Difference between revisions of "1992 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | 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 22:00, 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?
Solution
If twenty percent of the objects in the urn are beads, then the other 
% of the objects must be coins. Since 
% of the coins are silver, then 
% of the coins must be gold. Therefore, % of % of the objects in the urn are gold coins. Since 
, 48 percent of the objects in the urn must be gold coins. Thus the answer is 
.
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
