1992 AHSME Problems/Problem 2

Revision as of 13:52, 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. Unknown error_msg) of the objects must be coins. Since $40%$ (Error compiling LaTeX. Unknown error_msg) of the coins are silver, then $60%$ (Error compiling LaTeX. Unknown error_msg) of the coins must be gold. Therefore, $60%$ (Error compiling LaTeX. Unknown error_msg) of $80%$ (Error compiling LaTeX. Unknown error_msg) 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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