Breaking Down a GMATPrep Rate Problem

Stacey Koprince —  May 3, 2010 — 8 Comments

This week, we’re going to tackle a challenging GMATPrep problem solving question from the topic of Rates & Work.

Let’s start with the problem. Set your timer for 2 minutes. and GO!

*Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute, and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than gear P?

(A) 6

(B) 8

(C) 10

(D) 12

(E) 15

Given info about two different gears, P and Q, we have to figure out something about how quickly they move relative to each other. In particular, we’re supposed to figure out when this is true: (# of Gear Q revolutions) = (# of Gear P revolutions) + 6.

Quick Quiz:

Name something critical that you MUST do on a problem solving question after reading the question but before starting to write down information or solve the problem.

Brainstorm a little bit. There are multiple good answers to the quiz, but I’m looking for one in particular, so come up with a few ideas to see if you can hit the one that I want.

Here’s the answer: it’s critical to look at the answer choices before starting to solve. Why? Because the answer choices can actually help you decide the best way to solve the problem. In this problem, when I glanced at the answers, I immediately noticed that they were small, whole numbers. Great! I could actually test some of the answers to help me solve. I might start with the lowest number or I might start with one of the middle numbers “ I’m not sure about that until I get a bit further into the problem.

Back to the problem. Before we dive in, a word of advice: I strongly recommend drawing out what is happening. Draw two little circles for gears, label them and write the rates in. Draw a timeline that shows, with tick marks, when each one completes a full revolution, and so on. Make rate or work problems as visual and concrete as you can.

Okay, now that you’ve done that, you also noted that the first sentence gives us the rates in revolutions per minute, while the second sentence asks us to calculate something in seconds, right?. We’re going to have to do some conversions in order to solve, so let’s start there.

The answer choices are represented in seconds, so we could convert those to minutes; that would involve converting five numbers and those numbers would all be fractions. That’s kind of annoying. Alternatively, the problem contains two rates that are given in terms of revolutions per minute. Converting two is easier than converting five, so let’s convert the numbers in the problem.

Next, what do we want to convert to? Do we need to know the rate in terms of revolutions per second? Or do we want to know how many seconds it takes for each gear to rotate once?

Remember that timeline we drew earlier? We want something that we could just add to that timeline; that is, we could start drawing tick marks for Gear P and say, okay, after x seconds, it has rotated once, and then after 2x seconds, it has rotated twice, and so on. It’s often a lot easier to do rate problems this way, especially because it’s so easy to make a mistake with the algebra. So I’m going to start with figuring out how many seconds it takes for each gear to rotate once.

Gear Q’s rate is 40 revolutions per 60 seconds. 60 seconds / 40 revolutions = 6/4 = 3/2 = 1.5 seconds for one revolution. After 1.5 seconds, Gear Q has rotated once. After 3 seconds, Gear Q has rotated twice.

Gear P’s rate is 10 revolutions per minute. There are 60 seconds in a minute, so it rotates 10 times in 60 seconds. 60 seconds /10 revolutions = 6 seconds for one revolution. Okay, so after 6 seconds, Gear P has rotated once. After 12 seconds, it has rotated twice.

From here, I’m going to give you two different ways to finish this problem off. The long way is up first:

Hmm. Those two numbers, 6 and 12, are jogging my memory. Those are two of the answer choices! Earlier, I noted that it would be easy to try the answer choices in the problem because they’re small integers. Let’s try 6 and 12.

I already know that Gear P will have rotated once after 6 seconds and twice after 12. What about Gear Q? After 6 seconds, Gear Q will have rotated 6/1.5 = 4 times. Okay, so once and four times does that fit the question? (When will Gear Q have rotated 6 more times than Gear P?) Nope. Cross off A.

After 12 seconds, Gear Q will have rotated 12/1.5 = 8 times. Gear P will have rotated twice at this point. Does this fit the question? Yes! 8 = 2+6! Gear Q has rotated 6 more times than Gear P. We didn’t even have to use algebra or write any complicated equations (and,  yes, this was the long way).

Here’s the short way: if you did draw out a timeline as I suggested earlier, start placing tick marks on your timeline, every 6 seconds for Gear P and every 1.5 seconds for Gear Q. Wherever they overlap (12, in this case!) is our answer. (Note that you already know you won’t have to extend the timeline very far to find the overlap because the largest number in the answer choices is 15. If the largest number had been 145, then the longer way, above, might be a better bet.)

The correct answer is D.

Key Takeaways for Problem Solving Rate Problems:

(1) Determine that you have a rate (or work) problem: this occurs when some type of rate is discussed, possibly a rate for the movement of some object or possibly a rate for an amount of work that is completed.

(2) Before you start solving, glance at the answer choices for all problem solving problems. Does the format give you any clues about possible solution methods? If the answers consist of small, whole numbers, you can probably use them to help you solve. (If the numbers are very spread out, you can probably estimate. If there are variable expressions in the choices, you can try your own numbers. And so on!)

(3) Imagine you’re in the actual situation and draw it out. Make a number line: a timeline or a representation of distance (say, a 100-mile train track) and, if there’s more than one thing moving or doing work, put them both on that same number line. Draw tick marks and step through the problem: after 1 hour, this is where each train is on the 100-mile track; after two hours, this is where each one is; and so on.

(4) As always, write out all of your work and, when studying, analyze your work. Even when you answer correctly, there may still be an easier or more efficient way to do the problem!

* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.

Stacey Koprince

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Stacey Koprince is an Instructor and Trainer as well as the Director of Online Community for Manhattan Prep. She's also a management consultant who specializes in corporate strategy. She has been teaching various standardized tests for more than fifteen years and her entire teaching philosophy can be summed up in five words: teaching students how to think.

8 responses to Breaking Down a GMATPrep Rate Problem

  1. Can’t this also have been made even easier using relative rotation rates?

    Ie, the difference in rotation rates between Gear Q and P is 40 – 10 = 30 revs/min. Convert that to revolutions per second we get 0.5 revolutions per second. Now that we know that we want Gear Q to be 6 revolutions ahead of P, solve 6 revs / 0.5 = 12 seconds.

    To me this is even faster in a timed setting. What may confuse people with this relative approach is in the MGMAT books it shows up most often in a straight line – ie, Car A chasing Car B, not going around in circles.

  2. Hi Danny,
    Your approach is correct. we can do in that fashion also..to me your approach is perfect.
    Thanks !

  3. Hi Danny,

    THank you so much. it’s perfect.

  4. I have another ideal too. According Q and P at the same time, so P/10=P+6/40, then P=2, so that it’s P/10*60=2/10*60=12 Seconds

  5. Hi

    I did it in different way; the difference between works of the two gears equal 6.
    Thus, we have 2t/3 – t/6 = 6 . We solve for t and we gat 3t= 36 which gives us t= 12

    Bout I think that it is a little bit longer.

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