GRE Exam: A Simple Guide To Solving Rates Problems

Rate questions always come in one of two forms: distance or work.

In both cases, there will be a constant relationship between the rate, time, and work or distance:

The primary difference in the two equations is the following:
For distance questions, rate represents distance per unit of time and the result represents some distance traveled. For work questions, rate represents output per unit of time and the result represents the number of units produced (such as widgets, lawns mowed, papers, etc.).

Distance Problems: Rate × Time = Distance

The rate × time = distance formula is almost always the best way to solve distance equations. In simpler questions, you will be given two components of the formula and asked to solve for the third.

Example: If a car travels at a constant rate of 50 miles per hour, in how many hours will the car have traveled 325 miles?
Step 1: Set up the rate × time = distance chart:

Step 2: Solve for t:


Unit Conversions
When solving rate questions, be sure that you use the same unit throughout the question. If the rate is expressed per minute, then you must express time in minutes, not seconds or hours.

Example: Running at a constant rate, Bob travels 5 miles in 1 hour. If Bob travels for 20 minutes, how many miles does he travel?
Step 1: Set up the r × t = d chart:
rate (mi/hr) × time (hr) = distance (miles)

Bob’s rate is distance/time = 5 miles/1 hour, and his time is 20 minutes. It might be tempting to simply substitute these values into the original formula, but this would be incorrect. Why? Because the units do not match up. The rate is expressed per hour, whereas the time is expressed in minutes. Before plugging the values into the formula, you should convert the time from 20 minutes to hour.
Step 2: Once you have done so, you can plug the values into the formula:


Multiple Rates
For distance questions, the r × t = d table is particularly useful for problems that involve multiple rates or multiple travelers. In these situations, you will generally want to create two rows on the r × t = d table, fill them in with the appropriate values or variables, and use the resulting expressions to identify a relationship between the distances.

Example: Bob and Jack start at opposite ends of a 200-mile track. Bob travels at a constant rate of 50 miles per hour and Jack travels at a constant rate of 75 miles per hour. If they start traveling toward each other at the same time, in how many hours will they meet?
Step 1: Put the given information into the r × t = d table:

Note that there are two rows—one for Jack and one for Bob. You are asked to solve for the amount of time they travel, so let t represent time. Since Bob and Jack start and end at the same time, they will each have traveled for t hours. In terms of t, Bob travels 50t miles and Jack travels 75t miles.
Now you must identify the relationship between these distances. Since they are traveling toward each other, Bob will travel some of the 200 miles and Jack will travel the remaining distance. Thus the distances the two travel must add up to 200. Algebraically, you can represent this relationship as:

Step 2: Solve for t:

Example: Train A and train B start at the same point and travel in opposite directions. Train A travels at a constant rate of 80 miles per hour, and train B travels at a constant rate of 60 miles per hour. If train A starts traveling 2 hours before train B, how many miles will train A have traveled when the two trains are 720 miles apart?
Step 1: Put the given information into the r × t = d table. The rate for trains A and B are given as 80 miles per hour and 60 miles per hour, respectively. To solve for the number of miles Train A will have traveled, you need to determine how many hours Train A traveled. Let t = the number of hours Train A travels. Since Train A started 2 hours before Train B, and the two trains stop traveling at the same time, Train B must have traveled t – 2 hours.

Thus in terms of t, Train A’s distance is 80t and Train B’s distance is 60(t –2).

As in the previous example, it is essential to identify the relationship between the two trains’ distances. For the two trains to end up 720 miles apart, Train A must cover some portion of the 720 miles and Train B must cover the rest. Thus the sum of their distances is 720.

Expressed algebraically, the equation is:

Step 2: Now solve for t:

You are asked to solve for the number of miles Train A traveled, so plug 6 in for Train A’s time:

Train A’s distance is 80 × 6 = 480.

Don’t Average Rates!

Look at the following question:
Sarah goes on a 600-mile trip. She travels at a constant rate of 50 miles per hour for the first 300 miles of the trip, and at a constant rate of 100 miles per hour for the last 300 miles of the trip. What is Sarah’s average speed in miles per hour for the entire trip?

70
75
80
85

Solution: Many test-takers are tempted to average the rate of 50 and 100 to arrive at an average speed of 75. Though perhaps intuitive, this would be incorrect. Why? Because Sarah spent more time traveling at a rate of 50 miles per hour than at a rate of 100 miles per hour. Her overall rate will thus be closer to 50 than to 100. Based on this logic alone, you can immediately eliminate C, D, and E.

To actually calculate her average speed, you should use the rate formula: r × t = d formula to determine the time for each part of the trip:

Solve for x:

Solve for y:

The total time is x + y = 6 + 3 = 9. The average speed for the trip is thus .
The correct answer is Choice A.

Work Problems: Rate × Time = Work
The second type of rate problem involves work. In contrast to distance problems, work problems are concerned with some output per unit of time.

An output can be something produced (such as widgets, cups, cars, etc.) or a job done (such as mowing a lawn, cooking a meal, writing a paper, etc.). In both cases, you want to use the work formula, though as you will see, the way you express work will differ.

Let’s look at a simple work question:

If a machine produces pencils at a constant rate of 1,500 pencils per hour, in how many hours will the machine have produced 6,750 pencils?
SOLUTION: Put the given values into the rate × time = work (RTW) table. The machine’s rate is 1,500 pencils/hour, and its output (work) is 6,750 pencils.

Since you are trying to solve for time, assign a variable: t

Solve for t:

Let’s look at another example, this time with work represented as some job done instead of units produced:
Working at a constant rate, Bob can mow 3 same-sized lawns in 5 hours. How many hours will it take Bob to mow 2 same-sized lawns?

SOLUTION: Set up the RTW table. Remember that , so Bob’s rate will be .

Solve for t:


Combining Rates
In certain work questions, you will be given the individual rates of two or more elements and will be asked about what happens when they work together.
 

Example: Working alone at a constant rate, Bob can mow 1 lawn in 3 hours. Working alone at a constant rate, Jack can mow 1 same-sized lawn in 8 hours. If Bob and Jack work together but independently at their respective constant rates, how many hours will it take them to mow half a same-sized lawn?

SOLUTION: Bob’s rate is . Jack’s rate is .

To determine how long it takes them to mow the lawn when they work together, you need their combined rate.

The combined rate is the sum of their individual rates.

In this case, their combined rate is .

This means that, when they work together, Bob and Jack can mow of a lawn in 1 hour.

Now input this rate into the r × t = w formula:

Note that you input for the total work done, since the question is asking for the time necessary for them to mow half the lawn.

Solve for t: