Difference between revisions of "1992 AHSME Problems/Problem 2"

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== Solution ==
 
== Solution ==
If twenty percent of the objects in the urn are beads, then the other <math>80%</math> of the objects must be coins. Since <math>40%</math> of the coins are silver, then <math>60%</math> of the coins must be gold. Therefore, <math>60%</math> of <math>80%</math> of the objects in the urn are gold coins. Since <math>.6\times.8=.48</math>, 48 percent of the objects in the urn must be gold coins. Thus the answer is <math>\fbox{B}</math>.
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If twenty percent of the objects in the urn are beads, then the other <math>80</math>% of the objects must be coins. Since <math>40</math>% of the coins are silver, then <math>60</math>% of the coins must be gold. Therefore, <math>60</math>% of <math>80</math>% of the objects in the urn are gold coins. Since <math>.6\times.8=.48</math>, 48 percent of the objects in the urn must be gold coins. Thus the answer is <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 23:00, 19 October 2021

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 $80$% of the objects must be coins. Since $40$% of the coins are silver, then $60$% of the coins must be gold. Therefore, $60$% of $80$% 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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All AHSME Problems and Solutions

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