Problem Solving

BTGmoderatorDC wrote:Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and Y coordinates of P,Q and R are to be integers that satisfy the inequalities -4≤ X≤ 5 and 6≤ y≤ 16. How many different triangles with these properties could be constructed?

A. 110
B. 1100
C. 9900
D. 10000
E. 12100

OA C

Source: Official Guide

Take the task of building triangles and break it into stages.

Stage 1: Select any point where the right angle will be (point P).
The point can be selected from a 10x11 grid. So, there 110 points to choose from.
This means that stage 1 can be completed in 110 ways.

Stage 2: Select a point that is on the same horizontal line as the first point. This point will be point R.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the y-coordinate of point R.
In how many ways can we select the x-coordinate of point R?
Well, we can choose any of the 10 coordinates from -4 to 5 inclusive EXCEPT for the x-coordinate we chose for point P (in stage 1).
So, there are 9 coordinates to choose from.
This means that stage 2 can be completed in 9 ways.

Stage 3: Select a point that is on the same vertical line as the first point. This point will be point Q.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the x-coordinate of point Q.
In how many ways can we select the y-coordinate of point Q?
Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the y-coordinate we chose for point P (in stage 1).
So, there are 10 coordinates to choose from.
This means that stage 3 can be completed in 10 ways.

So, by the Fundamental Counting Principle (FCP), the total number of triangles = (110)(9)(10) = 9900

Answer: C
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Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent

BTGmoderatorDC wrote:Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and Y coordinates of P,Q and R are to be integers that satisfy the inequalities -4≤ X≤ 5 and 6≤ y≤ 16. How many different triangles with these properties could be constructed?

A. 110
B. 1100
C. 9900
D. 10000
E. 12100

OA C

Source: Official Guide

We can start by drawing the triangle in the xy-plane. This will help us visualize how we are to solve the problem.



(Move the R over to the right where it is actually at the vertex of the triangle.)

As we can see, the right triangle has a right angle at point P, and side PR is parallel to the x-axis; side PQ must be parallel to the y-axis.

To solve this question, we need to determine in how many ways we can construct points P, Q, and R, and then we will multiply those possibilities together. We are given that: -4 ≤ x ≤ 5 and 6 ≤ y ≤ 16. This means that there are 10 possible integer values for the x-coordinate and 11 possible integer values for the y-coordinate. Let's start by determining how many ways we can construct point P.

Since P is the first point we are trying to determine, we have all the options for the available x- and y-coordinates. Since there are 10 possible x-coordinates and 11 possible y-coordinates we have (10)(11) = 110 possible options. Next, we determine how many options we have for point R.

In determining point R, we must recognize that side PR of triangle PQR must remain parallel to the x-axis. This means that the y-coordinate of point R must match the y-coordinate of point P. Thus there is only 1 option for the y-coordinate of point R. However there are 9 total options for the x-coordinate of point R. Since point P has already exhausted 1 option out of the 10 total options, there are only 9 x-coordinates left for designating point R.
Finally, we determine how many ways in which we can construct point Q.

In a similar fashion to the method used for determining the number of ways to construct point R, we must remember that side PQ of triangle PQR must remain parallel to the y-axis. Thus, the x-coordinate of point Q must match the x-coordinate of point P. Therefore, there is only 1 option for the x-coordinate of point Q. However, there are 10 total options for the y-coordinate of point Q. Since point P has already exhausted 1 option out of the 11 total options, there are only 10 y-coordinates left for designating point Q.

In summary, we know the following:

There are 110 ways to select point P, 9 ways to select point R, and 10 ways to select point Q. Because we need to determine the total number of ways to create triangle PQR, we must use the fundamental counting principle. Thus, we multiply these three options together:

110 x 9 x 10 = 9,900

Answer: C