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M27-10

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M27-10  [#permalink]

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New post 16 Sep 2014, 01:27
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

72% (01:17) correct 28% (01:45) wrong based on 201 sessions

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Re M27-10  [#permalink]

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New post 16 Sep 2014, 01:27
3
2
Official Solution:


(1) \(x^2*y=3\). If \(x=1\), then \(y=z=3\) and \(yz=9\); but if \(x=3\), then \(y=\frac{1}{3}\), \(z=1\) and \(yz=\frac{1}{3}\). Not sufficient.

(2) \(\sqrt{x*y^2}=3\). Square this equation: \(x*y^2=9\), re-write as: \((xy)*y=9\). Now, since \(xy=z\) then: \(z*y=9\). Sufficient.


Answer: B
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Re: M27-10  [#permalink]

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New post 25 Jun 2016, 07:02
Bunuel wrote:
If \(x \gt 0\) and \(xy=z\), what is the value of \(yz\)?


(1) \(x^2*y=3\).

(2) \(\sqrt{x*y^2}=3\).



Given \(xy=z\) and asked us to find \(yz\).

For equation \(xy=z\) , multiplying y on both sides
=> \(xy^2\) =\(yz\) or \(yz\) = \(xy^2\).

Then we need to know the value of \(xy^2\) ---> eq 1.

Stat 1: \(x^2*y=3\)

=> y = 3 / \(x^2\) --> squaring... \(y^2\) = 9 / \(x^4\).

Then sub \(y^2\) value in eq 1 , we get the result as 9 / \(x^3\). We are not sure of x value. This is not sufficient.

Stat 2: Given \(\sqrt{x*y^2}\) = 3.

Squaring on both side we get \(xy^2\) = 9.

Then substituting in eq 1 , we get yz = 9. This is sufficient.

Hence correct answer is B.
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Re: M27-10  [#permalink]

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New post 16 Sep 2016, 19:20
stat 2
should't be y an incognito, since y2 is module? we don't know the sing of y?
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Re: M27-10  [#permalink]

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New post 17 Sep 2016, 00:14
mounirbr wrote:
stat 2
should't be y an incognito, since y2 is module? we don't know the sing of y?


Not sure I understand what you mean. Can you please tell me which step is unclear below?

(2) \(\sqrt{x*y^2}=3\). Square this equation: \(x*y^2=9\), re-write as: \((xy)*y=9\). Now, since \(xy=z\) then: \(z*y=9\). Sufficient.
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Re: M27-10  [#permalink]

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New post 09 Nov 2017, 13:25
What if for st.2 x=1/3; y=3? then z=1
OR
x=1. y=3 then z=3

I see multiple possible values, please help me understand.
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Re: M27-10  [#permalink]

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New post 09 Nov 2017, 21:18
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Re: M27-10  [#permalink]

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New post 30 Dec 2017, 04:23
Bunuel wrote:
If \(x \gt 0\) and \(xy=z\), what is the value of \(yz\)?


(1) \(x^2*y=3\).

(2) \(\sqrt{x*y^2}=3\).



Hi Bunuel.
Can we solve statement 2 in the way i mentioned below?

(2) \(\sqrt{x*y^2}=3\)

\(\sqrt{x}\) * \(\sqrt{y^2}\)= 3 --> \(\sqrt{x}\) * |y|= 3
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Re: M27-10  [#permalink]

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New post 30 Dec 2017, 05:11
sandysilva wrote:
Bunuel wrote:
If \(x \gt 0\) and \(xy=z\), what is the value of \(yz\)?


(1) \(x^2*y=3\).

(2) \(\sqrt{x*y^2}=3\).



Hi Bunuel.
Can we solve statement 2 in the way i mentioned below?

(2) \(\sqrt{x*y^2}=3\)

\(\sqrt{x}\) * \(\sqrt{y^2}\)= 3 --> \(\sqrt{x}\) * |y|= 3


No. You still need to square.
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Re: M27-10  [#permalink]

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New post 20 Sep 2018, 09:51
Bunuel wrote:
If \(x \gt 0\) and \(xy=z\), what is the value of \(yz\)?


(1) \(x^2*y=3\).

(2) \(\sqrt{x*y^2}=3\).



HI Bunnel, Is this 700 level question ? or 600-700 level question.
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Re: M27-10  [#permalink]

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New post 20 Sep 2018, 20:50
MaheshDuggirala wrote:
Bunuel wrote:
If \(x \gt 0\) and \(xy=z\), what is the value of \(yz\)?


(1) \(x^2*y=3\).

(2) \(\sqrt{x*y^2}=3\).



HI Bunnel, Is this 700 level question ? or 600-700 level question.


According to the stats, which you can check in the first post, the difficulty level of the question is 700+.
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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: M27-10 &nbs [#permalink] 20 Sep 2018, 20:50
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