Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jan 2511:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Jan 2410:00 AM EST
-12:00 PM EST
The Target Test Prep team is proud to introduce GMAT OnDemand, a video course created by Scott Woodbury-Stewart, founder and CEO, designed to provide you with everything you need to master the GMAT and achieve your score goals.
Jan 2410:00 AM EST
-11:59 PM EST
Through Friday, January 31, get an extra 25% off Target Test Prep GMAT plans. Just use the coupon code FLASH25 at checkout to save up to $400.- Jan 26
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease.
Updated on: 26 Mar 2014, 07:19
Originally posted by arunspanda on 26 Mar 2014, 06:17.
Last edited by Bunuel on 26 Mar 2014, 07:19, edited 1 time in total.
Last edited by Bunuel on 26 Mar 2014, 07:19, edited 1 time in total.
Renamed the topic and edited the question.
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
56% (02:17) correct
44%
(02:24)
wrong
based on 265
sessions
History
Date
Time
Result
Not Attempted Yet
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is
A. 5
B. 6
C. 7
D. 8
E. 10
A. 5
B. 6
C. 7
D. 8
E. 10
26 Mar 2014, 07:21
Kudos
Bookmarks
arunspanda
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is
A. 5
B. 6
C. 7
D. 8
E. 10
A. 5
B. 6
C. 7
D. 8
E. 10
In a plane if there are n points of which no three are collinear, then the number of triangles that can be formed by joining them is \(C^3_n\).
We are given that \(T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10\) --> \(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\) --> \(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\) --> \((n-1)n=20\) --> \(n=5\).
Answer: A.
Or: one additional point gives 10 more triangles, so when we add one point there are 10 different pairs of points which make triangles with that additional point --> \(C^2_n=10\) --> \((n-1)n=20\) --> \(n=5\).
Answer: A.
P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rule 3. Thank you.
General Discussion
26 Feb 2019, 07:44
Kudos
Bookmarks
Bunuel
arunspanda
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is
A. 5
B. 6
C. 7
D. 8
E. 10
A. 5
B. 6
C. 7
D. 8
E. 10
In a plane if there are n points of which no three are collinear, then the number of triangles that can be formed by joining them is \(C^3_n\).
We are given that \(T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10\) --> \(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\) -->\(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\) --> \((n-1)n=20\)--> \(n=5\).
Answer: A.
Or: one additional point gives 10 more triangles, so when we add one point there are 10 different pairs of points which make triangles with that additional point --> \(C^2_n=10\) --> \((n-1)n=20\) --> \(n=5\).
Answer: A.
Bunuel can you please expand upon the highlighted part? I get that we are cross multiplying for LCD, but I am getting lost.
Also, is there any way this is related to n-2 (number of triangles from a regular polygon)? I tried to use the idea that n(n-2) would give us the number of vertices*number of triangles formed from each.
Tn = n(n-2)
Tn+1 = n(n-2) + 10
... ?
