2013 AMC 8 Problems/Problem 11

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 is $\dfrac{4}{60}\times 60=\boxed{\textbf{(D)}\ 4}$

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.

Page

Toolbox

Search

2013 AMC 8 Problems/Problem 11

Problem

Solution

See Also