GMAT Math : Rate Problems

Study concepts, example questions & explanations for GMAT Math

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Example Questions

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Example Question #1 :Rate Problems

A group of students are making posters to advertise for a bake sale. 12 large signs and 60 small signs are needed. It takes 10 minutes to paint a small sign and 30 minutes to paint a large sign. How many students will be needed to paint all of the signs in 2 hours or less?

Possible Answers:

Correct answer:

Explanation:

In 2 hours, 1 student can paint 4 large signs or 12 small signs. Therefore, 3 students are required to paint the large signs () and 5 students are required to paint the small signs (). In total, 8 students are required.

Example Question #1 :How To Do Distance Problems

Jason is driving across the country. For the first 3 hours, he travels 60 mph. For the next 2 hours he travels 72 mph. Assuming that he has not stopped, what is his average traveling speed in miles per hour?

Possible Answers:

Correct answer:

Explanation:

In the first three hours, he travels 180 miles.

In the next two hours, he travels 144 miles.

for a total of 324 miles.

Divide by the total number of hours to obtain the average traveling speed.

Example Question #1 :Calculating Rate

Tom runs a 100m race in a certain amount of time. If John runs the same race, he takes 2 seconds longer. If John ran at 8m/s, approximately how fast did Tom run?

Possible Answers:

Correct answer:

Explanation:

Tom runs a 100m race in a certain amount of time. If John runs the same race, he takes 2 seconds longer. If John ran at 8m/s, how fast did Tom run?

Letdenote the amount of time that it took Tom to run the race. Then it took Johnseconds to run the same race going 8m/s. At 8m/s, it takes 12.5 seconds to finish a 100m race. This means it took Tom 10.5 seconds to finish. Running 100m in 10.5 seconds is the same as

Example Question #4 :Rate Problems

Jim and Julia, a married couple, work in the same building.

One morning, both left at 9:00, but in different cars. Jim arrived at 10:10; Julia arrived 10 minutes later. If Jim's average speed was 54 miles per hour, what was Julia's average speed (nearest whole number)?

Possible Answers:

Correct answer:

Explanation:

Jim arrived at the common destination in 70 minutes, orhours. His average speed was 54 miles per hour, so their workplace is

远离吉姆和茱莉亚的家里e.

Julia traveled those 63 miles in 80 minutes, orhours, so her average speed was

,

or, rounded, 47 miles per hour.

Example Question #5 :Rate Problems

Andy and his wife Donna both work at the same building.

One morning, Andy left home at 8:00; Donna left 5 minutes later. Each arrived at their common destination at 8:50. Andy drove at an average speed of 45 miles per hour; what was Donna's average speed, to the nearest mile per hour?

Possible Answers:

Correct answer:

Explanation:

Andy traveled for 50 minutes, orof one hour, at 45 miles per hour, so the office building ismiles from Andy and Donna's house. Donna traveled this distance for 45 minutes, orof one hour, so her speed was:

miles per hour.

Example Question #6 :Rate Problems

Kenny and Marie, a married couple, work in the same building.

One morning, both left at 9:00, but in different cars. Kenny arrived at 10:10; Marie arrived 10 minutes later. If Kenny's average speed was 6 miles per hour faster than Marie's, how far is their work place from their home (nearest whole mile)?

Possible Answers:

Correct answer:

Explanation:

Letbe the rate at which Kenny drove. Then Marie drove at a rate ofmiles per hour. The two drove the same distance, so, since Kenny drove 70 minutes, orhours, and Marie drove for 80 minutes, orhours, we can use the formulato set up the equation:

Since Kenny traveled at 48 miles per hour forhours, the distance each drove is

miles.

Example Question #7 :Rate Problems

Jerry took a car trip of 320 miles. The trip took a total of six hours and forty minutes; for the first four hours, his average speed was 60 miles per hour. What was his average speed for the remaining time?

Possible Answers:

Correct answer:

Explanation:

Jerry drove 60 miles per hour for 4 hours - that is,miles.

He drove the remainder of the distance, ormiles over a period ofhours, so his average speed was

miles per hour.

Example Question #1 :Calculating Rate

If it takes Sally 3 hours to drive\dpi{100} \small qmiles, how many hours will it take her to drive\dpi{100} \small rmiles at the same rate?

Possible Answers:

\dpi{100} \small \frac{r}{3q}

\dpi{100} \small \frac{3q}{r}

\dpi{100} \small \frac{3r}{q}

\dpi{100} \small \frac{qr}{3}

\dpi{100} \small \frac{3}{qr}

Correct answer:

\dpi{100} \small \frac{3r}{q}

Explanation:

If Sally drives q miles in 3 hours, her rate is 3/q miles per hour. Plug this rate into the distance equation and solve for the time:

\dpi{100} \small Distance = rate\times time

\dpi{100} \small r=\frac{q}{3}\times t

\dpi{100} \small t=\frac{3r}{q}

Example Question #1 :Rate Problems

A cat runs at a rate of 12 miles per hour. How far does he run in 10 minutes?

Possible Answers:

\dpi{100} 2\ miles

\dpi{100} 1\ mile

None of the other answers are correct.

\dpi{100} 12\ miles

\dpi{100} 10\ miles

Correct answer:

\dpi{100} 2\ miles

Explanation:

We need to convert hours into minutes and multiply this by the 10 minute time interval:

\small \frac{12\ miles}{1\ hour}x\frac{1\ hour}{60\ min}x\frac{10\ min}{1}=\frac{120\ miles}{60}=2\ miles

Example Question #2 :Calculating Rate

In order to qualify for the next heat, the race-car driver needs to average 60 miles per hour for two laps of a one mile race-track. The driver only averages 40 miles per hour on the first lap. What must be the driver's average speed for the second lap in order to average 60 miles per hour for both laps?

Possible Answers:

80 miles per hour

90 miles per hour

100 miles per hour

120 miles per hour

240 miles per hour

Correct answer:

80 miles per hour

Explanation:

If the driver needs to drive two laps, each one mile long, at an average rate of 60 miles per hour. To find the average speed, we need to add the speed for each lap together then divide by the number of laps. The equation would be as follows:

In our case we know lap one was driven atmiles per hour. We substitute this value in forand solve for.

Thus to averagemiles per hour for two laps with lap one beingmiles per hour, lap two would have to have a rate ofmiles per hour.

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