GMAT Prep Practice Test 1 order question

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davidnodine
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Topic: GMAT Prep Practice Test 1 order question
PostMon Aug 27, 2007 4:56 pm Reply with quote
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
e I, II and III

I understand how one works with fractions but I'm confused that the answer is D I and II. I'm sure its something simple that i'm over thinking. Thanks
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ratindasgupta
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PostMon Aug 27, 2007 11:06 pm Reply with quote
davidnodine 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
e I, II and III

I understand how one works with fractions but I'm confused that the answer is D I and II. I'm sure its something simple that i'm over thinking. Thanks
Are you sure II = x^2<1/x<2x ?

Coz if you take a fraction like 3/2,
x^2 = 9/4
1/x = 2/3
2x = 3

So i'm getting the expression as 1/x<x^2<2x.
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beny
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PostTue Aug 28, 2007 1:53 am Reply with quote
ratindasgupta wrote:
davidnodine 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
e I, II and III

I understand how one works with fractions but I'm confused that the answer is D I and II. I'm sure its something simple that i'm over thinking. Thanks
Are you sure II = x^2<1/x<2x ?

Coz if you take a fraction like 3/2,
x^2 = 9/4
1/x = 2/3
2x = 3

So i'm getting the expression as 1/x<x^2<2x.
Just because one example works, doesn't mean he typed the problem incorrectly...

x=.9 would satisfy criterion II.
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davidnodine
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PostTue Aug 28, 2007 2:23 pm Reply with quote
It is written correctly. Is there any way to look at this question without plugging in numbers?
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Guest
PostThu Aug 30, 2007 3:15 am Reply with quote
Lets take III and I , II can be tackled in same way ..

III ) 2x< x^2<1/x

if 2X < X^2 that means 2 < X ( Since X is positive so no sign change while cancelling X)

and if x^2< 1/X that means X^3 < 1 which means X< 1

so for III to be true we need a value of X which is > 2 and < 1 ... this is impossible ....

is u proceeed similarly for I and II you will find that they hold good for at least some value of X....
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uptowngirl92
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PostSat Nov 07, 2009 4:39 am Reply with quote
Source of the question is: GMAT Prep Test 1

If x is positive which of the following could be correct ordering of 1/x, 2x, x^2.
1) x^2 < 2x < 1/x

2) x^2 < 1/x < 2x

3) 2x < x^2 < 1/x

Choices are:

1) None
2) 1 Only
3) 3 Only
4) I and 2 only
5) 1, 2, 3

How to proceed without plugging in?I plugged in in the exam and got only choice 1 as correct and hence marke db which is wrong:(
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palvarez
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PostSat Nov 07, 2009 11:20 am Reply with quote
uptowngirl92 wrote:
Source of the question is: GMAT Prep Test 1

If x is positive which of the following could be correct ordering of 1/x, 2x, x^2.
1) x^2 < 2x < 1/x

2) x^2 < 1/x < 2x

3) 2x < x^2 < 1/x

Choices are:

1) None
2) 1 Only
3) 3 Only
4) I and 2 only
5) 1, 2, 3

How to proceed without plugging in?I plugged in in the exam and got only choice 1 as correct and hence marke db which is wrong:(
Hybrid approach to make things smooth

"Could be true" Find one positive instance for each case. Proving that there are no positive instances requires algebraic way.

"must be true" Find one negative instance for each case. Proving that there are all positive instances requires algebraic way.

Lets focus on algebraic way.

1. x^2 < 2x < 1/x
x^2 - 2x < 0 --> (0,2)
2x < 1/x
x^2 - (1/2) < 0 --> (-1/sqrt(2), 1/sqrt(2))

Combining together: (0, 1/sqrt(2))

The domain of this inequality is the above.

2. x^2 < 1/x < 2x

x^3 < 1
domain: (-inf, 1)

and x^2 > 1/2
domain: (-inf, 1/sqrt(2)) U (1/sqrt(2), +inf)

Find the intersection of both domains: (-inf, 1/sqrt(2). Note that x is positive.
(0, 1/sqrt(2)) is the domain.

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

x^2 - 2x > 0
domain: (-inf, 0) U (2, +inf)


x^3 < 1
domain: (-inf, 1)

Intersection: (-inf, 0)

But x is positive.


Some key things to remember:

(x-a)(x-b) > 0, (-inf, a) U (b, +inf), assumin a < b
(x-a)(x-b) < 0 (a, b), assumin a < b
(x^3-a^3) > 0 (a, +inf)
(x^3+a^3) < 0 (-inf, a)

Being cool and taking the restctions (like x being +ve) into account help your test.

Smart numbers help you to get rid of some choices, thereby ending up with fewer choices. There, focus on algebraic approach.
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