Welcome back to our series on GMAT quant rate problems. This article and the next will introduce concepts related to *average speed*. Though average speed is a simple concept, there are many ways for GMAT quant problems to “trick” test-takers with average speed scenarios.

## Official GMAT quant rate problems: Concepts related to average speed

Let’s start by clearing up a common misconception about average speed, with the help of **an example:**

Imagine that you decide to take a day trip to visit a historic town. First, you have to get out of your city, and because of heavy traffic, your average speed over the first half of the trip, by distance, is only 30 miles per hour. The second half of the trip is all highway, and you average 60 miles per hour. **What was your average speed for the whole trip?**

If you said 45 miles per hour, keep reading, because that’s incorrect. It’s true that 45 splits the difference between 30 and 60, but that isn’t how average speed works. Let’s assign a distance to the trip in the scenario above to see why:

Let’s say the trip to the town is 120 miles. You drove the first 60 miles of the trip at an average speed of 30 miles per hour, which took 2 hours. Then you drove the last 60 miles of the trip at an average speed of 60 miles per hour, which took 1 hour. In all, you covered 120 miles in 3 hours for an average speed of 40 miles per hour (since speed = distance/time).

Why is the average speed closer to 30 miles per hour than 60 miles per hour? Because average speed does not depend on the relative distances covered at different speeds; it depends on the relative *times spent traveling* at different speeds. Since you spent twice as much time driving at 30 mph (2 hours to 1 hour at 60 mph), your average speed is “twice as close” to 30 mph as to 60 mph (40 – 30 = 10, 60 – 40 = 20). If you had driven 90 miles, spending one hour averaging 30 mph and one hour averaging 60 mph, *then* your average speed would be 45 mph.

### Official GMAT problem involving average speed

**A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C at an average speed of 60 kilometers per hour. If the train did not stop at Station B, what was the average speed at which the train traveled from Station A to Station C?**

**The distance that the train traveled from Station A to Station B was 4 times the distance that the train traveled from Station B to Station C.****The amount of time that it took the train to travel from Station A to Station B was 3 times the amount of time that it took the train to travel from Station B to Station C.**

**(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.**

**(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.**

**(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.**

**(D) EACH statement ALONE is sufficient.**

**(E) Statements (1) and (2) TOGETHER are not sufficient.**

Given what we just observed about how average speed works, this question stem should lead you to think something like, “The average speed of the train depends on the ratio of travel times from Station A to Station B and from Station B to Station C.” T**his makes statement 2 look very promising, and in fact statement 2 is sufficient on its own.** You can check this by “plugging in” times of 3 hours and 1 hour or 6 hours and 2 hours. As long as you use times in the specified ratio of 3:1, the train’s average speed will come out to 75 kph.

**But if you dismissed statement 1 as the wrong kind of information (distance instead of time), then you got this one wrong.** Since we have the average speeds for each leg of the trip, knowing the ratio of distances for the two legs is sufficient to determine the overall average speed.

**Statement 1** tells us that the “A to B” distance is 4 times the “B to C” distance. Since we know from the question stem that the “A to B” average speed was 4/3 of the “B to C” average speed (80 kph to 60 kph), we can calculate that the A to B leg took three times as much time as the B to C leg.

- distance AB = 4(distance BC)
- speed AB = (4/3)(speed BC)
- Time = distance / speed
- Time AB = 4(distance BC) / (4/3)(speed BC)
- Time AB = 3(distance BC / speed BC)
- Time AB = 3(Time BC)

**So statement 1 ultimately provides the same data as statement 2:** the ratio of travel times for the AB and BC legs of the trip. Each statement on its own is sufficient, and** the correct answer is D**.

### Official GMAT data sufficiency average speed problem:

**On a certain nonstop trip, Marta averaged ***x*** miles per hour for 2 hours and ***y ***miles per hour for the remaining 3 hours. What was her average speed, in miles per hour, for the entire trip?**

**2***x***+ 3***y***= 280***y***=***x***+ 10**

**(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.**

**(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.**

**(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.**

**(D) EACH statement ALONE is sufficient.**

**(E) Statements (1) and (2) TOGETHER are not sufficient.**

If you said something like, “There are two variables *x* and *y*, so I will need two equations to solve the system,” and** went straight for answer choice C, you got this one wrong.** Counting variables is a good step, but it’s not foolproof.

