Speed · applications

Applications: Speed

16 problems

O-Level · SAT · GCSE Higher

01

A Car's Distance in 45 Minutes, and a Cyclist's Speed in Meters per Minute

heuristicCut the Hour into Quarters / Change Kilometers into Meters and Hours into Minutes

A car traveled from Town P to Town Q at a steady speed of 72 km/h. The trip took 45 minutes. (a) How far is it from Town P to Town Q, in kilometers? (b) A cyclist took 3 hours to ride the same road. What was the cyclist's speed in meters per minute (m/min)?

1 hour60 min · 72 km72 km/h
72 km/h is 72 km in every hour, and an hour is 60 minutes.
72 km/h means the car covers 72 km in every hour, and an hour is 60 minutes.
step 1 of 6

72 km/h means 72 km in every 60 minutes. Cut the hour into four quarters of 15 minutes to find the distance in 45 minutes. Then share the cyclist's distance, in meters, first over the hours and then over the minutes.

All steps
  1. 72 km/h means the car covers 72 km in every hour, and an hour is 60 minutes.
  2. Cut the hour into 4 equal parts of 15 minutes. In each 15 minutes the car covers 72 ÷ 4 = 18 km.
  3. (a) 45 minutes is 3 of those parts, so the car covers 3 × 18 = 54 km. It is 54 km from Town P to Town Q.
  4. The cyclist rides the same 54 km. One kilometer is 1000 meters, so that is 54 × 1000 = 54000 m.
  5. The cyclist takes 3 hours, so in each hour the cyclist rides 54000 ÷ 3 = 18000 m.
  6. (b) An hour is 60 minutes, so in each minute the cyclist rides 18000 ÷ 60 = 300 m. The speed is 300 m/min. Check: in 3 hours, which is 180 minutes, the cyclist rides 300 × 180 = 54000 m.

answer(a) 54 km; (b) 300 m/min

techniqueSpeed · Converting Compound Units

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Multiplying the speed by the minutes: 72 × 45 = 3240 km. The speed is in kilometers per hour, so the time must be in hours: 45 minutes is 34 of an hour, and 72 × 34 = 54 km.
  • Writing 45 minutes as 0.45 hours. An hour has 60 minutes, not 100, so 45 minutes is 45 ÷ 60 = 0.75 hours.
02

A Van's Trip in Two Legs at Two Different Speeds

heuristicFind the Time for Each Leg / Whole Distance ÷ Whole Time

A delivery van drove 80 km from Town A to Town B at an average speed of 40 km/h. It then drove 180 km from Town B to Town C at an average speed of 60 km/h. (a) How long did the whole trip from Town A to Town C take? (b) What was the van's average speed for the whole trip?

A to B40 km · 1 h40 km · 1 h80 km in 2 h
A to B: 80 km at 40 km/h. Each hour covers 40 km, so it takes 2 hours.
The first leg is 80 km at 40 km/h. Each hour covers 40 km, so the first leg takes 80 ÷ 40 = 2 hours.
step 1 of 5

Draw each leg as a bar cut into hours, each hour holding the distance the van covers in that hour. Count the hours, add the distances, and share the whole distance equally over the whole time.

All steps
  1. The first leg is 80 km at 40 km/h. Each hour covers 40 km, so the first leg takes 80 ÷ 40 = 2 hours.
  2. The second leg is 180 km at 60 km/h. Each hour covers 60 km, so the second leg takes 180 ÷ 60 = 3 hours.
  3. (a) The whole trip takes 2 + 3 = 5 hours.
  4. The whole distance is 80 + 180 = 260 km.
  5. (b) Share the 260 km equally over the 5 hours: 260 ÷ 5 = 52 km in each hour. The average speed is 52 km/h. Check: 52 km/h is between 40 and 60 km/h, and nearer 60, because the van spent more hours at 60 km/h.

answer(a) 5 hours; (b) 52 km/h

techniqueAverage Speed

examsO-Level · SAT

Common pitfalls

  • Taking the mean of the two speeds: (40 + 60) ÷ 2 = 50 km/h. The van drove for 2 hours at 40 km/h but for 3 hours at 60 km/h, so the faster speed lasted longer and pulls the average above 50 km/h.
  • Adding the two speeds: 40 + 60 = 100 km/h. The van never went faster than 60 km/h, so its average speed cannot be more than 60 km/h.
03

Two Cyclists Set Off Together at Different Speeds

heuristicDistances in the Ratio of the Speeds / The Gap Grows by the Same Amount Each Hour

Keith and Leonard set off on their bikes at the same time from the same gate and rode along a park trail in the same direction. Keith rode at 24 km/h and Leonard rode at 16 km/h. (a) How far apart were they after 45 minutes? (b) How long after they set off were they 10 km apart?

