Discussion

Question:

Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty in 8 hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?

Explanation:

Let the capacity of the tank be 24x litres.

Pipes A and B fill 3x and 2x litres per hour while pipe C empties 6x litres in an hour.

Let radius of the cone be r and height be h.

$\frac{1}{3}\pi r^{2}h=24x$

∴ πr²h = 72x

For first 19 hours, water inside the cone = 24x + 57x + 38x – 114x = 5x litres

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∆ABE ∼ ∆ACD

If AC = 2AB, CD = 2BE

∴ BE = r/2 and AB = h/2

After 50% reduction in the height of the water, volume

$=\frac{1}{3}\pi {(r/2)}^2(h/2)=\frac{\pi r^{2}h}{24}=\frac{72x}{24}=3x$

Option 1: Pipe A was open for 19 hours.

i.e., B and C were open for 1 more hour.

∴ 2x – 6x = –4x

The cone will have 5x – 4x = x litres of water.

∴ Option 1 is eliminated.

Option 2: Pipe A was open for 19 hours 30 minutes.

i.e., B and C were open for 1 more hour and A for 30 more minutes.

∴ 2x – 6x + 1.5x = –2.5x

The cone will have 5x – 2.5x = 2.5x litres of water

∴ Option 2 is eliminated.

Option 3: Pipe B was open for 19 hours 30 minutes.

i.e., A and C were open for 1 more hour and B for 30 more minutes.

∴ 3x – 6x + 1x = –2x

The cone will have 5x – 2x = 3x litres of water.

∴ Option 3 would be the possible option.

Hence, option (c).

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