Question

A man is 4 times as old as his son. 20 Years later he will be twice as old as his son. find their present ages.​

Answer 1

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Let the Man's age be x years and his son's age be y years

x = 4y -------------- (i)

After 20 years their ages will be

Man : x + 20

Son : y + 20

x + 20 = 2 ( y + 20 ) ----------------- (ii)  

Substituting value of x in equation (ii) with 4y

=> 4y + 20 = 2y + 40

=> 4y - 2y = 40 - 20  

=> 2y = 20

=> y = 10 years

Using in eq (i)

x = 4y

=> x = 40 years

Man's age = 40 yrs.

Son's age = 10 yrs.

Answer 2

SOLUTION :-

Let,

Present age of son = x

Present age of man = y

After 20 years,

Age of son = x + 20

Age of man = y + 20

According to the first condition,

$\longrightarrow \sf y = 4x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...........................(1)$

According to the second condition,

$\longrightarrow \sf 2(x+20= y+20)\\\\\longrightarrow 2x+40 = y+20\\\\\longrightarrow 2x-y = -20 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...........................(2)$

Substitute y = 4x in equation (2),

$\longrightarrow\sf 2x-4x= -20\\\\\longrightarrow -2x= -20\\\\\longrightarrow\bf x = 10$

Present age of son = 10 years

Present age of man = 10 × 4 = 40 years