Find all School-related info fast with the new School-Specific MBA Forum

It is currently 26 Jul 2014, 09:25

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

w, x, y, and z are integers. If z > y > x > w, is |w| > x^2

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
2 KUDOS received
Director
Director
avatar
Status: Preparing for the 4th time -:(
Joined: 25 Jun 2011
Posts: 563
Location: United Kingdom
Concentration: International Business, Strategy
GMAT Date: 06-22-2012
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 12

Kudos [?]: 273 [2] , given: 217

GMAT Tests User
w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 01 Dec 2011, 00:52
2
This post received
KUDOS
9
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

25% (08:29) correct 75% (02:22) wrong based on 507 sessions
w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?

(1) wx > yz
(2) zx > wy
[Reveal] Spoiler: OA

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610 :-(

VP
VP
avatar
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1365
Followers: 10

Kudos [?]: 131 [0], given: 10

GMAT Tests User
Re: w, x, y, and z are integers [#permalink] New post 02 Feb 2012, 00:09
check with substitutions.
a. let z,y,x,w be -1,-2,-8,-10 | -100
we get yes and no for w = -10 and -100 respectively.
Insufficient.

b. let z,y,x,w be 2,1|2,-8,-20
we get the same result for y= 1and 2 respectively.( y is positive otherwise the condition will not fulfill.
Sufficient.

Thus B it is.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!


Last edited by amit2k9 on 03 Feb 2012, 00:51, edited 2 times in total.
Senior Manager
Senior Manager
User avatar
Joined: 19 Apr 2011
Posts: 291
Schools: Booth,NUS,St.Gallon
Followers: 4

Kudos [?]: 64 [0], given: 51

GMAT Tests User
Re: w, x, y, and z are integers [#permalink] New post 02 Feb 2012, 04:24
Either statement alone is sufficient to solve the question
For statement 1)
2>1>-2>-3
where wx>yz.
The answer is true for |w| > x2 > |y| > z2 .
Sufficient.

Similarly for 2 assume some number such that the given condition is satisfied .
The answer is false for |w| > x2 > |y| > z2.

Hence each statement alone is enough.
_________________

+1 if you like my explanation .Thanks :)

Expert Post
16 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18746
Followers: 3251

Kudos [?]: 22454 [16] , given: 2619

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 02 Feb 2012, 06:26
16
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
rohitgoel15 wrote:
w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?

(1) wx > yz
(2) zx > wy


Good question. +1.

Answer to the question is neither E nor D, it's B.

w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?

Given: w, x, y, and z are integers and z>y>x>w: --w--x--y--z--
Question: is |w|>x^2>|y|>z^2?

Now, since z>y, then in order |y|>z^2 to hold true y must be negative and since y>x>w, then x and w must be negative. So in order the answer to be YES, y, x and w must be negative (notice that the reverse is not always true: they might be negative but the answer could be NO, but if answer is YES then they must be negative). So if we get that either of them is not negative then we could conclude that the answer to the question is NO.

(1) wx > yz --> the product of two smaller integers (w and x) is more than the product of to larger integers (y and z). It's possible for example that all but z are negative but even in this case we could have two different answers:
If (z=1)>(y=-2)>(x=-3)>(w=-4) --> in this case answer to the question "is |w|>x^2>|y|>z^2?" will be NO;
If (z=1)>(y=-2)>(x=-3)>(w=-10) --> in this case answer to the question "is |w|>x^2>|y|>z^2?" will be YES.
Not sufficient.

