1998 AJHSME Problems/Problem 9: Difference between revisions

Latest revision as of 18:49, 6 August 2020

Problem

For a sale, a store owner reduces the price of a $\$10$ scarf by $20\%$ . Later the price is lowered again, this time by one-half the reduced price. The price is now

$\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}$

Solution

Solution 1

$100\%-20\%=80\%$

$10\times80\%=10\times0.8$

$10\times0.8=8$

$\frac{8}{2}=4=\boxed{C}$

Solution 2

The first discount has percentage 20, which is then discounted again for half of the already discounted price.

$100-20=80$

$\frac{80}{2}=40$

$40\%\times10=10\times0.4=4=\boxed{C}$

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 ==Problem==
-For a sale, a store owner reduces the price of a <dollar/><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
+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>
@@ Line 31: / Line 31: @@
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{MAA Notice}}

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