LCM & HCF Quiz: Quantitative Aptitude Practice
10 timed LCM and HCF questions for RRB NTPC and SSC. Prime factorisation, remainder problems, bells and lights, fractions and the product rule.
LCM & HCF Quiz: Quantitative Aptitude Practice
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Answers to the starter quiz
Spoilers! These are the Starter quiz’s answers. Easy, Medium, Hard and Mixed rounds show their own answers when you finish.
What is the HCF of 84, 126 and 210?
- 42
- 21
- 14
- 84
Answer: 42. 84 = 2² × 3 × 7, 126 = 2 × 3² × 7, 210 = 2 × 3 × 5 × 7. Common primes with lowest powers: 2 × 3 × 7 = 42.
What is the LCM of 12, 18 and 30?
- 90
- 360
- 180
- 120
Answer: 180. Highest powers of every prime: 2² × 3² × 5 = 180.
The product of two numbers is 2,160 and their HCF is 12. What is their LCM?
- 180
- 160
- 216
- 120
Answer: 180. For two numbers, HCF × LCM = product, so LCM = 2,160 ÷ 12 = 180.
What is the least number which, when divided by 12, 15 and 20, leaves a remainder of 7 in each case?
- 127
- 57
- 77
- 67
Answer: 67. LCM of 12, 15 and 20 is 60. Add the common remainder: 60 + 7 = 67.
What is the greatest number that divides 245 and 1,029 leaving a remainder of 5 in each case?
- 8
- 16
- 24
- 32
Answer: 16. Remove the remainder: 240 and 1,024. HCF(240, 1,024) = 16.
Three traffic lights change every 24, 36 and 54 seconds. They all change together at 8:20:00 am. When will they next change together?
- 8:21:48 am
- 8:23:36 am
- 8:24:12 am
- 8:22:36 am
Answer: 8:23:36 am. LCM of 24, 36 and 54 is 216 seconds = 3 min 36 s. So the next common change is at 8:23:36 am.
What is the HCF of 2/3, 8/9 and 16/27?
- 16/3
- 2/3
- 1/27
- 2/27
Answer: 2/27. HCF of fractions = HCF of numerators / LCM of denominators = HCF(2, 8, 16)/LCM(3, 9, 27) = 2/27.
What is the greatest number that divides 43, 91 and 183 leaving the same remainder in each case?
- 4
- 8
- 12
- 2
Answer: 4. Take the differences: 91 - 43 = 48, 183 - 91 = 92, 183 - 43 = 140. HCF(48, 92, 140) = 4.
Two numbers are in the ratio 3:4 and their LCM is 180. What is the sum of the numbers?
- 75
- 135
- 105
- 90
Answer: 105. Let the numbers be 3h and 4h (3 and 4 are co-prime), so LCM = 12h = 180 and h = 15. The numbers are 45 and 60; sum 105.
What is the smallest perfect square that is divisible by each of 6, 8 and 15?
- 1,200
- 14,400
- 900
- 3,600
Answer: 3,600. LCM = 120 = 2³ × 3 × 5. Make every power even: 2^4 × 3² × 5² = 3,600 (= 60²).
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