Question

Two years ago a man’s age was three times the square of his son’s age. In three yrs time, his age Will be 4 times his son’s age. Find their present ages.​

Answer 1

Given,

• Two years ago a man’s age was three times the square of his son’s age
• In three year time, his age Will be 4 times his son’s age

To find,

• Present age of man's and his son

Solution,

Let us assume that, man's present age is x years and son's present age be y years.

So, we are given that,

⇒ 2 years ago, a man's age was three times the square of his son's age.

So,

$\sf \: → Man's \: age \: 2 \: years \: ago = x-2 \\ \\ \sf \: → Son's \: age \: 2 \: years \: ago = y-2$

Hence, according to the condition,

$\sf \: →x-2=3(y-2)² \\ \\ \sf→x=3(y-2)² + 2 - -(1)$

We are also given that,

→ In 3 years time, his age will be 4 times his son's age.

So,

$\sf→Man's \: age \: 3 \: years \: later = x + 3 \\ \\ \sf→ Son's \: age \: 3 \: years \: later = y + 3$

Hence, according to the condition,

$\sf \: ⇒ x + 3 = 4(y + 3) \\ \\ \sf⇒ x = 4(y + 3)-3 \\ \\ \sf \:⇒ x = 4y + 12-3 \\ \\ \sf⇒ x = 4y + 9 ----(2) \\ \\ \sf \: From \: Equating (1) \& (2), \\ \\ \\ \sf→3(y-2)² + 2 = 4y +9 \\ \\ \sf→ 3(y² - 4y + 4) + 2 = 4y +9 \\ \\ \sf→3y² - 12y + 12+2 = 4y +9 \: \\ \\ \sf→ 3y² - 12y + 14 = 4y + 9$

Transposing RHS to LHS,

$\sf→3y² - 12y + 14 - 4y - 9 = 0 \\ \\ \sf→ 3y² - 16y + 5 = 0 \\ \\ \sf→3y²-15y-y + 5 = 0 \\ \\ \sf→Зу(у - 5) - 1(y - 5) = 0 \\ \\ \sf →(y - 5)(3y - 1) = 0$

So,

$\sf \: ⇒ y = 5, y = \frac{1}{3}$

$\sf \: As, y = \frac{1}{3} \: isn't \: possible.$

Hence,

$\sf \: ⇒ y = 5$

Substituting this value in (2),

$\sf⇒ x = 4y +9 \\ \\ \sf⇒ x = 4(5) + 9\\ \\ \sf⇒ x = 20 +9 \\ \\ \sf⇒ x = 29$

Therefore, the present ages of man and his son is 29 years and 5 years respectively.

Answer 2

Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages. And, the present age of son = x = 5 years.