Rate and work · applications

Applications: Rate and Work

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

PSLE · O-Level · SAT · GCSE Higher

01

Cooperative Shared Task Completion

heuristicCommon Work Unit Normalization / Harmonic Rate Addition

Worker A can paint a school hall in 12 hours working alone. Worker B can paint the same hall in 15 hours working alone, and Worker C can paint it in 20 hours working alone. (a) What fraction of the hall can Worker A and Worker B paint together in 1 hour? (b) How many hours will it take for all three workers to paint the entire hall if they work together simultaneously?

Hall60 units
Call the hall 60 units, a number 12, 15 and 20 all divide.
Total work = 60 units.
step 1 of 9

Represent the hall as a block of 60 work units (LCM of 12, 15, and 20) and assign unit outputs per hour.

  1. Total work = 60 units.
  2. Worker A: 60 ÷ 12 = 5 units/hr.
  3. Worker B: 60 ÷ 15 = 4 units/hr.
  4. Worker C: 60 ÷ 20 = 3 units/hr.
  5. Workers A and B together do 5 + 4 = 9 units/hr.
  6. (a) Fraction in 1 hour = 960 = 320.
  7. All three workers together do 5 + 4 + 3 = 12 units/hr.
  8. Total time to complete 60 units: 60 ÷ 12 = 5 hours.
  9. (b) 5 hours.

answer(a) 320 of the hall; (b) 5 hours

techniqueRates · The Lowest Common Multiple

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the hours directly (12 + 15 + 20 = 47 hours) or averaging them, concluding it takes longer working together.
  • Taking the common denominator incorrectly when adding the three fractions.
02

Joint Work with Staggered Departures

heuristicAggregate Output Subtraction / Individual Remainder Allocation

Alice and Bob were assigned to digitize 600 pages of historical archives. Working alone, Alice can digitize all 600 pages in 20 hours. Alice and Bob began working together. After 4 hours of joint work, Bob had to leave for an urgent assignment. Alice continued alone and took another 10 hours to finish digitizing the remaining pages. (a) How many pages did Bob digitize during the 4 hours? (b) How many hours would Bob take to digitize all 600 pages working completely alone?

Alice303030303030303030303030303014 hourstogether 4 halone 10 h
Alice does 30 pages an hour.
Alice produces 600 ÷ 20 = 30 pages every hour.
step 1 of 6

Track Alice's contribution across the entire timeline: 14 hours at 30 pages/hr leaves 180 pages for Bob's 4-hour window.

  1. Alice produces 600 ÷ 20 = 30 pages every hour.
  2. Alice was present for 14 hours in total: 14 × 30 = 420 pages.
  3. The remaining pages were completed by Bob in his 4 hours: 600 − 420 = 180 pages.
  4. (a) Bob digitized 180 pages.
  5. Bob's rate: 180 ÷ 4 = 45 pages/hr.
  6. Time for Bob alone: 600 ÷ 45 = 1313 hours (13 hr 20 min).

answer(a) 180 pages; (b) 1313 hours, that is 13 h 20 min

techniqueRate, Total, Units · Rates

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Assuming Bob worked for 14 hours as well, rather than only during the first 4 hours.
  • Dividing 180 pages by 14 hours instead of Bob's actual working duration of 4 hours.
03

Machine Efficiency Ratios & Combined Output

heuristicRelative Efficiency Scaling / Composite Output Rate

A factory has two automated packaging machines, Machine X and Machine Y. Machine X is 1.5 times as fast as Machine Y. When both machines operate simultaneously, they can pack a shipment of 3600 gift hampers in 48 minutes. (a) How many hampers does Machine X pack per minute? (b) If Machine X starts alone and packs for 20 minutes, how many minutes will Machine Y take working alone to pack all the remaining hampers?

X3 partsY2 parts
X : Y = 1.5 : 1 = 3 : 2.
Ratio of rates: Machine X : Machine Y = 1.5 : 1 = 3 : 2.
step 1 of 9

Model Machine Y as 2 parts and Machine X as 3 parts (ratio 3 : 2). Total combined rate is 5 parts = 75 hampers/min.

