Three taps A, B and C fill a tank in 20 min, 15 min and 12 min, respectively. If all the taps are opened simultaneously, how long will they take to fill 40% of the tank?

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.

Let one pipe takes m h to fill the tank.
Then, the other pipe takes (m - 10) h.
According to the question.
? 1/m + 1/(m - 10) = 1/12
? (m - 10 + m)/{m(m - 10)} = 1/12
? 12m( m - 10 + m ) = ( m - 10 )
? m² - 34m + 120 = 0
? m² - 30m - 4m + 120 = 0
? ( m - 30 )( m - 4 ) = 0
? m = 30 or 4
? Faster pipe will take ( 30 - 10 ) h = 20 h to fill the tank.

Part filled by both in 2 min = 2 x ( 1/15 + 1/20 ) = 2 x ( 4 + 3)/60 = 7/30
Part unfilled = 1 - 7/30 = (30 - 7)/30 = 23/30
Now, B fills 1/20 part in 1 min.
? 23/30 part will be filled by B in (20 x 23)/30 min or in 46/3 min.
? Required time taken to fill the tank = 2 + 46/3 = 52/3 min.

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

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

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