AO3 · Fractions

Fraction Heuristics

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

01

Equal Fractions (Numerator Equating)

heuristicEquating Numerators

23 of Lucas's savings is equal to 47 of Nathan's savings. If Nathan has $48 more than Lucas, how much money do they have altogether?

Lucas2 of 3Nathan4 of 7
Two shaded parts of three, against four of seven. The shaded amounts are equal; the parts are not.
Rewrite 23 as an equivalent fraction with numerator 4: 23 = 46.
step 1 of 7

Equalize the numerators so that both models display the exact same number of equal shaded parts, allowing a direct comparison of total unit bars.

  1. Rewrite 23 as an equivalent fraction with numerator 4: 23 = 46.
  2. Now, 4 out of 6 parts of Lucas equals 4 out of 7 parts of Nathan.
  3. Draw Lucas's model with 6 equal units: [u][u][u][u][u][u] (where 4 units are shaded).
  4. Draw Nathan's model with 7 identical units: [u][u][u][u][u][u][u] (where 4 units are shaded).
  5. Compare the difference: 7u − 6u = u = $48.
  6. Total units: 6u + 7u = 13u.
  7. Calculate total: 13 × $48 = $624.

Common pitfalls

  • Confusing numerators with denominators, leading to setting L:N = 2:4 or 3:7.
  • Finding the LCM of the denominators (3 and 7) instead of the numerators (2 and 4).
  • Subtracting numerators (4 − 2) directly and equating 2u = $48 without converting to a common fractional baseline.
02

Fraction of a Remainder (Sequential Branching)

heuristicBranching Method / Remainder Unit Subdivision

Mrs. Tan had a sum of money. She spent 14 of it on groceries and 25 of the remainder on transport. She was left with $108. How much money did Mrs. Tan have at first?

Moneygroceriesremainder = 3 big units
One bar, four big units. Groceries take one; three remain.
Draw 1 long bar split into 4 equal big units. Groceries = 1 big unit; Remainder = 3 big units.
step 1 of 8

Cut the original bar into primary units, then subdivide the remainder box into new secondary units (or find a common denominator).

  1. Draw 1 long bar split into 4 equal big units. Groceries = 1 big unit; Remainder = 3 big units.
  2. Since the remainder (3 units) must be divided into 5 parts, find LCM of 3 and 5, which is 15. Subdivide each big unit into 5 sub-units.
  3. Total bar now consists of 4 × 5 = 20u.
  4. Groceries = 1 × 5 = 5u. Remainder = 15u.
  5. Transport = 25 × 15u = 6u.
  6. Units left = 15u − 6u = 9u.
  7. Set 9u = $108 ⟹ u = $12.
  8. Total at first = 20u = 20 × $12 = $240.

Common pitfalls

  • Adding fractions directly (14 + 25 = 1320) without recognizing that the second fraction applies only to the remainder, not the original whole.
  • Equating the final $108 to 35 of the whole rather than 35 of 34.
03

Fraction of Remainder with Fixed Offsets

heuristicWorking Backwards with Linear Units

A baker made some tarts. He sold 13 of them plus 6 tarts in the morning. In the afternoon, he sold 12 of the remaining tarts plus 4 tarts. In the end, he had 20 tarts left. How many tarts did he bake altogether?

Afternoonsold ½420 left
Start from the end. The afternoon remainder was halved; one half lost 4 more tarts and 20 were left.
Draw the afternoon remainder bar: split into 2 equal parts. One part has 4 extra tarts shaded out, leaving an unshaded box of 20.
step 1 of 7

Construct a two-tier bar model working strictly from the rightmost final segment back to the initial whole.

  1. Draw the afternoon remainder bar: split into 2 equal parts. One part has 4 extra tarts shaded out, leaving an unshaded box of 20.
  2. Observe that half of the remainder = 20 + 4 = 24.
  3. Therefore, the full afternoon remainder bar = 24 × 2 = 48.
  4. Draw the initial bar split into 3 units. Morning sale = 1 unit + 6.
  5. The remaining 2 units minus 6 tarts must equal 48: 2u − 6 = 48 ⟹ 2u = 54.
  6. Find u = 27.
  7. Total tarts = 3u = 3 × 27 = 81.

