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MY favourite XYZ

by goyalsau » Mon Oct 04, 2010 5:22 am
Is (x^7)(y^2)(z^3)>0 ?
(1) yz<0
(2) xz>0





[spoiler]I marked B but the official answer is C I am not able to understand How?[/spoiler]

Power of y does not matter because it is square

considering 2 statement
xz>0
it means whether they are both +ve or -ve

In any case result will be +ve because both x and z have odd power.
Please HELP me understand the solution.
Source is authentic
But OA C.
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by Rahul@gurome » Mon Oct 04, 2010 6:02 am
(1) xz>0 implies either x and z are both positive or both negative.
If x and z are positive then (x^7)(y^2)(z^3) > 0.
If x and z are negative then (x^7)(y^2)(z^3) < 0.
No unique answer.
So, (1) is NOT SUFFICIENT.

(2) yz < 0 implies that one of y and z is positive and the other one is negative.
If y = positive and z = negative then (y^2)(z^3) = negative, but now we don't have information about x, which can be positive or negative, and this information will change the value of (x^7)(y^2)(z^3).
So, definitely (2) is NOT SUFFICIENT.

Combining (1) and (2), we get the following cases:
Case 1: x = positive, y = negative, z = positive implies (x^7)(y^2)(z^3) > 0.
Case 2: x = negative, y = positive, z = negative implies (x^7)(y^2)(z^3) > 0.
Hence, (x^7)(y^2)(z^3) > 0.

[spoiler]The correct answer is (C).[/spoiler]

Does that help?
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by ramannjit » Mon Oct 04, 2010 6:19 am
goyalsau wrote:Is (x^7)(y^2)(z^3)>0 ?
(1) yz<0
(2) xz>0





[spoiler]I marked B but the official answer is C I am not able to understand How?[/spoiler]

Power of y does not matter because it is square

considering 2 statement
xz>0
it means whether they are both +ve or -ve

In any case result will be +ve because both x and z have odd power.
Please HELP me understand the solution.
Source is authentic
But OA C.
Each statement is as important as the other

(1) xz>0. States either x and z both +ve or both -ve. Consider x and z as +ve then (x^7)(y^2)(z^3) > 0. and now consider x and z are -ve then (x^7)(y^2)(z^3) < 0.so No definite answer. Nt Suff.


(2) yz < 0 states either y is +ve and z is -ve or the other way. Consider y = +ve and z = -ve; then (y^2)(z^3) = -ve. But nothing about "x" which can impact the value. So no definite answer, Nt Suff

(1) and (2),:
A: x = +ve, y = -ve, z = +ve => (x^7)(y^2)(z^3) > 0.
B: x = -ve, y = +ve, z = -ve => (x^7)(y^2)(z^3) > 0.
So, (x^7)(y^2)(z^3) > 0.

So I think C is correct.
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by kapil403 » Mon Oct 04, 2010 7:48 am
How can this be negative?

If x and z are negative then (x^7)(y^2)(z^3) < 0.
(-1)^7 2^2 (-1)^3 = +2
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by Rahul@gurome » Mon Oct 04, 2010 8:03 am
kapil403 wrote:How can this be negative?

If x and z are negative then (x^7)(y^2)(z^3) < 0.
(-1)^7 2^2 (-1)^3 = +2
U r right, here (x^7)(y^2)(z^3) > 0, but since we don't know y, so y can also be zero. In that case, it's not sufficient.
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by goyalsau » Mon Oct 04, 2010 9:04 am
If x and z are positive then (x^7)(y^2)(z^3) > 0.

If x and z are negative then (x^7)(y^2)(z^3) < 0.

I am not able to understand why you are changing the sign.
as i know x^7 will be -ve because x is -ve
and z^3 will be -ve because z is -ve

when -ve will be multiplied to -ve it will be +ve

I know there is something wrong with my concept
Please help me understand that.
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by shovan85 » Mon Oct 04, 2010 9:16 am
goyalsau wrote:If x and z are positive then (x^7)(y^2)(z^3) > 0.

If x and z are negative then (x^7)(y^2)(z^3) < 0.

I am not able to understand why you are changing the sign.
as i know x^7 will be -ve because x is -ve
and z^3 will be -ve because z is -ve

when -ve will be multiplied to -ve it will be +ve

I know there is something wrong with my concept
Please help me understand that.
:) Nothin wrong with your concept boss. Your approach is absolutely correct. The sole LOOPHOLE is you forgot the case y=0. Unless u take both options into account you cannot say anything about y.
U know y^2 >0 irrespective of y is +ve or -ve.
What about y = 0 the whole expression is getting 0 so u cannot say (x^7)(y^2)(z^3) > 0.

The job of first option is to let you know y is not equal to 0. :)

Hope I made my point
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by goyalsau » Mon Oct 04, 2010 9:21 am
:) Nothin wrong with your concept boss. Your approach is absolutely correct. The sole LOOPHOLE is you forgot the case y=0. Unless u take both options into account you cannot say anything about y.
U know y^2 >0 irrespective of y is +ve or -ve.
What about y = 0 the whole expression is getting 0 so u cannot say (x^7)(y^2)(z^3) > 0.

