Ratio and proportion · applications

Applications: Ratio and Proportion

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

O-Level · SAT · GCSE Higher

01

Direct Part-to-Part and Part-to-Whole Allocation

heuristicUnit Value Evaluation

The ratio of the number of boys to the number of girls in an art club was 4 : 7. There were 36 more girls than boys. (a) How many members were there in the art club altogether? (b) If each girl was given 3 paintbrushes and each boy was given 2 paintbrushes, how many paintbrushes were distributed in total?

Boysuuuu4u
Boys : girls = 4:7. Draw the boys as four equal units.
Draw Boys: 4 equal unit boxes [u][u][u][u].
step 1 of 8

Draw comparison unit bars for boys and girls, equate the excess units to the difference, and scale the units accordingly.

  1. Draw Boys: 4 equal unit boxes [u][u][u][u].
  2. Draw Girls: 7 identical unit boxes [u][u][u][u][u][u][u].
  3. Difference = 7 − 4 = 3 units = 36.
  4. Value of 1 unit = 36 ÷ 3 = 12.
  5. (a) Total units = 4 + 7 = 11 units = 11 × 12 = 132 members.
  6. (b) Paintbrushes for 4 units of boys = (4 × 12) × 2 = 96.
  7. Paintbrushes for 7 units of girls = (7 × 12) × 3 = 252.
  8. Total paintbrushes = 96 + 252 = 348.

answer(a) 132 members; (b) 348 paintbrushes

techniquePart-to-Whole Ratios · The Unitary Method

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Dividing 36 by the total number of units (11) rather than the difference in units (3).
  • Multiplying the unit value (12) directly by the combined brush rate (2 + 3 = 5) instead of multiplying boys and girls by their respective rates.
02

Repeated Identity (Common Term Bridging)

heuristicEqualizing the Shared Entity

The ratio of Alicia's stickers to Brenda's stickers was 3 : 5. The ratio of Brenda's stickers to Clara's stickers was 4 : 7. Alicia had 92 fewer stickers than Clara. How many stickers did the three girls have altogether?

Alicia3 unitsBrenda5 units3 : 5 row
Alicia : Brenda = 3:5. Draw Brenda as five units.
Draw Row 1: Alicia (3 units), Brenda (5 units).
step 1 of 8

Construct a two-row model for Brenda, subdividing the units until both rows contain identical unit counts.

  1. Draw Row 1: Alicia (3 units), Brenda (5 units).
  2. Draw Row 2: Brenda (4 units), Clara (7 units).
  3. Cut Brenda's 5 units into 4 smaller parts each (20 parts), and Brenda's 4 units into 5 parts each (20 parts).
  4. Alicia's bar becomes 3 × 4 = 12 units.
  5. Clara's bar becomes 7 × 5 = 35 units.
  6. Gap between Clara and Alicia = 35 − 12 = 23 units = 92.
  7. 1 unit = 92 ÷ 23 = 4.
  8. Total = (12 + 20 + 35) × 4 = 67 × 4 = 268 stickers.

answer268 stickers

techniqueCombining Two Ratios

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Combining the ratios directly as 3 : 5 : 7 without equalizing the common term Brenda.
  • Equating the difference of 92 to (7 − 3 = 4) units from the unscaled ratios.
03

Constant Total (Internal Transfer)

heuristicTotal Invariance

Daryl and Evan had some marbles in the ratio 7 : 5. After Daryl gave 18 marbles to Evan, the ratio of Daryl's marbles to Evan's marbles became 1 : 2. How many marbles did Daryl have at first?

BeforeDaryl 7uEvan 5u12u
Daryl : Evan = 7:5, so the total is 12 units.
Draw Total bar of 12 units: Daryl gets 7 units, Evan gets 5 units.
step 1 of 6

Draw a constant-length total bar split into 12 parts for the initial state and 3 big parts for the final state, then compare individual partitions.

