Question

One year ago a father was four times as old as his son .after 6 year his age exceed than twice of his son age by 9 year. Ratio of their present age is what

Answer 1

Lets denote the ages of the man and his son as M and S respectively.

One year ago their ages were M - 1 and S - 1 respectively.

So from the first statement, M - 1 = 4 * (S - 1) or

M - 4*S = -3 ← Equation 1

In 6 years time both would be older by 6 years, or M+6 and S+6.

From the second statement,

M + 6 = 2 * (S +6) + 9, or

M - 2*S = 15 ← Equation 2

Subtracting Equation 1 from Equation 2, we get:

M - 2*S - (M - 4*S) = 15 - (-3)

That is, 2*S = 18 , or S = 9. Therefore the Son’s present age is 9.

From Equation 2, M = 15 + 2*S or M = 15 + 18 = 33.

Therefore the Father’s present age is 33.

Answer 2

Son:Father = 19:73

You can find this out by using simple algebra. First make a column for 1 year ago and one for 6 years in the future:

1 year ago                                                                6 years in the future

Now, you know that one year ago, the father was 4 times the age of the son. So place that information into the table.

1 year ago
F=4S

Now, 7 years after 1 year ago, you know that the father's age is 29 years older plus twice as old as his son. So place that information in the table.

6 years in the future
F=2(s+7) +43

So now, place that into an equation.

4S+7=2S+43 (comparing the father's ages from 1 year ago to 6 years in the future)

Now, we solve it like a regular equation and get this:
S=18
But that is the son's age from one year ago. So we add one to it and then plug 18 in for the father's age, and add one and that is how you get 19 and 73.