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

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==Problem 9==
 
==Problem 9==
  
For a sale, a store owner reduces the price of a <dollar/>10 scarf by <math>20\% </math>.  Later the price is lowered again, this time by one-half the reduced price.  The price is now
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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
  
 
<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>

Revision as of 22:46, 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.

$100-20=80$

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

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

See also

1998 AJHSME (ProblemsAnswer KeyResources)
Preceded by
1997 AJHSME
Followed by
1999 AMC 8
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