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2013 AMC 8 Problems/Problem 11: Difference between revisions

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==Solution==
==Solution==
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>.
We use that fact that <math>d=rt</math>. In this case, let <math>x</math> represent the time.


Add the hours = <math>\frac{2}{5}  \text {hours}</math> + <math>\frac{2}{3}  \text {hours}</math> + <math>\frac{2}{4}  \text {hours}</math> = <math>\frac{94}{60} \text {hours}</math>.
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>.  


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 <math>\times</math> 3 days. x = <math>\frac{3}{2}  \text {hours}</math>.
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>.  


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.
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}  \text {hours}</math>. 
 
Adding up the hours yields <math>\frac{2}{5}  \text { hours}</math> + <math>\frac{2}{3}  \text { hours}</math> + <math>\frac{2}{4}  \text { hours}</math> = <math>\frac{94}{60}  \text { hours}</math>.
 
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>.
 
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, we multiply by <math>60</math>.


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

Revision as of 20:14, 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

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 (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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