Five friends recently visited a famous chocolatier, and collectively

Question

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.

gmat6nplus1 · Accepted Answer

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

Icerockboom · Answer

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.

Answer: B

chetan2u · Answer

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

viktorija · Answer

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) 
Two possible answers, hence insufficient.

(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)
Answer C

Bunuel · Answer

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.

mvictor · Answer

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

GMATGuruNY · Answer

Let M = the maximum weight purchased by any of the 5 friends.
Question stem, rephrased:
Is M>5?

Test the THRESHOLD.
Here, the threshold is M=5.
Constraint:
The sum of all 5 weights must be 16 pounds.

Statement 1: No two friends purchased the same amount of fudge. 
Case 1: M=5
Here, the other 4 weights must sum to 11.
One option for the other 4 weights: 4, 3, 2.5, 1.5.

Case 2: M=6
Here, the other 4 weights must sum to 10.
One option for the other 4 weights: 5, 4, 1, 0.

Since M=5 in Case 1 but M>5 in Case 2, INSUFFICIENT.

Statement 2: The minimum increment in which the chocolatier sells fudge is one pound. 
Case 1: M=5
Here, the other 4 weights must sum to 11.
One option for the other 4 weights: 5, 5, 1, 0.

Case 2: M=6
Here, the other 4 weights must sum to 10.
One option for the other 4 weights: 5, 5, 0, 0.

Since M=5 in Case 1 but M>5 in Case 2, INSUFFICIENT.

Statements combined: 
Case 1: M=5
Here, the other 4 weights must sum to 11.
Maximum possible values for the other 4 weights:
4, 3, 2, and 1, yielding a sum of 10.
Doesn't work.

Implication:
For the 5 weights to have a sum of 16, M>5.
SUFFICIENT.

ç

PRNDL · Answer

I have trouble when I am supposed to maximize the minimum value and minimize the maximum value.

Is there a method or shortcut to do that ?

Bunuel · Answer

14. Min/Max Problems 
     Theory   
 Maximizing/Minimizing Strategies on the GMAT - All in One Topic! 
     Questions   
 Worst Case Scenario Questions  
 DS Questions 
 PS Questions

For more check:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

PujaBhattacharya · Answer

I have a doubt

If 4 friends buy 1,2,3,4 pounds then 5th friend will buy 6 pounds----greater than 5 pounds

If 4 friends buy 1.5, 2.5, 3.5, 4.5 pounds then 5th friend will buy 4 pounds----less than 5 pounds

Then how's it answer C?

Posted from my mobile device

chetan2u · Answer

So what do the five friends get: 1.5, 2.5, 3.5, 4 and 4.5
But 3.5 to 4 or 4 to 4.5 does not meet the requirement of at least 1.
So this is not a valid distribution

TanushreeLahoti · Answer

Can anyone please explain the statement - The minimum increment in which the chocolatier sells fudge is one pound ?

TheVDR · Answer

If we have a question with balls or books or humans, we know these can't 1.2, 0.25, 3.65 as these have to be positive integers (or zero). Or imagine Haldiram selling Samosas. They may give 1 samosa, 2 samosas and so on but they won't sell 1.5 samosas or 2.25 samosas.

But chocolate fudge can be 1 pound, 1.3 pounds, 2.654 pounds etc.

Now it is said that "Minimum increment" - The minimum batch size or minimum quantity in which the fudge is sold is one pound. That means, we can have no fudge (0 pounds) OR 1 pound OR 2 pounds but not 1.3 pounds or 1.724 pounds as the batch size / plate size / portion size (whatever you want to call it) will increase only by 1 pound. It is the same as telling " The minimum increment in which Haldiram sells Samosas is one piece "

Five friends recently visited a famous chocolatier, and collectively

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