Common pitfalls

  • Subtracting the fixed amounts (e.g., 20 − 4 = 16) when reversing instead of adding them back.
  • Treating +6 and +4 as fractional units rather than absolute quantities.
  • Applying the initial fraction 13 to the final reversed amount.
04

Constant Part (Single Unchanged Quantity)

heuristicEqualizing the Unchanged Ratio/Part

At a library, 38 of the visitors were children and the rest were adults. After 45 more children entered the library and no adults left, 35 of the visitors were children. How many adults were in the library?

Beforeadults 5u3 : 5
Before: 38 children means children : adults = 3:5.
Before: Children : Adults = 3 : (8 − 3) = 3 : 5.
step 1 of 8

Convert fractions into part-to-part ratios and make the units of the unchanged identity identical.

  1. Before: Children : Adults = 3 : (8 − 3) = 3 : 5.
  2. After: Children : Adults = 3 : (5 − 3) = 3 : 2.
  3. Adults are unchanged. Find LCM of adult units (5 and 2), which is 10:
  4. Before scaled (× 2): Children : Adults = 6u : 10u.
  5. After scaled (× 5): Children : Adults = 15u : 10u.
  6. Change in children units: 15u − 6u = 9u.
  7. Equate unit change to actual change: 9u = 45 ⟹ u = 5.
  8. Adults = 10u = 10 × 5 = 50.

Common pitfalls

  • Equating the initial 3 units of children directly to the final 3 units of children because the numerators match.
  • Applying the increase of 45 to the total without adjusting the adult baseline units.
05

Constant Total (Internal Transfer)

heuristicFixing Total System Units

Container A and Container B held a total of 720 ml of oil. Initially, Container A was 512 full of the total oil. After 80 ml of oil was poured from Container B into Container A, what fraction of the total oil was in Container B?

Total720 ml
Draw the total as 12 units, because 512 names twelfths.
Draw a total bar of 12 units representing 720 ml.
step 1 of 6

Express both states in terms of a constant total unit bar of 12 parts.

  1. Draw a total bar of 12 units representing 720 ml.
  2. Compute value of u: 12u = 720 ⟹ u = 60 ml.
  3. Before: Container A = 5u = 300 ml, Container B = 7u = 420 ml.
  4. Since 80 ml moves from B to A, shift an 80 ml block visually from bar B to bar A.
  5. New volume of B = 420 − 80 = 340 ml.
  6. New fraction in B = 340720, divide top and bottom by 20 to simplify: 1736.

Common pitfalls

  • Assuming the total volume changes because an exchange occurred.
  • Adding 80 ml to A while forgetting to subtract 80 ml from B.
06

Constant Difference (Equal Reductions/Additions)

heuristicEqualizing Difference Units

Ethan had 25 as much pocket money as Fiona. After both of them spent $18 each on stationery, Ethan had 14 as much money as Fiona. How much money did Fiona have at first?

Ethan2uFiona5udifference 3u
Ethan : Fiona = 2:5. The gap is 3 units.
Before: Ethan : Fiona = 2 : 5 ⟹ Difference = 5 − 2 = 3 units.
step 1 of 6

Because both spend identical amounts, the gap (difference) between their amounts remains constant throughout.

  1. Before: Ethan : Fiona = 2 : 5 ⟹ Difference = 5 − 2 = 3 units.
  2. After: Ethan : Fiona = 1 : 4 ⟹ Difference = 4 − 1 = 3 units.
  3. Since the difference is already identical (3 units in both states), compare before and after directly.
  4. Ethan dropped from 2u to u: Change = 2u − u = u.
  5. Therefore: u = $18.
  6. Fiona at first = 5u = 5 × $18 = $90.

Common pitfalls

  • Equating (2 − 1) and (5 − 4) when the difference units are not aligned to a common multiple.
  • Subtracting 18 from only one of the parties in the model.
07

Fractional Comparison (More Than / Less Than)

heuristicDenominator Allocation to Reference Base

Store A sold 38 fewer laptops than Store B. Store C sold 14 more laptops than Store B. If Store A sold 180 fewer laptops than Store C, how many laptops did Store B sell?

