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H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 2)

HCF and LCM appear in nearly every aptitude paper. They look alike, so students mix them up.

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Q1. Two numbers are 48 and 82. If their H.C.F. is 2, what is their L.C.M.?

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Answer: B

Let the numbers be 48 and 82, with H.C.F. 2. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 48 x 82 = 3936. So L.C.M. = 3936 / 2 = 1968. Check: 2 x 1968 = 3936, which equals 48 x 82 = 3936.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q2. Two numbers are 67 and 33. If their H.C.F. is 1, what is their L.C.M.?

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Answer: D

Let the numbers be 67 and 33, with H.C.F. 1. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 67 x 33 = 2211. So L.C.M. = 2211 / 1 = 2211. Check: 1 x 2211 = 2211, which equals 67 x 33 = 2211.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q3. Two numbers are 66 and 26, with L.C.M. 858. What is their H.C.F.?

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Answer: C

Let the numbers be 66 and 26, with L.C.M. 858. For any two numbers, H.C.F. x L.C.M. = product of the numbers. Product: 66 x 26 = 1716. So H.C.F. = 1716 / 858 = 2. Check: 2 x 858 = 1716, which equals 66 x 26 = 1716.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q4. Traffic lights at 3 road crossings change after every 36, 72, 90 seconds respectively. If they all start together, after how many seconds will they occur simultaneously again?

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Answer: A

They all start together at time zero. Each one repeats after 36, 72, 90 seconds respectively. They next coincide after a time equal to the L.C.M. of these intervals. Break each into prime factors: 36 = 2 x 2 x 3 x 3; 72 = 2 x 2 x 2 x 3 x 3; 90 = 2 x 3 x 3 x 5. Taking every prime at its highest power gives L.C.M. = 360 seconds. Check: 360/36 = 10, 360/72 = 5, 360/90 = 4, so 360 seconds is exactly divisible by each interval.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q5. Three machines in a factory complete one full cycle every 60, 15, 20 seconds respectively. If they all start together, after how many seconds will they occur simultaneously again?

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Answer: B

They all start together at time zero. Each one repeats after 60, 15, 20 seconds respectively. They next coincide after a time equal to the L.C.M. of these intervals. Break each into prime factors: 60 = 2 x 2 x 3 x 5; 15 = 3 x 5; 20 = 2 x 2 x 5. Taking every prime at its highest power gives L.C.M. = 60 seconds. Check: 60/60 = 1, 60/15 = 4, 60/20 = 3, so 60 seconds is exactly divisible by each interval.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q6. Traffic lights at 3 road crossings change after every 15, 30, 45 seconds respectively. If they all start together, after how many seconds will they occur simultaneously again?

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Answer: D

They all start together at time zero. Each one repeats after 15, 30, 45 seconds respectively. They next coincide after a time equal to the L.C.M. of these intervals. Break each into prime factors: 15 = 3 x 5; 30 = 2 x 3 x 5; 45 = 3 x 3 x 5. Taking every prime at its highest power gives L.C.M. = 90 seconds. Check: 90/15 = 6, 90/30 = 3, 90/45 = 2, so 90 seconds is exactly divisible by each interval.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q7. Four machines in a factory complete one full cycle every 72, 48, 24, 45 seconds respectively. If they all start together, after how many seconds will they occur simultaneously again?

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Answer: A

They all start together at time zero. Each one repeats after 72, 48, 24, 45 seconds respectively. They next coincide after a time equal to the L.C.M. of these intervals. Break each into prime factors: 72 = 2 x 2 x 2 x 3 x 3; 48 = 2 x 2 x 2 x 2 x 3; 24 = 2 x 2 x 2 x 3; 45 = 3 x 3 x 5. Taking every prime at its highest power gives L.C.M. = 720 seconds. Check: 720/72 = 10, 720/48 = 15, 720/24 = 30, 720/45 = 16, so 720 seconds is exactly divisible by each interval.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q8. Bells at a temple toll at intervals of 20, 30, 15, 54 seconds respectively. If they all start together, after how many seconds will they occur simultaneously again?

