A tap can fill a tank in 6 hours. When the tank is half filled, three more similar taps are opened to fill water into the same tank. What is the total time taken to fill the tank completely by all the four pipes?
Answer: Option D
1 pipe can completely fill the tank in 6 hours
After first pipe fills 3 parts (half tank), 3 more pipes are added.
Now altogether four pipes are filling the remaining 3 parts (half tank)
3 parts of the tank can be filled in 3 hours (180 min) by each of the pipes
1 pipe -- 3 parts -- 180 min
4 pipes -- 3 parts -- X
So it takes 45 minutes to fill the remaining tank.
Hence total time taken by all the four pipes to fill the tank is,
3 hours by first pipe + 45 min by all four pipes together
Three pipes A, B and C can fill a tank in 6 hours. After three pipes worked for 2 hours C is closed. Then A and B filled the remaining part in 7 hours. calculate the number of hours taken by C alone to fill the tank is?
Answer: Option D
The capacity of the tank can be obtained by finding LCM of 6, 2 and 7 which is 42.
Total parts of the tank is 42
A, B, C together can fill the tank in 6 hours
A,B and C can fill 7 parts in 1 hour
From the question A, B and C together are filling only for 2 hours
Together they complete 14 parts in 2 hours
And the remaining 28 part will be filled by A and B in 7 hours
A and B together can fill 4 parts in 1 hour
From this we could find C’s filling parts per hour (A+B+C)-( A+B)=C
7-4= 3 parts
Total 42 parts of the tank will be filled by C alone
Two pipes A and B can separately fill a cistern in 60 min and 75 min respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 50 min. In how much time the third pipe alone can empty the cistern?
Answer: Option B
Let us assume the third pipe empties the cistern in x min.
Part of the cistern filled in 1 min when all the three pipes are opened simultaneously