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M20-21

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M20-21  [#permalink]

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New post 16 Sep 2014, 01:08
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A
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C
D
E

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  25% (medium)

Question Stats:

80% (01:07) correct 20% (00:38) wrong based on 74 sessions

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Re M20-21  [#permalink]

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New post 16 Sep 2014, 01:08
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Re: M20-21  [#permalink]

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New post 04 Jan 2015, 15:17
the equation can be simplified as : Is square(x*y).square(y*z) > 0 . If we combine both the equation we find that neither x, y,z are zero. And after , even if (x*y) or (y*z) is positive/negative , the overall equation will be always positive as it the square of two non zeros . Am i doing something wrong .. Can you please clarify ?
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M20-21  [#permalink]

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New post 05 Jan 2015, 05:50
jhasac wrote:
the equation can be simplified as : Is square(x*y).square(y*z) > 0 . If we combine both the equation we find that neither x, y,z are zero. And after , even if (x*y) or (y*z) is positive/negative , the overall equation will be always positive as it the square of two non zeros . Am i doing something wrong .. Can you please clarify ?


The point is that x^2*y^3*z^2 does NOT equal to (xy)^2*(yz)^2:

(xy)^2*(yz)^2 = x^2*y^4*z^2, not x^2*y^3*z^2.
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Re: M20-21  [#permalink]

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New post 16 Dec 2016, 20:40
I have a query

Equation breaks down as xy.xy.yz.

From 1 we know xy is positive. We don't know yz so NS

From 2 we know yz is negatve. We don't know xy so NS

COMBINING WE KNOW XY IS POSITIVE AND YZ IS NEGATIVE. SO POSITIVE X POSITIVE X NEGATIVE = NEGATIVE. We have answer as NO.


SHUDNT IT BE C ??
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Re: M20-21  [#permalink]

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New post 17 Dec 2016, 03:56
Omkar.kamat wrote:
I have a query

Equation breaks down as xy.xy.yz.

From 1 we know xy is positive. We don't know yz so NS

From 2 we know yz is negatve. We don't know xy so NS

COMBINING WE KNOW XY IS POSITIVE AND YZ IS NEGATIVE. SO POSITIVE X POSITIVE X NEGATIVE = NEGATIVE. We have answer as NO.


SHUDNT IT BE C ??


\(xy*xy*yz=x^2y^3z\). You are missing one z. It's \(X^2Y^3Z^2\)
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M20-21  [#permalink]

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New post 12 Sep 2017, 22:31
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Here's how I solved this.

STEM:

Is \(X^2Y^3Z^2 \gt 0\)

Since \(X^2\) and \(Z^2\) will ALWAYS be positive irrespective of their value, the stem can be Paraphrased as is - \(Y^3 \gt 0\). In plan english, What is the sign of Y?

S1:

\(XY \gt 0\)

Possible when both X&Y are positive or Both X&Y are negative. Meaning X can be either +ve or Negative. Insuff

S2:

\(YZ \lt 0\)

Same reasoning as S1. X could be +ve or -ve. Insuff

S1+S2:

Combining Both statements we get \(XY \gt YX\) => \(Y(X-Z) \gt 0\). This is still not enough to tell us tell us the exact sign of Y and Insuff

S1 Insuff. S2 Insuff. S1+S2 insuff. Final answer is E
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Re: M20-21  [#permalink]

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New post 05 Oct 2017, 00:27
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Buneul,

Kindly check on the below, we need to check whether \(x^2\) \(y^3\) \(z^2\) > 0 ?

Stat 1:
XY > 0 then we get two cases : case a) both positive and b) both negative

case a) When both +ve we get \(x^2\) \(y^3\) \(z^2\) as + + + > 0 (yes)
case b) when both -ve we get \(x^2\) \(y^3\) \(z^2\) as + - + < 0 (no)

Stat 2:
YZ < 0 then we get cases a) (y) +Ve and other -ve (z) and b) -ve (y)and other +ve(z)

case a) When y = +Ve and z = -ve we get \(x^2\) \(y^3\) \(z^2\) as + + + > 0 (yes)
case b) when y = -ve and z = +Ve we get \(x^2\) \(y^3\) \(z^2\) as + - + < 0 (no)

But in both statements when y is -ve then we get \(x^2\) \(y^3\) \(z^2\) < 0... right ? why this is not considered in official solution..plz let me know.
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Re: M20-21  [#permalink]

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New post 05 Oct 2017, 00:45
msk0657 wrote:
Buneul,

Kindly check on the below, we need to check whether \(x^2\) \(y^3\) \(z^2\) > 0 ?

Stat 1:
XY > 0 then we get two cases : case a) both positive and b) both negative

case a) When both +ve we get \(x^2\) \(y^3\) \(z^2\) as + + + > 0 (yes)
case b) when both -ve we get \(x^2\) \(y^3\) \(z^2\) as + - + < 0 (no)

Stat 2:
YZ < 0 then we get cases a) (y) +Ve and other -ve (z) and b) -ve (y)and other +ve(z)

case a) When y = +Ve and z = -ve we get \(x^2\) \(y^3\) \(z^2\) as + + + > 0 (yes)
case b) when y = -ve and z = +Ve we get \(x^2\) \(y^3\) \(z^2\) as + - + < 0 (no)

But in both statements when y is -ve then we get \(x^2\) \(y^3\) \(z^2\) < 0... right ? why this is not considered in official solution..plz let me know.


What are you implying?

For (1)+(2) x and y have the same sign and y and z have the opposite sign:

x > 0, y > 0 and z < 0, then x^2*y^3*z^2 = positive*positive*positive > 0;
x < 0, y < 0 and z > 0, then x^2*y^3*z^2 = positive*negative*positive > 0.
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Resources:
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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: M20-21  [#permalink]

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New post 20 Aug 2018, 16:50
Bunuel wrote:
Is \(X^2Y^3Z^2 \gt 0\) ?


(1) \(XY \gt 0\)

(2) \(YZ \lt 0\)


for \(X^2Y^3Z^2 \gt 0\) to be true one has to find whether \(y\) is positive or not, as \(x^2\) & \(z^2\) will always remain positive. Both statements (1) and (2) are insufficient to tell whether y is positive or negative; hence answer is E
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Re: M20-21 &nbs [#permalink] 20 Aug 2018, 16:50
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