  1. Ratio of rates: Machine X : Machine Y = 1.5 : 1 = 3 : 2.
  2. Together, they produce 3600 ÷ 48 = 75 hampers/min.
  3. 5 parts = 75 hampers/min ⟹ 1 part = 15 hampers/min.
  4. Machine X (3 parts) = 3 × 15 = 45 hampers/min.
  5. Machine Y (2 parts) = 2 × 15 = 30 hampers/min.
  6. (a) Machine X rate = 45 hampers/min.
  7. Machine X packs 20 × 45 = 900 hampers.
  8. Remaining = 3600 − 900 = 2700 hampers.
  9. Machine Y takes 2700 ÷ 30 = 90 minutes.

answer(a) 45 hampers per minute; (b) 90 minutes

techniqueRatio · Rates

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Setting the ratio of Machine X to Machine Y as 1 : 1.5 instead of 1.5 : 1.
  • Multiplying 20 minutes by the combined rate of 75 instead of Machine X's solo rate of 45.
04

Alternating Hourly Work Cycles

heuristicCycle Size Chunking / Remainder Interval Allocation

Darren and Ethan are clearing weeds from a community garden. Working alone, Darren can clear the plot in 8 hours, while Ethan can clear it in 12 hours. They agree to work alternately for 1 hour each, with Darren working in the 1st hour, Ethan working in the 2nd hour, Darren in the 3rd hour, Ethan in the 4th hour, and so on. (a) What fraction of the garden is cleared after the first 2 hours? (b) How many hours and minutes did it take altogether to clear the entire plot?

Plot24 unitsHoursD 3E 25 in 2 h
The plot is 24 units: Darren clears 3 an hour, Ethan 2.
Total plot = 24 units.
step 1 of 8

Draw a tape of 24 units. Group every 2 hours into a 5-unit block, then step through the final leftover units hour-by-hour.

  1. Total plot = 24 units.
  2. Darren clears 3 units in Hour 1; Ethan clears 2 units in Hour 2.
  3. (a) 2-hour block = 5 units out of 24 = 524.
  4. 4 blocks of 2 hours = 8 hours, clearing 4 × 5 = 20 units.
  5. Remaining to clear = 24 − 20 = 4 units.
  6. Hour 9 (Darren): clears 3 units, leaving 4 − 3 = 1 unit.
  7. Hour 10 (Ethan): Ethan needs 12 hour (30 min) to clear the last 1 unit.
  8. (b) Total time = 8 hr + 1 hr + 30 min = 9 hours 30 minutes.

answer(a) 524 of the garden; (b) 9 hours 30 minutes

techniqueRates · Remainders

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing 24 by 2.5 (average hourly rate) directly to get 9.6 hours, which fails to track discrete alternation and the correct worker on turn.
  • Letting Ethan work in the 9th hour instead of Darren.
05

Piece-Rate Compensation with Defect Penalties

heuristicSupposition on Output Rates / Total Loss per Defect

An artisan is paid $6.00 for every decorative glass lantern he successfully assembles. However, if a lantern is flawed, he is not paid and is penalized $9.00 for the wasted materials. Over a week, the artisan assembled 150 lanterns and received a net payment of $720.00. (a) How many flawed lanterns did the artisan assemble? (b) How much more money would he have received if none of the lanterns had been flawed?

If all good150 × $6 = $900
Suppose every lantern was good: $900.
Assume all 150 lanterns were good: Total = 150 × $6.00 = $900.00.
step 1 of 7

Assume all 150 lanterns were perfect. Every flawed lantern creates a gap of $6 (unearned) + $9 (penalty) = $15.

  1. Assume all 150 lanterns were good: Total = 150 × $6.00 = $900.00.
  2. Actual amount received = $720.00.
  3. Total shortfall = $900 − $720 = $180.00.
  4. Replacing 1 good lantern with 1 flawed lantern reduces earnings by: $6.00 (lost pay) + $9.00 (penalty) = $15.00.
  5. Number of flawed lanterns: 180 ÷ 15 = 12 flawed lanterns.
  6. (a) 12 flawed lanterns.
  7. (b) Shortfall is $180.00.

answer(a) 12 flawed lanterns; (b) $180

techniqueRate, Total, Units · Forming Equations

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Using only the $9 penalty as the difference per defective item, forgetting that the artisan also forfeits the $6 payment ($15 total difference).
  • Subtracting $9 from $6 ($3 difference) instead of adding.
06

Camp Rations / Resource Depletion with Group Changes

heuristicMan-Day Invariance / Consumption Rate Balancing

A youth camp stored enough food rations to feed 40 campers for 24 days, with each camper consuming an equal daily ration. After 6 days, 10 campers had to leave the camp due to bad weather. (a) For how many additional days will the remaining food rations last the remaining campers? (b) How many days longer did the food rations last in total compared to the original planned schedule?

