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

1. Find the LCM of 3^{5}*5^{7}*7^{2}, 3^{5}*5^{5}*7*11, 3^{6}*5^{3}*13 and 3^{7}*5^{2}*11*13.

a) 3^{7}*5^{7}*7^{2}*112*13

b) 3^{7}*5^{7}*7^{2}*11*13

c) 3^{6}*5^{7}*7^{2}*11*13

d) 3^{7}*5^{7}*7*11*13

View Answer

Explanation: The prime factors in the given numbers are 3, 5, 7, 11 and 13.

LCM = product of highest power of 3, 5, 7, 11, 13 = 3

^{7}*5

^{7}*7

^{2}*11*13.

2. Find the LCM of 16, 18, 24 and 30.

a) 720

b) 1440

c) 360

d) 540

View Answer

Explanation: 16 = 2

^{4}; 18 = 2*3

^{2}; 24 = 2

^{3}*3; 30 = 2*3*5.

Therefore, the LCM of 16, 18, 24 and 30 is 2

^{4}*3

^{2}*5 = 720.

3. Find the third term in the series of common multiple of 27, 81 and 135.

a) 1215

b) 2430

c) 810

d) 405

View Answer

Explanation: The series of common multiple is AP with first term and common difference equal to LCM.

The LCM of 27, 81 and 135 is 405.

The series of common multiple is 405, 810, 1215, 1620 ……

Therefore, the third term in the series of common multiple of 27, 81 and 135 is 1215.

4. Two numbers are in the ratio 4:5. Their LCM is 240. Find the sum of numbers.

a) 120

b) 108

c) 98

d) 132

View Answer

Explanation: Let the two numbers be 4x and 5x, then their LCM is 20x.

LCM 20x = 240, hence x=12.

The numbers are 4*12=48 and 5*12=60.

Therefore, sum of the numbers is 48+60 = 108.

5. Three runners running around a circular track can complete one revolution in 8, 12 and 18 minutes respectively. After how many minutes will they meet at the starting point?

a) 80

b) 84

c) 72

d) 60

View Answer

Explanation: This is a concept of LCM.

LCM of 8, 12 and 18 is 72.

Therefore, after 72 minutes they meet at the starting point.

6. Geeta made a necklace of either 20, 28 or 36 beads, not a single bead was left over. What could be the least number of beads Geeta had?

a) 1220

b) 1260

c) 1320

d) 1360

View Answer

Explanation: LCM of 25, 28 and 36 is 1260.

Therefore, least number of beads Geeta had is 1260.

7. Find the least number which is divisible by 22, 24, 38, 55, 72 and 76.

a) 75240

b) 75540

c) 72240

d) 72540

View Answer

Explanation: 22 = 2*11, 24 = 2

^{3}*3, 38 = 2*19, 55 = 5*11, 72 = 2

^{3}*3

^{2}, 76 = 2

^{2}*19.

LCM = 2

^{3}*3

^{2}*5*11*19 = 75240.

Therefore, the least number which is divisible by 22, 24, 38, 55, 72 and 76 is 75240.

8. Find the smallest number when decreased by 20 is divisible by 40, 50 and 70.

a) 1400

b) 1380

c) 1420

d) 1540

View Answer

Explanation: LCM of 40, 50 and 70 is 1400.

Therefore, the smallest number when decreased by 20 is divisible by 40, 50 and 70 is 1400+20 = 1420.

9. Which of the following number should be added to 1200 in order to make derived number divisible by each of 12, 26 and 27?

a) 204

b) 312

c) 192

d) 404

View Answer

Explanation: Let x be the number to be added.

LCM of 12, 26 and 27 is 702.

1200 + x = 2 * 702

x = 204.

Therefore, the number that should be added to 1200 in order to make derived number divisible by each of 12, 26 and 27 is 204.

10. The traffic signals at three different road crossings change after every 30, 45, 50 and 100 seconds respectively. If they all change simultaneously at 11:50 am, then at what time will they again change simultaneously?

a) 12:02 am

b) 12:10 pm

c) 12:15 am

d) 12:05 pm

View Answer

Explanation: LCM of 30, 45, 50 and 100 is 900.

900 seconds = 15 minutes.

Therefore, all traffic signals change simultaneously at 11:50 am + 15 minutes = 12:05 pm.

**More Aptitude Questions and Answers on LCM:**

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