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

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== Problem 7 ==
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{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #7]] and [[2010 AMC 10B Problems|2010 AMC 10B #10]]}}
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== Problem ==
 
Shelby drives her scooter at a speed of <math>30</math> miles per hour if it is not raining, and <math>20</math> 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 <math>16</math> miles in <math>40</math> minutes. How many minutes did she drive in the rain?
 
Shelby drives her scooter at a speed of <math>30</math> miles per hour if it is not raining, and <math>20</math> 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 <math>16</math> miles in <math>40</math> minutes. How many minutes did she drive in the rain?
  
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Let <math>x</math> be the time it is raining. Thus, the number of minutes it is not raining is <math>40-x</math> .  
 
Let <math>x</math> be the time it is raining. Thus, the number of minutes it is not raining is <math>40-x</math> .  
  
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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 \frac{1}{15}\
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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> per minute, and <math>\frac{1}{3}</math> per minute when it is not raining.
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Thus, we have the equation, <math>\frac{1}{2}</math> * x + <math>\frac{1}{3}</math> * (40-x) = 16
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Solving, gives <math>x</math> = <math>16</math> , so the amount of time it is not raining is <math>40</math> - <math>16</math> = <math>24</math>
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~coolmath2017
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==Video Solution==
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https://youtu.be/I3yihAO87CE?t=429
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~IceMatrix
  
 
== 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}}
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{{AMC10 box|year=2010|num-b=9|num-a=11|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:40, 15 February 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 it is raining. Thus, the number of minutes it is not raining 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 $\frac{1}{2}$ per minute, and $\frac{1}{3}$ per minute when it is not raining. Thus, we have the equation, $\frac{1}{2}$ * x + $\frac{1}{3}$ * (40-x) = 16

Solving, gives $x$ = $16$ , so the amount of time it is not raining is $40$ - $16$ = $24$

~coolmath2017

Video Solution

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

~IceMatrix

See also

2010 AMC 12B (Problems • Answer Key • Resources)
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 (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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