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# If x and y are integers and xy does not equal 0, is xy <

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Intern
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If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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09 Aug 2009, 05:02
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If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3

(2) x is to the right of 0 on the number line
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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09 Aug 2009, 05:12
lbsgmat wrote:
If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) x is to the right of 0 on the number line.

does xy have different signs

y = x^3(x-1), if x -ve then y is +ve , and if x is +ve we have 2 cases y can be either -ve or +ve (-ve if /x/<1)

from 2

x is +ve....insuff

both together

still insuff... E
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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09 Aug 2009, 05:18
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lbsgmat wrote:
If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) x is to the right of 0 on the number line.

St 1. x can be any number but -1 or 0 (according to the question stem). Any other interger +ve or -ve will produce a +ve y
however it is not sufficient since if x is +ve then xy>0, if x is -ve then xy<0 INSF

Statements combined yield only one answer: XY>0
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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09 Aug 2009, 06:39
my ans would be c too, pls post the ans
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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Updated on: 09 Aug 2009, 12:53
yezz wrote:
lbsgmat wrote:
If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) x is to the right of 0 on the number line.

does xy have different signs

y = x^3(x-1), if x -ve then y is +ve , and if x is +ve we have 2 cases y can be either -ve or +ve (-ve if /x/<1)

from 2

x is +ve....insuff

both together

still insuff... E

Marked in Red, since x and y are integers we have to ignore the condition |x|<1. So C(edited) should be the answer as y will always be +ve for any integer x other than 0.

Originally posted by Economist on 09 Aug 2009, 06:49.
Last edited by Economist on 09 Aug 2009, 12:53, edited 1 time in total.
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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09 Aug 2009, 07:04
Economist wrote:
yezz wrote:
lbsgmat wrote:
If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) x is to the right of 0 on the number line.

does xy have different signs

y = x^3(x-1), if x -ve then y is +ve , and if x is +ve we have 2 cases y can be either -ve or +ve (-ve if /x/<1)

from 2

x is +ve....insuff

both together

still insuff... E

Marked in Red, since x and y are integers we have to ignore the condition |x|<1. So A should be the answer as y will always be +ve for any integer x other than 0.

another silly mistake , thanks Econ....C it is
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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10 Aug 2009, 01:32
C it is...

On combining both the stmts you get an answer to the question asked i.e. xy>0....
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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20 Aug 2009, 11:35
If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3
(2) x is to the right of 0 on the number line.

We know xy not equals 0 , it can be +ve or -ve. Possible scenarios are X(+,+,-,-) , Y (+,-,-,+)

Now from Stmt 1 y = X^3(X-1) if X is +ve, then the question is X>1. If yes then Y will be +ve , if not then 0<x<1 then Y will be - ve . No information regarding value of x is given .

If X is -ve, then Y will be + ve irrespective of value of X . So there are two cases with this statement so statement is no sufficient.

From statement X is to right of 0 , and from question statement xy not equals 0 that means neither x nor y is zero.

Combining these two statement means X is positive and is greater than 1 , so xy will be positive

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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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21 Aug 2009, 17:50
1
C fo shiz.

I almost got tripped up in my own trickyness here - need to note that X and Y are integers, therefore 0 < x <1 is not possible. Once you have eliminated this condition, the two statements are sufficient.

These are key things to look out for:

If X and Y are integers
If X and Y are positive integers
If X and Y and different integers

My big tip here is to pay attention to the question and go back and double check the parameters for X and Y.
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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24 Aug 2009, 18:37
1
1.can solve to y = x^4 - x^3 = x^3(x-1)..so xy = x^4(x-1)..whether xy is +ve or -ve depends on value of x,as x^ 4 will always be +ve
2.x is +ve,dont know anything about y

combining xy is > 0
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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20 Nov 2011, 15:10
I) x^4-x^3 is always greater than or equal to 0. But in both cases xy must be equal to 0. So this case is impossible. That means y is always greater than 0. INSUFF.
2) if x is at right of 0 that means x is greater than 0.INSUFF.
Both together, x>0 and y>0 xy>0. C
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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30 Nov 2011, 10:35
1
1) says y=X^4-x^3
or
y=x^3(x-1)
multiply both sides by x
xy=x^4(x-1)
since we do not know anything about x it can be positive or negative. x^4 is always +ve. hence insufficient
2) Says X is >=1 does not say anything about y. Hence insufficient

Both together,
xy=x^4(x-1)
x^4 is def. +ve
(x-1) >=0 if x>=1

hence xy>=0 hence it CANNOT be < 0

Hence C.
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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24 Sep 2013, 06:19
In statement one you forgot to put the "^" sign!!!
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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27 Aug 2014, 07:20
Quote:
I think the OA is incorrect here.

