A rate tells you how much changes during one unit of time. Start by naming what is moving or being produced, then decide whether the situation has one constant rate, several stages, or several workers operating at once.
Suggested pace: 10 minutes on examples · 10–15 minutes on questions · 5 minutes on review. Take a second session if a method needs more work.
Keep in mind: amount = rate × time; time = amount ÷ rate
Method 1 of 4
A speed in miles per hour needs a time in hours. Converting minutes to decimal hours means dividing by 60, not moving a decimal point. Writing the units beside the numbers often catches this error before any multiplication.
A vehicle travels at a constant 48 miles per hour for 35 minutes. How far does it travel?
Answer: 28 miles
Check it: Thirty minutes would cover 24 miles. Five more minutes covers 4 miles, for 28 total.
At 36 miles per hour, how far do you travel in 20 minutes?
12 miles. Twenty minutes is 1/3 hour, so distance = 36 × 1/3.
Common mistake: 35 minutes is not 0.35 hour. At this speed, 0.35 hour would be only 21 minutes.
Method 2 of 4
The average of two speed readings is not automatically the average speed for a trip. A slower leg may take longer, so it contributes more to total time. Add all distances and all times before dividing.
A driver covers 60 miles at 30 mph, then another 60 miles at 60 mph without stopping. What is the average speed for the entire trip?
Answer: 40 mph
Check it: The driver spends twice as long at 30 mph as at 60 mph, so the average is closer to 30.
Why can you not always average two speeds by adding them and dividing by two?
That gives equal weight to each speed. It is valid for equal time intervals, but average speed in general is total distance divided by total elapsed time.
Common mistake: The arithmetic mean of 30 and 60 is 45, but that would apply to equal amounts of time at each speed, not these equal-distance legs.
Method 3 of 4
If a machine completes a job in 6 hours, its rate is 1/6 of the job per hour. Rates can be added when machines work together on the same divisible job at unchanged rates. A leak or opposing flow subtracts from the useful rate.
One pump fills an empty tank in 6 hours and another in 3 hours. Working together at constant rates, how long do they take?
Answer: 2 hours
Check it: The pair must be faster than the 3-hour pump alone. In 2 hours they fill 2/6 + 2/3 = 1 tank.
Each of two identical machines completes a job in 8 hours. How long do they need together?
4 hours. Their combined rate is 1/8 + 1/8 = 1/4 of the job per hour, if both can work on it together.
Common mistake: Adding 6 and 3 gives 9 hours, which would make two contributing pumps slower than either pump alone.
Method 4 of 4
For objects moving toward each other, both motions reduce the gap, so add their speeds. For a faster object catching a slower one in the same direction, subtract the speeds. Any head start must first be converted into an initial distance.
A cyclist is 12 miles ahead of a second cyclist. They ride in the same direction at 10 mph and 14 mph respectively. How long until the second catches the first?
Answer: 3 hours
Check it: The second travels exactly 12 miles farther, eliminating the initial gap.
Two vehicles travel toward each other at 30 and 40 miles per hour. How quickly does the gap shrink?
70 miles per hour. Both vehicles reduce the same gap, so add their speeds.
Common mistake: Dividing by 14 ignores the distance the first cyclist continues to travel.
A units check and a direction check are different. Confirm the units cancel correctly, then ask whether the result should be larger, smaller, faster, or slower.
Work the first six questions with the explanations closed. Review any missed or guessed answers before continuing with the other six. If you cannot set up a problem, return to its method above and try again from a blank page.
Work through one question at a time. Explain your approach, check the worked answer, then move on. Use scratch paper for calculations.
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Progress is saved in this browser. These original study questions do not produce an official AFQT score.
Answer: B. 180 seconds
Three laps cover 3 × 240 = 720 meters. Time equals distance divided by speed, so 720 ÷ 4 = 180 seconds. The 720-second choice uses the distance as a time without dividing by speed; 120 seconds accounts for only two laps.
Answer: D. 30 seconds
The front must travel the bridge length plus the train length before the rear clears the far end: 420 + 180 = 600 meters. At 20 meters per second, that takes 600 ÷ 20 = 30 seconds. The 21-second choice covers only the front’s trip across the bridge.
Answer: B. 5 minutes
The first stage takes 120 ÷ 60 = 2 minutes. The second takes 240 ÷ 80 = 3 minutes. The total is 5 minutes. Dividing all 360 pages by 80 gives 4.5 minutes, but the printer uses the faster rate only in the second stage.
Answer: C. 22.5 minutes
Press A prints 10/40 = 1/4 of the required number, but only 4/5 of that output is acceptable. Its contribution is (1/4)(4/5) = 1/5 of the order. Press B therefore completes 4/5 in 18 minutes. A full order takes 18 ÷ (4/5) = 22.5 minutes. The 24-minute choice treats all of Press A's printed labels as acceptable.
Answer: D. 95 meters
Total distance counts every part of the walk regardless of direction: 60 + 20 + 15 = 95 meters. The signed change in position is 60 − 20 + 15 = 55 meters east, but that is displacement, not the total distance walked. Eighty meters omits the last segment.
Answer: D. 20 minutes
The inflator runs for 5 minutes, rests for 3, runs for 5, rests for 3, and then runs for the final 4. Total elapsed time is 14 + 2 × 3 = 20 minutes. Fourteen omits both rests, 17 includes only one, and 8 subtracts the rests from operating time instead of adding them.
Answer: A. 13.5 packages per hour
Output is 3 × 20 + 2 × 24 = 108 packages. All eight clock hours count, so the average is 108 ÷ 8 = 13.5 packages per hour. Dividing by only the five productive hours gives 21.6. Averaging the two production rates to get 22 ignores both unequal productive durations and setup.
Answer: C. 5 miles per hour
The downstream speed is 18 ÷ 3 = 6 miles per hour, and the upstream speed is 18 ÷ 4.5 = 4 miles per hour. If the still-water speed is v and the current is c, then v + c = 6 and v − c = 4. Adding gives 2v = 10, so v = 5. Four and six are the ground speeds, not the still-water speed.
Answer: B. 24 seconds
Five tones contain four intervals between them. Those intervals total 4 × 6 = 24 seconds. The tones occur at 0, 6, 12, 18 and 24 seconds. Thirty seconds counts five intervals and would reach the sixth tone.
Answer: B. 300 seconds
One lap is 2(180 + 120) = 600 meters. Two laps plus the final two sides total 1,200 + 180 + 120 = 1,500 meters. At 5 meters per second, time is 1,500 ÷ 5 = 300 seconds. The 240-second choice omits the last partial lap; 360 counts a third full lap.
Answer: A. 30 seconds
The walkway’s motion and Lee’s walking are in the same direction, so Lee’s floor-relative speed is 0.8 + 1.2 = 2 meters per second. Time is 60 ÷ 2 = 30 seconds. Fifty seconds ignores the walkway’s motion; 75 is the time for a standing rider, and 150 incorrectly subtracts the speeds.
Answer: C. 37.5 minutes
A reaches the delivery point 9 × 0.5 = 4.5 kilometers from the start at 8:30 and stays there until 8:45. B needs 4.5 ÷ 12 = 0.375 hour, or 22.5 minutes, to reach that point. Starting at 8:15 puts B there at 8:37:30, while A is still stopped. The answer is 37.5 minutes after 8:00; 22.5 counts only B’s travel time. Without A’s stop, the catch-up would take longer.
A remembered choice letter is different from a method you can reconstruct. Keep one correction in your error log and use it on your next attempt.