  1. Draw Total bar of 12 units: Daryl gets 7 units, Evan gets 5 units.
  2. Draw identical Total bar divided into 3 equal blocks: Daryl gets 1 block, Evan gets 2 blocks.
  3. Convert 1 block into units: 12 ÷ 3 = 4 units.
  4. Daryl's change: from 7 units down to 4 units = 3 units.
  5. 3 units = 18 ⟹ 1 unit = 6.
  6. Daryl at first = 7 × 6 = 42 marbles.

answer42 marbles

techniqueRatios with a Constant Total

Common pitfalls

  • Comparing the 'Before' units (7) and 'After' units (1) directly (7 − 1 = 6 units) without normalizing total units.
  • Adding 18 to Daryl instead of subtracting it from Daryl.
04

Constant One Part (Single Unchanged Quantity)

heuristicEqualizing the Unchanged Quantity

The ratio of the number of fiction books to non-fiction books on a shelf was 5 : 3. After the librarian added 54 non-fiction books to the shelf and no fiction books were added or removed, the ratio of fiction books to non-fiction books became 2 : 3. How many fiction books were on the shelf?

Beforefiction 5u3u5 : 3
Before: fiction : non-fiction = 5:3.
Before: Fiction has 5 units, Non-fiction has 3 units.
step 1 of 7

Keep the fiction bar length strictly constant while extending the non-fiction bar to reflect the addition.

  1. Before: Fiction has 5 units, Non-fiction has 3 units.
  2. After: Fiction has 2 units, Non-fiction has 3 units.
  3. Equalize Fiction bars by subdividing: 5 units × 2 = 10 units; 2 units × 5 = 10 units.
  4. Adjust Non-fiction bars accordingly: Before = 3 × 2 = 6 units; After = 3 × 5 = 15 units.
  5. Change in Non-fiction = 15 − 6 = 9 units = 54.
  6. 1 unit = 54 ÷ 9 = 6.
  7. Fiction books = 10 × 6 = 60.

answer60 fiction books

techniqueRatios with a Constant Part

examsSAT

Common pitfalls

  • Assuming non-fiction units did not change because both ratios show the digit '3'.
  • Adding 54 to the total of the initial ratio units (5 + 3 = 8) without equalizing the unchanged component.
05

Constant Difference (Equal Changes / Age Progression)

heuristicDifference Invariance

Five years ago, the ratio of Mr. Wong's age to his son's age was 7 : 2. In 10 years' time from now, the ratio of Mr. Wong's age to his son's age will be 2 : 1. (a) How old is Mr. Wong's son now? (b) What is the ratio of Mr. Wong's age to his son's age now in simplest form?

5 yrs agofather 7uson 2ugap 5u
Five years ago: father 7u, son 2u. The gap is 5 units.
Past Model: Father (7 units), Son (2 units). Difference = 5 units.
step 1 of 9

Draw comparison bars where the difference segment between Father and Son remains fixed in length, then observe the unit elongation.

  1. Past Model: Father (7 units), Son (2 units). Difference = 5 units.
  2. Future Model: Father (2 parts), Son (1 part). Difference = 1 part.
  3. To align difference, cut each part into 5 units: Father = 10 units, Son = 5 units.
  4. Each person gains: 10 − 7 = 3 units.
  5. Years passed = 5 + 10 = 15 years.
  6. 3 units = 15 ⟹ 1 unit = 5 years.
  7. Son's age 5 years ago = 2 × 5 = 10.
  8. (a) Son's age now = 10 + 5 = 15 years old.
  9. (b) Father now = 35 + 5 = 40. Ratio = 40 : 15 = 8 : 3.

answer(a) 15 years old; (b) 8 : 3

techniqueRatios with a Constant Difference

Common pitfalls

  • Calculating the time interval as 10 − 5 = 5 years instead of 5 + 10 = 15 years.
  • Forgetting to add 5 years back to find the present age, reporting the age from 5 years ago as the final answer.
06

Chained Three-Party Ratios with Sub-Proportions

heuristicSub-Unit Branching

At an electronics fair, the ratio of the number of laptops to tablets to smartphones was 3 : 4 : 8. 13 of the laptops were Brand X and the rest were Brand Y. Half of the tablets were Brand X and the rest were Brand Y. None of the smartphones were Brand X. If there were 180 more Brand Y devices than Brand X devices, how many total devices were at the fair?

LaptopsXYY
Laptops are 3 blocks; 13 of them Brand X, so 1 block X and 2 Y.
Draw Laptops as 3 blocks: 1 block Brand X, 2 blocks Brand Y.
step 1 of 9

Represent each device category as discrete unit blocks, shade the Brand X portions, and compare total shaded versus unshaded blocks.