Store B8ubase
Both fractions are of Store B, so B is the base. Eighths and quarters both fit in 8 units.
Draw Store B as a bar of 8u.
step 1 of 6

Make the base bar for Store B equal to the LCM of the denominators (8 and 4), which is 8 units.

  1. Draw Store B as a bar of 8u.
  2. Store A is 3u shorter than B: Store A = 8u − 3u = 5u.
  3. Store C is 14 of B more than B: 14 × 8u = 2u. Store C = 8u + 2u = 10u.
  4. Compare Store C and Store A: 10u − 5u = 5u.
  5. Equate unit gap: 5u = 180 ⟹ u = 36.
  6. Store B = 8u = 8 × 36 = 288.

Common pitfalls

  • Treating '38 fewer' as Store A being 38 of Store B, instead of 1 − 38 = 58.
  • Assigning the denominator of 14 to Store C rather than the reference entity Store B.
08

Simultaneous Units and Parts (Everything Changed)

heuristicUnits and Parts System / Cross-Multiplication

The number of red pens was 34 of the number of blue pens. After 30 red pens and 10 blue pens were sold, the number of red pens became 23 of the number of blue pens. Find the original number of blue pens.

Red3uBlue4u
Before: red 3u, blue 4u.
Before: Red = 3u, Blue = 4u.
step 1 of 10

Use Units (u) and Parts (p) to represent two distinct states, then equate parts through substitution.

  1. Before: Red = 3u, Blue = 4u.
  2. After: Red = 2p, Blue = 3p.
  3. Express equations in u and p:
  4. Equation 1: 3u − 30 = 2p
  5. Equation 2: 4u − 10 = 3p
  6. Scale Equation 1 by 3: 9u − 90 = 6p.
  7. Scale Equation 2 by 2: 8u − 20 = 6p.
  8. Since both equal 6p: 9u − 90 = 8u − 20.
  9. Solve: u = 70.
  10. Blue pens = 4u = 4 × 70 = 280.

Common pitfalls

  • Assuming the unit difference (4u − 3u) remains constant even though different quantities (30 and 10) were removed.
  • Mixing up units (u) and parts (p) as if they have the same size.
09

Fractional Grouping / Sets (Number × Value)

heuristicGrouping into 1 Composite Set

35 of the coins in a piggy bank were 50-cent coins and the rest were 20-cent coins. The total value of all the coins was $57.00. How many coins were in the piggy bank in total?

50¢50¢50¢20¢20¢1 group = 5 coins
35 are 50-cent coins, so 3 of every 5 coins. Take 5 coins as one group.
From the fraction 35, define 1 group of 5 coins:
step 1 of 5

Form a single base group containing 3 fifty-cent coins and 2 twenty-cent coins, then find how many such groups fit into the total sum.

  1. From the fraction 35, define 1 group of 5 coins:
  2. 1 Group contains: 3 fifty-cent coins and 2 twenty-cent coins.
  3. Calculate the value of 1 group: (3 × $0.50) + (2 × $0.20) = $1.50 + $0.40 = $1.90.
  4. Find total number of groups: $57.00$1.90 = 30 groups.
  5. Calculate total coins: 30 groups × 5 coins/group = 150 coins.

Common pitfalls

  • Dividing the total value directly by 5 (the total units of items) ignoring coin values: $57 ÷ 5.
  • Mixing dollar and cent units (e.g., 150k + 40k = 57 instead of 5700).
10

Geometric Fractional Overlap

heuristicCommon Area Equating

A rectangle and a triangle overlap as shown. The unshaded area of the rectangle is 45 of its total area. The unshaded area of the triangle is 67 of its total area. If the total area of the entire figure is 176 cm2, find the area of the overlapping shaded region.

rectangletriangleSOverlapu
Call the shaded overlap one unit. Both shapes are measured against it.
Overlap = 1 unit.
step 1 of 6

Equate the shaded portion as a common unit of 1, and express the whole shapes as unit bars.

  1. Overlap = 1 unit.
  2. Rectangle = 5 units (Overlap = u, Unshaded = 4u).
  3. Triangle = 7 units (Overlap = u, Unshaded = 6u).
  4. Total figure consists of: Unshaded Rect + Overlap + Unshaded Tri = 4u + u + 6u = 11u.
  5. Equate: 11u = 176 cm2 ⟹ u = 16 cm2.
  6. Overlap region = u = 16 cm2.

Common pitfalls

  • Adding the full rectangle and triangle without subtracting the overlap, effectively double-counting the intersection.
  • Equating unshaded portions rather than the shared intersection.
11

Dual Independent Remainder Branches with Merged Exit

heuristicTwo-Variable Branching with Linear Elimination

Chloe and Daniel each had some money. Chloe spent 13 of her money on clothes and 14 of the remainder on makeup. Daniel spent 15 of his money on games and 12 of the remainder on books. Chloe and Daniel spent a combined total of $290, and they had a combined total of $250 left. How much did Chloe have at first?

Chloeu spentu left2u
Chloe: 13 then 14 of the rest is 13+16=12 spent. Two equal units.
Simplify Chloe: Clothes = 13, Remainder = 23. Makeup = 14 × 23 = 16. Total Chloe Spent = 36 = 12. Left = 12. Chloe = [u spent][u left].
step 1 of 8

Convert Chloe's spending into simple units (u) and Daniel's into parts (p), then utilize model comparison to cancel identical blocks.

  1. Simplify Chloe: Clothes = 13, Remainder = 23. Makeup = 14 × 23 = 16. Total Chloe Spent = 36 = 12. Left = 12. Chloe = [u spent][u left].
  2. Simplify Daniel: Games = 15, Remainder = 45. Books = 25. Total Daniel Spent = 35, Left = 25. Daniel = [3p spent][2p left].
  3. Draw Total Spent Model: [u] + [p][p][p] = $290.
  4. Draw Total Left Model: [u] + [p][p] = $250.
  5. Compare the two models: the difference is exactly p.
  6. p = 290 − 250 = 40.
  7. Substitute back to find u: [u] + 2(40) = 250 ⟹ u = 170.
  8. Chloe at first = 2u = 2 × $170 = $340.

Common pitfalls

  • Using the same unit letter or block size for both Chloe and Daniel, assuming u = p.
  • Applying Daniel's book fraction to the whole sum of Daniel and Chloe combined.
12

Internal Transfer with Unbalanced Offsets

heuristicTotal Invariance with Algebraic Re-balancing

Afiq had 35 as many gaming cards as Bala. Bala gave 14 of his cards and another 14 cards to Afiq. As a result, Afiq had 3 times as many cards as Bala. How many cards did Afiq have at first?

BeforeAfiq 12uBala 20u32u
Afiq : Bala = 3:5. Scale to 32 units so 14 of Bala is whole: 12u and 20u.
Before: Afiq : Bala = 3 : 5 (Total = 8 units). Scale to 32u so Bala is 20u and Afiq is 12u.
step 1 of 8

Total is invariant. Express the final state as unit blocks of the constant total and balance the transferred chunk.

  1. Before: Afiq : Bala = 3 : 5 (Total = 8 units). Scale to 32u so Bala is 20u and Afiq is 12u.
  2. Total = 32u.
  3. After: Afiq has 3 times as many as Bala ⟹ Afiq : Bala = 3 : 1 (Total = 4 parts).
  4. Since total is invariant, scale final ratio to 32 units: Afiq = 24u, Bala = 8u.
  5. Compare Bala before and after: Bala lost 20u − 8u = 12u.
  6. The lost amount equals the transferred chunk: 12u = 5u + 14.
  7. Subtract: 12u − 5u = 14 ⟹ 7u = 14 ⟹ u = 2.
  8. Afiq at first = 12u = 12 × 2 = 24 cards.