The job of first option is to let you know y is not equal to 0. :)

Hope I made my point

Thanks Buddy,
Now i know that i need 1 statement just to make sure that y is not equal to zero.

but why Rahul and ramanjit are changing the sign.


If x and z are positive then (x^7)(y^2)(z^3) > 0.


If x and z are negative then (x^7)(y^2)(z^3) < 0.

They are not talking about zero they are just flipping the sigh... I want to know what is the reason behind this flipping for sign..
Saurabh Goyal
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by shovan85 » Mon Oct 04, 2010 9:27 am
goyalsau wrote:
Thanks Buddy,
Now i know that i need 1 statement just to make sure that y is not equal to zero.

but why Rahul and ramanjit are changing the sign.


If x and z are positive then (x^7)(y^2)(z^3) > 0.


If x and z are negative then (x^7)(y^2)(z^3) < 0.

They are not talking about zero they are just flipping the sigh... I want to know what is the reason behind this flipping for sign..
Sign cannot be flipped... I m sure of it -ve * -ve is +ve
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by GMATMadeEasy » Mon Oct 04, 2010 2:20 pm
@Originial poster:
If x and z are negative then (x^7)(y^2)(z^3) < 0.

They are not talking about zero they are just flipping the sigh... I want to know what is the reason behind this flipping for sign..
You are right, they have written by mistake.

If X and Z are negative, (x^7)(y^2)(z^3) > 0 .

YOu are correct in your reasoning.
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by Rahul@gurome » Mon Oct 04, 2010 3:27 pm
GMATMadeEasy wrote: You are right, they have written by mistake.

If X and Z are negative, (x^7)(y^2)(z^3) > 0 .

YOu are correct in your reasoning.
U r right, that's mistake. Thanks for pointing that out.
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by ramannjit » Mon Oct 04, 2010 4:46 pm
goyalsau wrote::) Nothin wrong with your concept boss. Your approach is absolutely correct. The sole LOOPHOLE is you forgot the case y=0. Unless u take both options into account you cannot say anything about y.
U know y^2 >0 irrespective of y is +ve or -ve.
What about y = 0 the whole expression is getting 0 so u cannot say (x^7)(y^2)(z^3) > 0.

The job of first option is to let you know y is not equal to 0. :)

Hope I made my point

Thanks Buddy,
Now i know that i need 1 statement just to make sure that y is not equal to zero.

but why Rahul and ramanjit are changing the sign.


If x and z are positive then (x^7)(y^2)(z^3) > 0.


If x and z are negative then (x^7)(y^2)(z^3) < 0.

They are not talking about zero they are just flipping the sigh... I want to know what is the reason behind this flipping for sign..


IT IS A MISTAKE :(
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by anantbhatia » Tue Oct 05, 2010 9:13 am
(x^7)(y^2)(z^3)
=(xyz)^2 (x^5) z
=(xyz)^2 (x^4) xz

So, the first two terms will always be postive if x,y,z are not= 0.

(1) is not sufficient to know what xz would be.
(2) would have been sufficient if we knew that x,y,z are not=0

combining, we get that none of x,y,z is equal to 0. and xz<0. So we know for sure that it will be a negative result.
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by diebeatsthegmat » Tue Oct 05, 2010 8:44 pm
Rahul@gurome wrote:(1) xz>0 implies either x and z are both positive or both negative.
If x and z are positive then (x^7)(y^2)(z^3) > 0.
If x and z are negative then (x^7)(y^2)(z^3) < 0.
No unique answer.
So, (1) is NOT SUFFICIENT.

(2) yz < 0 implies that one of y and z is positive and the other one is negative.
If y = positive and z = negative then (y^2)(z^3) = negative, but now we don't have information about x, which can be positive or negative, and this information will change the value of (x^7)(y^2)(z^3).
So, definitely (2) is NOT SUFFICIENT.

Combining (1) and (2), we get the following cases:
Case 1: x = positive, y = negative, z = positive implies (x^7)(y^2)(z^3) > 0.
Case 2: x = negative, y = positive, z = negative implies (x^7)(y^2)(z^3) > 0.
Hence, (x^7)(y^2)(z^3) > 0.

[spoiler]The correct answer is (C).[/spoiler]

Does that help?
sorry i dont understand the xz part
from the info we know y^2>0 with all y,
xz>0 so z,x must be both + or both (-)
if x<0 and z<0 z^3*x^7>0 so x^7*y^2*z^3>0 ( yes)
if x>0 and y>0 of course x^7y^2z^3>0
so i think this condition is sufficient....
why is it not/???? i dont get it, please give me a fast explanation
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by diebeatsthegmat » Tue Oct 05, 2010 8:46 pm
Rahul@gurome wrote:
GMATMadeEasy wrote: You are right, they have written by mistake.

If X and Z are negative, (x^7)(y^2)(z^3) > 0 .

YOu are correct in your reasoning.
U r right, that's mistake. Thanks for pointing that out.
now i get it....
if y=0 we can say nothing
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