  1. Draw Laptops as 3 blocks: 1 block Brand X, 2 blocks Brand Y.
  2. Draw Tablets as 4 blocks: 2 blocks Brand X, 2 blocks Brand Y.
  3. Draw Smartphones as 8 blocks: all 8 blocks Brand Y.
  4. Sum Brand X blocks: 1 + 2 = 3 blocks.
  5. Sum Brand Y blocks: 2 + 2 + 8 = 12 blocks.
  6. Difference in blocks = 12 − 3 = 9 blocks = 180.
  7. 1 block = 180 ÷ 9 = 20.
  8. Grand total blocks = 3 + 4 + 8 = 15 blocks.
  9. Total devices = 15 × 20 = 300.

answer300 devices

techniqueThree-Part Ratios · Combining Two Ratios

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Applying fractions across the entire total of 15 units rather than to their specific individual category blocks.
  • Excluding smartphones from the Brand Y total, forgetting that having zero Brand X makes all smartphones Brand Y.
07

Number × Value in Ratio Form (Composite Grouping)

heuristicGrouping into 1 Composite Unit

In a charity box, the ratio of the number of $2 notes to $5 notes was 5 : 3, and the ratio of the number of $5 notes to $10 notes was 2 : 1. The total value of all the notes in the box was $1280. (a) How many $5 notes were in the box? (b) What was the total value of the $10 notes?

$2$2$2$2$2$2$2$2$2$210 × $2$5$5$5$5$5$56 × $5$10$10$103 × $10$2 : $5 = 5 : 3 and $5 : $10 = 2 : 1 → $5 notes as 6
Two ratios share the $5 notes: 5:3 and 2:1. Make the $5 count the same in both, 6.
Equalize $5 notes to get 1 base set:
step 1 of 6

Form a single representative group containing 10 two-dollar notes, 6 five-dollar notes, and 3 ten-dollar notes, then find how many such groups exist.

  1. Equalize $5 notes to get 1 base set:
  2. 1 Set contains: 10 of $2 notes, 6 of $5 notes, 3 of $10 notes.
  3. Value of 1 set = (10 × $2) + (6 × $5) + (3 × $10) = $20 + $30 + $30 = $80.
  4. Total sets = $1280 ÷ $80 = 16 sets.
  5. (a) Number of $5 notes = 16 × 6 = 96 notes.
  6. (b) Total value of $10 notes = 16 × (3 × $10) = 16 × $30 = $480.

answer(a) 96 notes; (b) $480

techniqueThe Unitary Method · Bar Models for Ratio

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing $1280 by the sum of the ratio numbers (10 + 6 + 3 = 19) instead of their monetary values.
  • Multiplying the final set count by note denominations directly without scaling by the ratio quantities.
08

Simultaneous Units and Parts / Everything Changed

heuristicCross-Multiplication / Units and Parts System

The ratio of Zachary's savings to Hannah's savings was 4 : 5. After Zachary spent $45 and Hannah spent $100, the ratio of Zachary's remaining savings to Hannah's remaining savings became 3 : 2. Find Zachary's savings at first.

Zachary4uHannah5u
Before: 4u and 5u.
Before: Zachary = 4u, Hannah = 5u.
step 1 of 9

Represent before states with units (u) and after states with parts (p). Create simultaneous unit comparisons to eliminate parts.

  1. Before: Zachary = 4u, Hannah = 5u.
  2. After: Zachary = 3p, Hannah = 2p.
  3. Equation 1: 4u − 45 = 3p.
  4. Equation 2: 5u − 100 = 2p.
  5. Scale Eq 1 by 2: 8u − 90 = 6p.
  6. Scale Eq 2 by 3: 15u − 300 = 6p.
  7. Equate both expressions for 6p: 15u − 300 = 8u − 90.
  8. 7u = 210 ⟹ u = 30.
  9. Zachary at first = 4u = 4 × $30 = $120.

answer$120

techniqueBefore-and-After Ratio Problems · Simultaneous by Elimination

examsSAT · GCSE Higher

Common pitfalls

  • Assuming the unit difference (5u − 4u = u) corresponds to the difference in expenditure ($100 − $45 = $55).
  • Subtracting ratio numbers directly: (4 − 3) : (5 − 2) = 1 : 3 and attempting to equate units.
09

Three-Party Closed System Internal Transfer

heuristicMulti-Entity Total Invariance

Initially, the ratio of Alan's tokens to Bryan's tokens to Colin's tokens was 3 : 4 : 5. Alan gave 20 tokens to Bryan, and Colin gave 30 tokens to Alan. In the end, the ratio of Alan's tokens to Bryan's tokens to Colin's tokens became 2 : 3 : 1. How many tokens did the three of them have altogether?