(2) zx > wy --> y is positive, because if it's negative then x and w are also negative (since y>x>w) and in this case no matter whether z is positive or negative: zx<wy (if z>0 then zx<0<wy and if z<0 then the product of two "more negative" numbers w and y will be more then the product of two "less negative" numbers z and x, so again we would have: zx<wy). Thus we have that y must be positive, which makes |w|>x^2>|y|>z^2 impossible (as discussed), hence the answer to the question is NO. Sufficient,

Answer: B.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 11 Apr 2011
Posts: 50
Followers: 0

Kudos [?]: 1 [0], given: 3

GMAT Tests User
Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 09 Feb 2012, 18:44
this question is pretty tough. took too much time substituting numbers while staying within the statements. :shock:
Manager
Manager
avatar
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 51
Concentration: General Management, Leadership
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

Kudos [?]: 5 [0], given: 118

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 24 Oct 2012, 22:35
(2) zx > wy --> y is positive, because if it's negative then x and w are also negative (since y>x>w) and in this case no matter whether z is positive or negative: zx<wy (if z>0 then zx<0<wy and if z<0 then the product of two "more negative" numbers w and y will be more then the product of two "less negative" numbers z and x, so again we would have: zx<wy). Thus we have that y must be positive, which makes |w|>x^2>|y|>z^2 impossible (as discussed), hence the answer to the question is NO. Sufficient,

Answer: B.

Hope it's clear.[/quote]


sorry but just trying to understand more in this option :) as this is really trick for me

if we plug z=4,y=3,x=2,w=1 then zx>wy seems positive & also if we plug z=4,y=1,x=-2,w=-3 then zx>wy fails.. so how we could say y is positive & proceed with option B.

Please correct me if i am wrong :|
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18746
Followers: 3251

Kudos [?]: 22454 [1] , given: 2619

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 25 Oct 2012, 03:34
1
This post received
KUDOS
Expert's post
breakit wrote:
sorry but just trying to understand more in this option :) as this is really trick for me

if we plug z=4,y=3,x=2,w=1 then zx>wy seems positive & also if we plug z=4,y=1,x=-2,w=-3 then zx>wy fails.. so how we could say y is positive & proceed with option B.

Please correct me if i am wrong :|


From z > y > x > w (given) and zx > wy (given) it follows that y must be positive. But it does not mean that if z > y > x > w and y>0, then zx > wy must hold true.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

1 KUDOS received
Senior Manager
Senior Manager
avatar
Joined: 06 Aug 2011
Posts: 408
Followers: 2

Kudos [?]: 52 [1] , given: 82

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 26 Oct 2012, 07:44
1
This post received
KUDOS
Verrrry Very tricky..!!! ..got wrong :(. Thanks bunuel for such a nice explanation !!
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 175

Kudos [?]: 893 [0], given: 235

GMAT Tests User Reviews Badge
Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 26 Oct 2012, 09:02
I would have guessed it C on real Gmat and moved on :)..... very difficult to answer in 2 mins and not a real Gmat question. But Bunuel explained it very well .
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Manager
avatar
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 51
Concentration: General Management, Leadership
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

Kudos [?]: 5 [0], given: 118

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 26 Oct 2012, 11:21
Bunuel wrote:
breakit wrote:
(2) zx > wy --> y is positive, because if it's negative then x and w are also negative (since y>x>w) and in this case no matter whether z is positive or negative: zx<wy (if z>0 then zx<0<wy and if z<0 then the product of two "more negative" numbers w and y will be more then the product of two "less negative" numbers z and x, so again we would have: zx<wy). Thus we have that y must be positive, which makes |w|>x^2>|y|>z^2 impossible (as discussed), hence the answer to the question is NO. Sufficient,

Answer: B.

Hope it's clear.



sorry but just trying to understand more in this option :) as this is really trick for me

if we plug z=4,y=3,x=2,w=1 then zx>wy seems positive & also if we plug z=4,y=1,x=-2,w=-3 then zx>wy fails.. so how we could say y is positive & proceed with option B.

Please correct me if i am wrong :|


From z > y > x > w (given) and zx > wy (given) it follows that y must be positive. But it does not mean that if z > y > x > w and y>0, then zx > wy must hold true.

