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Quantitative Aptitude — LCM
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Quantitative Aptitude — LCM

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8 mins
Time
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10
Questions
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1
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Quantitative Aptitude — LCM — Questions with Answers Open after you finish the quiz — all 10 questions, with answers and explanations.
Q1. What is the Least Common Multiple (LCM) of 12 and 18?
  1. 6
  2. 12
  3. 18
  4. 36
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Answer: D. 36
The prime factorization of 12 is 2² × 3 and 18 is 2 × 3². The LCM is 2² × 3² = 36.
Q2. Find the LCM of 8, 12, and 15.
  1. 60
  2. 90
  3. 120
  4. 180
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Answer: C. 120
The prime factorizations are 8=2³, 12=2²×3, 15=3×5. The LCM is 2³ × 3 × 5 = 120.
Q3. What is the smallest number that is exactly divisible by 6, 9, and 15?
  1. 30
  2. 60
  3. 90
  4. 120
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Answer: C. 90
The smallest number divisible by 6, 9, and 15 is their LCM. LCM(6, 9, 15) = 90.
Q4. Find the LCM of the fractions ¾, ⁵⁄₆, and ⁷⁄₈.
  1. 105/2
  2. 35/2
  3. 105/4
  4. 70/3
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Answer: A. 105/2
LCM of fractions = LCM of numerators ÷ HCF of denominators. LCM(3,5,7)=105, HCF(4,6,8)=2. So, LCM = 105/2.
Q5. Four bells ring at intervals of 4, 6, 8, and 12 minutes respectively. If they start ringing together at 9:00 AM, at what time will they ring together again?
  1. 9:12 AM
  2. 9:24 AM
  3. 9:36 AM
  4. 9:48 AM
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Answer: B. 9:24 AM
The time they ring together again is the LCM of their intervals. LCM(4, 6, 8, 12) = 24 minutes. So, 9:00 AM + 24 minutes = 9:24 AM.
Q6. Find the smallest number which when divided by 15, 20, and 25 leaves a remainder of 5 in each case.
  1. 300
  2. 305
  3. 295
  4. 310
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Answer: B. 305
First find LCM(15, 20, 25) = 300. The required number is LCM + remainder = 300 + 5 = 305.
Q7. The HCF of two numbers is 6 and their product is 216. Find their LCM.
  1. 18
  2. 36
  3. 72
  4. 108
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Answer: B. 36
For any two numbers, LCM × HCF = Product of the numbers. So, LCM = 216 ÷ 6 = 36.
Q8. What is the LCM of (2³ × 3² × 5) and (2² × 3³ × 7)?
  1. 2² × 3²
  2. 2³ × 3³ × 5
  3. 2² × 3² × 5 × 7
  4. 2³ × 3³ × 5 × 7
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Answer: D. 2³ × 3³ × 5 × 7
The LCM is found by taking the highest power of each prime factor present in the numbers.
Q9. Three runners run around a circular track. They complete one round in 200s, 240s, and 300s respectively. If they start at the same point and time, after how many seconds will they meet again at the starting point?
  1. 1200s
  2. 900s
  3. 600s
  4. 1500s
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Answer: A. 1200s
They will meet again at the starting point after a time equal to the LCM of their individual lap times. LCM(200, 240, 300) = 1200 seconds.
Q10. What is the least number which when divided by 10, 12, 15, and 18 leaves a remainder of 3 in each case?
  1. 177
  2. 180
  3. 186
  4. 183
Show answer
Answer: D. 183
First find LCM(10, 12, 15, 18) = 180. The required number is LCM + remainder = 180 + 3 = 183.
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