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
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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

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
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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

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.
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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

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.


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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 ?

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
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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

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
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