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:

w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]

Show Tags

25 Jun 2016, 18:39

Took me over 12 mins to solve by substitution.. This teaches us how elegant some DS Questions can be on the GMAT and how substitution may not be the best strategy on exam day..

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]

Show Tags

18 Dec 2016, 18:08

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 \(z>y>x>w\): --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.

I believe your statement "Now, since \(z>y\), then in order \(|y|>z^2\) to hold true \(y\) must be negative.." is wrong.

if 0 < y < z < 1 , then these y, z also satisfy \(z>y\) and \(|y|>z^2\) (for example y = 0.5 and z = 0.6) so it is not necessary for y to be negative.
_________________

Do not pray for an easy life, pray for the strength to endure a difficult one - Bruce Lee

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.

I believe your statement "Now, since \(z>y\), then in order \(|y|>z^2\) to hold true \(y\) must be negative.." is wrong.

if 0 < y < z < 1 , then these y, z also satisfy \(z>y\) and \(|y|>z^2\) (for example y = 0.5 and z = 0.6) so it is not necessary for y to be negative.

You should read the question carefully: w, x, y, and z are integers.
_________________

w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]

Show Tags

14 Nov 2017, 03:19

this question is extremely hard. Ones will not solve such question within 2 mins even if the person knows the solution already. Nevertheless, there is still a good chance that this question will appear in the actual test.