1977 AMC 12 Problem 3

Attempt Problem 3 of the 1977 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1977 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

3.

A man has $2.73\$2.73 in pennies, nickels, dimes, quarters and half dollars. If he has an equal number of coins of each kind, then the total number of coins he has is
33
55
99
1010
1515
Answer: E
Concepts:moneylinear equation
Small hint:
Find the value of one coin of each of the five types
Big hint:
The five coins in one complete set are worth 9191 cents
Solution:
One coin of each kind is worth 1+5+10+25+50=911+5+10+25+50=91 cents. Since 27391=3,\frac{273}{91}=3, there are three coins of each of the five kinds, or 5⋅3=155\cdot3=15 coins.
Therefore, the correct answer is E.
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