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

It is currently 21 Oct 2019, 20:44

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

For a scholarship, at most n candidates out of 2n + 1 can be

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
CEO
CEO
avatar
Joined: 15 Aug 2003
Posts: 3223
For a scholarship, at most n candidates out of 2n + 1 can be  [#permalink]

Show Tags

New post 17 Sep 2003, 02:52
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (03:44) correct 0% (00:00) wrong based on 7 sessions

HideShow timer Statistics

For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

1. 3
2. 4
3. 2
4. 5
Intern
Intern
avatar
Joined: 17 Sep 2003
Posts: 19
Location: GMAT Maze, Chaos.
  [#permalink]

Show Tags

New post 18 Sep 2003, 07:11
1
Do u know the answer to the problem ?

I m not able to solve it, the qn is kinda confusing.
Intern
Intern
avatar
Joined: 28 Aug 2003
Posts: 33
Location: USA
  [#permalink]

Show Tags

New post 18 Sep 2003, 17:36
I could do this only by backsolving.

63 = C(2n+1,1) + C(2n+1,2) + .... + C(2n+1,n)

Question stem is asking for n.

A. n = 3 then 2n+1 = 7.

7C1 + 7C2 + 7C3 = 7 + 21 + 35 = 63.

Answer A.

There has to be a better way, I think. Anyone? Vicks? Praet?
CEO
CEO
avatar
Joined: 15 Aug 2003
Posts: 3223
  [#permalink]

Show Tags

New post 19 Sep 2003, 02:56
edealfan wrote:
I could do this only by backsolving.

63 = C(2n+1,1) + C(2n+1,2) + .... + C(2n+1,n)

Question stem is asking for n.

A. n = 3 then 2n+1 = 7.

7C1 + 7C2 + 7C3 = 7 + 21 + 35 = 63.

Answer A.

There has to be a better way, I think. Anyone? Vicks? Praet?


Honestly, i cant think of a better way.
Vicks , can you help with this.
Manager
Manager
avatar
Joined: 23 May 2013
Posts: 93
GMAT ToolKit User
Re: For a scholarship, at most n candidates out of 2n + 1 can be  [#permalink]

Show Tags

New post 26 Feb 2014, 10:18
Praetorian wrote:
For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

A. 3
B. 4
C. 2
D. 5


No of ways to select at least one candidate out of x candidates is 2^x - 1
2^x - 1 =63
2^x = 64
hence X= 6

But we are told that 2n+1 = x
hence n = 2.5 or >2. Hence we choose A
_________________
“Confidence comes not from always being right but from not fearing to be wrong.”
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: For a scholarship, at most n candidates out of 2n + 1 can be  [#permalink]

Show Tags

New post 26 Feb 2014, 19:58
ankur1901 wrote:
Praetorian wrote:
For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

A. 3
B. 4
C. 2
D. 5


No of ways to select at least one candidate out of x candidates is 2^x - 1
2^x - 1 =63
2^x = 64
hence X= 6

But we are told that 2n+1 = x
hence n = 2.5 or >2. Hence we choose A


This is not correct. Getting a fractional value for n should be the clue since n = 2.5 could mean n = 2 candidates.

Remember that at most n candidates can be selected. The number of ways you can select at least 1 candidate up to n candidates out of the total 2n+1 is given as 63.

\((2n+1)C0 + (2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) + (2n+1)C(n+1) + (2n+1)C(n+2) + ... + (2n+1)C(2n+1) = 2^{2n+1}\)

Note that you are given that \((2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) = 63\). We also know \((2n+1)C0 = 1.\)
Also recall that \((2n+1)C0 + (2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) = (2n+1)C(n+1) + (2n+1)C(n+2) + ... + (2n+1)C(2n+1)\)
since nCr = nC(n-r)

\(1 + 63 + 63 + 1 = 2^{2n+1}\)
\(n = 3\)

For non-engineering students, this can be solved by backsolving though that's usually time consuming so I doubt GMAT will test such a concept.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Senior Manager
Senior Manager
User avatar
P
Joined: 10 Apr 2018
Posts: 267
Location: United States (NC)
For a scholarship, at most n candidates out of 2n+1 can be selected  [#permalink]

Show Tags

New post 21 Sep 2018, 10:41
1
For a scholarship, at most n candidates out of 2n+1 can be selected . If the number of different ways of selection of at least one candidate is 63 , the maximum number of candidates that can be selected for the scholarship is

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
_________________
Probus

~You Just Can't beat the person who never gives up~ Babe Ruth
GMAT Club Bot
For a scholarship, at most n candidates out of 2n+1 can be selected   [#permalink] 21 Sep 2018, 10:41
Display posts from previous: Sort by

For a scholarship, at most n candidates out of 2n + 1 can be

  post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne