Question

There are two taps to fill a tank while a third to empty it. When the third tap is closed, they can fill the tank in 10 minutes and 12 minutes, respectively. If all the three taps be opened, the tank is filled in 15 minutes. If the first two taps are closed, in what time can the third tap empty the tank when it is full?

• A. 8 min and 34 sec
• B. 7 min
• C. 9 min and 32 sec
• D. 6 min

Answer 1

The Correct Option is A

Let the rates of the taps be

Tap 1 fills the tank in 10 minutes, so its rate is \(\frac{1}{10}\) of the tank per minute.
Tap 2 fills the tank in 12 minutes, so its rate is \(\frac{1}{12}\) of the tank per minute.
Let the rate of the third tap (emptying) be \(\frac{1}{x}\) of the tank per minute [where \(x\) is the time it takes to empty the full tank.]

When all taps are open, the combined rate fills the tank in 15 minutes, so the total rate is \(\frac{1}{15}\) of the tank per minute.

Hence,
\(\Rightarrow\;\)\(\frac{1}{10} + \frac{1}{12} - \frac{1}{x} = \frac{1}{15}\)

\(\Rightarrow\;\)\(\frac{1}{10} + \frac{1}{12} = \frac{3}{30} + \frac{2.5}{30} = \frac{11}{60}\)

\(\Rightarrow\;\)\(\frac{11}{60} - \frac{1}{x} = \frac{1}{15}\)

\(\Rightarrow\;\)\(\frac{11}{60} - \frac{1}{15} = \frac{1}{x}\)

\(\Rightarrow\;\)\(\frac{11}{60} - \frac{4}{60} = \frac{7}{60}\)

\(\Rightarrow\;\)\(\frac{1}{x} = \frac{7}{60} \quad \Rightarrow \quad x = \frac{60}{7}\)

Thus, the third tap can empty the tank in 8 minutes 34 seconds.