In this page we have ** NCERT book Solutions for Class 8th Maths:Linear equations** for
EXERCISE 2 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

**Question 1: **

If you subtract ½ from a number and multiply the result by 1/2, you get 1/8. What is the number?

**Answer:**

Let x be the number then as per the question

The LCM of denominator is 8, so multiplying by 8 on both sides

4x-2=1

Transposing 2 to R.H.S, we obtain

4x=3

Dividing 4 on both the sides

x=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? **

**Answer:**

Let the breadth be *x* m.

Then as per question the length will be (2*x* + 2) m.

Perimeter of swimming pool = 2(*l + b*) = 154 m

2(2*x* + 2 + *x*) = 154

2(3*x* + 2) = 154

Dividing both sides by 2

3*x *+ 2 = 77

Transposing 2 to R.H.S, we obtain

3*x* = 77 − 2

3*x* = 75

Dividing 3 on both the sides

*x* = 25

So

Breath is 25 m

Length =2*x* + 2 = 2 × 25 + 2 = 52m

Hence, the breadth and length of the pool are 25 m and 52 m respectively.

**Question 3: **

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

**Answer:**

Let the length of equal sides be *x* cm.

Perimeter = *x *cm *+ x *cm + Base

Substituting the values, we get

The LCM of the denominator is 15, so multiplying both the sides by 15

30x+ 20=62

Transposing 20 to R.H.S, we obtain

30x=42

Dividing by 30 on both the sides

x=42/30=7/5

Therefore, the length of equal sides is 7/5 cm.

**Question 4: **

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

**Answer:**

Let one number be* x*. Therefore, the other number will be *x* + 15.

According to the question,

*x + x* + 15 = 95

2*x* + 15 = 95

Transposing 15 to R.H.S, we obtain

2*x* = 95 − 15

2*x* = 80

Dividing both sides by 2, we obtain

*x* = 40

So the other number will be

*x* + 15 = 40 + 15 = 55

Hence, the numbers are 40 and 55.

**Question 5: **

Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

**Answer:**

**In this type of question, the strategy is to choose common ratio and solve the equation to find the exact numbers**

Let the common ratio between these numbers be *x*. Therefore, the numbers will be 5*x* and 3*x* respectively.

Difference between these numbers = 18

5*x* − 3*x* = 18

2*x* = 18

Dividing both sides by 2,

*x* = 9

First number = 5*x* = 5 × 9 = 45

Second number = 3*x* = 3 × 9 = 27

**Question 6: **

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

**Answer:**

Let three consecutive integers be* x*, *x* + 1, and* x* + 2.

Then as per question

Sum of these numbers* = x+ x* + 1 + *x* + 2 = 51

3*x* + 3 = 51

Transposing 3 to R.H.S, we obtain

3*x* = 51 − 3

3*x* = 48

Dividing both sides by 3

*x* = 16, so other numbers will be

*x* + 1 = 17

*x* + 2 = 18

Hence, the consecutive integers are 16, 17, and 18.

**Question 7: **

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

**Answer :**

Let the three consecutive multiples of 8 be 8*x*, 8(*x* + 1), 8(*x* + 2).

As per questions,

Sum of these numbers* =* 8*x *+ 8(*x* + 1) + 8(*x* + 2) = 888

8(*x* + *x* + 1 +* x *+ 2) = 888

8(3*x* + 3) = 888

Dividing both sides by 8

3*x* + 3 = 111

Transposing 3 to R.H.S, we obtain

3*x* = 111 − 3

3*x* = 108

Dividing both sides by 3, we obtain

*x* = 36, so the three multiples will be

First multiple = 8*x = *8 × 36 = 288

Second multiple = 8(*x* + 1) = 8 × (36 + 1) = 8 × 37 = 296

Third multiple = 8(*x* + 2) = 8 × (36 + 2) = 8 × 38 = 304

Hence, the required numbers 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.

**Answer:**

Let three consecutive integers be *x*, *x* + 1, *x* + 2.

According to the question,

2*x *+ 3(*x* + 1) + 4(*x* + 2) = 74

2*x* + 3*x* + 3 + 4*x* + 8 = 74

9*x* + 11 = 74

Transposing 11 to R.H.S, we obtain

9*x* = 74 − 11

9*x* = 63

Dividing both sides by 9, we obtain

*x* = 7, so the integers are

*x* + 1 = 7 + 1 = 8

*x* + 2 = 7 + 2 = 9

Hence, the numbers 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?

**Answer :**

**In this type of question, the strategy is to choose common ratio and solve the equation to find the exact numbers**

Let common ratio between Rahul’s age and Haroon’s age be *y*.

Then

Age of Rahul =5*y* years

Age of haroon = 7*y* years

So After 4 years, the age of Rahul and Haroon will be (5*y *+ 4) years and (7*y *+ 4) years respectively.

According to the given question, after 4 years, the sum of the ages of Rahul and Haroon is 56 years.

(5*y *+ 4) + (7*y *+ 4) = 56

12*y *+ 8 = 56

Transposing 8 to R.H.S, we obtain

12*y* = 56 − 8

12*y* = 48

Dividing both sides by 12

*y* = 4,so the ages will be

Rahul’s age = 5*y* years = (5 × 4) years = 20 years

Haroon’s age = 7*y* years = (7 × 4) years = 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?

