A swimming pool is fitted with three pipes. T | vedclass

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(B) Let the time taken by the first pipe be $x$ hours.Then,the time taken by the second pipe is $(x - 5)$ hours.The time taken by the third pipe is $(

Vedclass · Accepted Answer

$3.75$

Vedclass · Answer

(B) Let the time taken by the first pipe be $x$ hours.Then,the time taken by the second pipe is $(x - 5)$ hours.The time taken by the third pipe is $(x - 5) - 4 = (x - 9)$ hours.According to the problem,the first and second pipes together fill the pool in the same time as the third pipe alone:$\frac{1}{x} + \frac{1}{x - 5} = \frac{1}{x - 9}$$(x - 5 + x) / (x(x - 5)) = 1 / (x - 9)$$(2x - 5)(x - 9) = x(x - 5)$$2x^2 - 18x - 5x + 45 = x^2 - 5x$$x^2 - 18x + 45 = 0$$(x - 15)(x - 3) = 0$Since $x$ must be greater than $9$ (as the third pipe takes $x-9$ hours),we take $x = 15$.Time taken by the first pipe = $15$ hours,second pipe = $10$ hours,third pipe = $6$ hours.Combined rate of second and third pipes = $\frac{1}{10} + \frac{1}{6} = \frac{3 + 5}{30} = \frac{8}{30} = \frac{4}{15}$ pools per hour.Time taken = $\frac{15}{4} = 3.75$ hours.

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