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GMATinsight
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The correct answer, which is not available in the options, should be 4\(\sqrt{5}\).
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chetan2u
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If an ant, which is sitting at one of the corners on the top face of a cubical box full of sweets, travels on the outer surface of the box to reach the diagonally opposite corner of the box then what will be the shortest distance that ant will have to travel? The side of the box is 4 inches each.

A) 12
B) \(4+4\sqrt{2}\)
C) \(2\sqrt{17}\)
D) \(8\sqrt{2}\)
E) \(4\sqrt{17}\)

Source: www.GMATinsight.com


Hi,

I think it is highly unlikely that such a Q is asked in GMAT..

Shortest route within a cube that is diagonal can still be thought of..
In a cube of side 4 it becomes \(\sqrt{4^2+4^2+4^2}=4√3\)..

But let's make this Q a bit easier..
Just concentrate on two squares of side 4 placed perpendicular to each other..
The ant has to move from one corner to opposite corner...When you open these two squares as if two squares next to each other, it becomes a rectangle with one side of (4+4) and other of 4..
The shortest distance here is the diagonal of this rectangle..
So ans is \(\sqrt{8^2+4^2}=\sqrt{80}=4√5\)...
But the choices do not have the correct answer..

GMATinsight it's again got faulty choices unless I have made some error in a hurry..

Bunuel kudos. You are an encyclopedia on all the Q's in our forum. You never miss out a single similar Q.

Hi chetan2u,

I the same method as you - opened up the cube and drew a line from point to point. Which becomes the hypotenuse of triangle with height 4 and base 8. I think the answer is not in the option. Why do you think that this question will not be asked on the GMAT?
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as i understand, first the ant has to go 4 inch, then 4√2 inch. the answer should be 4+4√2!
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Bunuel, could you please see this question? I believe the OA has some error



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