Question : Pipe P alone can fill a tank in 18 hours and Pipe Q alone can fill the same tank in 24 hours. If both the pipes are opened on alternate hours one at a time and pipe Q is opened for the first hour, then in how much time the tank will be full?
Option 1: $\frac{64}{5}$ hours
Option 2: $\frac{78}{5}$ hours
Option 3: $20$ hours
Option 4: $\frac{62}{3}$ hours
Correct Answer: $\frac{62}{3}$ hours
Solution :
According to the question,
Pipe P fills the tank in 18 hours, so its rate = $\frac{1}{18}$ per hour
Pipe Q fill tank in 24 hours, so its rate = $\frac{1}{24}$ per hour
⇒ Combined rate $=\frac{1}{24} + \frac{1}{18} = \frac{7}{72}$
So, in 2 hours, if both work alternatively starting with Q, the amount of work done = $\frac{7}{72}$
Thus, in 20 hours, the amount of work done = $\frac{70}{72}$
⇒ Remaining work = 1 – $\frac{70}{72}$ = $\frac{2}{72}$ which is to be done by Q alone
⇒ Time taken by Q to finish the remaining work = $\frac{\frac{2}{72}}{\frac{1}{24}}$ = $\frac{2}{3}$ hours
So, the total time $=20 + \frac{2}{3} = \frac{62}{3}$ hours
Hence, the correct answer is $\frac{62}{3}$ hours.
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