Too lengthy

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A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t2 + 12t + 24)Ï€ m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3d meters of space remaining. What is the height of the tank?

Answer

B 62/9

[Matt@VeritasPrep]

Let's start with the volume of the tank: π * 3^2 * h, or 9*h*π.

Assuming that "vertical space" refers to volume -- this is really unclear -- then at t = 4 minutes we've filled 9*h*Ï€ - d meters of space, giving us this equation:

(-t^2 + 12t + 24)Ï€ = (9*h*Ï€ - d)
or
(-4^2 + 12*4 + 24)Ï€ = (9*h*Ï€ - d)
or
56Ï€ = 9*h*Ï€ - d

If at t = 6 minutes we've filled all but d/3 meters, we have the equation

(-6^2 + 12*6 + 24)Ï€ = (9*h*Ï€ - d/3)
or
60Ï€ = 9*h*Ï€ - d/3

Subtracting the first equation from second, we have 4pi = (2d)/3, or d = 6pi. Plugging that back into the first equation, we have 56π = 9hπ - 6π, or h = 62/9.

I wish I had a shortcut for you, but I don't see one at first glance. This seems like one of those "accounting" sort of problems - the steps are reasonably clear, but they are many and treacherous.

[GMATGuruNY]

A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t2 + 12t + 24)Ï€ m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3d meters of space remaining. What is the height of the tank?

An alternate approach:

At the 4-minute mark, the height of the UNUSED portion of the tank -- the remaining "vertical space" -- is equal to d meters.
At the 6-minute mark, the height of the unused portion of the tank is equal to (1/3)d meters -- a decrease of 2/3.
Since the HEIGHT of the unused portion of the tank decreases by 2/3, the VOLUME of the unused portion of the tank also decreases by 2/3.

Let U = the volume of the unused portion of the tank at the 4-minute mark.
Between the 4-minute mark and the 6-minute mark, the value of U decreases by 2/3.

Volume of the solution at the 4-minute mark:
V = (-t² + 12t + 24)π = (-4² + 12*4 + 24)π = 56π cm³.

Volume of the solution at the 6-minute mark:
V = (-t² + 12t + 24)π = (-6² + 12*6 + 24)π = 60π cm³.

Between the 4-minute mark and the 6-minute mark, the volume of the solution INCREASES by 4π cm³.
As noted above, these 4π cm³ represent a 2/3 DECREASE in the value of U.
Thus:
4Ï€ = (2/3)U
U = 6π cm³.

Since the volume of the solution at the 4-minute mark is 56 cm³, and the value of U at the 4-minute mark is 6π cm³, we get:
Total volume = 56π + 6π = 62π cm³.

Since the volume of the tank is 62π cm³, and the radius is 3 cm, we get:
π(3²)h = 62π
h = 62/9.

[send2dar]

Let H be height of the cylinder.

After 4 minutes:
Ï€9H-Ï€9d = (-16+48+24)Ï€ => 9H-9d = 56 -(1)
AFter 6 minues:
Ï€9H-Ï€9(d/3) = (-36+72+24)Ï€ => 9H-3d = 60 -(2)
3(2)-(1) => 27H-9d-(9H-9d) = 180-56 => 18H = 124 => H = 62/9

Hope it helps.

Regards
--------
Tyring for 750+