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Official Guide Explanation:Problem Solving #229

Background

This is just one of hundreds of free explanations I've created to the quantitative questions in The Official Guide for GMAT Review (12th ed.). Click the links on the question number, difficulty level, and categories to find explanations for other problems.

These are the same explanations that are featured in my "Guides to the Official Guide" PDF booklets. However, because of the limitations of HTML and cross-browser compatibility, some mathematical concepts, such as fractions and roots, do not display as clearly online.

Question: 229
Page: 185
Difficulty: 7 (Very Difficult)
Category 1: Algebra > Inequalities > other
Category 2: Geometry > Coordinate Geometry > Other
Category 3: Geometry > Triangles > other

Explanation: If only because of the unfamiliarity of this question, this is the sort of thing you might choose to skip on test day. You can expect to see at least a question or two that you're just not prepared for, either because you don't recognize it as being similar to something you've seen before, or just because the GMAT is mixing in slightly new material.

That said, this is a permutations question. Start with one point, P. P could be any point in the space defined by the given range of x and y. There 10 points between -4 and 5, inclusive, and 11 between 6 and 16, inclusive. That's 10(11) = 110 different points that P could be. Since PR is parallel to the x - axis, R must have the same y - coordinate and P, but not the same x - coordinate. Since there are 10 possible x - coordinates and one of them is taken, that's a total of 9 possible values for R. Similarly, Q must have the same x - coordinate but not the same y - coordinate as P, leaving 10 possible values for P. Given that each of the points could be any of those possibilities, the number of possible triangles is the product of those: 110(9)(10) = 9,900, choice (C).

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