1. ## Find Laverne's Rate

At 9am, Laverne started from home on a hike to a mountain 12 miles away. After reaching the mountain, he took 1 hour for lunch and then returned over the same route, arriving home at 5pm. If his average rate returning was 1 mph less that his rate going, find his rate on the return trip.

My set up:

8x + 8(x - 1) = 12

I got 24 mph but the answer is 3 mph. What did I do wrong? I guess my entire equation is wrong, right?

2. In your set-up, for what does "x" stand? How did you arrive at your equation?

Originally Posted by magentarita
At 9am, Laverne started from home on a hike to a mountain 12 miles away. After reaching the mountain, he took 1 hour for lunch and then returned over the same route, arriving home at 5pm. If his average rate returning was 1 mph less that his rate going, find his rate on the return trip.
First, note that the total time spent travelling is seven hours, due to the one-hour lunch.

Using the standard equation "d = rt", we get:

. . .going:
. . . . .distance: 12
. . . . .rate: r
. . . . .time: 12/r

. . .returning:
. . . . .distance: 12
. . . . .rate: r - 1
. . . . .time: 12/(r - 1)

Add the two "time" expressions, set equal to the given total travel time, and solve the resulting rational equation.

If you get stuck, please reply showing how far you have gotten. Thank you!

3. Originally Posted by magentarita
At 9am, Laverne started from home on a hike to a mountain 12 miles away. After reaching the mountain, he took 1 hour for lunch and then returned over the same route, arriving home at 5pm. If his average rate returning was 1 mph less that his rate going, find his rate on the return trip.

My set up:

8x + 8(x - 1) = 12

I got 24 mph but the answer is 3 mph. What did I do wrong? I guess my entire equation is wrong, right?
Hello again magentarita,

I believe if we take out the hour spent for lunch, Laverne only spent 7 hours actually walking.

Let r = her rate going
Let r - 1 = her rate returning

Now the distance is constant at 12 miles.

If $D=rt$, then $\frac{D}{r}=t$

The time going and retuning is what we're concerned about. But, we know the total time walking is 7 hours.

Let $\frac{12}{r}=$ time going

Let $\frac{12}{r-1}=$ time returning

Now add the two times to a total of 7.

$\frac{12}{r}+\frac{12}{r-1}=7$

$12(r-1)+12r=7r(r-1)$

When I work though all of that I found r = 4 mph. This is the rate going.

So the rate returning would be r - 1 = 4 - 1 = 3 mph.

4. ## I got.....

Originally Posted by stapel
In your set-up, for what does "x" stand? How did you arrive at your equation?

First, note that the total time spent travelling is seven hours, due to the one-hour lunch.

Using the standard equation "d = rt", we get:

. . .going:
. . . . .distance: 12
. . . . .rate: r
. . . . .time: 12/r

. . .returning:
. . . . .distance: 12
. . . . .rate: r - 1
. . . . .time: 12/(r - 1)

Add the two "time" expressions, set equal to the given total travel time, and solve the resulting rational equation.

If you get stuck, please reply showing how far you have gotten. Thank you!
I got lost when reading the 9am and 5pm and the hour he took for lunch. This data blew me away. I got 8 in my equation by subtracting 5pm from 9am.

5. ## I thank you..........

Originally Posted by masters
Hello again magentarita,

I believe if we take out the hour spent for lunch, Laverne only spent 7 hours actually walking.

Let r = her rate going
Let r - 1 = her rate returning

Now the distance is constant at 12 miles.

If $D=rt$, then $\frac{D}{r}=t$

The time going and retuning is what we're concerned about. But, we know the total time walking is 7 hours.

Let $\frac{12}{r}=$ time going

Let $\frac{12}{r-1}=$ time returning

Now add the two times to a total of 7.

$\frac{12}{r}+\frac{12}{r-1}=7$

$12(r-1)+12r=7r(r-1)$

When I work though all of that I found r = 4 mph. This is the rate going.

So the rate returning would be r - 1 = 4 - 1 = 3 mph.

I got lost when reading the 9am and 5pm and the hour he took for lunch. This data blew me away. I got 8 in my equation by subtracting 5pm from 9am.