Fractions · applications

Applications: Fractions

18 question types · Model Method and algebra, side by side

PSLE · GCSE Higher

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.

answer$624

techniqueEquivalent Fractions · Same Top Number

examsPSLE

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.

answer$240

techniqueA Fraction of an Amount · Fractions of a Changing Whole

examsPSLE · GCSE Higher

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 1/2420 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.

answer81 tarts

techniqueFractions of a Changing Whole · Multiplying and Dividing Undo Each Other

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.

answer50 adults

techniqueRatios with a Constant Part

examsSAT

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. At first, Container A held 512 of the 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.

answer1736 of the oil

techniqueRatios with a Constant Total

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.

answer$90

techniqueRatios with a Constant Difference

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.

answer288 laptops

techniqueA Fraction of an Amount · Ratio, Fraction, Percent

examsPSLE · GCSE Higher

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.

answer280 blue pens

techniqueBefore-and-After Ratio Problems · Simultaneous by Elimination

examsSAT · GCSE Higher

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.

answer150 coins

techniqueA Fraction of a Group · The Unitary Method

examsPSLE · GCSE Higher

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.

answer16 cm2

techniqueA Fraction of an Amount · Area of a Composite Figure

examsPSLE · GCSE Higher

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 $300, and they had a combined total of $260 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] = $300.
  4. Draw Total Left Model: [u] + [p][p] = $260.
  5. Compare the two models: the difference is exactly p.
  6. p = 300 − 260 = 40.
  7. Substitute back to find u: [u] + 2(40) = 260 ⟹ u = 180.
  8. Chloe at first = 2u = 2 × $180 = $360.

answer$360

techniqueFractions of a Changing Whole · Simultaneous by Elimination

examsO-Level · SAT · GCSE Higher

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.

answer24 cards

techniqueRatios After a Transfer · Ratios with a Constant Total

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 Standardization

Tank A and Tank B have capacities in the ratio 3 : 4. Tap X fills 12 of Tank A in 6 minutes. Tap Y would fill 23 of Tank A in 10 minutes. Both tanks start empty. Tap X is turned on at Tank A and Tap Y at Tank B at the same moment. What fraction of Tank B is filled when Tank A is completely full?

Tank A3 blocksTap XTank B4 blocksTap Y
Capacities 3:4: three blocks against four, all the same size.
Represent Tank A with 3 equal blocks and Tank B with 4 blocks of the same size.
step 1 of 5

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

  1. Represent Tank A with 3 equal blocks and Tank B with 4 blocks of the same size.
  2. Tap X fills half of Tank A in 6 min, so all 3 blocks in 6 × 2 = 12 minutes.
  3. Tap Y fills 23 of Tank A, which is 2 blocks, in 10 min: 1 block every 5 minutes, so all 4 blocks of Tank B in 20 minutes.
  4. When Tank A is full, exactly 12 minutes have passed.
  5. In 12 of the 20 minutes Tank B needs, it is 1220 = 35 full.

answer35 of Tank B

techniqueRates · A Fraction of an Amount

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Comparing filling times directly without accounting for the fact that Tank A and Tank B have different total capacities.
  • Reading Tap Y's 23 as two thirds of Tank B, which gives 45.
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.

answer90 cm2

techniqueEquivalent Fractions · A Common Denominator

examsPSLE

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.

answer$510

techniqueA Fraction of an Amount · The Unitary Method

examsPSLE · GCSE Higher

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.
16

Slices of Cake Eaten at a Party, as Whole Cakes and What They Cost

heuristicGroup the Slices into Wholes / Improper Fraction to Mixed Number

At a class party, every cake is cut into 8 equal slices. The children eat 43 slices, which is 438 cakes. (a) How many cakes do the children eat? Give the answer as a mixed number. (b) Each cake cost $16. What was the cost of the cake that the children ate?

Cakes 1-3cake 1cake 2cake 3Cakes 4-6cake 4cake 5cake 6
Six cakes of 8 equal slices each. One whole cake is 8 slices, 88.
Draw each cake as a bar cut into 8 equal slices. One whole cake is 8 slices, which is 88.
step 1 of 6

Draw the cakes as bars of 8 slices each and fill them 8 slices at a time. The full bars give the whole number and the slices in the last bar give the fraction. Then find the cost of the whole cakes and of the extra slices separately.

