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Right triangle PQR is to be constructed in the xy-plane so
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Updated on: 31 Mar 2012, 05:32

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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?

Could someone please explain how those numbers came out? Where did the 11, 10, and 9 came from? There is no specific point for them.

gettinit wrote:

yes can someone please explain in detail please? Thanks

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 inequalitites -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

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1; Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A); Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

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) 1,100 (C) 9,900 (D) 10,000 (E) 12,100

I just plotted 10 horizontal points and 11 vertical points and solved it using the following formula:

(11-1)*(10-1)*11*10 = 10*9*11*10=9900 Matched with Bunuel's approach and found the logic was pretty similar.

If you plot the points; you would see that you can make 90 right triangles at every point. Then, just multiply this figure with the total number of points.

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 is less than or equal to x which is less than or equal to 5 and 6 is less than or equal to y which is less than or equal to 16. How many different triangles with these properties could be constructed?

Re: Right triangle PQR is to be constructed in the xy-plane
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03 Jul 2013, 12:44

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akijuneja wrote:

Right triangle PQR is to be constructed in the xy-plane so that the right angle is at Pand PR is parallel to the x-axis. The x- and y-coordinates of P, Q, and Rare 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 (0 9,900 (D) 10,000 (E) 12,100

Merging topics. Please refer to the solutions above.

Re: Right triangle PQR is to be constructed in the xy-plane so
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24 Jan 2016, 06:42

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amitdgr 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?

Re: Right triangle PQR is to be constructed in the xy-plane so
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18 Apr 2016, 14:47

Bunuel wrote:

abdullaiq wrote:

Could someone please explain how those numbers came out? Where did the 11, 10, and 9 came from? There is no specific point for them.

gettinit wrote:

yes can someone please explain in detail please? Thanks

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 inequalitites -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

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1; Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A); Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

I was going to solve this with this approach but was hesitant because of the rules of the triangle..Wouldn't we want to eliminate certain cases which don't satisfy the "sides" rule ( sum of 2 greater than the 3rd....diff of 2 smaller than the 3rd?)

Re: Right triangle PQR is to be constructed in the xy-plane so
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25 Apr 2016, 09:36

Avinashs87 wrote:

Bunuel wrote:

abdullaiq wrote:

Could someone please explain how those numbers came out? Where did the 11, 10, and 9 came from? There is no specific point for them.

gettinit wrote:

yes can someone please explain in detail please? Thanks

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 inequalitites -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

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1; Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A); Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

I was going to solve this with this approach but was hesitant because of the rules of the triangle..Wouldn't we want to eliminate certain cases which don't satisfy the "sides" rule ( sum of 2 greater than the 3rd....diff of 2 smaller than the 3rd?)

You really don't have to worry about this case - you're physically constructing triangles so there are no "hypothetical" side lengths. The third side length (the hypotenuse) is simply whatever it should be given the lengths of the first two sides.

Re: Right triangle PQR is to be constructed in the xy-plane so
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27 Apr 2016, 09:54

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amitdgr 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?

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

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 how many ways we can construct points P, Q, and R, and then we will multiply those possibilities together. We are given that: \(-4\leq{x}\)\(\leq{5}\) and \(6\leq{y}\)\(\leq{16}\). This means that there are 10 possible integer values for the x-coordinate and 11 possible integer values for the y-coordinates.

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

The answer is C.
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

Re: Right triangle PQR is to be constructed in the xy-plane so
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14 Jul 2016, 17:17

Bunuel wrote:

abdullaiq wrote:

Could someone please explain how those numbers came out? Where did the 11, 10, and 9 came from? There is no specific point for them.

gettinit wrote:

yes can someone please explain in detail please? Thanks

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 inequalitites -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

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1; Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A); Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

10C1*11C1*9C1*10C1=9900.

Answer: C.

Hope it helps.

Hi Bunuel,

i understood the explanation given but I have a doubt. When selecting coordinates for point P, how can we allow the value of X coordinate to be equal to 5? If x of P is 5, then point R will never lie in the range x<= 5. Shouldn't we consider x range from -4 till 4 for point P i.e. 9 values instead of current 10 ?

I don't understand the OG explanation of the following problem. Is there anyone to help me out?

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) 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) = 9900 Answer:

Right triangle PQR is to be constructed in the xy-plane so
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30 Nov 2016, 05:31

Since -4<=x<=5 and 6<= y<= 16, we have a drawing area of 10 (5 -[-4] + 1) by 11 (16-6+1). PQ is parallel to the vertical axis and PR is parallel to the horizontal axis (x). Si since the right angle is at P, we have to determine the possible coordinates for point P and then we can determine the "x" coordinate of PR and then the "y" coordinate for PQ.

Q

*

*

P * * R

Let's say we first choose the coordinates for "P". We can do this in 10 x 11 ways. Then, let's choose the x coordinate for "R" since its "y" coordinate is the same as P. This can be done in 9 ways out of ten since one spot was taken for point "P". The same can be done for the "y" coordinate of "Q" giving us 10 since out of the eleven 1 was taken for point "P"

So we should get 10x11x9x10=9900, hence, answer should be C.

