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

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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   25 Jun 2016, 19:39
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..
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   17 Dec 2016, 16:22
Is there any other solution that can be applied to this problem within 2 minutes?
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   18 Dec 2016, 00:06
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anairamitch1804 wrote:
Is there any other solution that can be applied to this problem within 2 minutes?


Not every problem has silver bullet solution. It is indeed a difficult question.
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   18 Dec 2016, 19:08
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.
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   18 Dec 2016, 23:14
Expert Reply
leanhdung 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 \(z>y>x>w\): --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.
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   18 Dec 2016, 23:16
Many thanks for your reply [SMILING FACE WITH SMILING EYES]

Sent from my SM-J710F using GMAT Club Forum mobile app
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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   14 Nov 2017, 04:19
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.
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   25 Aug 2020, 03:57
HI,

In this question, why are we considering just positive and negative? for example, if z=1/4 and y = 1/6 then y<z but mod of y will be greater than Z^2. Shouldn't positive fractions be considered? Please help :)
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   01 Oct 2020, 03:37
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.


Hi, is there a certain trick to look at this? I started with picking numbers and after 5 mins realised that the #2 can work but never satisfy the question. But it took a lot of time when i started with picking numbers. Any alternate approach to start solving the problem?
Thanks.
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink] New post   20 Jan 2024, 09:15
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Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
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