Difference between revisions of "2012 AMC 10B Problems/Problem 5"
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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}} |
Latest revision as of 14:02, 1 May 2021
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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