Difference between revisions of "2017 AMC 12A Problems/Problem 13"

(Solution)
 
Line 11: Line 11:
  
 
Solving equation, we get <math>x=135</math> <math>\Longrightarrow \boxed{B}</math>.
 
Solving equation, we get <math>x=135</math> <math>\Longrightarrow \boxed{B}</math>.
 +
 +
==Video Solution==
 +
https://youtu.be/N4MC_a4Z_2k
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2017|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:26, 7 August 2020

Problem

Driving at a constant speed, Sharon usually takes $180$ minutes to drive from her house to her mother's house. One day Sharon begins the drive at her usual speed, but after driving $\frac{1}{3}$ of the way, she hits a bad snowstorm and reduces her speed by $20$ miles per hour. This time the trip takes her a total of $276$ minutes. How many miles is the drive from Sharon's house to her mother's house?

$\textbf{(A)}\ 132 \qquad\textbf{(B)}\ 135 \qquad\textbf{(C)}\ 138 \qquad\textbf{(D)}\ 141 \qquad\textbf{(E)}\ 144$

Solution

Let total distance be $x$. Her speed in miles per minute is $\tfrac{x}{180}$. Then, the distance that she drove before hitting the snowstorm is $\tfrac{x}{3}$. Her speed in snowstorm is reduced $20$ miles per hour, or $\tfrac{1}{3}$ miles per minute. Knowing it took her $276$ minutes in total, we create equation: \[\text{Time before Storm}\, + \, \text{Time after Storm} = \text{Total Time} \Longrightarrow\] \[\frac{\text{Distance before Storm}}{\text{Speed before Storm}} + \frac{\text{Distance in Storm}}{\text{Speed in Storm}} = \text{Total Time} \Longrightarrow \frac{\tfrac{x}{3}}{\tfrac{x}{180}} + \frac{\tfrac{2x}{3}}{\tfrac{x}{180} - \tfrac{1}{3}} = 276\]

Solving equation, we get $x=135$ $\Longrightarrow \boxed{B}$.

Video Solution

https://youtu.be/N4MC_a4Z_2k

~savannahsolver

See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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

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

Invalid username
Login to AoPS