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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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2. Linear Equations in One Variable

| Exercise 2.2 |

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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Study Materials List:

Solutions ⇒ Class 8th ⇒ Mathematics
1. Rational Numbers
2. Linear Equations in One Variable
3. Understanding Quadrilaterals
4. Practical Geometry
5. Data Handling
6. Squares and Square Roots
7. Cubes and Cube Roots
8. Comparing Quantities
9. Algebraic Expressions and Identities
10. Visualising Solid Shapes
11. Mensuration
12. Exponents and Powers
13. Direct and Inverse Proportions
14. Factorisation
15. Introduction to Graphs
16. Playing with Numbers

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