I bicycle across the bridge, ride a flat road, then up the hill to the top of the hill... (more)

Question

I bicycle across the bridge, ride a flat road, then up the hill to the top of the hill. At the top I turn around to go down the hill, over the flats, and back over the bridge. As I look at my speedometer, it appears that I go a steady 15 mph on the flats (including the bridge), 6 mph on the “up-hill” and 24 mph on the “down-hill”. The entire ride takes 1.5 miles and I travel 18 miles. How many ”flat” miles were in my ride? How many “up-hill” miles? How many “down-hill” miles?

Answer 1

This can be treated as a simple system of variables.  If you look at the path, there's only two sections - the flat area, and the hill.  Let's call the distance of the flat area x, and the of the hill, y.  Then, we can observe a few things:

First, since the entire journey covers 18 miles back and forth, the distance one way must be half of that, or nine miles.  Since the flat area and the hill comprise the entire distance, we know that x + y = 9.

 

In terms of flat, uphill and downhill, we have a distance of 2x that is flat (the flat area, there and back), a distance of y that is uphill (the hill on the way out), and a distance of y that is downhill (the hill on the way back).  Based our speeds, we can construct an equation for the total time that should be taken, remembering the speed equation time = distance/speed:

 

2x/15 + y/6 + y/24 = 1.5 hours

Combine like terms:  the LCM of 6 and 24 is 24

2x/15 + 4y/24 + y/24 = 1.5

2x/15 + 5y/24 = 1.5

 

Now we have two equations and two unknowns, which is sufficient to solve.  I would do so by using x = 9 - y

2(9-y)/15 + 5y/24 = 1.5

18/15 - 2y/15 + 5y/24 = 3/2

LCM of 15 and 24 is 120, of 15 and 2 is 30

36/30 - 16y/120 + 25y/120 = 45/30

Combine like terms

9y/120 = 9/30

y/120 = 1/30

y = 4

Since x = 9 - y, we then have x = 5.  We can then answer the question:  2x flat miles is 10 miles, and there are 4 of both the uphill and downhill miles.