Gaps and Intervals on an Open Line
A straight road measuring 480 m has trees planted at regular intervals of 15 m from one end of the road to the other end. (a) How many trees are planted along the road in total? (b) If lampposts are to be placed at every 20 m along the same road from end to end, at how many locations will a tree and a lamppost share the exact same spot?
Draw a segmented distance line to visualize gaps between planted anchors.
- Draw line of 480 m.
- Mark 1st tree at 0 m. Every gap of 15 m adds 1 tree.
- Number of gaps = 480 ÷ 15 = 32.
- Total trees = 32 gaps + 1 starting tree = 33 trees.
- Common milestones: list multiples of 15 and 20 ⟹ 60 m, 120 m, 180 m, …
- Number of 60 m gaps = 480 ÷ 60 = 8.
- Shared spots = 8 + 1 = 9 spots.
answer(a) 33 trees; (b) 9 locations
techniqueMultiples · The Lowest Common Multiple
examsPSLE · GCSE Higher
Common pitfalls
- Forgetting to add 1 for the first tree at the start: 480 ÷ 15 = 32.
- Omitting the starting shared spot at 0 m in part (b), answering 8 instead of 9.
Gaps and Intervals: Time Chimes / Strikes
A grandfather clock takes 6 seconds to strike 4 o'clock (4 strikes). How many seconds will the same clock take to strike 10 o'clock (10 strikes) at the same striking rate?
Draw a timeline bar showing strikes as tick marks and spaces between ticks as duration boxes.
- Draw 4 tick marks: [Strike 1] --- [Strike 2] --- [Strike 3] --- [Strike 4].
- Count the gaps between ticks: exactly 3 gaps.
- 3 gaps = 6 seconds ⟹ 1 gap = 2 seconds.
- For 10 strikes, draw 10 ticks: exactly 10 − 1 = 9 gaps.
- Total time = 9 × 2 = 18 seconds.
answer18 seconds
techniqueTime Durations · Multiples
examsPSLE · GCSE Higher
Common pitfalls
- Using direct unitary method on strikes: 64 × 10 = 15 seconds, ignoring the fact that the first strike occurs at time zero.
- Dividing 6 by 4 instead of 3 to determine the interval length.
Classic Supposition / Assumption (Chicken and Rabbit)
A science lab has 36 microscopes and electronic balances altogether. Each microscope requires 2 batteries to operate, and each electronic balance requires 4 batteries to operate. A total of 104 batteries were used to power all 36 devices. How many electronic balances are there in the lab?
Assume all devices are microscopes, calculate the resulting battery deficit, and swap devices one-by-one to absorb the shortfall.
- Step 1 (Assumption): Assume all 36 devices are microscopes.
- Step 2 (Assumed Total): 36 × 2 = 72 batteries.
- Step 3 (Total Shortfall): 104 − 72 = 32 batteries missing.
- Step 4 (Unit Difference): Replacing 1 microscope with 1 balance adds 4 − 2 = 2 batteries.
- Step 5 (Number of Swaps): 32 ÷ 2 = 16 swaps.
- Number of electronic balances = 16 (and microscopes = 36 − 16 = 20).
answer16 electronic balances
techniqueTwo-Step Word Problems · Forming Equations
examsO-Level · SAT · GCSE Higher · H2
Common pitfalls
- Dividing the total shortfall (32) by 4 instead of the difference per device (2).
- Assuming all items are balances and forgetting to subtract from 36 to identify which item count was solved.
Three-Category Supposition with Constraint Linking
A donation tin contained 54 coins consisting of 20-cent, 50-cent, and $1 coins. There were twice as many 20-cent coins as 50-cent coins. The total value of all the coins was $37.20. How many $1 coins were in the donation tin?
Group two 20-cent coins and one 50-cent coin into a single composite 'bundle' of 3 coins worth $0.90, then apply supposition against $1 coins.
- Define 1 bundle of small coins: 2 of 20-cent + 1 of 50-cent = 3 coins, value = $0.90.
- Assume all 54 coins are $1 coins: Assumed value = 54 × $1 = $54.00.
- Surplus over actual value: $54.00 − $37.20 = $16.80.
- Replacing 3 of the $1 coins (worth $3.00) with 1 bundle of 3 small coins (worth $0.90) reduces total value by: $3.00 − $0.90 = $2.10.
- Number of bundles needed: $16.80 ÷ $2.10 = 8 bundles.
- Coins used in bundles: 8 × 3 = 24 coins.
