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Nervous
 
 

If 0 < x < 1, which of the following inequalities

by Nervous Tue Dec 25, 2007 5:26 am

If 0 < x < 1, which of the following inequalities must be true?

I. x^5 < x^3
II. x^4 + x^5 < x^3 + x^2
III. x^4 - x^5 < x^3 - x^2

A) None
B) I only
C) II only
D) I and II only
E) I, II and III

OA is E (highlight to see).
I marked the answer as D, but turned out that was wrong. Can some1 please explain how to check for the 3rd statement above?

Thanks!
RonPurewal
Students
 
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Re: If 0 < x < 1, which of the following inequalities

by RonPurewal Tue Dec 25, 2007 6:24 am

please obey our forum's conventions for how to name these threads. i went ahead and fixed this one for you (merry christmas); the white-on-white was cool enough to justify the favor.

Nervous Wrote:Can some1 please explain how to check for the 3rd statement above?

Thanks!


method #1: plug in two or three numbers between 0 and 1 and watch what happens. (you probably didn't need us to tell you this)

method #2: factor each side
left side = x^4(1 - x), right side = x^2(1 - x)
...since x^2 is bigger than x^4, the inequality holds.
alternatively: factor out just enough from the left side to make it look like the right side:
left side = x^2(x^2 - x^3) = x^2(right side)
...since x^2 is less than 1, the left side is consequently smaller than the right side.
Nervous
 
 

by Nervous Tue Dec 25, 2007 6:57 am

Much thanks for the explanation And for editing the post as well. Merry Christmas to you too! :)

thanks
Guest
 
 

Re: If 0 < x < 1, which of the following inequalities

by Guest Thu Dec 27, 2007 1:43 am

RPurewal Wrote:please obey our forum's conventions for how to name these threads. i went ahead and fixed this one for you (merry christmas); the white-on-white was cool enough to justify the favor.

Nervous Wrote:Can some1 please explain how to check for the 3rd statement above?

Thanks!


method #1: plug in two or three numbers between 0 and 1 and watch what happens. (you probably didn't need us to tell you this)

method #2: factor each side
left side = x^4(1 - x), right side = x^2(1 - x)
...since x^2 is bigger than x^4, the inequality holds.
alternatively: factor out just enough from the left side to make it look like the right side:
left side = x^2(x^2 - x^3) = x^2(right side)
...since x^2 is less than 1, the left side is consequently smaller than the right side.


Are you sure answer is E??

x^4 - x^5 < x^3 - x^2

x^4 + x^2 <x^5 + x^3

x^2 ( x^2 +1) < x^3(x^2+1)


Now x^2 is definetly bigger than x^3

So the equation falls apart hence. E doesn't hold true.

Any flaw in my reasoning?

Saurabh Malpani
Saurabh Malpani
 
 

Re: If 0 < x < 1, which of the following inequalities

by Saurabh Malpani Thu Dec 27, 2007 1:50 am

RPurewal Wrote:.

method #2: factor each side
left side = x^4(1 - x), right side = x^2(1 - x) -----> I think this is wrong it should be -ve of x^2(1 - x)
...since x^2 is bigger than x^4, the inequality holds.
alternatively: factor out just enough from the left side to make it look like the right side:
left side = x^2(x^2 - x^3) = x^2(right side)
...since x^2 is less than 1, the left side is consequently smaller than the right side.


Ron are you sure on this?
GK
 
 

by GK Thu Dec 27, 2007 3:02 am

I'm getting D as well.

Let x = 1/2

I. 1/32 < 1/16? Yes
II. 1/16 + 1/32 < 1/8 + 1/4 or 3/32 < 3/8? Yes
III. 1/16 -1/32 < 1/8 - 1/4 or 1/32 < -1/8? No.

Alternatively:
For III:
x^4 - x^5 < x^3 - x^2
or x^4(1-x) < x^2(x-1)
Since 0<x<1, 1-x is +ve and x-1 is -ve. So we LHS above will be +ve (x^4 is +ve and 1-x is +ve) and RHS will be -ve (x^2 is +ve but x-1 is -ve)..
+ve < -ve? No.

Not sure how the answer is E.
StaceyKoprince
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Joined: Wed Oct 19, 2005 9:05 am
Location: Montreal
 

by StaceyKoprince Thu Dec 27, 2007 11:11 pm

Hmm, something's funny here. Can the initial poster please double check the problem (IN GMATPREP - not from some other post on some other site) and make sure you entered it completely accurately? (Esp. statement 3)

I don't mean to doubt you but, as I was just saying in another post, it's incredibly rare for GMATPrep to be wrong. This stuff has been tested and retested tens of thousands of times, literally. I suppose it's possible for whoever's inputting the problem into the software to key in the wrong answer, but that really doesn't happen much.

Yet, statement 3, as written, has to be false:
x^4 - x^5 < x^3 - x^2

x^4 is larger than x^5 (when 0<x<1), so x^4-x^5 is positive.
x^3 is smaller than x^2 (when 0<x<1), so x^3 - x^2 is negative.
A positive number is not smaller than a negative number.

Was the sign on this one supposed to go the other way? (That is, is it supposed to be x^4 - x^5 > x^3 - x^2?)
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
mclaren7
 
 

by mclaren7 Sat Mar 15, 2008 2:49 pm

Yes, there is an error in Option III.

It should be:

x^4 - x^5 < x^2 - x^3.

KH
StaceyKoprince
ManhattanGMAT Staff
 
Posts: 9360
Joined: Wed Oct 19, 2005 9:05 am
Location: Montreal
 

by StaceyKoprince Mon Mar 17, 2008 4:06 pm

thanks for looking that up mclaren7!
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep