2017 AMC 10B Problems/Problem 7
Contents
Problem
Samia set off on her bicycle to visit her friend, traveling at an average speed of 
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 
kilometers per hour. In all it took her 
minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
Solution 1
Let's call the distance that Samia had to travel in total as 
, 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 
, or 
.
![\[\]](http://latex.artofproblemsolving.com/5/7/f/57f1406b65f9f2e6a365b7356da731780b42cceb.png)
She bikes at a rate of kph, so she travels the distance she bikes in 
hours. She walks at a rate of kph, so she travels the distance she walks in 
hours.
The total time is 
. This is equal to 
of an hour. Solving for , we have:
![\[\frac{22x}{85} = \frac{11}{15}\]](http://latex.artofproblemsolving.com/1/0/4/1047df3d263df2f4101416c6bbd2350f5b3fe49f.png)
![\[\frac{2x}{85} = \frac{1}{15}\]](http://latex.artofproblemsolving.com/a/b/e/abea6c21f569b9e8506517f21781841f11a2cf64.png)
![\[30x = 85\]](http://latex.artofproblemsolving.com/7/d/d/7ddd1cf0135652196e90aa16b630babae2eb6214.png)
![\[6x = 17\]](http://latex.artofproblemsolving.com/6/0/8/608bd4ededb3cc5b4bdfb775e9ac91ab4633320a.png)
![\[x = \frac{17}{6}\]](http://latex.artofproblemsolving.com/d/9/3/d9360642a7e793f542984bb113dba97831767793.png)
Since 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 
.
Solution 2 (Quick, Make Approximations)
Notice that Samia walks 
times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is
![\[\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}\]](http://latex.artofproblemsolving.com/2/0/e/20ef889e66387dafdc5a570a359d26425e468eb9.png)
Then, multiply this by the time
![\[\frac{17}{22} \cdot 44 \text{minutes} = 34 \text{minutes}\]](http://latex.artofproblemsolving.com/6/e/9/6e92faf047295ad4b08ee0f4d1e5db8ae61e96ca.png)
34 minutes is a little greater than 
of an hour so Samia traveled
![\[\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}\]](http://latex.artofproblemsolving.com/7/2/b/72bf4251e2268ae4a7db802aa095ca134d25df0e.png)
The answer choice a little greater than 2.5 is . (Note that we could've multiplied 
by and gotten the exact answer as well)
Solution 3 (Harmonic Mean)
Since the distance between biking and walking is equal, we can the given rates' harmonic mean to find the average speed.
![\[\overline{s}=\frac{2ab}{a+b}\implies\overline{s}=\frac{2 \cdot 17 \cdot 5}{17+5}=\frac{85}{11}\text{kph}\]](http://latex.artofproblemsolving.com/4/a/f/4afe774bda6674c262e88c7b3c6df303be2ae0a7.png)
Then, multiplying by 
hours gives the overall distance 
kilometers. Samia only walks half of that, so 
kilometers.
~mn28407
See Also
| 2017 AMC 10B (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 10 Problems and Solutions | ||
| 2017 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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. 
