Q1. What is the HCF (Highest Common Factor) of 12 and 18?
(a) 2
(b) 3
(c) 6
(d) 12
Answer: (c)
Explanation: The factors of 12 are 1,2,3,4,6,12; of 18 are 1,2,3,6,9,18. The greatest common factor is 6.
Q2. What is the LCM (Least Common Multiple) of 4 and 6?
(a) 8 (b) 12
(c) 16
(d) 24
Answer: (b)
Explanation: Multiples of 4: 4,8,12,16…; of 6: 6,12,18… The smallest common multiple is 12.
Q3. If the product of two numbers is 240 and their HCF is 8, what is their LCM?
(a) 30
(b) 40
(c) 60
(d) 120
Answer: (a)
Explanation: For any two numbers, Product = HCF × LCM. So LCM = Product/HCF = 240/8 = 30.
Q4. Find the HCF of 15, 25 and 35.
(a) 5
(b) 10
(c) 15
(d) 25
Answer: (a)
Explanation: Prime factors: 15=3×5, 25=5², 35=5×7. The common factor is 5.
Q5. The LCM of 8, 12 and 20 is:
(a) 40
(b) 60 (c) 120
(d) 240
Answer: (c)
Explanation: Prime factorization: 8=2³, 12=2²·3, 20=2²·5. LCM takes highest powers: 2³·3·5 = 8·3·5 = 120.
Q6. Two numbers are in the ratio 3:4 and their HCF is 5. What are the numbers?
(a) 9,12
(b) 12,16
(c) 15,20
(d) 18,24
Answer: (c)
Explanation: Let numbers be 3k and 4k. HCF = k =5 → numbers = 3×5=15 and 4×5=20.
Q7. The HCF of two co‑prime numbers is always:
(a) 0
(b) 1
(c) the smaller number
(d) the larger number
Answer: (b)
Explanation: Co‑prime numbers have no common factor other than 1, so HCF = 1.
Q8. If the LCM of two numbers is 180 and their HCF is 15, what is the product of the numbers? (a) 2700
(b) 1800
(c) 900
(d) 450
Answer: (a)
Explanation: Product = HCF × LCM = 15 × 180 = 2700.
Q9. Find the smallest number which when divided by 6, 9 and 15 leaves remainder 2 in each case.
(a) 92
(b) 90
(c) 94
(d) 96
Answer: (a)
Explanation: Required number = LCM(6,9,15) + 2. LCM = 90 → 90+2 = 92.
Q10. The HCF of 2²·3·5 and 2·3²·7 is:
(a) 2·3
(b) 2²·3
(c) 2·3² (d) 2·3·5·7
Answer: (a)
Explanation: Common prime factors with lowest powers: 2¹·3¹ = 6.
Q11. The LCM of 5, 7 and 9 is: (a) 35
(b) 45
(c) 63
(d) 315
Answer: (d)
Explanation: Since they are pairwise co‑prime except 9=3², LCM = 5·7·9 = 315.
Q12. If two numbers are such that one is a multiple of the other, then their HCF equals:
(a) the smaller number
(b) the larger number
(c) the LCM
(d) 1 Answer: (a)
Explanation: When one number divides the other, the smaller number is the greatest common divisor.
Q13. Find the greatest number that will divide 43, 91 and 183 leaving the same remainder in each case.
(a) 4
(b) 7
(c) 9
(d) 13
Answer: (a)
Explanation: Differences: 91‑43=48, 183‑91=92, 183‑43=140. HCF of 48,92,140 = 4.
Q14. The product of two numbers is 1080 and their LCM is 180. Their HCF is:
(a) 4
(b) 6
(c) 8
(d) 10
Answer: (b)
Explanation: HCF = Product/LCM = 1080/180 = 6.
Q15. Which of the following pairs has LCM equal to their product?
(a) 4 and 6 (b) 6 and 8
(c) 7 and 9
(d) 8 and 12
Answer: (c)
Explanation: 7 and 9 are co‑prime, so LCM = 7×9 = 63 = product.
Q16. The HCF of 0 and any non‑zero integer n is:
(a) 0 (b) n
(c) 1
(d) undefined
Answer: (b)
Explanation: Every number divides 0, so the greatest common divisor of 0 and n is |n|.
Q17. Find the LCM of fractions 2/3, 4/5 and 6/7.
(a) 12/105
(b) 12/35
(c) 12/1
(d) 12
Answer: (d)
Explanation: LCM of fractions = LCM of numerators / HCF of denominators. LCM(2,4,6)=12; HCF(3,5,7)=1 → LCM =12.
Q18. The HCF of fractions 3/4, 9/16 and 15/8 is:
(a) 3/16
(b) 3/8
(c) 3/4 (d) 1/2
Answer: (a)
Explanation: HCF of fractions = HCF of numerators / LCM of denominators. HCF(3,9,15)=3; LCM(4,16,8)=16 → HCF = 3/16.
Q19. Three bells toll at intervals of 6, 8 and 12 seconds. After how many seconds will they toll together again?
(a) 24
(b) 36
(c) 48
(d) 72 Answer: (a)
Explanation: They toll together at LCM(6,8,12)=24 seconds.
Q20. The greatest number which divides 1657 and 2037 leaving remainders 6 and 5 respectively is: (a) 123
(b) 127
(c) 235
(d) 305
Answer: (b)
Explanation: Subtract remainders: 1657‑6=1651, 2037‑5=2032. Required number = HCF(1651,2032)=127.
Q21. If the HCF of two numbers is 12 and their LCM is 144, then one of the numbers could be:
(a) 24
(b) 36 (c) 48
(d) 96
Answer: (c)
Explanation: Let numbers be 12a and 12b with a,b co‑prime and LCM=12ab=144 → ab=12. Possible (a,b) pairs: (1,12),(2,6),(3,4). Corresponding numbers: 12,144; 24,72; 36,48. So 48 is valid.
Q22. The least number which when increased by 5 is exactly divisible by 24, 32 and 40 is:
(a) 475
(b) 480
(c) 485
(d) 490
Answer: (a)
Explanation: Required number = LCM(24,32,40) – 5. LCM = 480 → 480‑5 = 475.
Q23. Two numbers are in the ratio 5:7 and their LCM is 210. What is their HCF?
(a) 5
(b) 6
(c) 7
(d) 10
Answer: (b)
Explanation: Let numbers be 5k and 7k. Since 5 and 7 are co‑prime, LCM = 5·7·k = 35k =210 → k=6. HCF = k =6.
Q24. Find the smallest 4‑digit number divisible by 18, 24 and 30.
(a) 1080
(b) 1200
(c) 1440
(d) 1560
Answer: (a)
Explanation: LCM(18,24,30)=360. Smallest 4‑digit multiple of 360 is 360×3=1080.
Q25. If the HCF of (a,b) = 1 and LCM of (a,b) = 60, then the product a·b equals:
(a) 30
(b) 60
(c) 120
(d) 360
Answer: (b)
Explanation: For any two numbers, a·b = HCF × LCM. Here HCF=1, LCM=60 → product = 60.