Keith3 units24 km/hLeonard2 units16 km/h
Same start and same time, so the distances are in the ratio of the speeds: 24 : 16 = 3 : 2.
Keith and Leonard ride for the same time, so their distances are in the same ratio as their speeds: 24 : 16 = 3 : 2.
step 1 of 5

The two ride for the same time, so the distances they cover are in the ratio of their speeds. Draw Keith's distance as 3 units and Leonard's as 2 units; the gap is 1 unit, and 1 unit is how far apart they get in any time.

All steps
  1. Keith and Leonard ride for the same time, so their distances are in the same ratio as their speeds: 24 : 16 = 3 : 2.
  2. So at any moment Keith has ridden 3 units while Leonard has ridden 2 units. The gap between them is 1 unit.
  3. In 1 hour Keith rides 24 km, which is 3 units, so 1 unit is 24 ÷ 3 = 8 km. The gap grows by 8 km every hour.
  4. (a) 45 minutes is 34 of an hour, so after 45 minutes the gap is 34 × 8 = 6 km.
  5. (b) A gap of 10 km takes 10 ÷ 8 = 114 hours, which is 1 hour 15 minutes. Check: in 114 hours Keith rides 30 km and Leonard rides 20 km, and 30 − 20 = 10 km.

answer(a) 6 km; (b) 1 hour 15 minutes

techniqueSpeed · Direct Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Answering part (a) with Keith's distance alone: 24 × 34 = 18 km. That is how far Keith is from the gate. Leonard has also moved 16 × 34 = 12 km, so they are only 18 − 12 = 6 km apart.
  • Reading 1.25 hours as 1 hour 25 minutes. The 0.25 is a quarter of an hour, and a quarter of 60 minutes is 15 minutes.
04

The Same Road Driven There and Back at Two Speeds

heuristicFaster Speed, Less Time / Times in the Ratio 4 : 3

Mr. Lee drove from Town P to Town Q at an average speed of 60 km/h. He drove back along the same road at 80 km/h, and the trip back took 30 minutes less than the trip there. (a) How long did the trip from Town P to Town Q take? (b) What is the distance between Town P and Town Q?

Speed there60 km/h3 partsSpeed back80 km/h4 parts
Same road both ways. Speeds 60 : 80 = 3 : 4, so the times are the other way around, 4 : 3.
The road is the same both ways, so speed × time is the same both ways. The speeds are in the ratio 60 : 80 = 3 : 4, so the times are the other way around, 4 : 3.
step 1 of 5

Draw the time for each trip as a bar of equal units. The faster trip takes fewer units. The difference of 1 unit is the 30 minutes saved.

All steps
  1. The road is the same both ways, so speed × time is the same both ways. The speeds are in the ratio 60 : 80 = 3 : 4, so the times are the other way around, 4 : 3.
  2. Draw the trip there as 4 units of time and the trip back as 3 units. The trip back takes 1 unit less.
  3. The trip back took 30 minutes less, so 1 unit is 30 minutes.
  4. (a) The trip there took 4 × 30 = 120 minutes, which is 2 hours.
  5. (b) The distance is 60 × 2 = 120 km. Check: the trip back took 3 × 30 = 90 minutes, which is 112 hours, and 80 × 112 = 120 km.

answer(a) 2 hours; (b) 120 km

techniqueInverse Proportion · Speed

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Giving the times the same ratio as the speeds, 3 : 4. The trip back is faster, so it must take less time, not more: the times are the other way around, 4 : 3.
  • Using 30 as the unit in hours: 4 × 30 = 120 hours, and 60 × 120 = 7200 km. The 30 is minutes, so 4 units are 120 minutes, which is 2 hours.
05

Two Cars Drive Toward Each Other from Two Towns

heuristicAdd the Speeds to Find How Fast the Gap Closes

Town A and Town B are 350 km apart. At 8:00 a.m., Car X left Town A and drove toward Town B at a steady 65 km/h. At the same time, Car Y left Town B and drove toward Town A at a steady 75 km/h. (a) How long did the cars take to meet? (b) How far was Car X from Town B when they met?