Planned6 days18 days × 40 = 720 units24 days
After 6 days, 18 days of food for 40 campers remain.
Remaining supply at Day 6: 18 days for 40 campers.
step 1 of 8

After 6 days, there were 18 days of rations left for 40 campers (40 × 18 = 720 units). Distribute these 720 units to 30 campers.

  1. Remaining supply at Day 6: 18 days for 40 campers.
  2. Total remaining units = 18 × 40 = 720 units.
  3. Now 30 campers share these 720 units.
  4. Days = 720 ÷ 30 = 24 days.
  5. (a) 24 additional days.
  6. Originally, the food was to last 18 more days.
  7. Difference = 24 − 18 = 6 days longer.
  8. (b) 6 days longer.

answer(a) 24 more days; (b) 6 days longer

techniqueInverse Proportion · Rates

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the total 960 camper-days by 30, forgetting that 240 camper-days were already consumed by 40 people.
  • Reporting 24 days as the extra duration instead of subtracting the planned 18 remaining days.
07

Tiered / Progressive Consumption Utility Billing

heuristicPiecewise Consumption Tiers / Step Function Boundary Inversion

A municipal utility company bills domestic water consumption on a tiered tariff: the first 20 m3 is billed at $1.20 per m3; the next 15 m3 (from 21 to 35 m3) is billed at $1.60 per m3; and every cubic meter above 35 m3 is billed at $2.40 per m3. (a) The Tan family consumed 32 m3 of water in March. What was their water bill? (b) The Lee family received a water bill of $72.00 in April. How many cubic meters of water did the Lee family consume?

32 m³20 × $1.20 = $2412 × $1.60 = $19.20
32 m³ is 20 in the first tier and 12 in the second.
Part (a): Split 32 m3 into 20 m3 (Block 1) and 12 m3 (Block 2).
step 1 of 8

Draw 3 stacked pricing blocks: Block 1 (up to 20 m³ at $1.20), Block 2 (next 15 m³ at $1.60), and Block 3 (excess at $2.40).

  1. Part (a): Split 32 m3 into 20 m3 (Block 1) and 12 m3 (Block 2).
  2. Cost = (20 × 1.20) + (12 × 1.60) = 24.00 + 19.20 = $43.20.
  3. (a) $43.20.
  4. Part (b): Block 1 takes $24.00 (20 m3). Leftover bill = 72 − 24 = $48.00.
  5. Block 2 takes $24.00 (15 m3). Leftover bill = 48 − 24 = $24.00.
  6. Block 3 rate is $2.40 per m3: 24.00 ÷ 2.40 = 10 m3.
  7. Total volume = 20 + 15 + 10 = 45 m3.
  8. (b) 45 m3.

answer(a) $43.20; (b) 45 m3

techniqueRate, Total, Units · Money as Decimals

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Multiplying the entire 32 m³ by $1.60, or dividing the entire $72 by $2.40.
  • Using 35 m³ in Tier 2 instead of the 15 m³ interval difference (35 − 20 = 15).
08

Cooperative Pair Work (Cycle System of 3 Workers)

heuristicCyclic Pair Addition / Simultaneous Elimination

Three painters, Ken, Liam, and Mark, were hired to paint a school: Ken and Liam working together take 12 days; Liam and Mark working together take 20 days; and Ken and Mark working together take 15 days. (a) How many days would it take for all three painters to paint the school working together? (b) How many days would Liam take to paint the entire school working completely alone?

K + L5 a dayL + M3 a dayK + M4 a day
The school is 60 units.
Set the whole school to 60 units of paint.
step 1 of 10

Double each pair's contribution so every painter is counted twice, then divide by 2 to get the combined daily team output.

