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Difference between revisions of "2013 AMC 8 Problems/Problem 11"

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==Solution==
 
==Solution==
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You have to find the hours spent on each day. For people who do not know how to do this, here is a slower way to do it (you will be able to do quickly once you learn it):
 
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On Monday, he was at a rate of 5 mph. So, 5x = 2 miles. x = <math>\frac{2}{5} \text {hours}</math>. For Wednesday, he jogged at a rate of 3 mph. Therefore, 3x = 2 miles. x = <math>\frac{2}{3}  \text {hours}</math>. On Friday, he jogged at a rate of 4 hours. So, 4x = 2 miles. x=<math>\frac{2}{4}  \text {hours}</math>.   
 
On Monday, he was at a rate of 5 mph. So, 5x = 2 miles. x = <math>\frac{2}{5} \text {hours}</math>. For Wednesday, he jogged at a rate of 3 mph. Therefore, 3x = 2 miles. x = <math>\frac{2}{3}  \text {hours}</math>. On Friday, he jogged at a rate of 4 hours. So, 4x = 2 miles. x=<math>\frac{2}{4}  \text {hours}</math>.   
  
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Once you find this, answer the actual question by finding the amount of time Grandfather would take by jogging at 4 mph per day. Set up the equation, 4x = 2 miles \times  3 days. x = <math>\frac{3}{2}  \text {hours}</math>.
 
Once you find this, answer the actual question by finding the amount of time Grandfather would take by jogging at 4 mph per day. Set up the equation, 4x = 2 miles \times  3 days. x = <math>\frac{3}{2}  \text {hours}</math>.
  
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To find the amount of time saved: <math>\frac{94}{60}  \text {hours}</math> - <math>\frac{3}{2}  \text {hours}</math> = <math>\frac{4}{60}  \text {hours}</math>. To convert this to minutes, use the conversion rate; multiply by 60.
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To find the amount of time saved, subtract the two amounts: <math>\frac{94}{60}  \text {hours}</math> - <math>\frac{3}{2}  \text {hours}</math> = <math>\frac{4}{60}  \text {hours}</math>. To convert this to minutes, use the conversion rate; multiply by 60.
  
 
Thus, the solution to this problem = <math>\boxed{\textbf{(D)}\ 4}</math>
 
Thus, the solution to this problem = <math>\boxed{\textbf{(D)}\ 4}</math>
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--[[User:Arpanliku|Arpanliku]] 17:15, 27 November 2013 (EST) I am not very good at LATEX, so this solution is a bit long.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=10|num-a=12}}
 
{{AMC8 box|year=2013|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:15, 27 November 2013

Problem

Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution

On Monday, he was at a rate of 5 mph. So, 5x = 2 miles. x = $\frac{2}{5} \text {hours}$. For Wednesday, he jogged at a rate of 3 mph. Therefore, 3x = 2 miles. x = $\frac{2}{3} \text {hours}$. On Friday, he jogged at a rate of 4 hours. So, 4x = 2 miles. x=$\frac{2}{4} \text {hours}$.

Add the hours = $\frac{2}{5} \text {hours}$ + $\frac{2}{3} \text {hours}$ + $\frac{2}{4} \text {hours}$ = $\frac{94}{60} \text {hours}$.

Once you find this, answer the actual question by finding the amount of time Grandfather would take by jogging at 4 mph per day. Set up the equation, 4x = 2 miles \times 3 days. x = $\frac{3}{2} \text {hours}$.

To find the amount of time saved, subtract the two amounts: $\frac{94}{60} \text {hours}$ - $\frac{3}{2} \text {hours}$ = $\frac{4}{60} \text {hours}$. To convert this to minutes, use the conversion rate; multiply by 60.

Thus, the solution to this problem = $\boxed{\textbf{(D)}\ 4}$

--Arpanliku 17:15, 27 November 2013 (EST) I am not very good at LATEX, so this solution is a bit long.

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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