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

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

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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?

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

Source : OG 11 (PS 248)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 31 Mar 2012, 05:32, edited 1 time in total.
Edited the question and added the OA
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Re: PS : RIGHT TRIANGLE PQR [#permalink]

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Is it C?

P can take a total of 10*11 co-ordinates.
for a given P, R can take a total of 9 co-ordinates and Q can take a total of 10 co-ordinates.

Hence, total = 110*9*10 = 9900
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Re: PS : RIGHT TRIANGLE PQR [#permalink]

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New post 09 Dec 2010, 20:57
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.
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Re: PS : RIGHT TRIANGLE PQR [#permalink]

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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:
arithmetic-og-question-88380.html?hilit=constructed#p790651
geometry-and-combinations-100675.html?hilit=constructed#p777591
700-question-94644.html?hilit=constructed#p818546

Hope it helps.
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Re: PS : RIGHT TRIANGLE PQR [#permalink]

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amitdgr wrote:
Source : OG 11 (PS 248)

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.

Ans: "C"
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Difficult Geometry Question [#permalink]

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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 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?

(a) 110
(b) 1,100
(c) 9,900
(d) 10,000
(e) 12,100
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Re: Right triangle PQR is to be constructed in the xy-plane [#permalink]

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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.

Similar questions to practice:
triangle-abc-will-be-constructed-in-a-xy-plane-according-to-132573.html
right-triangle-pqr-is-to-be-constructed-in-the-xy-plane-so-88380.html
right-triangle-abc-is-to-be-drawn-in-the-xy-plane-so-that-88958.html
a-right-triangle-abc-has-to-be-constructed-in-the-xy-plane-94644.html
a-right-triangle-abc-has-to-be-constructed-in-the-xy-plane-100675.html
right-triangle-rst-can-be-constructed-in-the-xy-plane-such-137129.html

All OG13 questions with solutions: the-official-guide-quantitative-question-directory-143450.html

Hope it helps.
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Re: Difficult Geometry Question [#permalink]

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

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New post 24 Jan 2016, 06:42
Expert's post
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)


MATH REVOLUTION VIDEO SOLUTION:


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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 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).

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

Answer: C.

Other discussion with alternate solutions:
arithmetic-og-question-88380.html?hilit=constructed#p790651
geometry-and-combinations-100675.html?hilit=constructed#p777591
700-question-94644.html?hilit=constructed#p818546

Hope it helps.


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

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New post 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).

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

Answer: C.

Other discussion with alternate solutions:
arithmetic-og-question-88380.html?hilit=constructed#p790651
geometry-and-combinations-100675.html?hilit=constructed#p777591
700-question-94644.html?hilit=constructed#p818546

Hope it helps.


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

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New post 27 Apr 2016, 09:54
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)


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

Image

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