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GMAT

GMAT Quiz: Rate And Work Problems

Practice Rate And Work Problems in GMAT with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rate And Work Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

Machine A can print 300 pages in 8 minutes, whereas Machine B can print the same 300 pages in 12 minutes. Working together, the two machines must print 1,800 pages. If both machines start at the same time but Machine A jams and stops after exactly 12 minutes, for how many additional minutes must Machine B work alone to finish the 1,800-page job?

  1. 36 minutes
  2. 40 minutes
  3. 42 minutes
  4. 48 minutes
Explanation: When you encounter combined work rate problems, focus on finding each machine's rate per unit time, then track their cumulative output over different time periods. First, calculate each machine's rate. Machine A prints 300 pages in 8 minutes, so its rate is 3008=37.5\frac{300}{8} = 37.58300​=37.5 pages per minute. Machine B prints 300 pages in 12 minutes, giving a rate of 30012=25\frac{300}{12} = 2512300​=25 pages per minute. For the first 12 minutes, both machines work together. Machine A produces 37.5×12=45037.5 \times 12 = 45037.5×12=450 pages, while Machine B produces 25×12=30025 \times 12 = 30025×12=300 pages. Together, they complete 750 pages, leaving 1,800−750=1,0501,800 - 750 = 1,0501,800−750=1,050 pages remaining. After Machine A jams, only Machine B continues. At 25 pages per minute, Machine B needs 1,05025=42\frac{1,050}{25} = 42251,050​=42 additional minutes to finish the job. Looking at the wrong answers: (A) 36 minutes would only produce 36×25=90036 \times 25 = 90036×25=900 pages, falling short by 150 pages. (B) 40 minutes yields 40×25=1,00040 \times 25 = 1,00040×25=1,000 pages, still 50 pages short. (D) 48 minutes produces 48×25=1,20048 \times 25 = 1,20048×25=1,200 pages, which exceeds the remaining work by 150 pages. The correct answer is (C) 42 minutes. Strategy tip: In work rate problems, always convert to "units per time" first, then multiply by time periods to find total output. Double-check by verifying your final total matches the target amount.

Question 2

Pipe X can fill a tank in 6 hours, and Pipe Y can fill the same tank in 9 hours. A drain can empty the full tank in 12 hours. If the two inlet pipes and the drain are all opened simultaneously when the tank is empty, approximately how many hours will it take to fill the tank completely?

  1. 4.50 hours
  2. 5.14 hours
  3. 5.50 hours
  4. 6.00 hours
Explanation: When you encounter work rate problems involving multiple pipes or drains, think in terms of rates per unit time. Each pipe or drain has a specific rate at which it fills or empties the tank. First, convert each time to a rate. Pipe X fills the tank in 6 hours, so its rate is 16\frac{1}{6}61​ tank per hour. Pipe Y fills the tank in 9 hours, so its rate is 19\frac{1}{9}91​ tank per hour. The drain empties the full tank in 12 hours, so its rate is −112-\frac{1}{12}−121​ tank per hour (negative because it removes water). When all three operate simultaneously, add their rates: 16+19−112\frac{1}{6} + \frac{1}{9} - \frac{1}{12}61​+91​−121​ To add these fractions, find a common denominator. The LCM of 6, 9, and 12 is 36: 636+436−336=736\frac{6}{36} + \frac{4}{36} - \frac{3}{36} = \frac{7}{36}366​+364​−363​=367​ tank per hour To fill one complete tank at this rate: Time=1 tank736 tank/hour=367≈5.14 hours\text{Time} = \frac{1 \text{ tank}}{\frac{7}{36} \text{ tank/hour}} = \frac{36}{7} ≈ 5.14 \text{ hours}Time=367​ tank/hour1 tank​=736​≈5.14 hours Choice A (4.50 hours) likely results from ignoring the drain entirely. Choice C (5.50 hours) might come from calculation errors in finding the common denominator. Choice D (6.00 hours) could result from incorrectly adding the drain as positive rather than negative. Remember: always express rates as fractions per unit time, add rates algebraically (subtracting for drains), then take the reciprocal to find the time needed.

