CBSE Class 8 Mathematics Factors and Multiples MCQs Set 17

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Multiple Choice Questions

Question 1: The H.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the numbers is 77, find the other.
(a) 121
(b) 88
(c) 63
(d) 99
Show Answer & Explanation

Answer: (d) 99

Explanation:
1. Product of the two numbers = HCF × LCM = 11 × 693 = 7623.
2. Other number = 7623 / 77 = 99.

Teacher's Note:
1. For two numbers, product = HCF × LCM.
2. Check: HCF(77, 99) = 11 and LCM = 693.

Question 2: What is the greatest number of 4 digits that, when divided by any of the numbers 6, 9, 12 and 17, leaves a remainder 1?
(a) 9801
(b) 9792
(c) 9794
(d) 9793
Show Answer & Explanation

Answer: (d) 9793

Explanation:
1. LCM(6, 9, 12, 17) = 612.
2. Greatest multiple of 612 below 10000 is 612 × 16 = 9792.
3. Number = 9792 + 1 = 9793.

Teacher's Note:
1. Find the greatest multiple of the LCM that is a four-digit number, then add the common remainder.
2. 612 × 17 = 10404 has five digits.

Question 3: A, B and C start together from the same point and run in the same direction around a circular stadium. A completes a round in 252 s, B in 308 s and C in 198 s. After what time will they meet again at the starting point?
(a) 3696 s
(b) 1848 s
(c) 1386 s
(d) 2772 s
Show Answer & Explanation

Answer: (d) 2772 s

Explanation:
1. The time is the LCM of 252, 308 and 198.
2. 252 = 2² × 3² × 7, 308 = 2² × 7 × 11, 198 = 2 × 3² × 11.
3. LCM = 2² × 3² × 7 × 11 = 2772 s = 46 min 12 s.

Teacher's Note:
1. Meeting again at the start means a whole number of rounds for each runner.
2. So use the LCM.

Question 4: The LCM of 5, 8, 12 and 20 will not be a multiple of
(a) 5
(b) 9
(c) 3
(d) 8
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. LCM = 2³ × 3 × 5 = 120.
2. 120 is divisible by 3, 8 and 5, but not by 9 (it has only one factor 3).

Teacher's Note:
1. Any multiple of the LCM is divisible by each number, but not necessarily by their powers.
2. Compute the LCM then test.

Question 5: Find the greatest possible length of planks into which three pieces of timber 42 m, 49 m and 63 m long can be cut with no piece left over.
(a) 63 m
(b) 8 m
(c) 49 m
(d) 7 m
Show Answer & Explanation

Answer: (d) 7 m

Explanation:
1. Required length = HCF(42, 49, 63).
2. 42 = 2·3·7, 49 = 7², 63 = 3²·7, so HCF = 7.

Teacher's Note:
1. The greatest equal length that divides all lengths is the HCF.
2. Pieces: 6 + 7 + 9 = 22 planks.

Question 6: The H.C.F. of two numbers is 12 and their difference is 12. The numbers are
(a) 66, 78
(b) 84, 96
(c) 70, 82
(d) 94, 106
Show Answer & Explanation

Answer: (b) 84, 96

Explanation:
1. Both numbers must be multiples of 12 differing by 12, i.e. 12m and 12(m + 1) with m and m + 1 coprime.
2. Checking options: HCF(84, 96) = 12 and 96 − 84 = 12; HCF(66, 78) = 6, HCF(70, 82) = 2, HCF(94, 106) = 2.

Teacher's Note:
1. The HCF must divide the difference, and both numbers must be multiples of the HCF.
2. Test each option's HCF.

Question 7: The L.C.M. of two prime numbers x and y (x > y) is 161. The value of 3y − x is
(a) −1
(b) −2
(c) 1
(d) 2
Show Answer & Explanation

Answer: (b) −2

Explanation:
1. LCM of two primes is their product: xy = 161 = 7 × 23.
2. x = 23, y = 7, so 3y − x = 21 − 23 = −2.

Teacher's Note:
1. For two distinct primes LCM = product.
2. Factorise 161.

