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If x 0, what is the value of sq. root (x^u/x^v) ?

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If x 0, what is the value of sq. root (x^u/x^v) ? [#permalink] New post 11 Jun 2010, 08:32
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A
B
C
D
E

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  25% (medium)

Question Stats:

81% (01:48) correct 19% (00:49) wrong based on 30 sessions
If x ≠ 0, what is the value of sq. root (x^u/x^v) ?

(1) u = v
(2) x is a perfect square.
[Reveal] Spoiler: OA

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Kudos [?]: 3 [0], given: 3

Re: problem [#permalink] New post 11 Jun 2010, 08:59
IMO E

A is INSUFFICIENT.
Equation becomes sqrt (x^(u-v)). If u-v = 0, then the value of the equation is +1 or -1.

B is INSUFFICIENT. We only know that x is a perfect square. Also, we know nothing about u & v.

A and B together dont provide us with any new information.
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Re: problem [#permalink] New post 11 Jun 2010, 09:46
Expert's post
hardnstrong wrote:
If x ≠ 0, what is the value of sq. root (x^u/x^v) ?
(1) u = v
(2) x is a perfect square.

OA will be posted later


\sqrt{\frac{x^u}{x^v}}=?

(1) u=v --> \sqrt{\frac{x^u}{x^v}}=\sqrt{\frac{x^u}{x^u}}=\sqrt{1}=1. Sufficient.

(2) x is a perfect square --> Clearly not sufficient.

Answer: A.
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Re: problem [#permalink] New post 11 Jun 2010, 20:41
Bunuel wrote:
hardnstrong wrote:
If x ≠ 0, what is the value of sq. root (x^u/x^v) ?
(1) u = v
(2) x is a perfect square.

OA will be posted later


\sqrt{\frac{x^u}{x^v}}=?

(1) u=v --> \sqrt{\frac{x^u}{x^v}}=\sqrt{\frac{x^u}{x^u}}=\sqrt{1}=1. Sufficient.

(2) x is a perfect square --> Clearly not sufficient.

Answer: A.


my question is sq. root of 1 can be +1 or -1 (like sq, root of 4 can be +2 or -2)
How can it be only 1?
please explain

OA is A
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Re: problem [#permalink] New post 12 Jun 2010, 03:33
Expert's post
hardnstrong wrote:
Bunuel wrote:
hardnstrong wrote:
If x ≠ 0, what is the value of sq. root (x^u/x^v) ?
(1) u = v
(2) x is a perfect square.

OA will be posted later


\sqrt{\frac{x^u}{x^v}}=?

(1) u=v --> \sqrt{\frac{x^u}{x^v}}=\sqrt{\frac{x^u}{x^u}}=\sqrt{1}=1. Sufficient.

(2) x is a perfect square --> Clearly not sufficient.

Answer: A.


my question is sq. root of 1 can be +1 or -1 (like sq, root of 4 can be +2 or -2)
How can it be only 1?
please explain

OA is A


Square root function can not give negative result.

Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root.

That is, \sqrt{25}=5, NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4.

Hope it helps.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: problem [#permalink] New post 12 Jun 2010, 21:18
Didnt know that
Thanks for the precious info.
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Re: problem   [#permalink] 12 Jun 2010, 21:18
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