Difference between revisions of "2017 AMC 10B Problems/Problem 7"

m
(Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
<math>\boxed{\textbf{(C)}\ 2.8}</math>
+
Let's call the distance that Samia had to travel in total as <math>2x</math>, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both <math>\frac{2x}{2}</math>, or <math>x</math>.
 
+
<cmath></cmath>
 +
She bikes at a rate of <math>17</math> kph, so she travels the distance she bikes in <math>\frac{x}{17}</math> hours. She walks at a rate of <math>5</math> kph, so she travels the distance she walks in <math>\frac{x}{5}</math> hours.
 +
<cmath></cmath>
 +
The total time is <math>\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}</math>. This is equal to <math>\frac{44}{60} = \frac{11}{15}</math> of an hour. Solving for <math>x</math>, we have:
 +
<cmath></cmath>
 +
<cmath>\frac{22x}{85} = \frac{11}{15}</cmath>
 +
<cmath>\frac{2x}{85} = \frac{1}{15}</cmath>
 +
<cmath>30x = 85</cmath>
 +
<cmath>6x = 17</cmath>
 +
<cmath>x = \frac{17}{6}</cmath>
 +
<cmath></cmath>
 +
Since <math>x</math> is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about <math>\boxed{\bold{(C) 2.8}}</math>.
 
{{AMC10 box|year=2017|ab=B|num-b=6|num-a=8}}
 
{{AMC10 box|year=2017|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:18, 16 February 2017

Problem

Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?

$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$

Solution

Let's call the distance that Samia had to travel in total as $2x$, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both $\frac{2x}{2}$, or $x$. \[\] She bikes at a rate of $17$ kph, so she travels the distance she bikes in $\frac{x}{17}$ hours. She walks at a rate of $5$ kph, so she travels the distance she walks in $\frac{x}{5}$ hours. \[\] The total time is $\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}$. This is equal to $\frac{44}{60} = \frac{11}{15}$ of an hour. Solving for $x$, we have: \[\] \[\frac{22x}{85} = \frac{11}{15}\] \[\frac{2x}{85} = \frac{1}{15}\] \[30x = 85\] \[6x = 17\] \[x = \frac{17}{6}\] \[\] Since $x$ is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about $\boxed{\bold{(C) 2.8}}$.

2017 AMC 10B (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 10 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