Difference between revisions of "1998 AJHSME Problems/Problem 9"

Revision as of 23:59, 9 June 2011

Problem 9

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 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
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

@@ Line 26: / Line 26: @@
 == See also ==
-{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
+{{AJHSME box|year=1998|num-b=8|num-a=10}}
 * [[AJHSME]]
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]

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Difference between revisions of "1998 AJHSME Problems/Problem 9"

Revision as of 23:59, 9 June 2011

Contents

Problem 9

Solution 1

Solution 2

See also