Difference between revisions of "2010 AMC 12B Problems/Problem 7"

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== Solution ==
 
== Solution ==
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Let <math>x</math> be the time it is not raining, and <math>y</math> be the time it is raining, in hours.
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We have the system: <math>30x+20y=16</math> and <math>x+y=2/3</math>
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Solving gives <math>x=\frac{4}{15}</math> and <math>y=\frac{2}{5}</math>
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We want <math>y</math> in minutes, <math>\frac{2}{5}*60=24 \Rightarrow C</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=6|num-a=8|ab=B}}
 
{{AMC12 box|year=2010|num-b=6|num-a=8|ab=B}}

Revision as of 19:20, 28 January 2011

Problem 7

Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$

Solution

Let $x$ be the time it is not raining, and $y$ be the time it is raining, in hours.

We have the system: $30x+20y=16$ and $x+y=2/3$

Solving gives $x=\frac{4}{15}$ and $y=\frac{2}{5}$

We want $y$ in minutes, $\frac{2}{5}*60=24 \Rightarrow C$

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 12 Problems and Solutions
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