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

10/3 mph

Step-by-step 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).

d no solution

step-by-step explanation: