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# Compare for x > 0

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

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

Compare for x > 0 [#permalink]  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.

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

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

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

Re: Compare for x > 0 [#permalink]  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
Joined: 23 Jan 2016
Posts: 137
Followers: 3

Kudos [?]: 108 [0], given: 15

Re: Compare for x > 0 [#permalink]  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
Intern
Joined: 26 Dec 2016
Posts: 4
Followers: 0

Kudos [?]: 7 [2] , given: 0

Re: Compare for x > 0 [#permalink]  20 Jan 2017, 15:38
2
KUDOS
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.
GRE Instructor
Joined: 10 Apr 2015
Posts: 1177
Followers: 44

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

Re: Compare for x > 0 [#permalink]  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

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

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

Re: Compare for x > 0 [#permalink]  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

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