Rate and work · applications

Applications: Rate and Work

18 problems

PSLE · O-Level · SAT · GCSE Higher

01

Three Workers Paint One Hall: Two of Them for an Hour, Then All Three Together

heuristicMake the Hall a Number of Units That 12, 15 and 20 All Divide / Add the Units Each Worker Paints in an Hour

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 all three workers take to paint the whole hall if they work together at the same time?

Hall60 units
Call the hall 60 units: 60 is the lowest common multiple of 12, 15 and 20.
Call the whole hall 60 units. 60 is the lowest common multiple of 12, 15 and 20, so each worker paints a whole number of units in an hour.
step 1 of 6

Cut the hall into 60 equal units, because 60 is the lowest common multiple of 12, 15 and 20. Then each worker paints a whole number of units in an hour, and the units add up.

All steps
  1. Call the whole hall 60 units. 60 is the lowest common multiple of 12, 15 and 20, so each worker paints a whole number of units in an hour.
  2. In one hour, Worker A paints 60 ÷ 12 = 5 units, Worker B paints 60 ÷ 15 = 4 units and Worker C paints 60 ÷ 20 = 3 units.
  3. In one hour, Workers A and B together paint 5 + 4 = 9 units.
  4. (a) 9 units out of 60 is 960 = 320 of the hall.
  5. All three workers together paint 5 + 4 + 3 = 12 units in one hour.
  6. (b) The 60 units take them 60 ÷ 12 = 5 hours. Check: in 5 hours Worker A paints 25 units, Worker B 20 and Worker C 15, and 25 + 20 + 15 = 60.

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

techniqueRates · The Lowest Common Multiple

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the three times: 12 + 15 + 20 = 47 hours. Three workers painting at once finish sooner than any one of them alone, so the answer must be less than 12 hours, the fastest worker's time.
  • Adding the fractions by adding the tops and the bottoms: 112 + 115 = 227. That is less than the 112 Worker A paints alone, so it cannot be right: first write both fractions as sixtieths, 560 + 460 = 960.
02

Two People Scan 600 Pages Together Until One of Them Leaves

heuristicFind Every Page Alice Scanned / The Rest Are Bob's, Scanned in His 4 Hours

Alice and Bob were asked to scan 600 pages of old records. Working alone, Alice can scan all 600 pages in 20 hours. Alice and Bob began working together. After 4 hours, Bob left for another job. Alice carried on alone and took another 10 hours to scan the pages that were left. (a) How many pages did Bob scan in his 4 hours? (b) How long would Bob take to scan all 600 pages working alone?

Alice303030303030303030303030303030 an hour
Alice scans 600 ÷ 20 = 30 pages an hour.
Alice scans all 600 pages in 20 hours, so she scans 600 ÷ 20 = 30 pages an hour.
step 1 of 6

Draw Alice's 14 hours as 14 equal parts of 30 pages. The pages she did not scan are Bob's, and he scanned them in the first 4 hours only.

All steps
  1. Alice scans all 600 pages in 20 hours, so she scans 600 ÷ 20 = 30 pages an hour.
  2. Alice worked 4 hours beside Bob and then 10 hours alone, which is 4 + 10 = 14 hours. In that time she scanned 14 × 30 = 420 pages.
  3. (a) Bob scanned the rest: 600 − 420 = 180 pages.
  4. Bob scanned those 180 pages in 4 hours, so he scans 180 ÷ 4 = 45 pages an hour.
  5. Alone, Bob needs 600 ÷ 45 = 1313 hours for all 600 pages.
  6. (b) 13 of an hour is 60 ÷ 3 = 20 minutes, so Bob would take 13 hours 20 minutes. Check: in 13 hours he scans 13 × 45 = 585 pages, and in 20 more minutes 45 ÷ 3 = 15 pages, which makes 600.

answer(a) 180 pages; (b) 1313 hours, which is 13 hours 20 minutes

techniqueRate, Total, Units · Rates

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Sharing Bob's 180 pages over 14 hours to find his rate. Bob left after 4 hours, so he scanned his 180 pages in 4 hours: 180 ÷ 4 = 45 pages an hour.
  • Counting only Alice's 10 hours alone: 10 × 30 = 300 pages, which leaves 300 for Bob. Alice also worked in the first 4 hours, beside Bob, so she worked 14 hours in all.
03

Two Packing Machines, One of Them 1.5 Times as Fast as the Other

heuristicWrite 1.5 : 1 as 3 : 2 and Share the Joint Rate in Parts / Find What X Leaves for Y

A factory has two packing machines, Machine X and Machine Y. Machine X is 1.5 times as fast as Machine Y. Working at the same time, the two machines pack 3600 gift hampers in 48 minutes. (a) How many hampers does Machine X pack in one minute? (b) Machine X starts alone and packs for 20 minutes. How many minutes will Machine Y take, working alone, to pack all the hampers that are left?