Common pitfalls

  • Forgetting to subtract the fixed offset of 14 from Bala when transferring to Afiq.
  • Failing to recognize that the total sum of cards remains unchanged before and after the transfer.
13

Fractional Rate-Based Flow / Shared Tasks

heuristicCommon Rate Standardisation

Tank A and Tank B have capacities in the ratio 3 : 4. Tap X fills 12 of Tank A in 6 minutes. Tap Y fills 23 of Tank B in 10 minutes. If both taps are turned on together, what fraction of Tank B will be filled when Tank A is completely full?

Tank A3 blocksTap XTank B4 blocksTap Y
Capacities 3:4: three blocks against four.
Represent Tank A capacity with 3 big blocks and Tank B with 4 big blocks.
step 1 of 5

Convert tank capacities into common blocks, then draw unit-per-minute bars.

  1. Represent Tank A capacity with 3 big blocks and Tank B with 4 big blocks.
  2. Find time for Tap X to fill 1 full Tank A: 6 min × 2 = 12 minutes.
  3. Find time for Tap Y to fill 1 full Tank B: 10 min × 32 = 15 minutes.
  4. When Tank A is full, exactly 12 minutes have elapsed.
  5. In 12 minutes, the fraction of Tank B filled by Tap Y is: 12 minutes15 minutes = 45.

Common pitfalls

  • Comparing filling times directly without accounting for the fact that Tank A and Tank B have different total capacities.
  • Multiplying rates directly with ratios instead of establishing a common time metric.
14

Multi-Tiered Overlapping Figures with Dependent Fractions

heuristicSequential Base Anchor / Common Denominator Scaling

The figure shows three overlapping shapes: A, B, and C. The overlap between A and B is 14 of the area of A. The same overlap is 16 of the area of B. The overlap between B and C is 15 of the area of C, and it is also 14 of the remaining unshaded area of B after excluding the overlap with A. If this remaining unshaded area of B is 72 cm2, find the total area of Shape C.

AB unshadedCUnshaded B72 cm²
Anchor on the one area the question gives in cm²: the unshaded part of B, drawn as 4 units.
Draw Unshaded Area of B as 4 equal units = 72 cm2.
step 1 of 5

Anchor the model on the specified remaining unshaded area of B and scale the target shape C accordingly.

  1. Draw Unshaded Area of B as 4 equal units = 72 cm2.
  2. Each unit = 72 ÷ 4 = 18 cm2.
  3. The overlap of B and C is 1 of these units: Overlap BC = 18 cm2.
  4. Shape C has this overlap as 15 of its total: Draw C with 5 identical units of 18 cm2.
  5. Area of C = 5 × 18 = 90 cm2.

Common pitfalls

  • Assuming the overlap between B and C is part of the overlap between A and B.
  • Applying the fraction 14 to the total area of B instead of the 'remaining unshaded area of B'.
15

Fractional Supposition with Leakage (Advanced Assumption)

heuristicComposite Unit Batching with Leakage Rate

A baker packed 240 pastries. 38 of them were egg tarts and the rest were cream puffs. He sold 56 of the egg tarts and 45 of the cream puffs. He sold each egg tart for $2 and each cream puff for $3. How much money did the baker collect in total?

240 pastriesegg tarts 3ucream puffs 5u8u
240 in eighths: u = 30. Egg tarts 3u, cream puffs 5u.
Draw a 240-unit bar split into 8 units: u = 240 ÷ 8 = 30.
step 1 of 5

Divide the initial bar into fractional sets, then subdivide each component into sold and unsold unit segments.

  1. Draw a 240-unit bar split into 8 units: u = 240 ÷ 8 = 30.
  2. Egg tarts = 3u = 90. Cream puffs = 5u = 150.
  3. Split Egg tarts into 6 segments: 90 ÷ 6 = 15 per segment. Sold = 5 segments = 75.
  4. Split Cream puffs into 5 segments: 150 ÷ 5 = 30 per segment. Sold = 4 segments = 120.
  5. Calculate monetary collection: (75 × $2) + (120 × $3) = $150 + $360 = $510.

Common pitfalls

  • Calculating total sales by applying a combined average fraction to the entire 240 pastries.
  • Multiplying the unsold fraction by the prices instead of the sold fraction.