Question

Father present age is 30 years more than son’s present age . Eight years later product of their ages will be 400.Find their present ages

Answer 1

Given :

• Father present age is 30 years more than son’s present age.

• Eight years later product of their ages will be 400.

To find :

• Their present ages =?

Step-by-step explanation :

Let, the present age of the son's be x.

Then, the present age of father's be x + 30.

Eight years later son's age will be, x + 8

Eight years later father's age will be, x + 30 + 8.

According to the question :

➮ (x +8) (x + 30 + 8) = 400

➮ (x + 8) (x + 38) = 400

➮ x(x + 38) 8(x + 38) = 400

➮ x² + 38x + 8x + 304 = 400

➮ x² + 46x = 400 - 304

➮ x² + 46x = 96

➮ x² = 46x - 96 = 0

➮ x² + 48x - 2x - 96 = 0

➮ x(x + 48) - 2(x + 48) = 0

➮ (x + 48) (x - 2) = 0

Therefore, We got the value of, x = - 48 Or 2.

Negative value cannot be taken. So,

The value of, x = 2.

Hence,

The present age of son's, x = 2 years.

Then, the present age of father's, x + 30 = 2 + 30 = 32 years

Answer 2

GIVEN,

• Father's age is 30 years more that his son's present age.
• Product of their ages = 400

LET

• Son's age be x

THEN,

Father's age = ( x + 30 )years ------(1)

$\sf{According\:to\:the\:question}$

$\bigstar{\sf{After\:8\:years}}$

Son's age = x + 8

Father's age = x + 30 + 8 = x + 38

$\longrightarrow \sf\: (x + 8) \times ( x + 38 ) = 400 \\ \\ \longrightarrow \sf\: {x}^{2} + 38x + 8x + 304 = 400 \\ \\ \longrightarrow \sf\: {x}^{2} + 46x - 96 = 0 \\ \\ \longrightarrow \sf\: {x}^{2} + 46x - 2x - 96 = 0 \\ \\ \longrightarrow \sf\: x ( x + 48 ) - 2 ( x + 48 ) = 0 \\ \\ \longrightarrow \sf\: ( x - 2 )( x + 48 ) = 0 \\ \\ \longrightarrow \sf\: x = 2\:or\:-48 years$

We know that age can not be negative so,

Age of son = 2 years

Putting this value in (1)

$\longrightarrow \sf\: Father's\:age\:=x + 30\\ \\ \longrightarrow \sf\: Father's\:age\:=32\:years$

So,

Son's age = $\boxed{2\:years}$

Father's age = $\boxed{32\:years}$