GMAT Prep Question

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shrikantkamble: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Jan 24, 2008 2:24 pm; Location: USA; Thanked: 1 times

by shrikantkamble » Fri Jul 25, 2008 6:41 am

Hi,

I am unable to find correct justification for this answer.
Let us assume if x = 2 ?

I selected None as answer but GMAT Prep says it should be option 4. Can you please explain me this ?

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Thanks & Regards,
Shrikant

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Source: Beat The GMAT — Problem Solving | Fri Jul 25, 2008 6:41 am

nitingupta5: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Fri Jul 11, 2008 8:40 am; Thanked: 2 times

by nitingupta5 » Fri Jul 25, 2008 6:45 am

Have u tried using fractions.
The question doesn't say they have to be positive integers

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shrikantkamble: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Jan 24, 2008 2:24 pm; Location: USA; Thanked: 1 times

by shrikantkamble » Fri Jul 25, 2008 6:59 am

Hi,

I can use fraction as well as integers. But these conditions should hold true for all values of x. According to my understanding the relation depends upon value of x. if value of x changes.... 0.1 or 2 or 1 then these conditions are not true.

Can anyone please help me out ?

Thanks & Regards,
Shrikant

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pepeprepa: Legendary Member; Posts: 661; Joined: Tue Jul 08, 2008 12:58 pm; Location: France; Thanked: 48 times

by pepeprepa » Fri Jul 25, 2008 7:47 am

x^2<2x<1/x
If we want it to be true we need x<2 and 2*x^2<1 so x<sqrt(1/2)
so solutions are 0<x<sqrt(1/2)

x^2<1/x<2x
If we want it to be true we need x^3<1 and 1<2*x^2 so sqrt(1/2)<x
so solutions exist

2x<x^2<1/x
If we want it to be true 2<x and x^3<1 so x under 2
so it is impossible

We have 1 and 2 which can exist

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fajoni: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Mon Mar 10, 2008 1:13 pm; Thanked: 2 times

by fajoni » Fri Jul 25, 2008 3:24 pm

Picking values for X that are less than 1 works well to illustrate the point made by the previous post.

If X=0.5 then:

1/X = 2

2X = 1

X^2 = .25

Thus, X^2< 2X <1/X

However, if X = 0.9:

1/X = 1.11

2X = 1.8

X^2 = .81

Thus, X^2 < 1/X < 2X

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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

by raunekk » Sat Jul 26, 2008 6:52 am

plugging numbers will b time consuming...

Is there a quicker method for this..??

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shrikantkamble: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Jan 24, 2008 2:24 pm; Location: USA; Thanked: 1 times

by shrikantkamble » Sun Jul 27, 2008 3:44 am

Hi ,

Here how we can assume x is in between 0 and 1. there is no such condition mentioned in the question. And it should hold true for all values of x irrespective of 0 to 1 or any other value.

I feel either question is wrong or Answer choice.

Thanks & Regards,
Shrikant

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

by parallel_chase » Sun Jul 27, 2008 3:54 am

shrikantkamble wrote:Hi ,

Here how we can assume x is in between 0 and 1. there is no such condition mentioned in the question. And it should hold true for all values of x irrespective of 0 to 1 or any other value.

I feel either question is wrong or Answer choice.

Read the question carefully, it says "what could be the correct ordering"

Therefore only way the options given could be possible is when x is between 0 and 1.

Hope this helps.

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