A group of four pumps is designed to fill a tank such that each of the
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02 Feb 2025, 06:46
OE
Let the speed of the largest pump be 's', so the speed of the each of the three pumps would be $\(\frac{2}{3}\)$ of the largest pump's speed $\(=\frac{2 s}{3}\)$.
So, the combined speed of all 4 pumps comes out to be $\(s+3 ( \frac{2s}{3} )\)$ which gives the time taken by all 4 pumps together to fill the tank as $\(\frac{1}{3 s}\)$ Work done $=$ Speed $\times$ Time $)$
The time taken by the largest pump alone to fill the tank is $\(\frac{1}{s}\)$
Finally the time taken by all 4 pumps together to fill the $\( \frac{\frac{1}{3s}}{\frac{1}{s}}=\frac{1}{3}\)$ of the time taken by the largest pump alone to fill the tank.
Hence the answer is (C).