Difference between revisions of "1998 AJHSME Problems/Problem 9"
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| − | ==Problem | + | ==Problem== |
| − | For a sale, a store owner reduces the price of a | + | For a sale, a store owner reduces the price of a <math>\$10</math> scarf by <math>20\% </math>. Later the price is lowered again, this time by one-half the reduced price. The price is now |
<math>\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}</math> | <math>\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}</math> | ||
| − | ==Solution 1== | + | ==Solution== |
| + | ===Solution 1=== | ||
<math>100\%-20\%=80\%</math> | <math>100\%-20\%=80\%</math> | ||
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<math>\frac{8}{2}=4=\boxed{C}</math> | <math>\frac{8}{2}=4=\boxed{C}</math> | ||
| − | ==Solution 2== | + | ===Solution 2=== |
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. | ||
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:49, 6 August 2020
Problem
For a sale, a store owner reduces the price of a 
scarf by 
. Later the price is lowered again, this time by one-half the reduced price. The price is now
Solution
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 Problem 8 |
Followed by Problem 10 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