X3 partsY2 parts
X : Y = 1.5 : 1 = 3 : 2, so X has 3 parts and Y has 2.
Machine X is 1.5 times as fast as Machine Y, so their rates are in the ratio 1.5 : 1 = 3 : 2. Give Machine X 3 parts and Machine Y 2 parts.
step 1 of 6

1.5 times as fast means the rates are in the ratio 1.5 : 1, which is 3 : 2. Draw Machine X's rate as 3 parts and Machine Y's as 2 parts, and share the joint rate among the 5 parts.

All steps
  1. Machine X is 1.5 times as fast as Machine Y, so their rates are in the ratio 1.5 : 1 = 3 : 2. Give Machine X 3 parts and Machine Y 2 parts.
  2. Together they pack 3600 ÷ 48 = 75 hampers a minute. That is 3 + 2 = 5 parts, so 1 part is 75 ÷ 5 = 15 hampers a minute.
  3. (a) Machine X packs 3 × 15 = 45 hampers a minute.
  4. Machine Y packs 2 × 15 = 30 hampers a minute.
  5. Machine X works alone for 20 minutes and packs 20 × 45 = 900 hampers, so 3600 − 900 = 2700 hampers are left.
  6. (b) Machine Y packs the 2700 hampers in 2700 ÷ 30 = 90 minutes. Check: 900 + 90 × 30 = 900 + 2700 = 3600 hampers.

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

techniqueRatio · Rates

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Giving Machine X 2 parts and Machine Y 3 parts. Machine X is the faster machine, so it gets the bigger share: 1.5 : 1 = 3 : 2 puts the 3 parts on Machine X.
  • Working out the first 20 minutes with the joint rate: 20 × 75 = 1500 hampers. Only Machine X was packing then, so it packed 20 × 45 = 900 hampers.
04

Two Gardeners Take Turns, One Hour Each, to Clear a Plot

heuristicMake the Plot 24 Units and Count 5 Units for Each Pair of Hours / Follow the Last Few Hours One at a Time

Darren and Ethan are clearing weeds from a community garden. Working alone, Darren can clear the plot in 8 hours, and Ethan can clear it in 12 hours. They take turns of 1 hour each: Darren works in the 1st hour, Ethan 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 long does it take altogether to clear the whole plot, in hours and minutes?

Plot24 unitsHoursD 3E 2
The plot is 24 units: Darren clears 3 an hour, Ethan 2.
Call the plot 24 units, the lowest common multiple of 8 and 12. Darren clears 24 ÷ 8 = 3 units an hour, and Ethan clears 24 ÷ 12 = 2 units an hour.
step 1 of 6

Cut the plot into 24 units, because 24 is the lowest common multiple of 8 and 12. Each pair of hours, one Darren's and one Ethan's, clears the same number of units, so count whole pairs first and then follow the last hours one at a time.

All steps
  1. Call the plot 24 units, the lowest common multiple of 8 and 12. Darren clears 24 ÷ 8 = 3 units an hour, and Ethan clears 24 ÷ 12 = 2 units an hour.
  2. (a) In the first 2 hours, Darren works one hour and Ethan works one hour: 3 + 2 = 5 units, which is 524 of the garden.
  3. Every pair of hours clears 5 units. Four pairs, which is 8 hours, clear 4 × 5 = 20 units, and 24 − 20 = 4 units are left. That is less than another 5, so follow the hours one at a time now.
  4. The 9th hour is Darren's turn. He clears 3 units, and 4 − 3 = 1 unit is left.
  5. The 10th hour is Ethan's turn. He clears 2 units in an hour, so 1 unit takes him half an hour, which is 30 minutes.
  6. (b) Altogether it takes 8 + 1 + 12 = 912 hours, which is 9 hours 30 minutes. Check: Darren worked 5 hours and cleared 5 × 3 = 15 units; Ethan worked 412 hours and cleared 412 × 2 = 9 units; 15 + 9 = 24.

