Question

A father is 3 times as old as his son. In 10 years’ time, his age will be double his son’s age. Find their present ages.



a. 10 years, 30 years

b. 12 years, 36 years

c. 15 years, 45 years

d. 11 years, 33 years

​

Answer 1

Given: The father's age is 3 times the age of his son. & In 10 years, his age will be double his son's age.

❍ Let's Consider, the present age of son be n years and his father's age be 3n years.

$\bigstar\;\underline{\boldsymbol{According\: to \:the\: Question :}}$

• In 10 years, the age of father will be double of his son's age.
• & Ten years later their ages will be: (n + 10) and (3n + 10).

⠀

$:\implies\sf Father_{\;(age)} = 2(Son_{\:(age)})$

$:\implies\sf (3n + 10) = 2(n + 10)$

$:\implies\sf 3n + 10 = 2n + 20$

$:\implies\sf 3n - 2n = 20 - 10$

$:\implies{\pmb{\sf n = 10}}$

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Therefore,

• Son's age, n = 10
• Father's age, 3n = 3(10) = 30

⠀

$\therefore\:\underline{\textsf{Hence, their present ages are \textbf{Option a) 10 years} and \textbf{30 years} }}.$

Answer 2

Given :-

A father is 3 times as old as his son. In 10 years’ time, his age will be double his son’s age.

To Find :-

Present age

Solution :-

Let

Age of father = x

Age of son = y

x = 3y (i)

After 10 years

x + 10 = 2(y + 10)

x + 10 = 2y + 20

x - 2y = 20 - 10

x - 2y = 10

3y - 2y = 10

y = 10

x = 3y

x = 3(10)

x = 30 years

Hence

Age of son = y = 10 years

Age of father = x = 30 years