Hope it's clear.[/quote]




thanks a lot :)
Manager
Manager
User avatar
Joined: 02 Sep 2010
Posts: 50
Location: India
Followers: 0

Kudos [?]: 41 [0], given: 17

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 27 Oct 2012, 05:46
Bunuel wrote:
breakit wrote:
(2) zx > wy --> y is positive, because if it's negative then x and w are also negative (since y>x>w) and in this case no matter whether z is positive or negative: zx<wy (if z>0 then zx<0<wy and if z<0 then the product of two "more negative" numbers w and y will be more then the product of two "less negative" numbers z and x, so again we would have: zx<wy). Thus we have that y must be positive, which makes |w|>x^2>|y|>z^2 impossible (as discussed), hence the answer to the question is NO. Sufficient,

Answer: B.

Hope it's clear.



sorry but just trying to understand more in this option :) as this is really trick for me

if we plug z=4,y=3,x=2,w=1 then zx>wy seems positive & also if we plug z=4,y=1,x=-2,w=-3 then zx>wy fails.. so how we could say y is positive & proceed with option B.

Please correct me if i am wrong :|


From z > y > x > w (given) and zx > wy (given) it follows that y must be positive. But it does not mean that if z > y > x > w and y>0, then zx > wy must hold true.

Hope it's clear.[/quote]
Amazing explanation again, Bunnel.
But can this be a real GMAT question?
_________________

The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!

Intern
Intern
avatar
Joined: 06 Apr 2011
Posts: 13
Followers: 0

Kudos [?]: 0 [0], given: 292

GMAT Tests User
Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 18 Nov 2012, 11:38
thanks Bunuel for excellent exp.
But how realistic the question is from GMAT perspective?
I re-attempted the question after reviewing your solution and still took me slightly more than 4 mins.

Thanks:)
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18746
Followers: 3251

Kudos [?]: 22454 [0], given: 2619

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 14 Jul 2013, 23:40
Expert's post
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 476
Followers: 1

Kudos [?]: 44 [0], given: 134

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 15 Jul 2013, 12:22
w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?


(1) wx > yz

If wx>yz then w and x must both be negative and their absolute values must be larger than the values of y and z respectively. If w is the smallest number in the group (i.e. -10) but it's absolute value is the greatest (as it must be because wx>yz then |x|>|y| Also, x^2 and y^2 will both be positive.

However, we can't be sure if |w| > x^2 > |y| > z^2? For example:

z > y > x > w
4 > 3 > -6 > -10

|w| > x^2 > |y| > z^2
|-10| > -6^2 > |3| > 4^2
10 > 36 > 3 > 16
For obvious reasons, that isn't valid.

z > y > x > w
1 > -2 > -3 > -14

|w| > x^2 > |y| > z^2
|-14| > -3^2 > |-2| > 1^2
14 > 9 > 2 > 1
This is valid. So, y could be positive or negative but w and y must be negative and their absolute values must be greater than xy.
INSUFFICIENT

(2) zx > wy

z > y > x > w
Values are not all negative

z > y > x > w
5 > 4 > 3 > 2
zx > wy
15 > 8 VALID

z > y > x > w
5 > 4 > 3 > -2
zx > wy
15 > -8 VALID

z > y > x > w
5 > 4 > -2 > -3
-10 > -8 INVALID

Either all z, y, x, w are positive or z, y, x are positive and w is negative.

is |w| > x^2 > |y| > z^2?

If z, y, x, w are all positive then |w| > x^2 > |y| > z^2? cannot be true
If z, y, x are positive and w is negative then |w| > x^2 > |y| > z^2? still cannot be true:
z > y > x > w
7 > 6 > 5 > -20

|w| > x^2 > |y| > z^2?
|-20| > 5^2 > |6| > 7^2
20 > 25 > 6 > 49

(B)

While I did get the question right, it took a while to do and I am not 100% positive if my methodology is correct. Could someone tell me if the way I solved it was correct?
1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 476
Followers: 1

Kudos [?]: 44 [1] , given: 134

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 20 Jul 2013, 13:12
1
This post received
KUDOS
Different way of solving, not by plugging in numbers.

w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?

(1) wx > yz

As opposed to plugging in numbers (which would take a very long time) we should try and analyze |w| > x^2 > |y| > z^2 to see where it may or may not hold true.

