BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

how many triangles?

Expert replies
by shipra » Sun Jun 29, 2008 2:52 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 x axis. The x and y coordinated of P,Q & R are to be integers that satisfy the inequalities -4<=x<=5 & 6<=y<=16.
How many different triangles wth these properties are possible?

A. 110
B. 1100
C. 9900
D. 10000
E. 12100
Join the discussion
Source: — Problem Solving |

how many triangles?

by simpdimp » Mon Jun 30, 2008 6:01 pm
X coordinate has 10 possible values between -4 and 5
Y coordinate has 11 possible values between 6 and 16

1. Fix P first, P can have 10 values across X axis and 11 across Y axis. So, that's 11*10 = 110

2. Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (10-1) values left along X axis. So far 110*9 = 990

3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (11-1) values left. So, 990*10 = 9900

The answer is C - 9900
Join the discussion

by shipra » Mon Jun 30, 2008 11:04 pm
Thanx a lot...
really helped...
Join the discussion

Re: how many triangles?

by rabab » Sat Oct 18, 2008 7:36 am
simpdimp wrote:Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (10-1) values left along X axis.

3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (11-1) values left.
Could you pls explain these two lines? Why does R have 9 values and Q have 10 values left?
Join the discussion

by cmoney » Sun Oct 19, 2008 11:46 am
I think that the point is that to make a triangle, the the points need to exist on different values. An example would be a triangle where the base points are the same...this creates a line, so it's impossible. Sine R cannot have the same value as P, you have subtract one from the possible locations. (10-1).

I hope that helped. Anyone feel free to correct me if I'm wrong.
Join the discussion

by rah_pandey » Tue Jun 16, 2009 10:15 am
I was trying to solve the problem like this

Lets choose points P and R on a straight line
no of ways of choosing 2 points out of possible 10 points(since -4<=x<=5) on a line parallel to X axis=10c2*2!=90------(A)

Please note 2! is used since P and R position can be interchanged and the resulting triangle will be different(mirror image)

now since triangle is right angled at P therefore no of possible points available for Q =11-1=10------(B)
(since 6<=y<=16) and also P has been fixed with the y coordinate satisfying the given relation. Please note Q can only have same x coordinate as P.
using (A) and (B) we get
90*10=900 combinations for point P&R such that 6<=y1<=16
where y1 is fixed. Now there are 11 possible values of y1

thus we get 900*11=9900
Join the discussion

by bhumika.k.shah » Sun Feb 07, 2010 9:42 pm
I am sowree i still dint understand how R is (10,-1)

Can someone please elaborate.
cmoney wrote:I think that the point is that to make a triangle, the the points need to exist on different values. An example would be a triangle where the base points are the same...this creates a line, so it's impossible. Sine R cannot have the same value as P, you have subtract one from the possible locations. (10-1).

I hope that helped. Anyone feel free to correct me if I'm wrong.
Join the discussion

by sars72 » Sun Feb 07, 2010 9:52 pm
bhumika.k.shah wrote:I am sowree i still dint understand how R is (10,-1)

Can someone please elaborate.
cmoney wrote:I think that the point is that to make a triangle, the the points need to exist on different values. An example would be a triangle where the base points are the same...this creates a line, so it's impossible. Sine R cannot have the same value as P, you have subtract one from the possible locations. (10-1).

I hope that helped. Anyone feel free to correct me if I'm wrong.
it's not (10,-1). It is 10-1 = 9 possible values for R. Since R and P cannot have the same value, we subtract 1 from possible number of values for R
Join the discussion

by bhumika.k.shah » Sun Feb 07, 2010 9:56 pm
Awrite. In that case Q has 10 values . right?

But what exactly are we trying to do here? ???

2. Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (10-1) values left along X axis. So far 110*9 = 990

3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (11-1) values left. So, 990*10 = 9900

sars72 wrote:
bhumika.k.shah wrote:I am sowree i still dint understand how R is (10,-1)

Can someone please elaborate.
cmoney wrote:I think that the point is that to make a triangle, the the points need to exist on different values. An example would be a triangle where the base points are the same...this creates a line, so it's impossible. Sine R cannot have the same value as P, you have subtract one from the possible locations. (10-1).

I hope that helped. Anyone feel free to correct me if I'm wrong.
it's not (10,-1). It is 10-1 = 9 possible values for R. Since R and P cannot have the same value, we subtract 1 from possible number of values for R
Join the discussion

by missrochelle » Tue Aug 24, 2010 4:59 am
simpdimp wrote:X coordinate has 10 possible values between -4 and 5
Y coordinate has 11 possible values between 6 and 16

1. Fix P first, P can have 10 values across X axis and 11 across Y axis. So, that's 11*10 = 110

2. Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (10-1) values left along X axis. So far 110*9 = 990

3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (11-1) values left. So, 990*10 = 9900

The answer is C - 9900
Can someone explain the 2nd part of the process? I understand why R has 9 values left on the X-axis, since you can't overlap -- but why don't R and Q also have 11 points on the y axis? Meaning if the triangles are small, you can have multiple triangles along the x axis?

