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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 25 Jun 2016, 19:39
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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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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 17 Dec 2016, 16:22
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Is there any other solution that can be applied to this problem within 2 minutes?
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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 18 Dec 2016, 00:06
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anairamitch1804
Is there any other solution that can be applied to this problem within 2 minutes?
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Not every problem has silver bullet solution. It is indeed a difficult question.
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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 18 Dec 2016, 19:08
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Bunuel
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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.
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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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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 18 Dec 2016, 23:14
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Bunuel
rohitgoel15
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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You should read the question carefully: w, x, y, and z are integers.
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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 18 Dec 2016, 23:16
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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 14 Nov 2017, 04:19
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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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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 25 Aug 2020, 03:57
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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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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 01 Oct 2020, 03:37
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Bunuel
rohitgoel15
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.
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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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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 01 May 2026, 09:14
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Experts, Is this question GMAT focus relevant?
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Bunuel
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w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 01 May 2026, 09:17
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imaityaroy
Experts, Is this question GMAT focus relevant?
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Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.­
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