Difference between revisions of "2012 AMC 10B Problems/Problem 5"

(Solutions)
(Solutions)
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== Solutions ==
 
== Solutions ==
  
Let X be the cost of her dinner.  
+
Let x be the cost of her dinner.  
  
<math>27.50=X+\frac{1}{10}*X+\frac{3}{20}*X</math>
+
<math>27.50=x+\frac{1}{10}*x+\frac{3}{20}*x</math>
  
<math>27+\frac{1}{2}=\frac{5}{4}*X</math>
+
<math>27+\frac{1}{2}=\frac{5}{4}*x</math>
  
<math>\frac{55}{2}=\frac{5}{4}X</math>
+
<math>\frac{55}{2}=\frac{5}{4}x</math>
  
<math>\frac{55}{2}*\frac{4}{5}=X</math>
+
<math>\frac{55}{2}*\frac{4}{5}=x</math>
  
<math>\boxed{22=X}</math>
+
<math>\boxed{22=x}</math>
  
 
'''OR'''
 
'''OR'''

Revision as of 02:30, 7 January 2016

Problem 5

Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the sales tax is added, and the tax is calculated on the pre-tip amount. She spends a total of 27.50 dollars for dinner. What is the cost of her dinner without tax or tip in dollars?

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$


Solutions

Let x be the cost of her dinner.

$27.50=x+\frac{1}{10}*x+\frac{3}{20}*x$

$27+\frac{1}{2}=\frac{5}{4}*x$

$\frac{55}{2}=\frac{5}{4}x$

$\frac{55}{2}*\frac{4}{5}=x$

$\boxed{22=x}$

OR

$\textbf{(D)}$

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 AMC 10 Problems and Solutions

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