In a 2-digit number, the digit at the unit's place is 5 less than the digit at the ten's place. The product of the digits is 36. Find the number.
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
Let the digit at the ten's place = $x$.
Then, digit at unit's place = $x - 5$ (given: unit's digit is 5 less than ten's digit).
Product of digits = $x(x - 5) = 36$
$$x^2 - 5x - 36 = 0$$
Splitting the middle term:
$$x^2 - 9x + 4x - 36 = 0$$
$$x(x - 9) + 4(x - 9) = 0$$
$$(x + 4)(x - 9) = 0$$
So, $x = -4$ or $x = 9$.
Since $x$ is a digit, $x$ cannot be negative. Therefore, $x = 9$.
Ten's digit = 9, Unit's digit = $9 - 5 = 4$.
The required number is 94.
Source: Chapter 4, Section 4.3 (Solution by Factorisation)
---
Explanation
- Examiners expect you to define the variable clearly, form the quadratic equation, and solve by factorisation (as this is the method taught in Ex. 4.2).
- Rejecting the negative value is essential — always state why (a digit cannot be negative). Missing this step can cost half a mark.
- State the final answer explicitly; writing just $x = 9$ without forming the number loses the conclusion mark.