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08 Aug 2020, 19:38
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Given that X ≠ 0, is ∛ < 5√ (should be the fifth root but I'm not sure how to type it or what the alt code is)
Statement 1: x <1
Statement 2: X > -1
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
Statement 1: x <1
Statement 2: X > -1
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
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09 Aug 2020, 02:12
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Weezy85 - Please follow the rules for posting in a specific forum. Also, I am moving this post to the appropriate forum: Data Sufficiency (DS).
12 Rules for Posting - Please Read this Before Posting New Question
12 Rules for Posting - Please Read this Before Posting New Question
09 Aug 2020, 04:44
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Weezy85
The question is
Given that \(X ≠ 0\)
IS \(\sqrt[3]{x} < \sqrt[5]{x} \)?
1) Statement 1: \(x <1\)
2) Statement 2: \(X > -1\)
to answer, lets multiply both sides by \(x^6\)
so the question will be, is \(x^3 < x\)
lets rearrange it a little bit:
\(x^3-x < 0\)
\(x(x^2-1) < 0\)
\(x (x-1)(x+1) < 0\)
there are 4 intervals to think about:
\(x>-1 \) --> will cause equation with positive sign (rejected)
\(-1>x>0\) --> will cause equation with positive sign (accepted)
\(0<x<1\) --> will cause equation with positive sign (rejected)
\(1<x\) --> will cause equation with positive sign (accepted)
so the question can be rewritten to:
does x have the intervals : \(-1>x>0\) or \(1<x\)
From statement (1),
we can't tell because it can be \(0<x<1\) which is rejected or \(-1>x>0\) which is accepted
From statement (2),
we can't tell because it can be \(0<x<1\) which is rejected or \(-1>x>0\) which is accepted
By combining (1) & (2),
Still we can't tell for the same reason
so E