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

techniqueRates · Remainders

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Using the mean rate: (3 + 2) ÷ 2 = 2.5 units an hour, so 24 ÷ 2.5 = 9.6 hours. The mean holds only over whole pairs of hours. After 8 hours the 4 units left are cleared by Darren's full hour at 3 units and half of Ethan's hour, not at 2.5 units an hour, so the last turns must be followed hour by hour.
  • Giving the 9th hour to Ethan. The turns go Darren, Ethan, Darren, and so on, so every odd hour, the 9th included, is Darren's.
05

A Craftsman Paid for Each Good Lantern and Fined for Each Flawed One

heuristicSuppose Every Lantern Was Good / Each Flawed Lantern Costs the Pay and the Fine

A craftsman is paid $6.00 for every glass lantern he makes without a flaw. For a lantern with a flaw he is paid nothing, and he is fined $9.00 for the wasted glass. In one week he made 150 lanterns and received $720.00 in all. (a) How many lanterns had a flaw? (b) How much more money would he have received if none of the lanterns had a flaw?

If all good150 × $6 = $900
Suppose every lantern was good: $900.
Suppose all 150 lanterns were good. He would receive 150 × $6 = $900.
step 1 of 5

Suppose all 150 lanterns were good, and compare what he would have received with what he did receive. Each flawed lantern accounts for the same part of the gap.

All steps
  1. Suppose all 150 lanterns were good. He would receive 150 × $6 = $900.
  2. He received $720, which is $900 − $720 = $180 less.
  3. Each lantern with a flaw loses him the $6 he would have been paid, and he pays the $9 fine as well, so each one costs him $6 + $9 = $15.
  4. (a) The $180 gap is made of $15 losses, so 180 ÷ 15 = 12 lanterns had a flaw.
  5. (b) With no flawed lanterns he would have received the $900, which is $180 more than he did. Check: 150 − 12 = 138 good lanterns pay 138 × 6 = $828, and 828 − 12 × 9 = 828 − 108 = $720.

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

techniqueRate, Total, Units · Forming Equations

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking $9 as the difference each flawed lantern makes: 180 ÷ 9 = 20. A flawed lantern also loses him the $6 he would have been paid, so each one makes a difference of $6 + $9 = $15.
  • Taking the difference $9 − $6 = $3: 180 ÷ 3 = 60. The $6 is money he does not get and the $9 is money he pays out, so the two losses add up; they do not cancel.
06

Food Stored for a Camp, and Some Campers Leave Early

heuristicCount the Food Left in Rations / Share Them Among the Campers Who Stay

A youth camp stored enough food to feed 40 campers for 24 days, and each camper eats the same amount each day. After 6 days, 10 campers had to leave the camp because of bad weather. (a) For how many more days will the food that is left feed the campers who stay? (b) How many days longer did the food last in all than was first planned?

Planned6 days18 days for 4024 days
After 6 days, the food left would feed 40 campers for 18 more days.
After 6 days, the food left would have fed the 40 campers for 24 − 6 = 18 more days.
step 1 of 5

After 6 days, the food left was planned to feed 40 campers for 18 more days. Count it in rations, one ration for one camper for one day, and share it among the 30 campers who stay.

All steps
  1. After 6 days, the food left would have fed the 40 campers for 24 − 6 = 18 more days.
  2. That is 40 × 18 = 720 rations, one ration for each camper for each day.
  3. After 10 campers leave, 40 − 10 = 30 campers stay, and they eat 30 rations a day.
  4. (a) The 720 rations last them 720 ÷ 30 = 24 more days.
  5. (b) The plan had 18 days of food left, and the food now lasts 24 more days, so it lasts 24 − 18 = 6 days longer. Check: 6 + 24 = 30 days in all, against the 24 days planned.