BeforeAlan 3uBryan 4uColin 5u12u
Before: 3:4:5, a total of 12 units.
Before model: Alan (3u), Bryan (4u), Colin (5u) → Total = 12u.
step 1 of 7

Draw total invariant bars of 12 units for both states. Trace the entity that experienced only a single one-way transfer to determine the unit value.

  1. Before model: Alan (3u), Bryan (4u), Colin (5u) → Total = 12u.
  2. After model: Alan (2p), Bryan (3p), Colin (p) → Total = 6p.
  3. Standardize After bar into 12 units by multiplying each part by 2:
  4. After: Alan (4u), Bryan (6u), Colin (2u).
  5. Inspect Colin's bar: dropped from 5u to 2u (a loss of 3u).
  6. Since Colin only gave away 30 tokens: 3u = 30 ⟹ u = 10.
  7. Total tokens = 12u = 12 × 10 = 120.

answer120 tokens

techniqueRatios After a Transfer · Three-Part Ratios

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Attempting to track Alan first, which involves two separate transactions (+30 and -20), increasing the likelihood of arithmetic errors.
  • Failing to standardize the total units before comparing individual changes.
10

Two-Way Offset Balancing (Opposite Direction Changes)

heuristicLinear Proportion Re-balancing

At a warehouse, the ratio of cartons of milk to cartons of juice was 5 : 2. After 35 cartons of milk were dispatched and 10 cartons of juice were added, the ratio of milk cartons to juice cartons became 3 : 2. How many cartons of milk were there at first?

Milk5uJuice2u
Before: milk 5u, juice 2u.
Before: Milk = 5u, Juice = 2u.
step 1 of 8

Draw bars showing milk decreasing and juice increasing, scale the final ratio parts to find common comparisons.

  1. Before: Milk = 5u, Juice = 2u.
  2. After: Milk = 3p = 5u − 35, Juice = 2p = 2u + 10.
  3. Multiply Juice equation by 3 and Milk equation by 2 to make parts equal to 6p:
  4. 2(5u − 35) = 6p ⟹ 10u − 70 = 6p.
  5. 3(2u + 10) = 6p ⟹ 6u + 30 = 6p.
  6. Equate: 10u − 70 = 6u + 30.
  7. 4u = 100 ⟹ u = 25.
  8. Milk at first = 5u = 5 × 25 = 125.

answer125 cartons of milk

techniqueBefore-and-After Ratio Problems

examsSAT · GCSE Higher

Common pitfalls

  • Treating both changes as reductions (writing 2u − 10 instead of 2u + 10).
  • Sign errors during transposition (e.g., writing 30 − 70 = −40 instead of 30 + 70 = 100).
11

Fractional Subsets and Residual Ratio Comparison

heuristicCommon Sub-Unit Partitioning

Lucas and Marcus had pocket money in the ratio 7 : 4. Lucas spent 35 of his money, and Marcus spent 12 of his money. Lucas had $48 more left than Marcus. (a) How much money did Lucas have at first? (b) What was the ratio of the total amount spent to the total amount left?

Lucas7 blocksMarcus4 blocks
Lucas : Marcus = 7:4, seven blocks against four.
Represent Lucas with 7 blocks and Marcus with 4 blocks.
step 1 of 10

Construct branching bars with subdivisions matching the fraction denominators, then compare the residual blocks.

  1. Represent Lucas with 7 blocks and Marcus with 4 blocks.
  2. Subdivide Lucas's 7 blocks into 5 parts each = 35 small units.
  3. Subdivide Marcus's 4 blocks into 5 parts each = 20 small units.
  4. Lucas leaves 25 × 35 = 14 units.
  5. Marcus leaves 12 × 20 = 10 units.
  6. Surplus of Lucas over Marcus = 14 − 10 = 4 units = $48.
  7. 1 unit = $48 ÷ 4 = $12.
  8. (a) Lucas at first = 35 × $12 = $420.
  9. (b) Spent units = (35 − 14) + (20 − 10) = 21 + 10 = 31 units.
  10. Left units = 14 + 10 = 24 units. Ratio = 31 : 24.

answer(a) $420; (b) 31 : 24

techniqueRatio, Fraction, Percent · Combining Two Ratios

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Applying Marcus's 12 to Lucas's money or to the total money.
  • Subtracting the fractions directly: 35 − 12 = 110 and equating that to $48.
12

Dual Mutual Exchange (Two-Way Sequential Pouring)

heuristicWorking Backwards / Multi-Stage Tracking

Container X and Container Y contained water in the ratio 5 : 3. First, 14 of the water in Container X was poured into Container Y. Then, 15 of the new amount of water in Container Y was poured back into Container X. In the end, Container X had 144 ml more water than Container Y. How much water was in Container X at first?

