AO3 · Whole numbers

Whole-Number Heuristics

10 question types · PSLE Paper 2 · Model Method and algebra, side by side

01

Gaps and Intervals on an Open Line

heuristicFence-Post Principle (Interval Counting: Objects = Gaps + 1)

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?

0 m480 ma 480 m road
The road, end to end.
Draw line of 480 m.
step 1 of 7 Practice whole numbers in the app

Draw a segmented distance line to visualize gaps between planted anchors.

  1. Draw line of 480 m.
  2. Mark 1st tree at 0 m. Every gap of 15 m adds 1 tree.
  3. Number of gaps = 480 ÷ 15 = 32.
  4. Total trees = 32 gaps + 1 starting tree = 33 trees.
  5. Common milestones: list multiples of 15 and 20 ⟹ 60 m, 120 m, 180 m, …
  6. Number of 60 m gaps = 480 ÷ 60 = 8.
  7. Shared spots = 8 + 1 = 9 spots.

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.
02

Gaps and Intervals: Time Chimes / Strikes

heuristicInterval-to-Event Translation

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?

4 strikes3 gaps = 6 s
Four strikes are four ticks. The seconds sit between them.
Draw 4 tick marks: [Strike 1] --- [Strike 2] --- [Strike 3] --- [Strike 4].
step 1 of 5 Practice whole numbers in the app

Draw a timeline bar showing strikes as tick marks and spaces between ticks as duration boxes.

  1. Draw 4 tick marks: [Strike 1] --- [Strike 2] --- [Strike 3] --- [Strike 4].
  2. Count the gaps between ticks: exactly 3 gaps.
  3. 3 gaps = 6 seconds ⟹ 1 gap = 2 seconds.
  4. For 10 strikes, draw 10 ticks: exactly 10 − 1 = 9 gaps.
  5. Total time = 9 × 2 = 18 seconds.

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.
03

Classic Supposition / Assumption (Chicken and Rabbit)

heuristicAssumption / Supposition Method

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?

36 microscopes · 0 balances222222222222222222222222222222222222Batteries36 × 2 = 7272
Suppose all 36 devices are microscopes.
Step 1 (Assumption): Assume all 36 devices are microscopes.
step 1 of 6 Practice whole numbers in the app

Assume all devices are microscopes, calculate the resulting battery deficit, and swap devices one-by-one to absorb the shortfall.

  1. Step 1 (Assumption): Assume all 36 devices are microscopes.
  2. Step 2 (Assumed Total): 36 × 2 = 72 batteries.
  3. Step 3 (Total Shortfall): 104 − 72 = 32 batteries missing.
  4. Step 4 (Unit Difference): Replacing 1 microscope with 1 balance adds 4 − 2 = 2 batteries.
  5. Step 5 (Number of Swaps): 32 ÷ 2 = 16 swaps.
  6. Number of electronic balances = 16 (and microscopes = 36 − 16 = 20).

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.
04

Three-Category Supposition with Constraint Linking

heuristicConsolidated Supposition / Composite Item Batching

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?

$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$1$154 one-dollar coins · 0 bundles of 3 · worth $54.001 bundle = 2 × 20¢ + 1 × 50¢ = 3 coins worth $0.90
Twice as many 20¢ as 50¢ coins: tie them into bundles of two 20¢ and one 50¢, three coins worth $0.90.
Define 1 bundle of small coins: 2 of 20-cent + 1 of 50-cent = 3 coins, value = $0.90.
step 1 of 7 Practice whole numbers in the app

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.

  1. Define 1 bundle of small coins: 2 of 20-cent + 1 of 50-cent = 3 coins, value = $0.90.
  2. Assume all 54 coins are $1 coins: Assumed value = 54 × $1 = $54.00.
  3. Surplus over actual value: $54.00 − $37.20 = $16.80.
  4. 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.
  5. Number of bundles needed: $16.80 ÷ $2.10 = 8 bundles.
  6. Coins used in bundles: 8 × 3 = 24 coins.
  7. Number of $1 coins remaining: 54 − 24 = 30 coins.

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.
05

Excess and Shortage (Opposite Directions)

heuristicGap and Difference (Total Gap / Unit Difference)

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?

6 each16 × 6 = 9614 left8 each110 markers18 short14 + 18 = 32 markers
From 14 left over to 18 short is a swing of 14 + 18 = 32 markers.
Find Total Gap: To move from a surplus of 14 to a deficit of 18 requires 14 + 18 = 32 markers.
step 1 of 5 Practice whole numbers in the app

Draw two distribution bars: the distance between having 14 extra and being 18 short represents the total gap to bridge.

  1. Find Total Gap: To move from a surplus of 14 to a deficit of 18 requires 14 + 18 = 32 markers.
  2. Find Unit Difference: Each student receives 8 − 6 = 2 extra markers.
  3. (a) Number of students = Total GapUnit Difference = 32 ÷ 2 = 16 students.
  4. (b) Find total markers using Case 1: 16 × 6 + 14 = 96 + 14 = 110 markers.
  5. Check using Case 2: 16 × 8 − 18 = 128 − 18 = 110 markers.

