The hardest part of an HCF or LCM word problem is rarely the calculation. It is deciding which one the story is asking for. This guide is built around that decision, with 16 worked problems.
Our HCF & LCM article covered how to calculate both. This guide is about something different: reading a real-life story problem and knowing straight away which one it is asking for. The calculation is usually the easy part. The hard part is the decision, and that is a skill you can learn with a short checklist and plenty of examples.
Every HCF or LCM word problem gives itself away through a pattern in the question, once you know what to look for.
| Signal Words / Situation | Use |
|---|---|
| “Largest” or “greatest” possible size, splitting into equal groups, dividing evenly with nothing left over | HCF |
| “Smallest” or “least” amount, things happening together again, repeating events lining up | LCM |
The core idea: HCF problems break something bigger into the largest possible equal pieces. LCM problems find the soonest point where repeating events line up again.
Try each problem on paper first, then tap Reveal Solution to compare.
Two ropes measuring 72 metres and 96 metres need to be cut into equal pieces, as long as possible, with none left over. How long should each piece be?
Why HCF: “as long as possible” with “none left over” signals the largest equal split.
Solution: Find HCF(72, 96). 72 = 23 × 32 and 96 = 25 × 3. Shared primes with the smallest powers: 23 × 3 = 24.
Answer: Each piece should be 24 metres long. (72 ÷ 24 = 3 pieces and 96 ÷ 24 = 4 pieces.)
A teacher has 48 pencils and 60 erasers. She wants to make identical gift kits with no items left over, using the greatest number of kits possible. How many kits can she make?
Why HCF: “greatest number of kits” with identical, evenly divided contents signals the HCF.
Solution: Find HCF(48, 60). Common factors of 48 and 60 include 1, 2, 3, 4, 6 and 12. The highest is 12.
Answer: She can make 12 kits, each with 48 ÷ 12 = 4 pencils and 60 ÷ 12 = 5 erasers.
Three containers hold 120 litres, 180 litres and 240 litres of water. What is the largest measuring container that can measure each amount exactly, with nothing left over?
Why HCF: “largest” size that divides every amount exactly. HCF works the same way with three numbers as with two.
Solution: 120 = 23 × 3 × 5, 180 = 22 × 32 × 5 and 240 = 24 × 3 × 5. Shared: 22 × 3 × 5 = 60.
Answer: The largest measuring container is 60 litres. It is used 2, 3 and 4 times.
Choosing HCF or LCM should never depend on how big the numbers look. It depends only on what the question is asking for.
The fix: always find the action phrase. “Cut into equal pieces, as long as possible” is a splitting situation, so it is HCF, however large or small the numbers are. Re-read the question’s action, not its numbers, before deciding.
A rectangular floor is 450 cm long and 300 cm wide. It is to be covered with identical square tiles, with no cutting. What is the largest possible tile, and how many are needed?
Why HCF: “largest possible” square that fits exactly in both directions.
Solution: HCF(450, 300): 450 = 2 × 32 × 52 and 300 = 22 × 3 × 52. Shared: 2 × 3 × 52 = 150. Tiles along the length: 450 ÷ 150 = 3. Along the width: 300 ÷ 150 = 2.
Answer: Tiles of side 150 cm, and 3 × 2 = 6 tiles are needed.
A rectangular field is 60 m by 84 m. Fence posts are to be placed around it, with a post at every corner and equal spacing between posts, as far apart as possible. How far apart are the posts, and how many posts are needed?
Why HCF: the spacing must divide both side lengths exactly, and “as far apart as possible” means the largest such spacing.
Solution: HCF(60, 84) = 12 (60 = 22 × 3 × 5 and 84 = 22 × 3 × 7). The perimeter is 2 × (60 + 84) = 288 m. Number of posts = 288 ÷ 12 = 24.
Answer: Posts are 12 m apart, and 24 posts are needed.
A teacher has 210 pencils and 140 notebooks to share equally among as many students as possible, with nothing left over. How many students are there, and what does each get?
