The tank is filled by three pipes with different uniform flow rates. While the first two pipes are operating simultaneously, they fill the tank in the same duration that the third pipe takes to fill it alone. The second pipe can fill the tank 5 hours quicker than the first pipe, yet 4 hours slower than the third pipe. What is the time (in hours) needed for the first pipe to fill the tank?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 15.
Let the time taken by the first pipe to fill the tank be x hours
Time taken by the second pipe to fill the tank = x - 5 hours
Time taken by third pipe to fill the tank = (x - 5 - 4) = x - 9 hours
Rate of first pipe = 1/x
Rate of second pipe = 1/(x - 5)
Rate of third pipe = 1/(x - 9)
According to the question;
The first two pipes operate simultaneously, they fill the tank in the same duration that the third pipe takes to fill it alone
⇒ 1/x + 1/(x - 5) = 1/(x - 9)
⇒ (x - 5)(x - 9) + x(x - 9) = x(x - 5)
Now,
⇒ 2x2 - 23x + 45 = x2 - 5x
⇒ x2 - 18x + 45 = 0
⇒ (x - 15)(x - 3)
Hence, x = 15 and 3
Since the time must be positive and reasonable for filling the tank,
The time taken by the first pipe to fill the tank is 15 hours.