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GRE Rates / Rate Questions | TTP GRE Blog

One of the major GRE quantitative reasoning topics you’re likely to see on test day centers on GRE rate questions. While rate questions come in many shapes and sizes, two popular types of rate questions are rate-time-distance questions and rate-time-work questions. In this article, we will cover both of those types of rate problems and how the GRE can test you on them.

Here are the topics we’ll cover:

Before getting into the details of specific types of GRE rate and work problems, let’s first discuss the concept of “rate.”

What Is a Rate?

In short, a rate is a ratio or fraction that compares two different units of measurement. Some more common rates that you are likely familiar with, from driving, are miles per hour (miles/hour) and kilometers per hour (kilometers/hour). But rates can be just about anything. For example:

  • Seltzers consumed per week: (seltzers/week)
  • Sneakers made per day: (sneakers/day)
  • Calories burned per mile: (calories/mile)


A rate is a ratio that compares two different units of measurement.

Now that we have a good idea of what a rate actually is, we can discuss a specific type of rate problem: rate-time-distance.


One of the more common types of rate questions involves the following formula:

Rate x Time = Distance

This formula can also be expressed as:

  • Time = Distance / Rate
  • Rate = Distance / Time

In general, you will know that you are working with a rate-time-distance problem when you’re asked things such as:

  • How fast was she driving?
  • What was the speed of the train?
  • What time did he arrive?
  • How long did the trip take?
  • How many miles did the plane travel?
  • What is the distance to and from work?


To solve distance problems, use the following formula: rate x time = distance.

When solving these types of math problems, we will use some version of the rate-time-distance formula. So, of course, it’s ideal to have this formula memorized.

To start, let’s do a few very basic rate problems, so we can practice using the formula. We will do a problem each asking for a rate, a time, or a distance.

Rate Example 1: Distance (Easy)

Martha drives from home to work at a constant rate of 40 miles per hour. If it takes her 1.5 hours to get to work, how many miles does she drive from home to work?


We are asked to determine how many miles she drives, so this is a question about distance. Thus, we will use the distance formula, d = r x t, where the rate is 40 miles per hour and time is 1.5 hours:

d = r x t

d = 40 x 1.5

d = 60 miles

Answer: E

Rate Example 2: Rate (Easy)

If an ant walks at a constant rate of 6 kilometers in a two-day period, what is the ant’s rate, in kilometers per day?


We need to determine the rate of the ant in kilometers per day. Thus, we use the formula for rate, r = d / t . We know that the distance is 6 kilometers and the time is 2 days.

r = d / t

r = 6 / 2

r = 3 kilometers per day

Answer: B

Rate Example 3: Time (Easy)

Vladimir walks at a constant rate of 100 steps per minute. If he walks to a store 500 steps away, how many minutes does it take him to walk to the store?


Our goal here is to calculate the time it took Vladimir to walk to the store. Thus, we use the formula t = d / r. We are given that the distance he travels is 500 steps, and his rate is 100 steps per minute.

t = d / r

t = 500 / 100

t = 5 minutes

Answer: B

Let’s try another, slightly more difficult example and employ the tips and tricks we’ve just learned to solve it.

Rate Example 4: Time (Medium)

Greta drove 300 miles from Pittsburgh to Philadelphia. If she drove at an average speed of 45 miles per hour for the first 3 hours of the trip and at an average speed of 55 miles per hour for the remainder of the trip, for how many hours did Greta drive at an average speed of 55 miles per hour?


Since Greta drove 45 mph for the first 3 hours of the trip, she drove a distance of 45 × 3 = 135 miles. Since the total distance from Pittsburgh to Philadelphia is 300 miles, the remaining distance is 300 – 135 = 165 miles.

Since we know that she drove the remaining distance of 165 miles at 55 mph, we can determine the time she drove at that average rate. Because we want to calculate the time, we use the formula t = d / r:

t = 165 / 55

t = 3

Greta drove for 3 hours at an average speed of 55 miles per hour.

Answer: D

So, now that we have practiced a few basic rate-time-distance questions, let’s look at how the GRE question-writers can throw in a wrinkle by requiring us to convert units in order to arrive at the correct answer.

Make Sure That Units Match When Performing Rate Calculations

For the problems above, we did not have to adjust any of the units of rate, time, or distance. However, in more challenging rate questions, we may need to adjust units before doing our calculations.

For example, we could be given a question that asks for the number of miles traveled but provides a rate of 30 miles per hour and a time of 20 minutes. In this case, we must recognize that we cannot multiply a rate in miles per hour by a time in minutes. Thus, before performing the multiplication, we must convert the rate to miles per minute or the time to hours. The easier of the two options is converting the time to hours.

Since there are 60 minutes in an hour, we convert 20 minutes to hours by dividing by 60, so we have 20/60 = 1/3 hours. Now that we have a rate of 30 miles per hour and a time of 1/3 hour, we have agreement of units, and we can calculate the distance:

distance = rate x time

distance = 30 x 1/3 = 10 miles


Before performing any rate calculations, ensure that the units match.

Let’s practice with an example.

Rate Example 5: Units (Medium)

A walrus swims at a constant rate of 7 miles per hour. If it travels at this rate for 12 minutes, how far does the walrus travel?

  • 1.4 miles
  • 1.71 miles
  • 7.2 miles
  • 8.4 miles
  • 14 miles


First, we note that the question is asking us to compute the distance traveled by the walrus. Thus, we know that we will use the formula distance = rate x time.

The walrus’ rate is 7 miles per hour, and its time is 12 minutes. Thus, the time units don’t match. We have two options: (1) convert the rate of 7 miles per hour to miles per minute or (2) convert 12 minutes to hours. It would appear that the second option is more straightforward.

We can convert 12 minutes to hours by using the fact that there are 60 minutes in 1 hour. We can set up a ratio to solve for the number of hours:

1 hour / 60 minutes = y hours / 12 minutes

Cross-multiplying, we have:

60 y = 12

We divide both sides by 60 to solve for y:

y = 12 / 60 = 1/5 hour

Now we know the walrus’ rate is 7 miles per hour and its time is 1/5 hour. We can now use the distance formula to determine the walrus’ traveling distance:

distance = rate x time

distance = 7 x 1 / 5

distance = 7/5 = 1.4 miles

Answer: A

Some of the More Common Types of Rate-Time-Distance Problems on the GRE

While rate-time-distance questions (otherwise known as speed-distance-time questions) can take on many forms, there are several types of rate questions that are more challenging than the ones we have covered up to this point. We will cover one more type here, but if you want additional help with this category of questions, check out the Target Test Prep GRE self-study course, where we cover the following speed problems in greater detail, including the specific math formulas you need to know to solve them:

  • Average Rates
  • Converging Rate Distance Problems
  • Diverging Rate Problems
  • Round Tip Rate Problems
  • Catch-Up Rate Problems
  • Catch-Up and Pass Rate Problems

As you study GRE rates, you’ll need to begin to recognize when each type of GRE rate problem is tested. For example, when two cars are driving toward each other, you should recognize that you have a converging rate question.

Now, let’s discuss average rates.

Average Rates on the GRE

Average rates are a specific type of rate question that you might encounter on your GRE. To solve for an average rate, you need to know the total distance traveled by an object and the total time it takes to travel that distance. With that, we can determine the average rate, using the following formula:

Average rate = (total distance) / (total time)


We determine the average rate by dividing the total distance by the total time.

Usually, when dealing with average rate questions, we are given two values for distance and two for time, which we plug into the average rate formula. So, we can express the average rate formula as:

Average rate = (distance one + distance two) / (time one + time two)

Now that we are clear on the average rate formula, let’s practice with an example.

Rate Example 6: Average Rate (Medium)

Rosa travels from home to work and then from work to home each day, taking the same 10-mile route each way. If she travels to work at a constant rate of 30 miles per hour and home from work at a constant rate of 40 mph, what is Rosa’s average speed, in miles per hour, when traveling to and from work?


