Practice Questions Class 8 NCERT Chapter 2 Linear Equations in One Variable Set 2

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 – 1/2) × 1/2 = 1/8

a/2 – 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) × 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 × (length + breadth)

So, 

2(2a + 2 + a) = 154

2(3a + 2) = 154

3a + 2 = 154/2

3a = 77 – 2

3a = 75

a = 75/3

a = 25

Therefore, Breadth = 25 m

Length = 2a + 2

= (2 × 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 – 4/3)

2a = (62 – 20)/15

2a = 42/15 

a = (42/30) × (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 – 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 – 3a = 18

2a = 18

a = 18/2

a = 19

Thus,

The  first numbers is (5a) = 5 × 9 = 45

And another number (3a) = 3 × 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 – 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 – 3

3a = 108

a = 108/3

a = 36

Thus, the three consecutive multiples of 8 are:

First no. = 8a = 8 × 36 = 288

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

Third No. = 8(a + 2) = 8 × (36 + 2) = 8 × 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 – 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 – 8

12a = 48

a = 48/12

a = 4

Therefore, Present age of Rahul = 5a = 5 × 4 = 20

And, present age of Haroon = 7a = 7 × 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 – 5a = 8

2a = 8

a = 8/2

a = 4

Therefore, Number of boys = 7 × 4 = 28

And, Number of girls = 5 × 4 = 20

Total number of students = 20 + 28 = 48

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 age of Baichung’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 – 29 = 135

3a – 3 = 135

3a = 135 + 3

3a = 138

a = 138/3

a = 46

Age of Baichung’s father = a = 46

Age of Baichung’s grandfather = (a + 26) = 46 + 26 = 72

Age of Baichung = (a – 29) = 46 – 29 = 17

Question 12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’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 – 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 × (5/2) + 2/3 = -7/12

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

5(a/2) = -7/12 – 2/3

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

5(a/2) = -15/12

5a/2 = -5/4

a = (-5/4) × (2/5)

a = – 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 ₹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 numbers of notes of ₹100, ₹50 and ₹10 be ‘2a’ , ‘3a’ and ‘5a’ respectively.

Value of ₹100 = 2a × 100 = 200a

Value of ₹50 = 3a × 50 = 150a

Value of ₹10 = 5a × 10 = 50a According to the question,

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

400a = 400000

a = 400000/400

a = 1000

Numbers of ₹100 notes = 2a = 2000

Numbers of ₹50 notes = 3a = 3000

Numbers of ₹10 notes = 5a = 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 number of ₹5 coins be ‘a’.

Then,

Number ₹2 coins = 3a

and, number of ₹1 coins = (160 – 4a) Now,

Value of ₹5 coins= a × 5 = 5a

Value of ₹2 coins = 3a × 2 = 6a

Value of ₹1 coins = (160 – 4a) × 1 = (160 – 4a)

According to the question,

Given, 5a + 6a + (160 – 4a) = 300

11a + 160 – 4a = 300

7a = 140

a = 140/7

a = 20

Number of ₹5 coins =  a = 20

Number of ₹2 coins = 3a = 60

Number of ₹1 coins = (160 – 4a) = 160 – 80 = 80

Question 16. The organizers 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 ₹3,000. 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 didn’t win will be (63 – a)

Total money given to the winner = a × 100 = 100a

Total money given to participant who didn’t win = 25 × (63 – a)

According to the question,

Given, 100a + 25 × (63 – a) = 3000

100a + 1575 – 25a = 3000

75a = 3000 – 1575

75a = 1425

a = 1425/75

a = 19

So, the number of winners are 19.