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w, x, y, and z are integers. If z > y > x > w, is w > x^2 [#permalink]
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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
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Re: w, x, y, and z are integers [#permalink]
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Updated on: 03 Feb 2012, 01:51
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 2,12,8,20 we get the same result for y= 1and 2 respectively.( y is positive otherwise the condition will not fulfill. Sufficient. Thus B it is.
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Re: w, x, y, and z are integers [#permalink]
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02 Feb 2012, 05:24
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.
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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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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.
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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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24 Oct 2012, 23:35
(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



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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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25 Oct 2012, 04:34
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.
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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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Verrrry Very tricky..!!! ..got wrong . Thanks bunuel for such a nice explanation !!
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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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26 Oct 2012, 10:02
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 .
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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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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



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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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27 Oct 2012, 06:46
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?
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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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18 Nov 2012, 12:38
thanks Bunuel for excellent exp. But how realistic the question is from GMAT perspective? I reattempted the question after reviewing your solution and still took me slightly more than 4 mins.
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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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15 Jul 2013, 13:22
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?



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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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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?)



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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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02 Jun 2014, 13:15
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 <wxyz> 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.



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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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03 Jun 2014, 09:00
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 <wxyz> 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.
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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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enigma123 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 Statement 1. Let's line up inequalities w<z x<y, if they are all positive we could get wx<yz by multiplying, but we have reverse that means that some of them negative or all negative. If all negative we should know the difference between integers. If w=20 and x=3, then w>x^2 but if w=4 the result will be reverse. INSUFFICIENT Statement 2. Let's line up inequalities z>y x>w, so we get zx>wy meaning that all numbers are positive. So we get that w > x^2 > y > z^2 is impossible. SUFFICIENT B



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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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enigma123 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 Given: z > y > x > w possible cases(as on a number line) for the above statement to be true wxyz0 (1)wxy0z (2)wx0yz (3)w0xyz (4)0wxyz (5) ASKED: Is w > x^2 > y > z^2? if y > z^2 then "y" must be negative{if it is positive then the distance[absolute value] of y cannot cannot be greater than z^2 } so if we can determine that only (1) and (2) are possible OR they are not possible we can get our answer as "y" is negative in only (1),(2) and positive in (3),(4),(5)STATEMENT 1 (1) wx > yz according to this (1),(2),(3) are possible and (4),(5) are not possible in order to get a definitive answer only (1),(2) can be possible or not be possible.As here (1),(2),(3) are possible so we cannot get a answer. INSUFFICIENT STATEMENT 2 (2) zx > wy according to this statement (1),(2) cannot be possible so this statement gives us a definitive answer as a NO that w > x^2 > y > z^2? is not possible. SUFFICIENT HENCE ANSWER is B please correct me if I'm wrong.



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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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21 Jan 2016, 11:25
Finally, i figured it out in 10 minutes.... It is a real matter of "negative, positive judgement of inequation. And it also strongly tests your logical judgement. Very good one.



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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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03 Feb 2016, 05:25
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enigma123 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 Responding to a pm: When you see z > y > x > w, you could consider a number line: wxyz The four integers divide the number line into five parts. 0 could be anywhere on the five parts. Question: is \(w > x^2 > y > z^2\)? (1) wx > yz Think of the number line. When will the product of the 2 smallest numbers be more than the two largest numbers? Say, when all numbers are negative.  (100)  (8)  (5)  (2)  Here, wx > yz Is \(w > x^2 > y > z^2\)? Yes. Now we can make a small change to this and get a No as answer.  (10)  (8)  (5)  (2)  Is \(w > x^2 > y > z^2\)? No. So this statement is clearly not sufficient. (2) zx > wy Now when will the product of the largest and third largest number be greater than the product of the second largest and the smallest number? That's easy, right? One case we can think of right away is when all numbers are positive. wxyz 1234 Is w > x^2 > y > z^2? No. Let's now try to get a Yes case. When will \(w > x^2\)? w must be negative with a large absolute value. If x is 2, w must be 5 or smaller. Let's go on. Now \(x^2\) is already greater than y. But can y be greater than \(z^2\) in this case? No. For y to be greater than \(z^2\), y should be negative with a large absolute value (compared with z). So is it possible that w, x and y are all negative and z is positive or 0? No. In that case zx will be negative (or 0) while wy will be positive. Hence, it is not possible to get a "Yes" case. Sufficient. Answer (B)
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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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03 Feb 2016, 10:32
newconcept123 wrote: Finally, i figured it out in 10 minutes.... It is a real matter of "negative, positive judgement of inequation. And it also strongly tests your logical judgement. Very good one. All said and done... I feel this question is not a GMAT like questions and is strictly a very difficult question even for practice of hard question of GMAT. in my opinion "A waste of time"
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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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