StartX 100uY 60u160u
Water only moves between X and Y, so the total bar never changes length.
Total volume is invariant throughout.
step 1 of 9

Work backwards from the final state using the constant total volume of the closed two-container system.

  1. Total volume is invariant throughout.
  2. End state: Container X has 144 ml more than Container Y.
  3. Before the second transfer, Container Y gave 15 of its water to X, leaving 45 in Y.
  4. Step forward systematically with scaled units: let initial X = 100u, Y = 60u.
  5. After first pour: X = 75u, Y = 85u.
  6. After second pour: Y = 85u × 45 = 68u, X = 160u − 68u = 92u.
  7. Difference = 92u − 68u = 24u = 144 ml.
  8. u = 6 ml.
  9. Container X at first = 100 × 6 = 600 ml.

answer600 ml

techniqueMultiplying and Dividing Undo Each Other · Ratios After a Transfer

Common pitfalls

  • Taking 15 of Container Y's *initial* volume (60u) instead of its *new* volume after receiving water from X (85u).
  • Assuming the final difference is simply the difference between the two transferred amounts.
13

Inverse Proportionality and Product Invariance

heuristicEqual Product Equating

Gear A, Gear B, and Gear C are meshed together in a row. The ratio of the number of teeth on Gear A to Gear B is 2 : 3, and the ratio of the number of teeth on Gear B to Gear C is 4 : 5. When Gear A makes 90 complete revolutions, how many complete revolutions does Gear C make?

A8 teethB12 teethC15 teethA: 90 rev B: 60 rev C: 48 rev
Teeth 2:3 and 4:5 share B; make B 12: A = 8, B = 12, C = 15 teeth.
Harmonize teeth count: A = 8 units, C = 15 units.
step 1 of 5

Set up an inverse ratio: the gear with more teeth turns fewer times in exact inverse proportion.

  1. Harmonize teeth count: A = 8 units, C = 15 units.
  2. Because teeth meshed is constant: RevA × TeethA = RevC × TeethC.
  3. Ratio of revolutions RevA : RevC is the inverse of teeth ratio: 15 : 8.
  4. 15 units of revolutions = 90 revs ⟹ 1 unit = 90 ÷ 15 = 6 revs.
  5. Revolutions for Gear C = 8 units = 8 × 6 = 48 revolutions.

answer48 revolutions

techniqueInverse Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Treating revolutions as directly proportional to teeth: setting RevC90 = 158, which would mean larger gears turn faster.
  • Multiplying revolutions across intermediate gears without accounting for common contact points.
14

Weighted Multi-Category Supposition in Ratio Form

heuristicUnit Batch Value Evaluation

A bookstore sold highlighters, pens, and markers. The ratio of the number of highlighters sold to pens sold was 3 : 4, and the ratio of pens sold to markers sold was 2 : 3. A highlighter cost $1.50, a pen cost $2.00, and a marker cost $3.50. The shopkeeper collected $65 more from the sale of markers than from the sale of pens. (a) How many pens were sold? (b) How much money was collected from all three items altogether?

1.501.501.503 highlighters2.002.002.002.004 pens3.503.503.503.503.503.506 markers
Pens appear in both ratios (3:4 and 2:3); make them 4 in both. One set: 3 highlighters, 4 pens, 6 markers.
1 Set contains: 3 highlighters, 4 pens, 6 markers.
step 1 of 8

Define 1 set containing 3 highlighters, 4 pens, and 6 markers. Find the revenue difference within 1 set and scale up to find the number of sets.