Town ATown B350 kmX 162.5 kmY 187.5 km10:30 a.m. · 2 h 30 min after 8:00they meet 162.5 km from A, 187.5 km from B
Each hour Car X drives 65 km in from the left and Car Y drives 75 km in from the right.
The cars drive toward each other. Each hour Car X covers 65 km of the road between them, and Car Y covers 75 km.
step 1 of 6

Draw the 350 km road with Car X starting at one end and Car Y at the other. Each hour the two cars together cover 140 km of the road between them.

All steps
  1. The cars drive toward each other. Each hour Car X covers 65 km of the road between them, and Car Y covers 75 km.
  2. Together they close 65 + 75 = 140 km of the gap every hour.
  3. In 2 hours they close 2 × 140 = 280 km, which leaves 350 − 280 = 70 km. That is half of 140 km, so it takes another half hour.
  4. (a) The cars meet after 212 hours, which is 2 hours 30 minutes, at 10:30 a.m.
  5. When they meet, Car X's distance from Town B is the road Car Y has already driven from Town B.
  6. (b) Car Y has driven 75 × 212 = 187.5 km, so Car X is 187.5 km from Town B. Check: Car X has driven 65 × 212 = 162.5 km, and 162.5 + 187.5 = 350 km.

answer(a) 2.5 hours (2 hours 30 minutes); (b) 187.5 km

techniqueSpeed · Rates

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Subtracting the speeds: 75 − 65 = 10 km/h. The cars drive toward each other, so both close the gap at once, and their speeds add: 65 + 75 = 140 km/h.
  • Giving 162.5 km for part (b). That is how far Car X is from Town A. The question asks how far it still is from Town B: 350 − 162.5 = 187.5 km.
06

A Motorcycle Catches Up with a Truck That Left Earlier

heuristicFind the Head Start / The Gap Closes by the Difference in Speed

At 9:00 a.m., a truck left a warehouse and drove along a highway at a steady 50 km/h. At 10:30 a.m., a motorcycle left the same warehouse and followed the same road at 80 km/h. (a) At what time did the motorcycle catch up with the truck? (b) How far had the motorcycle driven when it caught up with the truck?

1:00 p.m. · 2 h 30 min after 10:30Truck75 km head start125 km200 kmMotorcycle200 km200 kmcaught up: 200 km each
By 10:30 the truck has driven 112 hours at 50 km/h: a 75 km head start.
From 9:00 a.m. to 10:30 a.m. is 112 hours. By 10:30 a.m. the truck has driven 50 × 112 = 75 km, so it has a 75 km head start.
step 1 of 5

Draw both distances from the warehouse as bars, starting at 10:30 a.m., when the truck already has a head start. Each hour the motorcycle's bar grows 30 km more than the truck's, until the two bars are equal.

All steps
  1. From 9:00 a.m. to 10:30 a.m. is 112 hours. By 10:30 a.m. the truck has driven 50 × 112 = 75 km, so it has a 75 km head start.
  2. Both drive in the same direction. Each hour the motorcycle drives 80 km and the truck drives 50 km, so the gap shrinks by 80 − 50 = 30 km.
  3. To close the 75 km gap takes 75 ÷ 30 = 212 hours.
  4. (a) 212 hours after 10:30 a.m. is 1:00 p.m.
  5. (b) The motorcycle has driven 80 × 212 = 200 km. Check: the truck has driven from 9:00 a.m. to 1:00 p.m., which is 4 hours, and 50 × 4 = 200 km.

answer(a) 1:00 p.m.; (b) 200 km

techniqueSpeed · Rate, Total, Units

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the 212 hours to 9:00 a.m. and answering 11:30 a.m. The 212 hours are counted from when the motorcycle set off, at 10:30 a.m.
  • Dividing the 75 km head start by the motorcycle's speed: 75 ÷ 80. The truck keeps moving away, so the motorcycle gains only 80 − 50 = 30 km each hour.
07

Two Vans Drive Toward Each Other, One Setting Off Later

heuristicTake Away What the First Van Drove Alone / Then Close the Gap Together

Town M and Town N are 380 km apart. At 8:00 a.m., Van A left Town M and drove toward Town N at 60 km/h. At 8:30 a.m., Van B left Town N and drove toward Town M at 80 km/h. (a) How far apart were the two vans at 8:30 a.m.? (b) At what time did the two vans pass each other?

