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

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Solving gives <math>x=\frac{4}{15}</math> and <math>y=\frac{2}{5}</math>
 
Solving gives <math>x=\frac{4}{15}</math> and <math>y=\frac{2}{5}</math>
  
We want <math>y</math> in minutes, <math>\frac{2}{5}*60=24 \Rightarrow C</math>
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We want <math>y</math> in minutes, <math>\frac{2}{5} \cdot 60=24 \Rightarrow C</math>
  
 
== Solution 2  ==
 
== Solution 2  ==
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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 <math>\frac{1}{2}</math> miles per minute, and <math>\frac{1}{3}</math> miles per minute when it is raining.  
 
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 <math>\frac{1}{2}</math> miles per minute, and <math>\frac{1}{3}</math> miles per minute when it is raining.  
Thus, we have the equation, <math>\frac{1}{2} * (40-x) + \frac{1}{3} * x = 16</math>  
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Thus, we have the equation, <math>\frac{1}{2} \cdot (40-x) + \frac{1}{3} \cdot 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>
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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
 +
 +
==Video Solution==
 +
https://youtu.be/7GezNKKIEl4
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 +
~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 15:49, 23 November 2023

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

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, $\frac{1}{2} \cdot (40-x) + \frac{1}{3} \cdot x = 16$

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

~coolmath2017

Video Solution

https://youtu.be/7GezNKKIEl4

~Education, the Study of Everything

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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