It is currently 18 Mar 2018, 06:56

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Five friends recently visited a famous chocolatier, and collectively p

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44290
Five friends recently visited a famous chocolatier, and collectively p [#permalink]

### Show Tags

23 Jan 2015, 07:44
Expert's post
3
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

54% (01:25) correct 46% (01:18) wrong based on 126 sessions

### HideShow timer Statistics

Five friends recently visited a famous chocolatier, and collectively purchased a total of 16 pounds of fudge. Did any one friend purchase more than 5 pounds of fudge?

(1) No two friends purchased the same amount of fudge.

(2) The minimum increment in which the chocolatier sells fudge is one pound.

Kudos for a correct solution.
[Reveal] Spoiler: OA

_________________
Manager
Joined: 04 Oct 2013
Posts: 176
GMAT 1: 590 Q40 V30
GMAT 2: 730 Q49 V40
WE: Project Management (Entertainment and Sports)
Re: Five friends recently visited a famous chocolatier, and collectively p [#permalink]

### Show Tags

23 Jan 2015, 08:12
1
KUDOS
Bunuel wrote:
Five friends recently visited a famous chocolatier, and collectively purchased a total of 16 pounds of fudge. Did any one friend purchase more than 5 pounds of fudge?

(1) No two friends purchased the same amount of fudge.

(2) The minimum increment in which the chocolatier sells fudge is one pound.

Kudos for a correct solution.

Let's calculate the average pound per person --> 3,2.

1) set up two sets S=(a; b; c; d; e) and set Q=(f; g; h; i; j). No two friends purchased the same amount of fudge. We expect $$a+b+c+d+e = f+g+h+i+j=16$$. Assume S (1; 2; 3; 4; 6) the answer to our question is yes. Assume Q (3,2; 3,3; 3,4; 3,5; 2,6) the answer to our question is no.

2) set up two sets S and Z. Assume S (1; 2; 3; 4; 6) the answer to our question is yes; assume Z (3; 3; 3; 3; 4) the answer to our question is no.

1+2) minimum increment is one. Set Q is not acceptable, thus the answer is C.

hope it helps
_________________

learn the rules of the game, then play better than anyone else.

Intern
Joined: 27 Mar 2014
Posts: 26
Re: Five friends recently visited a famous chocolatier, and collectively p [#permalink]

### Show Tags

23 Jan 2015, 10:07
The two statement is contradict. So the answer must A, B, D, E.

S1/ The average fudge/people is 16/5 = 3.2
All can less than 5, regardless of any equal value.

s2/ Suppose we don't have any person buy more than 5. So the most fudge bought by a person must be 5.
The increment is at least 1. So the other value could be 4, 3, 2, 1. Total is 15 => Supposing statement is incorrect. There must be one person bought more than 5 pounds.

Some thing like Dirichlet distribution.

Math Expert
Joined: 02 Aug 2009
Posts: 5719
Re: Five friends recently visited a famous chocolatier, and collectively p [#permalink]

### Show Tags

23 Jan 2015, 10:16
Expert's post
1
This post was
BOOKMARKED
Icerockboom wrote:
The two statement is contradict. So the answer must A, B, D, E.

S1/ The average fudge/people is 16/5 = 3.2
All can less than 5, regardless of any equal value.

s2/ Suppose we don't have any person buy more than 5. So the most fudge bought by a person must be 5.
The increment is at least 1. So the other value could be 4, 3, 2, 1. Total is 15 => Supposing statement is incorrect. There must be one person bought more than 5 pounds.

Some thing like Dirichlet distribution.

ans C ...you are missing the point where some of them buy same no of fudge for eg 4,4,4,2,2, STATEMENT 2 ALONE WILL NOT BE SUFFICIENT
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

GMAT online Tutor

Manager
Joined: 14 Oct 2014
Posts: 66
Location: United States
GMAT 1: 500 Q36 V23
Re: Five friends recently visited a famous chocolatier, and collectively p [#permalink]

### Show Tags

24 Jan 2015, 20:18
1
KUDOS
Sum of 5 friends=16 pounds
Question: Did any one of the friends buy >5 pounds?

(1) No two same amounts:
Answer to the question is YES: (6, 1,2,3,4)
Answer to the question is NO: (1.5, 2.5, 3.5, 4, 4.5)

(2) The minimum amount is 1 pound.
YES: (6, 1,2,3,4)
NO: (4,4,4,2,2)
Insufficient.

(1)+(2) The only case that satisfy both statements is (6, 1,2,3,4)
Math Expert
Joined: 02 Sep 2009
Posts: 44290
Re: Five friends recently visited a famous chocolatier, and collectively p [#permalink]

### Show Tags

26 Jan 2015, 04:47
Bunuel wrote:
Five friends recently visited a famous chocolatier, and collectively purchased a total of 16 pounds of fudge. Did any one friend purchase more than 5 pounds of fudge?

(1) No two friends purchased the same amount of fudge.

(2) The minimum increment in which the chocolatier sells fudge is one pound.

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

In this Min/Max problem, your goal is to determine whether the minimum value of the highest total can or must be greater than 5. Statement 1 is not sufficient largely because the question (as statement 2 hints) does not prohibit fractional pounds, so the 16 total pounds can be distributed with all values under 5 (for example, 1, 2, 3, 4.4, and 4.6) or with one value well above 5 (0.1, 0.2, 0.3, 0.4, and the rest all going to the large value).

Statement 2 serves as a check against statement 1 ("Why Are You Here?") but is not sufficient on its own. But taken together, the statements tell you that the five values must be unique integers. To get to 15 total (one less than the question's mandated 16 pounds), there are really two ways:

Minimize the low value at 0, which guarantees that one value has to be above 5 (0, 1, 2, 3 for the lower 4, or 0, 2, 3, 4 all require that the high value is over 5).

or

Maximize the low value at 1, which gives the distribution 1, 2, 3, 4, 5. But since there is a 16th pound, and by adding one to any of the values 1-4 you'd match one of the higher values, the extra pound must go to the high value, meaning that even if you maximize the lower values, the minimum high value must be 6. Accordingly, the statements together guarantee the answer "yes" and the answer is C.
_________________
Board of Directors
Joined: 17 Jul 2014
Posts: 2752
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Five friends recently visited a famous chocolatier, and collectively p [#permalink]

### Show Tags

01 Aug 2017, 15:21
I chose A till i read the second statement...clearly the answer is C, as the only option that could be: 1,2,3,4,6
Five friends recently visited a famous chocolatier, and collectively p   [#permalink] 01 Aug 2017, 15:21
Display posts from previous: Sort by