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
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
Part of tank filled by first tap in 1 h = 1/3
Part of tank filled by second tap in 1 h = 1/4
Part of tank emptied by third tap in 1 h = 1/5
Part of the tank filled by all pipes opened simultaneously in 1 h
= 1/3 + 1/4 - 1/5 = (20 + 15 - 12)/60 = 23/60
Time taken by all the taps to fill the tank when it is empty = 23/60 h = 214/23 h
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
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
Here, a = 16 and = 9
Required time = ?ab
?16 x 9 = 4 x 3 = 12 min
1st pipe takes 1 h to fill 1/2 part of the tank.
So, time taken to fill the whole tank (m) = 2 h
2nd pipe takes 1 h to 1/3 part of the tank
So, time taken to fill the whole tank (n) = 3 h
Let 3rd pipe takes P h to empty the tank = x
? 1/m + 1/n - 1/x = 7/12 ? 1/2 + 1/3 - 1/x = 7/12
? 1/x = (6 + 4 - 7)/12 = 3/12 = 1/4
? x = 4 h
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