A cistern has three pipe A ,B,C. A and B can fill it in 3 hours and 4 hours respectively. A and C can empty it in 1 hours. If the pipe are opened at 3pm, 4pm, 5pm respectively on the same day. The cistern will empty at?
Solution (By Examveda Team)
A can fill whole cistern in 3 hours.
Work rate of A = 100/3 = 33.33% per hour.
B can fill cistern in 4 hours.
Work rate of B = 100/4 = 25% per hour.
Now, A and C can empty the cistern in 1 hours.
Work rate of (A +C) = 100% per hours.
But C is doing negative work i.e emptieng the cistern.
So, work rate of C = - 100 - 33.33% = - 133.33% (Negative sign shows C is empting the cistern).
Now,
From 3 pm to 4 PM, cistern will be filled by 33.33%.
Next hour, cistern will get water = 33.33 +25 % = 58.33.
From now from next hours, the work rate of A, B and C = 33.33 +25 - 133.33% = -75
Now combinning 3 hours work, cistern will be filled
= 33.33 + 33.33 +25 - 75% = 16.66%
Now, System will take another,
= 16.66 *60/75 = 13.32 minutes to get cistern empty.
So, cistern will be emptied at 6 : 13:32 PM.
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It's properly
My suggestion to examveda team is to read the problem prooerly
A and B can fill 2/3+1/4=11/12 part of cistern in 2 hours.
A,B, and C can empty 1/3+1/4-1= -5/12 part of cistern in 1 hour
5/12 part is emptied in 1 hour
11/12 part will be emptied in 1/(5/12)*(11/12)= 2.2 hr
Thus at 7hr 12 min(5+2.2 hrs) cistern will be empty.
Cistern filled in 1 hour by pipe A alone = 1/3,
Cistern filled in 1 hour by pipes A and B together = 1/3 + 1/4 = 7/12
Hence cistern filled by 5pm = 1/3 + 7/12 = 11/12.
Let cistern be emptied in t hours after 5pm when pipe C is also opend.
Therefore 11/12 + (7/12)t - t = 0
This gives t = 11/5 hours
A,B fill the cistern in 2 hours= 2/4+1/3= 11/12
A,B,C empty the cistern in 1 hour= 1/4+1/3-1= -5/12
5/12 part of tank was emptied in 1 hour
11/12 part of tank would require time= 1/(5/12)*(11/12)= 2.2 hr
Thus at Time= 5+2.2= 7hr 20 min cistern will be empty.