GRE Prep 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Number of different triangles possible using the given points
[#permalink]
31 Oct 2022, 03:16
31 Oct 2022, 03:16
1
Expert Reply
1
Bookmarks
00:00
Question Stats:
60% (00:50) correct
40% (00:57) wrong based on 20 sessions
Hide Show timer Statistics
Attachment:
screenshot.1053.jpg [ 11.53 KiB | Viewed 2547 times ]
Quantity A |
Quantity B |
Number of different triangles possible using the given points as vertices |
42 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
Re: Number of different triangles possible using the given points
[#permalink]
15 Nov 2022, 14:58
15 Nov 2022, 14:58
Expert Reply
The brute force approach is to count all the triangles literally inside the circle you can draw
Just take one point as a reference and then join with another two points. Then the point after until you map the entire circle.
It will not take that much time but is tedious. It is not the way the question is conceived to be solved.
Instead, this is a combinatorics question in disguise
You have seven points and three sides for each triangle
\(\frac{7!}{3!4!}= \frac{7*6*5*4*3*2*1}{3*2*1-4*3*2*1}=35\)
B is the answer
Just take one point as a reference and then join with another two points. Then the point after until you map the entire circle.
It will not take that much time but is tedious. It is not the way the question is conceived to be solved.
Instead, this is a combinatorics question in disguise
You have seven points and three sides for each triangle
\(\frac{7!}{3!4!}= \frac{7*6*5*4*3*2*1}{3*2*1-4*3*2*1}=35\)
B is the answer
gmatclubot
Re: Number of different triangles possible using the given points [#permalink]
15 Nov 2022, 14:58
Moderators:
