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Manager
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w, x, y, and z are integers. If z > y > x > w, is w > x^2
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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
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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
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18 Dec 2016, 00:06
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
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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 \(z>y>x>w\): wxyzQuestion: 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
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18 Dec 2016, 23:14
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 \(z>y>x>w\): wxyzQuestion: 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
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18 Dec 2016, 23:16
Many thanks for your reply [SMILING FACE WITH SMILING EYES] Sent from my SMJ710F 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
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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
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20 Apr 2018, 12:47
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 \(z>y>x>w\): wxyzQuestion: 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. brunelhow to attempt such problems involving 4 variable?Because finding combinations that will make answer YES AND NO can be tedious sometimes.any tips to approach such problems?



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Joined: 25 Jun 2018
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w, x, y, and z are integers. If z > y > x > w, is w > x^2
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18 Oct 2018, 05:39
How is statement 2 sufficient for the values z=3, y=1, x=0 and w=4?
here, z>y>x>w and all the assumed values are integers. Please help




w, x, y, and z are integers. If z > y > x > w, is w > x^2
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18 Oct 2018, 05:39



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