  1. Set the whole school to 60 units of paint.
  2. Ken + Liam paint 5 units per day.
  3. Liam + Mark paint 3 units per day.
  4. Ken + Mark paint 4 units per day.
  5. Combining the three pairs counts 2 Kens, 2 Liams, and 2 Marks: 5 + 3 + 4 = 12 units/day.
  6. 1 Ken + 1 Liam + 1 Mark = 12 ÷ 2 = 6 units/day.
  7. All three together: 60 ÷ 6 = 10 days.
  8. (a) 10 days.
  9. Since Ken and Mark together paint 4 units/day, Liam alone paints 6 − 4 = 2 units/day.
  10. Liam alone: 60 ÷ 2 = 30 days.

answer(a) 10 days; (b) 30 days

techniqueRates · Simultaneous by Elimination

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Forgetting to divide the sum of the rates by 2, assuming the combined rate is 12 units/day instead of 6 units/day.
  • Averaging the three pair days ((12 + 20 + 15)/3 = 15.67 days).
09

Rate with Continuous Leakage / Depletion

heuristicMulti-Channel Rate Balance / Net Replenishment Evaluation

A community water reservoir with a total capacity of 36000 liters currently holds 24000 liters of water. An intake pipe replenishes the reservoir at a rate of 100 liters/min. At the same time, the reservoir supplies water to a village at 70 liters/min, and an undetected crack leaks water at 10 liters/min. (a) How many hours will it take to fill the reservoir completely to its capacity? (b) If the intake pipe is closed the moment the reservoir is completely full, how many hours and minutes will it take for the reservoir to empty completely while supplying the village and leaking at the same rates?

Reservoir24000 in36000
12000 liters short of full.
Capacity deficit = 36000 − 24000 = 12000 liters.
step 1 of 6

Model the reservoir as an accumulating unit bucket gaining 20 liters/min, and emptying at 80 liters/min when the tap shuts.

  1. Capacity deficit = 36000 − 24000 = 12000 liters.
  2. Every minute: +100 ℓ in, −70 ℓ to village, −10 ℓ leak ⟹ +20 ℓ/min net gain.
  3. Fill time: 12000 ÷ 20 = 600 min = 10 hours.
  4. (a) 10 hours.
  5. When tap shuts: loss is 70 + 10 = 80 ℓ/min.
  6. Empty time from full (36000 ℓ): 36000 ÷ 80 = 450 minutes = 7 hr 30 min.

answer(a) 10 hours; (b) 7 hours 30 minutes

techniqueRates · Rate, Total, Units

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Using the full 36,000 liters instead of the 12,000 liter deficit to find the filling time.
  • Subtracting the 10 liter leak from the 70 liter supply instead of adding them together to find total outflow.
10

Variable Production Rate Across Intervals

heuristicPiecewise Timeline Production / Acceleration Re-balancing

An industrial 3D printing workshop was contracted to produce 720 precision gears. The printing facility operated at its standard rate of 40 gears per hour for the first 5 hours. Due to machine overheating, the printing speed was throttled down by 25% for the next 4 hours. Following cooling, a second printer was activated alongside the first, doubling the standard production rate to 80 gears per hour until the order was completed. (a) How many gears were produced in total during the first 9 hours? (b) How many hours did the entire production run take from start to finish?

Hours5 h × 40Gears200
Phase 1: 5 hours at 40 an hour, 200 gears.
Phase 1: 5 hr × 40 = 200 gears.
step 1 of 8

Plot production along a 720-unit bar: Block 1 (200 gears), Block 2 (120 gears), and Block 3 (remaining 400 gears).

  1. Phase 1: 5 hr × 40 = 200 gears.
  2. Phase 2: 25% off 40 = 10 reduction ⟹ 30 gears/hr.
  3. Phase 2 output: 4 hr × 30 = 120 gears.
  4. First 9 hours output: 200 + 120 = 320 gears.
  5. (a) 320 gears.
  6. Remaining bar: 720 − 320 = 400 gears.
  7. Phase 3 speed is 80 gears/hr: 400 ÷ 80 = 5 hours.
  8. Total hours: 5 + 4 + 5 = 14 hours.

answer(a) 320 gears; (b) 14 hours

techniqueRate, Total, Units · Rates

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Subtracting 25% from 5 hours instead of reducing the production rate of 40 gears/hour.
  • Reporting only the 5 hours of Phase 3 instead of the total 14 hours for the entire job.
11

Two Bag Sizes of Coffee Beans: The Price per 100 g, and a Café's Weekly Saving

heuristicUnit Price per 100 g / Counting Whole Bags for the Weekly Amount

A shop sells the same coffee beans in two bag sizes: a 250 g bag for $6.00 and a 400 g bag for $8.80. (a) Find the price per 100 g of each bag. Which bag is the better buy? (b) A café uses exactly 2 kg of these beans each week. How much does it save each week by buying only the better bags instead of only the other bags?

