2012 AMC 10B Problems/Problem 5: Difference between revisions
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== | == Solution == | ||
Let | Let <math>x</math> be the cost of her dinner. | ||
<math>27.50= | <math>27.50=x+\frac{1}{10}*x+\frac{3}{20}*x</math> | ||
<math>27+\frac{1}{2}=\frac{5}{4}* | <math>27+\frac{1}{2}=\frac{5}{4}*x</math> | ||
<math>\frac{55}{2}=\frac{5}{4} | <math>\frac{55}{2}=\frac{5}{4}x</math> | ||
<math>\frac{55}{2}*\frac{4}{5}= | <math>\frac{55}{2}*\frac{4}{5}=x</math> | ||
<math> | <math>x=22</math> | ||
<math>\boxed{\textbf{(D)}}</math> | |||
==See Also== | |||
{{AMC10 box|year=2012|ab=B|num-b=4|num-a=6}} | |||
{{MAA Notice}} | |||
[[Category: Introductory Algebra Problems]] | |||
Latest revision as of 17:47, 19 October 2025
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?
Solution
Let 
be the cost of her dinner.
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| 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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