Lesson 5 of 35 · Quantitative Reasoning: arithmetic

Rates, work and travel

Use distance equals rate times time, average speeds correctly, relative speed for meetings and chases, and add work rates rather than times.

About 22 minutes

In this lesson

  • Set up any travel question with distance equals rate times time, keeping units consistent.
  • Find an average speed from total distance and total time.
  • Solve meeting and catch-up problems with relative speed.
  • Combine work rates, including a drain or a worker who leaves part-way.

The idea

One relationship, three forms

Travel questions rest on distance=rate×time\text{distance} = \text{rate} \times \text{time}, and work questions on work=rate×time\text{work} = \text{rate} \times \text{time}. Every question gives two of the three quantities, sometimes in disguise, and asks for the third. A small table with a row for each traveller or worker and columns for rate, time and distance keeps the facts straight.

Units first

Rates in kilometres per hour and times in minutes do not mix. Convert before computing: 45 minutes is 34\frac{3}{4} hour, and 1.9 hours is 1 hour and 0.9×60=540.9 \times 60 = 54 minutes, not 1 hour 9 minutes. When answer choices are clock times, convert the decimal part of an hour carefully at the end.

Average speed is total distance over total time

average speed=total distancetotal time\text{average speed} = \frac{\text{total distance}}{\text{total time}}

It is not the average of the speeds, unless equal times are spent at each speed. Driving 120 km out at 60 km/h and back at 40 km/h takes 2 hours plus 3 hours, so the average speed is 2405=48\frac{240}{5} = 48 km/h, not 50. More time is spent at the slower speed, which pulls the average towards it. For two equal distances at speeds aa and bb there is a shortcut:

average speed=2aba+b\text{average speed} = \frac{2ab}{a + b}

and 2×60×40100=48\frac{2 \times 60 \times 40}{100} = 48 agrees.

Relative speed

When two objects move towards each other, or away from each other, the gap between them changes at the sum of their speeds. When one follows the other in the same direction, the gap changes at the difference. So two cyclists 63 km apart riding towards each other at 15 and 13 km/h close the gap at 28 km/h and meet after 6328=2.25\frac{63}{28} = 2.25 hours.

For a chase, first find the head start: the distance the leader covers before the follower sets off. The catch-up time is the head start divided by the closing speed.

Work: add rates, not times

If a job takes a worker aa hours, the worker's rate is 1a\frac{1}{a} of the job per hour. Rates of people working together add; times do not. A drain, or anyone undoing the work, contributes a negative rate. For two workers,

1T=1a+1b,T=aba+b\frac{1}{T} = \frac{1}{a} + \frac{1}{b}, \qquad T = \frac{ab}{a + b}

A useful device without a calculator: pretend the job has a size equal to the LCM of the individual times. If one printer takes 10 hours and another 15, call the job 30 pages; the printers do 3 and 2 pages an hour, 5 together, so the job takes 6 hours. Fractions become whole numbers.

Sanity checks

  • Working together must be faster than the fastest worker alone and, with two equal workers, exactly half of one worker's time.
  • An average speed lies between the slowest and fastest speeds.
  • When two travellers set off towards each other at the same moment, they meet closer to the slower one's start.

These checks take seconds and often eliminate two or three answer choices before any arithmetic.

Worked examples

Example 1: Average speed on a round trip

A van drives 120 km to a depot at 60 km/h and returns along the same road at 40 km/h. What is its average speed for the round trip?

  1. Find each time: 12060=2\frac{120}{60} = 2 hours out and 12040=3\frac{120}{40} = 3 hours back.

  2. Total distance over total time: 2402+3=48\frac{240}{2 + 3} = 48 km/h.

  3. Check with the equal-distance shortcut: 2×60×4060+40=4800100=48\frac{2 \times 60 \times 40}{60 + 40} = \frac{4800}{100} = 48.

Answer

48 km/h. The tempting 50 km/h averages the speeds, which would be right only if the van spent equal times, not equal distances, at each speed.

