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Re: w, x, y, and z are integers [#permalink]
02 Feb 2012, 04:24

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.
_________________

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
02 Feb 2012, 06:26

12

This post received KUDOS

Expert's post

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,

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
24 Oct 2012, 22:35

(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.

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
25 Oct 2012, 03:34

1

This post received KUDOS

Expert's post

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.

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
26 Oct 2012, 09:02

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 .
_________________

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
26 Oct 2012, 11:21

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.

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
27 Oct 2012, 05:46

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?
_________________

The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
18 Nov 2012, 11:38

thanks Bunuel for excellent exp. But how realistic the question is from GMAT perspective? I re-attempted the question after reviewing your solution and still took me slightly more than 4 mins.

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
15 Jul 2013, 12:22

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:

|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

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?

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2 [#permalink]
20 Jul 2013, 13:12

1

This post received KUDOS

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?)

gmatclubot

Re: w, x, y, and z are integers. If z > y > x > w, is |w| > x^2
[#permalink]
20 Jul 2013, 13:12