Re: Right triangle PQR is to be constructed in the xy-plane so
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20 May 2017, 08:31

Here we have right triangle QPR. <P is 90 degrees.

-4 ≤ x ≤ 5 6 ≤ y ≤ 16

The question is asking how many possible triangles we can locate on xy axis with given coordinates.

Lets figure out how many places we can put point P. We know that PR is parallel to x axis and y is parallel to y axis. For coordinate x point P we have range from -4 to 5 including: {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5} => we have 10 different coordinates here, meaning coordinate x of point P or P(x) can be located in 9 different places P(x) = 10.

For coordinate y point P => P(y) we have range from 6 to 16 including {6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} – here we have 11 different coordinated, meaning coordinate y for point P can be located in 11 different places P(y) = 11

Therefore we can calculate 10x11 = 110 => 110 different places where we can locate point P (Pxy) = 110.

Let’s find out how many places we can locate point R. Since R is on the same line as P (line is parallel to x axis) => whatever P(y) value is, R(y) =has to be the same. So we need to find out what R(x) value might be. So again we have coordinate x for point R we have range of 10 different values, and ONE of this values is taken for coordinate x point P => (Px). Meaning there is 9 options for coordinate x for point R. Meaning we can place R(x) – in 9 different locations.

R(x) = 9.

Now let’s find out, how many places we can put point Q. Since Q is on the same line as P (and line is parallel to y axis), x value must be the same => whatever P(x) value is, P(y) must be the same. So we need to find out how many places we can put y coordinate of point Q or in another words => whatever Q(y) value might be. Here again we have range or 11 y values possible => {6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}, but ONE of them is taken by point P. => P(y), meaning there is only 10 places possible we can locate coordinate y for point Q.

Q(y) = 10

Therefore, there is 110 potential places to put P, 9 potential places to put R and 10 potential places to put Q. => 110x9x10 => 9900 potential places to put triangle PQR. And that is answer choice B.

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18 Jul 2017, 09:13

1) Find total number of integers for x and y that fit in range.

x: (5-(-4)) + 1 = 10 y: (16-6) + 1 = 11

2) Draw a right triangle with right angle at P. It helps to visualize so I made P(0,0) Q(3,0) and R(0,4)

3) Start with P. P can be anything from x and y range, so possibilities are (10 x values)(11 y values) = 110 possibilities. 4) Next is R. Keep in mind: R's x-values can differ, but y cannot because PR is parallel to x-axis. R can only have 9 x-values because you used one x-value for P. So possibilities are (9 x values)(1 y value) = 9 possibilities. 5) Last is Q. Keep in mind: Q's y- values can differ, but x cannot because PQ will be parallel to y-axis. Q can only have 10 y-values because you used one y-value for P. So possibilities are (1 x value)(10 y value) = 10 possibilities.

Re: Right triangle PQR is to be constructed in the xy-plane so
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08 Jun 2018, 21:21

Bunuel wrote:

abdullaiq wrote:

Could someone please explain how those numbers came out? Where did the 11, 10, and 9 came from? There is no specific point for them.

gettinit wrote:

yes can someone please explain in detail please? Thanks

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 inequalitites -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

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1; Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A); Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

Bunuel Can you please brief how do we can randomly select any point for the x coordinate for vertex R and the y coordinate for vertex Q? How did you evaluate that for every point it will prove the triangular inequality correct?

Re: Right triangle PQR is to be constructed in the xy-plane so
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08 Jun 2018, 22:19

teaserbae wrote:

Bunuel wrote:

abdullaiq wrote:

Could someone please explain how those numbers came out? Where did the 11, 10, and 9 came from? There is no specific point for them.

gettinit wrote:

yes can someone please explain in detail please? Thanks

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 inequalitites -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

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1; Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A); Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

Bunuel Can you please brief how do we can randomly select any point for the x coordinate for vertex R and the y coordinate for vertex Q? How did you evaluate that for every point it will prove the triangular inequality correct?

Well, any three points on a plane can form a triangle, no?
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Re: Right triangle PQR is to be constructed in the xy-plane so
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08 Jun 2018, 23:11

Bunuel wrote:

teaserbae wrote:

Bunuel wrote:

Could someone please explain how those numbers came out? Where did the 11, 10, and 9 came from? There is no specific point for them.

Bunuel Can you please brief how do we can randomly select any point for the x coordinate for vertex R and the y coordinate for vertex Q? How did you evaluate that for every point it will prove the triangular inequality correct?

Well, any three points on a plane can form a triangle, no?

Any 3 points on a plane can't form a triangle. But in this question, any 3 points seem to make a triangle, I just wanted to know the proof behind that

gmatclubot

Re: Right triangle PQR is to be constructed in the xy-plane so &nbs
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