Two pipes A and B can fill a cistern in 37½ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:

(A + B)'s 1 hour's work =	❨	1	+	1	❩	=	9	=	3	.
		12		15			60		20

(A + C)'s hour's work =	❨	1	+	1	❩	=	8	=	2	.
		12		20			60		15

Part filled in 2 hrs =	❨	3	+	2	❩	=	17	.
		20		15			60

Part filled in 6 hrs =	❨	3 x	17	❩	=	17	.
			60			20

Remaining part =	❨	1 -	17	❩	=	3	.
			20			20

Now, it is the turn of A and B and	3	part is filled by A and B in 1 hour.
	20

∴ Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.

Part filled by A in 1 min =	1	.
	20

Part filled by B in 1 min =	1	.
	30

Part filled by (A + B) in 1 min =	❨	1	+	1	❩	=	1	.
		20		30			12

∴ Both pipes can fill the tank in 12 minutes.

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

∴	1	+	1	=	1
	x		(x - 5)		(x - 9)

⟹	x - 5 + x	=	1
	x(x - 5)		(x - 9)

⟹ (2x - 5)(x - 9) = x(x - 5)

⟹ x² - 18x + 45 = 0

(x - 15)(x - 3) = 0

⟹ x = 15.    [neglecting x = 3]

Part filled in 2 hours =	2	=	1
	6		3

Remaining part =	❨	1 -	1	❩	=	2	.
			3			3

∴ (A + B)'s 7 hour's work =	2
	3

(A + B)'s 1 hour's work =	2
	21

∴ C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=	❨	1	-	2	❩	=	1
		6		21			14

∴ C alone can fill the tank in 14 hours.

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

∴	1	+	1	=	1
	x		(x + 6)		4

⟹	x + 6 + x	=	1
	x(x + 6)		4

⟹ x² - 2x - 24 = 0

⟹ (x -6)(x + 4) = 0

⟹ x = 6.     [neglecting the negative value of x]

Part filled by (A + B + C) in 3 minutes = 3	❨	1	+	1	+	1	❩	=	❨	3 x	11	❩	=	11	.
		30		20		10					60			20

Part filled by C in 3 minutes =	3	.
	10

∴ Required ratio =	❨	3	x	20	❩	=	6	.
		10		11			11

Work done by the waste pipe in 1 minute =	1	-	❨	1	+	1	❩
	15			20		24

=	❨	1	-	11	❩
		15		120

= -	1	.    [-ve sign means emptying]
	40

∴ Volume of	1	part = 3 gallons.
	40

Volume of whole = (3 x 40) gallons = 120 gallons.

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =	❨	4 x	1	❩	=	2	.
			6			3

Remaining part =	❨	1 -	1	❩	=	1	.
			2			2

∴	2	:	1	:: 1 : x
	3		2

⟹ x =	❨	1	x 1 x	3	❩	=	3	hours i.e., 45 mins.
		2		2			4

So, total time taken = 3 hrs. 45 mins.

Net part filled in 1 hour	❨	1	+	1	-	1	❩	=	17	.
		5		6		12			60

∴ The tank will be full in	60	hours i.e., 3	9	hours.
	17		17

Work done by the leak in 1 hour =	❨	1	-	3	❩	=	1	.
		2		7			14

∴ Leak will empty the tank in 14 hrs.