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# PS: Correct ordering

Author Message
Senior Manager
Joined: 16 Jan 2009
Posts: 342
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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07 Jun 2009, 11:27
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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If x is positive , which of the following could be correct ordering of $$1/x,2x and x^2$$?
I.$$x^2<2x<1/x$$
II.$$x^2<1/x<2x$$
III.$$2x<x^2<1/x$$

A.none.
B.I
C.III
D.I and II
E.All
D

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Lahoosaher

Director
Joined: 03 Jun 2009
Posts: 756
Location: New Delhi
WE 1: 5.5 yrs in IT

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07 Jun 2009, 11:53
I. x^2 < 2x < 1/x
lets evaluate both parts seperately
x^2 < 2x
or x < 2 (as x is +ve)

2x < 1/x
or x^2 < 0.5
or x < 0.7

Together for x <0.7

II. x^2 < 1/x < 2x
x^2 < 1/x
or x^3 < 1
or x < 1

1/x < 2x
or 1/2 < x^2
or 0.7 < x

Together, possible for 0.7 < x < 1

III. 2x < x^2 < 1/x
2x < x^2
or 2 < x (dividing x from both sides, since x is +ve)

x^2 < 1/x
or x^3 < 1
or x < 1

together, x>2 and x<1, not possible

Hence answer I and II or "D"
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Manager
Joined: 14 May 2009
Posts: 188

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08 Jun 2009, 03:30
1
[quote="amolsk11"]If x is positive , which of the following could be correct ordering of $$1/x,2x and x^2$$?
I.$$x^2\frac{1}{\sqrt[3]{2}}$$
$$1&2&3 \longrightarrow \frac{1}{\sqrt[3]{2}} 2$$
$$2: \longrightarrow x^2<1/x \longrightarrow x^3<1 \longrightarrow x<1$$

Impossible

D 1&2

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: PS: Correct ordering   [#permalink] 08 Jun 2009, 03:30
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