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 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 <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> | + | 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 <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 == |
Revision as of 22:33, 4 October 2016
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 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 , 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.