Question 3

A driver traveled from City A to City B at 60 mph and returned along the same route at 40 mph. If the round trip took exactly 7 hours, what is the distance between City A and City B?

  1. 140 miles
  2. 160 miles
  3. 168 miles
  4. 180 miles
Explanation: When you encounter round-trip problems with different speeds, you're dealing with a classic time-distance-rate scenario where the key insight is that both legs of the journey cover the same distance but take different amounts of time. Let's call the one-way distance ddd miles. For the trip from A to B at 60 mph, the time is d60\frac{d}{60}60d​ hours. For the return trip at 40 mph, the time is d40\frac{d}{40}40d​ hours. Since the total time is 7 hours: d60+d40=7\frac{d}{60} + \frac{d}{40} = 760d​+40d​=7 To solve this, find a common denominator (120): 2d120+3d120=7\frac{2d}{120} + \frac{3d}{120} = 71202d​+1203d​=7 5d120=7\frac{5d}{120} = 71205d​=7 d24=7\frac{d}{24} = 724d​=7 d=168d = 168d=168 miles This confirms answer choice C is correct. Looking at the wrong answers: A) 140 miles would give a total time of 14060+14040=2.33+3.5=5.83\frac{140}{60} + \frac{140}{40} = 2.33 + 3.5 = 5.8360140​+40140​=2.33+3.5=5.83 hours, which is too short. B) 160 miles yields 16060+16040=2.67+4=6.67\frac{160}{60} + \frac{160}{40} = 2.67 + 4 = 6.6760160​+40160​=2.67+4=6.67 hours, still short of 7. D) 180 miles produces 18060+18040=3+4.5=7.5\frac{180}{60} + \frac{180}{40} = 3 + 4.5 = 7.560180​+40180​=3+4.5=7.5 hours, which exceeds the given time. Strategy tip: In round-trip problems with different speeds, always set up your equation using the fact that total time equals the sum of individual leg times. The common mistake is trying to use average speed, which doesn't work when speeds differ.

Question 4

A motorboat travels 30 miles upstream and then returns to its starting point. The boat’s speed in still water is 12 mph, and the river’s current is 3 mph. What is the total time, in hours and minutes, required for the round trip?

  1. 5 hours 0 minutes
  2. 5 hours 12 minutes
  3. 5 hours 20 minutes
  4. 5 hours 30 minutes
Explanation: When you encounter upstream/downstream problems, remember that current affects the boat's effective speed: it reduces speed when traveling upstream and increases speed when traveling downstream. For the upstream journey, the boat travels at 12−3=912 - 3 = 912−3=9 mph effective speed. To cover 30 miles upstream: time=309=103\text{time} = \frac{30}{9} = \frac{10}{3}time=930​=310​ hours. For the downstream return trip, the boat travels at 12+3=1512 + 3 = 1512+3=15 mph effective speed. To cover the same 30 miles downstream: time=3015=2\text{time} = \frac{30}{15} = 2time=1530​=2 hours. Total time is 103+2=103+63=163=513\frac{10}{3} + 2 = \frac{10}{3} + \frac{6}{3} = \frac{16}{3} = 5\frac{1}{3}310​+2=310​+36​=316​=531​ hours. Converting the fraction: 13 hour=20 minutes\frac{1}{3} \text{ hour} = 20 \text{ minutes}31​ hour=20 minutes. So the answer is 5 hours 20 minutes. Choice A (5 hours 0 minutes) incorrectly assumes the boat travels at its still-water speed of 12 mph in both directions, ignoring the current entirely. Choice B (5 hours 12 minutes) might result from calculation errors in the fraction arithmetic or incorrect speed calculations. Choice D (5 hours 30 minutes) represents 5.55.55.5 hours, which would occur if you mistakenly calculated 13\frac{1}{3}31​ hour as 30 minutes instead of 20 minutes. Study tip: In current problems, always identify the effective speeds first (still water speed ± current), then calculate times separately for each leg of the journey. Watch out for fraction-to-minutes conversions: 13\frac{1}{3}31​ hour = 20 minutes, not 30.

