Atank is filled by three pipes with uniform flow.The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by third pipe alone.The second pipe fills the tank 5hours faster than the first pipe and 4hours slower than the third pipe .The time required by the first pipe is:
Answers
Answer:
15 hrs.
Step-by-step explanation:
Let t hours be time taken by first pipe.
The second pipe is 5 hrs faster than first pipe => (t - 5) hrs.
The third pipe is 4 hrs slower than 2nd pipe => (t -5) - 4 = (t -9) hrs.
It is given the first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by third pipe alone.
1/t + 1/t-5 = 1/t-9
[t - 5 + t]/t(t-5) = 1/t-9
2t-5 / t² - 5t = 1/t-9
(2t-5)(t-9) = (t²-5t)1
2t² - 18t - 5t + 45 = t² - 5t
t² - 18t + 45 = 0
t² - 15t - 3t + 45 = 0
t(t-15) - 3(t-15) = 0
(t - 3)(t-15) = 0
=> t = 3 or t = 15.
If first pipe takes 3 hrs, then 2nd and 3rd will take - 2hrs and -6 hrs. since these cannot be negative, the time required by first pipe cannot be 3 hrs.
If first pipe takes 15 hrs,
The 2nd pipe will take (15-5) = 10 hrs and 3rd pipe will take (15 - 9) = 6 hrs.
Thus time taken by first pipe will be 15 hrs.