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Difference between revisions of "2017 AMC 10B Problems/Problem 2"
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Sofia ran <math>5</math> laps around the <math>400</math>-meter track at her school. For each lap, she ran the first <math>100</math> meters at an average speed of <math>4</math> meters per second and the remaining <math>300</math> meters at an average speed of <math>5</math> meters per second. How much time did Sofia take running the <math>5</math> laps?
 
Sofia ran <math>5</math> laps around the <math>400</math>-meter track at her school. For each lap, she ran the first <math>100</math> meters at an average speed of <math>4</math> meters per second and the remaining <math>300</math> meters at an average speed of <math>5</math> meters per second. How much time did Sofia take running the <math>5</math> laps?
  
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<math>\textbf{(A)}\ 5</math> minutes and <math>35</math> seconds <math>\qquad\textbf{(B)}\ 6</math> minutes and <math>40</math> seconds <math>\qquad\textbf{(C)}\ 7</math> minutes and <math>5</math> seconds <math>\qquad\textbf{(D)}\ 7</math> minutes and <math>25</math> seconds <math>\qquad\textbf{(E)}\ 8</math> minutes and <math>10</math> seconds
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<math>\textbf{(A)}\ \text{5 minutes and 35 seconds}\qquad\textbf{(B)}\ \text{6 minutes and 40 seconds}\qquad\textbf{(C)}\ \text{7 minutes and 5 seconds}\qquad\textbf{(D)}\ \text{7 minutes and 25 seconds}</math><math>\qquad\textbf{(E)}\ \text{8 minutes and 10 seconds}</math>
  
 
==Solution==
 
==Solution==
  
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If Sofia ran the first <math>100</math> meters of each lap at <math>4</math> meters per second and the remaining <math>300</math> meters of each lap at <math>5</math> meters per second, then she took <math>\frac{100}{4}+\frac{300}{5}=25+60=85</math> seconds for each lap. Because she ran <math>5</math> laps, she took a total of <math>5 \cdot 85=425</math> seconds, or <math>7</math> minutes and <math>5</math> seconds. The answer is <math>\boxed{\textbf{(C)} }</math>.
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If Sofia ran the first <math>100</math> meters of each lap at <math>4</math> meters per second and the remaining <math>300</math> meters of each lap at <math>5</math> meters per second, then she took <math>\frac{100}{4}+\frac{300}{5}=25+60=85</math> seconds for each lap. Because she ran <math>5</math> laps, she took a total of <math>5 \cdot 85=425</math> seconds, or <math>7</math> minutes and <math>5</math> seconds. The answer is <math>\boxed{\textbf{(C)}\ \text{7 minutes and 5 seconds}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=1|num-a=3}}
 
{{AMC10 box|year=2017|ab=B|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:52, 16 February 2017

Problem

Sofia ran $5$ laps around the $400$-meter track at her school. For each lap, she ran the first $100$ meters at an average speed of $4$ meters per second and the remaining $300$ meters at an average speed of $5$ meters per second. How much time did Sofia take running the $5$ laps?

$\textbf{(A)}\ \text{5 minutes and 35 seconds}\qquad\textbf{(B)}\ \text{6 minutes and 40 seconds}\qquad\textbf{(C)}\ \text{7 minutes and 5 seconds}\qquad\textbf{(D)}\ \text{7 minutes and 25 seconds}$$\qquad\textbf{(E)}\ \text{8 minutes and 10 seconds}$

Solution

If Sofia ran the first $100$ meters of each lap at $4$ meters per second and the remaining $300$ meters of each lap at $5$ meters per second, then she took $\frac{100}{4}+\frac{300}{5}=25+60=85$ seconds for each lap. Because she ran $5$ laps, she took a total of $5 \cdot 85=425$ seconds, or $7$ minutes and $5$ seconds. The answer is $\boxed{\textbf{(C)}\ \text{7 minutes and 5 seconds}}$.

See Also

2017 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

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