  1. 1 Set contains: 3 highlighters, 4 pens, 6 markers.
  2. Cost of pens in 1 set = 4 × $2.00 = $8.00.
  3. Cost of markers in 1 set = 6 × $3.50 = $21.00.
  4. Difference in 1 set = $21.00 − $8.00 = $13.00.
  5. Number of sets = $65 ÷ $13 = 5 sets.
  6. (a) Pens sold = 5 sets × 4 pens/set = 20 pens.
  7. Total money in 1 set = (3 × 1.50) + 8.00 + 21.00 = 4.50 + 8.00 + 21.00 = $33.50.
  8. (b) Grand total revenue = 5 × $33.50 = $167.50.

answer(a) 20 pens; (b) $167.50

techniqueThe Unitary Method · Bar Models for Ratio

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Equating the $65 difference directly to the quantity difference (6 − 4 = 2) units without factoring in item prices.
  • Omitting the price of highlighters when calculating total collection.
15

Multi-Tiered Dual-Subgroup Demographic Shift

heuristicHierarchical Units and Remainder Re-scaling

At a funfair, the ratio of the number of adults to children was 3 : 5. Among the children, the ratio of the number of boys to girls was 1 : 3. Later, 33 adults and 7 boys entered the funfair, while 6 girls left the funfair. In the end, the ratio of the number of adults to children became 1 : 1, and the ratio of the number of boys to girls became 1 : 2. How many children were at the funfair at first?

Adults3 unitsChildren5 units
Adults : children = 3:5.
Draw Adults as 3 units and Children as 5 units.
step 1 of 8

Construct a two-tier bar model: Tier 1 partitions the whole into adults and children; Tier 2 subdivides the children bar. Solve using the isolated subgroup change before confirming against the overall total.

  1. Draw Adults as 3 units and Children as 5 units.
  2. Subdivide each unit into 4 mini-units:
  3. Adults = 12 mini-units, Children = 20 mini-units.
  4. Children bar is split into Boys (5 mini-units) and Girls (15 mini-units).
  5. Focus on children's final ratio: End Boys : End Girls = p : 2p.
  6. Set up relationship: 2 × (5u + 7) = 15u − 6.
  7. 10u + 14 = 15u − 6 ⟹ 5u = 20 ⟹ u = 4.
  8. Calculate total initial children: 20u = 20 × 4 = 80 children.

answer80 children

techniqueRatios with a Constant Part · Combining Two Ratios

examsSAT

Common pitfalls

  • Applying the initial 1 : 3 boy-to-girl ratio to the entire population of adults and children combined.
  • Ignoring the change in girls (subtracting 6) and adding all changes together.
  • Failing to scale the primary ratio (3 : 5) by 4, leading to cumbersome fractional units (1.25u and 3.75u).
16

A Van's Fuel Log: The Fuel Used for Each Kilometer, and How Far the Tank Lasts

heuristicDivide Each Output by Its Input, Then Use the Constant

On three trips, a delivery van used 4.8 liters of fuel to travel 60 km, 12 liters to travel 150 km and 19.2 liters to travel 240 km. (a) Show that the fuel used is in direct proportion to the distance, and find the constant of proportionality in liters per kilometer. (b) The van starts its next trip with 36 liters of fuel in the tank. The driver refuels when 4 liters are left. How far can the van travel on this trip before she must refuel?

km60150240liters4.81219.210 km blocks61524
Each distance is a whole number of 10 km blocks: 6, 15 and 24 blocks.
Every distance is a whole number of 10 km blocks: 60 km is 6 blocks, 150 km is 15 blocks and 240 km is 24 blocks.
step 1 of 5

Cut every distance into blocks of 10 km. If each block uses the same amount of fuel on every trip, the fuel is in direct proportion to the distance, and that amount gives the constant.

  1. Every distance is a whole number of 10 km blocks: 60 km is 6 blocks, 150 km is 15 blocks and 240 km is 24 blocks.
  2. Share each trip's fuel over its blocks: 4.8 ÷ 6 = 0.8, 12 ÷ 15 = 0.8 and 19.2 ÷ 24 = 0.8 liters. Every 10 km block uses the same 0.8 liters, so the fuel is in direct proportion to the distance.
  3. (a) One kilometer uses 0.8 ÷ 10 = 0.08 liters. The constant of proportionality is 0.08 liters per kilometer.
  4. (b) The driver keeps 4 liters in the tank, which leaves 36 − 4 = 32 liters to use.
  5. 32 liters is 32 ÷ 0.8 = 40 blocks of 10 km, so the van can travel 40 × 10 = 400 km. Check: 400 × 0.08 = 32 liters.

