Difference between revisions of "2018 AMC 10B Problems/Problem 2"
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+ | {{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #2]] and [[2018 AMC 10B Problems|2018 AMC 10B #2]]}} | ||
+ | |||
+ | == 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> | ||
− | + | == Solutions == | |
− | ==Solution== | + | === 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. | + | 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>\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. |
+ | |||
+ | === Solution 2 (Faster) === | ||
+ | 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>. | ||
+ | ~coolmath_2018 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/77dDIzKprzA | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2018|ab=B|num-b=1|num-a=3}} | ||
+ | {{AMC12 box|year=2018|ab=B|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} |
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?
Solutions
Solution 1
Let Sam drive at exactly mph in the first half hour, mph in the second half hour, and mph in the third half hour.
Due to , and that min is half an hour, he covered miles in the first mins.
SImilarly, he covered miles in the nd half hour period.
The problem states that Sam drove miles in min, so that means that he must have covered miles in the third half hour period.
, so .
Therefore, Sam was driving miles per hour in the third half hour.
Solution 2 (Faster)
The average speed for the total trip is Therefore the average speed for the total trip is the average of the average speeds of the three intrevals. So we have and solving for . So the answer is . ~coolmath_2018
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.