The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find their present ages.
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
Let the present age of son = x years.
Then, present age of man = $2x^2$ years.
Eight years hence:
- Son's age = $(x + 8)$ years
- Man's age = $(2x^2 + 8)$ years
According to the condition:
$$2x^2 + 8 = 3(x + 8) + 4$$
$$2x^2 + 8 = 3x + 24 + 4$$
$$2x^2 - 3x - 20 = 0$$
Factorising:
$$2x^2 - 8x + 5x - 20 = 0$$
$$2x(x - 4) + 5(x - 4) = 0$$
$$(x - 4)(2x + 5) = 0$$
So, $x = 4$ or $x = -\dfrac{5}{2}$
Since age cannot be negative, $x = 4$.
Present age of son = 4 years
Present age of man = $2 \times 4^2 = 32$ years
Source: Chapter 4, Exercise 4.2
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Explanation
- Examiners award marks for: correct variable assumption (1), forming the equation (1), simplifying to standard form (1), correct factorisation (1), and stating both answers with rejection of negative value (1).
- Always reject the negative root when the variable represents age and explicitly state why.
- Check: Eight years hence — man is 40, son is 12; $3 \times 12 + 4 = 40$ ✓