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-4 ≤ x ≤ 5 and 6 ≤ y ≤ 16

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by sanju09 » Sat Dec 07, 2013 2:51 am
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, 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. 1,100
C. 9,900
D. 10,000
E. 12,100

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Source: — Problem Solving |

by Uva@90 » Sat Dec 07, 2013 5:13 am
Hi,
The range of X-axis is 10 and
the range of Y-axis is 11.

So, Point P can lie anywhere among the points mentioned above. So P can have 10*11 possibilities.
Point R, which lies on X axis, can have 9 possibilities(10-1)
Point Q,which lies on Y axis, can have 10 possibilities(11-1)

So, total possibilities = 10*11*9*10 = 9900

Answer is C

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Uva.
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by theCodeToGMAT » Sat Dec 07, 2013 5:37 am
Total ways when P is on left of R = 10*9*11 + 10*8*11 + ..... +10*1*11 = 10*11 (9)(10)/2 = 4950

Total ways when P is on right of R = 4950

Total ways = 4950 + 4950 = 9900
[spoiler]
{C}[/spoiler]
R A H U L
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by Brent@GMATPrepNow » Sat Dec 07, 2013 6:52 am
sanju09 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, 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. 1,100
C. 9,900
D. 10,000
E. 12,100
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) = [spoiler]9900 = C[/spoiler]


If you're interested, here's another 700+ level question involving counting triangles in the coordinate plane: https://www.beatthegmat.com/how-many-tri ... 28974.html

Cheers,
Brent

Aside: Here's a free video on the FCP: https://www.gmatprepnow.com/module/gmat-counting?id=775
Brent Hanneson - Creator of GMATPrepNow.com
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