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# M20-21

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Math Expert
Joined: 02 Sep 2009
Posts: 46167

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16 Sep 2014, 01:08
00:00

Difficulty:

25% (medium)

Question Stats:

75% (01:04) correct 25% (00:33) wrong based on 59 sessions

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Is $$X^2Y^3Z^2 \gt 0$$ ?

(1) $$XY \gt 0$$

(2) $$YZ \lt 0$$

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Math Expert
Joined: 02 Sep 2009
Posts: 46167

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16 Sep 2014, 01:08
Official Solution:

Statements (1) and (2) combined are insufficient. Consider:

$$X$$ is positive, $$Y$$ is positive, $$Z$$ is negative - the answer is "yes";

$$X$$ is negative, $$Y$$ is negative, $$Z$$ is positive - the answer is "no".

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Intern
Joined: 20 Aug 2014
Posts: 5

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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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 46167

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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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Intern
Joined: 04 Mar 2016
Posts: 46
Location: India

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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 ??
Math Expert
Joined: 02 Sep 2009
Posts: 46167

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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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Manager
Joined: 12 Jun 2016
Posts: 221
Location: India
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12 Sep 2017, 22:31
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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Retired Moderator
Joined: 26 Nov 2012
Posts: 599

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05 Oct 2017, 00:27
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.
Math Expert
Joined: 02 Sep 2009
Posts: 46167

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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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Re: M20-21   [#permalink] 05 Oct 2017, 00:45
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# M20-21

Moderators: chetan2u, Bunuel

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