One way for wx to be greater than yz (despite w and x being less than y and z on the number line) is for BOTH w and x to be negative. y could be positive or negative but z has to be positive. The other way is for all w, x, y, z to be negative.

|y| has to be greater than z^2 for the question to be true. If y is positive (and therefore, z is positive and greater than y) this isn't possible because a positive number (y) cannot be greater than the square of a larger number (Z) (assuming all values are integers which they are) If y is positive, the answer to the question is "NO"

If y is negative, (in which case z could be negative or positive) the question is possible. For example, if y=-3 and z=1, z would be greater than y but |y| would be greater than 1^2. This means the answer to the question could be "YES" In other words, it's possible that |y|>|z| but y<z.
INSUFFICIENT

(2) zx > wy

We know that all z,y,x,w cannot be negative because zx < wy as opposed to the other way around. The only way for zx > wy is for all but w to be positive so:

If z is positive and we know that z,y,x are positive and x is negative then we know that |w| > x^2 > |y| > z^2 never holds true. If z is greater than the other two positives (y and z) then there is no way z>2 will be less than |y| (because z cannot be a fraction as we are told all variables are integers) Therefore, this inequality fails every time regardless of values of y,x,w.
SUFFICIENT

(B)

(p.s. are there any problems similar to this which I can practice on?)
Intern
Intern
avatar
Joined: 19 Apr 2014
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 80

GMAT ToolKit User
Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 02 Jun 2014, 12:15
Bunuel wrote:
rohitgoel15 wrote:
w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?

(1) wx > yz
(2) zx > wy


Good question. +1.

Answer to the question is neither E nor D, it's B.

w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?

Given: w, x, y, and z are integers and z>y>x>w.
Question: is |w|>x^2>|y|>z^2?

Now, since z>y, then in order |y|>z^2 to hold true y must be negative and since y>x>w, then x and w must be negative. So in order the answer to be YES, y, x and w must be negative (notice that the reverse is not always true: they might be negative but the answer could be NO, but if answer is YES then they must be negative). So if we get that either of them is not negative then we could conclude that the answer to the question is NO.

(1) wx > yz --> the product of two smaller integers (w and x) is more than the product of to larger integers (y and z). It's possible for example that all but z are negative but even in this case we could have two different answers:
If (z=1)>(y=-2)>(x=-3)>(w=-4) --> in this case answer to the question "is |w|>x^2>|y|>z^2?" will be NO;
If (z=1)>(y=-2)>(x=-3)>(w=-10) --> in this case answer to the question "is |w|>x^2>|y|>z^2?" will be YES.
Not sufficient.

(2) zx > wy --> y is positive, because if it's negative then x and w are also negative (since y>x>w) and in this case no matter whether z is positive or negative: zx<wy (if z>0 then zx<0<wy and if z<0 then the product of two "more negative" numbers w and y will be more then the product of two "less negative" numbers z and x, so again we would have: zx<wy). Thus we have that y must be positive, which makes |w|>x^2>|y|>z^2 impossible (as discussed), hence the answer to the question is NO. Sufficient,

Answer: B.

Hope it's clear.




Hi Bunuel,
It would be great if you could help me with this one thing. I love to solve question of inequalities using a number line approach. I plotted the number line thus

<----w-------x-------y--------z------> and read the question. For me the question translated to basically asking where is the 0 on this on number line in respect of y?
Because |y| > z^2 only if y<0.


After that I simply told myself if I get options that state definitely where the zero is Ill chose it without definitely knowing the traditional solution. So I look at the options do a little imagination of the following sort :

<----- (-6) --- (-3) ---- (-2) ---- 0 ----(2)----(3) -----(6)------>

1) wx>yz : This statement tells me (i) <------(w)---(x)--------------------------------------------------(0)---(y)--(z)--> (kind of to scale)
or (ii) <-----(w)---(x)-----(y)---0----(z)--------->

Therefore, INSUFFICIENT.

2) zx>wy: (i)<---------(w)------(x)-------(0)-------(y)------(z)

and I just know that B is sufficient to tell me where the zero is.

A lot of this is based on intuition. I know it might not be the best strategy. However, it works for me.

