In a 21-meter race between a tortoise and a hare, the tortoise leaves 9 minutes before the hare. The hare, by running at an average speed of 0.5 meter per hour faster than the toroise, corsses the finish line 3 minutes before the tortoise. What are the average speeds of the tortoise and the hare?
To find the average speeds of the tortoise and the hare, we can set up a system of equations using the given information.
Let's start by assigning variables:
- Let x be the average speed of the tortoise in meters per minute.
- Let x + 0.5 be the average speed of the hare in meters per minute.
Given that the tortoise leaves 9 minutes before the hare, we can assume that the hare takes 9 minutes less time to complete the race. This means that the time taken by the hare to finish the race is 21 - 9 = 12 minutes.
We also know that the hare crosses the finish line 3 minutes before the tortoise. This means that the time taken by the tortoise to finish the race is 12 + 3 = 15 minutes.
Using the formula speed = distance / time, we can now set up the following equations:
For the hare:
(x + 0.5) = (21 / 12)
For the tortoise:
x = (21 / 15)
Now we can solve these equations to find the values of x (tortoise speed) and x + 0.5 (hare speed).
Hare's speed:
(x + 0.5) = (21 / 12)
x + 0.5 = 1.75 (simplifying the fraction 21 / 12)
x = 1.25 (subtracting 0.5 from both sides)
Tortoise's speed:
x = (21 / 15)
x = 1.4 (simplifying the fraction 21 / 15)
So, the average speed of the tortoise is 1.4 meters per minute, and the average speed of the hare is 1.25 meters per minute.
time = distance/speed, if the hare's speed is x m/min, then since 0.5 m/hr = 1/120 m/min, and the hare's time is 3 min less than the tortoise's
21/x = 21/(x - 1/120) - 3
Yet this has no rational solutions.
And since when does a hare go only 1/120 meter/min faster than a tortoise? I suspect a typo somewhere, but fix the data if necessary, and the above logic should work for you.