Question

40,2 years ago a father was 6 times older than his son. After 18 years he will become double as old as his son. Find the present age​

Answer 1

Answer:

Baby shark, doo, doo, doo, doo, doo, doo

Baby shark, doo, doo, doo, doo, doo, doo

Baby shark

Mommy shark, doo, doo, doo, doo, doo, doo

Mommy shark, doo, doo, doo, doo, doo, doo

Answer 2

Let :

• Present age of father = x years
• Present age of son = y years

Given :

• 2 years ago , father was 6 times older than his son
• After 18 years he will become double as old as his son

To find :

• Present age of father
• Present age of son

Solution :

2 years ago -

• Age of father = x-2
• Age of son = y-2

Also , Age of father = 6×Age of son

(x-2) = 6(y-2)

x-2 = 6y - 12

x = 6y - 12 + 2

x = 6y - 10 ....... equation 1

18 years later -

• Age of father = x+18
• Age of son = y+18

Also , Age of father = 2×(age of son)

(x+18) = 2×(y+18)

x + 18 = 2y + 36

x = 2y + 36 - 18

x = 2y + 18 .......... equation 2

Putting value of, x from equation 1 into equation 2,

we get;

6y - 10 = 2y + 18

6y - 2y = 18 + 10

4y = 28

y = 28 ÷ 4

y = 7

Putting value of y in equation 1, we get ;

x = 6(7) - 10

x = 42 - 10

x = 32

ANSWER :

• Present age of father = 32 years
• Present age of son = 7 years