Question

4 years ago, a father was 6 times as old as his son. 10 years later, the father will be 5/2 times as old as his son. Find the present age of the father present age of the son.

Answer 1

Let the present age of son be x yrs

Present age of father = y yrs

4 years ago,

Age of son = x-4

Age of father = y -4

According to question ,

y-4 = 6(x-4)

y-4 = 6x – 24

y – 6x = -24+4

y – 6x = -20 ………………………..(1)

10 years later,

Age of son = x+ 10

Age of father = y+ 10

Y+10 = (x+10)

2y + 20 = 5x+ 50

2y – 5x = 50-20

2y – 5x = 30 ……………………….(2)

Multiply eqn(1) by 2

We get

2y – 12x = -40

2y – 5x = 30

(-) (+) (-)

-7x = -70

X = 70/7 = 10

Put the value of x in eqn(1)

y-6*10 = -20

y – 60 = -20

y = -20+60

y = 40

so , the present age of son = 10 yrs

present age of father = 40 yrs