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]]}} | |
− | <math>\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68</math> | + | == Problem == |
+ | Sam drove <math>96</math> miles in <math>90</math> minutes. His average speed during the first <math>30</math> minutes was <math>60</math> mph (miles per hour), and his average speed during the second <math>30</math> minutes was <math>65</math> mph. What was his average speed, in mph, during the last <math>30</math> minutes? | ||
+ | |||
+ | <math> | ||
+ | \textbf{(A) } 64 \qquad | ||
+ | \textbf{(B) } 65 \qquad | ||
+ | \textbf{(C) } 66 \qquad | ||
+ | \textbf{(D) } 67 \qquad | ||
+ | \textbf{(E) } 68 | ||
+ | </math> | ||
+ | |||
+ | == Solution 1 == | ||
+ | Suppose that Sam's average speed during the last <math>30</math> minutes was <math>x</math> mph. | ||
+ | |||
+ | 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*} | ||
+ | 30+32.5+x\cdot0.5&=96 \\ | ||
+ | x\cdot0.5&=33.5 \\ | ||
+ | x&=\boxed{\textbf{(D) } 67}. | ||
+ | \end{align*}</cmath> | ||
+ | ~Haha0201 ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 2 == | ||
+ | Suppose that Sam's average speed during the last <math>30</math> minutes was <math>x</math> mph. | ||
+ | |||
+ | 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 duration (<math>30</math> minutes each), his average speed during the entire trip was the average of <math>60</math> mph, <math>65</math> mph, and <math>x</math> mph. We have | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{60+65+x}{3}&=64 \\ | ||
+ | 60+65+x&=192 \\ | ||
+ | x&=\boxed{\textbf{(D) } 67}. | ||
+ | \end{align*}</cmath> | ||
+ | ~coolmath_2018 ~MRENTHUSIASM | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/77dDIzKprzA | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
+ | https://youtu.be/VN88m4xUHM0 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==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}} |
Latest revision as of 02:37, 28 May 2023
- The following problem is from both the 2018 AMC 12B #2 and 2018 AMC 10B #2, so both problems redirect to this page.
Contents
Problem
Sam drove miles in
minutes. His average speed during the first
minutes was
mph (miles per hour), and his average speed during the second
minutes was
mph. What was his average speed, in mph, during the last
minutes?
Solution 1
Suppose that Sam's average speed during the last minutes was
mph.
Recall that a half hour is equal to minutes. Therefore, Sam drove
miles during the first half hour,
miles during the second half hour, and
miles during the last half hour. We have
~Haha0201 ~MRENTHUSIASM
Solution 2
Suppose that Sam's average speed during the last minutes was
mph.
Note that Sam's average speed during the entire trip was mph. Since Sam drove at
mph,
mph, and
mph for the same duration (
minutes each), his average speed during the entire trip was the average of
mph,
mph, and
mph. We have
~coolmath_2018 ~MRENTHUSIASM
Video Solution
~savannahsolver
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
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.