- Number of $1 coins remaining: 54 − 24 = 30 coins.
answer30 one-dollar coins
techniqueForming Equations · Counting Coins
examsO-Level · SAT · GCSE Higher · H2
Common pitfalls
- Treating 20-cent and 50-cent coins as independent entities during supposition rather than bundling them according to their fixed 2 : 1 constraint.
- Subtracting 8 from 54 directly, forgetting that each bundle consumes 3 coins.
Excess and Shortage (Opposite Directions)
A teacher wants to distribute markers equally to a class of art students. If she gives each student 6 markers, she will have 14 markers left over. If she gives each student 8 markers, she will be short of 18 markers. (a) How many students are there in the art class? (b) How many markers does the teacher have?
Draw two distribution bars: the distance between having 14 extra and being 18 short represents the total gap to bridge.
- Find Total Gap: To move from a surplus of 14 to a deficit of 18 requires 14 + 18 = 32 markers.
- Find Unit Difference: Each student receives 8 − 6 = 2 extra markers.
- (a) Number of students = Total GapUnit Difference = 32 ÷ 2 = 16 students.
- (b) Find total markers using Case 1: 16 × 6 + 14 = 96 + 14 = 110 markers.
- Check using Case 2: 16 × 8 − 18 = 128 − 18 = 110 markers.
answer(a) 16 students; (b) 110 markers
techniqueRemainders · Forming Equations
examsPSLE
Common pitfalls
- Subtracting the shortage from the excess (18 − 14 = 4) instead of adding them to find the total gap between surplus and deficit.
- Dividing the total gap (32) by the sum of allocations (6 + 8 = 14) instead of the difference (8 − 6 = 2).
Excess and Excess / Shortage and Shortage (Same Direction)
Uncle Raymond packed oranges into gift boxes. If he packed 5 oranges into each box, he had 48 oranges left unpacked. If he packed 9 oranges into each box, he still had 8 oranges left unpacked. (a) How many gift boxes did Uncle Raymond have? (b) How many oranges did he have altogether?
Compare the remaining loose oranges in both scenarios to identify how many surplus oranges were absorbed into larger packs.
- Case 1 leaves 48 loose oranges; Case 2 leaves 8 loose oranges.
- Number of loose oranges packed into existing boxes: 48 − 8 = 40 oranges.
- Extra oranges packed per box: 9 − 5 = 4 oranges.
- (a) Number of boxes: 40 ÷ 4 = 10 boxes.
- (b) Total oranges: 10 × 9 + 8 = 98 oranges.
answer(a) 10 gift boxes; (b) 98 oranges
techniqueRemainders · Forming Equations
examsPSLE
Common pitfalls
- Adding the two excesses (48 + 8 = 56) as if one was a shortage, obtaining 56 ÷ 4 = 14 boxes.
- Confusing total boxes with total oranges and answering 10 for both parts.
Container Mass with Partial Liquid / Content
A metal drum filled completely with oil has a total mass of 24.8 kg. When 34 of the oil is pumped out, the remaining oil and drum have a mass of 8.6 kg. (a) What is the mass of the empty metal drum? (b) What was the mass of the oil when the drum was completely full?
Represent the total mass as 1 drum block plus 4 oil units, then match the removed mass directly to 3 oil units.
- Full model: [Drum] + [u][u][u][u] = 24.8 kg.
- Partially empty model: [Drum] + [u] = 8.6 kg.
- The mass lost corresponds exactly to the 3 oil units removed:
- 3u = 24.8 − 8.6 = 16.2 kg.
- Value of u = 16.2 ÷ 3 = 5.4 kg.
- (b) Full oil mass = 4u = 4 × 5.4 = 21.6 kg.
- (a) Mass of empty drum = 8.6 − 5.4 = 3.2 kg.
answer(a) 3.2 kg; (b) 21.6 kg
techniqueForming Equations · Grams and Kilograms
examsO-Level · SAT · GCSE Higher · H2
Common pitfalls
- Assuming the drum mass is also reduced by 34 (e.g., calculating 24.8 × 14 = 6.2 kg).
- Subtracting 16.2 kg directly from 8.6 kg and obtaining a negative mass.
Working Backwards (Multi-Stage Transfer)
A box contained some game tokens. Sean took half of the tokens and 4 more tokens. Then, Terry took 13 of the remaining tokens and 6 more tokens. Finally, Uma took 14 of what was left and 3 more tokens. In the end, there were 15 tokens left in the box. How many tokens were in the box at first?
