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DaniyalAlwani
IanStewart

I know this question should be super easy but please explain when I went wrong.

Target question: Possible values of x: 0<=x<=1

Statement 1: 1<=x<=3
0 is not part of this; hence false

Statement 2: -1<=x<=2
Both 0 and 1 are part of this range, hence B should be the answer.

Please, correct me if I am wrong.

I think there are two things wrong with your approach. First, I believe you're assuming x is an integer, and in pure algebra questions, you should never assume (unless told) that your unknowns are integers. It's possible, using Statement 1, that x = 2.5, for example.

But it also appears that you're approaching the question 'backwards', if I understand correctly how you solved. It seems to me you assumed x was either 0 or 1, and then tested whether each Statement was true. That's doing things the wrong way around. In DS, the Statements are facts. They cannot be wrong, so you never want to test whether a Statement is true -- the Statements are automatically true. It's the question itself that we don't know the answer to. So here, we aren't sure, before using the Statements, if x is between -1 and 2. When we look at Statement 2 alone, we know for an absolute fact that x is between -2 and 3, but that's all we know. So it's possible that x = 1, say, and then the answer to the question "Is -1 < x < 2?" is "yes". But it's also possible that x = 2.5, or x = -1.5, and then the answer to the question "Is -1 < x < 2?" is "no". Since we can get two different answers to the question, "yes" and "no", Statement 2 is not sufficient alone. And as in my post above, even using both Statements we can't be certain x is between -1 and 2, so the answer is E.
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Bunuel
Is \(-1 < x < 2\) ?


(1) \(0 < x < 4\)

(2) \(-2 < x < 3\)



DS21188

We need the range to be within -1 < x < 2 for sufficiency.

Statement 1:
We can select x = 1 or x = 3.5, x = 1 would be within (-1, 2) but x = 3.5 would be outside. Insufficient.

Statement 2:
We can select x = 1 or x = -1.5, x = 1 would be within (-1, 2) but x = -1.5 would be outside. Insufficient.

Combined:
Combined we can reduce the range to 0 < x < 3. However, this does not fully incorporate -1 < x < 2 so this is insufficient. E.g. x = 2.5 is outside of range.

Ans: E
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