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).
06

Excess and Excess / Shortage and Shortage (Same Direction)

heuristicNet Gap Evaluation

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?

5 each10 × 5 = 5048 loose9 each10 × 9 = 90
Both ways leave oranges loose: 48 the first way, 8 the second.
Case 1 leaves 48 loose oranges; Case 2 leaves 8 loose oranges.
step 1 of 5 Practice whole numbers in the app

Compare the remaining loose oranges in both scenarios to identify how many surplus oranges were absorbed into larger packs.

  1. Case 1 leaves 48 loose oranges; Case 2 leaves 8 loose oranges.
  2. Number of loose oranges packed into existing boxes: 48 − 8 = 40 oranges.
  3. Extra oranges packed per box: 9 − 5 = 4 oranges.
  4. (a) Number of boxes: 40 ÷ 4 = 10 boxes.
  5. (b) Total oranges: 10 × 9 + 8 = 98 oranges.

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.
07

Container Mass with Partial Liquid / Content

heuristicElimination of Constant Tare Mass

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?

Fulldrumuuuu24.8 kg
Full: the drum plus four units of oil, 24.8 kg.
Full model: [Drum] + [u][u][u][u] = 24.8 kg.
step 1 of 7 Practice whole numbers in the app

Represent the total mass as 1 drum block plus 4 oil units, then match the removed mass directly to 3 oil units.

  1. Full model: [Drum] + [u][u][u][u] = 24.8 kg.
  2. Partially empty model: [Drum] + [u] = 8.6 kg.
  3. The mass lost corresponds exactly to the 3 oil units removed:
  4. 3u = 24.8 − 8.6 = 16.2 kg.
  5. Value of u = 16.2 ÷ 3 = 5.4 kg.
  6. (b) Full oil mass = 4u = 4 × 5.4 = 21.6 kg.
  7. (a) Mass of empty drum = 8.6 − 5.4 = 3.2 kg.

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.
08

Working Backwards (Multi-Stage Transfer)

heuristicInverse Operations / Flow Reversal

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?

End15 leftBefore Umauuu34u = 2415 + 3 = 18 = 3 of 4 units → u = 6
Undo Uma: put back her 3 extra tokens, 18. That was 3 of 4 units, so a unit is 6 and the box held 24.
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.
step 1 of 3 Practice whole numbers in the app

Build step-by-step reverse unit bars from the final 15 tokens back to the initial total.

  1. 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.
  2. 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.
  3. Stage 1 (Sean): Add back 4 tokens ⟹ 45 + 4 = 49. This represents half the box ⟹ Initial tokens: 49 × 2 = 98.

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.
09

Grouping with Promotional Discounts (Bundle Pricing)

heuristicComposite Set Grouping with Remainder Units

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?

5 files: 1 groups of 5 (4 paid + 1 free) and 0 loose1 group: 4 × $3.50 = $14.00 for 5 files
Buy 4, get 1 free: a group of 5 files costs 4 × $3.50 = $14.00.
Draw 1 Group of 5 files: [$3.50][$3.50][$3.50][$3.50][FREE] = 5 files for $14.00.
step 1 of 5 Practice whole numbers in the app

Draw representative 5-item blocks containing 4 paid units and 1 zero-cost unit, followed by loose unit blocks.

  1. Draw 1 Group of 5 files: [$3.50][$3.50][$3.50][$3.50][FREE] = 5 files for $14.00.
  2. Divide target files by group size: 43 ÷ 5 = 8 groups with 3 files left over.
  3. Cost of 8 groups: 8 × $14.00 = $112.00.
  4. Cost of 3 loose files: 3 × $3.50 = $10.50.
  5. Combine totals: $112.00 + $10.50 = $122.50.

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.
10

Simultaneous Elimination (Scale and Subtract)

heuristicCommon Coefficient Equalization

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?

Set 1$168
Set 1: 3 shirts and 4 pairs of shorts, $168.
Set 1: 3 shirts + 4 shorts = $168.
step 1 of 9 Practice whole numbers in the app

Draw two sets of comparison bars, double the second set to make shorts identical, and deduce the price of shirts from the excess length.

  1. Set 1: 3 shirts + 4 shorts = $168.
  2. Set 2: 5 shirts + 2 shorts = $182.
  3. Double Set 2: 10 shirts + 4 shorts = $182 × 2 = $364.
  4. Compare Doubled Set 2 with Set 1: the 4 shorts cancel out.
  5. Difference in shirts: 10 − 3 = 7 shirts.
  6. Price difference: $364 − $168 = $196 ⟹ 1 shirt = $196 ÷ 7 = $28.
  7. Find 4 shorts from Set 1: $168 − (3 × $28) = $168 − $84 = $84.
  8. (a) 1 pair of shorts = $84 ÷ 4 = $21.
  9. (b) 2 shirts + 3 shorts = (2 × $28) + (3 × $21) = $56 + $63 = $119.

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).