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M26-31

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M26-31  [#permalink]

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New post 16 Sep 2014, 00:26
1
15
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A
B
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D
E

Difficulty:

  55% (hard)

Question Stats:

56% (01:00) correct 44% (01:33) wrong based on 243 sessions

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M26-31  [#permalink]

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New post 16 Sep 2014, 00:26
2
3
Official Solution:

If \(xyz \ne 0\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2}) \lt 0\)?


\(xyz \ne 0\) means that neither of the unknowns is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2} \lt 0\)? Since \(x^4\) and \(z^2\) are positive numbers, then the question boils down to whether \(\sqrt[3]{y} \lt 0\), 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: \(\sqrt[3]{125} = 5\) and \(\sqrt[3]{-64} = -4\)).

(1) \(\sqrt[5]{y} \gt \sqrt[4]{x^2}\). As even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y} \gt \sqrt[4]{x^2} \gt 0\), (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.


Answer: D
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Re: M26-31  [#permalink]

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New post 30 Jun 2015, 02:05
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.


Answer: D
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Re: M26-31  [#permalink]

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New post 30 Jun 2015, 02:43
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.


Answer: D


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;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3.
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Re: M26-31  [#permalink]

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New post 30 Jun 2015, 03:06
OHH ......thanks a lot
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Re M26-31  [#permalink]

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New post 25 Jul 2016, 09:31
I think this is a high-quality question and I agree with explanation.
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Re: M26-31  [#permalink]

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New post 25 Aug 2017, 10:16
if \sqrt{x^2} = |x|

then why not \sqrt{[square_root]x^2}[/square_root] = |x|
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Re: M26-31  [#permalink]

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New post 16 Jul 2018, 07:55
Bunuel wrote:
Official Solution:

If \(xyz \ne 0\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2}) \lt 0\)?


\(xyz \ne 0\) means that neither of the unknowns is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2} \lt 0\)? Since \(x^4\) and \(z^2\) are positive numbers, then the question boils down to whether \(\sqrt[3]{y} \lt 0\), 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: \(\sqrt[3]{125} = 5\) and \(\sqrt[3]{-64} = -4\)).

(1) \(\sqrt[5]{y} \gt \sqrt[4]{x^2}\). As even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y} \gt \sqrt[4]{x^2} \gt 0\), (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.


Answer: D


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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Re: M26-31  [#permalink]

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New post 16 Jul 2018, 09:02
tejyr wrote:
Bunuel wrote:
Official Solution:

If \(xyz \ne 0\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2}) \lt 0\)?


\(xyz \ne 0\) means that neither of the unknowns is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2} \lt 0\)? Since \(x^4\) and \(z^2\) are positive numbers, then the question boils down to whether \(\sqrt[3]{y} \lt 0\), 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: \(\sqrt[3]{125} = 5\) and \(\sqrt[3]{-64} = -4\)).

(1) \(\sqrt[5]{y} \gt \sqrt[4]{x^2}\). As even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y} \gt \sqrt[4]{x^2} \gt 0\), (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.


Answer: D


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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M26-31  [#permalink]

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New post 29 Aug 2018, 13:52
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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New post 18 Sep 2018, 06:29
Deceptively tricky!
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Re: M26-31  [#permalink]

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New post 25 Nov 2018, 07:09
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 \(\sqrt[4]{x^2}\) part of the equation 1 becomes positive.

Next for the \(\sqrt[5]{y}\) part -
Note that odd roots retain the symbol.
Let \(y = 32\) So \(\sqrt[5]{y} = 2\)
Let \(y = -32\) So \(\sqrt[5]{y} = -2\)

Hence y has to be positive to satisfy the equation. If \(y\) is negative then \(\sqrt[5]{y}\) CANNOT BE GREATER THAN \(\sqrt[4]{x^2}\).
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Re: M26-31 &nbs [#permalink] 25 Nov 2018, 07:09
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