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

(Solution 2)
(Solution 2)
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Thus, we have the equation, <math>\frac{1}{2} * (40-x) + \frac{1}{3} * x = 16</math>  
 
Thus, we have the equation, <math>\frac{1}{2} * (40-x) + \frac{1}{3} * x = 16</math>  
  
Solving, gives <math>x</math> = <math>24</math> , so the amount of time it is not raining is <math>40</math> - <math>16</math> = <math>24</math>
+
Solving, gives <math>x</math> = <math>24</math> , so the amount of time she drove in the rain is <math>24</math> minutes.
  
 
~coolmath2017
 
~coolmath2017

Revision as of 12:01, 11 July 2021

The following problem is from both the 2010 AMC 12B #7 and 2010 AMC 10B #10, so both problems redirect to this page.

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

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, $\frac{2}{5}*60=24 \Rightarrow C$

Solution 2

Let $x$ be the time she drove in the rain. Thus, the number of minutes she did not drive in the rain is $40-x$ .

Since we are calculating the time in minutes, it is best to convert the speeds in minutes. Thus, the speed per minute when it is not raining is $\frac{1}{2}$ miles per minute, and $\frac{1}{3}$ miles per minute when it is raining. Thus, we have the equation, $\frac{1}{2} * (40-x) + \frac{1}{3} * x = 16$

Solving, gives $x$ = $24$ , so the amount of time she drove in the rain is $24$ minutes.

~coolmath2017

Video Solution

https://youtu.be/I3yihAO87CE?t=429

~IceMatrix

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
2010 AMC 10B (ProblemsAnswer KeyResources)
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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