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
Here, a = 16 and = 9
Required time = ?ab
?16 x 9 = 4 x 3 = 12 min
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
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
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.
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 1st pipe in 1 min = 1/20
Part filled by 2nd pipe in 1 min = 1/24
Part filled by all the pipes in 1 min = 1/15
Work done by the waste pipe 1 min
= 1/15 - (1/20 + 1/24 )
= 1/15 - 11/120
= (8 - 11)/120 = ( - 3/120 ) = ( -1/40 )
[-ve sign indicates emptying]
Now, volume of 1/40 part = 6 gallon
? Volume of whole tank = 40 x 6 = 240 gallon
Part of the tank filled by both taps in 5 min = 5 x (1/20 + 1/60)
= 5 x (6 +2 )/120 = 8/24 = 1/3
? Remaining part = (1 - 1/3) = 2/3
? 1/60 Part is now filled in 1 min.
? 2/3 Part is now filled in 60 x 2/3 = 40 min.
Let the cistern gets emptied in m h after 3:pm
Work done by A in m h, by B in (m - 1) h and by c in (m - 2) h= 0
? m/3 + (m - 1)/4 - (m - 2) = 0
? 4m + 3(m - 1) - 12(m - 2) = 0
? 5m = 21
? m = 21/5 = 4.2
? m = 4 h 12 min
? Required time = 7 : 12 pm.
Time taken to fill the tank by the tap having 2 cm diameter = 61 min
? Time taken to fill the tank by the tap having 1 cm diameter
= 61 x (2/1)2 = 244 min
Similarly, time taken to fill the tank by the tap having 4/3 cm diameter
= 61 x [(2 x 3)/4]2 = 61 x 9/4 = 549/4 min.
? Part of the tank filled by all the three pipes in 1 min
= 1/61 + 1/244 + 1/(549/4)
= (36 + 9 + 16)/2196 = 61/2196 = 1/36
Hence, required time taken = 36 min
Part filled by 1st pipe in 1 h = 1/14
Part filled by 2nd pipe in 1 h = 1/16
Part filled by the two pipes in 1 h
= ( 1/14 + 1/16 ) = (8 + 7)/112 = 15/112
? Time taken by these two pipes to fill the cistern = 112/15 h = 7 h 28 min
Due to leakage, the time taken 7 h 28 min + 92 min = 9 h
&thee4; Work done by ( two pipes + leak ) in 1 h = 1/9
Work done by the leak in 1 h = 1/9 - 15/112 = (112 - 135)/1008
= - 23/1008
? Time taken by leak to empty the full cistern
= 1008/23 = 4319/23 h
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