Build step-by-step reverse unit bars from the final 15 tokens back to the initial total.
- Stage 3 (Uma): Add back 3 tokens ⟹ 15 + 3 = 18. This represents 3 out of 4 units ⟹ 1 unit = 6 ⟹ Before Uma: 6 × 4 = 24.
- Stage 2 (Terry): Add back 6 tokens ⟹ 24 + 6 = 30. This represents 2 out of 3 units ⟹ 1 unit = 15 ⟹ Before Terry: 15 × 3 = 45.
- Stage 1 (Sean): Add back 4 tokens ⟹ 45 + 4 = 49. This represents half the box ⟹ Initial tokens: 49 × 2 = 98.
answer98 tokens
techniqueMultiplying and Dividing Undo Each Other · A Fraction of an Amount
examsPSLE · GCSE Higher
Common pitfalls
- Subtracting the constant amounts during back-calculation (15 − 3 = 12) instead of adding them back.
- Multiplying by the fraction taken rather than dividing by the fraction remaining.
Grouping with Promotional Discounts (Bundle Pricing)
A bookstore sells files at $3.50 each. A special discount is offered: 'Buy 4 files and get 1 additional file free'. Mrs. Lee needs 43 files for her company. What is the least amount of money she needs to spend?
Draw representative 5-item blocks containing 4 paid units and 1 zero-cost unit, followed by loose unit blocks.
- Draw 1 Group of 5 files: [$3.50][$3.50][$3.50][$3.50][FREE] = 5 files for $14.00.
- Divide target files by group size: 43 ÷ 5 = 8 groups with 3 files left over.
- Cost of 8 groups: 8 × $14.00 = $112.00.
- Cost of 3 loose files: 3 × $3.50 = $10.50.
- Combine totals: $112.00 + $10.50 = $122.50.
answer$122.50
techniqueRemainders · Money as Decimals
examsPSLE
Common pitfalls
- Dividing 43 by 4 instead of 5, assuming free files do not count toward the target quantity.
- Applying the 'buy 4' promo to the remaining 3 files, giving a discount that was not earned.
Simultaneous Elimination (Scale and Subtract)
3 shirts and 4 pairs of shorts cost $168. 5 shirts and 2 pairs of shorts cost $182. (a) What is the cost of 1 pair of shorts? (b) How much do 2 shirts and 3 pairs of shorts cost?
Draw two sets of comparison bars, double the second set to make shorts identical, and deduce the price of shirts from the excess length.
- Set 1: 3 shirts + 4 shorts = $168.
- Set 2: 5 shirts + 2 shorts = $182.
- Double Set 2: 10 shirts + 4 shorts = $182 × 2 = $364.
- Compare Doubled Set 2 with Set 1: the 4 shorts cancel out.
- Difference in shirts: 10 − 3 = 7 shirts.
- Price difference: $364 − $168 = $196 ⟹ 1 shirt = $196 ÷ 7 = $28.
- Find 4 shorts from Set 1: $168 − (3 × $28) = $168 − $84 = $84.
- (a) 1 pair of shorts = $84 ÷ 4 = $21.
- (b) 2 shirts + 3 shorts = (2 × $28) + (3 × $21) = $56 + $63 = $119.
answer(a) $21; (b) $119
techniqueSimultaneous by Elimination · Solving Simultaneous Equations by Scaling
examsO-Level · SAT · GCSE Higher
Common pitfalls
- Doubling the items in Set 2 without doubling the money on the right-hand side.
- Subtracting equations without equalizing one of the item types first (5s − 3s and 4h − 2h simultaneously without eliminating either variable).
A Library Order of Dictionaries and Atlases
A primary school buys 48 dictionaries for its library at $27 each. It also buys some atlases at $15 each. The whole bill is $1776. (a) How much do the 48 dictionaries cost? (b) How many atlases does the school buy?
Draw the bill as one bar in two parts, the dictionaries and the atlases. Long multiplication gives the first part, and the second part is what is left of the bill.
- The bill of $1776 is one bar with two parts: the 48 dictionaries at $27 each, and an unknown number of atlases at $15 each.
- Find 48 × 27 by long multiplication. Multiply by the ones digit first: 48 × 7 = 336.
- Now multiply by the tens digit. The 2 in 27 stands for 2 tens, so write a 0 in the ones place, then 48 × 2 = 96: the second row is 48 × 20 = 960.
