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.

OG#228

Expert replies
by tapanmittal » Fri Jul 03, 2015 9:55 pm
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 coordinates of P,Q and R are to be integers that satisfy the following inequalities
i. x greater than equal to -4 & less than equal to 5
ii. y is greater than equal to 6 & less than equal to 16.

How many different triangles can be constructed with these properties?

Ans-9,900
Join the discussion
Source: — Problem Solving |

by talaangoshtari » Fri Jul 03, 2015 10:53 pm
Since x and y should be integers,

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

we have 10 options for x and 11 options for y. The total number of ways that we can put p on xy plane according to the given ranges for x and y is:

10 × 11 = 110

Since p and r have the same y, once we put p on xy plane, we are no longer concerned about the y of r, we know that yp = yr. So we should only choose x for the point r. Since we selected one x for point p before, now we have one less options for x. In other words, 10 - 1 = 9.

And point q should have the same x as p. So we have 11 - 1 = 10 options.

=> the total triangles = 110 × 9 × 10 = 9900
Join the discussion

by Brent@GMATPrepNow » Sat Jul 04, 2015 1:15 pm
The question should read:
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 can be constructed?
(A) 110
(B) 1100
(C) 9900
(D) 10000
(E) 12100

C
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) = [spoiler]9900 = C[/spoiler]

------------------------------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by nikhilgmat31 » Thu Jul 09, 2015 3:39 am
simply awesome question - seriously.

P = 11*10 ways
R = 9 ways of selecting x
Q = 10 ways of selecting y

11*10*9*10

9900
Join the discussion