Question

The age of a man is twice than s son eight yearso hence their ages will be in the ratio of 7:4 find their present age

Answer 1

Heya !!

Here's your answer..
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Let the age of man be ( x ) and age of his son be ( y ).

Given that age of man is twice than his son.

x = 2y ---(1)

After 8 years their ages will be in the ratio of 7:4.

After 8 years age of man = x + 8
After 8 years ago of his son = y + 8

$\frac{x + 8}{ y + 8} = \frac{7}{4} \\ \\ 4x + 32 = 7y + 56$

from eq. (1) .. put the value of x

4( 2y ) + 32 = 7y + 56

8y + 32 = 7y + 56

8y - 7y = 56 - 32

y = 24

put value of y in eq.(1)

x = 2y
x = 2(24)
x = 48

Age of Man ( x ) is 48 years.
Age of his son ( y ) is 24 years.
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Hope it helps.
Thanks :)

Answer 2

Let,
Age of son = x
Age of man =>2x

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According to the question, after 8 years,

Son's age = ( x + 8) years
Father's age = (2x + 8) years

Ratio = 7:4

Then,

$\frac{2x + 8}{x + 8} = \frac{7}{4}$

$4(2x + 8) = 7(x + 8)$

$8x + 32 = 7x + 56$

8x- 7x = 56 - 32

x = 24

Present age of

• son = 24 years
• father =2(24) = 48 years.


I hope this will help you


-by ABHAY