GRE average speed problems trip up even well-prepared students because the correct answer almost never matches what you'd expect from simply averaging two speeds. Average speed equals total distance divided by total time — not the mean of the individual speeds — and the GRE uses that misconception as a deliberate trap. This guide breaks down every formula, shortcut, and strategy you need to solve these problems quickly and accurately on test day.
The formula is deceptively simple: Average Speed = Total Distance ÷ Total Time. That's it. No matter how many segments a journey has, you add up every mile traveled, add up every hour spent traveling, and divide. The word "average" in average speed does not mean arithmetic mean — it means the single constant speed that would cover the same total distance in the same total time.
You can expect approximately 1-2 rate-related questions per GRE quantitative section, and average speed problems are among the most common. These questions appear in multiple formats: multiple choice, numeric entry, and quantitative comparison.
Here's the trap that catches most students: if you drive 60 mph for one leg and 40 mph for another leg of equal distance, the average speed is not 50 mph. It's 48 mph. The reason? You spend more time traveling at the slower speed. At 60 mph, you cover 120 miles in 2 hours. At 40 mph, you cover 120 miles in 3 hours. That's 240 miles in 5 hours, giving an average of 48 mph.
The arithmetic mean (adding speeds and dividing by 2) only gives the correct average speed when you spend equal time at each speed — not equal distance. Since most GRE problems involve equal distances (like round trips), the simple average is almost always wrong.
| Speeds | Wrong Answer (Mean) | Correct Average Speed | Difference |
|---|---|---|---|
| 40 mph and 60 mph | 50 mph | 48 mph | 2 mph lower |
| 30 mph and 90 mph | 60 mph | 45 mph | 15 mph lower |
| 20 mph and 60 mph | 40 mph | 30 mph | 10 mph lower |
| 50 mph and 50 mph | 50 mph | 50 mph | 0 (equal speeds) |
Worked Example
A driver travels 120 miles at 60 mph, then 120 miles at 40 mph. What is the average speed for the entire trip?
Every GRE average speed problem builds on one relationship: Distance = Rate × Time. This formula has three forms depending on what you need to find:
For average speed problems, you'll most often use T = D ÷ R to calculate how long each segment of a journey takes, then sum those times to get total time.
| Scenario | Formula | When to Use |
|---|---|---|
| Basic distance-rate-time | D = R × T | Any problem involving distance, speed, and time |
| Find average speed | Total Distance ÷ Total Time | Any multi-segment journey |
| Round trip (equal distances) | 2UV ÷ (U + V) | Same distance at two different speeds |
| Find time from distance and rate | T = D ÷ R | When time is the unknown |
| Find rate from distance and time | R = D ÷ T | When speed is the unknown |
The hardest part of GRE rate problems isn't the math — it's translating the word problem into an equation. Start by identifying what the problem gives you and what it asks for. Create a simple table for each segment with three columns: Distance, Rate, and Time. Fill in what you know, then use D = R × T to find the missing value.
For example, "A train travels for 3 hours at 80 km/h" gives you R = 80 and T = 3, so D = 240 km. Once you have all segments filled in, finding average speed is straightforward addition and division.
Enter the distance and speed for each segment of a two-part journey. See the true average speed compared to the arithmetic mean trap.
When you travel the same distance at two different speeds — which is exactly what happens on a round trip — there's a powerful shortcut. Let D be the one-way distance, U be the speed going, and V be the speed returning:
The distance D cancels out completely, which means the average speed for any equal-distance round trip depends only on the two speeds — not on how far you actually traveled. This is the harmonic mean of the two speeds, and for equal distances it is always less than or equal to the arithmetic mean.
Use 2UV/(U+V) whenever a problem involves two legs of equal distance at different speeds. This covers round trips, there-and-back scenarios, and any problem that says "the same route" or "equal halves." Do not use it when the distances are unequal — for those problems, you must calculate total distance and total time separately.
Worked Example
Sam drives to work at 30 mph and returns home along the same route at 50 mph. What is Sam's average speed for the round trip?
This is the single most common trap in GRE average speed problems. The test makers know that your instinct is to add the two speeds and divide by 2. They deliberately include that wrong answer among the choices. If you see an average speed question with speeds U and V traveling equal distances, expect (U+V)/2 to be sitting right there as a tempting but incorrect option.
The correct answer is always lower than the arithmetic mean (for equal-distance trips). Use this as a quick sanity check: if your answer equals the simple average, you probably made the classic mistake.
GRE speed distance time problems sometimes mix units to create confusion. You might be given a speed in miles per hour but a time in minutes, or a distance in kilometers with speed in meters per second. Before plugging numbers into any formula, verify that all your units match. The most common conversion: divide minutes by 60 to convert to hours when working with mph.
Some GRE problems never tell you the actual distance — they just give you two speeds and ask for average speed. When no distance is provided, pick a smart number. The best choice is the LCM (Least Common Multiple) of the two speeds, because it gives you clean integer times for each segment. For speeds of 40 mph and 60 mph, use LCM(40, 60) = 120 miles. Time at 40 mph = 3 hours, time at 60 mph = 2 hours. No messy fractions.
Follow this process for every GRE rate problem and you'll avoid the common pitfalls:
When a problem gives two speeds but no distance, the LCM method eliminates fractions and speeds up your calculation. Here's how it works: if the two speeds are 50 km/h and 75 km/h, find LCM(50, 75) = 150. Assume each leg is 150 km. Time at 50 = 3 hours, time at 75 = 2 hours. Total = 300 km in 5 hours = 60 km/h. Clean arithmetic, no calculator needed.
| Problem Type | Difficulty | Frequency | Key Formula |
|---|---|---|---|
| Basic D = R × T | Easy | Common | D = R × T |
| Average speed (two segments) | Medium | Common | Total D ÷ Total T |
| Round trip average speed | Medium | Common | 2UV ÷ (U + V) |
| Multi-segment journey | Medium–Hard | Less common | Total D ÷ Total T |
| Relative speed (same direction) | Hard | Rare | Speed₁ − Speed₂ |
| Work rate problems | Medium–Hard | Common | R × T = W |
Worked Example
A cyclist rides 18 km at 36 km/h, then rides the remaining 72 km at 72 km/h. What is the average speed for the entire 90 km trip?
Test your understanding with these GRE-format questions. Each one targets a different aspect of average speed problem-solving. Select your answer and check the explanation.