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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?

- $6$ hours
- $10$ hours
- $15$ hours
- $30$ hours

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Answer is C.

Given A = B + 5 and C = B – 4, and let B = x hours, so,

A = (x+5) hours and C = (x-4) hours

Condition, given in the question, A + B = C

$\frac{1}{x+5} + \frac{1}{x} = \frac{1}{x-4}$

$x^{2} -8x-20=0$

On solving the above quadratic equation, we will get value of x = -2, 10.

Acceptable value of x = **10**

Hence the time required by the first pipe A = x + 5 = **15 hours**