MY favourite XYZ

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

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.

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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Source: Beat The GMAT — Data Sufficiency | Mon Oct 04, 2010 5:22 am

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

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?

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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ramannjit: Senior | Next Rank: 100 Posts; Posts: 47; Joined: Sun May 09, 2010 7:21 am; Location: NewDelhi; Thanked: 1 times

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.

Ramannjit

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kapil403: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Tue Jul 20, 2010 10:51 am

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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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

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.

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

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.

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

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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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Mon Oct 04, 2010 9:21 am

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
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

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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GMATMadeEasy: Legendary Member; Posts: 768; Joined: Mon Nov 30, 2009 3:46 amFollowed by: 7 members; Thanked: 21 times

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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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

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.

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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ramannjit: Senior | Next Rank: 100 Posts; Posts: 47; Joined: Sun May 09, 2010 7:21 am; Location: NewDelhi; Thanked: 1 times

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

Ramannjit

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anantbhatia: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Sat Feb 06, 2010 11:56 pm; Thanked: 5 times

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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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

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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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

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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