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

Redeem

OG13 | Pand C

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 206
Joined: Sun Jun 24, 2012 5:44 pm
Thanked: 5 times
Followed by:3 members

OG13 | Pand C

by [email protected] » Mon Feb 09, 2015 6:41 pm
Right triangle PQR is to be constructed in the xy-plane
so that the right angle is at P and PRis 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. Howmany different triangles with these
properties could be constructed?

I am not able to understand the coordinate geometry method. Is there any other way this question can be solved through P and C. Please help.
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 am
Location: Boston, MA
Thanked: 1153 times
Followed by:128 members
GMAT Score:770

by DavidG@VeritasPrep » Mon Feb 09, 2015 6:58 pm
Best to consider the number of options point by point.

We'll start with P:

Well, for the x-coordinate, we can pick anything between -4 and 5 inclusive. That gives us 10 options. For the y-coordinate, we can pick anything between 6 and 16 inclusive. That gives us 11 options. So we have a total of 10*11 = 110 options for point P.

Now we'll set R. Well, we know that PR is parallel to the x axis, which means that R will have the same y coordinate as P. That means we just need to see how many options we have for x. Clearly, it can't share an x with P, or it would be the same point, so if there were 10 options for P's x coordinate, there are 9 left for R.

Next, we'll set Q. If PQ is perpendicular to PR, then Q will have the same x coordinate as P. That means we just need to find how many options we have for y. Again, it can't have the same value as P, or they'd be the same point, so if there were 11 'y' options for P, there must be 10 'y' options left for Q.

In sum: we've got 110 options for P, 9 options left for R's x-coordinate, and 10 options left for Q's y-coordinate, giving us 110 * 9 * 10 = 9900 options in all.
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course
Top
User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 am
Location: Boston, MA
Thanked: 1153 times
Followed by:128 members
GMAT Score:770

by DavidG@VeritasPrep » Mon Feb 09, 2015 7:01 pm
Note: the initial problem wasn't transcribed 100% correctly. Instead of -4<x<5, it should read -4≤x≤5. Similarly, it should read 6≤y≤16.
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Feb 09, 2015 7:02 pm
I posted a solution here:

https://www.beatthegmat.com/how-many-tri ... 13159.html
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
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Mon Feb 09, 2015 7:03 pm
[email protected] wrote:Right triangle PQR is to be constructed in the xy-plane
so that the right angle is at P and PRis 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. Howmany different triangles with these
properties could be constructed?

I am not able to understand the coordinate geometry method. Is there any other way this question can be solved through P and C. Please help.
The question was not correctly transcribed. It 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
Top
Post new topic Post Reply
• Page 1 of 1