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Answer: C

They all start together at time zero. Each one repeats after 20, 30, 15, 54 seconds respectively. They next coincide after a time equal to the L.C.M. of these intervals. Break each into prime factors: 20 = 2 x 2 x 5; 30 = 2 x 3 x 5; 15 = 3 x 5; 54 = 2 x 3 x 3 x 3. Taking every prime at its highest power gives L.C.M. = 540 seconds. Check: 540/20 = 27, 540/30 = 18, 540/15 = 36, 540/54 = 10, so 540 seconds is exactly divisible by each interval.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q9. Bells at a temple toll at intervals of 72, 48, 15, 40 seconds respectively. If they all start together, after how many seconds will they occur simultaneously again?

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Answer: A

They all start together at time zero. Each one repeats after 72, 48, 15, 40 seconds respectively. They next coincide after a time equal to the L.C.M. of these intervals. Break each into prime factors: 72 = 2 x 2 x 2 x 3 x 3; 48 = 2 x 2 x 2 x 2 x 3; 15 = 3 x 5; 40 = 2 x 2 x 2 x 5. Taking every prime at its highest power gives L.C.M. = 720 seconds. Check: 720/72 = 10, 720/48 = 15, 720/15 = 48, 720/40 = 18, so 720 seconds is exactly divisible by each interval.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q10. Traffic lights at 4 road crossings change after every 60, 24, 20, 72 seconds respectively. If they all start together, after how many seconds will they occur simultaneously again?

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Answer: B

They all start together at time zero. Each one repeats after 60, 24, 20, 72 seconds respectively. They next coincide after a time equal to the L.C.M. of these intervals. Break each into prime factors: 60 = 2 x 2 x 3 x 5; 24 = 2 x 2 x 2 x 3; 20 = 2 x 2 x 5; 72 = 2 x 2 x 2 x 3 x 3. Taking every prime at its highest power gives L.C.M. = 360 seconds. Check: 360/60 = 6, 360/24 = 15, 360/20 = 18, 360/72 = 5, so 360 seconds is exactly divisible by each interval.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q11. Bells at a temple toll at intervals of 54, 20, 15, 48 seconds respectively. If they all start together, after how many seconds will they occur simultaneously again?

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Answer: C

They all start together at time zero. Each one repeats after 54, 20, 15, 48 seconds respectively. They next coincide after a time equal to the L.C.M. of these intervals. Break each into prime factors: 54 = 2 x 3 x 3 x 3; 20 = 2 x 2 x 5; 15 = 3 x 5; 48 = 2 x 2 x 2 x 2 x 3. Taking every prime at its highest power gives L.C.M. = 2160 seconds. Check: 2160/54 = 40, 2160/20 = 108, 2160/15 = 144, 2160/48 = 45, so 2160 seconds is exactly divisible by each interval.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q12. Find the least 4-digit number which is divisible by 21, 24, 15.

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Answer: D

Let L be the L.C.M. of 21, 24, 15. L.C.M.(21, 24, 15) = 840. The least 4-digit number is 1000. Dividing 1000 by 840 leaves remainder 160. Rounding up to the next multiple of 840 needs 680 more: 1000 + 680 = 1680. Check: 1680/840 = 2, a whole number, and 1680 >= 1000.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q13. Find the least 5-digit number which is divisible by 32, 35, 28.

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Answer: C

Let L be the L.C.M. of 32, 35, 28. L.C.M.(32, 35, 28) = 1120. The least 5-digit number is 10000. Dividing 10000 by 1120 leaves remainder 1040. Rounding up to the next multiple of 1120 needs 80 more: 10000 + 80 = 10080. Check: 10080/1120 = 9, a whole number, and 10080 >= 10000.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q14. Find the least 4-digit number which is divisible by 40, 21.

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Answer: D

Let L be the L.C.M. of 40, 21. L.C.M.(40, 21) = 840. The least 4-digit number is 1000. Dividing 1000 by 840 leaves remainder 160. Rounding up to the next multiple of 840 needs 680 more: 1000 + 680 = 1680. Check: 1680/840 = 2, a whole number, and 1680 >= 1000.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q15. Find the least 4-digit number which is divisible by 24, 32, 36.

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Answer: A

Let L be the L.C.M. of 24, 32, 36. L.C.M.(24, 32, 36) = 288. The least 4-digit number is 1000. Dividing 1000 by 288 leaves remainder 136. Rounding up to the next multiple of 288 needs 152 more: 1000 + 152 = 1152. Check: 1152/288 = 4, a whole number, and 1152 >= 1000.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q16. Find the least 5-digit number which is divisible by 28, 45, 15.