Example 2: A courier catching up

One courier leaves a depot at 9:00 riding at 12 km/h. A second courier leaves the same depot at 9:30 along the same road at 18 km/h. At what time does the second courier catch the first?

  1. Head start: in the half hour before 9:30 the first courier covers 12×0.5=612 \times 0.5 = 6 km.

  2. Same direction, so the gap closes at the difference of the speeds: 18−12=618 - 12 = 6 km/h.

  3. Catch-up time: 66=1\frac{6}{6} = 1 hour after 9:30, so at 10:30.

  4. Check: by 10:30 the first courier has ridden 1.5 hours, 12×1.5=1812 \times 1.5 = 18 km; the second has ridden 1 hour, 18×1=1818 \times 1 = 18 km.

Answer

At 10:30. Adding the speeds instead would give a closing speed of 30 km/h and a catch-up after 12 minutes, which is impossible for riders going the same way.

Example 3: Two pipes and a drain

Pipe A fills a tank in 6 hours and pipe B fills it in 9 hours. A drain empties a full tank in 12 hours. If the tank starts empty and all three are open, how long does it take to fill?

  1. Pick a tank size equal to lcm(6,9,12)=36\text{lcm}(6, 9, 12) = 36 units. Pipe A adds 6 units per hour, pipe B adds 4, and the drain removes 3.

  2. Net rate: 6+4−3=76 + 4 - 3 = 7 units per hour, so filling takes 367=517\frac{36}{7} = 5\frac{1}{7} hours.

  3. Check with fractions: 16+19−112=6+4−336=736\frac{1}{6} + \frac{1}{9} - \frac{1}{12} = \frac{6 + 4 - 3}{36} = \frac{7}{36} of the tank per hour, and the reciprocal is 367\frac{36}{7}.

Answer

5175\frac{1}{7} hours, about 5 hours 9 minutes. Without the drain the pipes alone need 3610=3.6\frac{36}{10} = 3.6 hours, so the drain adds time, as it should.

Example 4: Finishing a job alone

Ana can paint a fence in 10 days and Ben in 15 days. Ana works alone for 4 days, then stops, and Ben finishes the fence alone. How many days does Ben work?

  1. Call the job 30 units, the LCM of 10 and 15. Ana paints 3 units a day and Ben 2.

  2. In 4 days Ana paints 12 units, leaving 18.

  3. Ben needs 182=9\frac{18}{2} = 9 days.

  4. Check with fractions: Ana does 410\frac{4}{10} of the job and Ben 915=610\frac{9}{15} = \frac{6}{10}, which add to 1.

Answer

Ben works 9 days.

Common traps

  • Averaging the speeds

    The average speed is total distance over total time. Over equal distances it is always below the simple average of the speeds, because more time is spent going slowly. Expect the simple average among the choices.

  • Adding times instead of rates

    Two machines taking 4 and 6 hours do not take 10 hours together, or 5. Convert each to a rate, add the rates, and turn the total back into a time. The combined time must be shorter than either machine's time alone.

  • Decimal hours read as minutes

    1.9 hours is 1 hour 54 minutes, not 1 hour 9 minutes. Multiply the decimal part by 60 whenever the answer is a clock time or a number of minutes.

  • Forgetting the head start

    When travellers set off at different times, the early one has already covered some distance. Work out where everyone is when the last one starts, then apply relative speed from that moment.

Check yourself

Answer each question before you open its answer.

Question 1: Average speed over two halves

Problem Solving

A runner covers the first half of a 12-kilometre course at 8 km/h and the second half at 12 km/h. What is her average speed for the whole course?

  1. A.4.8 km/h
  2. B.9.6 km/h
  3. C.10 km/h
  4. D.10.4 km/h
  5. E.20 km/h
Show the answer

Answer: B. 9.6 km/h

Working

Times: 68=0.75\frac{6}{8} = 0.75 hour and 612=0.5\frac{6}{12} = 0.5 hour, a total of 1.25 hours. Average speed: 121.25=9.6\frac{12}{1.25} = 9.6 km/h. Shortcut for equal distances: 2×8×128+12=19220=9.6\frac{2 \times 8 \times 12}{8 + 12} = \frac{192}{20} = 9.6.

Why the other options are wrong

  • A. This divides only one half of the course, 6 km, by the total time.
  • C. This is the simple average of the two speeds; she spends longer at the slower speed, so the true average is lower.
  • D. This weights each speed by the other half's time (8×0.5+12×0.758 \times 0.5 + 12 \times 0.75 over 1.25 hours), swapping the times.
  • E. This adds the speeds, as for the closing speed of two runners approaching each other.

What carries over

Average speed is total distance over total time, never the average of the speeds over equal distances.

Question 2: Filling against a drain

Problem Solving

Pipe P fills an empty tank in 3 hours and pipe Q fills it in 5 hours. A drain, opened at the same moment, empties a full tank in 6 hours. With P, Q and the drain all open, how long does it take to fill the empty tank?

  1. A.1130\frac{11}{30} hour
  2. B.1371\frac{3}{7} hours
  3. C.1781\frac{7}{8} hours
  4. D.2 hours
  5. E.28112\frac{8}{11} hours
Show the answer

Answer: E. 28112\frac{8}{11} hours

Working

Net rate: 13+15−16=10+6−530=1130\frac{1}{3} + \frac{1}{5} - \frac{1}{6} = \frac{10 + 6 - 5}{30} = \frac{11}{30} of the tank per hour, so the time is 3011=2811\frac{30}{11} = 2\frac{8}{11} hours. With a 30-unit tank: P adds 10 units per hour, Q adds 6 and the drain removes 5, a net 11 units per hour, and 3011\frac{30}{11} hours again. It is longer than the 1781\frac{7}{8} hours the pipes need without the drain, as it must be.

Why the other options are wrong

  • A. 1130\frac{11}{30} is the fraction of the tank filled per hour, a rate; the time is its reciprocal.
  • B. This adds the drain's rate instead of subtracting it: 13+15+16=2130\frac{1}{3} + \frac{1}{5} + \frac{1}{6} = \frac{21}{30}.
  • C. This is the time for the two pipes alone; the drain was left out.
  • D. This combines the times directly, 3+5−63 + 5 - 6; times do not add or subtract, rates do.

What carries over

Add rates with signs, then take the reciprocal to get the time.

Question 3: Two trains, different start times

Problem Solving

Two stations are 345 km apart. Train A leaves the first station at 8:00 and travels towards the second at 60 km/h. Train B leaves the second station at 9:00 and travels towards the first at 90 km/h. At what time do the trains meet?

  1. A.10:09
  2. B.10:18
  3. C.10:54
  4. D.11:18
  5. E.11:54
Show the answer

Answer: C. 10:54

Working

By 9:00 train A has covered 60 km, leaving 285 km between the trains. They close the gap at 60+90=15060 + 90 = 150 km/h, so they meet 285150=1.9\frac{285}{150} = 1.9 hours after 9:00, which is 1 hour 54 minutes later, at 10:54. Check: train A has run 2.9 hours, 2.9×60=1742.9 \times 60 = 174 km; train B has run 1.9 hours, 1.9×90=1711.9 \times 90 = 171 km; 174+171=345174 + 171 = 345.

Why the other options are wrong

  • A. This reads 1.9 hours as 1 hour 9 minutes; 0.9 hour is 54 minutes.
  • B. This starts both trains at 8:00, ignoring that train B sets off an hour later.
  • D. This starts both trains at 9:00 with the full 345 km between them, forgetting the 60 km train A has already covered.
  • E. This counts train A's extra hour twice: it is already used up in the 60 km head start, so it must not be added to the meeting time.

What carries over

Place everyone at the moment the last traveller starts, then divide the gap by the closing speed.

Key points

  • Distance = rate × time; convert units before computing.

  • Average speed is total distance over total time.

  • Towards or apart, add speeds; same direction, subtract them.

  • Work rates add (drains subtract); times never add.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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