a certain tank can be filled by two pipes in 80 minutes. The larger pipe by itself can fill the tank in 2 hours less than the smaller pipe by itself. How long does it take each pipe to fill the tank on its own?
Let the rate of filling of the larger pipe be L tank/hour, and let the rate of filling of the smaller pipe be S tank/hour.
We are given that the tank can be filled by two pipes in 80 minutes. In other words, in 1 hour and 20 minutes, the two pipes together can fill the tank.
The rate of filling by the two pipes together is 1 tank/80 minutes. We need to convert this rate to tanks per hour.
Since 1 hour = 60 minutes, we have
1 tank/80 minutes = (1 tank/80 minutes) * (60 minutes/1 hour) = 60/80 = 3/4 tank/hour.
We can write the rate equation for the two pipes together as:
L + S = 3/4.
We are also given that the larger pipe can fill the tank in 2 hours less than the smaller pipe can fill the tank by itself. In other words, L is S - 2.
Substituting this into the first equation, we have:
(S - 2) + S = 3/4.
Combining like terms, we obtain:
2S - 2 = 3/4.
Adding 2 to both sides, we have:
2S = 3/4 + 2 = 3/4 + 2 * 4/4 = 3/4 + 8/4 = 11/4.
Dividing both sides by 2, we obtain:
S = (11/4)/2 = 11/8.
Substituting this back into the equation L = S - 2, we have L = 11/8 - 16/8 = -5/8.
Since L represents the filling rate of the larger pipe, it must be positive. Therefore, the solution L = -5/8 is extraneous (not valid).
Thus, there is no valid solution for the question as stated.