Expert: Steve Holleran - 3/23/2008

Question
On account of repairs, the average speed of a train on a 200km trip is reduced by 5km an hour. This makes the train 2 hours late. What is its normal speed?

Answer
Hi Anna,

Let's say the train's normal rate is r and normal time for the trip is t.  Then 200 = r * t.

Now, if its speed is reduced by 5, it would be r-5, and its time increased by 2 becomes t+2 for the same 200 mile trip:

               200 = (r - 5)(t + 2) = rt + 2r - 5t -10

since rt = 200 from above, we can substitute:

               200 = 200 + 2r - 5t - 10

                0 = 2r - 5t - 10

               5t + 10 = 2r  so r = (5/2)t + 5

Now sub back into  the first equation:

                (5/2 t + 5)(t) = 200

                 5/2 t^2 + 5t - 200 = 0

Mult by 2:        5t^2 + 10t - 400 = 0

Div by 5:          t^2 + 2t - 80 = 0

                  (t + 10)(t - 8) = 0   so t = 8.

Then r = (5/2)(8) + 5 = 20 + 5 = 25 km/hr.


Hope this is what you wanted.
Steve