There are three taps of diameters 1cm, \( \Large \Large \frac{4}{3} \) cm and 2 cm, respectively. The ratio of the water flowing through them is equal to the ratio of the square of their diameters. The biggest tap can fill the tank alone in 61 min. If all the taps are opened simultaneously, how long will the tank take to be filled?
Correct Answer: Description for Correct answer:
Time taken to fill the tank by the tap having 2 cm diameter = 61 min.
Therefore, Time taken to fill the tank by the tap having 1 m diameter
= \( \Large 61 \times \left(\frac{2}{1}\right)^{2} \) = 244 min.
Similarly, time takn to fill the tank by the tap having \( \Large \frac{4}{3} \)cm diameter
= \( \Large 61 \times \left(\frac{2}{\frac{4}{3}}\right)^{2}=61 \times \frac{9}{4}=\frac{549}{4} \) min.
Therefore, Part of the tank filled by all the three pipes in 1 min
= \( \Large \frac{1}{61}+\frac{1}{244}+\frac{1}{ 549/4} \)
=\( \Large \frac{36+9+16}{2196} = \frac{61}{2196} = \frac{1}{36} \)
Hence, required time taken = 36 min.
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