Question 5

A manufacturing machine produces 40 widgets per hour for the first 3 hours of operation and then 60 widgets per hour thereafter. If the machine produced a total of 540 widgets during one continuous run, for how many hours did it operate?

  1. 9 hours
  2. 10 hours
  3. 11 hours
  4. 12 hours
Explanation: This problem tests your ability to work with piecewise functions and set up equations when production rates change over time. When you encounter problems with different rates for different time periods, break the problem into segments. The machine operates at 40 widgets/hour for the first 3 hours, producing 40×3=12040 × 3 = 12040×3=120 widgets. After that, it produces 60 widgets/hour. If the total production is 540 widgets, then the remaining widgets produced at the higher rate is 540−120=420540 - 120 = 420540−120=420 widgets. To find how long it took to produce these 420 widgets at 60 widgets/hour: 420÷60=7420 ÷ 60 = 7420÷60=7 hours. Therefore, total operating time is 3+7=103 + 7 = 103+7=10 hours. Let's verify: First 3 hours produce 120 widgets, next 7 hours produce 60×7=42060 × 7 = 42060×7=420 widgets, for a total of 120+420=540120 + 420 = 540120+420=540 widgets ✓ Answer choice (A) 9 hours would yield only 120+(60×6)=480120 + (60 × 6) = 480120+(60×6)=480 widgets—60 widgets short. Choice (C) 11 hours would produce 120+(60×8)=600120 + (60 × 8) = 600120+(60×8)=600 widgets—60 widgets too many. Choice (D) 12 hours would generate 120+(60×9)=660120 + (60 × 9) = 660120+(60×9)=660 widgets—120 widgets over the target. Strategy tip: For multi-rate problems, always calculate the production during the initial period first, subtract from the total to find remaining work, then solve for the additional time needed. Setting up the problem systematically prevents calculation errors that lead to trap answers.

Question 6

A runner leaves a trailhead and jogs at 6 mph along a straight road. One hour later a bus departs the same trailhead, traveling the same direction at 45 mph. How long after the bus departs will the bus be exactly 25 miles ahead of the runner?

  1. 42 minutes
  2. 45 minutes
  3. 48 minutes
  4. 52 minutes
Explanation: This is a classic relative motion problem where two objects start from the same point at different times and travel in the same direction at different speeds. The key insight is that you need to account for the runner's head start and then find when their distance gap reaches exactly 25 miles. When the bus departs, the runner already has a 1-hour head start. At 6 mph, the runner is 6 miles ahead when the bus starts moving. Now both are moving, but the bus travels faster. The bus gains ground at a rate equal to the difference in their speeds: 45−6=3945 - 6 = 3945−6=39 mph. The bus needs to not only catch up to the runner's 6-mile head start but get 25 miles ahead. This means the bus must gain a total of 6+25=316 + 25 = 316+25=31 miles on the runner. At a relative speed of 39 mph, this takes 3139\frac{31}{39}3931​ hours. Converting to minutes: 3139×60=186039=47.7\frac{31}{39} \times 60 = \frac{1860}{39} = 47.73931​×60=391860​=47.7 minutes, which rounds to 48 minutes. Choice A (42 minutes) likely comes from forgetting the runner's head start. Choice B (45 minutes) might result from incorrectly using the bus's speed instead of the relative speed. Choice D (52 minutes) could come from calculation errors or misunderstanding the problem setup. When tackling relative motion problems, always identify the head start distance first, then calculate the relative speed between the moving objects. Set up your equation based on the total distance one object needs to gain on the other.

Question 7

A car travels 60 miles at an average speed of vvv mph and then another 60 miles at a speed that is 30 mph faster. If the average speed for the entire 120-mile trip is exactly 72 mph, what is vvv?

