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1998 AJHSME Problems/Problem 9: Difference between revisions

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==Problem 9==
==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>
<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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== See also ==
== 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]]
* [[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 $\$10$ 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.

$100-20=80$

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

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

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

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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