1998 AJHSME Problems/Problem 9

Problem

For a sale, a store owner reduces the price of a <dollar/> $10$ 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

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1998 AJHSME Problems/Problem 9

Contents

Problem

Solution

Solution 1

Solution 2

See also