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2010 AMC 10B Problems/Problem 10: Difference between revisions

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<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30</math>
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30</math>


==Solution==
==Solution 1==
We know that <math>d = vt</math>
We know that <math>d = vt</math>


Line 23: Line 23:
So, our answer is <math> \boxed{\textbf{(C)}\ 24} </math>
So, our answer is <math> \boxed{\textbf{(C)}\ 24} </math>


==Solution 2==
We let <math>t</math> be the time Shelby drives in the rain. This gives us the equation <math> 20t + 30(\frac{2}{3} - t) = 16</math>. Expanding and rearranging gives us <math>10t = 4</math>, or <math>t = 0.4</math> hours. we multiply <math>0.4</math> by <math>60</math>, which gives us <math> \boxed{\textbf{(C)}\ 24} </math>.
==See Also==
==See Also==
{{AMC10 box|year=2010|ab=B|num-b=9|num-a=11}}
{{AMC10 box|year=2010|ab=B|num-b=9|num-a=11}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 01:21, 5 January 2018

Problem

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 1

We know that $d = vt$

Since we know that she drove both when it was raining and when it was not and that her total distance traveled is $16$ miles.

We also know that she drove a total of $40$ minutes which is $\dfrac{2}{3}$ of an hour.

We get the following system of equations, where $x$ is the time traveled when it was not raining and $y$ is the time traveled when it was raining:

$\left\{\begin{array}{ccc} 30x + 20y & = & 16 \\x + y & = & \dfrac{2}{3} \end{array} \right.$

Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:

$-10y = -4 \Leftrightarrow y = \dfrac{2}{5}$

We know now that the time traveled in rain was $\dfrac{2}{5}$ of an hour, which is $\dfrac{2}{5}*60 = 24$ minutes

So, our answer is $\boxed{\textbf{(C)}\ 24}$

Solution 2

We let $t$ be the time Shelby drives in the rain. This gives us the equation $20t + 30(\frac{2}{3} - t) = 16$. Expanding and rearranging gives us $10t = 4$, or $t = 0.4$ hours. we multiply $0.4$ by $60$, which gives us $\boxed{\textbf{(C)}\ 24}$.

See Also

2010 AMC 10B (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 AMC 10 Problems and Solutions

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