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

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==Problem==
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{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #2]] and [[2018 AMC 10B Problems|2018 AMC 10B #2]]}}
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== 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?
 
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?
  
 
<math>\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68</math>
 
<math>\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68</math>
  
 
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== Solutions ==
==Solution==
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=== Solution 1 ===
 
 
 
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.
 
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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SImilarly, he covered <math>\frac{65}{2}</math> miles in the <math>2</math>nd half hour period.  
 
SImilarly, he covered <math>\frac{65}{2}</math> miles in the <math>2</math>nd half hour period.  
  
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 - (30 + \frac{65}{2}) = 33 \frac{1}{2}</math> miles in the third half hour period.
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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.
  
 
<math>rt = d</math>, so <math>x \cdot \frac{1}{2} = 33 \frac{1}{2}</math>.
 
<math>rt = d</math>, so <math>x \cdot \frac{1}{2} = 33 \frac{1}{2}</math>.
  
 
Therefore, Sam was driving <math>\boxed{\textbf{(D) } 67}</math> miles per hour in the third half hour.
 
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
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== Video Solution ==
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https://youtu.be/77dDIzKprzA
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 14:21, 19 January 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?

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

Solutions

Solution 1

Let Sam drive at exactly $60$ mph in the first half hour, $65$ mph in the second half hour, and $x$ mph in the third half hour.

Due to $rt = d$, and that $30$ min is half an hour, he covered $60 \cdot \frac{1}{2} = 30$ miles in the first $30$ mins.

SImilarly, he covered $\frac{65}{2}$ miles in the $2$nd half hour period.

The problem states that Sam drove $96$ miles in $90$ min, so that means that he must have covered $96 - \left(30 + \frac{65}{2}\right) = 33 \frac{1}{2}$ miles in the third half hour period.

$rt = d$, so $x \cdot \frac{1}{2} = 33 \frac{1}{2}$.

Therefore, Sam was driving $\boxed{\textbf{(D) } 67}$ miles per hour in the third half hour.

Solution 2 (Faster)

The average speed for the total trip is \[\text{avg. speed} = \frac{96}{\frac{3}{2}} = 64.\] Therefore the average speed for the total trip is the average of the average speeds of the three intrevals. So we have $64 = \frac{60 + 65 + x}{3}$ and solving for $x = 67$. So the answer is $\boxed{\textbf{(D) } 67}$. ~coolmath_2018

Video Solution

https://youtu.be/77dDIzKprzA

~savannahsolver

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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