Question

A man is twice as old as twice as his son.20 years ago ,the age of the man was 12 times the age of the son Find their present ages

Answer 1

let n be how old he is today. n-20=s/12 let s be his sons age 20 yrs ago. S=(20/12)+20. 20/12=1.666667, so the son is 21.6666667. The father is twice as old, so n=43.333334. Hope this helps! pls mark brainliest!

Answer 2

Given :-

A man is twice as old as his son. 20 years ago, the age of man was 12 times the age of the son

To Find :-

What are their present ages

Solution :-

Let, the present age of son be x years

And, the present age of man will be 2x years

According to the question :

⇒ $\sf 12(x - 20) =\: 2x - 20$

⇒ $\sf 12x - 240 =\: 2x - 20$

⇒$\sf 12x - 2x =\: - 20 + 240$

⇒ $\sf 10x =\: 220$

⇒$\sf x =\: \dfrac{\cancel{220}}{\cancel{10}}$

➠$\sf\bold{\purple{x =\: 22\: years}}$

Hence, the required ages are,

✧ Present age of son = x = 22 years

✧ Present age of man = 2x = 2 × 22 = 44 years

∴ The present age of son is 22 years and the present age of man is 44 years.

${\red{ \boxed{\large {\bold{VERIFICATION :-}}}}}$

$↦ \sf 12(x - 20) =\: 2x - 20$

$↦ \sf 12x - 240 =\: 2x - 20$

By putting x = 22 we get,

↦ $\sf 12(22) - 240 =\: 2(22) - 20$

↦$\sf 12 \times 22 - 240 =\: 2 \times 22 - 20$

↦ $\sf 264 - 240 =\: 44 - 20$

↦$\sf 24 =\: 24$

➦ LHS = RHS

Hence, Verified ✔