Question

13. 5 years ago a man was 7 times as old
as his son. After 5 years he will be
thrice as old as his son. Find their
present
ages.​

Answer 1

Answer:

Answer

Let Five years ago the age of son be x years and age of father be 7x years

Present age of son =x+5

Present age of father =7x+5

5 years later their age will (x+10) and (7x+10)

∴7x+10=3(x+10)

7x−3x=20

4x=20

x=5

So, the present age of son=x+5=5+5=10years

and the present age of father=7x+5=35+5=40years

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Answer 2

Answer:

Their present age of the man & his son is 40 years and 10 years respectively.

Explanation:

Question says that,

5 years ago a man was 7 times his son's age. 5 years hence the man becomes 3 times his son's age.

Find their present age.

If we assume the age of the son 5 years ago was $x$ years.

Then, the man's age 5 year ago was 7 times $x$

Or, $7x$ years.

Therefore, present age of

The man will be $(7x + 5)$ years.

His son will be $(x + 5)$ years.

Now, 5 years later

The man's age $(7x + 10)$ years.

Son's age $(x + 10)$ years.

Given,

The man's age = 3× the age of his son.

Or, we can say that

$\implies (7x + 5) = 3(x + 5)$

Solving for $\boldsymbol x$

$\implies 7x + 10 = 3(x + 10)$

$\implies 7x + 10 = 3x + 30$

$\implies 7x - 3x = 30 - 20$

$\implies 4x = 20$

$\implies x = \dfrac{20}{4}$

${\therefore{ \underline {\sf {The \: value \: of \: { \boldsymbol{x}} \: is \: 5 .}}}}$

Hence, their present age of

➪ The man is $(7x + 5) = \bf 40 \: years.$

➪ His son is $(x + 5) = \bf 10 \: years.$