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20 Jul 2014, 11:23
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
55% (01:53) correct
45%
(01:56)
wrong
based on 1007
sessions
History
Date
Time
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Not Attempted Yet
If \(xy\neq{0}\), is \(x^y>0\)?
(1) \(\sqrt[x]{32}=\frac{1}{2}\)
(2) \(\sqrt[-y]{16}=\frac{1}{2}\)
(1) \(\sqrt[x]{32}=\frac{1}{2}\)
(2) \(\sqrt[-y]{16}=\frac{1}{2}\)
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
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20 Jul 2014, 11:23
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SOLUTION
If \(xy\neq{0}\), is \(x^y>0\)?
(1) \(\sqrt[x]{32}=\frac{1}{2}\).
Rewrite as \(32^{\frac{1}{x}}=2^{-1}\);
Since \(32=2^5\), then we'd have that \(2^{\frac{5}{x}}=2^{-1}\);
Equate the powers of 2 on both sides: \(\frac{5}{x}=-1\).
Solving gives \(x=-5\).
Now, if \(y\) is even, then \(x^y=(-5)^{even}>0\) but if \(y\) is not even, say if it's 1, then \(x^y=(-5)^1<0\). Not sufficient.
(2) \(\sqrt[-y]{16}=\frac{1}{2}\)
Rewrite as \(16^{\frac{1}{-y}}=2^{-1}\);
Since \(16=2^4\), then we'd have that \(2^{\frac{4}{-y}}=2^{-1}\);
Equate the powers of 2 on both sides: \(\frac{4}{-y}=-1\).
Solving gives \(y=4\).
Since given that \(x\) is not zero (from \(xy\neq{0}\)), then \(x^y=x^{4}>0\). Sufficient.
Answer: B.
Try NEW Exponents and Roots PS question.
If \(xy\neq{0}\), is \(x^y>0\)?
(1) \(\sqrt[x]{32}=\frac{1}{2}\).
Rewrite as \(32^{\frac{1}{x}}=2^{-1}\);
Since \(32=2^5\), then we'd have that \(2^{\frac{5}{x}}=2^{-1}\);
Equate the powers of 2 on both sides: \(\frac{5}{x}=-1\).
Solving gives \(x=-5\).
Now, if \(y\) is even, then \(x^y=(-5)^{even}>0\) but if \(y\) is not even, say if it's 1, then \(x^y=(-5)^1<0\). Not sufficient.
(2) \(\sqrt[-y]{16}=\frac{1}{2}\)
Rewrite as \(16^{\frac{1}{-y}}=2^{-1}\);
Since \(16=2^4\), then we'd have that \(2^{\frac{4}{-y}}=2^{-1}\);
Equate the powers of 2 on both sides: \(\frac{4}{-y}=-1\).
Solving gives \(y=4\).
Since given that \(x\) is not zero (from \(xy\neq{0}\)), then \(x^y=x^{4}>0\). Sufficient.
Answer: B.
Try NEW Exponents and Roots PS question.
20 Jul 2014, 12:12
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If \(xy\neq{0}\), is \(x^y>0\)?
(1) \(\sqrt[x]{32}=\frac{1}{2}\)
(2) \(\sqrt[-y]{16}=\frac{1}{2}\)
Sol: Given \(xy\neq{0}\)
St 1 \(\sqrt[x]{32}=\frac{1}{2}\) can be written as (2)^5/x= 2^-1
or x=-5
Now if y=2 then x^y>0 but if y=1 then x^y =-5<0. Since 2 answers are possible so St 1 is not sufficient
St2 can written as (2)^-4/y=2^-1
or y=4.
Now we know \(x\neq{0}\) and y=4 so for any value of x, x^y >0
Ans is B
(1) \(\sqrt[x]{32}=\frac{1}{2}\)
(2) \(\sqrt[-y]{16}=\frac{1}{2}\)
Sol: Given \(xy\neq{0}\)
St 1 \(\sqrt[x]{32}=\frac{1}{2}\) can be written as (2)^5/x= 2^-1
or x=-5
Now if y=2 then x^y>0 but if y=1 then x^y =-5<0. Since 2 answers are possible so St 1 is not sufficient
St2 can written as (2)^-4/y=2^-1
or y=4.
Now we know \(x\neq{0}\) and y=4 so for any value of x, x^y >0
Ans is B