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Answer: B

Let L be the L.C.M. of 28, 45, 15. L.C.M.(28, 45, 15) = 1260. The least 5-digit number is 10000. Dividing 10000 by 1260 leaves remainder 1180. Rounding up to the next multiple of 1260 needs 80 more: 10000 + 80 = 10080. Check: 10080/1260 = 8, a whole number, and 10080 >= 10000.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q17. Find the least 4-digit number which is divisible by 24, 36, 12.

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Answer: C

Let L be the L.C.M. of 24, 36, 12. L.C.M.(24, 36, 12) = 72. The least 4-digit number is 1000. Dividing 1000 by 72 leaves remainder 64. Rounding up to the next multiple of 72 needs 8 more: 1000 + 8 = 1008. Check: 1008/72 = 14, a whole number, and 1008 >= 1000.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q18. Find the least 5-digit number which is divisible by 35, 45.

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Answer: D

Let L be the L.C.M. of 35, 45. L.C.M.(35, 45) = 315. The least 5-digit number is 10000. Dividing 10000 by 315 leaves remainder 235. Rounding up to the next multiple of 315 needs 80 more: 10000 + 80 = 10080. Check: 10080/315 = 32, a whole number, and 10080 >= 10000.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q19. Find the least 5-digit number which is divisible by 32, 15, 12.

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Answer: A

Let L be the L.C.M. of 32, 15, 12. L.C.M.(32, 15, 12) = 480. The least 5-digit number is 10000. Dividing 10000 by 480 leaves remainder 400. Rounding up to the next multiple of 480 needs 80 more: 10000 + 80 = 10080. Check: 10080/480 = 21, a whole number, and 10080 >= 10000.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q20. Find the greatest 4-digit number which is divisible by 15, 28.

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Answer: B

Let L be the L.C.M. of 15, 28. L.C.M.(15, 28) = 420. The greatest 4-digit number is 9999. Dividing 9999 by 420: 9999/420 = 23 remainder 339. Subtracting that remainder gives the greatest 4-digit multiple of 420: 9999 – 339 = 9660. Check: 9660/420 = 23, a whole number, and 9660 <= 9999.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q21. Find the greatest 4-digit number which is divisible by 45, 35, 24.

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Answer: C

Let L be the L.C.M. of 45, 35, 24. L.C.M.(45, 35, 24) = 2520. The greatest 4-digit number is 9999. Dividing 9999 by 2520: 9999/2520 = 3 remainder 2439. Subtracting that remainder gives the greatest 4-digit multiple of 2520: 9999 – 2439 = 7560. Check: 7560/2520 = 3, a whole number, and 7560 <= 9999.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q22. Find the greatest 5-digit number which is divisible by 24, 32.

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Answer: A

Let L be the L.C.M. of 24, 32. L.C.M.(24, 32) = 96. The greatest 5-digit number is 99999. Dividing 99999 by 96: 99999/96 = 1041 remainder 63. Subtracting that remainder gives the greatest 5-digit multiple of 96: 99999 – 63 = 99936. Check: 99936/96 = 1041, a whole number, and 99936 <= 99999.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q23. Find the greatest 4-digit number which is divisible by 28, 45.

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Answer: B

Let L be the L.C.M. of 28, 45. L.C.M.(28, 45) = 1260. The greatest 4-digit number is 9999. Dividing 9999 by 1260: 9999/1260 = 7 remainder 1179. Subtracting that remainder gives the greatest 4-digit multiple of 1260: 9999 – 1179 = 8820. Check: 8820/1260 = 7, a whole number, and 8820 <= 9999.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q24. Find the greatest 4-digit number which is divisible by 24, 21.

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Answer: D

Let L be the L.C.M. of 24, 21. L.C.M.(24, 21) = 168. The greatest 4-digit number is 9999. Dividing 9999 by 168: 9999/168 = 59 remainder 87. Subtracting that remainder gives the greatest 4-digit multiple of 168: 9999 – 87 = 9912. Check: 9912/168 = 59, a whole number, and 9912 <= 9999.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

Q25. Find the greatest 5-digit number which is divisible by 12, 15.

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Answer: D

Let L be the L.C.M. of 12, 15. L.C.M.(12, 15) = 60. The greatest 5-digit number is 99999. Dividing 99999 by 60: 99999/60 = 1666 remainder 39. Subtracting that remainder gives the greatest 5-digit multiple of 60: 99999 – 39 = 99960. Check: 99960/60 = 1666, a whole number, and 99960 <= 99999.

Related topics: H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples

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