To further improve my skills w.r.t. this method please help me out by finding weaknesses in the reasoning and ways to improve it. It would be great.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18746
Followers: 3251

Kudos [?]: 22454 [0], given: 2619

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post 03 Jun 2014, 08:00
Expert's post
prashantgupta23 wrote:
Bunuel wrote:
rohitgoel15 wrote:
w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?

(1) wx > yz
(2) zx > wy


Good question. +1.

Answer to the question is neither E nor D, it's B.

w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 > |y| > z^2?

Given: w, x, y, and z are integers and z>y>x>w.
Question: is |w|>x^2>|y|>z^2?

Now, since z>y, then in order |y|>z^2 to hold true y must be negative and since y>x>w, then x and w must be negative. So in order the answer to be YES, y, x and w must be negative (notice that the reverse is not always true: they might be negative but the answer could be NO, but if answer is YES then they must be negative). So if we get that either of them is not negative then we could conclude that the answer to the question is NO.

(1) wx > yz --> the product of two smaller integers (w and x) is more than the product of to larger integers (y and z). It's possible for example that all but z are negative but even in this case we could have two different answers:
If (z=1)>(y=-2)>(x=-3)>(w=-4) --> in this case answer to the question "is |w|>x^2>|y|>z^2?" will be NO;
If (z=1)>(y=-2)>(x=-3)>(w=-10) --> in this case answer to the question "is |w|>x^2>|y|>z^2?" will be YES.
Not sufficient.

(2) zx > wy --> y is positive, because if it's negative then x and w are also negative (since y>x>w) and in this case no matter whether z is positive or negative: zx<wy (if z>0 then zx<0<wy and if z<0 then the product of two "more negative" numbers w and y will be more then the product of two "less negative" numbers z and x, so again we would have: zx<wy). Thus we have that y must be positive, which makes |w|>x^2>|y|>z^2 impossible (as discussed), hence the answer to the question is NO. Sufficient,

Answer: B.

Hope it's clear.




Hi Bunuel,
It would be great if you could help me with this one thing. I love to solve question of inequalities using a number line approach. I plotted the number line thus

<----w-------x-------y--------z------> and read the question. For me the question translated to basically asking where is the 0 on this on number line in respect of y?
Because |y| > z^2 only if y<0.


After that I simply told myself if I get options that state definitely where the zero is Ill chose it without definitely knowing the traditional solution. So I look at the options do a little imagination of the following sort :

<----- (-6) --- (-3) ---- (-2) ---- 0 ----(2)----(3) -----(6)------>

1) wx>yz : This statement tells me (i) <------(w)---(x)--------------------------------------------------(0)---(y)--(z)--> (kind of to scale)
or (ii) <-----(w)---(x)-----(y)---0----(z)--------->

Therefore, INSUFFICIENT.

2) zx>wy: (i)<---------(w)------(x)-------(0)-------(y)------(z)

and I just know that B is sufficient to tell me where the zero is.

A lot of this is based on intuition. I know it might not be the best strategy. However, it works for me.

To further improve my skills w.r.t. this method please help me out by finding weaknesses in the reasoning and ways to improve it. It would be great.


Yes, your logic makes sense. Notice that your solution is basically the same as mine.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2   [#permalink] 03 Jun 2014, 08:00
    Similar topics Author Replies Last post
Similar
Topics:
3 Experts publish their posts in the topic If w>x>y>z>0, is z>4? BN1989 2 01 Apr 2012, 02:47
23 Experts publish their posts in the topic w, x, y, and z are integers. If w >x>y>z>0, is y a common enigma123 9 26 Jan 2012, 21:19
1 w, x, y, and z are integers. If w > x > y > z > 0, is y a co ashiima 5 10 Dec 2011, 16:51
If x < y, y <z> w, which of the following andrehaui 1 30 May 2007, 05:11
w, x, y, and z are integers. If w > x > y > z > vshaunak@gmail.com 11 20 May 2007, 05:35
Display posts from previous: Sort by

w, x, y, and z are integers. If z > y > x > w, is |w| > x^2

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.