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Difference between revisions of "2018 AMC 10B Problems/Problem 2"

m (→‎Problem: LaTeX'ed the problem.)
(Made Sol 1 considerably more concise. Also, corrected typos in Sol 2 as well as mentioning an underlying assumption.)
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</math>
 
</math>
  
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== Solutions ==
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== Solution 1 ==
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=== Solution 1 ===
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Suppose that Sam's average speed during the last <math>30</math> minutes was <math>x</math> mph.
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Let Sam drive at exactly <math>60</math> mph in the first half hour, <math>65</math> mph in the second half hour, and <math>x</math> mph in the third half hour.
 
  
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Due to <math>rt = d</math>, and that <math>30</math> min is half an hour, he covered <math>60 \cdot \frac{1}{2} = 30</math> miles in the first <math>30</math> mins.
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Recall that a half hour is equal to <math>30</math> minutes. Therefore, Sam drove <math>60\cdot0.5=30</math> miles during the first half hour, <math>65\cdot0.5=32.5</math> miles during the second half hour, and <math>x\cdot0.5</math> miles during the last half hour. We have <cmath>\begin{align*}
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30+32.5+x\cdot0.5&=96 \\
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x\cdot0.5&=33.5 \\
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x&=\boxed{\textbf{(D) } 67}.
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\end{align*}</cmath>
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~Haha0201 ~MRENTHUSIASM
  
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SImilarly, he covered <math>\frac{65}{2}</math> miles in the <math>2</math>nd half hour period.  
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== Solution 2 ==
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Suppose that Sam's average speed during the last <math>30</math> minutes was <math>x</math> mph.
  
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The problem states that Sam drove <math>96</math> miles in <math>90</math> min, so that means that he must have covered <math>96 - \left(30 + \frac{65}{2}\right) = 33 \frac{1}{2}</math> miles in the third half hour period.
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Note that Sam's average speed during the entire trip was <math>\frac{96}{3/2}=64</math> mph. Since Sam drove at <math>60</math> mph, <math>65</math> mph, and <math>x</math> mph for the same amount of time (<math>30</math> minutes each), his average speed during the entire trip was the average of the speeds <math>60</math> mph, <math>65</math> mph, and <math>x</math> mph. We have
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<cmath>\begin{align*}
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\frac{60+65+x}{3}&=64 \\
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60+65+x&=192 \\
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x&=\boxed{\textbf{(D) } 67}.
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\end{align*}</cmath>
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~coolmath_2018 (Solution)
  
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<math>rt = d</math>, so <math>x \cdot \frac{1}{2} = 33 \frac{1}{2}</math>.
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~MRENTHUSIASM (Minor Edits)
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Therefore, Sam was driving <math>\boxed{\textbf{(D) } 67}</math> miles per hour in the third half hour.
 
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=== Solution 2 (Faster) ===
 
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The average speed for the total trip is <cmath>\text{avg. speed} = \frac{96}{\frac{3}{2}} = 64.</cmath> Therefore the average speed for the total trip is the average of the average speeds of the three intrevals. So we have <math>64 = \frac{60 + 65 + x}{3}</math> and solving for <math>x = 67</math>. So the answer is <math>\boxed{\textbf{(D) } 67}</math>.
 
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~coolmath_2018
 
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 12:35, 18 September 2021

The following problem is from both the 2018 AMC 12B #2 and 2018 AMC 10B #2, so both problems redirect to this page.

Problem

Sam drove $96$ miles in $90$ minutes. His average speed during the first $30$ minutes was $60$ mph (miles per hour), and his average speed during the second $30$ minutes was $65$ mph. What was his average speed, in mph, during the last $30$ minutes?

$\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68$

Solution 1

Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.

Recall that a half hour is equal to $30$ minutes. Therefore, Sam drove $60\cdot0.5=30$ miles during the first half hour, $65\cdot0.5=32.5$ miles during the second half hour, and $x\cdot0.5$ miles during the last half hour. We have \begin{align*} 30+32.5+x\cdot0.5&=96 \\ x\cdot0.5&=33.5 \\ x&=\boxed{\textbf{(D) } 67}. \end{align*} ~Haha0201 ~MRENTHUSIASM

Solution 2

Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.

Note that Sam's average speed during the entire trip was $\frac{96}{3/2}=64$ mph. Since Sam drove at $60$ mph, $65$ mph, and $x$ mph for the same amount of time ($30$ minutes each), his average speed during the entire trip was the average of the speeds $60$ mph, $65$ mph, and $x$ mph. We have \begin{align*} \frac{60+65+x}{3}&=64 \\ 60+65+x&=192 \\ x&=\boxed{\textbf{(D) } 67}. \end{align*} ~coolmath_2018 (Solution)

~MRENTHUSIASM (Minor Edits)

Video Solution

https://youtu.be/77dDIzKprzA

~savannahsolver

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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
2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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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