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

techniqueInverse Proportion · Rates

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Sharing all 960 rations among the 30 campers: 960 ÷ 30 = 32 days. The 40 campers had already eaten 240 rations in the first 6 days, so only 720 rations are left to share.
  • Giving 24 days as the answer to (b). The 24 days are the days the food lasts after day 6, and the plan had 18 days left at that point, so the food lasts only 24 − 18 = 6 days longer.
07

A Water Bill Charged in Three Price Bands

heuristicFill the Cheapest Band First / Take Away Each Full Band's Cost to Find the Water in the Top Band

A water company charges for water in three price bands. The first 20 m3 costs $1.20 per m³. The next 15 m3, from 20 m3 up to 35 m3, costs $1.60 per m³. Every cubic meter above 35 m3 costs $2.40. (a) The Tan family used 32 m3 of water in March. What was their water bill? (b) The Lee family's water bill for April was $72.00. How many cubic meters of water did the Lee family use?

32 m³20 × $1.20 = $2412 × $1.60 = $19.20
32 m³ is 20 m³ in the first band and 12 m³ in the second.
The Tan family's 32 m3 is the first 20 m3 at $1.20 and the next 32 − 20 = 12 m3 at $1.60.
step 1 of 6

Draw the water as three blocks, one for each price band. For part (a), fill the blocks from the cheapest. For part (b), take away the cost of each full block from the bill until only the top band is left.

All steps
  1. The Tan family's 32 m3 is the first 20 m3 at $1.20 and the next 32 − 20 = 12 m3 at $1.60.
  2. (a) Their bill is 20 × 1.20 + 12 × 1.60 = 24.00 + 19.20 = $43.20.
  3. The Lee family's first 20 m3 cost 20 × 1.20 = $24.00, which leaves 72.00 − 24.00 = $48.00 of the bill.
  4. Their next 15 m3 cost 15 × 1.60 = $24.00, which leaves 48.00 − 24.00 = $24.00 for the water above 35 m3.
  5. Water above 35 m3 costs $2.40 per m³, so $24.00 pays for 24.00 ÷ 2.40 = 10 m3.
  6. (b) The Lee family used 20 + 15 + 10 = 45 m3. Check: 24.00 + 24.00 + 10 × 2.40 = $72.00.

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

techniqueRate, Total, Units · Money as Decimals

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Charging all 32 m3 at one price, such as 32 × 1.60 = $51.20. Only the water above 20 m3 costs $1.60; the first 20 m3 always cost $1.20 per m³.
  • Dividing the whole bill by the top price: 72 ÷ 2.40 = 30 m3. Only the part of the bill above $48.00 buys water at $2.40; the first $48.00 paid for the first 35 m3 at the lower prices.
08

Three Painters Whose Times Are Known Only in Pairs

heuristicAdd the Three Pairs, Which Counts Each Painter Twice, and Halve / Take Away a Pair to Leave One Painter

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. Ken and Mark working together take 15 days. (a) How many days would all three painters take to paint the school working together? (b) How many days would Liam take to paint the whole school working alone?

The school is 60 units of painting.K + L12 daysL + M20 daysK + M15 days
Call the school 60 units, the lowest common multiple of 12, 20 and 15.
Call the whole school 60 units, the lowest common multiple of 12, 20 and 15.
step 1 of 6

Cut the school into 60 units, the lowest common multiple of 12, 20 and 15, and find each pair's units in a day. Putting the three pairs side by side shows two Kens, two Liams and two Marks.

All steps
  1. Call the whole school 60 units, the lowest common multiple of 12, 20 and 15.
  2. In one day, Ken and Liam paint 60 ÷ 12 = 5 units, Liam and Mark paint 60 ÷ 20 = 3 units, and Ken and Mark paint 60 ÷ 15 = 4 units.
  3. Put the three pairs together: 5 + 3 + 4 = 12 units a day. Each painter is in two of the pairs, so this is two Kens, two Liams and two Marks.
  4. So one Ken, one Liam and one Mark together paint 12 ÷ 2 = 6 units a day.
  5. (a) All three paint the 60 units in 60 ÷ 6 = 10 days.
  6. (b) Ken and Mark paint 4 of those 6 units, so Liam paints 6 − 4 = 2 units a day, and alone he needs 60 ÷ 2 = 30 days. Check: Ken paints 5 − 2 = 3 units a day and Mark 3 − 2 = 1, and 3 + 1 = 4, as Ken and Mark should.

