Find the greatest four-digit integer that leaves remainder 6 on division by 15, 20 and 24.

The number is 120k + 6. The greatest value at most 9999 is 120 × 83 + 6 = 9966.

Three ropes of lengths 8.4 m, 12.6 m and 18.9 m are cut into equal pieces of the greatest possible length. How many pieces are obtained?

In decimetres, HCF(84, 126, 189) = 21. Total pieces = (84 + 126 + 189) ÷ 21 = 19.

Three machines start a cycle together. Their cycle times are 2.4, 3.6 and 4.5 minutes. How many cycles does the fastest machine complete before all three first finish a cycle together again?

Scale by 10: LCM(24, 36, 45) = 360, so the common time is 36 minutes. The fastest machine completes 36 ÷ 2.4 = 15 cycles.

Data sufficiency: What is the HCF of positive integers a and b? I. a and b are consecutive integers. II. a × b = 420.

Statement I alone is sufficient; Statement II alone is not.. Consecutive integers always have HCF 1, so I alone is sufficient. II allows (20, 21) with HCF 1 and (10, 42) with HCF 2, so II alone is insufficient.

Two buses leave a depot together at 6:45 a.m. One leaves every 36 minutes and the other every 54 minutes. At what time will they next leave together?

LCM(36, 54) = 108 minutes = 1 hour 48 minutes. Thus the next common departure is at 8:33 a.m.

A divisor leaves remainder 25 on dividing 925 and remainder 35 on dividing 1295. How many positive integers can be such a divisor?

The divisor must be a factor of HCF(900, 1260) = 180 and exceed 35. The eligible factors are 36, 45, 60, 90 and 180, giving 5.

Two numbers have HCF 21 and LCM 1386. If one is 126, what is the other?

Other number = HCF × LCM ÷ known number = 21 × 1386 ÷ 126 = 231.

A courtyard 8.10 m long and 6.30 m wide is covered with the largest possible equal square slabs without cutting. What is the side of each slab in cm?

Convert to cm and take HCF(810, 630) = 90. This is the largest slab side.

What is the least non-negative amount, in ₹, to subtract from ₹5380 so that the balance can be divided exactly into shares of ₹45, ₹60 or ₹75?

LCM(45, 60, 75) = 900. The remainder on dividing 5380 by 900 is 880, so subtract ₹880.

A hall measuring 9.24 m by 6.72 m is tiled with the largest equal square tiles without cutting. Laying each tile costs ₹18. What is the total laying cost in ₹?

Largest tile side = HCF(924, 672) = 84 cm. There are 11 × 8 = 88 tiles, so the laying cost is 88 × 18 = ₹1584.

Two numbers are in the ratio 9:14. Their HCF and LCM have sum 1651. Find the larger number.

If the HCF is h, the LCM is 126h. Hence 127h = 1651, so h = 13 and the larger number is 14 × 13 = 182.

A gardener plants 156 rose plants and 234 marigold plants in rows. Each row has the same number of plants and only one kind. What is the greatest possible number of plants per row?

The row size must divide both totals. HCF(156, 234) = 78.

Find the least positive integer that leaves remainders 8, 14 and 20 when divided by 15, 21 and 27 respectively, and is divisible by 11.

The common deficit is 7, so N = 945k − 7. Since 945 leaves remainder 10 modulo 11, the least suitable k is 4; N = 3780 − 7 = 3773.

Which pair is co-prime?

(35, 64). 35 = 5 × 7 and 64 = 2^6 have no common prime factor, so their HCF is 1. The other pairs share 13, 17 and 19 respectively.

Find the smallest non-negative integer x such that 4321 + x is divisible by 24, 32 and 40.

LCM(24, 32, 40) = 480. Since 4321 = 480 × 9 + 1, add 480 − 1 = 479.

Find the HCF of 7/15 and 14/25.

Both fractions are reduced. HCF = HCF(7, 14) ÷ LCM(15, 25) = 7/75.

Packets containing the same number of sweets leave 3 sweets unused whether each packet holds 14, 21 or 35 sweets. If there are more than 100 sweets, what is the least possible total?

The total is 210k + 3 because LCM(14, 21, 35) = 210. The least value above 100 is 213.

What is the least positive length, in metres, that is an exact multiple of 1.2 m, 1.8 m and 2.7 m?

Scale all lengths by 10: LCM(12, 18, 27) = 108. Divide by 10 to get 10.8 m.

Find the least integer greater than 30 that leaves remainder 5 when divided by 9, 12 and 18.

LCM(9, 12, 18) = 36. The first suitable number above 30 is 36 + 5 = 41.

Find LCM ÷ HCF for the fractions 6/14, 9/21 and 15/28.

First reduce them to 3/7, 3/7 and 15/28. Their HCF is 3/28 and LCM is 15/7, so the ratio is 20.

For positive integers a and b, HCF(a, b) = 24 and a × b = 69120. What is their LCM?

For two integers, HCF × LCM = their product. Thus LCM = 69120 ÷ 24 = 2880.

The greatest integer dividing 486, 666 and 966 with the same remainder is d. What is the remainder?

d = HCF(180, 300) = 60. Since 486 = 60 × 8 + 6, the common remainder is 6.

How many unordered pairs of positive integers have HCF 14 and LCM 5880?

After dividing by 14, the co-prime factors have product 420 = 2² × 3 × 5 × 7. Assign each of the 4 prime-power blocks to one factor: 2^4 ÷ 2 = 8 unordered pairs.

Find the greatest divisor that leaves remainders 17 and 23 on dividing 1037 and 1451 respectively.

Subtract the remainders: 1020 and 1428. Their HCF is 204, which exceeds both remainders.

Two numbers have ratio 8:15 and product 17280. Find their HCF.

Write the numbers as 8h and 15h since 8 and 15 are co-prime. Then 120h² = 17280, giving h = √144 = 12.

Topic

HCF and LCM of Numbers MCQ – Part 2 - Topic Guide and MCQs

Part of Quantitative Aptitude > 02 HCF And LCM Of Numbers - 25 questions, about 25 min, with an explanation after every answer.

Question count

about 25 min

Estimated time

Mixed

Difficulty

03 Nov 2026

Last updated

Quick answer

HCF and LCM of Numbers MCQ – Part 2 is a topic inside Quantitative Aptitude (02 HCF And LCM Of Numbers). It runs 25 untimed MCQs - roughly 25 min - and the review explains every answer once you submit.

Topic introduction

What HCF and LCM of Numbers MCQ – Part 2 covers

The shape of the set before you start it.

HCF and LCM of Numbers MCQ – Part 2 sits inside Quantitative Aptitude, under the 02 HCF And LCM Of Numbers lane, and the portal keeps it as one self-contained set: 25 questions, four options each, no clock and a written explanation behind every answer.

Attempt it once to find the gaps, read why each distractor is wrong, then retake the same set a few days later. The questions you miss twice are the ones worth writing down.

Curated by Gurpreet Singh · Punjab Police SI, Rank 2 (713/800) · Founder, House of Aspirants · Read his story →

Question bank

HCF and LCM of Numbers MCQ – Part 2: all 25 questions

Read the question, pick an option, then open the answer.

Find the greatest four-digit integer that leaves remainder 6 on division by 15, 20 and 24.
1. 9996
2. 9846
3. 9960
4. 9966
Show answer

Answer: (D) 9966

Explanation: The number is 120k + 6. The greatest value at most 9999 is 120 × 83 + 6 = 9966.
Three ropes of lengths 8.4 m, 12.6 m and 18.9 m are cut into equal pieces of the greatest possible length. How many pieces are obtained?
1. 19
2. 21
3. 39
4. 9
Show answer

Answer: (A) 19

Explanation: In decimetres, HCF(84, 126, 189) = 21. Total pieces = (84 + 126 + 189) ÷ 21 = 19.
Three machines start a cycle together. Their cycle times are 2.4, 3.6 and 4.5 minutes. How many cycles does the fastest machine complete before all three first finish a cycle together again?
1. 15
2. 8
3. 10
4. 30
Show answer

Answer: (A) 15

Explanation: Scale by 10: LCM(24, 36, 45) = 360, so the common time is 36 minutes. The fastest machine completes 36 ÷ 2.4 = 15 cycles.
Data sufficiency: What is the HCF of positive integers a and b? I. a and b are consecutive integers. II. a × b = 420.
1. Each statement alone is sufficient.
2. Statement II alone is sufficient; Statement I alone is not.
3. Both statements together are sufficient, but neither alone is sufficient.
4. Statement I alone is sufficient; Statement II alone is not.
Show answer

Answer: (D) Statement I alone is sufficient; Statement II alone is not.

