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Compare for x > 0

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GMAT Club Legend
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Joined: 07 Jun 2014
Posts: 4710
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1612 [0], given: 375

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Compare for x > 0 [#permalink] New post 17 May 2016, 18:05
Expert's post
00:00

Question Stats:

83% (00:34) correct 16% (00:40) wrong based on 53 sessions
\(x > 0\)

Quantity A
Quantity B
\(\frac{1}{x}\)
\(\frac{(1 + x)}{x^2}\)


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.




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Question: 2
Page: 81
[Reveal] Spoiler: OA

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Sandy
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GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4710
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 91

Kudos [?]: 1612 [0], given: 375

CAT Tests
Re: Compare for x > 0 [#permalink] New post 17 May 2016, 18:11
Expert's post
Explanation

Note that Quantity B, \(\frac{(1 + x)}{x^2}\) can be expressed as \(\frac{1}{x^2}+\frac{x}{x^2}\), which can be simplified to \(\frac{1}{x^2}+\frac{1}{x}\).

Note that Quantity A is \(\frac{1}{x}\), and for all nonzero values of \(x\), \(\frac{1}{x^2} >0\). It follows that \(\frac{1}{x^2}+\frac{1}{x} > \frac{1}{x}\); that is, Quantity B is greater than Quantity A. Thus the correct answer is Choice B.
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Sandy
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Manager
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Joined: 23 Jan 2016
Posts: 137
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Kudos [?]: 108 [0], given: 15

Re: Compare for x > 0 [#permalink] New post 18 May 2016, 00:14
As x > 0, no matter what 1/x^2 + 1/x will always be great than 1/x, Hence B
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Intern
Intern
Joined: 26 Dec 2016
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Kudos [?]: 7 [2] , given: 0

Re: Compare for x > 0 [#permalink] New post 20 Jan 2017, 15:38
2
This post received
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Quantity A Quantity B
1/X (1+X)/X^2
Multiply both by X
1 (1+X)/X
1 1/X + X/X
1 1/X + 1
Since X Always >0 , Then Quantity B is always greater.
Answer is B.
GRE Instructor
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Joined: 10 Apr 2015
Posts: 1177
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Kudos [?]: 1053 [0], given: 6

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Re: Compare for x > 0 [#permalink] New post 25 Jan 2017, 11:10
Expert's post
sandy wrote:
\(x > 0\)

Quantity A
Quantity B
\(\frac{1}{x}\)
\(\frac{(1 + x)}{x^2}\)




Another approach is to use Matching Operations

Given:
Quantity A: 1/x
Quantity B: (1 + x)

Since x ≠ 0, we can be certain that x² is POSITIVE
So, we can safely multiply both quantities by x² to get:
Quantity A: x²/x
Quantity B: (1 + x)

Simplify to get:
Quantity A: x
Quantity B: 1 + x

Subtract x from both sides to get:
Quantity A: 0
Quantity B: 1

Answer: B

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GRE Instructor
User avatar
Joined: 10 Apr 2015
Posts: 1177
Followers: 44

Kudos [?]: 1053 [0], given: 6

CAT Tests
Re: Compare for x > 0 [#permalink] New post 29 Sep 2018, 07:53
Expert's post
sandy wrote:
\(x > 0\)

Quantity A
Quantity B
\(\frac{1}{x}\)
\(\frac{(1 + x)}{x^2}\)




Given:
Quantity A: 1/x
Quantity B: (1 + x)/x²

USEFUL FRACTION PROPERTY: (a + b)/c = a/b + a/c

So, we can rewrite Quantity B as follows:
Quantity A: 1/x
Quantity B: 1/x² + x/x²

Simplify Quantity B:
Quantity A: 1/x
Quantity B: 1/x² + 1/x

Subtract 1/x from both quantities to get:
Quantity A: 0
Quantity B: 1/x²

Since x > 0, we know that x² > 0, which means 1/x² > 0

Answer: B

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Re: Compare for x > 0   [#permalink] 29 Sep 2018, 07:53
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