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

techniqueRates · Simultaneous by Elimination

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking 12 units a day as the three painters' rate, which gives 60 ÷ 12 = 5 days. Adding the three pairs counts each painter twice, so the three together paint only 12 ÷ 2 = 6 units a day.
  • Taking the mean of the three times: (12 + 20 + 15) ÷ 3 ≈ 15.7 days. Three painters working at once finish sooner than any pair of them, so the answer must be less than 12 days.
09

A Reservoir Filled by One Pipe While Water Leaves It Two Ways

heuristicFind the Water Gained Each Minute, In Less Out / With the Intake Shut, Only the Water Out Counts

A village reservoir holds 36000 liters when it is full, and it now holds 24000 liters. An intake pipe fills it at 100 liters a minute. At the same time, the reservoir sends 70 liters a minute to the village, and a crack leaks 10 liters a minute. (a) How many hours will it take to fill the reservoir? (b) The intake pipe is closed the moment the reservoir is full. The village supply and the leak carry on at the same rates. How long will it take for the reservoir to empty, in hours and minutes?

Reservoir24000 in12000 to fill36000
The reservoir is 36000 − 24000 = 12000 liters short of full.
The reservoir needs 36000 − 24000 = 12000 liters more to be full.
step 1 of 6

Draw the reservoir as a bar with the water it holds and the space still empty. Each minute, compare the water that comes in with the water that goes out.

All steps
  1. The reservoir needs 36000 − 24000 = 12000 liters more to be full.
  2. Each minute, 100 liters come in, and 70 + 10 = 80 liters go out to the village and through the crack.
  3. So each minute the reservoir gains 100 − 80 = 20 liters.
  4. (a) 12000 liters at 20 liters a minute take 12000 ÷ 20 = 600 minutes, which is 600 ÷ 60 = 10 hours.
  5. Once the intake is closed, nothing comes in, and 80 liters go out each minute from the full 36000 liters.
  6. (b) The reservoir empties in 36000 ÷ 80 = 450 minutes. 450 = 7 × 60 + 30, so that is 7 hours 30 minutes. Check: 450 × 80 = 36000 liters.

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

techniqueRates · Rate, Total, Units

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the whole 36000 liters by 20 to find the filling time: 1800 minutes. The reservoir already holds 24000 liters, so only the 12000 liters still missing have to come in.
  • Taking the leak away from the village supply: 70 − 10 = 60 liters a minute out. The water sent to the village and the water lost through the crack both leave the reservoir, so together 70 + 10 = 80 liters leave it each minute.
10

A Workshop Makes Gears at Three Different Rates

heuristicRate × Hours for Each Stretch of Time / Divide the Gears Left by the Last Rate

A workshop has to make 720 gears. For the first 5 hours, its machine made 40 gears an hour. The machine then grew too hot, so for the next 4 hours it made 25% fewer gears each hour. After that, a second machine was switched on beside the first, and together they made 80 gears an hour until the order was finished. (a) How many gears were made in the first 9 hours? (b) How many hours did the whole order take, from start to finish?

Hours5 h × 40Gears200
The first 5 hours: 5 × 40 = 200 gears.
In the first 5 hours the workshop made 5 × 40 = 200 gears.
step 1 of 6

Cut the order into three stretches of time, each at its own rate. Rate × hours gives the gears made in each stretch, and the gears left give the hours of the last stretch.

All steps
  1. In the first 5 hours the workshop made 5 × 40 = 200 gears.
  2. 25% fewer gears an hour means 14 of 40 fewer, which is 40 ÷ 4 = 10 fewer, so the hot machine made 40 − 10 = 30 gears an hour.
  3. In the next 4 hours it made 4 × 30 = 120 gears.
  4. (a) In the first 5 + 4 = 9 hours the workshop made 200 + 120 = 320 gears.
  5. That left 720 − 320 = 400 gears, which the two machines made at 80 an hour in 400 ÷ 80 = 5 hours.
  6. (b) The whole order took 5 + 4 + 5 = 14 hours. Check: 200 + 120 + 5 × 80 = 720 gears.

