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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. The formula is d = r \times t, where d is distance, r is rate, and t is time.
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We use that fact that <math>d=rt</math>. Let d= distance, r= rate or speed, and t=time. In this case, let <math>x</math> represent the time.
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On Monday, he was at a rate of <math>5 \text{ m.p.h}</math>. So, <math>5x = 2 \text{ miles}\implies x = \frac{2}{5} \text { hours}</math>.
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For Wednesday, he walked at a rate of <math>3 \text{ m.p.h}</math>. Therefore, <math>3x = 2 \text{ miles}\implies x = \frac{2}{3}  \text { hours}</math>.
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On Friday, he walked at a rate of <math>4 \text{ m.p.h}</math>. So, <math>4x = 2 \text{ miles}\implies x=\frac{2}{4}=\frac{1}{2}  \text {hours}</math>. 
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Adding up the hours yields <math>\frac{2}{5}  \text { hours}</math> + <math>\frac{2}{3}  \text { hours}</math> + <math>\frac{1}{2}  \text { hours}</math> = <math>\frac{47}{30}  \text { hours}</math>.
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We now find the amount of time Grandfather would have taken if he walked at <math>4 \text{ m.p.h}</math> per day. Set up the equation, <math>4x = 2 \text{ miles} \times 3 \text{ days}\implies x = \frac{3}{2}  \text { hours}</math>.
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To find the amount of time saved, subtract the two amounts: <math>\frac{47}{30}  \text { hours}</math> - <math>\frac{3}{2}  \text { hours}</math> = <math>\frac{1}{15}  \text { hours}</math>. To convert this to minutes, we multiply by <math>60</math>.
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Thus, the solution to this problem is <math>\dfrac{1}{15}\times 60=\boxed{\textbf{(D)}\ 4}</math>
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==Video Solution==
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https://youtu.be/b3z2bfTLk4M ~savannahsolver
  
 
==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}}

Latest revision as of 16:57, 5 May 2022

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

We use that fact that $d=rt$. Let d= distance, r= rate or speed, and t=time. In this case, let $x$ represent the time.

On Monday, he was at a rate of $5 \text{ m.p.h}$. So, $5x = 2 \text{ miles}\implies x = \frac{2}{5} \text { hours}$.

For Wednesday, he walked at a rate of $3 \text{ m.p.h}$. Therefore, $3x = 2 \text{ miles}\implies x = \frac{2}{3} \text { hours}$.

On Friday, he walked at a rate of $4 \text{ m.p.h}$. So, $4x = 2 \text{ miles}\implies x=\frac{2}{4}=\frac{1}{2} \text {hours}$.

Adding up the hours yields $\frac{2}{5} \text { hours}$ + $\frac{2}{3} \text { hours}$ + $\frac{1}{2} \text { hours}$ = $\frac{47}{30} \text { hours}$.

We now find the amount of time Grandfather would have taken if he walked at $4 \text{ m.p.h}$ per day. Set up the equation, $4x = 2 \text{ miles} \times 3 \text{ days}\implies x = \frac{3}{2} \text { hours}$.

To find the amount of time saved, subtract the two amounts: $\frac{47}{30} \text { hours}$ - $\frac{3}{2} \text { hours}$ = $\frac{1}{15} \text { hours}$. To convert this to minutes, we multiply by $60$.

Thus, the solution to this problem is $\dfrac{1}{15}\times 60=\boxed{\textbf{(D)}\ 4}$

Video Solution

https://youtu.be/b3z2bfTLk4M ~savannahsolver

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

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