Explanation: Consecutive integers always have HCF 1, so I alone is sufficient. II allows (20, 21) with HCF 1 and (10, 42) with HCF 2, so II alone is insufficient.
Two buses leave a depot together at 6:45 a.m. One leaves every 36 minutes and the other every 54 minutes. At what time will they next leave together?
1. 8:15
2. 7:39
3. 8:33
4. 9:27
Show answer

Answer: (C) 8:33

Explanation: LCM(36, 54) = 108 minutes = 1 hour 48 minutes. Thus the next common departure is at 8:33 a.m.
A divisor leaves remainder 25 on dividing 925 and remainder 35 on dividing 1295. How many positive integers can be such a divisor?
1. 5
2. 18
3. 6
4. 8
Show answer

Answer: (A) 5

Explanation: The divisor must be a factor of HCF(900, 1260) = 180 and exceed 35. The eligible factors are 36, 45, 60, 90 and 180, giving 5.
Two numbers have HCF 21 and LCM 1386. If one is 126, what is the other?
1. 231
2. 11
3. 252
4. 462
Show answer

Answer: (A) 231

Explanation: Other number = HCF × LCM ÷ known number = 21 × 1386 ÷ 126 = 231.
A courtyard 8.10 m long and 6.30 m wide is covered with the largest possible equal square slabs without cutting. What is the side of each slab in cm?
1. 45
2. 90
3. 30
4. 180
Show answer

Answer: (B) 90

Explanation: Convert to cm and take HCF(810, 630) = 90. This is the largest slab side.
What is the least non-negative amount, in ₹, to subtract from ₹5380 so that the balance can be divided exactly into shares of ₹45, ₹60 or ₹75?
1. 180
2. 880
3. 20
4. 80
Show answer

Answer: (B) 880

Explanation: LCM(45, 60, 75) = 900. The remainder on dividing 5380 by 900 is 880, so subtract ₹880.
A hall measuring 9.24 m by 6.72 m is tiled with the largest equal square tiles without cutting. Laying each tile costs ₹18. What is the total laying cost in ₹?
1. 1584
2. 792
3. 1440
4. 2376
Show answer

Answer: (A) 1584

Explanation: Largest tile side = HCF(924, 672) = 84 cm. There are 11 × 8 = 88 tiles, so the laying cost is 88 × 18 = ₹1584.
Two numbers are in the ratio 9:14. Their HCF and LCM have sum 1651. Find the larger number.
1. 182
2. 117
3. 252
4. 169
Show answer

Answer: (A) 182

Explanation: If the HCF is h, the LCM is 126h. Hence 127h = 1651, so h = 13 and the larger number is 14 × 13 = 182.
A gardener plants 156 rose plants and 234 marigold plants in rows. Each row has the same number of plants and only one kind. What is the greatest possible number of plants per row?
1. 39
2. 78
3. 156
4. 26
Show answer

Answer: (B) 78

Explanation: The row size must divide both totals. HCF(156, 234) = 78.
Find the least positive integer that leaves remainders 8, 14 and 20 when divided by 15, 21 and 27 respectively, and is divisible by 11.
1. 1883
2. 3773
3. 2828
4. 938
Show answer

Answer: (B) 3773

Explanation: The common deficit is 7, so N = 945k − 7. Since 945 leaves remainder 10 modulo 11, the least suitable k is 4; N = 3780 − 7 = 3773.
Which pair is co-prime?
1. (57, 76)
2. (39, 65)
3. (35, 64)
4. (51, 85)
Show answer

Answer: (C) (35, 64)

Explanation: 35 = 5 × 7 and 64 = 2^6 have no common prime factor, so their HCF is 1. The other pairs share 13, 17 and 19 respectively.
Find the smallest non-negative integer x such that 4321 + x is divisible by 24, 32 and 40.
1. 319
2. 1
3. 479
4. 159
Show answer

Answer: (C) 479

Explanation: LCM(24, 32, 40) = 480. Since 4321 = 480 × 9 + 1, add 480 − 1 = 479.
Find the HCF of 7/15 and 14/25.
1. 7/5
2. 1/75
3. 14/5
4. 7/75
Show answer

