GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 22:11 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Let T_n be the number of all possible triangles formed by

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Manager  Joined: 04 Oct 2013
Posts: 150
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Let T_n be the number of all possible triangles formed by  [#permalink]

Show Tags

2
11 00:00

Difficulty:   85% (hard)

Question Stats: 49% (02:23) correct 51% (02:16) wrong based on 156 sessions

HideShow timer Statistics

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

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.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 58428
Re: Let T_n be the number of all possible triangles formed by  [#permalink]

Show Tags

arunspanda wrote:
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

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 $$T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10$$ --> $$C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10$$ --> $$\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10$$ --> $$(n-1)n=20$$ --> $$n=5$$.

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 --> $$C^2_n=10$$ --> $$(n-1)n=20$$ --> $$n=5$$.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rule 3. Thank you.
_________________
Senior Manager  P
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 435
Re: Let T_n be the number of all possible triangles formed by  [#permalink]

Show Tags

Bunuel wrote:
arunspanda wrote:
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

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 $$T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10$$ --> $$C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10$$ -->$$\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10$$ --> $$(n-1)n=20$$--> $$n=5$$.

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 --> $$C^2_n=10$$ --> $$(n-1)n=20$$ --> $$n=5$$.

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
... ?
Math Expert V
Joined: 02 Sep 2009
Posts: 58428
Re: Let T_n be the number of all possible triangles formed by  [#permalink]

Show Tags

energetics wrote:
Bunuel wrote:
arunspanda wrote:
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

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 $$T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10$$ --> $$C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10$$ -->$$\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10$$ --> $$(n-1)n=20$$--> $$n=5$$.

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 --> $$C^2_n=10$$ --> $$(n-1)n=20$$ --> $$n=5$$.

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
... ?

$$C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10$$;

Express (n+1)! as (n-2)!*(n-1)n(n+1) and n! as (n-3)!*(n-2)(n-1)n:

$$\frac{(n-2)!*(n-1)n(n+1)}{3!(n-2)!}-\frac{(n-3)!*(n-2)(n-1)n}{3!(n-3)!}=10$$

$$\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10$$

$$\frac{(n-1)n(n+1)-(n-2)(n-1)n}{6}=10$$

$$\frac{(n-1)n(n+1-n+2)}{6}=10$$

$$\frac{(n-1)n*3}{6}=10$$

$$(n-1)n=20$$

$$n=5$$.
_________________
Manager  S
Joined: 17 May 2018
Posts: 140
Location: India
Re: Let T_n be the number of all possible triangles formed by  [#permalink]

Show Tags

We could also solve this by directly inputting the numbers. 6c3-5c3=10.Hence value of n is 5

Posted from my mobile device Re: Let T_n be the number of all possible triangles formed by   [#permalink] 26 Feb 2019, 11:08
Display posts from previous: Sort by

Let T_n be the number of all possible triangles formed by

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  