M03-23

Official Solution:

If one of two parallel lines has 7 points on it and the other has 8 points, how many distinct triangles can be formed using these points as vertices?

A. 168
B. 196
C. 316
D. 364
E. 455


Note that this is not a Geometry question. While it uses basic knowledge of lines and figures, it is actually a probability and combinatorics question. There are 8 questions within GMAT Prep Focus Edition that use similar principles. Here is one example.

Approach #1:

There are two types of triangles that can be formed:

1. Triangles with two vertices on the line with 8 points and the third vertex on the line with 7 points: \(C^2_8*C^1_7=28*7=196\);

2. Triangles with two vertices on the line with 7 points and the third vertex on the line with 8 points: \(C^2_7*C^1_8=21*8=168\);

Total number of triangles: \(196+168=364\).

Approach #2:

Any three distinct points chosen from the total of \(8+7=15\) points will form a triangle, except for the cases where the three points are collinear.

Therefore, the total number of triangles can be calculated as \(C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364\). Here, \(C^3_8\) and \(C^3_7\) represent the number of different sets of 3 collinear points possible from the line with 8 points and the line with 7 points, respectively.

Note: ­The area of a square, rectangle, the volume of a cube or a rectangular solid, and the Pythagorean theorem are not considered by the GMAT as specific geometry knowledge and can still be tested on the exam. There are several questions involving this in the GMAT Prep Focus mocks. Thus, the question above is not about geometry; it's rather on percents.


Answer: D