ORIGINAL WORKED EXAMPLE
Apply the idea
A courier travels 135 kilometres at a constant 45 kilometres per hour. How long does the trip take?
- Use time=distance/rate.
- 135 km divided by 45 km/h leaves hours.
- 135/45=3.
Answer 3 hours.
GMAT · Applied problem solving · TOPIC LESSON
Motion, average speed and combined work
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
BY THE END, YOU CAN
- Explain and apply distance-rate-time.
- Explain and apply opposite and same-direction motion.
- Explain and apply delayed starts.
- Explain and apply average speed versus mean speeds.
- Explain and apply combined work rates.
- Explain and apply changing rates.
- Explain and apply filling and emptying.
- Explain and apply consistent time units.
BEFORE YOU START
- Fractions, unit conversions and linear equations.
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GMAT study through an editorial skill map. It does not reproduce official questions, predict exam content or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
Original GMAT-style worked examples and practice
All passages, examples and exercises here are original. No official GMAC questions, choices or explanations are reproduced. This independent preparation resource is not affiliated with or endorsed by GMAC. GMAT is a trademark of the Graduate Management Admission Council.
01 / UNDERSTAND & APPLY
Distance equals rate times elapsed time
For constant speed, distance=rate×time. Label the units and isolate the requested quantity: time=distance/rate and rate=distance/time. A table of each traveler’s rate, time and distance can keep relationships separate even when no drawing is needed.
A changing-speed trip must be divided into intervals over which the rate is known. The total distance is the sum of segment distances and the total time is the sum of segment times. A single average rate is computed from those totals, not assumed.
ORIGINAL PRACTICE · 1
A runner travels 12 kilometres in 80 minutes. What is the average speed in kilometres per hour?
Reveal answer & reasoning
9 km/h.
80 minutes=4/3 hour, so speed=12÷(4/3)=9. Dividing by 80 without converting gives kilometres per minute, not per hour.
02 / UNDERSTAND & APPLY
Closing speed depends on direction
For two travelers moving toward each other on a straight route, the gap closes at the sum of their speeds. For one traveler catching another in the same direction, the gap closes at the difference, provided the pursuer is faster.
Relative speed describes the change in separation, not either traveler’s own distance. Use the initial gap divided by closing speed to find meeting time, then use an individual speed to answer where they meet. State a common time origin.
ORIGINAL WORKED EXAMPLE
Apply the idea
Two cyclists start 84 kilometres apart and ride toward each other at 12 and 16 km/h. When do they meet, and how far has the slower cyclist ridden?
- Closing speed is 12+16=28 km/h.
- Time is 84/28=3 hours.
- The slower cyclist rides 12×3=36 kilometres.
Answer After 3 hours; 36 kilometres for the slower cyclist.
ORIGINAL PRACTICE · 1
A car traveling 70 km/h is 30 kilometres behind a bus traveling 55 km/h in the same direction. How long to catch it?
Reveal answer & reasoning
2 hours.
The gap closes at 70−55=15 km/h, so 30/15=2 hours. Using the sum 125 describes opposite-direction closing and is inappropriate here.
03 / UNDERSTAND & APPLY
Separate elapsed times from the clock
A later starter has less travel time at any shared clock time. You can either calculate the early traveler’s head start and use relative speed, or assign separate elapsed-time expressions. Do not give both travelers the same duration when their start times differ.
A pursuer moving no faster than the leader cannot close a positive head start under constant speeds. Check this feasibility before solving an equation that might produce a negative time.
ORIGINAL WORKED EXAMPLE
Apply the idea
A bus leaves at 9:00 at 40 km/h. A car follows the same route at 10:00 at 60 km/h. At what time does it catch the bus?
- At 10:00 the bus has a 40-kilometre head start.
- The car closes the gap at 60−40=20 km/h.
- It catches up 40/20=2 hours after 10:00.
Answer 12:00 noon.
ORIGINAL PRACTICE · 1
A walker starts at 4 km/h. A runner leaves the same point 30 minutes later at 10 km/h, following the walker along the same route in the same direction. How many minutes after the runner starts is the catch?
Reveal answer & reasoning
20 minutes.
The head start is 4×1/2=2 kilometres. Closing speed is 6 km/h, so time=2/6=1/3 hour=20 minutes. Thirty minutes is the delay, not the catch-up duration.
04 / UNDERSTAND & APPLY
Average speed weights by time
Average speed equals total distance divided by total elapsed travel time. The arithmetic mean of speeds is valid for equal times, because each speed receives equal time weight. For equal distances at speeds u and v, the average is 2uv/(u+v), not (u+v)/2.
If the problem includes rest time in the elapsed interval, include it in total time even though it contributes no distance. Distinguish average speed from average velocity: a return trip may have zero net displacement but positive total distance.
ORIGINAL WORKED EXAMPLE
Apply the idea
A vehicle travels 60 kilometres at 30 km/h and returns the same distance at 60 km/h. What is its average speed over the entire trip?
- Outbound time is 60/30=2 hours.
- Return time is 60/60=1 hour.
- Total distance is 120 kilometres and time is 3 hours.
Answer 40 km/h.
