Question

You ride your bike to campus a distance of 3 miles and return home on the same route. Going to campus you ride mostly downhill and average 5 miles per hour faster than on your trip home. If the round trip takes 54 minutes what is your average rate on the return trip

Answer 1

10/3 mph

Explanation:

Obviously, (time going) + (time returning) = (total time spent en route) = 54 min. Since time = distance / rate,

3 miles 3 miles

----------------------- + --------------------- = 54 min

downhill speed uphill speed

Let u = uphill speed and d = downhill speed; then d = u + 5 (all in mph)

Then we have:

3 miles 3 miles

----------------------- + --------------------- = 54 min

u + 5 u

and our task here is to determine the uphill speed, u.

The LCD is u(u + 5). Thus we have:

3u 3(u + 5) miles

----------------------- + ----------------------- = 54 min = 0.9 hr

u(u + 5) u(u + 5)

so that:

6u + 15

----------------------- = 0.9 hr or 6u + 15 = 0.9(u)(u + 5), or

u(u + 5)

6u + 15 = 0.9u² + 4.5u

Combining the u terms, we get:

15 = 0.9u² + 4.5u, or 0.9u² + 1.5u - 15 = 0

Eliminating the fractions, we get 9u² + 15u - 150, or

3u^2 + 5u - 50 = 0

This factors into (3u - 10)(u + 5) = 0. The only positive root is u = 10/3.

Your average rate on the return trip (uphill) is 10/3 mph (3 1/3 mph).