We need to calculate Rosa’s average rate (speed) for her trip to and from work. We will use the formula for average rate:

average rate = (distance one + distance two) / (time one + time two)

The route she takes is the same distance (10 miles) for each leg of her trip. Thus, we know that distance 1 = distance 2 = 10 miles.

We have to calculate each of the times individually for each leg of her trip. Her trip to work is a distance of 10 miles, and her rate is 30 mph, so her home-to-work time is:

t = d / r

t = 10 / 30

t = 1/3 hour

Now let’s calculate the time for her return trip, from work to home. Again, the distance is 10 miles, but now her rate is 40 mph. Thus, her work-to-home time is:

t = d / r

t = 10 / 40

t = 1/4

Now that we know the two times of 1/3 hour and 1/4 hour, we can use the average rate formula to calculate the average rate of Rosa’s round trip.

average rate = (distance one + distance two) / (time one + time two)

average rate = (10 + 10) / (1/3 + 1/4)

average rate = 20 / (4/12 + 3/12)

average rate = 20 / 7/12

average rate = 240 / 7 = 34 2/7

Answer: B

Next, let’s discuss one other type of rate question: work problems.

Rate-Time-Work Problems

So, I have some good news! Work problems are quite similar to distance problems. The only difference is that instead of calculating distance with a rate and a time, we calculate work. We can consider “work” some sort of job or task that is completed. For example:

  • a hose filling a pool
  • a person shoveling a walkway
  • a person painting a house
  • a machine completing a job.

All of the above scenarios (and many more) would constitute work.

Let’s review the rate-time-work formula.

The Rate-Time-Work Formula

We use the following formula for work:

Rate x Time = Work

This formula can also be expressed as:

  • Time = Work/Rate
  • Rate = Work/Time


To solve work problems, use the formula rate x time = work.

Next, let’s discuss an important yet tricky component of all work questions: determining an object’s work-rate.

Determining the Work-Rate of an Object

We already know that the work-rate of an object is always expressed as work/time. In some cases, it will be straightforward to determine an object’s work, and in other cases, less so. For example:

  • Tom can wash 12 plates every 5 hours.

Thus, Tom’s work is 12, and his time is 5, for a rate of 12/5 plates/hour.

  • A machine can produce 15 toys in 1 day. Thus, the work is 15 and the time is 1, for a rate of 15/1 = 15 toys/day.

While determining the work rate for the above scenarios was pretty straightforward, doing so gets a little more challenging when the work value is implied as one job. For example:

  • A particular hose can fill a pool in 3 hours.

While we are not told specifically that work is equal to 1, it’s implied that “filling a pool” is 1 job. Thus, work is 1 and time is 3, for a rate of 1/3 pool/hour.

Now that we are familiar with the rate-time-work formula and can see how to determine an object’s work-rate, let’s practice a basic work-rate problem.

Work Example 1: Time (Easy)

A soda machine working at a constant rate can fill 90 cans of soda in 2 hours. In how many hours will the soda machine fill 270 cans?


For this work problem, we are given that the rate is 90 cans / 2 hours, which can be reduced to 45 cans per hour. We are asked to determine the time it takes to perform the work of filling 270 cans. Using the formula for time, we have:

time = work / rate

time = 270 / 45

time = 6 hours

Answer: D

Let’s try another example.

Work Example 2: Time (Medium)

Eric can solve 3 homework problems in 25 minutes. He works at a constant rate and takes no breaks. The number of hours it will take him to solve 26 problems is closest to which of the following?


We see that this is a work problem, and we are asked to determine the time. Thus, we will use the formula time = work / rate.

We see that the rate is 3/25 problems / minute. The total work is 26 problems. We substitute these values into the formula and solve for time:

time = work / rate

time = 26 / (3/25)

time = 26 x (25/3)

time = 650 / 3 minutes

Notice that the question is asking the time in hours to solve 26 problems. So, we have to convert 650 / 3 minutes to hours by dividing by 60. Thus, we have:

Time = (650 / 3) / 60

Time = (650 / 3) x (1 / 60)

Time = 65 / 18, or about 3.8 hours.

Answer: B

Next, let’s discuss a popular type of work problem: the combined worker problem.

