Sam took a short hike with his friends up to Mt. Wayshett. They hiked up at a steady pace of 1.5 mph, and downhill at a rate of 4 mph. If the climb to climb up the mountain took an hour more than the time to hike down, how long was the entire hike?

Answer 1

Let \( t \) be the time in hours Sam took to hike down the mountain. Therefore, the time he took to hike up the mountain would be \( t + 1 \) hours.

We can express the distances for both hikes (up and down) with the following equations:

• Distance going up: This is given by the formula \( \text{distance} = \text{rate} \times \text{time} \). Sam hikes up at 1.5 mph for \( t + 1 \) hours, so the distance up is: \[ d = 1.5(t + 1) \]

• Distance going down: Sam hikes down at 4 mph for \( t \) hours, so the distance down is: \[ d = 4t \]

Since the distance up and the distance down are the same, we can set these two expressions equal to each other: \[ 1.5(t + 1) = 4t \]

Now, we'll solve for \( t \):

1. Expand the left-hand side: \[ 1.5t + 1.5 = 4t \]

2. Rearranging the equation gives: \[ 1.5 = 4t - 1.5t \] \[ 1.5 = 2.5t \]

3. Divide both sides by 2.5 to isolate \( t \): \[ t = \frac{1.5}{2.5} = 0.6 \text{ hours} \]

Now that we have \( t \), we can find the time taken for the ascent: \[ t + 1 = 0.6 + 1 = 1.6 \text{ hours} \]

Now, to find the total time of the hike, we add the time going up and the time going down: \[ \text{Total time} = t + (t + 1) = 0.6 + 1.6 = 2.2 \text{ hours} \]

Thus, the duration of the entire hike was \(\boxed{2.2}\) hours.