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Re: Let u, v, and w denote the lengths of three separate line segments. In [#permalink]
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For U,V,W to become the lengths of the sides of the triangle, they must satisfy the two conditions

1). The sum of any two sides of a triangle is always greater than the third side
2). The difference between any two sides of a triangle is less than the third side

statement 1:

u,v,w ==> u,u+5, v

5< v< 2u+5

So, V is always greater than an unique value (5). SUFFICIENT

Statement2:

7-2<v<7+2

5<v<9

So, V is always greater than an unique value (5). SUFFICIENT

Both are sufficient alone. Hence,D
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Re: Let u, v, and w denote the lengths of three separate line segments. In [#permalink]
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Bunuel wrote:
Let u, v, and w denote the lengths of three separate line segments. In order for u, v, and w to represent the lengths of three sides of a triangle, what value must v exceed?

(1) w = u + 5

(2) u = 2 and w = 7


The sum of two sides of a triangle must exceed the third side. So if one side is u, and the other is u+5, the third side will clearly need to be larger than 5, so that when we add u and the third side, we get something that exceeds u + 5. So I imagine the intended answer is D here.

But that value 5 is not at all unique. Certainly v must exceed 5, but it also must exceed 4, or √2, or 0.01. For information to be sufficient in a DS question asking for a value, you need to be able to find a unique numerical answer for that value, and you can't do that here. I see a lot of prep company DS questions that ask, in various ways, for minimum or maximum values, and that type of question logically doesn't make sense as a DS question. There's a reason no official DS questions are set up this way.

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Re: Let u, v, and w denote the lengths of three separate line segments. In [#permalink]
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