A cistern can be filled by one of two pipes in 52 minutes and by the other pipe in 60 minutes. Both pipes are opened together for a certain time but being particularly clogged only $$\frac{13}{20}$$​ of the full quantity of water flows through the first pipe and only $$\frac{3}{4}$$​ of the full quantity of water flows through the second pipe. The obstructions, however, were suddenly removed in both the pipes after some time and the cistern got filled in $$15\frac{3}{5}$$​ minutes from that moment. For how many minutes did the pipes remain clogged?

Correct option is C

Given:

First pipe fills the tank in 52 minutes.

Second pipe fills the tank in 60 minutes.

Formula Used:

Total work = Efficiency $$\times$$ Total times

According to the question:

780 = $$\left[15\times \frac{13}{20} + 13 \times \frac{3}{4}\right]t$$ + (15 + 13) $$\frac{78}{5}$$

780 = $$\frac{39}{2}$$ t + 28 $$\times$$ $$\frac{78}{5}$$

780 = 39$$\left[\frac{t}{2} + \frac{56}{5}\right]$$​​

20 = $$\frac{t}{2}$$ + $$\frac{56}{5}$$

$$\frac{t}{2} = 20 - \frac{56}{5}$$

t = $$\frac{88}{5}$$

t = 17$$\frac{3}{5}$$ minutes

Thus, the correct option is (c)

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