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

techniqueRate, Total, Units · Rates

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking 25% off the 4 hours instead of off the rate. The machine worked all 4 hours; what fell by 25% was the number of gears it made each hour, from 40 to 30.
  • Giving 5 hours as the answer to (b). That is only the time the two machines worked together; the order had already taken 9 hours before that, so it took 9 + 5 = 14 hours in all.
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 bag$1.20$1.20$1.20$1.20$1.20$6.00100 g: $2.40
The 250 g bag is 5 lots of 50 g, each 6.00 ÷ 5 = $1.20, so each 100 g costs $2.40.
The 250 g bag is 5 lots of 50 g, so each 50 g costs 6.00 ÷ 5 = $1.20, and each 100 g costs 2 × 1.20 = $2.40.
step 1 of 6

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.

All steps
  1. The 250 g bag is 5 lots of 50 g, so each 50 g costs 6.00 ÷ 5 = $1.20, and each 100 g costs 2 × 1.20 = $2.40.
  2. The 400 g bag is 4 lots of 100 g, so each 100 g costs 8.80 ÷ 4 = $2.20.
  3. (a) The 400 g bag is the better buy: each 100 g costs 2.40 − 2.20 = $0.20 less.
  4. 2 kg is 2000 g, which is 2000 ÷ 400 = 5 bags of 400 g: 5 × 8.80 = $44.00.
  5. In 250 g bags, 2000 g is 2000 ÷ 250 = 8 bags: 8 × 6.00 = $48.00.
  6. (b) The café saves 48.00 − 44.00 = $4.00 a week. Check: 2000 g is 20 lots 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$800Euros9090909090909090720 euros
Cut $800 into 8 units of $100; each unit buys 100 × 0.90 = 90 euros, so Lena received 8 × 90 = 720 euros.
Cut $800 into 8 units of $100. At $1 = €0.90, each unit buys 100 × 0.90 = 90 euros, so Lena received 8 × 90 = 720 euros.
step 1 of 6

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.

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

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.

All steps
  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. (a) The six blocks have a mass of 6 × 9000 = 54000 g, which is 54 kg.
  4. 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.
  5. Each cubic centimeter of the stopper has a mass of 60 ÷ 50 = 1.2 g.
  6. (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 meterBoots12 000boots: 2 × 0.03 = 0.06 m2720 N on 0.06 m2: 12 000 N per m2
In boots, 720 N presses on 2 × 0.03 = 0.06 m2: 720 ÷ 0.06 = 12000 N/m2.
In boots, the hiker's 720 N presses on 2 × 0.03 = 0.06 m2 of snow, so the pressure is 720 ÷ 0.06 = 12000 N/m2.
step 1 of 6

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.

All steps
  1. In boots, the hiker's 720 N presses on 2 × 0.03 = 0.06 m2 of snow, so the pressure is 720 ÷ 0.06 = 12000 N/m2.
  2. In snowshoes, 720 N presses on 2 × 0.12 = 0.24 m2, so the pressure is 720 ÷ 0.24 = 3000 N/m2.
  3. (a) 12000 N/m2 in boots and 3000 N/m2 in snowshoes: four times the area gives a quarter of the pressure.
  4. At 3500 N/m2, the 0.24 m2 under the snowshoes holds up at most 3500 × 0.24 = 840 N.
  5. The hiker is 720 N of that, so 840 − 720 = 120 N is left for the pack.
  6. (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.
15

Juice Boxes in Packs of 6 and 10: The Price of One Box, and the Most a Budget Buys

heuristicUnit Price: Divide Each Pack's Price by Its Boxes / Count Only the Whole Packs the Money Pays For

For a class party, Maya compares two packs of the same juice boxes. A pack of 6 boxes costs $4.20, and a pack of 10 boxes costs $7.50. The boxes are sold only in whole packs. (a) Find the price of one juice box in each pack. Which pack is the better buy? (b) Maya has $21 to spend. How many juice boxes can she buy if she buys only packs of 6? How many can she buy if she buys only packs of 10?

The price of each boxPack of 6$0.70$0.70$0.70$0.70$0.70$0.70$4.20Pack of 10$7.50$4.20 ÷ 6 = $0.70 a box
Share $4.20 among the 6 boxes: 4.20 ÷ 6 = $0.70 a box.
Share $4.20 among the 6 boxes in the small pack: 4.20 ÷ 6 = $0.70 a box.
step 1 of 6

Share each pack's price equally among its boxes to find the price of one box, and compare. Then count how many whole packs of each size $21 pays for.

