The Correct Order

[Atonio79]

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

[med00]

I think ans should be A

[Anson800]

IMO D. Take 0.3 and 0.7 and check

[GmatG]

Originally Posted by Anson800

IMO D. Take 0.3 and 0.7 and check

Agree.. we have to chk for both integer and decimal values
IMO D

[NishantG]

D for me too

[Atonio79]

D is the correct answer... Did yall figure that out by picking numbers?

Thanks

[GmatG]

Originally Posted by Atonio79

D is the correct answer... Did yall figure that out by picking numbers?

Thanks

Yes we will have to pik the approprate no. one integer and other decimail (such as 3 and 1/3)

[mals24]

Can you explain how you got D with a worked out example.
I feel the answer is B.

For instance if we take x as 7 and 0.7
i) x=7
1/x = 1/7 = 0.143
x^2 = 49
2x = 14
i.e. x^2>2x>1/x.

ii) x = 0.7
1/x = 1.43
x^2 = 0.49
2x = 1.4
i.e. 1/x>2x>x^2 which is statement I.

[Makumajon]

I. Is x^2 < 2x < 1/x possible?
Is x<2<1/x^2 possible. Yes, but x's value must be a very small proper fraction, say x=0.1.

II. Is x^2 < 1/x < 2x possible?
Is x<1/x^2<2 possible? Yes, but x must still be a proper fraction and close to 1, say x=0.9

III. Is 2x < x^2 < 1/x possible?
Is 2<x<1/x^2. Nope, since x>2, 1/x^2 's value will no longer be greater than than x?

[gmat731]

Terrific Makumajon!!!! Clear and concise. How does it strike you? The method to divide makes it so clear and easy to approach.