A tank 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 the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
A.6 hours
B.10 hours
C.15 hours
D.30 hours
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Suppose, first pipe alone takes x hours to fill the tank .
Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.
As per question, we get
\begin{aligned}
\frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9} \\
=> \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\\
=> (2x - 5)(x - 9) = x(x - 5)\\
=> x^2 - 18x + 45 = 0
\end{aligned}
After solving this euation, we get
(x-15)(x+3) = 0,
As value can not be negative, so x = 15
Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.
As per question, we get
\begin{aligned}
\frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9} \\
=> \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\\
=> (2x - 5)(x - 9) = x(x - 5)\\
=> x^2 - 18x + 45 = 0
\end{aligned}
After solving this euation, we get
(x-15)(x+3) = 0,
As value can not be negative, so x = 15
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