Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

man, I'm quite weak at number properties as well I just found out, trying this one... well lets give it a shot:

xy != 0 means x and y can be +ve or -ve but none is zero

(1) the ^4 neutralizes the sign and says that the absolute value of y > x
The true value cannot be know.
- both could be +ve then y>x
- both could be -ve then y<x
---> INSUFF

(2) rewrite to 1/(x^3) < 1/(x^3), and signs play a role
For this statement to be true, the true values could be:
- both could be +ve then x>y
- both could be -ve then x<y
---> INSUFF

(1)&(2)
On the premise, that the absolute value of y > x from (1),
in order for (2) to be true, x must be -ve.
Hence, y>x
SUFF.

Re: number properties 3 [#permalink]
04 Nov 2005, 08:31

jdtomatito wrote:

If xy is different from 0, is x < y?

(1) x^4 < y^4

(2) x^(-3) < y^(-3)

(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0
since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0
We have to cases:
case1 : x-y<0 and x+y>0 ---> -y<x<y
case2: x-y>0 and x+y<0 ---> y<x<-y
---> insuff

(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0
--> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0
1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0
---> 1/x-1/y < 0 --> (y-x)/xy<0
we have two cases:
case1 : y-x<0 and xy>0 ---> y<x and xy>0
for example: x=3, y=2 we have 1/3-1/2<0
case2: y-x>0 and xy<0 ---> y>0>x
--->insuff

Last edited by laxieqv on 04 Nov 2005, 08:42, edited 1 time in total.

Re: number properties 3 [#permalink]
04 Nov 2005, 09:25

laxieqv wrote:

jdtomatito wrote:

If xy is different from 0, is x < y?

(1) x^4 < y^4

(2) x^(-3) < y^(-3)

(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff

(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff

Re: number properties 3 [#permalink]
04 Nov 2005, 10:00

laxieqv wrote:

jdtomatito wrote:

If xy is different from 0, is x < y?

(1) x^4 < y^4

(2) x^(-3) < y^(-3)

(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff

(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff

good explanation; but the bold part does not satisfy statement 2. pick some numbers and try
Answer is good: B is suff

Re: number properties 3 [#permalink]
04 Nov 2005, 10:05

gotoknow3 wrote:

laxieqv wrote:

jdtomatito wrote:

If xy is different from 0, is x < y?

(1) x^4 < y^4

(2) x^(-3) < y^(-3)

(1) x^4<y^4 --> x^4-y^4<0 ---> (x^2-y^2) ( x^2+y^2) <0 since x^2+y^2> 0 ( coz no x , y equal to 0 so the sum can't be 0)---> x^2-y^2<0 ---> (x-y)*(x+y) <0 We have to cases: case1 : x-y<0 and x+y>0 ---> -y<x<y case2: x-y>0 and x+y<0 ---> y<x<-y ---> insuff

(2) x^(-3)< y^(-3) --> 1/x^3< 1/y^3 --> 1/x^3-1/y^3<0 --> ( 1/x-1/y) * (1/x^2+ 1/xy+ 1/y^2) <0 1/x^2+ 1/y^2 >= | 2/xy|>= 2/xy> 1/xy, thus 1/x^2+1/xy+1/y^2>0 ---> 1/x-1/y < 0 --> (y-x)/xy<0 we have two cases: case1 : y-x<0 and xy>0 ---> y<x and xy>0 for example: x=3, y=2 we have 1/3-1/2<0 case2: y-x>0 and xy<0 ---> y>0>x --->insuff

good explanation; but the bold part does not satisfy statement 2. pick some numbers and try Answer is good: B is suff

Actually there are three questions I think have a wrong OA. Questions are from Princeton +48 Gmat math.

OE: 1 is insuf because negative values for the variables give you the opposite value than negative values.
In statement 2, negative numbers retain their sign but we need to look closely at how fractions react. If you plug in x=2 and y=1 that gives you a "no". If you plug in x=1/2 and y=1/4 you also get a "no". Therefore statement 2 is sufficient and answer is B.

But if you plug in x=-1 y=1 you get a yes with statement 2, thus OA/OE is wrong.