Town MTown N380 kmVan A 180 kmVan B 200 km11:00 a.m. · 2 h 30 min after 8:30A alone: 30 kmthey pass at 11:00 a.m.
From 8:00 to 8:30 only Van A moves: 60 × 12 = 30 km.
From 8:00 a.m. to 8:30 a.m. only Van A is moving. In that half hour it drives 60 × 12 = 30 km.
step 1 of 5

Draw the 380 km road. Mark off the part Van A drives before Van B sets off; the two vans close the rest of the road together.

All steps
  1. From 8:00 a.m. to 8:30 a.m. only Van A is moving. In that half hour it drives 60 × 12 = 30 km.
  2. (a) At 8:30 a.m. the vans are 380 − 30 = 350 km apart.
  3. From 8:30 a.m. both vans drive toward each other, so the gap closes by 60 + 80 = 140 km every hour.
  4. Closing 350 km takes 350 ÷ 140 = 212 hours.
  5. (b) 212 hours after 8:30 a.m. is 11:00 a.m., so the vans pass each other at 11:00 a.m. Check: by then Van A has driven for 3 hours, 60 × 3 = 180 km, and Van B for 212 hours, 80 × 212 = 200 km, and 180 + 200 = 380 km.

answer(a) 350 km; (b) 11:00 a.m.

techniqueSpeed · Rate, Total, Units

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the whole 380 km by 140 km/h. For the first 30 minutes only Van A was moving, so the two vans close the gap together only over the 350 km left at 8:30 a.m.
  • Adding the 212 hours to 8:00 a.m. and answering 10:30 a.m. The 212 hours start at 8:30 a.m., when both vans are moving.
08

Two Drivers Pass Each Other 30 km from the Halfway Point

heuristicThe Faster Driver Has Driven 60 km More / Divide by the Difference in Speed

Amy left Town X and Ben left Town Y at the same time, driving toward each other along the same road. Amy drove at 70 km/h and Ben drove at 50 km/h. When they passed each other, they were 30 km from the point halfway between the two towns. (a) How far apart are Town X and Town Y? (b) After passing Amy, how long did Ben take to reach Town X?

Town XTown YAmy 210 kmBen 150 kmM3 h of driving3030
Mark the halfway point M. Amy, the faster driver, is 30 km past M when they pass; Ben is 30 km short of it.
Mark the halfway point M of the road. Amy is faster, so when they pass each other she has gone 30 km past M, and Ben is 30 km short of M.
step 1 of 6

Draw the road with its halfway point M. Amy's bar reaches 30 km past M and Ben's stops 30 km short of M, so the difference between the two bars is 60 km.

All steps
  1. Mark the halfway point M of the road. Amy is faster, so when they pass each other she has gone 30 km past M, and Ben is 30 km short of M.
  2. So Amy has driven 30 + 30 = 60 km more than Ben.
  3. Each hour Amy drives 70 − 50 = 20 km more than Ben, so it takes 60 ÷ 20 = 3 hours for her to be 60 km ahead. They pass each other after 3 hours.
  4. In 3 hours Amy drives 70 × 3 = 210 km and Ben drives 50 × 3 = 150 km.
  5. (a) The towns are 210 + 150 = 360 km apart. Check: M is 360 ÷ 2 = 180 km from each town, and Amy's 210 km is 30 km past it.
  6. (b) After they pass, Ben still has to drive the 210 km that Amy has already driven. That takes 210 ÷ 50 = 4.2 hours, which is 4 hours 12 minutes.

answer(a) 360 km; (b) 4.2 hours (4 hours 12 minutes)

techniqueSpeed · Ratios with a Constant Difference

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Taking the difference in distance as 30 km. Amy is 30 km past the halfway point and Ben is 30 km short of it, so she has driven 30 + 30 = 60 km more than he has.
  • Answering part (b) with the 150 km Ben has already driven: 150 ÷ 50 = 3 hours. After passing Amy, Ben still has to drive the part of the road on Amy's side, which is the 210 km she drove.
09

A Journey Slowed by Traffic After the First Third

heuristicFind the Time for the First Third / Share What Is Left of the Distance over What Is Left of the Time

Mr. Tan drove 360 km from City A to City B. For the first 13 of the distance, he drove at an average speed of 80 km/h. Heavy traffic then slowed him down for the rest of the way. The whole journey took 512 hours. (a) How long did the first 13 of the journey take? (b) What was his average speed for the rest of the journey?

