Question

Tap A is 50% more efficient than Tap B. Both the taps together can fill the tank in (48/5) hours. Find the time taken by Tap C which is 60%more efficient than Tap B, to fill the tank alone.

Answer 1

Time taken by Tap C alone to fill the tank is 48 hours.

Step-by-step explanation:

Let the time taken by Tap A to fill the tank alone is x hours.

Given:

• Tap A is 50% more efficient than Tap B.

$= > A = x \: hrs \\ = > \: B = \frac{x}{0.5} hrs \\ = > B = 2x \: hrs$

• Both the taps together can fill the tank in (48/5) hours.

$= > \frac{1}{A} + \frac{1}{B} = \frac{5}{48} \\ = > \frac{1}{x} + \frac{1}{2x} = \frac{5}{48} \\ = > \frac{3}{2x} = \frac{5}{48} \\ = > x = \frac{72}{5} hrs$

• Tap C which is 60%more efficient than Tap B.

$= > B = 2x \\ = > B = \frac{144}{5} hrs \\ = > C = \frac{B}{0.6} hrs \\ = > C = \frac{5B}{3} hrs \\ = > C = \frac{5}{3} \times \frac{144}{5} hrs \\ = > C = 48 \: hrs$

Time taken by Tap C alone to fill the tank is 48 hours.