**Problem 1 :**

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

**Problem 2 :**

The ages of John and David are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages ?

**Problem 3 :**

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 ?

**Problem 4 :**

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?

**Problem 5 :**

The organizers of an essay competition decide that a winner in the competition gets a prize of $100 and a participant who does not win gets a prize of $25. The total prize money distributed is $3000. Find the number of winners, if the total number of participants is 63.

**Problem 1 :**

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

**Solution :**

Let the three consecutive multiples of 8 be

x, (x + 8) and (x + 16)

**Given :** The sum of three consecutive multiples of 8 is 888.

Then, we have

x + (x + 8) + (x + 16) = 888

3x + 24 = 888

Subtract 24 from each side.

3x = 864

Divide each side by 24.

x = 288

Then,

x + 8 = 288 + 8 = 296

x + 16 = 288 + 16 = 304

So, the three consecutive multiples of 8 are 288, 296, and 304.

**Problem 2 :**

The ages of John and David are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages ?

**Solution :**

**Given :** The ages of John and David are in the ratio 5 : 7.

Then, the present ages of John and David are

5x and 7x

**Given :** Four years later the sum of their ages will be 56 years.

Then, we have

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

Simplify and solve for x.

12x + 8 = 56

Subtract 8 from each side.

12x = 48

Divide each side by 12.

x = 4

Age of John is

5x = 5(4)

5x = 20

Age of David is

7x = 7(4)

7x = 28

So, the present ages of John and David are 20 years and 28 years respectively.

**Problem 3 :**

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 :**

**Given :** The number of boys and girls in a class are in the ratio 7 : 5.

Then,

Number of boys = 7x

Number of girls = 5x

**Given :** The number of boys is 8 more than the number of girls.

Then, we have

7x = 5x + 8

Subtract 5x from each side.

2x = 8

Divide each side by 2.

x = 4

Number of boys is

7x = 7(4)

7x = 28

Number of girls is

5x = 5(4)

5x = 20

Total number of students in the class is

= 28 + 20

= 48

So, the total class strength is 48.

**Problem 4 :**

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 x be the age Baichung's father.

Then,

Age of Baichung = x - 29

Age of Baichung's grandfather = x + 26

**Given :** The sum of the ages of all the three is 135 years.

Then, we have

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

3x - 3 = 135

Add 3 to each side.

3x = 138

Divide each side by 3.

x = 46

Age of Baichung's father is is 46 years.

x - 29 = 46 - 29

x - 29 = 17

Age of Baichung is 20 years.

x + 26 = 46 + 26

x + 26 = 72

Age of Baichung's grand father is 72 years.

**Problem 5 :**

The organizers of an essay competition decide that a winner in the competition gets a prize of $100 and a participant who does not win gets a prize of $25. The total prize money distributed is $3000. Find the number of winners, if the total number of participants is 63.

**Solution :**

Let x be the number of winners.

Then, the number of participants do not win is (63 - x).

**Given :** Winner gets $100, loser gets $25 and total prize money distributed is $3000.

Then, we have

100x + 25(63 - x) = 3000

100x + 1575 - 25x = 3000

75x + 1575 = 3000

75x = 1425

Divide each side by 75.

x = 19

So, the number of winners is 19.

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