1) $$y = x^4 - x^3$$. Thus, we know that $$xy = x(x^4 - x^3) = x^5 - x^4 = x^4(x - 1)$$. Our question is then, is $$x^4(x-1)<0?$$

Since we have no idea about x this is clearly insufficient.

2) If y is positive, then NO, but if y is negative then YES. Insufficient.

Taking the two statements together. We need to find out if $$x^4(x - 1)<0$$.

To find where this function changes signs, we set it equal to zero and then test values between our critical points. $$x^4(x-1)=0$$ gives us x = 0 and x = 1.

Testing on our number line: for x<0 we see that $$xy = x^4(x-1)$$ is negative; for 0<x<1, we see that xy is STILL negative; for x>1 we see that xy IS POSITIVE.

Therefore, simply knowing that x is positive does not provide us with enough information. We still need to know whether x>1 or x<1.

Edit: Just saw that x and y are integers! Very sneaky! I'm leaving this post because I think it's valuable to see this thought process anyway.
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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08 Sep 2015, 08:52
Question Stem - Does X and Y are opposite signs ?

Statement 1 - Y = (X^4 - X^3)

X^3(X - 1) = X cannot take the value 1 as Y not equal to zero.

Whether X is negative or positive Y will always be positive. X = 2 then Y = 8. If X = -2 then Y = 8. Therefore not suff...

Statement 2- X is positive
Nothing on Y therefore Not Suff...

Combined Y is always positive and x is positive , therefore suff...
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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15 Sep 2015, 03:36
here X and Y are integers. , xy doesn't equal to 0 .
It means either of the integers is not 0.
It is xy<0.

Two possibilities are there
1: x>0 and Y<0.
2; X<0 and Y>0.

Statement 1
y=X^4-X^3.
Minimum value of X we can take
in positive integers is 2 and in negative integer is -1.

if we take X=1, then Y=0. it is not acceptable.

if X=2,X>0. Y=16-8=8.

take If X=-1 then, Y=1-(-1)=1+1+2.

In both cases Y>0. we dont know x is +ve or -ve.so not sufficient.

Statement 2
X is right to 0 on number line . it means X>0 is +ve . but we dont know value of y.

Both statements combine X>0 and Y>0. so XY>0

So option c is correct.
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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28 May 2017, 02:44
lbsgmat wrote:
If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3

(2) x is to the right of 0 on the number line

1 - y= x^3(x-1) since question says x and y are integers and different from 0 you have 2 possible solution
if x<0, y>0 so xy<0
if x>0 (or x>=1) , y>0 so xy>0
1 NOT SUFF

2- CLEARLY NOT SUFF

1&2 -2 force the solution --> x>0, y<0 and so xy < 0
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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28 May 2017, 19:55
lbsgmat wrote:
If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3

(2) x is to the right of 0 on the number line

The goal is to determine if xy < 0 or if xy => 0.

Statement 1) y = x^3(x-1)

Now test cases for x

x = -1, y = -1(-2) = 2, xy = -2

x = 2, y = 8(1) = 8, xy = 16.

Not sufficient.

Statement 2) Rephrased it states that x > 0 or x is positive. However, we do not know the value of y. Insufficient.

Statements 1+2) Now we know that the cases in which x < 0 are invalid, so now y = x^3(x-1), and x != 1, so this solution is always positive. Sufficient.
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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27 Jul 2018, 07:29
Can someone help with a clearer solution with explanation?
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Re: If x and y are integers and xy does not equal 0, is xy <  [#permalink]

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02 Oct 2018, 02:10
shivangis wrote:
Can someone help with a clearer solution with explanation?

We have to check xy < 0.

Given xy != 0 and Integers.

So x!= 0 and y!= 0

Cond-1 : y = x^4 -x^3

One point to note is y != 0 Hence x != 1

As x!= 0 we can safely multiply x on both the sides.

xy= x^5 - x^3 = x^4( x - 1 )

Now as X^4 is always positive we can focus on (x-1)

Thus sign of xy depends on sign of ( x - 1 )
So if x<1 => xy < 1 and if x> 1 => xy > 1

Insufficient

Cond-2 : x in on the right side of the 0 i.e. x > 0

But nothing is said about y and hence xy can be less than or greater than 0

Insufficient

On Mixing both of them:-

x > 0 ( Given by Cond-2 ) and x > 1 ( Deduced by Cond-1 ), we can say x > 1 ( always )

Hence xy = x^4 ( x -1 ) shall always be positive. Hence both of the conditions are sufficient to answer the question.
Re: If x and y are integers and xy does not equal 0, is xy <   [#permalink] 02 Oct 2018, 02:10
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