**Answer :**

Let the common ratio between the number of boys and numbers of girls be *y*

Number of boys = 7y

Number of girls = 5*y*

As per the question,

Number of boys = Number of girls + 8

7*y* = 5*y* + 8

Transposing 5*y* to L.H.S, we obtain

7*y* – 5*y* = 8

2*y* = 8

Dividing both sides by 2

*y* = 4

So the strength of boys and girls are

Number of boys = 7*y* = 7 × 4 = 28

Number of girls = 5*y*= 5 × 4 = 20

Hence, 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?

**Answer :**

Let us assume Baichung’s father’s age be *x* years.

Therefore, Baichung’s age and Baichung’s grandfather’s age will be (*x* − 29) years and (*x* + 26) years respectively.

As per the question,

The sum of the ages of these 3 people is 135 years.

So

*x +( x* – 29) +( *x* + 26) = 135

3*x* − 3 = 135

Transposing 3 to R.H.S, we obtain

3*x* = 135 + 3

3*x* = 138

Dividing both sides by 3

*x* = 46, then the ages of three people will be

Baichung’s father’s age = *x* years = 46 years

Baichung’s age = (*x* − 29) years = (46 − 29) years = 17 years

Baichung’s grandfather’s age = (*x* + 26) years = (46 + 26) years = 72 years

**Question 12 : **

Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

**Answer :**

Let Ravi’s present age be *x* years.

Fifteen years later, Ravi’s age = 4 × His present age

*x* + 15 = 4*x*

Transposing *x* to R.H.S, we obtain

15 = 4*x* − *x*

15 = 3*x*

Dividing both sides by 3, we obtain

5 = *x*

Hence, Ravi’s present age = 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 What is the number?

**Answer:**

Let x be the rational number ,then

The LCM of 2,3 and 12 is 12,so multiplying by 12

30x+8=-7

Transposing 8 to R.H.S, we obtain

30x=-15

Dividing by 30 on both the sides

x=-1/2

**Question 14 : **

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4, 00,000. How many notes of each denomination does she have**? **

**Answer :**

Let the common ratio between the numbers of notes of different denominations be *y*. Therefore, numbers of Rs 100 notes, Rs 50 notes, and Rs 10 notes will be 2*y*, 3*y*, and 5*y * respectively.

Amount of Rs 100 notes = Rs (100 × 2y) = Rs 200*y*

Amount of Rs 50 notes = Rs (50 × 3*y*)= Rs 150*y*

Amount of Rs 10 notes = Rs (10 × 5*y*) = Rs 50*y*

As per question the total amount is Rs 400000.

200*x* + 150*x* + 50*x* = 400000

400*x* = 400000

Dividing both sides by 400, we obtain

*x* = 1000,so number of notes of each denominations

Number of Rs 100 notes = 2*x* = 2 × 1000 = 2000

Number of Rs 50 notes = 3*x* = 3 × 1000 = 3000

Number of Rs 10 notes = 5*x* = 5 × 1000 = 5000

**Question 15: **

I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

**Answer :**

Let the number of Rs 5 coins be *x*.

Then Number of Rs 2 coins = 3*x*

*Now it is given that that total number of coins is 160,so*

Number of Re 1 coins = 160 − (Number of coins of Rs 5 and of Rs 2)

= 160 − (3*x* + *x*) = 160 − 4*x*

Amount of Re 1 coins = Rs [1 × (160 − 4*x*)] = Rs (160 − 4*x*)

Amount of Rs 2 coins = Rs (2 × 3*x*)= Rs 6*x*

Amount of Rs 5 coins = Rs (5 × *x*) = Rs 5*x*

As per question total amount is Rs 300.

160 − 4*x* + 6*x* + 5*x* = 300

160 + 7*x* = 300

Transposing 160 to R.H.S, we obtain

7*x* = 300 − 160

7*x* = 140

Dividing both sides by 7, we obtain

*x* = 20, so the number of coins of each denominations will be

Number of Re 1 coins = 160 − 4*x* = 160 − 4 × 20 = 160 − 80 = 80

Number of Rs 2 coins = 3*x* = 3 × 20 = 60

Number of Rs 5 coins = *x* = 20

**Question 16 : **

The organizers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3000. Find the number of winners, if the total number of participants is 63.

**Answer:**

Let the number of winners be *x*.

Therefore, the number of participants who did not win will be 63 − *x*.

Amount given to the winners = Rs (100 × *x*) = Rs 100*x*

Amount given to the participants who did not win = Rs [25(63 − *x*)]

= Rs (1575 − 25*x*)

As per the question,

100*x* + 1575 − 25*x *= 3000

Transposing 1575 to R.H.S, we obtain

75*x *= 3000 − 1575

75*x *= 1425

Dividing both sides by 75, we obtain

*x* = 19

Hence, number of winners = 19

Download this assignment as pdf

Go Back to Class 8 Maths Home page Go Back to Class 8 Science Home page

- Biology Foundation Course for AIPMT/Olympiad : Class 9
- Science FR Ninth Class Part-3 (Biology) (PB)
- Physics Part 1 Class - 9
- Chemistry Part 2 Class - 9
- The IIT Foundation Series Physics Class 9
- The IIT Foundation Series Chemistry Class 9
- Oswaal CBSE CCE Question Banks With Complete Solution For Class 9 Term-II (October To March 2016) Science
- x am idea science class 9 term -2
- NCERT Exemplar Problems: Solutions Science Class 9
- NCERT Solutions - Science for Class IX