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

Redeem

help me with this question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 512
Joined: Mon Jun 18, 2012 11:31 pm
Thanked: 42 times
Followed by:20 members

help me with this question

by sana.noor » Tue Jul 30, 2013 11:36 pm
How many disciples does Tiffany have?

(1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her disciples.

(2) there are 28 ways that she could decide which 2 disciples she will recommend promoting.

OA is D
Work hard in Silence, Let Success make the noise.

If you found my Post really helpful, then don't forget to click the Thank/follow me button. :)
Top
Source: — Data Sufficiency |
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Fri Jun 21, 2013 2:46 am
Location: India
GMAT Score:740

by gmatsid » Wed Jul 31, 2013 1:49 am
Statement 1 seems incomplete. How many people on each list ? If nit equal, what sort of distribution. With the current information, you could have infinite no of atleast 2 disciples distributed on 200-500 lists.

Statement 2: nC2=28 => n!/(n-2)!2! = 28 => n(n-1)=56 => n^2-n-56=0 => n^2-8n+7n-56=0
=> n(n-8)+7(n-8)=0 =>(n+7)(n-8)=0

n=-7 not possible. So n=8.

Hence B
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 » Wed Jul 31, 2013 3:01 am
sana.noor wrote:How many disciples does Tiffany have?

(1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her disciples.

(2) there are 28 ways that she could decide which 2 disciples she will recommend promoting.

OA is D
Statement 2: There are 28 ways that she could decide which 2 disciples she will recommend promoting.
Number of ways to choose 2 people from n choices = (n)(n-1) / (2*1).
Since there are 28 ways, we get:
(n)(n-1) / (2*1) = 28.
(n)(n-1) = 56.
Since n must be a positive integer, n=8, implying that there are 8 disciples.
SUFFICIENT.

Statement 1: There are between 200 and 500 lists she could make consisting of the names of at least 2 of her disciples.
Since the two statements cannot contradict each other, it must be possible that 8 disciples will satisfy the constraint in statement 1.
The question at hand is whether there can be MORE THAN 8 or FEWER THAN 8 disciples.

Case 1: 9 disciples
For each disciple, there are two options: to be included on a list or NOT to be included.
Since there are 2 options for each of the 9 disciples, we get:
2*2*2*2*2*2*2*2*2 = 512.
Of these 512 lists, the following cases violate the constraint that at least 2 disciples must be included:
Number of ways to choose NO disciples = 1.
Number of ways to choose exactly ONE disciple = 9.
Thus:
Total number of possible lists = 512-1-9 = 502.
Since 502 is not between 200 and 500, there cannot be 9 disciples.

Case 2: 7 disciples
Applying the reasoning used in Case 1, the total number of possible lists ≈ 2*2*2*2*2*2*2 = 128.
Since the total number of possible lists will not be between 200 and 500, there cannot be 7 disciples.

Thus, just like statement 2, statement 1 requires that the number of disciples = 8.
SUFFICIENT.

The correct answer is D.
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 Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Wed Jul 31, 2013 3:57 pm
We're getting into deep waters here, but I wanted to explain the trick used for the first statement and make it a little easier to work with.

If I have a group of x disciples to choose from, the sum of the ways to choose all the subgroups from 0 of the disciples to all x of the disciples is 2^x.

(This formula is intuitive enough: for every member of the group, there are two choices - SELECTED or NONSELECTED - and since we have x disciples to consider, our formula is 2^x. A proper proof is here.)

Since we want to know the number of ways of choosing at least 2 of the members, we want to subtract the number of ways of choosing 0 disciples and the number of ways of choosing 1 disciple.

x choose 0 = 1 (there's only one way to choose 0 disciples)
x choose 1 = x (also pretty straightforward)

So the number of ways of choosing at least 2 of x disciples is 2^x - x - 1.

Statement 1 gives us the inequality

200 < 2^x - x - 1 < 500
201 < 2^x - x < 501

The only power of 2 in this range at all is 2^8, so x = 8, and S1 is SUFFICIENT.
Top
Post new topic Post Reply
• Page 1 of 1