1992 AHSME Problems/Problem 2

Revision as of 14:52, 17 October 2021 by Arcticturn (talk | contribs) (Solution)

Problem

An urn is filled with coins and beads, all of which are either silver or gold. Twenty percent of the objects in the urn are beads. Forty percent of the coins in the urn are silver. What percent of objects in the urn are gold coins?

$\text{(A) } 40\%\quad \text{(B) } 48\%\quad \text{(C) } 52\%\quad \text{(D) } 60\%\quad \text{(E) } 80\%$

Solution

If twenty percent of the objects in the urn are beads, then the other $80%$ (Error compiling LaTeX. ! Missing $ inserted.) of the objects must be coins. Since $40%$ (Error compiling LaTeX. ! Missing $ inserted.) of the coins are silver, then $60%$ (Error compiling LaTeX. ! Missing $ inserted.) of the coins must be gold. Therefore, $60%$ (Error compiling LaTeX. ! Missing $ inserted.) of $80%$ (Error compiling LaTeX. ! Missing $ inserted.) of the objects in the urn are gold coins. Since $.6\times.8=.48$, 48 percent of the objects in the urn must be gold coins. Thus the answer is $\fbox{B}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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