answer(a) Fuel divided by distance is 0.08 for every trip, so the fuel is in direct proportion to the distance and the constant is 0.08 liters per km; (b) 400 km

techniqueThe Constant of Proportionality · Direct Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing for one trip only. Any single pair of numbers gives a quotient; the fuel is in direct proportion to the distance only because all three trips give the same one, so every trip must be checked.
  • Dividing the whole 36 liters by 0.08 and answering 450 km. The 4 liters the driver keeps in the tank are not used on the trip, so only 32 liters are shared out.
17

Apple Pickers in an Orchard: A Bigger Team, Then a Bigger Order

heuristicFind What One Picker Fills in One Hour, Then Scale

In an orchard, 6 pickers fill 90 crates of apples in 3 hours. Every picker works at the same steady rate. (a) How many crates do 9 pickers fill in 3 hours? (b) The orchard has an order for 400 crates, which must be filled in 5 hours. How many pickers are needed?

6 pickers, 3 hours: 90 crates
One box for each picker in each hour: 6 × 3 = 18 boxes, holding 90 crates between them.
Draw one box for each picker in each hour: 6 pickers for 3 hours make 6 × 3 = 18 boxes, and together they hold 90 crates.
step 1 of 5

Draw one box for each picker in each hour. The boxes all hold the same number of crates, so find what one box holds and count boxes.

  1. Draw one box for each picker in each hour: 6 pickers for 3 hours make 6 × 3 = 18 boxes, and together they hold 90 crates.
  2. One box, one picker for one hour, holds 90 ÷ 18 = 5 crates.
  3. (a) 9 pickers for 3 hours make 9 × 3 = 27 boxes, so they fill 27 × 5 = 135 crates.
  4. (b) 400 crates need 400 ÷ 5 = 80 boxes.
  5. The 80 boxes are spread over 5 hours, so each hour needs 80 ÷ 5 = 16 boxes, one for each picker: 16 pickers are needed. Check: 16 × 5 × 5 = 400 crates.

answer(a) 135 crates; (b) 16 pickers

techniqueProportion by Unit Value · Direct Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Scaling (a) by the time as well as by the team. The time is still 3 hours; only the number of pickers changes, from 6 to 9, so the crates change in the same way: 90 × 96 = 135.
  • Treating a longer time as needing more pickers. Each picker fills more crates in 5 hours than in 3, so a longer time needs fewer pickers; dividing the 80 picker-hours by the 5 hours is what gives the team.
18

Paint Mixed Too Dark: White Added Until the Shade Is Lighter

heuristicRewrite Both Ratios So the Unchanged Share Matches

A painter mixes 14 liters of paint from blue paint and white paint in the ratio 3 : 4. The shade is too dark, so she adds white paint only, until the ratio of blue paint to white paint is 2 : 5. (a) How many liters of white paint does she add? (b) White paint is sold only in 2-liter cans, at $9.50 a can. White paint is sold only in whole cans. How much does she pay for the cans of white paint she needs?

Beforeblue 3 partswhite 4 parts14 LAfterblue 2 partswhite 5 parts
Before, blue : white = 3 : 4; after, 2 : 5. No blue is added, so the blue bar is the same length in both.
No blue paint is added, so the blue share is the same before and after. Before, blue : white = 3 : 4; after, blue : white = 2 : 5.
step 1 of 5

The blue share is the same paint before and after, so rewrite both ratios with the same number of units for blue. Then one unit means the same amount of paint in both, and the white units can be compared.

  1. No blue paint is added, so the blue share is the same before and after. Before, blue : white = 3 : 4; after, blue : white = 2 : 5.
  2. Rewrite both ratios so the blue share matches. The least common multiple of 3 and 2 is 6: 3 : 4 = 6 : 8 and 2 : 5 = 6 : 15.
  3. Before, the 14 liters are 6 + 8 = 14 units, so one unit is 14 ÷ 14 = 1 liter.
  4. (a) The white goes from 8 units to 15 units, so she adds 15 − 8 = 7 units, which is 7 liters.
  5. (b) 3 cans hold only 6 liters, so she buys 4 cans and pays 4 × $9.50 = $38.00. Check: blue 6 liters, white 15 liters, and 6 : 15 = 2 : 5.

