SSC MTS MATHS QUIZ

Three taps A, B and C can fill a tank in 50,60 and 30 hours respectively. How long (in hours) would the three taps take to fill the tank if all of them are together?

Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?

A tank is filled by tap A in 30 minutes and by tap B in 45 minutes. When the tank is full, tap C can empty it in 90 minutes. If all the taps are opened together, then the tank will be full:

Pipe A can fill the tank in 12 hours and pipe B can fill the tank in 20 hours. A third pipe C empties the tank in 30 hours. If all pipes are opened together then find the time taken to fill the tank completely.

Two inlet taps A and B can fill a tank in 36 minutes and 60 minutes respectively. Find the time taken by both the taps together to fill $\frac{1}{6}$ th of the tank.

Pipes A and B can fill a tank in $12 \mathrm{~h}$ and $18 \mathrm{~h}$, respectively, whereas pipe $\mathrm{C}$ can empty the full tank in $8 \mathrm{~h}$. A and B are opened for $4 \frac{1}{2} h$ and then closed. C alone will empty the tank in (in h):

Three taps A ,B and C can fill a tank in 12 , 15 and 20 hours respectively. If A is open for all the time and B and C are open for one hour each alternatively, the tank will be full in 

A pipe can fill a cistern in 12 minutes and another pipe can fill it in 15 minutes, but a third pipe can empty it in 6 minutes. The first two pipes are kept open for 5 minutes in the beginning and then the third pipe is also opened. Number of minutes taken to empty the cistern is:

 $\frac{3}{7}$ th part of a tank is filled in 15 minutes, then find in how many minutes the remaining part will be filled.

Pipe $A$ and $B$ running together can fill the cistern in 6hours. A takes 5 hours more than B to fill the cistern when opened separately. In how much time pipe B can fill the cistern alone?