Difference between revisions of "2003 AMC 8 Problems/Problem 11"

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On Friday, the shoes would cost <math> 40 \cdot 1.1= 44 </math> dollars. Then on Monday, the shoes would cost <math> 44- \frac{44}{10}=44-4.4=\boxed{\textbf{(B)}\ 39.60} </math>.
 
On Friday, the shoes would cost <math> 40 \cdot 1.1= 44 </math> dollars. Then on Monday, the shoes would cost <math> 44- \frac{44}{10}=44-4.4=\boxed{\textbf{(B)}\ 39.60} </math>.
  
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==See Also==
 
{{AMC8 box|year=2003|num-b=10|num-a=12}}
 
{{AMC8 box|year=2003|num-b=10|num-a=12}}

Revision as of 02:48, 24 December 2012

Problem

Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by $10%$ (Error compiling LaTeX. Unknown error_msg). Over the weekend, Lou advertises the sale: Ten percent off the listed price. Sale starts Monday." How much does a pair of shoes cost on Monday that cost $40$ dollars on Thursday?

$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44$

Solution

On Friday, the shoes would cost $40 \cdot 1.1= 44$ dollars. Then on Monday, the shoes would cost $44- \frac{44}{10}=44-4.4=\boxed{\textbf{(B)}\ 39.60}$.

See Also

2003 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AJHSME/AMC 8 Problems and Solutions