Difference between revisions of "2011 AMC 8 Problems/Problem 10"
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<math> \textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75 </math> | <math> \textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75 </math> | ||
| − | ==Solution== | + | ==Solution 1 (Algebra)== |
Let <math>x</math> be the number of miles you ride. The number of miles you ride after the first half mile is <math>x-0.5.</math> We can write this equation: | Let <math>x</math> be the number of miles you ride. The number of miles you ride after the first half mile is <math>x-0.5.</math> We can write this equation: | ||
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x &= \boxed{\textbf{(C)}\ 3.3}\end{align*}</cmath> | x &= \boxed{\textbf{(C)}\ 3.3}\end{align*}</cmath> | ||
| − | ==Solution 2== | + | ==Solution 2 (Table)== |
| − | |||
| − | 0 | + | \begin{array}{|c|c|c|} |
| − | + | Miles & Money & Remark\\ | |
| − | 0.5 | + | \hline |
| − | + | 0 & 8 & \text{2 for tip} \\ | |
| − | 3.3 | + | \hline |
| + | 0.5 & 5.6 & \text{-2.4 for first 0.5 mi} \\ | ||
| + | \hline | ||
| + | 3.3 & 0 & \text{-5.6 for 2.8 mi} | ||
| + | \end{array} | ||
Miles travelled = <math>\boxed{\textbf{(C)}\ 3.3}</math> | Miles travelled = <math>\boxed{\textbf{(C)}\ 3.3}</math> | ||
| − | + | Solution by Anshulb, formatted by Cheetahboy93 | |
| − | |||
| − | |||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 22:20, 18 February 2024
Problem
The taxi fare in Gotham City is $2.40 for the first 
mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
Solution 1 (Algebra)
Let 
be the number of miles you ride. The number of miles you ride after the first half mile is 
We can write this equation:
Solution 2 (Table)
\begin{array}{|c|c|c|} Miles & Money & Remark\\ \hline 0 & 8 & \text{2 for tip} \\ \hline 0.5 & 5.6 & \text{-2.4 for first 0.5 mi} \\ \hline 3.3 & 0 & \text{-5.6 for 2.8 mi} \end{array}
Miles travelled = 
Solution by Anshulb, formatted by Cheetahboy93
Video Solution
~savannahsolver
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
