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Question from GMAT Prep - I can't figure it out!

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werdwerd: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Mon Mar 09, 2009 12:22 pm

by werdwerd » Tue Mar 10, 2009 5:29 pm

Hey everyone, I have two questions I'm stuck on:

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

From my testing of numbers, statement I is fine. But the answer is both statements I and II. Could someone explain to me how statement II might hold true? I've tried fractions, whole numbers...can't figure it out, and it's driving me nuts

and second:

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is:

A – between 2 and 10
B – between 10 and 20
C – between 20 and 30
D – between 30 and 40
E – greater than 40

I have no idea how to approach this second one...

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Source: Beat The GMAT — Problem Solving | Tue Mar 10, 2009 5:29 pm

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Re: Question from GMAT Prep - I can't figure it out!

by Ian Stewart » Tue Mar 10, 2009 11:23 pm

werdwerd wrote:Hey everyone, I have two questions I'm stuck on:

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

From my testing of numbers, statement I is fine. But the answer is both statements I and II. Could someone explain to me how statement II might hold true? I've tried fractions, whole numbers...can't figure it out, and it's driving me nuts

To see how II might be true, you could rewrite the inequalities. Since we know x is positive, we can multiply or divide by x here without worrying about whether we need to reverse the inequality:

If x^2 < 1/x < 2x, then:

1/x < 2x
1/2 < x^2
sqrt(1/2) < x

We also know that x^2 < 1/x, so x < 1. That is, II should be true for any value of x between sqrt(1/2) (which is roughly 0.7) and 1.

As for your second question, if you search the forum, you should find several solutions; I've certainly solved it a few times here.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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werdwerd: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Mon Mar 09, 2009 12:22 pm

by werdwerd » Wed Mar 11, 2009 4:41 am

That explanation was great - thank you! And sorry for refraining from the search for the second question, I'm new here. Much more clarity now.

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sumanr84: Legendary Member; Posts: 758; Joined: Sat Aug 29, 2009 9:32 pmFollowed by: 2 members; Location: Bangalore,India; Thanked: 67 times

by sumanr84 » Mon Jul 19, 2010 10:13 am

Ian Stewart wrote:
werdwerd wrote:Hey everyone, I have two questions I'm stuck on:

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

From my testing of numbers, statement I is fine. But the answer is both statements I and II. Could someone explain to me how statement II might hold true? I've tried fractions, whole numbers...can't figure it out, and it's driving me nuts
To see how II might be true, you could rewrite the inequalities. Since we know x is positive, we can multiply or divide by x here without worrying about whether we need to reverse the inequality:

If x^2 < 1/x < 2x, then:

1/x < 2x
1/2 < x^2
sqrt(1/2) < x

We also know that x^2 < 1/x, so x < 1. That is, II should be true for any value of x between sqrt(1/2) (which is roughly 0.7) and 1.

As for your second question, if you search the forum, you should find several solutions; I've certainly solved it a few times here.

Nice approach suggested by Ian as this is huge time blocker question ..
To see how III. 2x < x^2 < 1/x, could not be true on a similar line of reasoning,

2x < x^2 and so 2 < x
x^2 < 1/x and so x^3 < 1 or x < 1
It's clear to see that x > 2 but x < 1 is not a possible scenario. Hence, III is not right.

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