Question

Four years ago a mean was six times as old as his son. In five years time, he will be only three times as old his son. Find the difference in their ages in ten years time

Answer 1

Answer:Let the present ages of father and son be x and y, respectively.

As per the problem, six years ago,

(x−6)=3(y−6)

Six years later, (x+6)=2(y+6)

Solving these two equations, we get father's age is 42 years, and son's age is 18 years.

Step-by-step explanation:42;18

Answer 2

Answer:

I hope it's helpful

Step-by-step explanation:

Let us assume that the father’s age is x and the son’s age is y.

According to the question, a father is now three times as old as his son,

so, x = 3y

five years before their ages would have been, x-5 and y-5. And the question says the father was four times as old as his son,

x-5 = 4( y -5)

x-5= 4y -20

x-4y + 15 = 0

3y - 4y + 15 =0 , we already know x = 3y

y = 15 and x = 45

The father is presently 45 and the son is 15

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