12. A cyclist travels 40 km at a speed of x km/h. Find the time taken in terms of x. Find the time taken when his speed is
reduced by 2 km/h. If the difference between the times is
1 hour, find the original speed x.

Step-by-step explanation:

distance = rate times time, so

           distance

time =

               rate

Here the distance is 40 km and the rate is x.  Then

              40 km

time =  =  (40/x) hr

             x km/hr

If the speed is reduced by 2 km/hr, we get:

              40 km                                            40 km                  40

time =  or time = = hr

             (x km/hr - 2 km/hr)                      (x - 2) (km/hr)       x - 2

The difference between the times is:

 40          40

- = 1 hr          Solve this for the original speed, x

x - 2         x

40x - 40x + 80                                80

 =  1 hr     or     = 1     or 80 = x^2 - 2x

     x(x - 2)                                     x(x - 2)

Rewriting this in standard quadratic form:  x^2 - 2x - 80 = 0

Here a = 1, b = -2 and c = -80, and so the discriminant b^2 - 4ac is:

(-2)^2 - 4(1)(-80) = 324

The solutions are:

      -(-2) ± √324       2 ± 18

x = = = 1 ± 9, or 10 (discard the other root)

              2                      2

The original speed was 10 km/hr

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step-by-step explanation:

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