250 g bag100 g100 g50 g$6.00400 g bag100 g100 g100 g100 g$8.80
Cut each bag into 100 g parts: the 250 g bag is 212 parts and the 400 g bag is 4 parts.
Cut each bag into 100 g parts: the 250 g bag is 212 parts and the 400 g bag is 4 parts.
step 1 of 7

Cut each bag into 100 g parts and find the price of one part. Then count the bags of each size that make 2 kg.

  1. Cut each bag into 100 g parts: the 250 g bag is 212 parts and the 400 g bag is 4 parts.
  2. One part of the 250 g bag costs 6.00 ÷ 2.5 = $2.40.
  3. One part of the 400 g bag costs 8.80 ÷ 4 = $2.20.
  4. (a) The 400 g bag is the better buy: each 100 g costs 2.40 − 2.20 = $0.20 less.
  5. 2 kg is 2000 g, which is 5 bags of 400 g: 5 × 8.80 = $44.00.
  6. In 250 g bags, 2000 g is 8 bags: 8 × 6.00 = $48.00.
  7. (b) The café saves 48.00 − 44.00 = $4.00 a week. Check: 2000 g is 20 parts of 100 g, and 20 × 0.20 = $4.00.

answer(a) $2.40 and $2.20 per 100 g, so the 400 g bag is the better buy; (b) $4.00 a week

techniqueComparing Unit Rates · Money as Decimals

examsSAT · GCSE Higher

Common pitfalls

  • Calling the cheaper bag, $6.00 against $8.80, the better buy: the bags hold different amounts of coffee, so only the prices for the same mass, 100 g, can be compared.
  • Giving $0.20 as the weekly saving: that is the saving on each 100 g, and 2 kg holds 20 lots of 100 g.
12

Dollars Changed into Euros for a Trip, and the Leftover Euros Changed Back

heuristicConverting Both Ways / Each Rate in Its Own Direction

Before a trip to Italy, Lena changed $800 into euros at the rate $1 = €0.90. She spent €630 on the trip. When she came home, her bank bought back all her leftover euros at the rate €1 = $1.05. (a) How many euros did Lena have left at the end of the trip? (b) How many dollars did the bank pay her for them, and how many dollars less was that than those euros had cost her before the trip?

Dollars$100$100$100$100$100$100$100$100$800Euros9090909090909090
Cut $800 into 8 units of $100; each unit buys 100 × 0.90 = 90 euros.
Cut $800 into 8 units of $100. At $1 = €0.90, each unit buys 100 × 0.90 = 90 euros.
step 1 of 7

Cut the $800 into 8 units of $100. Each unit becomes €90, so the euros are 8 equal units too, and the unit left over is followed back to the bank.

  1. Cut $800 into 8 units of $100. At $1 = €0.90, each unit buys 100 × 0.90 = 90 euros.
  2. So Lena received 8 × 90 = 720 euros.
  3. She spent €630, which is 630 ÷ 90 = 7 units, so 1 unit is left.
  4. (a) €90 were left.
  5. The bank pays 90 × 1.05 = $94.50 for that unit.
  6. The same unit of euros had cost her $100 before the trip.
  7. (b) The bank paid $94.50, which is 100 − 94.50 = $5.50 less. Check: 94.50 + 5.50 = 100.

answer(a) €90; (b) $94.50, which is $5.50 less than the $100 those euros had cost

techniqueExchange Rates · Money as Decimals

examsPSLE · GCSE Higher

Common pitfalls

  • Changing the €90 back with the first rate, to get 90 ÷ 0.90 = $100: the bank uses its own rate for the way back, €1 = $1.05.
  • Multiplying €90 by 0.90 to change it back: 0.90 is the number of euros for each dollar, so it changes dollars into euros, never euros into dollars.
13

A Tank of Cooking Oil: Its Mass, and Which of Two Objects Floats in It

heuristicMass = Density × Volume / Comparing Densities

In a science lab, a tank 50 cm long, 40 cm wide and 45 cm tall holds cooking oil to a depth of 30 cm. The density of the oil is 0.9 g/cm3. An object floats in a liquid when its density is less than the liquid's, and sinks when its density is greater. (a) What is the mass of the oil, in kilograms? (b) A pine block measuring 20 cm by 10 cm by 10 cm has a mass of 1200 g, and a rubber stopper with a volume of 50 cm3 has a mass of 60 g. Find the density of each. Which of them floats in the oil?

