Problem:
Joe has a collection of coins, consisting of -cent coins, -cent coins, and -cent coins. He has more -cent coins than -cent coins, and the total value of his collection is cents. How many more -cent coins does Joe have than -cent coins?
Answer Choices:
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Solution:
Let be the number of -cent coins Joe has, and let be the requested value - the number of -cent coins Joe has minus the number of -cent coins he has. Then Joe has -cent coins and -cent coins. The given information leads to the equations
These equations simplify to and . Solving these equations simultaneously yields and . Joe has more -cent coins than -cent coins. Indeed, Joe has -cent coins, -cent coins, and -cent coins.
The problems on this page are the property of the MAA's American Mathematics Competitions