Question
Mathematics Question on Time and Work
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?
8 min and 34 sec
7 min
9 min and 32 sec
6 min
8 min and 34 sec
Solution
Let the rates of the taps be
Tap 1 fills the tank in 10 minutes, so its rate is 101 of the tank per minute.
Tap 2 fills the tank in 12 minutes, so its rate is 121 of the tank per minute.
Let the rate of the third tap (emptying) be x1 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 151 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.