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Right triangle PQR is to be constructed in the xy-plane so

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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 09 Jun 2018, 04:54
teaserbae wrote:
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


Hey teaserbae

You seemed to have misunderstood a plane for a line. Any 3 points on a line cannot
form a triangle. On the contrary, any 3 points on a plane can form a triangle.

Definition of a plane:
In mathematics, a plane is a flat, two-dimensional surface that extends infinitely far. A plane is
the two-dimensional analogue of a point(zero dimensions), a line (one dimension) and three-dimensional space.


Hope this helps you!
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 04 Sep 2018, 07:33
Could we not just see that we have 9 possibilities for R when we constrain P and Q, and since this is a combinatorics problem, understand that the answer will be derived from a multiplicative relationship? Therefore, the answer must be a multiple of 9 - hence, C.
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Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 06 Sep 2018, 03:54
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.

Other discussion with alternate solutions:
http://gmatclub.com/forum/arithmetic-og ... ed#p790651
http://gmatclub.com/forum/geometry-and- ... ed#p777591
http://gmatclub.com/forum/700-question- ... ed#p818546

Hope it helps.

Hi Bunuel i would to take this question as an opportunity to ask you few questions. Today i have formally started my preparation for quant i have read all MGMT maths books and Gmat club math book my concepts are fine and this is my first 700+ and oG tag question but the real problem i am facing is the time management such as in this question i was unable to find a concrete procedure and it took me 5 min to sort the matter out and find discreetly what the question was asking. i would like to know that is there any specific pattern in quants that keep on repeating in a different creative way so that one dont waste time sorting out the method but rather recognize the pattern and slove it. just like the chess puzzle or i would put it in another way, how solving this question is helping me in quant practice. Furthermore should i practice 700+ questions only or should i go for 600-700 range as well.
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 02 Jun 2019, 18:58
why we cannot calculate the rule of foundation of triangle = i mean if 2 side greater than other...... ?!!!
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 20 Sep 2019, 06:47
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?

(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100

Source : OG 11 (PS 248)


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I did not find the answers very detailed and had to spend a lot of time figuring it out. I've broken down every step into the most simplified manner and this will be the only solution that you need to look at if you are stuck :)

1. Make an x-y plane.
2. Draw a right angle triangle anywhere in the upper half of the plane.
3. I named it as QPR where P is the right angle, PR is the line parallel to the x axis and QP is the line parallel to the y-axis. Draw this out somewhere.
4. Now name P (x1, y1)
5. Name R (x2, y1) as it shares the same y coordinate with P being parallel to the x axis. (Draw the diagram)
6. Q (x1, y2)
7. Let's start with the cominations now :)
How many points can the x-coordinate have? -4 to 5 is a total of 10 points.
How many points can the y-coordinate have? 6 to 16 is a total of 11 points.
Total combinations = no. of ways P can be filled* R can be filled* Q can be filled.
: P can be filled in (10 ways *11 ways)
: R can be filled in (9 ways *1 way); why? 9 because it can be filled in 10-1 ways, we lost 1 to P, the x-coordinate of R cannot possibly be the same as P or it won't be a triangle anymore. We have already established that y coordinate is the same as P, so there is only one way to fill it= which is the same as P's Y coordinate.
: Q can be filled in (1 way *10 ways): try to understand this on your own by applying the same approach as R's solution.
Therefore, Total combinations = (10*11) * (9*1) * (1*10)
= 110*90 = 990

Phew.. Hope this helps.


I need 10 kudos points to be able to add error log into pdf format on gmat club, so please help if this answer provided you with some clarity. Thanks
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 13 May 2020, 06:35
Can we look at the problem statement as this
out of 10 point on the horizontal line - we have to select 2 points & its a permutation. So 10P2
Similarly on the vertical line - we need to select 2 points & its a permutation so 11P2

Now- total triangle is 10P2 * 11 P2 - 9900
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 22 May 2020, 13:14
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?

(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100

Source : OG 11 (PS 248)


A quick solution to solve this problem is to look at the x coordinates, which have 10 options for P and Q to select. And given P and Q cannot be at the same point, the combination of PQ = 10*9

Then go through all options, you will notice only option C has a factor of 9. So the correct answer is C.
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 26 May 2020, 07:37
RaghavSingla 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).

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 ?

Please help me understand where am I wrong.


Same question please. Are we considering all possible shapes of a right angle triangle here where one side is parallel to the x-axis ?
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 03 Jun 2020, 03:02
When we consider the fact that P has the right angle, and that PR is parallel to X axis, we can say that the value for the X coordinate will be the same for both of these points.
Next, when we know that PQ is perpendicular to PR, we must be able to tell that Q will lie on the same vertical line as P, therefore having the same y coordinate.
So from the above understanding, we can say that there are a total of two different x coordinates (i.e for point P. and for point R) that are required to be selected, and similarly there are two different y coordinates( i.e. for point P and for point Q) to be selected.
From the given range of x coordinates we have 10 values in it's range, and 11 values possible for y coordinates.
Selecting 2 values from 10 is 10P2, as the order of selection matters i.e 90
And Selecting 2 values from 11, 11P2 is 110.
Combining both we can conclude that the no. of points , and hence the no. of triangles possible will be 9900.(C)
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 21 Jun 2020, 07:57
altairahmad wrote:
RaghavSingla 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).

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 ?

Please help me understand where am I wrong.


Same question please. Are we considering all possible shapes of a right angle triangle here where one side is parallel to the x-axis ?

altairahmad RaghavSingla
Yes we are considering all shapes.(mirror images of all triangle you already thought :))
As per your doubt if we take x=5, then also triangle can be formed by taking x=4,3,2 etc
Still there will 10 possible values of x you can take to make a triangle.
I hope it helps :)
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Re: Right triangle PQR is to be constructed in the xy-plane so  [#permalink]

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New post 02 Jul 2020, 05:15
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.

Other discussion with alternate solutions:
http://gmatclub.com/forum/arithmetic-og ... ed#p790651
http://gmatclub.com/forum/geometry-and- ... ed#p777591
http://gmatclub.com/forum/700-question- ... ed#p818546

Hope it helps.


What about other the sideLength constraints of the triangle?
And Pythagorean Theorem constraint?

Are we really supposed to blindly apply this Combinatorics formula?
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Re: Right triangle PQR is to be constructed in the xy-plane so   [#permalink] 02 Jul 2020, 05:15

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