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31 Aug 2015, 10:39
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Knewton Brutal Challenge
A vertically oriented cylindrical tank, a rectangular solid tank, and a cubic tank sit flat on the floor of a factory. Three pumps are turned on simultaneously, each of which pumps water at a different, constant rate, one pump from the cylindrical tank to the rectangular solid tank, one pump from the rectangular solid tank to the cubic tank, and one pump from the cubic tank to the cylindrical tank. The pumping mechanisms are already filled with water before pumping begins, so that there is no change in the amount of water contained within the pumping mechanisms. If the area of the base of the cylindrical tank is equal to the area of one face of the cubic tank, how many feet will the water level (the height which the water reaches in a tank) have risen in the cylindrical tank after 45 minutes? Assume that none of the tanks overflows or completely drains in 45 minutes.
(1) After 30 minutes, the water level of the cubic tank has decreased by 2 feet, and the water level of the rectangular solid tank, whose bottom face has a length and width that are respectively twice and three times as long as an edge of the cubic tank, has decreased by 3 feet.
(2) After 30 minutes, the volume of water in the cubic tank has decreased by 800 cubic feet, and the volume of the rectangular solid tank has decreased by 7,200 cubic feet.
31 Aug 2015, 23:15
Kudos
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IMO : A
Given that no water overflows --> It means the volume of the total water present in all 3 tanks will always be constant.
i.e. Volume of water in Cylindrical tank + Rectangular Tank + Cubical Tank = const.
Question Stem:
Let "r" be the radius of the circular base of cylinder. And "a" be the side of the cube.
Given \(r^2 = a^2\) --(i)
St 1:
length of rectangular tank L = 2a
breadth of rectangular tank B = 3a
Thus Area of base = \(6a^2\)
Volume of Water lost from Rectangular tank + water lost from Cubical tank = water gained by cylindrical tank
\(3* 6a^2 * 3 + 2* a^2 = k * r^2\) (where k = height gained by cylindrical tank)
Thus we can get height gained for 30 min.
Hence from this we can find out height gained for 45 min. Since it is DS question we need not calculate the exact value.
Hence Suff
St 2:
volume of water in the cubic tank has decreased by 800 =\(h * a^2\) (where h = height gained by cubic tank)
volume of the rectangular solid tank has decreased by 7,200
Total volume gained by cylindrical tank = 8000 in 30 min
For 45 min = 12000 = \(k * a^2\) (where k = height gained by cylindrical tank)
Here we cannot determine the value of k or h.
Hence not suff
Given that no water overflows --> It means the volume of the total water present in all 3 tanks will always be constant.
i.e. Volume of water in Cylindrical tank + Rectangular Tank + Cubical Tank = const.
Question Stem:
Let "r" be the radius of the circular base of cylinder. And "a" be the side of the cube.
Given \(r^2 = a^2\) --(i)
St 1:
length of rectangular tank L = 2a
breadth of rectangular tank B = 3a
Thus Area of base = \(6a^2\)
Volume of Water lost from Rectangular tank + water lost from Cubical tank = water gained by cylindrical tank
\(3* 6a^2 * 3 + 2* a^2 = k * r^2\) (where k = height gained by cylindrical tank)
Thus we can get height gained for 30 min.
Hence from this we can find out height gained for 45 min. Since it is DS question we need not calculate the exact value.
Hence Suff
St 2:
volume of water in the cubic tank has decreased by 800 =\(h * a^2\) (where h = height gained by cubic tank)
volume of the rectangular solid tank has decreased by 7,200
Total volume gained by cylindrical tank = 8000 in 30 min
For 45 min = 12000 = \(k * a^2\) (where k = height gained by cylindrical tank)
Here we cannot determine the value of k or h.
Hence not suff
16 Oct 2016, 08:54
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