Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 2)
 
(15 intermediate revisions by 8 users not shown)
Line 1: Line 1:
−
== Problem 7 ==
+
{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #7]] and [[2010 AMC 10B Problems|2010 AMC 10B #10]]}}
 +
 
 +
== 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?
  
Line 12: Line 14:
 
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>
+
We want <math>y</math> in minutes, <math>\frac{2}{5} \cdot 60=24 \Rightarrow C</math>
  
 
== Solution 2  ==
 
== Solution 2  ==
−
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 she drove in the rain. Thus, the number of minutes she did not drive in the rain is <math>40-x</math> .
 +
 
 +
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} \cdot (40-x) + \frac{1}{3} \cdot x = 16</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
 +
 
 +
==Video Solution==
 +
https://youtu.be/7GezNKKIEl4
 +
 
 +
~Education, the Study of Everything
  
−
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.  
+
==Video Solution==
−
Thus, we have the equation, <math>\frac{1}{2}</math> * x + <math>\frac{1}{3}</math> * (40-x) = 16
+
https://youtu.be/I3yihAO87CE?t=429
  
−
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>
+
~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}}
 +
{{AMC10 box|year=2010|num-b=9|num-a=11|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_7&oldid=205397"