Combined Worker Problems

Many work problems involve two or more objects working together. We refer to these problems as combined worker problems. Some of the most commonly tested GRE combined work-time problems involve two objects or workers working together to complete a job. The formula that we use to solve these problems is quite intuitive:

work of worker one + work of worker two = total work

In some cases, the total work will be greater than one, and in other cases, it will be one if the total work is one job. Also, while the objects work together in combined worker problems, sometimes, they work for the same amount of time, and other times they do not.

Let’s practice with one of each scenario below.

Work Example 3: Combined Worker (Medium)

Working at a constant rate, Tyrone can paint a fence in 4 hours. Working at a constant rate, his brother can paint the same fence in 2 hours. How long will it take them to paint the fence if they both work together at their respective constant rates?


Tyrone’s rate is 1 fence in 4 hours, or 1/4 fence per hour. His brother’s rate is 1 fence in 2 hours, or 1/2 fence per hour. Because the two brothers are working together, their combined rate is the sum of their individual rates:

combined rate = 1/4 + 1/2

combined rate = 1/4 + 2/4 = 3/4 fence per hour

We know that the total work is 1 fence painted. We need to determine the total time it takes to complete the work, so we will use the formula time = work / rate:

time = work / rate

time = 1 / (3/4)

time = 4/3 hours

Answer: B

Now, let’s practice an example in which the workers do not work for the same amount of time.

Work Example 4: Combined Worker (Hard)

Working alone at a constant rate, machine A takes 2 hours to build a car. Working alone at a constant rate, machine B takes 3 hours to build the same car. If they work together for 1 hour at their respective constant rates, and then machine B breaks down, how much additional time will it take machine A to finish the car by itself?


Let’s first determine the rates of machine A and machine B and their combined rate. First, machine A can build 1 car in 2 hours, so A’s rate is 1/2 car per hour. Machine B can build 1 car in 3 hours, so B’s rate is 1/3 car per hour.

The combined rate is the sum of their respective rates:

combined rate = 1/2 + 1/3

combined rate = 3/6 + 2/6

combined rate = 5/6 car per hour.

We are told that the two machines worked together for 1 hour. So, the total work is:

work = rate x time

work = 5/6 x 1

work = 5/6 car

So, for the 1 hour they worked together, machines A and B produced 5/6 of a car.

After machine B broke down, machine A had to finish building the car by itself. Thus, machine A had 1 – 5/6 = 1/6 of a car to build, working by itself. Let’s find out how much time it took for machine A to do this work.

time = work / rate

time = (1/6) / (1/2)

time = (1/6) x 2

time = 2/6 = 1/3 hour

Answer: C


The GRE often tests rates, so it’s important to be familiar with what rate problems look like and learn the appropriate math tips and problem-solving strategies to help you solve those problems.

A rate is a fraction or ratio that compares two different units of measurement. Examples include miles per hour, cars sold per month, and pages read per minute.

When you are solving rate practice problems, it’s important to make sure that the units match before you finish the problem. For example, a question could provide you a rate in gallons per minute, but the answer choices give you the time in hours. Unit conversion is necessary.

Rate-time-distance problems require that you use the formula rate x time = distance to solve problems involving those three units.

Rate-time-work problems are similar to rate-time-distance problems, and the formula is similar: rate x time = work. But, in this case, you are dealing with situations in which work is performed. Examples of work include filling a pool, picking apples, or manufacturing a particular number of widgets.

An advanced type of rate problem is a combined worker problem, which involves two or more workers (people, machines, etc.) that have individual rates but work in concert to finish a particular job. The key to being successful with this type of problem is to combine the rates by adding them, and then using the appropriate formula to obtain the answer.

What’s Next?

The rates discussed in this article are examples of ratios, a topic that is also a highly tested one on the GRE. You can benefit from a detailed review of ratios and more math practice by reading our article about part-to-part and part-to-whole ratios.

Rizwan Ahmed
Rizwan Ahmed
AuditStudent.com, founded by Rizwan Ahmed, is an educational platform dedicated to empowering students and professionals in the all fields of life. Discover comprehensive resources and expert guidance to excel in the dynamic education industry.


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