  1. Draw each cake as a bar cut into 8 equal slices. One whole cake is 8 slices, which is 88.
  2. Fill the bars 8 slices at a time. Five bars take 5 × 8 = 40 slices, and the other 43 − 40 = 3 slices go into a sixth bar.
  3. (a) Five bars are full and the sixth bar has 3 of its 8 slices filled, so the children eat 538 cakes.
  4. The 5 whole cakes cost 5 × $16 = $80.
  5. One slice is 18 of a cake, so it costs $16 ÷ 8 = $2, and the 3 slices cost 3 × $2 = $6.
  6. (b) The cake the children ate cost $80 + $6 = $86. Check: 43 slices at $2 each cost 43 × $2 = $86.

answer(a) 538 cakes; (b) $86

techniqueImproper Fractions · Mixed Numbers and Improper Fractions · A Fraction of an Amount

examsPSLE

Common pitfalls

  • Writing 438 as 438 or as 535. The whole number is how many times 8 goes into 43, which is 5 times, and the remainder is still counted in eighths: 3 slices out of 8 is 38.
  • Answering (b) with 5 × $16 = $80, the cost of the whole cakes only. The children also ate 38 of a sixth cake, and that part costs 38 of $16, which is $6 more.
17

A Ribbon Cut into Quarter-Meter Pieces, and the Length Not Used

heuristicCount Unit Pieces / Mixed Number to Improper Fraction and Back

Mrs. Diaz has a ribbon 534 m long. She cuts all of it into pieces that are each 14 m long. (a) How many pieces does she cut? (b) She uses 10 of the pieces to tie bows on gift boxes. What is the total length of the pieces that she does not use? Give the answer as a mixed number.

Ribbon1 m1 m1 m1 m1 m3/4 m5 and 3/4 m
The ribbon: 5 whole meters and 34 m. Every meter is marked into quarters, the length of one piece.
Draw the ribbon as 5 whole meters and 34 of a meter. Each piece is 14 m long, so mark every meter into 4 quarters.
step 1 of 6

Draw the ribbon as 5 whole meters and 34 of a meter, and mark every quarter of a meter. Counting the quarters turns the mixed number into an improper fraction; putting the unused quarters back into whole meters turns it into a mixed number again.

  1. Draw the ribbon as 5 whole meters and 34 of a meter. Each piece is 14 m long, so mark every meter into 4 quarters.
  2. Each whole meter gives 4 pieces: 5 × 4 = 20 pieces. The 34 m gives 3 more pieces.
  3. (a) She cuts 20 + 3 = 23 pieces. This shows that 534 = 234.
  4. She uses 10 pieces, so 23 − 10 = 13 pieces are not used. Together they measure 134 m.
  5. Put the quarters back into whole meters, 4 quarters to a meter: 13 = 3 × 4 + 1, so 13 quarters make 3 whole meters and 1 quarter of a meter.
  6. (b) The unused pieces measure 314 m. Check: the 10 pieces used measure 104 = 224 m, and 534 − 224 = 314.

answer(a) 23 pieces; (b) 314 m

techniqueMixed Numbers and Improper Fractions · Improper Fractions

examsPSLE · GCSE Higher

Common pitfalls

  • Writing 534 as 84 or 154 by adding or multiplying the wrong numbers. Each of the 5 whole meters is 4 quarters, so the whole meters give 5 × 4 = 20 quarters, and the 3 quarters are added to that to make 234.
  • Giving (b) as 13 or as 134. The 13 is a number of pieces, not a length, and the question asks for the length as a mixed number: 134 m is 314 m.
18

Pizzas Shared Equally at Two Picnic Tables, Compared

heuristicA Division Written as a Fraction / Common Denominator to Compare

At a school picnic, all the pizzas are the same size. At the first table, 3 pizzas are shared equally among 4 children. At the second table, 2 pizzas are shared equally among 3 children. (a) What fraction of a pizza does each child at the first table get? (b) Does each child at the first table or each child at the second table get more pizza, and how much more?

Table 14 children
Table 1: three pizzas, each cut into 4 equal parts, one part for each of the 4 children.
First table: cut each of the 3 pizzas into 4 equal parts, one part for each of the 4 children. Each part is 14 of a pizza.
step 1 of 5

Cut every pizza into as many equal parts as there are children, and give each child one part of every pizza. Then draw the two shares on bars of the same length and cut both into twelfths to compare them.

  1. First table: cut each of the 3 pizzas into 4 equal parts, one part for each of the 4 children. Each part is 14 of a pizza.
  2. (a) Each child takes one part from every pizza, so each child gets 3 quarters: 3 ÷ 4 = 34 of a pizza.
  3. Second table: cut each of the 2 pizzas into 3 equal parts. Each child takes one part from each pizza, so each child gets 2 thirds: 2 ÷ 3 = 23 of a pizza.
  4. Draw the two shares on bars one pizza long and cut both bars into twelfths: 34 = 912 and 23 = 812.
  5. (b) The first share is one twelfth longer, so each child at the first table gets 912 − 812 = 112 of a pizza more. Check: 4 × 34 = 3 pizzas and 3 × 23 = 2 pizzas.

answer(a) 34 of a pizza; (b) each child at the first table, by 112 of a pizza

techniqueFractions as Division · A Common Denominator

examsPSLE

Common pitfalls

  • Writing the share at the first table as 43, children over pizzas. The amount being shared, 3 pizzas, is divided by the number of children, 4, so it goes on top: 3 ÷ 4 = 34. With fewer pizzas than children, each share must be less than one pizza.
  • Deciding that each child at the second table gets more because there are fewer children there. There are also fewer pizzas; only the two shares, written over the same denominator, show which is larger.
Mr. Chalk Drill fraction arithmetic