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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}</math> * x + <math>\frac{1}{3}</math> * (40-x) = 16  
 
Thus, we have the equation, <math>\frac{1}{2}</math> * x + <math>\frac{1}{3}</math> * (40-x) = 16  
  
−
Solving, gives <math>x</math> = 16, so the amount of time it is not raining is <math>40</math>-<math>16</math> = <math>24</math>
+
Solving, gives <math>x</math> = 16, so the amount of time it is not raining is <math>40</math> - <math>16</math> = <math>24</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:26, 25 April 2020

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 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 $40$ - $16$ = $24$

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

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