Answer: (D) 7/75

Explanation: Both fractions are reduced. HCF = HCF(7, 14) ÷ LCM(15, 25) = 7/75.
Packets containing the same number of sweets leave 3 sweets unused whether each packet holds 14, 21 or 35 sweets. If there are more than 100 sweets, what is the least possible total?
1. 108
2. 210
3. 213
4. 423
Show answer

Answer: (C) 213

Explanation: The total is 210k + 3 because LCM(14, 21, 35) = 210. The least value above 100 is 213.
What is the least positive length, in metres, that is an exact multiple of 1.2 m, 1.8 m and 2.7 m?
1. 5.4
2. 10.8
3. 0.108
4. 108
Show answer

Answer: (B) 10.8

Explanation: Scale all lengths by 10: LCM(12, 18, 27) = 108. Divide by 10 to get 10.8 m.
Find the least integer greater than 30 that leaves remainder 5 when divided by 9, 12 and 18.
1. 77
2. 31
3. 41
4. 36
Show answer

Answer: (C) 41

Explanation: LCM(9, 12, 18) = 36. The first suitable number above 30 is 36 + 5 = 41.
Find LCM ÷ HCF for the fractions 6/14, 9/21 and 15/28.
1. 5
2. 10
3. 60
4. 20
Show answer

Answer: (D) 20

Explanation: First reduce them to 3/7, 3/7 and 15/28. Their HCF is 3/28 and LCM is 15/7, so the ratio is 20.
For positive integers a and b, HCF(a, b) = 24 and a × b = 69120. What is their LCM?
1. 1658880
2. 2880
3. 5760
4. 1440
Show answer

Answer: (B) 2880

Explanation: For two integers, HCF × LCM = their product. Thus LCM = 69120 ÷ 24 = 2880.
The greatest integer dividing 486, 666 and 966 with the same remainder is d. What is the remainder?
1. 60
2. 54
3. 12
4. 6
Show answer

Answer: (D) 6

Explanation: d = HCF(180, 300) = 60. Since 486 = 60 × 8 + 6, the common remainder is 6.
How many unordered pairs of positive integers have HCF 14 and LCM 5880?
1. 16
2. 12
3. 8
4. 4
Show answer

Answer: (C) 8

Explanation: After dividing by 14, the co-prime factors have product 420 = 2² × 3 × 5 × 7. Assign each of the 4 prime-power blocks to one factor: 2^4 ÷ 2 = 8 unordered pairs.
Find the greatest divisor that leaves remainders 17 and 23 on dividing 1037 and 1451 respectively.
1. 102
2. 204
3. 408
4. 34
Show answer

Answer: (B) 204

Explanation: Subtract the remainders: 1020 and 1428. Their HCF is 204, which exceeds both remainders.
Two numbers have ratio 8:15 and product 17280. Find their HCF.
1. 144
2. 6
3. 24
4. 12
Show answer

Answer: (D) 12

Explanation: Write the numbers as 8h and 15h since 8 and 15 are co-prime. Then 120h² = 17280, giving h = √144 = 12.

Why it matters

Why HCF and LCM of Numbers MCQ – Part 2 is worth the hours

Where the marks actually come from.

HCF and LCM of Numbers MCQ – Part 2 is factual rather than analytical: you either recall it in seconds or you lose the mark to somebody who did. That makes it one of the cheapest sections to convert - a handful of short sittings usually settles the whole cluster.

Exam relevance

Exams where HCF and LCM of Numbers MCQ – Part 2 matters

Derived from the subject and lane this topic belongs to.

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Expected questions

Expected questions from HCF and LCM of Numbers MCQ – Part 2

What an examiner usually reaches for in this cluster.

Expect one-line recall questions - definitions, stand-for expansions, identifiers and the ordering of things. Practise the wording as well as the concept: these papers test terminology, and the set above uses the same four-option shape they do.

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Previous year questions

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How this topic behaves in real papers.

Past papers reuse clusters like this almost verbatim across years and posts. Run the same set until every distractor is familiar - previous-year style questions stop being recall and start being free marks.

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How long is the HCF and LCM of Numbers MCQ – Part 2 quiz?

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The set answers that: each question carries four options and a full explanation, so you learn the wording an examiner uses, not just the fact behind it.

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