Quantitative Aptitude

LCM and HCF: Practice Questions with Answers

14 practice questions on LCM and HCF, with answers; most also have an explanation or a worked solution. Try each one before you open its answer.

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1. Q.1

   Find the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively.

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   Answer: 13

   Solution:

   1. The required number is the HCF of (70 − 5) and (125 − 8), i.e., HCF(65, 117).
   2. 65 = 5 × 13. 117 = 3² × 13. HCF = 13.
   3. Check: 70 ÷ 13 = 5 remainder 5. 125 ÷ 13 = 9 remainder 8 — both match.

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2. Q.2

   Find the smallest number which, when divided by 20, 25 and 30, leaves remainders 17, 22 and 27 respectively.

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   Answer: 297

   Solution:

   1. Check the "k less than divisor" pattern: 20−17=3, 25−22=3, 30−27=3 — consistent, so k = 3.
   2. LCM(20, 25, 30): 20 = 2² × 5, 25 = 5², 30 = 2 × 3 × 5 → LCM = 2² × 3 × 5² = 300.
   3. Required number = LCM − k = 300 − 3 = 297.
   4. Check: 297 ÷ 20 = 14 remainder 17; 297 ÷ 25 = 11 remainder 22; 297 ÷ 30 = 9 remainder 27 — all match.

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3. Q.3

   Two numbers are in the ratio 3 : 4 and their LCM is 48. Find the HCF and the two numbers.

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   Answer: HCF = 4, numbers are 12 and 16

   Solution:

   1. Let the numbers be 3x and 4x. Since 3 and 4 share no common factor, x itself is the HCF, and the LCM is 3 × 4 × x = 12x.
   2. 12x = 48 → x = 4.
   3. The numbers are 3 × 4 = 12 and 4 × 4 = 16. HCF = x = 4. Check: HCF(12, 16) = 4 and LCM(12, 16) = 48 — matches.

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4. Q.4

   Find the HCF and LCM of 24, 36, and 60.

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   Answer: HCF = 12, LCM = 360

   Solution:

   1. 24 = 2³ × 3. 36 = 2² × 3². 60 = 2² × 3 × 5.
   2. HCF = common factors (2 and 3 appear in all three; 5 doesn’t) at their lowest power across all three: 2² × 3¹ = 4 × 3 = 12.
   3. LCM = all factors at their highest power: 2³ × 3² × 5¹ = 8 × 9 × 5 = 360.

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5. Q.5

   Three bells ring at intervals of 15, 20 and 30 minutes respectively. If they all ring together at 9:00 AM, when will they next ring together?

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   Answer: 10:00 AM

   Solution:

   1. LCM(15, 20, 30): 15 = 3 × 5, 20 = 2² × 5, 30 = 2 × 3 × 5 → LCM = 2² × 3 × 5 = 60.
   2. They next ring together after 60 minutes, i.e. at 10:00 AM.

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6. Q.6

   Find the smallest number which, when divided by 12, 16 and 24, leaves a remainder of 7 in each case.

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   Answer: 55

   Solution:

   1. First find LCM(12, 16, 24): 12 = 2² × 3, 16 = 2⁴, 24 = 2³ × 3 → LCM = 2⁴ × 3 = 48.
   2. Smallest number leaving the same remainder R = LCM + R = 48 + 7 = 55.

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7. Q.7

   Find the HCF and LCM of 12 and 18.

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   Answer: HCF = 6, LCM = 36

   Solution:

   1. 12 = 2² × 3. 18 = 2 × 3².
   2. HCF = common factors at their lowest power = 2¹ × 3¹ = 6.
   3. LCM = all factors at their highest power = 2² × 3² = 4 × 9 = 36.
   4. Check: HCF × LCM = 6 × 36 = 216, and 12 × 18 = 216 — matches.

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8. Q.8

   Find the HCF and LCM of 8, 12, and 20.

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   Answer: HCF = 4, LCM = 120

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9. Q.9

   Find the smallest number which, when divided by 15, 20 and 25, leaves remainders 11, 16 and 21 respectively.

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   Answer: 296. k = 4 in each case (15−11=20−16=25−21=4); LCM(15,20,25) = 300; 300 − 4.

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10. Q.10

    Find the LCM of 7 and 11.

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    Answer: 77. 7 and 11 are co-prime (HCF = 1), so LCM = 7 × 11 directly.

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11. Q.11

    Three bells ring at intervals of 6, 8 and 12 minutes. If they ring together at 12:00 noon, when will they next ring together?

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    Answer: 12:24 PM. LCM(6,8,12) = 24 minutes.

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12. Q.12

    Find the smallest number which, when divided by 10, 15 and 20, leaves a remainder of 4 in each case.

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    Answer: 64. LCM(10,15,20) = 60; 60 + 4.

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13. Q.13

    Find the HCF and LCM of 15 and 20.

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    Answer: HCF = 5, LCM = 60

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14. Q.14

    The HCF of two numbers is 6 and their LCM is 90. If one of the numbers is 18, find the other.

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    Answer: 30. HCF × LCM = product of numbers = 540; 540/18.

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