# Linear Equations in One Variable : Exercise 2.2 (Mathematics NCERT Class 8th)

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

**Let the number be**

*Sol.**x*.

Given,

Dividing both the sides by , we get,

Transposing to RHS, we get,

**Q.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?
**

**Let the breadth be**

*Sol.**x*m. Therefore, the length will be (2

*x*+ 2) m.

Now, given perimeter of swimming pool = 2(length + breadth) = 154 m.

2 (2

*x*+ 2 +

*x*) = 154

2 (3

*x*+ 2) = 154

Dividing both the sides by 2, we get,

(3

*x*+ 2) = 77

Transposing 2 from LHS to RHS, we get,

3

*x*= 77 â€“ 2

3

*x*= 75

Dividing by 3 on both the sides, we get,

*x*= 25

Therefore, (2

*x*+ 2) = (2Ã—25 + 2) = 52

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

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

**Let the length of equal sides be**

*Sol.**x*cm.

Now, perimeter of triangle =

*x + x*+ base =

Transposing from LHS to RHS, we get,

Dividing both the sides by 2, we get,

Therefore, the length of equal sides is cm.

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

**Let one number be**

*Sol.**x*. Thus, the other number will be

*x*+ 15.

Given,

*x*+

*x*+ 15 = 95

2

*x*+ 15 = 95

Transposing 15 from LHS to RHS, we get,

2

*x*= 95 â€“ 15

2

*x*= 80

Dividing both the sides by 2, we get,

*x*= 40

*x*+ 15 = 40 + 15 = 55

Thus, the numbers are 40 and 55.

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

**Let the first number be 5**

*Sol.**x*and second number be 3

*x*.

Given, 5

*x*3

*x*= 18

2

*x*= 18

Dividing by 2 on both the sides, we get,

*x*= 9

Since, first number = 5

*x*= 5 Ã— 9 = 45

Second number = 3

*x*= 3 Ã— 9 = 27

Therefore, the numbers are 45 and 27.

**Q.6 Three consecutive integers add up to 51. What are these integers?
**Sol. Let the three consecutive integers be

*x*,

*x*+ 1 and

*x*+ 2.

Given,

*x*+

*x*+ 1 +

*x*+ 2 = 51

3

*x*+ 3 = 51

Transposing 3 from LHS to RHS, we get,

3

*x*= 51 â€“ 3

3

*x*= 48

Dividing by 3 on both the sides, we get,

*x*= 16

Therefore,

*x*+ 1 = 17 and

*x*+ 2 = 18.

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

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

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

*Sol.*Given, 8a + 8(a + 1) + 8(a + 2) = 888

8(a + a + 1 + a + 2) = 888

8(3a + 3) = 888

Dividing by 8 on both the sides, we get,

3a + 3 = 111

Transposing 3 from LHS to RHS, we get,

3a = 111 â€“ 3

3a = 108

Dividing by 3 on both the sides, we get,

a = 36

Therefore, 8a = 8 Ã— 36 = 288,

8(a + 1) = 8(36 + 1) = 8 Ã— 37 = 296

8(a + 2) = 8(36 + 2) = 8 Ã— 38 = 304

Thus, the required numbers are 288, 296 and 304.

**Q.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.
**

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

*Sol.*Given, 2a + 3(a + 1) + 4(a + 2) = 74

2a + 3a + 3 + 4a + 8 = 74

9a + 11 = 74

Transposing 11 from LHS to RHS, we get,

9a = 74 â€“ 11

9a = 63

Dividing by 9 on both the sides, we get,

a = 7

Therefore, a + 1 = 7 + 1 = 8, a + 2 = 7 + 2 = 9

Therefore, the required numbers are 7, 8 and 9.

**Q.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?
**

**Let the age of Rahul be 5**

*Sol.**x*and the age of Haroon be 7

*x*.

After 4 years, the age of Rahul will be (5

*x*+ 4) and that of Haroon will be (7

*x*+ 4)

Given, (5

*x*+ 4) + (7

*x*+ 4) = 56

12

*x*+ 8 = 56

Transposing 8 from LHS to RHS, we get,

12

*x*= 56 â€“ 8

12

*x*= 48

Dividing by 12 on both the sides, we get,

*x*= 4

Therefore, age of Rahul = 5

*x*= 5 Ã— 4 = 20 years

Age of Haroon = 7

*x*= 7 Ã— 4 = 28 years

**Q.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?
**

**Let the number of boys be 7**

*Sol.**x*and number of girls be 5

*x*.

