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by nakul_anand » Sat Nov 28, 2009 8:32 am
If X is positive, which one of the following could be the correct order 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
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by papgust » Sat Nov 28, 2009 7:44 pm
IMO B

Consider three cases here to test the order - Integer, Proper fraction and Improper fraction.

If x=1/2,
x^2 < 2x < 1/x (Satisfies I)

If x=3,
1/x < 2x < x^2 (No options to satisfy)

If x=3/2,
1/x < x^2 < 2x ( " )

Hence B.
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by nakul_anand » Tue Dec 01, 2009 4:35 am
Hello Papgust....

I did pretty much the same thing and got the same answer (which is B).

But the OA is D.

Wasn't able to figure out why II could be correct.
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by raghavakumar85 » Tue Dec 01, 2009 10:13 am
you must consider all the possible values of x ie., if x is 1, >1 and is a fraction (0<x<1). then you will get the right option D. this question tests the knowledge of number properties.

consider x = 3 then I holds good

consider x = 3/4; then 4/3;6/4;(3/4)^2 => x^2 < 1/x < 2x

thanks!
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by papgust » Tue Dec 01, 2009 8:57 pm
raghavakumar85 wrote: consider x = 3 then I holds good
How does x when equal to 3 satisfy I?

x=3, 3^2 > 2*3 > 1/3 (which is the reverse order of I)
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by Testluv » Tue Dec 01, 2009 9:27 pm
x = 1/2 satisfies I

x = 3/4 satisfies II

If x = 3/4, then x^2 = 9/16, and 1/x = 4/3, and 2x = 3/2

9/16 < 4/3 < 3/2

So, both I and II are possible orders. If x is positive, however, it is impossible that x^2 < 1/x while simultaneously 2x < x^2.

Choose D.
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by [email protected] » Tue Dec 11, 2012 5:37 pm
Testluv,

how do I know that I have to choose 3/4? I tested numbers and I chose 2 (Integer) and 1/2 ( Non-Integer) and I got therefore B. So what is your key to pick the right numbers?

Thanks!
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by GMATGuruNY » Tue Dec 11, 2012 7:41 pm
[email protected] wrote:Testluv,

how do I know that I have to choose 3/4? I tested numbers and I chose 2 (Integer) and 1/2 ( Non-Integer) and I got therefore B. So what is your key to pick the right numbers?

Thanks!
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