Question

A man is twice as old 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

Please mark this as 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,

$⇒ \sf 12(x - 20) =\: 2x - 20$

$⇒ \sf 12x - 240 =\: 2x - 20$

$⇒ \sf 12x - 2x =\: - 20 + 240$

$⇒ \sf 10x =\: 220$

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

$➠ \sf\bold{\red{x =\: 22\: years}}$

Hence, the required ages are,

• Present age of son = x = 22 years

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

$\therefore$ The present age of son is 22 years and the present age of man is 44 years.

$\begin{gathered}\\\end{gathered}{\blue{\boxed{\large{\bold{VERIFICATION :-}}}}}$

$↦ \sf 12(x - 20) =\: 2x - 20$

$↦ \sf 12x - 240 =\: 2x - 20$

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 ✔

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