Q1. [1] § 1.4 Summary
If $x = ab^3$ and $y = a^3b$, where $a$ and $b$ are prime numbers, then $[\text{HCF}(x,y) - \text{LCM}(x,y)]$ is equal to :
- (a) $1 - a^3b^3$
- (b) $ab(1 - ab)$
- (c) $ab - a^4b^4$
- (d) $ab(1 - ab)(1 + ab)$
Previously asked in CBSE board exam
2025 30/4/1 Q1
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
(d) $ab(1-ab)(1+ab)$
Since $x = ab^3 = a^1b^3$ and $y = a^3b = a^3b^1$, HCF $= a^1b^1 = ab$ and LCM $= a^3b^3$. So HCF $-$ LCM $= ab - a^3b^3 = ab(1-a^2b^2) = ab(1-ab)(1+ab)$.
Explanation
- HCF = product of lowest powers of common prime factors = $a^1 \cdot b^1 = ab$.
- LCM = product of highest powers of all prime factors = $a^3 \cdot b^3$.
- Factorise $ab - a^3b^3 = ab(1 - a^2b^2)$, then use the difference-of-squares identity: $1 - a^2b^2 = (1-ab)(1+ab)$.
- Examiner expects you to show the HCF and LCM step clearly before computing the difference.
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