Learn to master this common type of GMAT word problem!
Speed
Rate is another word for speed, and the equation for this is D = RT (which some people remember as the “dirt” equation). D = distance, R = rate (a.k.a. speed, a.k.a. velocity), and T = time. One common source of errors is that all three variables have to be in the same units. If you travel at 30 mph for 10 minutes, you do not go 30*10 = 300 miles!
It’s also helpful to remember, as etymology would suggest, that a “rate” is a “ratio”, which in term is a fraction. With rates, we can always use ratios and always set up proportions. One way to find how far one goes at 30 mph in 10 minutes is to say
Crossmultiply, and you get 10 = 2x. So, x must equal 5 miles.
Average Speed
Many trickier rate questions ask about “average speed” or “average velocity” (for GMAT purposes, those two are identical). The formula for average speed is:
For a single trip at one speed, there’s nothing particularly mysterious about this question. This concept becomes much trickier in twoleg trips, especially trips in which the car travels at one speed in one leg, and at another speed in another leg. You can never simply average the two velocities given, and that will always be a tempting incorrect choice on the GMAT. You always need to apply D = RT separately in each leg of the trip, and then you need to add results from the individual legs to find the total distance and the total time.
Practice Questions
For practice, here are some average velocity questions in TwoLeg trips. The last two are challenging.
1) A car drives 300 miles at 30 mph, and then 300 miles at 60 miles per hour. What is the car’s average speed, in mph?

 (A) 36

 (B) 40

 (C) 42

 (D) 45
 (E) 57
2) A car drives for 3 hours at 40 mph, and then drives 300 miles at 60 mph. What is the car’s average speed, in mph?

 (A) 45

 (B) 47.5

 (C) 50

 (D) 52.5
 (E) 55
3) For the first 150 miles of a trip, a car drives at v mph. For the next 200 miles, the car drives at (v + 25) mph. The average speed of the whole trip is 35 mph. Find the value of v.

 (A) 20

 (B) 25

 (C) 30

 (D) 35
 (E) 40
4) A car travels at one speed for 4 hours, and then at twice that speed for 6 hours. The average velocity for the whole 10 hour trip is 40 mph. Find the initial speed in mph.

