It is currently 23 Jun 2017, 07:13

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

VP
Joined: 30 Jun 2008
Posts: 1034
Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

13 Oct 2008, 03:13
5
KUDOS
94
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

64% (02:53) correct 36% (02:14) wrong based on 1405 sessions

### HideShow timer Statistics

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

_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Last edited by Bunuel on 31 Mar 2012, 05:32, edited 1 time in total.
Edited the question and added the OA
SVP
Joined: 17 Jun 2008
Posts: 1548
Re: PS : RIGHT TRIANGLE PQR [#permalink]

### Show Tags

13 Oct 2008, 05:20
9
KUDOS
1
This post was
BOOKMARKED
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
Intern
Joined: 09 Dec 2010
Posts: 1
Re: PS : RIGHT TRIANGLE PQR [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 39598
Re: PS : RIGHT TRIANGLE PQR [#permalink]

### Show Tags

10 Dec 2010, 01:01
25
KUDOS
Expert's post
41
This post was
BOOKMARKED
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:

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.

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.
_________________
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010
Re: PS : RIGHT TRIANGLE PQR [#permalink]

### Show Tags

29 May 2011, 07:34
1
KUDOS
1
This post was
BOOKMARKED
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"
_________________
Manager
Joined: 16 Sep 2010
Posts: 220
Location: United States
Concentration: Finance, Real Estate
GMAT 1: 740 Q48 V42

### Show Tags

23 Aug 2011, 20:26
1
This post was
BOOKMARKED
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
Math Expert
Joined: 02 Sep 2009
Posts: 39598
Re: Right triangle PQR is to be constructed in the xy-plane [#permalink]

### Show Tags

03 Jul 2013, 12:44
2
KUDOS
Expert's post
7
This post was
BOOKMARKED
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15921

### Show Tags

15 Jul 2014, 14:53
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 39598
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

24 Jan 2016, 06:42
4
KUDOS
Expert's post
3
This post was
BOOKMARKED
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:

_________________
Manager
Joined: 16 Dec 2013
Posts: 52
Location: United States
GPA: 3.7
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

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:

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.

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?)
Manager
Joined: 23 May 2013
Posts: 191
Location: United States
Concentration: Technology, Healthcare
Schools: Stanford '19 (A)
GMAT 1: 760 Q49 V45
GPA: 3.5
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

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:

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.

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.
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1044
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

27 Apr 2016, 09:54
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.

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

_________________

Jeffery Miller

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

Manager
Joined: 27 Jan 2013
Posts: 103
Location: India
GMAT 1: 700 Q49 V35
GPA: 3.7
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

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:

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.

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 ?

Intern
Joined: 29 Sep 2011
Posts: 10
Concentration: Technology, General Management
Schools: University of Dhaka - Class of 2009
GMAT Date: 08-14-2012
GPA: 3.47
WE: Education (Education)

### Show Tags

26 Aug 2016, 08:04
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
SVP
Joined: 12 Sep 2015
Posts: 1554

### Show Tags

26 Aug 2016, 08:22
1
KUDOS
Top Contributor
1
This post was
BOOKMARKED
emahmud wrote:
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
[Reveal] Spoiler:
C

RELATED VIDEO

_________________

Brent Hanneson – Founder of gmatprepnow.com

Intern
Joined: 29 Sep 2011
Posts: 10
Concentration: Technology, General Management
Schools: University of Dhaka - Class of 2009
GMAT Date: 08-14-2012
GPA: 3.47
WE: Education (Education)

### Show Tags

26 Aug 2016, 22:31
Thanks. Now I got the point.
Math Expert
Joined: 02 Sep 2009
Posts: 39598
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

27 Aug 2016, 02:11
emahmud wrote:
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

Merging topics.

_________________
Intern
Joined: 16 Oct 2015
Posts: 21
Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

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.
Intern
Joined: 02 Jan 2017
Posts: 15
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

27 Apr 2017, 09:41
X coordinates: -4 TO 5=Total==10 points
Y == 6 TO 16==11 POINTS

Now use Combinations.
For any Triangle PQR Coordinates will be like P(X1,Y1), Q(X2,Y1),R(X1,Y3)

So For P,Q, R possible combination OF POINTS == (10*11) * (9*1) * (1*10) ==9900
Intern
Joined: 16 May 2017
Posts: 22
Re: Right triangle PQR is to be constructed in the xy-plane so [#permalink]

### Show Tags

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.
Attachments

Problem 228.docx [213.08 KiB]

Re: Right triangle PQR is to be constructed in the xy-plane so   [#permalink] 20 May 2017, 08:31
Similar topics Replies Last post
Similar
Topics:
1 Right triangle RST can be constructed in the xy-plane such 2 17 Aug 2014, 23:52
18 Right triangle RST can be constructed in the xy-plane such 6 12 Sep 2016, 11:24
50 A right triangle ABC has to be constructed in the xy-plane 13 21 May 2017, 14:03
44 A right triangle ABC has to be constructed in the xy-plane 10 29 Sep 2016, 15:15
104 Right triangle PQR is to be constructed in the xy-plane so 23 07 Jan 2017, 15:19
Display posts from previous: Sort by