Why HCF: “as many as possible” with an equal share of every item and none left over.
Solution: HCF(210, 140) = 70 (210 = 2 × 3 × 5 × 7 and 140 = 22 × 5 × 7, so the shared part is 2 × 5 × 7 = 70).
Answer: 70 students. Each gets 210 ÷ 70 = 3 pencils and 140 ÷ 70 = 2 notebooks.
Two bells ring every 15 minutes and every 20 minutes. If they both ring together right now, how soon will they next ring together?
Why LCM: “next ring together” describes two repeating events lining up again.
Solution: Find LCM(15, 20). Multiples of 15: 15, 30, 45, 60. Multiples of 20: 20, 40, 60. The first shared value is 60.
Answer: The bells next ring together in 60 minutes.
A gardener has two plants. One blooms every 4 days and the other every 6 days. If both bloom today, in how many days will they bloom together again?
Why LCM: “bloom together again” is another repeating-event alignment.
Solution: LCM(4, 6): multiples of 4 are 4, 8, 12; multiples of 6 are 6, 12. The LCM is 12.
Answer: They will bloom together again in 12 days.
Three runners complete a lap of a track in 8 minutes, 12 minutes and 16 minutes. If they all start together, after how many minutes will they all be back at the starting point at the same time?
Why LCM: “all back at the same time” is three repeating events lining up again.
Solution: LCM(8, 12, 16): 8 = 23, 12 = 22 × 3 and 16 = 24. Largest powers: 24 × 3 = 48.
Answer: All three runners are at the start together after 48 minutes. By then the runners have done 6, 4 and 3 laps.
Multiplying the numbers always gives a common multiple, but not necessarily the first one.
The fix: the bells already ring together at 60 minutes, so 300 minutes is only the fifth time they meet. The LCM is the smallest common multiple: 60. Multiplying works as a shortcut only when the numbers share no common factor.
Two buses leave a station at 7:00 am. Bus A leaves every 40 minutes and bus B every 60 minutes. At what time do they next leave together?
Why LCM: “next leave together” is two repeating events lining up.
Solution: LCM(40, 60): 40 = 23 × 5 and 60 = 22 × 3 × 5, so LCM = 23 × 3 × 5 = 120 minutes = 2 hours.
Answer: They next leave together at 9:00 am. (The LCM gives a time gap. You must add it to the start time.)
The LCM is a gap of time, and the question asks for a clock time.
The fix: convert 120 minutes to 2 hours and add it to the start time: 7:00 am + 2 hours = 9:00 am. Always re-read the last sentence of the problem to see what form of answer it wants (a time, a number of days, a length, a count).
Two security guards work night shifts. One works every 4th day and the other every 6th day. They are both on duty on Monday. On which day of the week are they next on duty together?
Why LCM: repeating cycles that line up again.
Solution: LCM(4, 6) = 12 days. Count the days of the week: 12 = 7 + 5, so 12 days after Monday is 5 days after Monday.
Answer: They are next on duty together on Saturday.
Traffic light A is green for a moment every 40 seconds and traffic light B every 60 seconds. They are both green together now. How many times in the next 10 minutes are they both green together (not counting now)?
Why LCM: two repeating events lining up, then a count of how many times that happens.
Solution: LCM(40, 60) = 120 seconds = 2 minutes. In 10 minutes = 600 seconds, the number of times is 600 ÷ 120 = 5.
Answer: They are green together 5 more times in 10 minutes (at 2, 4, 6, 8 and 10 minutes).
Tiles shaped like rectangles 12 cm by 18 cm are laid side by side, all facing the same way, to make the smallest possible solid square. What is the side of the square, and how many tiles are needed?
Why LCM: the side of the square must be a multiple of both 12 and 18, and “smallest” asks for the lowest such multiple.
Solution: LCM(12, 18) = 36. The square has a side of 36 cm. Tiles across: 36 ÷ 12 = 3. Tiles down: 36 ÷ 18 = 2.
Answer: A square of side 36 cm, made from 3 × 2 = 6 tiles.
