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

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(→‎Solution 3 (Harmonic Mean))
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==Solution 3 (Harmonic Mean)==
 
==Solution 3 (Harmonic Mean)==
  
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Since the distance that both rates travel are equivalent, we can find the harmonic mean of the two rates to find the average speed. Therefore, let <math>a=17</math> and <math>b=5</math>
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Since the distance between biking and walking is equal, we can the given rates' harmonic mean to find the average speed.
  
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<cmath>\overline{s}=\frac{2ab}{a+b}\implies\overline{s}=\frac{2 \cdot 17 \cdot 5}{17+5}=\frac{85}{11}</cmath>
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<cmath>\overline{s}=\frac{2ab}{a+b}\implies\overline{s}=\frac{2 \cdot 17 \cdot 5}{17+5}=\frac{85}{11}\text{kph}</cmath>
  
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Now, we can find the overall distance by multiply the average speed by the time(hours), 44 minutes. Let <math>s=\frac{11}{15}</math>
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Then, multiplying by <math>\frac{11}{15}</math> hours gives the overall distance <math>\frac{17}{3}</math> kilometers. Samia only walks half of that, so  <math>\frac{17}{3}\approx \boxed{\textbf{(C)}2.8}</math> kilometers.
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<cmath>d=\overline{s}\cdot t\implies d=\frac{85}{11}\cdot\frac{11}{15}=\frac{17}{3}</cmath>
 
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Because the distance Samia walks is half the total distance, the distance she walks is <math>\frac{17}{6}\text{kilometers}</math>. With some knowledge of sixths, one finds that Samia walked, in kilometers, <math>2.8\overline{3}\approx \boxed{2.8\implies\textbf{(C)}}</math>
 
  
 
~mn28407
 
~mn28407

Revision as of 01:50, 1 June 2020

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 1

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 $\boxed{\bold{(C)} 2.8}$.

Solution 2 (Quick, Make Approximations)

Notice that Samia walks $\frac {17}{5}$ 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 \[\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}\] Then, multiply this by the time \[\frac{17}{22} \cdot 44 \text{minutes} = 34 \text{minutes}\] 34 minutes is a little greater than $\frac {1}{2}$ of an hour so Samia traveled \[\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}\] The answer choice a little greater than 2.5 is $\boxed{\bold{(C)} 2.8}$. (Note that we could've multiplied $\frac {34}{60}=\frac {17}{30}$ by $5$ 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.

\[\overline{s}=\frac{2ab}{a+b}\implies\overline{s}=\frac{2 \cdot 17 \cdot 5}{17+5}=\frac{85}{11}\text{kph}\]

Then, multiplying by $\frac{11}{15}$ hours gives the overall distance $\frac{17}{3}$ kilometers. Samia only walks half of that, so $\frac{17}{3}\approx \boxed{\textbf{(C)}2.8}$ 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

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