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payalkhndlwl
Will at least three identical cubes fit inside a cylindrical shipping container?

(1) The edge of one of the cubical boxes is 4/5 as wide as the diameter of the cylinder.

(2) If the cylinder had 50% more capacity, seven of the identical cubical boxes could fit inside the cylinder


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Hello! Can anyone dumb the solution down?

Got lost in : why is the diagonal of largest edge (square) has to be less than d*sqrt 2
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payalkhndlwl
Will at least three identical cubes fit inside a cylindrical shipping container?

(1) The edge of one of the cubical boxes is 4/5 as wide as the diameter of the cylinder.

(2) If the cylinder had 50% more capacity, seven of the identical cubical boxes could fit inside the cylinder

(1) The edge of one of the cubical boxes is 4/5 as wide as the diameter of the cylinder.

Let the diameter of cylinder = 5
--> Edge of cube = 4/5*5 = 4

For the cube to fit in the cylinder, the diagonal of the cube should be at least equal to diameter
--> Diagonal = \(\sqrt{2}\)*edge
--> Diagonal = 4\(\sqrt{2}\) = 5.65 (approx)

So, cube will definitely not fit in the cylinder - A Definite NO

Sufficient

(2) If the cylinder had 50% more capacity, seven of the identical cubical boxes could fit inside the cylinder

Nothing can be deduced about the radius and height of the cylinder with the given volume.

Insufficient

Option A
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Quote:
1) The edge of one of the cubical boxes is 4/5 as wide as the diameter of the cylinder.
The diagonal of the square will be the diameter of the circle (base of the cylinder). diameter = s*(2)^0.5
Edge of the cube is (4/5)*d
(4/5)*d * 2‾√2 is greater than d, so doesnt make sense.

This is enough to prove that the cubes cannot fit in the cylinder.

But in statement 1, we do not know the height so then we cannot say for sure that the cubes will not fit in the cylinder. Can someone please explain why this is sufficient.
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