Question

2.
2 years ago, my age was 4 times the age of my son then. 6 years ago, my age

was twice the square of the age of my son then. Find the present age of my son.​

Answer 1

Answer:

The present age of my son is 10 years.

Let the present age of me be x and my son be y.

Two years ago,

My age was x-2 and

my son's age was y-2

It is given that two years ago my age was 4 and half times the age of my son. Or in other words, my age was 2 times the age of my son.

So, (x-2) = 4.5(y-2)

⇒ x = (9/2)y - 7          ...(1)

Six years ago,

My age was x-6 and

My son's age was y-6

It is given that my age was twice the square of the age of my son.

So, (x-6) = 2×(y-6)²

Substituting the value of x from (1), we get:

(9/2)y - 7  - 6 = 2(y-6)²

⇒ (9/2)y - 13  = 2(y-6)²

⇒ (9/2)y - 13  = 2y² + 72 - 24y

⇒ (9/2)y = 2y² - 24y + 85

⇒ 9y = 4y² - 48y + 170

⇒ 4y² - 57y + 170 = 0

⇒ 4y² - 40y - 17y + 170 = 0

⇒ 4y(y - 10) - 17(y - 10) = 0

⇒ (y-10)(4y-17)

By zero product rule,

we get y = 10 and y = 17/4 = 4.25

y = 4.25 is discarded as age cannot be fractional.

So, present age of son is y = 10 years.

Step-by-step explanation: