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Re: The greatest possible (straight line) distance, in inches, between any [#permalink]
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The greatest possible (straight line) distance, in inches, between any two points on a certain cube is 10.

The greatest distance between 2 points in a cube is the main diagonal which passes through the center of the cube. The diagonal of the face of the cube is not the main diagonal.

The length of the main diagonal =\( \sqrt{l^2 + b^2 + h^2}\)
Let's say, the side of a cube is a .
We can say that in a cube, l = b= h = a
So the length of the main diagonal = \( \sqrt{a^2 + a^2 + a^2}\) = \(\sqrt{3 a^2}\) = \(a\sqrt{3}\)
Its given that greatest possible length is 10. So we can equate that
\(a\sqrt{3}\) = 10
\(a = 10/\sqrt{3}\)


If the cube is modified so that its length is doubled and its width is doubled while its height remains unchanged, then what is the greatest possible (straight line) distance, in inches, between any two points on the modified box?

Then l = 2* a
b = 2* a
h = a.
The length of the main diagonal =\( \sqrt{l^2 + b^2 + h^2}\) = \( \sqrt{4*a^2 + 4*a^2 + a^2}\)
=\(\sqrt{9*a^2}\) = \(3a = 3* 10/\sqrt{3}\) = \(10\sqrt{3}\)

Option B is the answer.

Thanks,
Clifin J Francis,
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Re: The greatest possible (straight line) distance, in inches, between any [#permalink]
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Kudos
Answer is B.

But I selected D by mistake. Since I wanted to make the diagonal length of the 2nd cube GREATEST, I wanted to make the diagonal of the 1st cube the LEAST. So, I assumed the side length as 10inch. In the process, I missed the line "The greatest possible (straight line) distance, in inches, between any two points".

Had the question said, "The (straight line) distance, in inches, between two points", then D would have been the correct option.

Hope you don't make the same mistake I made.
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Re: The greatest possible (straight line) distance, in inches, between any [#permalink]
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Kudos
Quote:
The greatest possible (straight line) distance, in inches, between any two points on a certain cube is 10. If the cube is modified so that its length is doubled and its width is doubled while its height remains unchanged, then what is the greatest possible (straight line) distance, in inches, between any two points on the modified box?

A. 10√2
B. 10√3
C. 20
D. 30
E. 30√3

The longest distance in a cube is a diagonal connecting opposite faces, it's given by √3a, where a is the side of the cube.
So, it's given to us that √3a = 10 for a cube, which means, the side of the cube is a = 10/√3.
Now length and width of the cube are doubled.
Dimension of the new cuboid formed are
l = 20/√3, w = 20/√3, h = 10/√3

Now, similar formula for the longest straight line distance (let it be d) between any two points in a cuboid is
d = √(l^2 + w^2 + h^2), substituting values for l, w and h, we get
d = √(400/3 + 400/3 + 100/3)
d = √(900/3)
d = √300 = 10√3
Hence, correct answer is B) 10√3
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Re: The greatest possible (straight line) distance, in inches, between any [#permalink]
Bunuel wrote:
The greatest possible (straight line) distance, in inches, between any two points on a certain cube is 10. If the cube is modified so that its length is doubled and its width is doubled while its height remains unchanged, then what is the greatest possible (straight line) distance, in inches, between any two points on the modified box?


A. 10√2
B. 10√3
C. 20
D. 30
E. 30√3

Longest straight line of a cube is a*rt3 a being the side of the cube

then a becomes 10/rt3

Then the diagnol of the cube = (10/rt3)^2 + (2*20/rt3)^2
=>9/3 *100 = sq*length

Therefore the lenth equals rt3*10

Therefore IMO B
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Re: The greatest possible (straight line) distance, in inches, between any [#permalink]
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