**Never forget the fundamental relationships between speed, distance, and time!** If, as we are told, Marta averaged *x* miles per hour for 2 hours, then 2*x* represents the time of 2 hours multiplied by the speed of *x* miles per hour! Since speed * time = distance, 2*x* is the distance covered during the first 2 hours of the trip. By the same reasoning, 3*y* is the distance covered during the final 3 hours of the trip. So together, 2*x* + 3*y* represents the total distance for Marta’s trip.

Therefore **statement 1** provides us with a total trip distance of 280 miles. Since we know from the question stem that the trip took a total of 5 hours (2 + 3), we can use this distance of 280 miles to set up Marta’s average speed as 280 miles / 5 hours = (280 / 5) miles per hour. **Statement 1 on its own is sufficient.**

**Statement 2** tells us, algebraically, that Marta’s average speed for the last 3 hours was 10 miles per hour faster than her average speed for the first 2 hours. This is insufficient on its own, because we don’t know whether the speeds *x* and *y* are 10 mph and 20 mph or 110 mph and 120 mph!

This problem is an excellent demonstration of an important data sufficiency principle: on an “applied” DS problem with a real-world scenario like this one, *think about what the variables in the equations mean in terms of the real-world elements***,** like speed, distance, and time. Divorcing the algebra from the real-world scenario it represents increases your chances of missing something.

### Official GMAT problem:

**Khalil drove 120 kilometers in a certain amount of time. What was his average speed, in kilometers per hour, during this time?**

**If Khalil had driven at an average speed that was 5 kilometers per hour faster, his driving time would have been reduced by 20 minutes.****If Khalil had driven at an average speed that was 25% faster, his driving time would have been reduced by 20%.**

**(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.**

**(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.**

**(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.**

**(D) EACH statement ALONE is sufficient.**

**(E) Statements (1) and (2) TOGETHER are not sufficient.**

This is a tricky one. If you’ve really absorbed the content in this article series about the inverse relationship between speed and time, the values in **statement 2 might jump out at you.** Let’s translate those percentage changes to fractional factors.

An average speed increase of 25% means an average speed multiplication by a factor of 5/4. And a driving time reduction by 20% means a driving time multiplication by a factor of 4/5. Notice that these fractions are reciprocals. **So statement 2 tells us nothing that we don’t already know** as part of our fundamental knowledge base for speed, distance, and time! If a variable like speed is multiplied by a given factor, an inversely-related variable like time is multiplied by the reciprocal of that factor.

**Statement 2 will not contribute anything to answering the question** about Khalil’s average speed. Either statement 1 is sufficient on its own (answer A), or it’s not, and statement 2 doesn’t help (answer E). The usefulness of the data in statement 1 might be hard to recognize, but we can turn it into an equation:

Distance = speed * time

120 = (*s* + 5)(*t* – ⅓)

The distance of 120 was given in the question stem. If the variables *s* and *t* are the actual speed and time for Khalil’s trip, (*s* + 5) and (*t* – ⅓) represent the hypothetical speed and time given in statement 1. (The ⅓ is a conversion into hours of the 20 minutes from the statement.) Of course, we can also set up the simple equation implied by the question stem itself:

Distance = speed * time

120 = *s* * *t*

Now we have a system of two equations with two variables, so the whole thing is solvable. We could express *s* in terms of *t* or *t* in terms of *s* and then substitute this expression into the other equation in the system. All of this came from the question stem and from statement 1, so** the correct answer is A.**

This concludes our introduction to average speed. The next article will make an important distinction between average speed and instantaneous speed.

If you are looking for extra help in preparing for the GMAT, we offer extensive one-on-one GMAT tutoring. You can schedule a complimentary 30-minute consultation call with one of our tutors to learn more!

**Contributor: ***Elijah Mize (Apex GMAT Instructor)*