Question

Five years ago, a man was 7 times as old as his son. Five years hence, he will be 3 times as old
as his son. Find their present ages.​

Answer 1

GIVEN:

• Five years ago, a man was 7 times as old as his son.
• Five years hence, he will be 3 times as old as his son.

To Find:

• Their present ages

Solution:

• Let present age of son be x
• Five years ago, son's age was (x – 5) years.
• Five years ago father's age was 7(x – 5) years.

$\sf \: Father's \: Present \: age \: = \: [ 7(x \: – \: 5) + 5] years \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \sf \: = 7x \: – \: 35 +5 \: years \\ \sf \: \: \: \: \: \: \: \: = (7x \: – \: 30) years$

• Five years hence, son's age will be (x + 5) years.

According to the question,

$\sf \: Father's \: age = 3 × son's \: age \\ \sf \implies (7x \:– \:30) + 5 – 3 (x + 5) \\ \sf \implies 7x \:–\: 30 + 5 = 3x + 15 \\ \implies \: \sf \: 7x\: – \:3x = 15 + 30\: –\: 5 \\ \sf \implies \: 4x = 40 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies x = 10 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Hence, Son's present age = 10 years
• Father's present age = (7 x 10 – 30) = 40 years.