- (a) Add the two rows: 336 + 960 = 1296. The dictionaries cost $1296.
- The rest of the bill pays for the atlases: $1776 − $1296 = $480.
- (b) Each atlas costs $15, so the school buys 480 ÷ 15 = 32 atlases. Check: 1296 + 32 × 15 = 1296 + 480 = 1776.
answer(a) $1296; (b) 32 atlases
techniqueTwo Digits by Two Digits · Two-Step Word Problems
examsPSLE · GCSE Higher
Common pitfalls
- Writing 48 × 2 = 96 as the second row without the 0 in the ones place. The 2 in 27 is 2 tens, so the row is 960, and 336 + 96 = 432 is far too small for 48 books at $27.
- Dividing the whole bill by the price of an atlas, 1776 ÷ 15. The bill also pays for the dictionaries, so their $1296 must be taken away first.
School Fair Money Shared Equally Among the Classes
At a school fair, the parents' association raises $3024. The money is for class outings and is shared equally among the school's 18 classes. (a) How much does each class receive? (b) A local business then offers to pay for two classes' outings, so the whole $3024 is shared equally among the other 16 classes instead. How much more does each of those 16 classes receive than before?
Draw the money as one bar cut into 18 equal shares. When two classes no longer need their shares, those two shares are cut into 16 equal pieces, one more piece for each of the other classes.
- Draw the $3024 as one bar cut into 18 equal shares, one for each class. To find one share, divide 3024 by 18 by long division.
- Since 3 is less than 18, start with 30: 30 ÷ 18 = 1 remainder 12. Bring down the 2 to make 122.
- 18 × 6 = 108, so 122 ÷ 18 = 6 remainder 122 − 108 = 14. Bring down the 4 to make 144.
- (a) 18 × 8 = 144, so 144 ÷ 18 = 8 with nothing left over. Each class receives $168.
- The two shares no longer needed are 2 × $168 = $336, and they are shared among the other 16 classes.
- (b) Long division gives 336 ÷ 16 = 21, so each of the 16 classes receives $21 more. Check: 16 × (168 + 21) = 16 × 189 = 3024.
answer(a) $168; (b) $21 more
techniqueLong Division · Two-Step Word Problems
examsPSLE · GCSE Higher
Common pitfalls
- Answering $189 for part (b). That is the new share; the question asks how much more it is than the old share of $168.
- Stopping the long division after two digits and answering $16. Every digit of 3024 must be brought down, so there is a third round, 144 ÷ 18 = 8, and the share is $168.
Coaches and Minibuses for a School Trip
A school is taking 468 students and teachers on a trip to the science center. A coach seats 44 people. (a) How many coaches must the school hire so that everyone has a seat? (b) Instead, the school hires only as many coaches as it can fill completely, and everyone left over travels in minibuses that seat 12 people each. How many minibuses does it need?
Draw the 468 people as one bar and cut it into coach loads of 44. The quotient counts the full coaches and the remainder counts the people left over; each part of the question reads that remainder in its own way.
- Draw the 468 people as one bar and cut it into coach loads of 44. Long division tells us how many loads there are.
- Since 4 is less than 44, start with 46: 46 ÷ 44 = 1 remainder 2. Bring down the 8 to make 28. Since 28 is less than 44, the next digit of the quotient is 0, so 468 ÷ 44 = 10 remainder 28.
- (a) Ten coaches carry 10 × 44 = 440 people, and the 28 left over still need seats. Round up: the school must hire 10 + 1 = 11 coaches.
- Part (b) reads the same division another way. Only full coaches go, so the school hires 10 coaches, and the remainder, 28 people, travels in minibuses.
- 28 ÷ 12 = 2 remainder 4. Two minibuses carry 24 people, and the last 4 still need seats, so round up again.
- (b) The school needs 2 + 1 = 3 minibuses. Check: 10 × 44 + 28 = 440 + 28 = 468, and 3 minibuses have 3 × 12 = 36 seats for the 28 people.
answer(a) 11 coaches; (b) 3 minibuses
techniqueInterpreting the Remainder · Long Division · Remainders
examsPSLE · GCSE Higher
Common pitfalls
- Answering 10 coaches for part (a) because the quotient is 10. The 28 people in the remainder still have to travel, so one more coach is needed.
- Starting part (b) from the 11 coaches of part (a), which leaves nobody for the minibuses. Part (b) hires full coaches only, so it keeps the quotient, 10, and the minibuses carry the remainder, 28.