Taps A, B and C are attached with a tank and velocity of water coming through them are 42 L/h, 56 L/h and 48 L/h, respectively. A and B are inlets and C is outlet. If all the taps are opened simultaneously, tank is filled in 16 h. What is the capacity of the tank?

Let time taken by B to fill the tank = T h.
? Time taken by A to fill the tank = T + (T x 80)/100 = 9T/5 h

According to the formula,
Time taken by both the taps to fill the tank = ab /(a + b)
? 45 = (T x 9T/5)/(T + 9T/5)
? 45 x 14T/5 = 9T²/5
? T = 45 x 14/9 = 70 h

Part of the tank filled in 1 min by A, B and C = 1/20 + 1/15 +1/12
= (3 + 4 + 5)/60 = 12/60 = 1/5
? Time taken by A, B and C to fill the tank = 5 min
? Time taken by A, B and C to fill 40% of the tank = 40% of 5 = (40/100) x 5 = 2 min.

Let number of outlets be x
? Number of inlets = ( 7 - x )
Time taken to fill the tank when all the pipes are opened = 30/11h
Part of tank filled in 1 h when all the pipes are opened = 11/30 h

According to the question,
(7 - x)/10 - x/15 = 11/30
? {3(7 - x) - 2x}/30 = 11/30
? 21 - 3x - 2x = 11
? 5x = 10
? x = 2
Hence, number of outlets = 2 and number of inlets = 7 - 2 = 5

Total quantity of solution P, Q and R from A, B and C receptively, after 3min
= 3/30 + 3/20 + 3/10 = 3 x (2 + 3 + 6)/60 = 11/20
Quantity of solution R in liquid in 3 min = 3/10
? part of solution R = [3/10] / [11/20 ] = (3 x 20)/(10 x 11) = 6/11

According to the question,
Double the 2.5 min = 5 min
Now, part filled in 5 min = 5 x ( 1/12 + 1/15 ) = [5 x ( 5 + 4)]/60 = (5 x 9)/60 = 3/4
Part emptied in 1 min when P, R and M, all are opened. = 1/6 - ( 1/12 + 1/5 ) = 1/6 - (5 + 4)/60 = 1/60
One - sixtieth part is emptied in 1 min.
? Three - fourth part will be emptied in 60 x (3/4) = 15 x 3 = 45 min.

Part of the tank filled with A and B in
1 min = 1/25 x 5/6 + 1/20 x 2/3 = 1/30 + 1/30
= 2/30 = 1/15
Hence, Time taken to fill the tank = 15 min

Part of the tank filled by the taps A, B and C in 3 min = 1/20 + 1/30 - 1/15 = (3 + 2 - 4)/60 = 1/60
? Time taken to fill [ 1 - ( 1/20 + 1/30 )] or 55/60th part of the tank = 3 x 55 = 165 min
Remaining part of the tank = 1 - 55/60 = 5/60 = 1/12
Tap A fill 1/0 part in 1 min, then
Remaining part = 1/12 - 1/20 = (5 -3)/60 = 2/60 = 1/30
i.e, 1/30th part is filled by B in 1 min
Hence, required time to fill the whole tank = (165 + 1 +1 ) min = 167 min

Work done by the inlet in 1 h = (1/8 - 1/24) = 1/24
Work done by the inlet in 1 min = 1/24 x 1/60 = 1/1440
? Volume of 1/1440 part = 3 L
? Volume of the whole = 3 x 1440 = 4320 L

Time taken by A to fill the tank, m = 15 min
? Time taken by B to fill the tank, n = 15 x 4 = 60
? Required time taken m x n/(m + n)
= (15 x 60)/(15 + 60) = (15 x 60)/75 = 12 min

Part filled by A in 1 h = 1/16
Part fill by B in 1 h = 1/10
Part filled by ( A + B ) in 2 h = 1/16 + 1/10 = 13/80
? Part filled by ( A + B ) in 12 h = 6 x 13/80 = 78/80
? Remaining part = 1 - 78/80 = 2/80 = 1/40
Now, it is the turn of A
Time taken by A to fill 1/40 part of the tank = (1/40) x 16 = 2/5 h
? Total time taken ( 12 + 2/5) h = 12²/₅ h