2013 AMC 8 Problems/Problem 11

Revision as of 20:14, 27 November 2013 by Fadebekun (talk | contribs) (Solution)

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 $d=rt$. 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, $4x = 2 \text{ miles}\implies x=\frac{2}{4}  \text {hours}$.

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

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: $\frac{94}{60}  \text { hours}$ - $\frac{3}{2}  \text { hours}$ = $\frac{4}{60}  \text { hours}$. To convert this to minutes, we multiply by $60$.

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

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AJHSME/AMC 8 Problems and Solutions

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