Two taps A and B can fill an overhead tank in 10 hours and 15 hours re

1.	Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank?
Answer» Given, Tap A can fill a tank in= 10 hours Tap B can fill the tank in = 15 hours Work done by Tap A in 1 hour \(\frac{1}{10}\) Work done by tap B in 1 hour \(=\frac{1}{15}\) ∴ Work done by both tap in 1 hour \(=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}\)part Work done by both tap in 4 hours \(={4}\times\frac{1}{6}=\frac{2}{3}\)part Remaining part \(={1}-\frac{2}{3}=\frac{1}{3}\)part Hence, time taken by A to fill remaining part \(=\frac{\frac{1}{3}}{\frac{1}{10}}=\frac{10}{3}={3}\frac{1}{3}\)hours

Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank?

Answer»

Given,

Tap A can fill a tank in= 10 hours

Tap B can fill the tank in = 15 hours

Work done by Tap A in 1 hour \(\frac{1}{10}\)

Work done by tap B in 1 hour \(=\frac{1}{15}\)

∴ Work done by both tap in 1 hour \(=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}\)part

Work done by both tap in 4 hours \(={4}\times\frac{1}{6}=\frac{2}{3}\)part

Remaining part \(={1}-\frac{2}{3}=\frac{1}{3}\)part

Hence, time taken by A to fill remaining part \(=\frac{\frac{1}{3}}{\frac{1}{10}}=\frac{10}{3}={3}\frac{1}{3}\)hours