Question

A man is three times as old as his son. Eight years ago the man was 7 times as old as his son. find their present ages?

Answer 1

Hi there !
Here's the answer :


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Let the present age of Man's son be x
and the present age of Man = 3x
( °•° Man's age = 3× his son's age)


Eight years ago(before),
the age of Man's son = x-8
the age of man = 3x-8

Given,
Man's age = 7 × His son's age
=> 3x-8 = 7( x-8)
=> 3x-8 = 7x-56
=> 4x = 56-8
=> 4x = 48
=> x = 12

•°• Present age of Man's son x = 12 years
& Present age of Man 3x = 3×12 = 36 years

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Hope it helps

Answer 2

let the man age is x
and his son age is y

according to the question
x= 3y. -(1)

and

Eight years ago the man was 7 times as old as his son.
then (x+8) = 7(y-8)
x-8=7y-56 (2)

put value of x in (2) equation

3y-8=7y-56
7y-3y=56-8
4y=48
y=12 years

put value of y in (1) equation

x=3×12
x=36 years


hence age of men is 36 years and age of son is 12 years

I hope it is helped you

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