Question 8: Given that \( 2^h\times 2^3=2^9 \), find the value of h.
(a) 12
(b) 3
(c) 8
(d) 6
Show Answer & Explanation

Answer: (d) 6

Explanation:
1. Add exponents: h + 3 = 9.
2. So h = 6.

Teacher's Note:
1. Same base, multiply, add the exponents.
2. 2⁶ × 2³ = 2⁹.

Question 9: Which of the following statements are true?
1. The H.C.F. of two co-prime numbers is 1.
2. If a and b are co-prime, their L.C.M. is ab.
3. 385 and 462 are co-prime.
4. The product of two numbers is 2160 and their H.C.F. is 12; their L.C.M. is 180.
(a) 1, 2 and 4 only
(b) 3 and 4 only
(c) 1, 2 and 3 only
(d) 1 and 3 only
Show Answer & Explanation

Answer: (a) 1, 2 and 4 only

Explanation:
1. 1 and 2 are definitions/properties of co-prime numbers.
2. 385 = 5·7·11 and 462 = 2·3·7·11 share 7 and 11, so they are not co-prime (HCF 77).
3. LCM = 2160 / 12 = 180, so 4 is true.

Teacher's Note:
1. Co-prime means HCF = 1.
2. LCM × HCF = product of the numbers.

Question 10: In 65*5, which digit should replace * so that the number is divisible by 9?
(a) 3
(b) 1
(c) 2
(d) 4
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Sum of digits = 6 + 5 + * + 5 = 16 + *.
2. For divisibility by 9 the sum must be 18, so * = 2: 6525 = 9 × 725.

Teacher's Note:
1. A number is divisible by 9 if its digit sum is.
2. Only one digit works since 16 + * must equal 18.

Question 11: The least number divisible by 15, 20, 24, 32 and 36 is
(a) 360
(b) 1440
(c) 720
(d) 2880
Show Answer & Explanation

Answer: (b) 1440

Explanation:
1. 15 = 3·5, 20 = 2²·5, 24 = 2³·3, 32 = 2⁵, 36 = 2²·3².
2. LCM = 2⁵ × 3² × 5 = 1440.

Teacher's Note:
1. Take the highest power of each prime.
2. 1440 / 32 = 45, so it is divisible by 32.

Question 12: Which of the following statements are true?
1. The least number of five digits exactly divisible by 16, 18, 24 and 30 is 10080.
2. The greatest number of five digits exactly divisible by 9, 12, 18 and 24 is 99720.
3. The least number divisible by 15, 20, 24, 32 and 36 is 1440.
(a) 2 and 3 only
(b) 1, 2 and 3
(c) 1 and 2 only
(d) 1 and 3 only
Show Answer & Explanation

Answer: (d) 1 and 3 only

Explanation:
1. LCM(16, 18, 24, 30) = 720 and 720 × 14 = 10080, so 1 is true.
2. LCM(9, 12, 18, 24) = 72; the greatest five-digit multiple is 72 × 1388 = 99936, not 99720, so 2 is false.
3. LCM(15, 20, 24, 32, 36) = 1440, so 3 is true.

Teacher's Note:
1. Find the LCM, then the nearest multiple.
2. Statement 2 fails because 99936 is larger than 99720.

Question 13: If GCF(a, b) = 2² × 7 × 11, LCM(a, b) = 2⁵ × 3² × 5 × 7³ × 11² and b = 2⁵ × 3² × 5 × 7 × 11, then a is
(a) 2² × 7 × 11
(b) 2² × 3² × 7³
(c) 2² × 7³ × 11²
(d) 2⁵ × 7³ × 11²
Show Answer & Explanation

Answer: (c) 2² × 7³ × 11²

Explanation:
1. a = GCF × LCM / b.
2. Exponent of 2: 2 + 5 − 5 = 2; of 3: 0 + 2 − 2 = 0; of 5: 0; of 7: 1 + 3 − 1 = 3; of 11: 1 + 2 − 1 = 2.
3. So a = 2² × 7³ × 11².

Teacher's Note:
1. Add exponents of GCF and LCM then subtract exponents of b.
2. Check that GCF has min exponents: 2², 7, 11.

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Factors and Multiples Objective Questions & Solutions for Class 8 Mathematics

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