GRE rate problems are some of the most challenging and perplexing questions on the GRE Quantitative Reasoning section. They test your ability to apply mathematical concepts such as distance, time, and speed to real-world scenarios. While GRE rate problems can be daunting, they are also some of the most rewarding to solve on the GRE exam. Once you master the basics of rate problems, you will be well on your way to conquering the GRE Quantitative Reasoning section.

In this blog:

## What are GRE Rate Problems?

GRE rate problems are typically word problems that involve two or more objects moving at different speeds. The goal of the problem is to determine the distance travelled by one or more of the objects, the time it takes them to travel that distance, or their relative speed. GRE rate problems can be very challenging, but they are also very fair. The test makers always provide you with all the information you need to solve the problem, but it is up to you to figure out how to use that information correctly.

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## Shorter GRE Rate-Time-Work Formula

Here are some fundamental GRE rate-time-work formulas, that are essential tools for solving various rate-related problems on the GRE. Including those involving vehicles travelling at different speeds, objects moving in opposite directions, and other GRE rate problems. Knowing how to apply these formulas will help you navigate rate-related questions with ease.

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## New GRE Rate-Time-Distance Problems

**Question:**A boat travels upstream for 5 hours and then returns downstream in 3 hours. If the speed of the current is 2 miles per hour, what is the speed of the boat in still water?

**Solution:**

Let the speed of the boat in still water be “B” miles per hour.

When travelling upstream, the boat’s effective speed is (B – 2) miles per hour, and when travelling downstream, it’s (B + 2) miles per hour.

Using the formula: Distance = Speed × Time

Upstream distance = (B – 2) × 5 = 5B – 10 miles

Downstream distance = (B + 2) × 3 = 3B + 6 miles

Since the distances are the same (going upstream and coming downstream), we can set them equal to each other:

5B – 10 = 3B + 6

Solving for B:

2B = 16

B = 8 miles per hour

So, the speed of the boat in still water is 8 miles per hour.

**2. Question: **Two cyclists start from the same point and travel in opposite directions. One cyclist travels at a speed of 12 miles per hour, while the other travels at a speed of 15 miles per hour. How far apart are they after 2 hours?

**Solution: **

Since the cyclists are travelling in opposite directions, you can add their speeds to find the relative speed at which they are moving apart:

Relative speed = 12 mph + 15 mph = 27 mph

Now, use the formula: Distance = Speed × Time

Distance = 27 mph × 2 hours = 54 miles

So, the two cyclists are 54 miles apart after 2 hours.

**3. Question: **A car and a bus start from the same point and travel in the same direction. The car travels at a speed of 60 miles per hour, while the bus travels at a speed of 45 miles per hour. If the bus departs 3 hours later than the car, how long does it take for the bus to catch up to the car?

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**Solution: **

First, calculate the distance the car travels in the 3-hour head start:

Distance = Speed × Time = 60 mph × 3 hours = 180 miles

Now, the bus needs to cover this 180-mile distance to catch up to the car. The relative speed between the bus and the car is:

Relative speed = 60 mph (car’s speed) – 45 mph (bus’s speed) = 15 mph

Now, use the formula: Time = Distance / Speed

Time = 180 miles / 15 mph = 12 hours

So, it takes the bus 12 hours to catch up to the car.

**4. Question: **A plane flies from City A to City B at an average speed of 400 miles per hour and returns from City B to City A at an average speed of 500 miles per hour. If the total time for the round trip is 5 hours, what is the distance between the two cities?

**Solution: **Let D be the distance between the two cities in miles.

To find the time it takes to fly from A to B and back, we can use the formula: Time = Distance / Speed.

Time for the first leg (A to B): D / 400

Time for the second leg (B to A): D / 500

The total time for the round trip is given as 5 hours:

D / 400 + D / 500 = 5

To solve for D, we need to find a common denominator:

(5D/2000) + (4D/2000) = 5

Combine the fractions:

(9D/2000) = 5

Now, solve for D:

9D = 5 * 2000

9D = 10,000

D = 10,000 / 9

D ≈ 1,111.11 miles

So, the distance between the two cities is approximately 1,111.11 miles.

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**5. Question:** A train leaves City X for City Y, travelling at a constant speed. At the same time, another train leaves City Y for City X, also travelling at a constant speed. The two trains meet 6 hours later, and the train from City X has travelled 360 miles farther than the train from City Y. If the train from City X travels at 60 miles per hour, what is the speed of the train from City Y?

**Solution: **

Let the speed of the train from City Y be “Y” miles per hour.

The train from City X travels 360 miles farther, so it covers 360 additional miles in the same time. The time both trains travel is 6 hours.

Using the formula: Distance = Speed × Time

For the train from City X:

Distance = 60 mph × 6 hours = 360 miles

For the train from City Y:

Distance = (Y mph) × 6 hours

Given that the train from City X has travelled 360 miles more than the train from City Y, we can set up the equation:

360 = 6Y

Now, solve for Y:

Y = 360 / 6

Y = 60 miles per hour

So, the speed of the train from City Y is also 60 miles per hour.

**6. Question: **A car travels from Town A to Town B at a speed of 40 miles per hour and returns from Town B to Town A at a speed of 60 miles per hour. If the total travel time is 5 hours, what is the distance between the two towns?

**Solution: **

Let D be the distance between the two towns in miles.

To find the time it takes to travel from A to B and back, we can use the formula: Time = Distance / Speed.

Time for the first leg (A to B): D / 40

Time for the second leg (B to A): D / 60

The total time for the round trip is given as 5 hours:

D / 40 + D / 60 = 5

To solve for D, we need to find a common denominator:

(3D/120) + (2D/120) = 5

Combine the fractions:

(5D/120) = 5

Now, solve for D:

5D = 5 * 120

5D = 600

D = 600 / 5

D = 120 miles

So, the distance between the two towns is 120 miles.

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