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

Latest revision as of 01:37, 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.

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 duration ( $30$ minutes each), his average speed during the entire trip was the average of $60$ 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 ~MRENTHUSIASM

Video Solution

https://youtu.be/77dDIzKprzA

~savannahsolver

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/VN88m4xUHM0

~Education, the Study of Everything

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.

@@ Line 2: / Line 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 <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>
-== Solutions ==
+== Solution 1 ==
-=== Solution 1 ===
+Suppose that Sam's average speed during the last <math>30</math> minutes was <math>x</math> mph.
-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.
-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.
+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*}
++32.5+x\cdot0.5&=96 \\
+x\cdot0.5&=33.5 \\
+x&=\boxed{\textbf{(D) } 67}.
+\end{align*}</cmath>
+~Haha0201 ~MRENTHUSIASM
-SImilarly, he covered <math>\frac{65}{2}</math> miles in the <math>2</math>nd half hour period.
+== Solution 2 ==
+Suppose that Sam's average speed during the last <math>30</math> minutes was <math>x</math> mph.
-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.
+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*}
-<math>rt = d</math>, so <math>x \cdot \frac{1}{2} = 33 \frac{1}{2}</math>.
+\frac{60+65+x}{3}&=64 \\
++65+x&=192 \\
-Therefore, Sam was driving <math>\boxed{\textbf{(D) } 67}</math> miles per hour in the third half hour.
+x&=\boxed{\textbf{(D) } 67}.
+\end{align*}</cmath>
-=== Solution 2 (Faster) ===
+~coolmath_2018 ~MRENTHUSIASM
-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 ==
@@ Line 28: / Line 37: @@
 ~savannahsolver
+==Video Solution (HOW TO THINK CRITICALLY!!!)==
+https://youtu.be/VN88m4xUHM0
+~Education, the Study of Everything
 ==See Also==

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