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Difference between revisions of "2011 AMC 8 Problems/Problem 10"

m (→‎Solution)
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==Solution==
 
==Solution==
 
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:
 +
 +
  
 
<cmath>\begin{align*} 10 &= 2.4 + 0.2 \times \frac{x-0.5}{0.1} + 2\\
 
<cmath>\begin{align*} 10 &= 2.4 + 0.2 \times \frac{x-0.5}{0.1} + 2\\
Line 11: Line 13:
 
2.8 &= x-0.5\\
 
2.8 &= x-0.5\\
 
x &= \boxed{\textbf{(C)}\ 3.3}\end{align*}</cmath>
 
x &= \boxed{\textbf{(C)}\ 3.3}\end{align*}</cmath>
 +
 +
==Solution 2==
 +
Miles | Money | Remark
 +
0    | 8    | 2 for tip
 +
0.5  | 5.6  | -2.4 for first 0.5 mi
 +
3.3  | 0    | -5.6 for 2.8 mi
 +
 +
Miles travelled = \boxed{\textbf{(C)}\ 3.3}\end{align*}<math></math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=9|num-a=11}}
 
{{AMC8 box|year=2011|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:34, 31 December 2022

Problem

The taxi fare in Gotham City is $2.40 for the first $\frac12$ 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?

$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$

Solution

Let $x$ be the number of miles you ride. The number of miles you ride after the first half mile is $x-0.5.$ We can write this equation:


\begin{align*} 10 &= 2.4 + 0.2 \times \frac{x-0.5}{0.1} + 2\\ 5.6 &= 2(x-0.5)\\ 2.8 &= x-0.5\\ x &= \boxed{\textbf{(C)}\ 3.3}\end{align*}

Solution 2

Miles | Money | Remark 0 | 8 | 2 for tip 0.5 | 5.6 | -2.4 for first 0.5 mi 3.3 | 0 | -5.6 for 2.8 mi

Miles travelled = \boxed{\textbf{(C)}\ 3.3}\end{align*}$$ (Error compiling LaTeX. Unknown error_msg)

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. AMC logo.png

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