Each block holds 10 000 cubic centimeters of oil.Oil10 00010 00010 00010 00010 00010 00050 × 40 × 30 = 60 000 cm3
The oil fills 50 × 40 × 30 = 60000 cm3: 6 blocks of 10000 cm3.
The oil fills 50 × 40 × 30 = 60000 cm3, which is 6 blocks of 10000 cm3.
step 1 of 7

A density of 0.9 g/cm3 means that each cubic centimeter of oil has a mass of 0.9 g. Split the oil into equal blocks, find the mass of one, and compare the mass in each cubic centimeter of the two objects with the oil's.

  1. The oil fills 50 × 40 × 30 = 60000 cm3, which is 6 blocks of 10000 cm3.
  2. Each cubic centimeter has a mass of 0.9 g, so one block has a mass of 10000 × 0.9 = 9000 g.
  3. Six blocks: 6 × 9000 = 54000 g.
  4. (a) 54000 g = 54 kg.
  5. The pine block has a volume of 20 × 10 × 10 = 2000 cm3, so each cubic centimeter of it has a mass of 1200 ÷ 2000 = 0.6 g.
  6. Each cubic centimeter of the stopper has a mass of 60 ÷ 50 = 1.2 g.
  7. (b) 0.6 g is less than the oil's 0.9 g, so the pine block floats; 1.2 g is more, so the stopper sinks, even though it is the lighter of the two objects.

answer(a) 54 kg; (b) the block's density is 0.6 g/cm3 and the stopper's is 1.2 g/cm3, so the pine block floats and the stopper sinks

techniqueDensity · Volume of a Cuboid

examsSAT · GCSE Higher

Common pitfalls

  • Deciding by mass that the light stopper floats and the heavy block sinks: what decides it is the mass in each cubic centimeter, the density, not the whole mass.
  • Using the tank's height of 45 cm for the volume of the oil: the oil reaches a depth of only 30 cm.
14

A Hiker on Snow in Boots and in Snowshoes: The Pressure Each Way, and the Heaviest Pack

heuristicPressure as Force over Area / Working Back from the Snow's Limit

A hiker weighs 720 N and stands on snow with his weight spread evenly over both feet. Each of his boots has a sole of area 0.03 m2, and each of his snowshoes has an area of 0.12 m2. (a) What pressure does he put on the snow in boots, and what pressure in snowshoes, in N/m2? (b) The snow holds any pressure up to and including 3500 N/m2 without a foot sinking in. What is the heaviest pack, in newtons, that he can carry on snowshoes without sinking? Ignore the weight of the snowshoes.

Pressure on the snow, in newtons on each square meterboots: 2 × 0.03 = 0.06 m2
In boots, 720 N presses on 2 × 0.03 = 0.06 m2.
In boots, the hiker's 720 N presses on 2 × 0.03 = 0.06 m2 of snow.
step 1 of 7

Pressure is the force on each square meter. Share the weight over the area in contact with the snow, then work back from the snow's limit to the greatest total weight it holds up.

  1. In boots, the hiker's 720 N presses on 2 × 0.03 = 0.06 m2 of snow.
  2. The pressure is 720 ÷ 0.06 = 12000 N/m2.
  3. In snowshoes, 720 N presses on 2 × 0.12 = 0.24 m2, so the pressure is 720 ÷ 0.24 = 3000 N/m2.
  4. (a) 12000 N/m2 in boots and 3000 N/m2 in snowshoes: four times the area gives a quarter of the pressure.
  5. At 3500 N/m2, the 0.24 m2 under the snowshoes holds up at most 3500 × 0.24 = 840 N.
  6. The hiker is 720 N of that, so 840 − 720 = 120 N is left for the pack.
  7. (b) The heaviest pack is 120 N. Check: (720 + 120) ÷ 0.24 = 3500 N/m2, which is the limit exactly.

answer(a) 12000 N/m2 in boots and 3000 N/m2 in snowshoes; (b) 120 N

techniquePressure · Rate, Total, Units

examsGCSE Higher

Common pitfalls

  • Using the area of one boot or one snowshoe: he stands on both feet, so his weight is spread over twice that area.
  • Giving 840 N as the answer: that is the most the snow holds up in total, and the hiker's own 720 N is part of it.
Mr. Chalk Read the guide