Question

Please help me see the attachment

Answer 1

Let the present age of son be x

Given, man is twice the age of son. So, 2x

Eight years hence, age of son = x + 8

And age of man = 2x + 8

After eight years their ages in the ratio 7 : 4 will be 7x and 4x



According to the question,

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

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

=> 8x + 32 = 7x + 56

=> 8x - 7x = 56 - 32

=> x = 24



$\therefore{}$ The age of son => x

=> $\underline{24 \: years}$


$\therefore{}$ The age of man => 2x

=> 2 × 24

=> $\underline{48 \: years }$



$\therefore{}$ The ages of son and the man is 24 years and 48 years respectively.

Answer 2

$\huge\bf\green {Hey \: there}$


Question - The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.


Solution :-


Let the present age of son be x years


Present age of the man = 2x years


After 8 years :


Age of son = ( x + 8 ) years


Age of man = ( 2x + 8 ) years


Ratio of their ages after 8 years = 7 : 4


=> ( 2x + 8 ) / ( x + 8 ) = 7 / 4


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


=> 8x + 32 = 7x + 56


=> 8x - 7x = 56 - 32


=> x = 24 years


∴ Present age of son = 24 years


∴ Present age of the man = 2 × 24 = 48 years



$\huge\bf\pink {Hope \: It \: Helps}$


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