Length of the rectangle = 2a + 2 We know that,

Perimeter = 2 Ã— (length + breadth)

So,

2(2a + 2 + a) = 154

2(3a + 2) = 154

3a + 2 = 154/2

3a = 77 â€“ 2

3a = 75

a = 75/3

a = 25

Length = 2a + 2

= (2 Ã— 25) + 2

= 50 + 2

Length = 52 m

### Question 3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 62/15 cm. What is the length of either of the remaining equal sides?

Solution:

Base of isosceles triangle = 4/3 cm

Perimeter of triangle = 62/15

Let the length of equal sides of triangle be ‘a’.

So,

2a = (62/15 â€“ 4/3)

2a = (62 â€“ 20)/15

2a = 42/15

a = (42/30) Ã— (1/2)

a = 42/30

a = 7/5

So, length of either of the remaining equal sides are 7/5 cm each.

### Question 4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution:

Let one of the numbers be ‘a’.

Then, the other number becomes (a + 15) Given in the question,

Also given that,

a + (a + 15) = 95

2a + 15 = 95

2a = 95 â€“ 15

2a = 80

a = 80/2

a = 40

So, First number = 40

And, other number is = (a + 15) = 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 numbers be ‘5a’ and ‘3a’.  So, according to the question,

5a â€“ 3a = 18

2a = 18

a = 18/2

a = 19

Thus,

The  first numbers is (5a) = 5 Ã— 9 = 45

And another number (3a) = 3 Ã— 9 = 27.

### Question 6. Three consecutive integers add up to 51. What are these integers?

Solution:

Let the three consecutive integers be ‘a’, ‘a + 1’ and ‘a + 2’. So, according to the question,

a + (a + 1) + (a + 2) = 51

3a + 3 = 51

3a = 51 â€“ 3

3a = 48

a = 48/3

a = 16

So, the integers are

First integer will be (a) = 16

Second integer will be (a + 1) = 17

& third integer will be (a + 2) = 18

### Question 7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Solution:

Let the three consecutive multiples of 8 be ‘8a’, ‘8(a+1)’ and ‘8(a+2)’. According to the question,

Given,

8a + 8(a + 1) + 8(a + 2) = 888

8 (a + a + 1 + a + 2) = 888 (Taking 8 as common)

8 (3a + 3) = 888

3a + 3 = 888/8

3a + 3 = 111

3a = 111 â€“ 3

3a = 108

a = 108/3

a = 36

Thus, the three consecutive multiples of 8 are:

First no. = 8a = 8 Ã— 36 = 288

Second no. = 8(a + 1) = 8 Ã— (36 + 1) = 8 Ã— 37 = 296

Third No. = 8(a + 2) = 8 Ã— (36 + 2) = 8 Ã— 38 = 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 are ‘a’, ‘a+1’ and ‘a+2’. According to the question,

Given,

2a + 3(a + 1) + 4(a + 2) = 74

2a + 3a +3 + 4a + 8 = 74

9a + 11 = 74

9a = 74 â€“ 11

9a = 63

a = 63/9

a = 7

Thus, the numbers are:

First integer. = a = 7

Second integer = a + 1 = 8

and Third integer = a + 2 = 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 ages of Rahul and Haroon be ‘5a’ and ‘7a’.

Four years later,

The ages of Rahul and Haroon will be (5a + 4) and (7a + 4) respectively. According to the question,

Given, (5a + 4) + (7a + 4) = 56

5a + 4 + 7a + 4 = 56

12a + 8 = 56

12a = 56 â€“ 8

12a = 48

a = 48/12

a = 4

Therefore, Present age of Rahul = 5a = 5 Ã— 4 = 20

And, present age of Haroon = 7a = 7 Ã— 4 = 28

### 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 be ‘7a’ and girls be ‘5a’.

According to the question,

Given, 7a = 5a + 8

7a â€“ 5a = 8

2a = 8

a = 8/2

a = 4

Therefore, Number of boys = 7 Ã— 4 = 28

And, Number of girls = 5 Ã— 4 = 20

Total number of students = 20 + 28 = 48

### Question 11. Baichungâ€™s father is 26 years younger than Baichungâ€™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 age of Baichungâ€™s father be ‘a’.

