Capacity of tap B is 80% more than that of A. If both the taps are opened simultaneously, they take 45 h to fill the tank. How long will B take to fill the tank alone?
Correct Answer: Description for Correct answer:
Let time taken by B to fill the tank, a = xh.
Therefore, Time taken by A to fill the tank
b = \( \Large x+\frac{x \times 80}{100}=\frac{9x}{5}h \)
According to the formula,
Time taken by both the taps to fill the tank
\( \Large t = \frac{ab}{a+b} \)
=> \( \Large 45 = \frac{x \times \frac{9x}{5}}{x+\frac{9x}{5}} \)
=> \( \Large 45 \times \frac{14x}{5}=\frac{9x^{2}}{5} \)
Therefore, x = \( \Large \frac{45 \times 14}{9} \)
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