Find the smallest number that leaves a remainder of 5 when divided by 12, by 15 or by 20.
Why LCM: we need a number that is a common multiple, plus the remainder.
Solution: LCM(12, 15, 20): 12 = 22 × 3, 15 = 3 × 5 and 20 = 22 × 5, so LCM = 22 × 3 × 5 = 60. The number is 60 + 5 = 65.
Answer: 65. Check: 65 ÷ 12 = 5 remainder 5, 65 ÷ 15 = 4 remainder 5, 65 ÷ 20 = 3 remainder 5. ✓
Some questions add an extra step. Use the same decision (splitting or repeating), then handle the extra detail.
Find the greatest number that divides 126 and 94 and leaves a remainder of 6 in each case.
Why HCF: we want the greatest divisor. First remove the remainder so that each number divides exactly.
Solution: Subtract the remainder: 126 − 6 = 120 and 94 − 6 = 88. HCF(120, 88) = 8 (120 = 23 × 3 × 5 and 88 = 23 × 11). The divisor must also be bigger than the remainder, and 8 is bigger than 6.
Answer: 8. Check: 126 ÷ 8 = 15 remainder 6, and 94 ÷ 8 = 11 remainder 6. ✓
The HCF of two numbers is 8 and their LCM is 120. One number is 24. Find the other.
Why the product rule: for two numbers, HCF × LCM = the product of the numbers.
Solution: Other number = (8 × 120) ÷ 24 = 960 ÷ 24 = 40.
Answer: The other number is 40. Check: HCF(24, 40) = 8 and LCM(24, 40) = 120. ✓
Packaging companies use HCF thinking to find the largest box size that fits different products without wasted space. Bus and train timetables, factory shifts and machine servicing schedules use LCM thinking to work out when different cycles will line up again.
Traffic engineers use similar timing ideas to design “green wave” corridors, where lights on a major road are timed so that cars travelling at the right speed meet a series of green lights in a row. The timing of signals along a route is planned so that their cycles line up in the way the traveller needs.
Read each scenario and decide: does it need HCF or LCM? Your score is tracked as you go, and your best score is saved on this device.
1. HCF(56, 84) = 28. Pieces: 56 ÷ 28 = 2 and 84 ÷ 28 = 3, so 5 pieces in all, each 28 m long.
2. LCM(12, 18) = 36 seconds.
3. LCM. “Smallest number” that is divisible by 6, 8 and 10 is the LCM: 120 biscuits.
1. HCF(540, 360) = 180 (540 = 22 × 33 × 5, 360 = 23 × 32 × 5, shared: 22 × 32 × 5 = 180). Tile side 180 cm. Tiles = (540 ÷ 180) × (360 ÷ 180) = 3 × 2 = 6.
2. LCM(12, 18, 24) = 23 × 32 = 72 minutes = 1 hour 12 minutes. 9:00 am + 1 h 12 min = 10:12 am.
3. LCM(6, 9, 15) = 2 × 32 × 5 = 90. The number is 90 + 2 = 92.
4. Other number = (12 × 360) ÷ 72 = 60. Check: 72 = 23 × 32 and 60 = 22 × 3 × 5, so HCF = 22 × 3 = 12 and LCM = 23 × 32 × 5 = 360. ✓
Students who can calculate an HCF and LCM still stall on word problems because they guess. Teach the decision before the calculation: ask “is it splitting or repeating?” and ask them to say why. Let learners act out the bells or buses with counters on a number line. Ask for an estimate before the calculation (should the answer be bigger or smaller than the given numbers?). Encourage them to write the answer as a full sentence with units. A good routine is to have students write their own HCF and LCM story problem and swap with a partner.
Word problems are where mathematics is actually used. Recognising an HCF or LCM situation hidden inside a real scenario is a far more valuable skill than calculating either one in isolation. The same skill, spotting which tool a problem is really asking for, carries into every other area of mathematics, from algebra word problems to planning and engineering.
12 questions. Select an answer for each, then submit to see your score instantly.