For example,
triangle 1: P= (-4,6) R=(-3,6) Q=(-4,7)
then moving up the y axis
triangle 2: P=(-4,7) Q=(-3,7) R=(-4,8)
Join the discussion

by GMATGuruNY » Tue Aug 24, 2010 5:41 am
When a question asks for the number of triangles that can be constructed, it's not a geometry question but a combinations question. Why? Because a triangle is a combination of 3 points.

We need to determine how many ways we can combine P, Q and R to form a triangle. For each point, we need to choose an x value and a y value.

Point P:
x value: -4≤x≤5, giving us 10 choices.

y value: 6≤y≤16, giving us 11 choices.

Now we have to combine the number of choices for x with the number of choices for y. It's as though we have 10 shirts and 11 ties, and we need to determine how many outfits can be made:

(number of choices for x)*(number of choices for y)=10*11=110 choices for P.

Point Q:
x value: In order to construct a right triangle, Q has to have the same x coordinate as P (so that Q is directly above P and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for P's x value.

y value: If P and Q have the same x value, they can't have the same y value, or they will be the same point. We used 1 of our 11 choices for y when we chose P, so we have 11-1=10 choices for Q's y value.

(number of choices for x)*(number of choices for y)=1*10=10 choices for Q.

Point R:
y value: For PR to be parallel to the x axis, P and R have to share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for P's y value.

x value: If P and R have the same y value, they can't have the same x value, or they will be the same point. We used 1 of our 10 choices for x when we chose P, so we have 10-1=9 choices for R's x value.

(number of choices for x)*(number of choices for y)=9*1=9 choices for R.

So we have 110 choices for P, 10 choices for Q, and 9 choices for R. We need to determine how many ways we can combine P, Q and R to make a triangle. It's as though we have 110 shirts, 10 ties, and 9 pairs of pants, and we need to determine the number of outfits that can be made:

(number of choices for P)*(number of choices for Q)*(number of choices for R) = 110*10*9 = 9900.

The correct answer is C.

Hope this helps!
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by ajayiitr » Wed Oct 06, 2010 7:05 pm

My confusion is that to draw a line PR, we have 10 points and we have to select 2. So, I did 10C2 = 10*9/2.

Now, I know my answer will not match to the correct answer because of this denominator 2. Why should we multiply 10C2 by 2?


My understanding is that a line can be drawn by just selecting 2 points out of available points and way you do it is combination, since order does not matter, i.e. (picking 2 and 3 is same as picking 3 and 2 = same line.)

Anyways, from there, to pick y value of line PR there are 11 values. So, 10C2*11. Then, to pick Q, there could be 10 values of y left. So, 10C2*11*10 = (10*9/2)*11*10 = 9900/2, which is the wrong answer.

Can please someone clear this confusion that what is wrong in using combination to draw a line here.

Thanks.
Join the discussion

by ajayiitr » Thu Oct 14, 2010 12:44 pm
Can someone please answer my question?

Thanks.
ajayiitr wrote:
My confusion is that to draw a line PR, we have 10 points and we have to select 2. So, I did 10C2 = 10*9/2.

Now, I know my answer will not match to the correct answer because of this denominator 2. Why should we multiply 10C2 by 2?


My understanding is that a line can be drawn by just selecting 2 points out of available points and way you do it is combination, since order does not matter, i.e. (picking 2 and 3 is same as picking 3 and 2 = same line.)

Anyways, from there, to pick y value of line PR there are 11 values. So, 10C2*11. Then, to pick Q, there could be 10 values of y left. So, 10C2*11*10 = (10*9/2)*11*10 = 9900/2, which is the wrong answer.

Can please someone clear this confusion that what is wrong in using combination to draw a line here.

Thanks.
Join the discussion

by GMATGuruNY » Thu Oct 14, 2010 1:22 pm
ajayiitr wrote:Can someone please answer my question?

Thanks.
ajayiitr wrote:
My confusion is that to draw a line PR, we have 10 points and we have to select 2. So, I did 10C2 = 10*9/2.

Now, I know my answer will not match to the correct answer because of this denominator 2. Why should we multiply 10C2 by 2?


My understanding is that a line can be drawn by just selecting 2 points out of available points and way you do it is combination, since order does not matter, i.e. (picking 2 and 3 is same as picking 3 and 2 = same line.)

Anyways, from there, to pick y value of line PR there are 11 values. So, 10C2*11. Then, to pick Q, there could be 10 values of y left. So, 10C2*11*10 = (10*9/2)*11*10 = 9900/2, which is the wrong answer.

Can please someone clear this confusion that what is wrong in using combination to draw a line here.

Thanks.
In your approach, order does matter when counting the number of choices for line segment PR. Since Q must have the same x coordinate as P, if we place P at (-4,6) and R at (5,6), we'll be able to construct different triangles than if we place R at (-4,6) and P at (5,6).

Thus, the number of choices for line segment PR = 10P2*11. Since this would leave us 10 choices for Q's y value, the number of possible triangles would be 10P2*11*10 = 9900.

Hope this clears your confusion.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by ajayiitr » Thu Oct 14, 2010 1:30 pm
Awesome, Thanks so much.
Join the discussion