Question

The product of the age of Punita 2 years ago and her age after 4 years from now is 1 year more than twice her present age. Find her present age.

Hint:

To solve age-related problems, translate the given information into algebraic expressions and solve the resulting quadratic equation.

Answer 1

Let Punita's present age be \( x \).
- Her age 2 years ago is \( x - 2 \).
- Her age after 4 years is \( x + 4 \).
According to the given information: \[ (x - 2)(x + 4) = 2x + 1. \] Step 1: Expand the left-hand side
Expand the product \( (x - 2)(x + 4) \): \[ x^2 + 4x - 2x - 8 = 2x + 1, \] \[ x^2 + 2x - 8 = 2x + 1. \] Step 2: Simplify the equation
Now, simplify the equation by subtracting \( 2x \) from both sides: \[ x^2 - 8 = 1. \] Next, add 8 to both sides: \[ x^2 = 9. \] Step 3: Solve for \( x \)
Taking the square root of both sides: \[ x = 3 \quad \text{or} \quad x = -3. \] Since age cannot be negative, we conclude that Punita's present age is: \[ x = 3 \, \text{years}. \]
Conclusion:
Punita's present age is 3 years.