Distance120 km120 km120 km360 km
Three equal units of 360 ÷ 3 = 120 km.
Cut the 360 km into 3 equal units. Each unit is 360 ÷ 3 = 120 km.
step 1 of 5

Draw the 360 km as 3 equal units, one for the first third and two for the rest. Under it, draw the time: the first unit's time, and the time left over for the other two units.

All steps
  1. Cut the 360 km into 3 equal units. Each unit is 360 ÷ 3 = 120 km.
  2. (a) He drove the first 120 km at 80 km/h, which took 120 ÷ 80 = 112 hours.
  3. The rest of the journey is the other 2 units: 2 × 120 = 240 km.
  4. The whole journey took 512 hours, so the rest took 512 − 112 = 4 hours.
  5. (b) His average speed for the rest was 240 ÷ 4 = 60 km/h. Check: 112 hours at 80 km/h and 4 hours at 60 km/h cover 120 + 240 = 360 km.

answer(a) 1.5 hours; (b) 60 km/h

techniqueAverage Speed · A Fraction of an Amount

examsO-Level · SAT

Common pitfalls

  • Taking 13 of the time instead of 13 of the distance: 512 ÷ 3 hours at 80 km/h. The question says the first 13 of the distance, so it is 120 km, and its time is 120 ÷ 80 = 112 hours.
  • Taking the 112 hours away from 5 hours instead of 512 hours, and getting 312 hours for the rest. The whole journey took 512 hours, so the rest took 4 hours.
10

A Cyclist Catches Up and Then Pulls Ahead

heuristicAdd the Distance Behind and the Distance Ahead / Divide by the Difference in Speed

Cyclist A and Cyclist B rode along a straight road in the same direction. Cyclist A rode at a steady 25 km/h and Cyclist B at a steady 15 km/h. At first, Cyclist A was 8 km behind Cyclist B. (a) How long did it take until Cyclist A was 12 km ahead of Cyclist B? (b) How far did Cyclist A ride in that time?

-10-8-6-4-20+2+4+6+8+10+12+14BA +12 km2 h · measured from B, at 0
Measure from B. At the start A is 8 km behind: at −8.
Measure from Cyclist B. At the start, Cyclist A is 8 km behind B.
step 1 of 6

Measure where Cyclist A is from Cyclist B, on a line with B at 0. A starts at 8 km behind and ends at 12 km ahead, and moves 10 km closer to the end every hour.

All steps
  1. Measure from Cyclist B. At the start, Cyclist A is 8 km behind B.
  2. At the end, Cyclist A is 12 km ahead of B.
  3. To get there, A first makes up the 8 km to catch up with B, and then pulls 12 km ahead. So A gains 8 + 12 = 20 km on B.
  4. Each hour A rides 25 − 15 = 10 km more than B.
  5. (a) Gaining 20 km takes 20 ÷ 10 = 2 hours.
  6. (b) In 2 hours Cyclist A rides 25 × 2 = 50 km. Check: B rides 15 × 2 = 30 km, and A started 8 km behind B, so A ends 50 − 8 − 30 = 12 km ahead.

answer(a) 2 hours; (b) 50 km

techniqueSpeed · Rate, Total, Units

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Subtracting the two distances: 12 − 8 = 4 km. A must first make up the 8 km to catch up, and only then open the 12 km lead, so the two distances add: 8 + 12 = 20 km.
  • Dividing the 20 km by Cyclist A's speed: 20 ÷ 25. B is moving too, so each hour A gains only 25 − 15 = 10 km on B.
11

Two Walkers Meet, Turn Around at the Ends, and Meet Again

heuristicThree Lengths of the Path Together by the Second Meeting / Each Walker Goes Three Times as Far

John and Peter started walking toward each other at the same time along a straight path, John from Point X and Peter from Point Y. They met for the first time 700 m from Point X. Each walked on to the far end, turned around at once, and walked back at the same speed as before. They met for the second time 400 m from Point Y. (a) How far had John walked by the time they met for the second time? (b) What is the distance between Point X and Point Y?