Given, 7

*x*â€“ 8 = 5

*x*

Transposing 7

*x*from LHS to RHS, we get,

- 8 = 5

*x*â€“ 7

*x*

- 8 = â€“ 2

*x*

8 = 2

*x*

Dividing both the sides by 2, we get,

4 =

*x*

*x*= 4

Therefore, number of boys = 7

*x*= 7 Ã— 4 = 28

Number of boys = 5

*x*= 5 Ã— 4 = 20

Hence, total class strength = 28 + 20 = 48 students.

**Q.11 Baichungâ€™s father is 26 years younger than Baichungâ€™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?
**

**Let Baichungâ€™s fatherâ€™s age be a years.**

*Sol.*Hence, age of Baichungâ€™s father and Baichungâ€™s grandfather will be (a â€“ 29) and (a + 26) respectively.

Given, a + (a â€“ 29) + (a + 26) = 135

3a â€“ 3 = 135

Transposing 3 from LHS to RHS, we get,

3a = 135 + 3

3a = 138

Dividing by 3 on both the sides, we get,

a = 46

Thus, age of Baichungâ€™s father = a years = 46 years

Age of Baichung = (a â€“ 29) = (46 â€“ 29) = 17 years

Age of Baichungâ€™s Grandfather = (a + 26) = (46 + 26) = 72 years.

**Q.13 A rational number is such that when you multiply it by ****and add **** to the product, you get****. What is the number?
**

**Let the number be a.**

*Sol.*Given,

Transposing from LHS to RHS, we get,

Multiplying by on both the sides, we get,

Thus, the required rational number is .

**Q.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?
**

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

*Sol.*Therefore, amount of Rs 100 notes = Rs (100 Ã— 2x) = Rs 200x

Amount of Rs 100 notes = Rs (50 Ã— 3x) = Rs 150x

Amount of Rs 100 notes = Rs (10 Ã— 5x) = Rs 50x

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

400x = 400000

Dividing by 400 on both the sides, we get,

x = 1000

Thus, number of Rs 100 notes = 2x = 2 Ã— 1000 = 2000

Number of Rs 50 notes = 3x = 3 Ã— 1000 = 3000

Number of Rs 10 notes = 5x = 5 Ã— 1000 = 5000

**Q.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?
**

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

*Sol.*Number of Rs 2 coins = 3 x number of Rs 5 coins = 3a

Number of Rs 1 coins = 160 â€“ (number of coins of Rs 5 and of Rs 2) = 160 â€“ (3a + a) = 160 â€“ 4a.

Now, amount of Rs 1 coins = Rs [1 x (160 â€“ 4a)] = Rs (160 â€“ 4a)

Amount of Rs 2 coins = Rs (2 x 3a) = Rs 6a

Amount of Rs 5 coins = Rs (5 x a) = Rs 5a

Given, 160 â€“ 4a + 6a + 5a = 300

160 + 7a = 300

Transposing 160 from LHS to RHS, we get,

7a = 300 â€“ 160

7a = 140

Dividing by 7 on both the sides, we get,

a = 20

Thus, number of Rs 5 coins = a = 20

Number of Rs 2 coins = 3a = 3 x 20 = 60

Number of Rs 1 coins = 160 â€“ 4a = 160 - 4 x 20 = 160 â€“ 80 = 80

**Q.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 of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63.
**

**Let the number of winners be y.**

*Sol.*Hence, the number of participants who did not win will be 63 â€“ y.

Amount given to the winners = Rs (100 x y) = Rs 100y

Amount given to the participants who did not win = Rs [25 (63 â€“ y)] = Rs (1575 â€“ 25y)

Given, 100y + 1575 â€“ 25y = 3000

Transposing 1575 from LHS to RHS, we get,

75y = 3000 â€“ 1575

75y = 1425

Dividing by 75 on both the sides, we get,

y = 19.

Thus, number of winners are 19.