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]]}} | {{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #2]] and [[2018 AMC 10B Problems|2018 AMC 10B #2]]}} | ||
− | ==Problem== | + | == 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 <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> | + | <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 == | |
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− | ==Video Solution== | ||
https://youtu.be/77dDIzKprzA | https://youtu.be/77dDIzKprzA | ||
Latest revision as of 20:27, 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 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
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.