XY1st meeting · 700 m from XTogether1 lengthat the 1st meeting
At the first meeting they have walked the path once between them, and John's part is 700 m.
At the first meeting, John and Peter have walked the whole path once between them. John's part of it is 700 m.
step 1 of 6

Draw the lengths of the path the two walkers cover together: 1 length at the first meeting and 3 lengths at the second. Then draw John's distance, 3 times his first 700 m, and mark it on the path.

All steps
  1. At the first meeting, John and Peter have walked the whole path once between them. John's part of it is 700 m.
  2. After that, each walks on to the far end, turns around and walks back until they meet. Together they walk the path once more to reach the two ends, and once more to meet again.
  3. So by the second meeting they have walked 1 + 2 = 3 lengths of the path together, 3 times as much as at the first meeting.
  4. (a) Their speeds do not change, so each has walked 3 times as far as before. John has walked 3 × 700 = 2100 m.
  5. John's 2100 m is the whole path from X to Y, and then 400 m back from Y.
  6. (b) So the path is 2100 − 400 = 1700 m long. Check: Peter walked 1700 − 700 = 1000 m by the first meeting, so 3 × 1000 = 3000 m by the second. That is 1700 m to X and 1300 m back, which leaves him 1700 − 1300 = 400 m from Y.

answer(a) 2100 m; (b) 1700 m

techniqueSpeed · Ratio

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Thinking they have walked the path twice together by the second meeting. After the first meeting, they walk one more length together to reach the two ends, and one more to meet again, so it is 3 lengths in all, not 2.
  • Adding the 400 m: 2100 + 400 = 2500 m. John's 2100 m is the whole path and then 400 m back toward X, so the path is 400 m shorter than 2100 m.
12

Two Cars Meet, Then Each Drives the Other's Part of the Road

heuristicCompare the Two Parts of the Road for Each Car / Guess and Check the Meeting Time

Car P left Town A and Car Q left Town B at the same time, driving toward each other at steady speeds. They met at Point M. After passing M, Car P took 2 hours to reach Town B, and Car Q took 412 hours to reach Town A. (a) How many hours after they set off did the cars meet at M? (b) Find the ratio of Car P's speed to Car Q's speed.

Car PAM: ? hMB: 2 h
Both cars drive for the same time until they meet. Car P drives AM in that time, then MB in 2 hours.
Both cars drive for the same time until they meet at M; call it the meeting time. Car P drives AM in the meeting time, and then MB in 2 hours.
step 1 of 5

Draw each car's time as a bar in two parts, one for AM and one for MB. Both cars drive for the same time until they meet. For each car, the times on AM and MB are in the ratio of the two parts of the road, so the two cars must give the same ratio. A longer meeting time makes the first ratio bigger and the second smaller, so only one time makes them equal.

All steps
  1. Both cars drive for the same time until they meet at M; call it the meeting time. Car P drives AM in the meeting time, and then MB in 2 hours.
  2. Car P keeps one speed, so the two parts of the road are in the ratio of its times: AM : MB = meeting time : 2.
  3. Car Q drives MB in the meeting time, and then AM in 412 hours, so AM : MB = 412 : meeting time.
  4. (a) Both ratios describe the same two parts of the road, so they are equal. Try 3 hours: 3 : 2, and 412 : 3 = 9 : 6 = 3 : 2. They match, so the cars met after 3 hours.
  5. (b) On AM, Car P took 3 hours and Car Q took 412 hours. On the same road the faster car takes less time, so the speeds are in the ratio of the times turned around: 412 : 3 = 9 : 6 = 3 : 2. Check: on MB, Car P took 2 hours and Car Q took 3 hours, and 3 : 2 is the same ratio again.

answer(a) 3 hours; (b) 3 : 2

techniqueSpeed · Inverse Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Taking the mean of the two times after M: (2 + 412) ÷ 2 = 314 hours. The meeting time must give both cars the same ratio AM : MB, and 314 : 2 is not equal to 412 : 314.
  • Writing the speeds in the same order as the times on AM: 3 : 412 = 2 : 3. Car P took less time on the same road, so Car P is the faster car, and the speeds are 3 : 2.
13

A Rabbit Naps During a Race Against a Tortoise

heuristicFind Each Animal's Time from Its Speed / The Nap Is the Whole Time Minus the Running Time

A rabbit and a tortoise ran a 1200 m race. The rabbit ran at 150 m/min, and the tortoise crawled at 20 m/min. The rabbit ran for 4 minutes, stopped for a nap, and then ran on at the same speed to the finish line. The tortoise crawled without stopping and finished 2 minutes ahead of the rabbit. (a) For how many minutes did the rabbit run in all? (b) How long was the rabbit's nap?

