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license

by j_shreyans » Thu Mar 12, 2015 9:11 am
To receive a driver license, sixteen year-olds at Culliver High School have to pass both a written and a practical driving test. Everyone has to take the tests, and no one failed both tests. If 30% of the 16 year-olds who passed the written test did not pass the practical, how many sixteen year-olds at Culliver High School received their driver license?

(1) There are 188 sixteen year-olds at Culliver High School.

(2) 20% of the sixteen year-olds who passed the practical test failed the written test.

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by DavidG@VeritasPrep » Thu Mar 12, 2015 9:57 am
Matrix!

Let's start by inserting all the info we're given in the question stem. We know 30% of those who passed the written test did not pass the practical. So we'll call the number who passed the written 'w,' which means .3w passed the written, but did not pass the practical. Consequently, if 30% of those who passed the written failed the practical, 70% of those who passed the written, passed the practical as well. Call that cell .7w. We want to know how many students passed both tests, so we're looking for 7w. Put another way, if we have 'w' we have sufficiency. (Also, we know that no one failed both tests, so we can insert 0 into that cell.) This gives us the following matrix:

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S1: This gives us the total students. Our table now looks like this:

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No way this gives us w. Not Sufficient

S2: Let's call the total who passed the practical 'p.' If 20% of those failed the written test, we can designate the Passed Practical/Failed Written as .2p. And then we know that 80% of those who passed the practical, also passed the written. So let's call that .8p. (Notice this is the same cell that we've already designated as .7w, so those quantities are equal.) Now we have this:

Image


Well, we know that .7w = .8p, but there's still no way to solve for w or p. Not Sufficient.


Together, we'll have the following matrix:

Image


Now we have .7w = .8p. But if we add the bottom row, we'd also have w + .2p = 188; 2 linear equations/2 variables, so sufficient. (Anytime we have at least as many unique linear equations as we have variables, we know that we can solve for all the variables without actually doing so.) Answer is C
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by Brent@GMATPrepNow » Thu Mar 12, 2015 10:54 am
It's worth noting that David's Double Matrix approach can be used for most questions featuring a population in which each member has two characteristics associated with it.
Here, we have a population of sixteen-year-olds at Culliver High School, and the two characteristics are:
- passed the written test or did not pass the written test
- passed the practical driving test or did not pass the practical driving test

This question type is VERY COMMON on the GMAT, so be sure to master the technique.

To learn more about the Double Matrix Method, watch our free video: https://www.gmatprepnow.com/module/gmat- ... ems?id=919

Once you're familiar with this technique, you can attempt these additional practice questions:

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Difficult Problem Solving questions
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Easy Data Sufficiency questions
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Medium Data Sufficiency questions
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Difficult Data Sufficiency questions
- https://www.beatthegmat.com/double-set-m ... 71423.html
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- https://www.beatthegmat.com/mba/2011/05/ ... question-3

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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