SOLUTION: I am having trouble figuring out how to solve this problem and any help would be appreciated. This problem is not from a text book and is one that my professor create himself.

Question 214298: I am having trouble figuring out how to solve this problem and any help would be appreciated. This problem is not from a text book and is one that my professor create himself.
A speedy tortoise can run with a speed of 10 cm/sec and a hare can run 20 times as fast. If they both start at the same time, but the hare stops for 2 minutes to rest, the tortoise wins by 20cm.
1. How long does the race take?
2. What is the length of the race?
Found 2 solutions by stanbon, scott8148:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A speedy tortoise can run with a speed of 10 cm/sec and a hare can run 20 times as fast. If they both start at the same time, but the hare stops for 2 minutes to rest, the tortoise wins by 20cm.
1. How long does the race take?
Tortoise DATA:
rate = 10 cm/sec ; distance = x cm ; time = x/10 seconds
---
Hare DATA:
rate = 200 cm/sec ; distance = (x-20) cm ; time = (x-20)/200 sec
---------------------------------------
Equation
Hare time + 120 seconds = tortoise time
(x-20)/200 + 120 = x/10
Multiply thru by 200 to get:
x-20 + 200*120 = 20x
19x = 200*120 -20
x = 1262.11 cm (length of the race)
x/10 = 126.21 seconds (time of the race)
----
Cheers,
Stan H.
2. What is the length of the race?

Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
1. t * 10 = [(t - 120) * 200] + 20 ___ time is in seconds

10t = 200t - 23980 ___ 23980 = 190t ___ 126.2 = t (approx)

2. d = 10 * t = 10 * 126.2 = 1262 cm
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