Question

Jacy went on a bike ride of 60 miles. He realized that if he had gone 10 mph faster, he would have arrived 8 hours sooner. How fast did he actually ride?

Answer 1

Let r be the rate at which he travels and t be the time.

Then, rt = 60

If he travels 8 mph faster, he arrives 10 hours sooner, so

(r+8)(t - 10) = 60

Multiplying, rt - 10r + 8t - 80 = 60

Substituting rt = 60, we get

60 - 10r + 8t - 80 = 60

Then, 8t = 10r + 80

t = 5/4r + 10

Substitute this into rt = 60

r(5/4r + 10) = 60

5/4r^2 + 10r - 60 = 0

Multipling by 4/5

r^2 + 8r - 48 = 0

(r + 12)(r - 4) = 0

r = -12, 4

A negative solution does not make sense

Thus, r = 4 4 mph is the solution.

Checking, rt = 60, so 4t = 60, and t = 15

Showing that this meets the other condition,

(r+8)(t - 10) = (4+8)(15-10) = 12 * 5 = 60

It checks.