ORIGINAL PRACTICE · 1
A vehicle travels for 1 hour at 40 km/h and 1 hour at 70 km/h. What is its average speed?
Reveal answer & reasoning
55 km/h.
Total distance is 40+70=110 kilometres over 2 hours, giving 55. The arithmetic mean works here because the durations are equal; it would not generally work for equal distances.
05 / UNDERSTAND & APPLY
Add rates, not completion times
Choose one entire job as the unit. A worker completing it alone in a hours has rate 1/a job per hour. Independent constant rates add, so two workers together have rate 1/a+1/b. The completion time is the reciprocal of the combined rate.
This model assumes simultaneous work can be combined without interference and the rates remain constant. Adding individual completion times would describe serial work on separate whole jobs, not collaboration on one job.
ORIGINAL WORKED EXAMPLE
Apply the idea
One machine finishes a batch in 6 hours and another in 9 hours. How long together at constant independent rates?
- Rates are 1/6 and 1/9 batch per hour.
- Combined rate is 3/18+2/18=5/18.
- Time for one batch is 1÷(5/18)=18/5 hours.
Answer 3.6 hours, or 3 hours 36 minutes.
ORIGINAL PRACTICE · 1
Two workers together finish a task in 4 hours. One alone takes 12 hours. How long does the other take alone?
Reveal answer & reasoning
6 hours.
The other rate is 1/4−1/12=1/6 job per hour, so its time is 6 hours. Subtracting completion times, 12−4=8, does not subtract rates.
06 / UNDERSTAND & APPLY
Partition the work at every change
When a worker joins, leaves or changes speed, create separate intervals. In each interval, completed fraction=rate×time. Add completed fractions and equate to one whole job. The final interval may be shorter than the time required for an entire job at that rate.
A convenient work-unit total can remove fractions: if individual completion times are 6 and 8 hours, use 24 work units so rates become 4 and 3 units per hour. This is a scaling choice, not a change to the actual job.
ORIGINAL WORKED EXAMPLE
Apply the idea
A completes a job alone in 6 hours and B in 12. A works alone for 2 hours, then B joins. How much additional time is needed?
- A completes 2/6=1/3 of the job.
- The remaining fraction is 2/3. Together the rate is 1/6+1/12=1/4.
- Additional time=(2/3)/(1/4)=8/3 hours.
Answer 2 hours 40 minutes additional.
ORIGINAL PRACTICE · 1
A machine makes 20 parts per hour for 3 hours, then 30 per hour for 2 hours. What is its average production rate?
Reveal answer & reasoning
24 parts per hour.
It makes 60+60=120 parts in 5 hours, giving 24 per hour. The mean of 20 and 30 is 25, but the durations are unequal.
07 / UNDERSTAND & APPLY
Outflow is a negative contribution to net rate
For a tank, a filling pipe adds positive volume per time and a drain removes it. Subtract outflow from inflow to get the net rate. If net rate is zero, the level remains constant; if it is negative, a partially filled tank empties rather than fills.
The model assumes stated rates continue over the relevant interval. An empty tank cannot lose water below zero, and a full tank may overflow; use physical boundaries rather than extending a linear model beyond its meaning.
ORIGINAL WORKED EXAMPLE
Apply the idea
An inlet fills an empty tank in 8 hours. A drain empties a full tank in 12 hours. With both open and constant rates, how long to fill from empty?
- Inflow is 1/8 tank per hour and outflow is 1/12.
- Net rate is 3/24−2/24=1/24.
- Time for one tank is 24 hours.
Answer 24 hours.
ORIGINAL PRACTICE · 1
A tank is half full. Inflow is 6 litres per minute and outflow is 10 litres per minute. If capacity is 80 litres, how long until empty?
Reveal answer & reasoning
10 minutes.
Initial water is 40 litres. Net loss is 10−6=4 litres per minute, so 40/4=10 minutes. Adding the rates ignores their opposite effects.
08 / UNDERSTAND & APPLY
Convert before combining rates
Rates can be combined only after using compatible units. A rate per minute and a rate per hour cannot be added numerically as written. Convert to a common time unit, and preserve fractions until the final conversion.
A decimal part of an hour is not a number of minutes. For example, 0.4 hour means 0.4×60=24 minutes, not 40 minutes. Read whether the question asks total elapsed time, additional time or a clock time.
ORIGINAL WORKED EXAMPLE
Apply the idea
One scanner processes 12 pages per minute and another 480 pages per hour. How many minutes do they need together for 1,000 pages?
- 480 pages per hour=8 pages per minute.
- Together they process 12+8=20 pages per minute.
- Time=1000/20.
Answer 50 minutes.
ORIGINAL PRACTICE · 1
A job takes 2.75 hours. Express this as hours and minutes.
Reveal answer & reasoning
2 hours 45 minutes.
The fractional hour is 0.75×60=45 minutes. Writing 2 hours 75 minutes both misinterprets the decimal and uses an unnormalized minute count.
KEEP THE BIG IDEAS
Quick revision
- Use total distance divided by total time for average speed.
- Compute gaps with relative speed and keep delayed start times distinct.
- Add work rates and take the reciprocal for a whole-job time.
- Split at every rate change and convert time units consistently.
READ WITH CONTEXT
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