Question

One year ago, a man was 8 times as old as his daughter. Now his age is equal to the square of his daughter's age. The present age of the daughter is:

• A. 5 years
• B. 6 years
• C. 7 years
• D. 8 years

Hint:

Age problems often reduce to forming quadratic equations — assign variables and work step by step.

Answer 1

The Correct Option is C

Let daughter’s current age be \( x \).
Then 1 year ago, daughter’s age = \( x - 1 \), man’s age = \( 8(x - 1) \) Now man's current age = \( 8(x - 1) + 1 = 8x - 7 \) Given: \[ 8x - 7 = x^2 \Rightarrow x^2 - 8x + 7 = 0 \] Solving the quadratic: \[ x = \frac{8 \pm \sqrt{64 - 28}}{2} = \frac{8 \pm \sqrt{36}}{2} = \frac{8 \pm 6}{2} \Rightarrow x = 7 \text{ (valid)}, x = 1 \text{ (too young)} \] Daughter’s present age = \fbox{7 years}