Two pipes A and B can separately fill a tank in 36 minutes and 45 minutes respectively

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Two pipes A and B can fill a tank in 36 minutes and 45 minutes respectively. Another pipe C can empty the tank in 30 minutes. First A and B are opened. After 7 minutes, C is also opened. The tank is filled up in

1. 39 minutes
2. 46 minutes
3. 40 minutes
4. 45 minutes

As per the given in question , we have

Part of the tank filled by pipes A and B in 1 minute =	1	+	1	=	5 + 4	=	9	=	1
36	45	180	180	20

Part of the tank filled by these pipes in 7 minutes =	7
20

Remaining unfilled part = 1 –	7	=	20 - 7	=	13
20	20	20

When all three pipes are opened.

=	1	-	1	=	3 - 2	=	1
20	30	60	60

∴ Time taken in filling	13	part	=	13	× 60 = 39 minutes
20	20

Required time = 39 + 7 = 46 minutes

Correct Answer:

Description for Correct answer:

(A + B)'s 7 minutes filling =\( \Large \left(5+4\right) \times 7 = 63 \) units Remaining capacity =\( \Large 180 - 63 = 117 \) units Now C is opened, it empties by 6 units/min.So total units filled in tank is =\( \Large \left(5+4\right) -6 = 3 \) units/min Now tank can be filled in=\( \Large \frac{117}{3} =39\)minTank is filled up in

= \( \Large 7 + 39\ minutes = 46 \) min

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