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A vending machine accepts only $0.25 coins and $0.10 coins. A customer pays an exact total using fewer than 12 coins. How many $0.10 coins did the customer use?
(1) The total paid is $2.05.
(2) The customer used more $0.25 coins than $0.10 coins.
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Correct answer
A
Let Q be the number of $0.25 coins and D the number of $0.10 coins, both nonnegative integers with Q + D < 12.
Statement (1): 25Q + 10D = 205, i.e. 5Q + 2D = 41. The total 205 ends in 5, an odd multiple of 5; quarters contribute a units digit of 5 each and dimes contribute 0, so the count of quarters must be odd. Checking odd Q against the under-12-coins bound:
Q = 1 → D = 18 → Q + D = 19 (too many) Q = 3 → D = 13 → Q + D = 16 (too many) Q = 5 → D = 8 → Q + D = 13 (too many) Q = 7 → D = 3 → Q + D = 10 (valid) Q = 9 → D negative (invalid)
Only (Q, D) = (7, 3) works, so D = 3. Sufficient.
Statement (2): Q > D with no total leaves many options, (2,1), (3,1), (5,2), and so on, so D is not determined. Not sufficient.
The trap assumes (2) is needed to narrow the Diophantine solutions; in fact the under-12 bound plus the parity move already pins a unique pair. Skipping the parity move forces a long enumeration and invites a slip.
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