chocolatier

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chocolatier

by j_shreyans » Tue May 19, 2015 9:09 am
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.

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by GMATGuruNY » Tue May 19, 2015 9:12 am
j_shreyans 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.
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.

The correct answer is C.
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