Then, age of Baichung’s grandfather = (a + 26)

and, Age of Baichung = (a – 29) According to the question,

Given, a + (a + 26) + (a – 29) = 135

3a + 26 â€“ 29 = 135

3a â€“ 3 = 135

3a = 135 + 3

3a = 138

a = 138/3

a = 46

Age of Baichungâ€™s father = a = 46

Age of Baichungâ€™s grandfather = (a + 26) = 46 + 26 = 72

Age of Baichung = (a – 29) = 46 â€“ 29 = 17

### Question 12. Fifteen years from now Raviâ€™s age will be four times his present age. What is Raviâ€™s present age?

Solution:

Let the present age of Ravi be ‘a’.

Fifteen years later, Ravi age will be (a+15) years. According to the question,

Given, a + 15 = 4a

4a â€“ a = 15

3a = 15

a = 15/3

a = 5

Therefore, Present age of Ravi = 5 years.

### Question 13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?

Solution:

Let the rational be ‘a’.

According to the question,

Given, a Ã— (5/2) + 2/3 = -7/12

5(a/2) + 2/3 = -7/12

5(a/2) = -7/12 â€“ 2/3

5(a/2) = (-7- 8)/12

5(a/2) = -15/12

5a/2 = -5/4

a = (-5/4) Ã— (2/5)

a = â€“ 10/20

a = -1/2

Therefore, the rational number will be -1/2.

### Question 14. Lakshmi is a cashier in a bank. She has currency notes of denominations â‚¹100, â‚¹50 and â‚¹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is â‚¹4,00,000. How many notes of each denomination does she have?

Solution:

Let the numbers of notes of â‚¹100, â‚¹50 and â‚¹10 be ‘2a’ , ‘3a’ and ‘5a’ respectively.

Value of â‚¹100 = 2a Ã— 100 = 200a

Value of â‚¹50 = 3a Ã— 50 = 150a

Value of â‚¹10 = 5a Ã— 10 = 50a According to the question,

Given, 200a + 150a + 50a = 400000

400a = 400000

a = 400000/400

a = 1000

Numbers of â‚¹100 notes = 2a = 2000

Numbers of â‚¹50 notes = 3a = 3000

Numbers of â‚¹10 notes = 5a = 5000

### Question 15. I have a total of â‚¹300 in coins of denomination â‚¹1, â‚¹2 and â‚¹5. The number of â‚¹2 coins is 3 times the number of â‚¹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Solution:

Let number of â‚¹5 coins be ‘a’.

Then,

Number â‚¹2 coins = 3a

and, number of â‚¹1 coins = (160 â€“ 4a) Now,

Value of â‚¹5 coins= a Ã— 5 = 5a

Value of â‚¹2 coins = 3a Ã— 2 = 6a

Value of â‚¹1 coins = (160 â€“ 4a) Ã— 1 = (160 â€“ 4a)

According to the question,

Given, 5a + 6a + (160 â€“ 4a) = 300

11a + 160 â€“ 4a = 300

7a = 140

a = 140/7

a = 20

Number of â‚¹5 coins =  a = 20

Number of â‚¹2 coins = 3a = 60

Number of â‚¹1 coins = (160 â€“ 4a) = 160 â€“ 80 = 80

### Question 16. The organizers of an essay competition decide that a winner in the competition gets a prize of â‚¹100 and a participant who does not win gets a prize of â‚¹25. The total prize money distributed is â‚¹3,000. Find the number of winners, if the total number of participants is 63.

Solution:

Let the numbers of winner be ‘a’

Then, the number of participant who didnâ€™t win will be (63 â€“ a)

Total money given to the winner = a Ã— 100 = 100a

Total money given to participant who didnâ€™t win = 25 Ã— (63 – a)

According to the question,

Given, 100a + 25 Ã— (63 – a) = 3000

100a + 1575 â€“ 25a = 3000

75a = 3000 â€“ 1575

75a = 1425

a = 1425/75

a = 19

So, the number of winners are 19.

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