Question

One year ago, a man was 8 times as old as his son. Now his age is equal to the

square of his son’s age. Their present ages are

(a) 7 years, 49 years

(b) 5 years, 25 years

(c) 1 years, 50 years

(d) 6 years, 49 years​

Answer 1

Step-by-step explanation:

let present age of his son x

present age of man is y

1 year ago

son's age = x-1

man's age = y-1

y-1 = 8(x-1)

y-1 = 8x-8

8x-y = 7.............1

now

y = x^2..............2

place the value of 2 in 1

8x-x^2=7

taking minus common

x^2-8x+7=0

x^2-(7+1)x+7=0

x^2-7x-x+7=0

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

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

x=7years. x=1 not possible

y = x^2

y = 7^2

y =49 years

so option a is correct