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?

Answer 1

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.