HCF and LCM Questions and Answers - Algebraic Variables

This set of Aptitude Questions and Answers (MCQs) focuses on “Algebraic Variables”.

1. Find the HCF of polynomials (x+1)²(x-2)(x+3) and (x+1)(x+2)(x+3)².
a) (x+1)(x+2)(x+3)²
b) (x+1)(x-2)(x+3)²
c) (x+1)(x-3)
d) (x+1)(x+3)
View Answer

Answer: d
Explanation: The common factors between (x+1)²(x-2)(x+3) and (x+1)(x+2)(x+3)² is (x+1)(x+3).
Hence, HCF is (x+1)(x+3).

2. Find the HCF of 16(x⁵+x⁴-6x³) and 40(x⁴+4x³+3x²).
a) 8(x+3)
b) 8x(x+3)
c) 8x(x+3)²
d) 8x²(x+3)
View Answer

Answer: d
Explanation: Let p(x) = 16(x⁵+x⁴-6x³) = 16x³(x-2)(x+3) and q(x) = 40(x⁴+4x³+3x²) = 40x²(x+1)(x+3).
The common factors are 8x²(x+3).
Therefore, HCF is 8x²(x+3).

3. Find the HCF of 48x⁵y⁷z³, 18x⁶y⁴z⁵ and 54x⁷y²z⁷.
a) 6x⁵y²z³
b) 6x⁵y³z³
c) 8x⁵y²z³
d) 6x⁵y⁴z³
View Answer

Answer: a
Explanation: The common factors between 48x⁵y⁷z³, 18x⁶y⁴z⁵ and 54x⁷y²z⁷ is 6x⁵y²z³.
Therefore, HCF is 6x⁵y²z³.

4. Find the LCM of 8a⁴b⁵c⁶, 10a⁶b²c³ and 15a⁵b⁶c⁴.
a) 240a⁶b⁶c⁶
b) 120a⁶b⁶c⁶
c) 120a⁵b⁶c⁶
d) 120a⁶b⁷c⁶
View Answer

Answer: b
Explanation: LCM of 8, 10 and 15 is 120.
We have to choose the highest power of a, b and c to find its LCM.
Therefore, the LCM of 8a⁴b⁵c⁶, 10a⁶b²c³ and 15a⁵b⁶c⁴ is 120a⁶b⁶c⁶.

5. Find the LCM of (x+4)(x²-4x+4), (x-2)(x²-16) and (x-4)(x+3).
a) (x-2)(x²-16) (x+3)
b) (x+2)(x²-16) (x+3)
c) (x-2)²(x²-16) (x+3)
d) (x-2)²(x²-16) (x-3)
View Answer

Answer: c
Explanation: Let p(x) = (x+4)(x²-4x+4) = (x+4)(x-2)², q(x) = (x-2)(x²-16) = (x-2)(x+4)(x-4) and r(x) = (x-4)(x+3).
Therefore, LCM is (x-2)²(x²-16) (x+3).

6. Find the LCM of 40(2x⁶+3x⁵-2x⁴) and 60(2x⁵+x⁴-x³).
a) 120(2x-1)(x²+3x+2)
b) 120(2x-1)(x²-3x+2)
c) 120(2x-1)²(x²+3x+2)
d) 120(2x-1)(x-1)(x+2)
View Answer

Answer: a
Explanation: Let p(x) = 40(2x⁶+3x⁵-2x⁴) = 2³*5x⁴(x+2)(2x-1) and q(x) = 60(2x⁵+x⁴-x³)
= 2²*3*5x³(x+1)(2x-1).
Therefore, LCM is 120(2x-1)(x+1)(x+2) = 120(2x-1)(x²+3x+2).

7. The HCF and LCM of two polynomials is (x+2) and (x+1)(x+2)(x-4). If one of the polynomial is (x+2)(x-4), find the other.
a) (x+1)
b) (x+1)(x+2)
c) (x+1)(x+2)(x-4)
d) (x+1)(x-2)
View Answer

Answer: b
Explanation: We know that, product of two polynomials = HCF * LCM.
Other polynomial = \(\frac{(x+2)*(x+1)(x+2)(x-4)}{(x+2)(x-4)}\)=(x+1)(x+2).

8. If LCM of two variables is p²q³, then which of the following could not be its HCF?
a) p²
b) p²q²
c) p³q³
d) q²
View Answer

Answer: c
Explanation: LCM is a multiple of HCF.
p²q³ is not a multiple of p³q³.
Hence, p³q³ could not be its HCF.

9. What is the LCM of x³y-xy³, x³y²+x²y³ and x²y+xy²?
a) x²y(x²-y²)
b) x²y²(x²-y²)
c) xy(x²-y²)
d) x²y²(x²+y²)
View Answer

Answer: b
Explanation: p(x) = x³y-xy³ = xy(x²-y²), q(x) = x³y²+x²y³ = x²y²(x+y) and r(x) = x²y+xy² = xy(x+y).
Therefore, LCM is x²y²(x²-y²).

10. Find the HCF of p⁶-q⁶ and p⁸-q⁸.
a) p-q
b) p+q
c) p²+q²
d) p²-q²
View Answer

Answer: d
Explanation: Let p(x) = p⁶-q⁶ = (p²-q²)(p⁴+q⁴+p²q²) and q(x) = p⁸-q⁸ = (p⁴+q⁴)(p²+q²)(p²-q²).
Therefore, HCF is p²-q².

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