There are two inlets A and B connected to a tank. A and B can fill the tank in 16 h and 10 h, respectively. If both the pipes are opened alternately for 1 h, starting from A, then how much time will the tank take to filled?

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

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

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

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

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 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

Area of tap ? Work done by pipe.
When diameter is doubled, area will be four times. so, it will work four times faster.
Hence, required time taken to empty the tank = 40 x 1/4 = 10 min.

Here, A = 24 min, B = 32, T = 9 min
? Required time = B(1 - T/A )
= 32 ( 1 - 9/24 )
= 32 x 15/24 = 20 min.

Let the number of minutes taken to empty the cistern be T min.
According to the questions,
T/6 - (T + 5)/12 - (T + 5)/15 = 0
? T/6 - T/12 - 5/12 - T/12 - T/12 - T/15 = 0
? T/6 - T/12 - T/15 = 5/12 + 5/15
? (10T - 5T - 4T)/ 60 = (25 + 20)/60
? T/60 = 45/60
? T = 45 min.

Work done by both the taps in 5 min
= 5 (1/20 +1/25) = (5 x 9)/100 = 9/20
Remaining part = (1 - 9/20) = 11/20
Now, 1/20 part is filled in 1 min.
So, 11/20 part will be filled in 11 min.
Hence, the tank will be full in 11 more min.