Difference between revisions of "1998 AJHSME Problems/Problem 9"
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(→Solution 1: Fixed a LaTeX problem) |
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| Line 9: | Line 9: | ||
<math>100\%-20\%=80\%</math> | <math>100\%-20\%=80\%</math> | ||
| − | <math>10\times80\%= | + | <math>10\times80\%=10\times0.8</math> |
| − | < | + | <math>10\times0.8=8</math> |
| − | < | + | <math>\frac{8}{2}=4=\boxed{C}</math> |
==Solution 2== | ==Solution 2== | ||
| Line 19: | Line 19: | ||
The first discount has percentage 20, which is then discounted again for half of the already discounted price. | The first discount has percentage 20, which is then discounted again for half of the already discounted price. | ||
| − | < | + | <math>100-20=80</math> |
| − | < | + | <math>\frac{80}{2}=40</math> |
| − | |||
| − | </math> | ||
| + | <math>40\%\times10=10\times0.4=4=\boxed{C}</math> | ||
== See also == | == See also == | ||
Revision as of 22:45, 9 June 2011
Contents
Problem 9
For a sale, a store owner reduces the price of a <dollar/>10 scarf by 
. Later the price is lowered again, this time by one-half the reduced price. The price is now
Solution 1
Solution 2
The first discount has percentage 20, which is then discounted again for half of the already discounted price.
See also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by 1997 AJHSME |
Followed by 1999 AMC 8 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
