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Director
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DS- GMATPrep [#permalink] New post 05 Jun 2007, 02:33
00:00
A
B
C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
If x is positive, which of the following could be the 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 only
c. III only
d. I and II only
e. I, II and III
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 [#permalink] New post 05 Jun 2007, 04:02
as far as I can see only I is possible

Say if x were .5
.25 < 1 < 2 which holds valid (this also eliminates option III since 2x seems greater than x^2 for all 0<x<1>= 1 and found that it does not work out.

So my answer is Option B.

But since I am very very sleepy and because DS is my weak point in quant, I would like to know the OA.
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Re: DS- GMATPrep [#permalink] New post 05 Jun 2007, 16:49
vshaunak@gmail.com wrote:
If x is positive, which of the following could be the 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 only
c. III only
d. I and II only
e. I, II and III



I. x^2 < 2x < 1/x
.5^2= .25 < .5x2=1 < 1/.5=2 so I. is possible

II. x^2 < 1/x < 2x
.8^2=.64 < 1/.8=1.25 < 2*.8=1.6 so II. is possible

III. 2x < x^2 <1/x
Not possible

So I say D
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 [#permalink] New post 05 Jun 2007, 21:01
Yup .. well spotted guys. I was really sleepy then :)
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 [#permalink] New post 05 Jun 2007, 21:40
May be I have not understood the Q correctly.

Are we not supposed to find the equation which is true for all values of x rather than some specific values?
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 [#permalink] New post 06 Jun 2007, 00:50
OA is 'D'. Thanks guys.
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 [#permalink] New post 06 Jun 2007, 01:41
There are three possibilities for value of x here,
i) 0<x>1 & x=1
ii)1<x>2 & x=2
iii)x>2
Now for i) taking from previous examples x=0.5
then the ordering is x^2<2x<1/x (this is I) & for x=1, 1/x=x^2<2x

For ii) lets say x=1.5
the ordering is 1/x<x^2<2x & for x=2, 1/x< 2x=x^2

for iii) lets say x=3
the ordering is 1/x<2x<x^2

SO in my opinion, the answer is "b"
Tell me plz how its d and not b
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 [#permalink] New post 08 Jun 2007, 07:11
shahrukh wrote:
There are three possibilities for value of x here,
i) 0<x>1 & x=1
ii)1<x>2 & x=2
iii)x>2
Now for i) taking from previous examples x=0.5
then the ordering is x^2<2x<1/x (this is I) & for x=1, 1/x=x^2<2x

For ii) lets say x=1.5
the ordering is 1/x<x^2<2x & for x=2, 1/x< 2x=x^2

for iii) lets say x=3
the ordering is 1/x<2x<x^2

SO in my opinion, the answer is "b"
Tell me plz how its d and not b


From my post above.

II. x^2 < 1/x < 2x
.8^2=.64 < 1/.8=1.25 < 2*.8=1.6 so II. is possible

You only have to prove it can work for a single case not that it is always true.
  [#permalink] 08 Jun 2007, 07:11
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