Tortoise60 min · 20 m/min · 1200 m
The tortoise never stops: 1200 ÷ 20 = 60 minutes.
The tortoise never stops, so it takes 1200 ÷ 20 = 60 minutes to finish.
step 1 of 4

Draw each animal's time as a bar. The tortoise's bar is all crawling. The rabbit's bar is 2 minutes longer, and only a small part of it is running: 4 minutes at the start and the rest of the running at the end.

All steps
  1. The tortoise never stops, so it takes 1200 ÷ 20 = 60 minutes to finish.
  2. The rabbit finished 2 minutes after the tortoise, so from start to finish the rabbit took 60 + 2 = 62 minutes.
  3. (a) At 150 m/min the rabbit needs only 1200 ÷ 150 = 8 minutes of running for the 1200 m: 4 minutes before its nap and 8 − 4 = 4 minutes after it.
  4. (b) The rest of the 62 minutes is the nap: 62 − 8 = 54 minutes. Check: 4 + 54 + 4 = 62 minutes.

answer(a) 8 minutes; (b) 54 minutes

techniqueSpeed · Time Durations

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Taking the rabbit's time as 60 − 2 = 58 minutes. The tortoise finished 2 minutes ahead, so the rabbit finished 2 minutes later, after 62 minutes.
  • Taking away only the first 4 minutes of running: 62 − 4 = 58 minutes. After the nap the rabbit still had to run the rest of the race, which took another 4 minutes, so it ran for 8 minutes in all.
14

A Car and a Truck Leave Together, and a Bus Comes the Other Way

heuristicLook at the Road at 10:00 a.m. / Close the Gap Between the Truck and the Bus

At 9:00 a.m., a car and a truck left Town P and drove toward Town Q. The car drove at 75 km/h and the truck at 45 km/h. At the same time, a bus left Town Q and drove toward Town P at 75 km/h. The car passed the bus at 10:00 a.m. (a) What is the distance between Town P and Town Q? (b) At what time did the truck pass the bus?

Town PTown Qcar 75→truck 45→bus 75 from Q←10:00 a.m.
At 10:00 a.m., one hour in, the car is 75 km from Town P.
Look at the road at 10:00 a.m., one hour after the start. The car has driven 75 km from Town P.
step 1 of 6

Draw the road at 10:00 a.m., one hour after the start, with the car, the truck and the bus on it. The car and the bus are at the same place, and the truck is behind the car.

All steps
  1. Look at the road at 10:00 a.m., one hour after the start. The car has driven 75 km from Town P.
  2. The bus passes the car at 10:00 a.m., so the bus is also 75 km from Town P. It has driven 75 km from Town Q.
  3. (a) The towns are 75 + 75 = 150 km apart.
  4. At 10:00 a.m. the truck has driven 45 km from Town P. It is 75 − 45 = 30 km behind the car, so it is 30 km from the bus.
  5. The truck and the bus drive toward each other, so they close the 30 km at 45 + 75 = 120 km/h. That takes 30 ÷ 120 = 14 hour, which is 15 minutes.
  6. (b) The truck passes the bus 15 minutes after 10:00 a.m., at 10:15 a.m. Check: in 114 hours the truck drives 56.25 km and the bus drives 93.75 km, and 56.25 + 93.75 = 150 km.

answer(a) 150 km; (b) 10:15 a.m.

techniqueSpeed · Rate, Total, Units

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the 15 minutes to 9:00 a.m. and answering 9:15 a.m. The 30 km gap is measured at 10:00 a.m., so the 15 minutes are counted from 10:00 a.m.
  • Dividing the 30 km gap by the truck's speed alone: 30 ÷ 45. The bus is also driving toward the truck, so the gap closes at 45 + 75 = 120 km/h.
15

A Driver Speeds Up for the Last Three-Fifths of the Road

heuristicFive Units of Road / 16 km More in Each Hour of the Faster Part

Mr. Ahmad drove from Town X to Town Y. He drove the first 25 of the distance in 2 hours. Then he drove 16 km/h faster for the rest of the way, which took him 212 hours. (a) What was his average speed for the first 25 of the journey? (b) What is the distance between Town X and Town Y?