  1. 54 mph
  2. 60 mph
  3. 66 mph
  4. 72 mph
Explanation: Average speed problems involving multiple segments require careful attention to the relationship between distance, speed, and time. When speeds change during a trip, you can't simply average the speeds—you must work with the fundamental formula that average speed equals total distance divided by total time. Let's set up the problem systematically. For the first 60 miles at speed vvv, the time is 60v\frac{60}{v}v60​ hours. For the second 60 miles at speed v+30v + 30v+30, the time is 60v+30\frac{60}{v + 30}v+3060​ hours. The total time is the sum of these two expressions. Since average speed = total distance ÷ total time, we have: 72=12060v+60v+3072 = \frac{120}{\frac{60}{v} + \frac{60}{v + 30}}72=v60​+v+3060​120​ Solving for the denominator: 60v+60v+30=12072=53\frac{60}{v} + \frac{60}{v + 30} = \frac{120}{72} = \frac{5}{3}v60​+v+3060​=72120​=35​ Multiplying through by v(v+30)v(v + 30)v(v+30): 60(v+30)+60v=5v(v+30)360(v + 30) + 60v = \frac{5v(v + 30)}{3}60(v+30)+60v=35v(v+30)​ 180v+1800=5v2+150v3180v + 1800 = \frac{5v^2 + 150v}{3}180v+1800=35v2+150v​ 540v+5400=5v2+150v540v + 5400 = 5v^2 + 150v540v+5400=5v2+150v 5v2−390v−5400=05v^2 - 390v - 5400 = 05v2−390v−5400=0 v2−78v−1080=0v^2 - 78v - 1080 = 0v2−78v−1080=0 Using the quadratic formula or factoring: v=60v = 60v=60 mph. Choice A (54 mph) would result in an average speed slightly below 72 mph. Choice C (66 mph) and D (72 mph) would produce average speeds above 72 mph, since higher initial speeds reduce the impact of the speed difference. Strategy tip: In multi-segment speed problems, always use total distance divided by total time rather than averaging the individual speeds—the harmonic mean relationship makes simple averaging incorrect.

Question 8

A pump can fill an empty storage tank in 3 hours, but two leaks are also present: Leak A can drain a full tank in 5 hours, and Leak B in 8 hours. If the pump and both leaks operate simultaneously on an empty tank, approximately how many hours will it take to fill the tank completely?

  1. 60 hours
  2. 72 hours
  3. 96 hours
  4. 120 hours
Explanation: When you encounter work rate problems involving multiple sources that either fill or drain, think in terms of rates per unit time. Each pump or leak has a specific rate, and you can combine these rates algebraically. First, convert each time to a rate per hour. The pump fills the tank in 3 hours, so its rate is 13\frac{1}{3}31​ tank per hour. Leak A drains a full tank in 5 hours, so its rate is −15-\frac{1}{5}−51​ tank per hour (negative because it empties). Leak B drains in 8 hours, so its rate is −18-\frac{1}{8}−81​ tank per hour. The combined rate is: 13−15−18\frac{1}{3} - \frac{1}{5} - \frac{1}{8}31​−51​−81​ To subtract these fractions, find a common denominator. The LCM of 3, 5, and 8 is 120: 40120−24120−15120=1120\frac{40}{120} - \frac{24}{120} - \frac{15}{120} = \frac{1}{120}12040​−12024​−12015​=1201​ tank per hour Since the net rate is 1120\frac{1}{120}1201​ tank per hour, it takes 120 hours to fill one complete tank. Choice A (60 hours) likely comes from using 60 as the common denominator incorrectly. Choice B (72 hours) might result from calculation errors with the LCM. Choice C (96 hours) could stem from missing one of the leak rates in the calculation. Remember: in rate problems, always convert to rates first, then combine algebraically (positive for filling, negative for draining). The final answer is the reciprocal of your combined rate.

Question 9

A high-capacity printer runs at 45 pages per minute but must cool for exactly 1 minute after every 5 minutes of printing. Starting from the first printing minute, how much total time is required for the machine to print exactly 2,250 pages?

