Three taps are fitted in a cistern. The empty cistern is filled by the first and the second taps in 3 and 4 h, respectively. The full cistern is emptied by the third tap in 5 h. If all three taps are opened simultaneously, the empty cistern will be filled up in?

Part filled by tap A in 1 min = 1/60
Let tap B fills the tank in x min
Then, Part filled by tap, B in 1 min = 1/x
According to the question,
1/60 + 1/x = 1/40
? 1/x = 1/40 - 1/60
&rArr ; 1/x = (3 - 2)/120
&rArr ; 1/x = 1/120
? Tap B can fill the tank in 120 min.

Here w = 2 litres per hour
? Required answer = {(15 x 10) / (15 - 10)} x 2 = 60 litres.

Let the first pipe be shut up after x minutes
Now, applying the given rule, we have 30{1 - (x + 10)/40 } = x
[Here t = (x + 10) minutes]
or x = 90/7 minutes

Time taken to fill the whole tank = (12 x 14 x 8) / (14 x 8 + 12 x 8 - 12 x 14) = 168/5 minutes
? In 7 minutes 5/168 x 7 = 5/24 part of the tank will be filled
? Requied answer = 1 - 5/24 = 19/24 part

Required answer = (25 x 40 x 30) / (40 x 30 + 25 x 30 - 25 x 40)
= 600 / 19 = 31¹¹/₁₉ minutes

Part filled by A in 1 min = 1/30
Part filled by B in 1 min = 1/10
Part emptied by C in 1 min = - 1/40
Net part filled in 1 h by ( A + B + C ) = ( 1/30 + 1/10 - 1/40 )
= (4 + 12 - 3)/120 = 13/120

? Required time to fill the tank = 120/13 = 9³/₁₃ h

Let the leak takes x h to empty the tank.
Now, part filled by inlet in 1 h = 1/8
part filled in 1 h when both tap and leak works together = 1/(8+2) = 1/10

According to the question.
1/x = 1/8 - 1/10 = (5 - 4) / 40 = 1/40
? x = 40 h

Work done by C in 1 min = (1/60 + 1/75 - 1/50 )
= (5 + 4 - 6)/300 = 3/300 = 1/100
Hence, C can empty the full tank in 100 min.

Part filled by ( A + B ) in 1 h = 1/6
Part filled by ( B + C ) in 1 h = 1/10
Part filled by ( A + C ) in 1 h = 1/12
? Part filled by 2 ( A + B + C ) in 1 h = 1/6 +1/10 + 1/12 = (10 + 6 + 5)/60 = 21/60 = 7/20
? Part filled by ( A + B + C ) in 1 h = 7/(2 x 20) = 7/40
? Required time 40/7 = 5⁵/₇ h

Part filled by pipe P in 1 min = 1/12
Part filled by pipe Q in 1 min = 1/15
Part filled by both pipes in 1 min = 1/12 + 1/15 + = (5 + 4)/60 = 9/60
Now, Part filled by both pipes in 3 min = (3 x 9)/60 = 27/60 = 9/20
? Remaining part = 1 - 9/20 = 11/20
Let the remaining part is filled by pipe Q in T min.
Then, T/15 = 11/20
T = (15 x 11)/20 = (3 x 11)/4
= 33/4 = 8⁴/₄ min