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 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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