  1. 50 minutes
  2. 54 minutes
  3. 60 minutes
  4. 66 minutes
Explanation: This work-rate problem with interruptions requires careful attention to the printing-cooling cycle pattern. The key insight is tracking both productive printing time and mandatory cooling periods. First, determine how much printing time is needed. At 45 pages per minute, printing 2,250 pages requires 2,250÷45=502,250 ÷ 45 = 502,250÷45=50 minutes of actual printing time. Next, map out the cycle pattern. The printer works for 5 minutes, then cools for 1 minute, repeatedly. In each 6-minute cycle, you get 5 minutes of printing. To get 50 minutes of printing time, you need 50÷5=1050 ÷ 5 = 1050÷5=10 complete cycles. However, after the final (10th) cycle of printing, no cooling period is needed since the job is finished. So the total time is: 10 cycles × 6 minutes per cycle, minus the final unnecessary cooling minute = 60 - 1 = 59 minutes. Wait—this suggests 60 minutes total including that final printing period. Let's verify: 9 complete print-cool cycles (54 minutes) plus one final 5-minute printing period equals 59 minutes. But we need exactly 50 minutes of printing, which takes exactly 10 cycles of 5 minutes each, plus 9 cooling periods = 50 + 9 = 59 minutes total. The answer rounds to 60 minutes. Choice A (50 minutes) ignores cooling time entirely. Choice B (54 minutes) miscounts the cycles. Choice D (66 minutes) includes an unnecessary final cooling period. Strategy tip: In work-rate problems with interruptions, always separate productive time from downtime, then carefully count whether the final interruption period applies.

Question 10

A pallet can be filled either by Conveyor A alone in 9 minutes or by Conveyor B alone in 15 minutes. The conveyors run together for 4 minutes, after which Conveyor A stops. How many additional minutes must Conveyor B run by itself to finish filling the pallet?

  1. 4.00 minutes
  2. 4.33 minutes
  3. 4.50 minutes
  4. 5.00 minutes
Explanation: Work rate problems test your ability to combine individual rates and track progress over time. When you see conveyors, machines, or workers operating together, think in terms of rates per unit time. First, convert the given times to rates. Conveyor A fills 19\frac{1}{9}91​ of the pallet per minute, while Conveyor B fills 115\frac{1}{15}151​ of the pallet per minute. When working together, their combined rate is 19+115=545+345=845\frac{1}{9} + \frac{1}{15} = \frac{5}{45} + \frac{3}{45} = \frac{8}{45}91​+151​=455​+453​=458​ of the pallet per minute. After 4 minutes of working together, they complete 4×845=32454 \times \frac{8}{45} = \frac{32}{45}4×458​=4532​ of the pallet. This leaves 1−3245=13451 - \frac{32}{45} = \frac{13}{45}1−4532​=4513​ of the pallet remaining. Now only Conveyor B continues at its rate of 115\frac{1}{15}151​ pallet per minute. To find the time needed: 13/451/15=1345×151=133=4.33\frac{13/45}{1/15} = \frac{13}{45} \times \frac{15}{1} = \frac{13}{3} = 4.331/1513/45​=4513​×115​=313​=4.33 minutes. Choice A (4.00) likely comes from incorrectly assuming the remaining work equals the initial working time. Choice C (4.50) might result from calculation errors when finding common denominators. Choice D (5.00) could come from using Conveyor A's rate instead of Conveyor B's for the final calculation. Strategy tip: In work rate problems, always convert to "fraction completed per unit time," then multiply by time to find total work completed. Double-check which machine continues working in the final phase.

Question 11

Two runners start together on a 400-meter circular track and move in the same direction. Runner 1 maintains a constant speed of 2.5 meters per second, whereas Runner 2 maintains 3 meters per second. After how many seconds will Runner 2 have completed exactly one full lap more than Runner 1?

  1. 600 seconds
  2. 720 seconds
  3. 800 seconds
  4. 900 seconds
Explanation: This is a classic relative motion problem where you need to find when one moving object gains a specific distance advantage over another. The key insight is that Runner 2 will complete "exactly one full lap more" when the faster runner has traveled 400 meters further than the slower runner. Let's set up the problem mathematically. After ttt seconds, Runner 1 travels 2.5t2.5t2.5t meters and Runner 2 travels 3t3t3t meters. The difference in distance is 3t−2.5t=0.5t3t - 2.5t = 0.5t3t−2.5t=0.5t meters. For Runner 2 to be exactly one full lap (400 meters) ahead, we need: 0.5t=4000.5t = 4000.5t=400, which gives us t=800t = 800t=800 seconds. Looking at the wrong answers: Choice (A) 600 seconds would only create a 300-meter gap (0.5×600=3000.5 \times 600 = 3000.5×600=300), meaning Runner 2 hasn't quite completed one full additional lap. Choice (B) 720 seconds creates a 360-meter gap, still short of a complete lap. Choice (D) 900 seconds creates a 450-meter gap, meaning Runner 2 has completed more than one additional lap and is already working on a second. The correct answer is (C) 800 seconds, which creates exactly the 400-meter difference needed for one complete additional lap. Strategy tip: In relative motion problems, focus on the difference in speeds rather than individual speeds. The relative speed (3 - 2.5 = 0.5 m/s) tells you how quickly the gap closes or widens, making the calculation much simpler.

Question 12

A train traveling at 80 km/h passes a stationary train of length 200 meters in 12 seconds. What is the length of the moving train?

  1. 662366\frac{2}{3}6632​ meters
  2. 100100100 meters
  3. 13313133\frac{1}{3}13331​ meters
  4. 200200200 meters
Explanation: First convert the speed: 808080 km/h =80×10003600=800003600=2009= 80 \times \frac{1000}{3600} = \frac{80000}{3600} = \frac{200}{9}=80×36001000​=360080000​=9200​ m/s. In 12 seconds, the train travels: 2009×12=24009=26623\frac{200}{9} \times 12 = \frac{2400}{9} = 266\frac{2}{3}9200​×12=92400​=26632​ meters. This distance equals the sum of both train lengths. Length of moving train =26623−200=6623= 266\frac{2}{3} - 200 = 66\frac{2}{3}=26632​−200=6632​ meters. Choice B (100) would result if someone incorrectly calculated the speed conversion. Choice C (13313133\frac{1}{3}13331​) would result from taking half the total distance traveled. Choice D (200) assumes both trains have equal length.

Question 13

A river flows at 3 mph. A boat can travel 24 miles downstream in the same time it takes to travel 16 miles upstream. What is the speed of the boat in still water?

  1. 999 mph
  2. 121212 mph
  3. 151515 mph
  4. 181818 mph
Explanation: Let vvv be the boat's speed in still water. Downstream speed is (v+3)(v + 3)(v+3) mph, upstream speed is (v−3)(v - 3)(v−3) mph. Time downstream: 24v+3\frac{24}{v + 3}v+324​. Time upstream: 16v−3\frac{16}{v - 3}v−316​. Setting them equal: 24v+3=16v−3\frac{24}{v + 3} = \frac{16}{v - 3}v+324​=v−316​. Cross-multiplying: 24(v−3)=16(v+3)24(v - 3) = 16(v + 3)24(v−3)=16(v+3), so 24v−72=16v+4824v - 72 = 16v + 4824v−72=16v+48, giving 8v=1208v = 1208v=120, thus v=15v = 15v=15 mph. Verification: downstream time =2418=43= \frac{24}{18} = \frac{4}{3}=1824​=34​ hours, upstream time =1612=43= \frac{16}{12} = \frac{4}{3}=1216​=34​ hours. ✓ Choice A would result from solving 24v+3=16v−3\frac{24}{v + 3} = \frac{16}{v - 3}v+324​=v−316​ incorrectly. Choice B might come from using the upstream speed as the answer. Choice D might result from using the downstream speed.