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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   55% (hard)

Question Stats: 57% (01:06) correct 43% (01:34) wrong based on 255 sessions

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If $$xyz \ne 0$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2}) \lt 0$$?

(1) $$\sqrt{y} \gt \sqrt{x^2}$$

(2) $$y^3 \gt \frac{1}{z{^4}}$$

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Official Solution:

If $$xyz \ne 0$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2}) \lt 0$$?

$$xyz \ne 0$$ means that neither of the unknowns is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2} \lt 0$$? Since $$x^4$$ and $$z^2$$ are positive numbers, then the question boils down to whether $$\sqrt{y} \lt 0$$, which is the same as whether $$y \lt 0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} = 5$$ and $$\sqrt{-64} = -4$$).

(1) $$\sqrt{y} \gt \sqrt{x^2}$$. As even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y} \gt \sqrt{x^2} \gt 0$$, (or which is the same as $$y \gt 0$$). Therefore, the answer to the original question is NO. Sufficient.

(2) $$y^3 \gt \frac{1}{z{^4}}$$. The same here as $$\frac{1}{z{^4}} \gt 0$$ then $$y^3 \gt \frac{1}{z{^4}} \gt 0$$, (or which is the same as $$y \gt 0$$). Therefore, the answer to the original question is NO. Sufficient.

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HI BUNUEL
FOR THE BEWLO QUESTION AND SOLUTION PROVIDED
FOR STTEMENT 1 LETS SUPPOSE X^2 =625 ....THEN 4 TH ROOT OF X^2 THAT IS (X^2)1/4 MEANS IT CAN BE +5 OR -5 RIGHT SO HOW (Y)^1/5 > (X^2)1/4 PROVES THAT Y IS GREATER THAN ZERO PLS ADVISE

QUESTION AND ANSWER AS POSTED ON THREAD
xyz≠0 means that neither of the unknowns is equal to zero. Next, (x−4)∗(3√y)∗(z−2)=3√yx4∗z2, so the question becomes: is 3√yx4∗z2<0? Since x4 and z2 are positive numbers, then the question boils down to whether 3√y<0, which is the same as whether y<0 (recall that odd roots have the same sign as the base of the root, for example: 3√125=5 and 3√−64=−4).

(1) 5√y>4√x2. As even root from positive number (x2 in our case) is positive then 5√y>4√x2>0, (or which is the same as y>0). Therefore, the answer to the original question is NO. Sufficient.

(2) y3>1z4. The same here as 1z4>0 then y3>1z4>0, (or which is the same as y>0). Therefore, the answer to the original question is NO. Sufficient.

Math Expert V
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2007a4ps351 wrote:
HI BUNUEL
FOR THE BEWLO QUESTION AND SOLUTION PROVIDED
FOR STTEMENT 1 LETS SUPPOSE X^2 =625 ....THEN 4 TH ROOT OF X^2 THAT IS (X^2)1/4 MEANS IT CAN BE +5 OR -5 RIGHT SO HOW (Y)^1/5 > (X^2)1/4 PROVES THAT Y IS GREATER THAN ZERO PLS ADVISE

QUESTION AND ANSWER AS POSTED ON THREAD
xyz≠0 means that neither of the unknowns is equal to zero. Next, (x−4)∗(3√y)∗(z−2)=3√yx4∗z2, so the question becomes: is 3√yx4∗z2<0? Since x4 and z2 are positive numbers, then the question boils down to whether 3√y<0, which is the same as whether y<0 (recall that odd roots have the same sign as the base of the root, for example: 3√125=5 and 3√−64=−4).

(1) 5√y>4√x2. As even root from positive number (x2 in our case) is positive then 5√y>4√x2>0, (or which is the same as y>0). Therefore, the answer to the original question is NO. Sufficient.

(2) y3>1z4. The same here as 1z4>0 then y3>1z4>0, (or which is the same as y>0). Therefore, the answer to the original question is NO. Sufficient.

You should read solutions more carefully: even roots from positive numbers are positive ONLY!

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root.

