Two taps A and B can fill a tank in 25 min and 20 min, respectively. But taps are not opened properly so the taps A and B allow 5/6th and 2/3rd part of water, respectively. How long will they take to fill the tank?

Quantity of water admitted by tap A in 1 h = 42 L
Quantity of water admitted by tap B in 1h = 56 L
Quantity of water removed by tap C in 1 h = 48 L
So, quantity of water filled in the tank in 1 h = ( 42 + 56 - 48 ) L = 50 L
? Quantity for water filled in 16 h = 16 x 50 = 800 L
Hence, capacity of tank = 800 L

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

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

Part filled by X in 1st min and Y in the 2nd min = ( 1/6 + 1/7) = 13/42
Part filled by ( X + Y ) working alternately in 6 min = 1/2 x 13/42 x 6 =13/14
? Remaining Part = ( 1- 13/14 ) = 1/14
Now, it is the turn of X, one-sixth part is filled in 1 min.
One-fourteenth part is filled in (6 x 1/14) min = 3/7
? Required time = ( 6 + 3/7 ) = 6³/₇ min