Road5 unitsFirst 2 h1 h1 h1 unit each hour
Five equal units of road. The first 2 units took 2 hours: 1 unit each hour.
Draw the whole road as 5 equal units. He drove the first 2 units in 2 hours, so at his first speed he drove 1 unit every hour.
step 1 of 6

Draw the road as 5 equal units. The first 2 units take 2 hours, so at his first speed he covers 1 unit an hour. Compare the faster part with what the first speed would have covered in the same time.

All steps
  1. Draw the whole road as 5 equal units. He drove the first 2 units in 2 hours, so at his first speed he drove 1 unit every hour.
  2. He drove the other 3 units in 212 hours, at 16 km/h faster.
  3. At his first speed, 212 hours would cover only 212 units. The extra 16 km in each of the 212 hours covered the last 12 unit: 16 × 212 = 40 km.
  4. So 12 unit is 40 km, and 1 unit is 2 × 40 = 80 km.
  5. (a) He drove 1 unit, 80 km, every hour, so his average speed for the first 25 was 80 km/h.
  6. (b) The whole road is 5 units: 5 × 80 = 400 km. Check: the rest is 3 × 80 = 240 km at 80 + 16 = 96 km/h, and 240 ÷ 96 = 212 hours.

answer(a) 80 km/h; (b) 400 km

techniqueAverage Speed · A Fraction of an Amount

examsO-Level · SAT

Common pitfalls

  • Taking the extra distance as just 16 km, so that 12 unit is 16 km and 1 unit is 32 km. 16 km/h faster means 16 km more in each hour, and he drove faster for 212 hours, so the extra distance is 16 × 212 = 40 km.
  • Answering part (a) with the faster speed, 80 + 16 = 96 km/h. Part (a) asks about the first 25 of the road, before he sped up, so the answer is 80 km/h.
16

A Runner's Speed in Meters per Second, Changed into Kilometers per Hour

heuristicChange One Unit at a Time / Seconds into Hours, Then Meters into Kilometers

At sports day, Zara runs 200 m in 40 seconds. Her brother Omar rides his bike to school at 15 km/h. (a) What is Zara's average speed in meters per second? (b) Change Zara's speed into kilometers per hour. Who is faster, and by how many kilometers per hour?

200 m in 40 s÷ 405 m in 1 s
In 1 second Zara runs 200 ÷ 40 = 5 m.
Zara runs 200 m in 40 seconds, so in 1 second she runs 200 ÷ 40 = 5 m.
step 1 of 5

5 m/s means 5 m in every second. Build up from one second to one minute and then to one hour, and change the meters into kilometers at the end.

All steps
  1. Zara runs 200 m in 40 seconds, so in 1 second she runs 200 ÷ 40 = 5 m.
  2. (a) Her average speed is 5 m/s.
  3. One minute is 60 seconds, so at this speed she covers 5 × 60 = 300 m in each minute.
  4. One hour is 60 minutes: 300 × 60 = 18000 m in each hour. That is 18000 ÷ 1000 = 18 km, so 5 m/s is 18 km/h.
  5. (b) Zara's 18 km/h is faster than Omar's 15 km/h, by 18 − 15 = 3 km/h. Check: 18 ÷ 3.6 = 5 m/s.

answer(a) 5 m/s; (b) 18 km/h, so Zara is faster, by 3 km/h

techniqueConverting Compound Units · Speed

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Changing only one unit: 5 × 3600 = 18000, and calling it 18000 km/h. The 18000 is the number of meters in an hour, so the meters must also be changed into kilometers: 18000 ÷ 1000 = 18.
  • Dividing by 3.6 instead of multiplying: 5 ÷ 3.6 is less than 2 km/h, slower than walking. The same speed always has a bigger number in km/h than in m/s, so going from m/s to km/h multiplies by 3.6.