That is:
$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3.
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OHH ......thanks a lot
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I think this is a high-quality question and I agree with explanation.
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if \sqrt{x^2} = |x|

then why not \sqrt{[square_root]x^2}[/square_root] = |x|
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Bunuel wrote:
Official Solution:

If $$xyz \ne 0$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2}) \lt 0$$?

$$xyz \ne 0$$ means that neither of the unknowns is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2} \lt 0$$? Since $$x^4$$ and $$z^2$$ are positive numbers, then the question boils down to whether $$\sqrt{y} \lt 0$$, which is the same as whether $$y \lt 0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} = 5$$ and $$\sqrt{-64} = -4$$).

(1) $$\sqrt{y} \gt \sqrt{x^2}$$. As even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y} \gt \sqrt{x^2} \gt 0$$, (or which is the same as $$y \gt 0$$). Therefore, the answer to the original question is NO. Sufficient.

(2) $$y^3 \gt \frac{1}{z{^4}}$$. The same here as $$\frac{1}{z{^4}} \gt 0$$ then $$y^3 \gt \frac{1}{z{^4}} \gt 0$$, (or which is the same as $$y \gt 0$$). Therefore, the answer to the original question is NO. Sufficient.

Hi Bunnel,
For second statement you considered 1/z^4 > 0 because (like 1st statement logic of even root->positive) z^1/4 >0
Let me know if my understanding is correct
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Math Expert V
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tejyr wrote:
Bunuel wrote:
Official Solution:

If $$xyz \ne 0$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2}) \lt 0$$?

$$xyz \ne 0$$ means that neither of the unknowns is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2} \lt 0$$? Since $$x^4$$ and $$z^2$$ are positive numbers, then the question boils down to whether $$\sqrt{y} \lt 0$$, which is the same as whether $$y \lt 0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} = 5$$ and $$\sqrt{-64} = -4$$).

(1) $$\sqrt{y} \gt \sqrt{x^2}$$. As even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y} \gt \sqrt{x^2} \gt 0$$, (or which is the same as $$y \gt 0$$). Therefore, the answer to the original question is NO. Sufficient.

(2) $$y^3 \gt \frac{1}{z{^4}}$$. The same here as $$\frac{1}{z{^4}} \gt 0$$ then $$y^3 \gt \frac{1}{z{^4}} \gt 0$$, (or which is the same as $$y \gt 0$$). Therefore, the answer to the original question is NO. Sufficient.

Hi Bunnel,
For second statement you considered 1/z^4 > 0 because (like 1st statement logic of even root->positive) z^1/4 >0
Let me know if my understanding is correct

No. $$\frac{1}{z{^4}} \gt 0$$ because a number in even power is non-negative.
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Hi Bunuel ,

I read the explanation for solution above on why 1 is sufficient and I still don't get how square or even roots of positive numbers are positive. Statement 1 is an equation so why would we not consider 2 solutions? I know you explained it again above but I have never seen a case like this before.

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Deceptively tricky!
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oasis90 wrote:
Hi Bunuel ,

I read the explanation for solution above on why 1 is sufficient and I still don't get how square or even roots of positive numbers are positive. Statement 1 is an equation so why would we not consider 2 solutions? I know you explained it again above but I have never seen a case like this before.

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Hey oasis90

For those who still have doubts. Here is a break-down
-

Firstly notice that all we need to know is that if y is negative.
If y is negative then the equation is less than zero. Otherwise it is not.

Try taking any number for x.
Let $$x = -3$$. So $$x^2 = 9$$.
Let $$x= 3$$ Again we have $$x^2 = 9$$.

So whatever the number x is. The square of x will be positive.
And isn't it true that square root of a positive number is always positive?
So $$\sqrt{x^2}$$ part of the equation 1 becomes positive.

Next for the $$\sqrt{y}$$ part -
Note that odd roots retain the symbol.
Let $$y = 32$$ So $$\sqrt{y} = 2$$
Let $$y = -32$$ So $$\sqrt{y} = -2$$

Hence y has to be positive to satisfy the equation. If $$y$$ is negative then $$\sqrt{y}$$ CANNOT BE GREATER THAN $$\sqrt{x^2}$$. Re: M26-31   [#permalink] 25 Nov 2018, 08:09
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