# Linear Equations in One Variable – Exercise 2.2 / Chapter 2

**CBSE CLASS 8 MATHEMATICS**

**Linear Equations in One Variable – Chapter 2**

**Exercise 2.2**

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

Solution:

Let the number be x.

Given (x – ½) ½ = 1/8.

½ x – ½ x ½ = 1/8

x/2 – ¼ = 1/8

x/2 = 1/8 + ¼ = 1/8 + 2/8 =3/8 (Convert to like fractions)

x = ¾.

**2. The perimeter of a rectangular swimming pool is 154m. Its length is 2m more than twice its breadth. What are the length and the breadth of the pool?**

Solution:

Let the breadth of the pool be b.

Length = 2 + 2b

Given perimeter = 154

We know that perimeter of a rectangle = 2 (l + b)

Therefore, 154 = 2(2 + 2b + b)

154 = 2(2 + 3b)

77 = 2 + 3b

77 – 2 = 3b

75 = 3b

Breadth = b = 75/3 = 25m

Length = 2 + 2 x 25 = 2 + 50 = 52m.

**3. The base of an Isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 2/15cm. What is the length of either of the remaining equal sides?**

Solution:

Let the length of equal sides be x.

Given perimeter = 4 2/15 cm = 62/15 cm

Therefore, x + x + 4/3 = 62/15

2x + 4/3 = 62/15

2x = 62/15 -4/3 = 62/15 – 20/15 = 42/15 = 14/5

x = 7/5 = 1 2/5 cm.

**4. Sum of two numbers be 95. If one exceeds the other by 15, find the numbers.**

Solution:

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

Given sum of two number = 95

i.e, x + x + 15 = 95

2x + 15 = 95

2x = 95 – 15 = 80

x = 80/2 = 40

So one number is 40 and the other number is x + 15 = 40 + 15 = 55.

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

Solution:

Given two numbers are in the ratio 5:3.

Let the two numbers be 5x and 3x.

Difference = 18 means 5x – 3x = 18

2x = 18

x = 9

The numbers are 5x = 45 and 3x = 27.

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

Solution:

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

Given x + x + 1+ x+ 2 = 51

3x + 3 = 51

3x = 51 – 3 = 48

x = 48/3 = 16.

So the three consecutive integers are 16, 17 and 18.

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

Solution:

Let the three consecutive multiples of 8 be 8x, 8x +8, 8x + 16.

Given 8x + 8x + 8+ 8x + 16 = 888

24x + 24 = 888

24(x + 1) = 888

x + 1 = 888/ 24 = 37

x = 37 -1 = 36.

So the three consecutive multiples of 8 are 8x = 8 x 36 = 288

8x +8 = 288 + 8 = 296

8x + 16 = 288 + 16 = 304.

**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 be x, x+1, x+2.

When it is multiplied by 2, 3 and 4 respectively, the numbers become 2x, 3(x+1) and 4(x+2).

Given 2x + 3(x+1) + 4(x+2) = 2x + 3x+3+4x+8 = 74

9x + 11 = 74

9x = 74 – 11 = 63

x = 63/9 = 7

So the three consecutive integers are 7, 8 and 9.

**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:

It is given that the ages are in the ratio 5:7.

So let the present ages be 5x and 7x.

After 4 years, their ages become 5x + 4 and 7x + 4.

Given 5x + 4+ 7x + 4 = 56

12x + 8 = 56

12x = 56 – 8 = 48

x = 48/12 = 4.

So the present age of Rahul is 5x = 5 x 4 = 20 years

Resent age of Haroon is 7x = 7 x 4 = 28 years.

**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:

It is given that the number of boys and girls are in the ratio 7:5.

Let the number of boys be 7x and the number of girls be 5x.

Given 7x = 8+5x

i.e, 7x – 5x = 8

2x = 8

x = 8/2 = 4

So the number of boys are 7x = 7 x 4 = 28

And number of girls = 5x = 5 x 4 = 20

Total class strength = 28 + 20 = 48.

**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 the age of Baichung be x years.

Age of Baichung’s father = x + 29

Age of Baichung’s grandfather = x + 29 + 26 = x + 55

Given x + x + 29 + x + 55 = 135

3x + 84 = 135

3x = 135 – 84 = 51

x = 51/3 = 17.

Therefore, age of Baichung = 17 years

Age of Baichung’s father = x + 29 = 17 + 29 = 46 years

Age of Baichung’s grandfather = 46 + 26 = 72 years.

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

Solution:

Let Ravi’s present age be x years.

After 15 years, Ravi’s age = x + 15.

Given, 4x = x + 15

4x – x = 15

3x = 15

x = 15/3 = 5.

So Ravi’s present age is 5 years.

**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 number be x.

It is given that 5/2 x + 2/3 = -7/12.

5x/2 = -7/12 – 2/3 = -7/12 – 8/12 (convert to like fractions)

5x/2 = -15/12

x = -1/2.

**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 these notes is 2:3:5. The total cash with Lakshmi is Rs 400000.How many notes of each denomination does she have?**

Solution:

Given ratio is 2:3:5.

Let the number of currency notes of Rs 100, 50 and 10 respectively be 2x, 3x and 5x .

Converting all the denominations into rupees,

So we get 100 x 2x = 200x, 50 x 3x = 150x, 10 x 5x = 50x.

Total cash = 400000.

Given, 200x + 150x + 50x = 400000.

400x = 400000

x = 400000/400 = 1000.

Therefore, number of 100 rupee notes = 2x = 2000

Number of 50 rupee notes = 3x = 3000

Number of 10 rupee notes = 5x = 5000.

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

Solution:

Let the number of 5 rupee coins be x.

Then given number of 2 rupee coins = 3x.

Total number of coins = 160.

Then number of 1 rupee coins = 160 –(x + 3x)

Converting all the denominations into rupees,

So we get x coins of 5 rupee becomes 5x, 3x coins of 2 rupee becomes 6x and 160 –(x +3x) coins of 1 rupee becomes 160 -4x.

Given 5x + 6x + 160 – 4x = 300

7x + 160 =300

7x = 140

x = 140/7 = 20.

Therefore, number of 5 rupee coins = 20

Number of 2 rupee coins = 3x = 60

Number of 1 rupee coins = 160 – 4x = 160 – 80 = 80.

**16. The organisers 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 Rs 25. The total prize money distributed is Rs 3000. Find the number of winners, if the total number of participants is 63.**

Solution:

Let the number of winners be x.

Number of participants who does not win = 63 –x.

Amount given to the winner = 100x

Amount given to the participant who does not win =25 (63-x)

Total prize money distributed = 3000.

i.e, 100x + 25(63 – x) = 100x + 1575 – 25x = 3000

75x = 1425

x = 1425/75 = 19.

So number of winners = 19.