Question

* one year ago a man was 8 times as ald as his son, Now
his age is equal to the square of his son's age, find the
present age​

Answer 1

Let present age of man be x years and present age of man (father) be y years.

One years ago a man was eight times aa old as his son.

$\bold{One\:year\:ago\:=>}\: \begin{cases} \text{Age of son = (x - 1) years} \\ \text{Age of father = (y - 1) years} \end{cases}$

According to question,

=> y - 1 = 8(x - 1)

=> y - 1 = 8x - 8

=> y = 8x - 8 + 1

=> y = 8x - 7 ___ (eq 1)

Now his age is equal to the square of his son's age.

$\bold{At\:present\:=>}\: \begin{cases} \text{Age of son = (x) years} \\ \text{Age of father = (y) years} \end{cases}$

According to question,

=> y = x²

=> 8x - 7 = x²

=> x² - 8x + 7 = 0

=> x² - 7x - x + 7 = 0

=> x(x - 7) -1(x - 7) = 0

=> x = 7, 1

Substitute value of x in (eq 1)

• If x = 7

=> y = 8(7) - 7

=> y = 56 - 7

=> y = 49

• If x = 1

=> y = 8(1) - 7

=> y = 8 - 7

=> y = 1 [Not possible]

•°• Age of father is 49 years and age of son is 7 years

Answer 2

Answer:

let son age=y

man age=z

1 yr ago

son age=y-1

man age =z-1

atq

z-1=8(y-1)

and

z=y^2

so

y^2-1=8y-8

y^2-8y+8-1=0

y^2-8y+7=0

y(y-7)-(y-7)=0

(y-1)(y-7)=0

y=7

y=1 (not valid)

son age=7yrs

man age=7^2=49 ans