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:

When dealing with age-related problems, set up an equation based on the given conditions and solve it.

Answer 1

The Correct Option is C

To solve the problem, let's denote the daughter's current age as \( x \). According to the problem:

• The man was 8 times as old as his daughter one year ago. Therefore, one year ago, the daughter's age was \( x - 1 \) and the man's age was \( 8(x - 1) \).
• Currently, the man's age is equal to the square of his daughter's age. Hence, the man's present age is \( x^2 \).

Setting up the equation from the above conditions:

• Since the man's age one year ago is \( x^2 - 1 \), we have:
  \( 8(x - 1) = x^2 - 1 \)

Expanding and simplifying the equation:

• \( 8x - 8 = x^2 - 1 \)
• Rearranging gives:
  \( x^2 - 8x + 7 = 0 \)

This is a quadratic equation which can be solved by factorization:

• The factors of 7 that add up to -8 are -7 and -1. Thus, the equation can be written as:
  \( (x - 7)(x - 1) = 0 \)

Solving for \( x \):

• \( x - 7 = 0 \) or \( x - 1 = 0 \)
• This gives \( x = 7 \) or \( x = 1 \)

Since it is not reasonable for the daughter, who had an age last year, to be \( 1 \) year old now, the only logical solution is:

• The current age of the daughter is \( 7 \) years.