answer(a) 7 liters; (b) $38.00

techniqueRewriting a Ratio to Match a Share · Ratios with a Constant Part

examsPSLE · O-Level

Common pitfalls

  • Adding 5 − 4 = 1 unit of white, 2 liters. The blue share is 3 units before and 2 units after, so a unit is a different amount of paint in the two ratios; they must be rewritten so the blue shares match before the white shares are compared.
  • Paying for 3.5 cans, or for 3 cans. Paint is sold only in whole cans, and 3 cans hold 6 liters, 1 liter short of the 7 liters she needs.
19

A Hiking Trail on Two Maps with Different Scales

heuristicMap to Ground by One Scale, Then Ground to Map by the Other

A hiking map is drawn to the scale 1 : 25000. On it, the trail from a parking lot to a waterfall measures 18.4 cm along its bends. (a) How long is the trail on the ground, in kilometers? (b) The same trail is drawn on a regional map on which 1 cm stands for 2 km. How long is the trail on the regional map, in centimeters?

Hiking map18.4 cm1 cm is 250 m
On the hiking map, 1 cm stands for 25000 cm, which is 250 m on the ground.
On the hiking map, 1 cm stands for 25000 cm on the ground, which is 250 m.
step 1 of 5

Find what one centimeter on the hiking map stands for on the ground, count the centimeters of trail, then share the ground distance into the regional map's centimeters.

  1. On the hiking map, 1 cm stands for 25000 cm on the ground, which is 250 m.
  2. The trail is 18.4 cm on the map, so on the ground it is 18.4 × 250 = 4600 m.
  3. (a) 4600 m = 4.6 km, so the trail is 4.6 km long.
  4. (b) On the regional map, every 2 km of ground is 1 cm. The trail's 4.6 km is 4.6 ÷ 2 = 2.3 lots of 2 km.
  5. So the trail is 2.3 cm long on the regional map. Check: 2.3 × 2 = 4.6 km, the same trail.

answer(a) 4.6 km; (b) 2.3 cm

techniqueMap Scales

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Writing 18.4 × 25000 = 460000 and calling it kilometers or meters. A scale written as a ratio has no unit, so the answer is in the map's unit, centimeters, and 100000 cm make one kilometer.
  • Scaling the 18.4 cm up for the regional map. Each centimeter of the regional map stands for more ground than a centimeter of the hiking map, so the same trail must be shorter there; going through the ground distance, 4.6 km, keeps the direction right.
20

An L-Shaped Living Room on a Floor Plan, and the Boxes of Flooring It Needs

heuristicScale Each Length, So the Area Scales by the Square

A floor plan of an apartment is drawn to the scale 1 : 50 on paper ruled in 1 cm squares. The L-shaped living room covers 108 of the squares. (a) What is the floor area of the real living room, in square meters? (b) Flooring is sold in boxes that each cover 2.2 m2, at $64 a box. How much does it cost to buy enough boxes to cover the living room floor? Ignore any waste from cutting.

1 square: 50 cm by 50 cm of floorwhich is 0.5 m by 0.5 m
One 1 cm square of the plan stands for 50 cm by 50 cm of floor: 0.5 m by 0.5 m.
One 1 cm square on the plan stands for a square of floor 50 cm by 50 cm, which is 0.5 m by 0.5 m.
step 1 of 5

Work out what one square of the plan stands for on the floor, as a length each way and then as an area, and count the squares.

  1. One 1 cm square on the plan stands for a square of floor 50 cm by 50 cm, which is 0.5 m by 0.5 m.
  2. That piece of floor has an area of 0.5 × 0.5 = 0.25 m2. Both of its sides were scaled by 50, so its area was scaled by 50 × 50 = 2500.
  3. (a) The room covers 108 squares, so its area is 108 × 0.25 = 27 m2.
  4. (b) 12 boxes cover 12 × 2.2 = 26.4 m2, which is less than 27 m2, so 13 boxes are needed.
  5. 13 boxes cost 13 × $64 = $832. Check: 13 × 2.2 = 28.6 m2, enough for the 27 m2 room.

answer(a) 27 m2; (b) $832

techniqueScaling Area · Map Scales

examsO-Level · SAT

Common pitfalls

  • Multiplying the plan area by 50 only: 108 × 50 = 5400 cm2, about half a square meter, far too small for a living room. The scale multiplies each length by 50, and an area is two lengths multiplied, so it is multiplied by 50 × 50 = 2500.
  • Rounding 12.27… boxes down to 12. Twelve boxes cover 26.4 m2 and leave 0.6 m2 of floor bare, so a thirteenth box must be bought.
Mr. Chalk Drill ratio and proportion