A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
एक नल एक टंकी को 6 घंटे में भर सकता है। आधा टैंक भरने के बाद, तीन और समान नल खोले जाते हैं। टैंक को पूरी तरह से भरने में लिया गया कुल समय कितना है?
A)
3 hrs 45 min
B)
4 hrs
C)
4 hrs 15 min
D)
5 hrs
Correct Answer :
3 hrs 45 min
Explanation :
Time taken by one tap to fill the half of the tank =3 hours
Part filled by the four taps in 1 hour=4/6=2/3
Remaining part=1 -1/2=1/2
2/3 of the tank is filled by four taps in 1 hour
let the time taken to fill Remaining part is x
Therefore, 2/3:1/2 :: 1:x
or x=(1/2)*1*(3/2)=3/4 hours
ie 45 min
So, the total time taken = 3 hrs + 45 min = 3 hrs 45 min
Three Pipes A,B and C together filled a tank in 6 hours
So 1 hour work of A,B and C =1/6
C closed after 2 hour's, Part filled in 2 hours working together(A,B and C) = 2/6 = 1/3
Remaining part = (1-1/3) = 2/3
A and B can fill remaining in 8 hours, (A + B)'s 8 hour's work = 2/3
As the pipes are operating alternatively for one minute, thus (A+B)'s 2 minutes job is =(1/4 +1/6) =5/12
in next, 2 minutes both pipe can fill another 5/12 part.
So in 4 minutes A and B are operating alternatively will fill 5/12 + 5/12 = 5/6
Remaining part of cistern= 1 - 5/6 =1/6
Pipe A can fill 1/4 part of the cistern in 1 minute,and can fill 1/6 part of the cistern in = 4*(1/6) =2/3 minutes
Total time taken to fill the Cistern = time taken to fill (5/6 part + 1/6 part)
=> 4 minutes + 2/3 minutes = 4(2/3) minutes Or, 4 minutes 40 seconds
Let a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), when opening both the pipes, the net part emptied in 1 hour = (1/y)-(1/x)
=> (1/8 - 1/16) = 1/16
Time taken to empty the full tank = 16 min.
Hence, time taken to empty the half tank = 8 min
Let a pipe can fill a tank in x hours, and another pipe can empty the full tank in y hours (where y> x), then on opening both the pipes, the net part filled in 1 hour = (1/x)-(1/y)
=> (1/2 - 1/3) = 1/6
Part filled by A in 1 hour=1/36
Part filled by B in 1 hour= 1/45
Part filled by (A+B)'s in 1 hour=1/36 +1/45
= 9/180
= 1/20
Hence, both the pipes together will fill the tank in 20 hours
Let the reservoir be filled by the 1st pipe in x hours
The second pipe will fill it in (x+10) hours
1/x + (1/(x+10))= 1/12
=> (2x+10)/((x)*(x+10))= 1/12
=> x2-14x-120=0
=> (x-20)(x+6)=0
=> x=20 (value can not be -ve )
So, the second pipe will take 30 hours to fill the reservoir
Work done by the two pipes in 1 hour= 1/14+1/16=15/112
Time taken by these two pipes to fill the tank=112/15 hrs
Due to leakage, time taken = 7 hrs 28 min + 32 min= 8 hours
Therefore, work done by (two pipes + leak) in 1 hr= 1/8
work done by leak n 1 hour=15/112 -1/8=1/112
Leak will empty full cistern n 112 hours