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Solutions ⇒ Class 8th ⇒ Mathematics ⇒ 2. Linear Equations in One Variable

# Solutions 2. Linear Equations in One Variable - Exercise 2.2 | Class 8 Mathematics - Toppers Study

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## Solutions 2. Linear Equations in One Variable - Exercise 2.2 | Class 8 Mathematics - Toppers Study

Exercise 2.2

Q1.   If you subtract   from a number and multiply the result by  , you get . What is the number?

Solution:

Let the number be x

so, the equation will be

Q2.   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 swimming pool x.

So, the equation will be

so, the length  of pool 2x + 2

= 2×25 + 2

= 50 + 2

= 52

and the breadth of pool x = 25

Q3. The base of an isosceles triangle   cm the perimeter of a triangle is 4  cm what is the length of either remaining equal sides?

Solution:

Let the length of one remaining equal sides x.So, equation wil

Q4.   Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution:

Let the second number is x.

So, the equation will be  x + x + 15 = 95

So, the first number will be x + 15

= 40 + 15

= 55

And second number: x = 40

Q5.   Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Solution:

Let,

Both numbers will be 5x and 3x

So, the equation will be 5x -  3x = 18

2x = 18

so, the first number will be 5x

= 5 × 9

= 45

and the second number will be =  3 × 9

=  27

Q6.   Three consecutive integers add up to 51. What are these integers?

Solution:

Let, all numbers x, x+1, x+2 respectively.

So, the equation will be x+x+1+x+2=51

⇒ 3x +3=51

⇒ 3x=51-3

So, the first number will be x = 16

, second number will be x + 1

= 16 + 1

= 17

and the third  number will be x + 2 =  18

Q7. The sum of three consecutive multiples of 8 is 888. Find the multiples?

Solution:

Let, three consecutive multiples of 8 is x, x+8, and x+16 respectively

So, the equations will be : x + x + 8 + x + 16 = 888

3x + 24 = 888

⇒ 3x = 888 - 24

So, the first multiple x = 288

Second multiple x + 8 = 288+8 = 296

Third multiple x + 16 = 288 + 16 = 304

Q8.   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, all numbers be x, x+1, x+2

So, the equation will be: 2(x) + 3(x+1) +4(x+2) =74

⇒ 2x+ 3x+3+ 4x+8 = 74

⇒ 9x+11=74

⇒ 9x=74-11

So, first number is x = 7

Second number is x + 1

= 7 + 1 = 8

Third number is x + 2

= 7 + 2

= 9

Q9.   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 ages of Rahul and Haroon be 5x and 7x respectively

So, the equation will be 5x + 4 + 7x + 4 = 56

⇒ 12x + 8 = 56

⇒ 12x = 56 - 8

So, the present age of Rahul: 5x = 5 × 4 = 20

The present age of Haroon: 7x = 7 × 4 = 28

Q10.   Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Solution:

Let, the age of Aman’s son x

5(x - 10) = 3x - 10

5x – 50 = 3x - 10

5x – 3x = -10 + 50

2x = 40

The present age of Aman: 3x = 3 × 20 = 60

The present age of his son: x = 20

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