Exercise 2.2 Page: 28

**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.

According to the question,

(x – 1/2) × ½ = 1/8

x/2 – ¼ = 1/8

x/2 = 1/8 + ¼

x/2 = 1/8 + 2/8

x/2 = (1+ 2)/8

x/2 = 3/8

x = (3/8) × 2

x = ¾

**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?**

Solution:

Given that,

Perimeter of rectangular swimming pool = 154 m Let the breadth of rectangle be = x

According to the question,

Length of the rectangle = 2x + 2 We know that,

Perimeter = 2(length + breadth)

⇒ 2(2x + 2 + x) = 154 m

⇒ 2(3x + 2) = 154

⇒ 3x +2 = 154/2

⇒ 3x = 77 – 2

⇒ 3x = 75

⇒ x = 75/3

⇒ x = 25 m

Therefore, Breadth = x = 25 cm

Length = 2x + 2

= (2 × 25) + 2

= 50 + 2

= 52 m

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

Solution:

Let one of the numbers be= x.

Then, the other number becomes x + 15 According to the question,

x + x + 15 = 95

⇒ 2x + 15 = 95

⇒ 2x = 95 – 15

⇒ 2x = 80

⇒ x = 80/2

⇒ x = 40

First number = x = 40

And, other number = x + 15 = 40 + 15 = 55

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

Solution:

Let the two numbers be 5x and 3x. According to the question,

5x – 3x = 18

⇒ 2x = 18

⇒ x = 18/2

⇒ x = 19

Thus,

The numbers are 5x = 5 × 9 = 45

And 3x = 3 × 9 = 27.

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

Solution:

Let the three consecutive integers be x, x+1 and x+2. According to the question,

x + (x+1) + (x+2) = 51

⇒ 3x + 3 = 51

⇒ 3x = 51 – 3

⇒ 3x = 48

⇒ x = 48/3

⇒ x = 16

Thus, the integers are

x = 16

x + 1 = 17

x + 2 = 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, 8(x+1) and 8(x+2). According to the question,

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

⇒ 8 (x + x+1 + x+2) = 888 (Taking 8 as common)

⇒ 8 (3x + 3) = 888

⇒ 3x + 3 = 888/8

⇒ 3x + 3 = 111

⇒ 3x = 111 – 3

⇒ 3x = 108

⇒ x = 108/3

⇒ x = 36

Thus, the three consecutive multiples of 8 are:

8x = 8 × 36 = 288

8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296

8(x + 2) = 8 × (36 + 2) = 8 × 38 = 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 are x, x+1 and x+2. According to the question,

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

⇒ 2x + 3x +3 + 4x + 8 = 74

⇒ 9x + 11 = 74

⇒ 9x = 74 – 11

⇒ 9x = 63

⇒ x = 63/9

⇒ x = 7

Thus, the numbers are:

x = 7

x + 1 = 8

x + 2 = 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:

Let the ages of Rahul and Haroon be 5x and 7x. Four years later,

The ages of Rahul and Haroon will be (5x + 4) and (7x + 4) respectively. According to the question,

(5x + 4) + (7x + 4) = 56

⇒ 5x + 4 + 7x + 4 = 56

⇒ 12x + 8 = 56

⇒ 12x = 56 – 8

⇒ 12x = 48

⇒ x = 48/12

⇒ x = 4

Therefore, Present age of Rahul = 5x = 5×4 = 20

And, present age of Haroon = 7x = 7×4 = 28

**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:

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

According to the question,

7x = 5x + 8

⇒ 7x – 5x = 8

⇒ 2x = 8

⇒ x = 8/2

⇒ x = 4

Therefore, Number of boys = 7×4 = 28

And, Number of girls = 5×4 = 20

Total number of students = 20+28 = 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’s father be x.

Then, the age of Baichung’s grandfather = (x+26)

and, Age of Baichung = (x-29) According to the question,

x + (x+26) + (x-29) = 135

⇒ 3x + 26 – 29 = 135

⇒ 3x – 3 = 135

⇒ 3x = 135 + 3

⇒ 3x = 138

⇒ x = 138/3

⇒ x = 46

Age of Baichung’s father = x = 46

Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72

Age of Baichung = (x-29) = 46 – 29 = 17

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

Solution:

Let the present age of Ravi be x.

Fifteen years later, Ravi age will be x+15 years. According to the question,

x + 15 = 4x

⇒ 4x – x = 15

⇒ 3x = 15

⇒ x = 15/3

⇒ x = 5

Therefore, Present age of Ravi = 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 be x.

According to the question,

x × (5/2) + 2/3 = -7/12

⇒ 5x/2 + 2/3 = -7/12

⇒ 5x/2 = -7/12 – 2/3

⇒ 5x/2 = (-7- 8)/12

⇒ 5x/2 = -15/12

⇒ 5x/2 = -5/4

⇒ x = (-5/4) × (2/5)

⇒ x = – 10/20

⇒ x = -½

Therefore, the rational number is -½.

**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?**

Solution:

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

Value of Rs 100 = 2x × 100 = 200x

Value of Rs 50 = 3x × 50 = 150x

Value of Rs 10 = 5x × 10 = 50x According to the question,

200x + 150x + 50x = 4,00,000

⇒ 400x = 4,00,000

⇒ x = 400000/400

⇒ x = 1000

Numbers of Rs 100 notes = 2x = 2000

Numbers of Rs 50 notes = 3x = 3000

Numbers of Rs 10 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 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?**

Solution:

Let the number of Rs 5 coins be x.

Then,

number Rs 2 coins = 3x

and, number of Rs 1 coins = (160 – 4x) Now,

Value of Rs 5 coins = x × 5 = 5x

Value of Rs 2 coins = 3x × 2 = 6x

Value of Rs 1 coins = (160 – 4x) × 1 = (160 – 4x)

According to the question,

5x + 6x + (160 – 4x) = 300

⇒ 11x + 160 – 4x = 300

⇒ 7x = 140

⇒ x = 140/7

⇒ x = 20

Number of Rs 5 coins = x = 20

Number of Rs 2 coins = 3x = 60

Number of Rs1 coins = (160 – 4x) = 160 – 80 = 80

**16. The organisers of an essay competition decide that a winner in the competition gets a prize of Rs100 and a participant who does not win gets a prize of Rs25. The total prize money distributed is Rs3,000. Find the number of winners, if the total number of participants is 63.**

Solution:

Let the numbers of winner be x.

Then, the number of participants who didn’t win = 63 – x

Total money given to the winner = x × 100 = 100x

Total money given to participant who didn’t win = 25×(63-x)

According to the question,

100x + 25×(63-x) = 3,000

⇒ 100x + 1575 – 25x = 3,000

⇒ 75x = 3,000 – 1575

⇒ 75x = 1425

⇒ x = 1425/75

⇒ x = 19

Therefore, the numbers of winners are 19.