 (A) 25

 (B) 35

 (C) 40

 (D) 50
 (E) 60
Practice Question Explanations
1) In order to figure out the average velocity, we need to know both the total distance and the total time. From the question, we know the total distance is 600 miles. We need to figure out the time of each leg separately. In the first leg, T = D/R = 300/30 = 10 hr. In the second leg, T = D/R = 300/60 = 5 hours. The total time is 10 + 5 = 15 hours. The average velocity, total distance divided by total time, is 600/15 = 40 mph. Answer = B.
2) In the first leg, we know time and rate, so find distance: D = RT = (3)*(40) = 120 miles. In the second leg, we know distance and rate, so find time: T = R/D = 300/60 = 5 hours. Total distance = 120 + 300 = 420 miles. Total time = 3 + 5 = 8. Average velocity = 420/8 = 210/4 = 105/2 = 52.5 mph. Answer = D.
3) The distance of the first leg is 150 miles, and the rate is v, so the time of the first leg is
The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is
The total distance was 350 miles, and the average speed was 35 mph, so the total time of the trip must have been T = D/R = 350/35 = 10 hours. At this point, the algebra becomes hairy, so I will just plug in numbers from the answer choices.
Choice A. If v = 20 mph, then v + 25 = 45 mph. The first leg takes 150/20 = 7.5 hours, and the last leg 200/45 takes way more than three hours, so this total time is well over 10 hours. This choice is not correct.
Choice B. If v = 25, then v + 25 = 50. The first leg takes 150/25 = 6 hours. The second leg takes 200/50 = 4 hours. The total is 10 hours, which is the correct value, so this is the correct answer choice. Answer = B.
4) If the average velocity for the 10 hour trip is 40 mph, that means the total distance is D = RT = (40)*(10) = 400 miles. The distance in the first leg is d_{1} = RT = 4v. The distance in the second leg is d_{2} = RT = (2v)*(6) = 12v. The total distance is the sum, 4v + 12v = 16 v. Set this equal to the numerical value of the total distance.
So the initial speed is v = 25 mph. Answer = A.
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Hi. You have a wonderful gift for explaining these problems. Thank you for sharing with us. Can you hear it? Do you know what’s coming? That’s right… But … in your explanation for problem number 3, when you state “The distance for the second leg is 200 and the rate is v so…” shouldn’t the “+25” be included? You do have it in the equation, just not in its description. Thanks. And a waytoofullofmyself “you’re welcome” in advance.
Hi John,
I think you had just the right level of self confidence to say “you’re welcome” in advance. 😉 Thanks for spotting that error. I’ve just edited the text so that it matches the word problem as well as the equation for leg 2 in the explanation.
Mike! You explained it so simply. I’ve been going through a lot of your blog posts lately. For someone who has always disliked math, you make learning math fun and intriguing! Thank you for that and keep posting more blogs!! 😀
Hi Ramy! Thanks for your kind words. 🙂 Happy studying!
I felt my firefighter aptitude test because I didnt know this simple and easy equations.
Thank you so much for explaining this.
Alex Rios
Hey,
Just a headsup, whenever I print this out the popup shows up on the printed pages asking me to enter my name for the newsletter. It shows up after I press “print.” Took three tries. Maybe have logic to have it only show up in the first five seconds or if someone navigates away? Super annoying 🙁
Thanks for the problems!
Hi Caroline,
Thanks for alerting us to this odd behavior – it’s annoying, we agree! We’re going to take a look at that plugin and see what can be done to make it not behave in that way.
All best,
Jessica
Hi Caroline!
We decided to move the popup, so you can print uninterrupted now! Thanks again for letting us know.
All best,
Jessica
Thanks!
Thanks!
Mike,
Thanks so much for a detailed post. Speed problems always trick me but this one sure boosts confidence.
Dear Omkar,
You are quite welcome, my friend. 🙂 I am very glad you found this helpful! Best of luck to you!
Mike 🙂
Mike, you are too awesome and so humble unlike Ron purewal from Manhattan GMAT prep. I think both you and Ron are great illustrators, gifted, and naturally genius.
Dear Abdul,
Thank you very much for your kind words. 🙂 My friend, I wish you the very best of good fortune in your studies.
Mike 🙂
Hi Mike ,
Are these the type pf questions that will be asked on the gmat ? Or is this just to get a hang of the topic?
Thanks in advance
Dear Ammara,
Any of the questions in this article could be a genuine GMAT question. The GMAT loves word problems on the D = RT theme. Does this make sense?
Mike 🙂
Thanks Mike for posting such good stuff.
I have a question.
While solving Question #3, I was not able to solve it within 23 minutes.
Most of my time I wasted in doing calculation to solve following equation
10=150/v + 200/(v+25)
When I saw solution provided by you, I realized that you didn’t spend time in solving above equation but rather checked which answer choice matches above equation.
So my question is how do we determine which approach to use (when we have approximately 23 minutes for each question) ? i.e. whether to solve equation completely to find final answer or whether to substitute all answer choices one by one in equation and see which one satisfies equation?
Appreciate your inputs.
Thanks
Peter,
That’s an excellent question. In part, it’s a matter of developing experience and intuition with your own practice, so that when you get to an equation such as the one you have here, you have an immediate gut sense, “Hmm, this looks like it’s going to take a bit of time to solve” and think to look around for alternatives.
Think about it. Basically every math question on the GMAT is designed to be solved in under 90 seconds. If you have hit on an approach that going to take 5 minutes, then there absolutely must be something more efficient that you are overlooking, and backsolving is often a likely candidate.
Whenever (a) the problem introduces a variable, (b) the problem states everything in the scenario in terms of that variable, (c) the problem asks for the value of the variable, and (d) all the answers are numerical, then that’s a situation crying out for backsolving. See this post for more on backsolving:
http://magoosh.com/gmat/2012/gmatplugginginstrategyalwaysstartwithanswerchoicec/
Does all this make sense?
Mike 🙂
Yes Mike. It makes sense. Thanks for your detailed explanation. I appreciate it.
Peter,
You are quite welcome. Best of luck to you!
Mike 🙂
keep up the great vvork Mike
Real gud stuff Mike.. keep up the gud vvork 🙂
Thank you for your kind vvords. 😛
Mike 🙂