All steps
  1. Share $4.20 among the 6 boxes in the small pack: 4.20 ÷ 6 = $0.70 a box.
  2. Share $7.50 among the 10 boxes in the big pack: 7.50 ÷ 10 = $0.75 a box.
  3. (a) One box costs 0.75 − 0.70 = $0.05 less in the pack of 6, so the smaller pack is the better buy.
  4. Packs of 6: 5 × 4.20 = $21.00, which is all of the $21. Five packs hold 5 × 6 = 30 boxes.
  5. Packs of 10: 2 × 7.50 = $15.00, and a third pack would bring the cost to $22.50, more than $21. Two packs hold 20 boxes, and $6.00 is left over.
  6. (b) Maya can buy 30 boxes in packs of 6, but only 20 boxes in packs of 10. Check: 30 × 0.70 = $21.00 and 20 × 0.75 = $15.00.

answer(a) $0.70 a box in the pack of 6 and $0.75 a box in the pack of 10, so the pack of 6 is the better buy; (b) 30 boxes in packs of 6, and 20 boxes in packs of 10

techniqueComparing Unit Rates · Rates

examsSAT · GCSE Higher

Common pitfalls

  • Deciding that the bigger pack must be the better buy. The pack of 10 holds more boxes, but each of its boxes costs $0.75, which is more than the $0.70 a box in the pack of 6. Only the price of one box can tell which pack is the better buy.
  • Working out 21 ÷ 0.75 = 28 boxes for the packs of 10. The boxes are sold only in whole packs, and 28 boxes would be 2.8 packs, so Maya can buy only 2 packs, which is 20 boxes.
16

A Skateboard Priced in Euros on a Trip and in Dollars at Home

heuristicChange Both Prices into One Currency / Multiply Toward Euros, Divide Back to Dollars

On a school trip to Spain, Leo sees a skateboard priced at 36 euros. At home, the same skateboard costs $40. The exchange rate is $1 = 0.8 euros. (a) How many euros is the home price of $40 worth? (b) How many dollars is the price in Spain worth? Where is the skateboard cheaper, and by how many dollars?

One unit$5= 4 eurosin euros4
At $1 = 0.8 euros, $5 buys 5 × 0.8 = 4 euros: one unit.
At $1 = 0.8 euros, $5 buys 5 × 0.8 = 4 euros. So one unit is $5, which is the same money as 4 euros.
step 1 of 5

At $1 = 0.8 euros, every $5 buys 5 × 0.8 = 4 euros. Cut both prices into equal units, each unit being $5 or 4 euros, and compare the numbers of units.

All steps
  1. At $1 = 0.8 euros, $5 buys 5 × 0.8 = 4 euros. So one unit is $5, which is the same money as 4 euros.
  2. The home price, $40, is 40 ÷ 5 = 8 units.
  3. (a) 8 units are 8 × 4 = 32 euros. Check: 40 × 0.8 = 32.
  4. The price in Spain, 36 euros, is 36 ÷ 4 = 9 units.
  5. (b) 9 units are 9 × 5 = $45. That is 45 − 40 = $5 more than at home, so the skateboard is cheaper at home, by $5. The gap is one unit, which is also 4 euros.

answer(a) 32 euros; (b) $45, so the skateboard is cheaper at home, by $5

techniqueExchange Rates

Common pitfalls

  • Comparing 36 with 40 and saying the skateboard is cheaper in Spain. The two numbers are in different currencies, so they cannot be compared until both prices are in dollars, or both are in euros. 36 euros is $45, which is more than $40.
  • Multiplying the euros by the rate: 36 × 0.8 = 28.8, and calling it $28.80. The rate $1 = 0.8 euros changes dollars into euros by multiplying. Going from euros back to dollars divides: 36 ÷ 0.8 = $45.
17

Two Blocks of the Same Size, Aluminum and Pine, Compared by Density

heuristicDensity = Mass ÷ Volume / Volume = Mass ÷ Density

In a science class, Sofia has two blocks of the same size. Each block is 10 cm long, 5 cm wide and 4 cm tall. The aluminum block has a mass of 540 g, and the pine block has a mass of 100 g. (a) Find the density of each block, in g/cm3. (b) Sofia wants a block of pine with the same mass as the aluminum block. What volume of pine does she need?

Two blocks of the same size, 200 cubic centimeters eachAluminum540 gPine100 g10 × 5 × 4 = 200 cm3each
Each block has a volume of 10 × 5 × 4 = 200 cm3.
Each block has a volume of 10 × 5 × 4 = 200 cm3.
step 1 of 6

The two blocks have the same volume, so compare the mass in each cubic centimeter of them. Then use the density of pine the other way around: count the cubic centimeters of pine that make 540 g.

All steps
  1. Each block has a volume of 10 × 5 × 4 = 200 cm3.
  2. The aluminum block's 540 g are shared over 200 cm3: 540 ÷ 200 = 2.7 g in each cubic centimeter.
  3. The pine block's 100 g are shared over 200 cm3: 100 ÷ 200 = 0.5 g in each cubic centimeter.
  4. (a) The density of the aluminum is 2.7 g/cm3, and the density of the pine is 0.5 g/cm3. The blocks are the same size, so the heavier one is the denser one here.
  5. Each cubic centimeter of pine holds 0.5 g, so 2 cubic centimeters hold 1 g. For 540 g, the pine must fill 540 × 2 = 1080 cm3.
  6. (b) She needs 1080 cm3 of pine, which is 1080 ÷ 200 = 5.4 times the volume of the aluminum block. Check: 1080 × 0.5 = 540 g.

answer(a) aluminum 2.7 g/cm3 and pine 0.5 g/cm3; (b) 1080 cm3

techniqueDensity

examsSAT · GCSE Higher

Common pitfalls

  • Multiplying in part (b): 540 × 0.5 = 270, and answering 270 cm3. That much pine has a mass of only 270 × 0.5 = 135 g. Each cubic centimeter of pine holds only half a gram, so 540 g of pine fills more than 540 cm3: divide the mass by the density.
  • Deciding that the new pine block is as dense as the aluminum block, because the two have the same mass. The pine needs 5.4 times the volume for that mass, so its density is still 540 ÷ 1080 = 0.5 g/cm3. Density belongs to the material, not to the size of the block.
18

A Dancer Standing Flat on Both Feet and on the Toes of One Foot

heuristicPressure = Force ÷ Area / The Same Force on a Smaller Area Gives a Bigger Pressure

A ballet dancer weighs 480 N. When she stands flat on both feet, each foot touches the floor over an area of 120 cm2. When she stands on the tips of the toes of one foot, only 8 cm2 of her shoe touches the floor. (a) What pressure does she put on the floor when she stands flat on both feet, in N/cm2? (b) What pressure does she put on the floor when she stands on the toes of one foot? How many times the pressure in part (a) is that?

Floor touched, in square centimetersBoth feet1201202402 × 120 = 240 cm2
Standing flat, both feet touch the floor: 2 × 120 = 240 cm2.
Standing flat, both of her feet touch the floor: 2 × 120 = 240 cm2.
step 1 of 5

Pressure is the force on each square centimeter. Share the dancer's weight over the area that touches the floor. Her weight is the same both times; only the area changes.

All steps
  1. Standing flat, both of her feet touch the floor: 2 × 120 = 240 cm2.
  2. (a) Her 480 N are shared over 240 cm2: 480 ÷ 240 = 2 N on each square centimeter. The pressure is 2 N/cm2.
  3. On the toes of one foot, the same 480 N are shared over only 8 cm2: 480 ÷ 8 = 60 N on each square centimeter.
  4. (b) The pressure is 60 N/cm2, and 60 ÷ 2 = 30, so it is 30 times the pressure when she stands flat.
  5. Check with the areas: 240 ÷ 8 = 30, so her two flat feet touch 30 times as much floor as her toes. The same force on a thirtieth of the area gives 30 times the pressure.

answer(a) 2 N/cm2; (b) 60 N/cm2, which is 30 times the pressure in part (a)

techniquePressure

examsGCSE Higher

Common pitfalls

  • Using the area of one foot when she stands flat: 480 ÷ 120 = 4 N/cm2. She stands on both feet, so her weight is spread over 2 × 120 = 240 cm2.
  • Thinking that the pressure stays the same because her weight does not change. Pressure is the force on each square centimeter, and on her toes the 480 N are shared among only 8 square centimeters, so each one takes much more.