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If x^3y^5 > 0, and x^2z^3 <0, which of the following must be
[#permalink]
13 Jul 2020, 09:59

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Question Stats:

If \(x^3y^5 > 0\), and \(x^2z^3 <0\), which of the following must be true ?

A. \(x>0\)

B. \(z<0\)

C. \(xy >0\)

D. \(yz<0\)

E. \(\frac{x^2}{z}<0\)

F. \(xyz<0\)

A. \(x>0\)

B. \(z<0\)

C. \(xy >0\)

D. \(yz<0\)

E. \(\frac{x^2}{z}<0\)

F. \(xyz<0\)

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If x^3y^5 > 0, and x^2z^3 <0, which of the following must be
[#permalink]
15 Jul 2020, 08:18

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Check out this post first to become familiar with Inequality Theory

\(x^3y^5 > 0\)

Product of two variables x and y with odd powers is > 0 => both the variables have the same sign

=> Either both x and y are positive or both are negative. (NOTE: We do not know about the exact signs of x and y, but we do know about their relative signs)

\(x^2z^3 <0\)

=> We know that \(x^2\) being square of a number will always be non-negative

=> \(z^3\) has to be < 0 then only \(x^2z^3 <0\)

=> z < 0 (\(z^3\)is odd power of z)

So, we know that x and y have the same sign and z < 0. Let's go through each and every option in detail now.

A. \(x>0\) => NOT Necessary as we know that x can be both positive or negative

B. \(z<0\) => TRUE, proved above

C. \(xy >0\) => TRUE, proved above. Both x and y have the same sign so xy > 0

D. \(yz<0\) => NOT Necessary as z < 0 and for yz < 0 y has to be positive, but we do not know that.

E. \(\frac{x^2}{z}<0\) => TRUE as for \(\frac{x^2}{z}<0\) z<0 which is true

F. \(xyz<0\) => TRUE as xy > 0 and z < 0 so xyz < 0

So, Answer will be B,C,E,F

Hope it helps!

Watch below video to know about the Basics of Inequalities

_________________

\(x^3y^5 > 0\)

Product of two variables x and y with odd powers is > 0 => both the variables have the same sign

=> Either both x and y are positive or both are negative. (NOTE: We do not know about the exact signs of x and y, but we do know about their relative signs)

\(x^2z^3 <0\)

=> We know that \(x^2\) being square of a number will always be non-negative

=> \(z^3\) has to be < 0 then only \(x^2z^3 <0\)

=> z < 0 (\(z^3\)is odd power of z)

So, we know that x and y have the same sign and z < 0. Let's go through each and every option in detail now.

A. \(x>0\) => NOT Necessary as we know that x can be both positive or negative

B. \(z<0\) => TRUE, proved above

C. \(xy >0\) => TRUE, proved above. Both x and y have the same sign so xy > 0

D. \(yz<0\) => NOT Necessary as z < 0 and for yz < 0 y has to be positive, but we do not know that.

E. \(\frac{x^2}{z}<0\) => TRUE as for \(\frac{x^2}{z}<0\) z<0 which is true

F. \(xyz<0\) => TRUE as xy > 0 and z < 0 so xyz < 0

So, Answer will be B,C,E,F

Hope it helps!

Watch below video to know about the Basics of Inequalities

_________________

Ankit

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Re: If x^3y^5 > 0, and x^2z^3 <0, which of the following must be
[#permalink]
20 Feb 2021, 07:40

1

Solution:

\(x^3\)\(y^5\)>0

As both the power of x & y are odd, for \(x^3\)\(y^5\)>0, both x &y should have same sign i.e. either positive or negative.

\(x^2\)\(z^3\)<0

Here, \(x^2 \)is positive but \(z^3\) is negative and thus, z is negative as well.

So, we know x & y both have same sign and z is negative.

A. x>0---> x can be positive or negative not necessarily true.

B. z<0---->Yes z is negative thus this has to true

C. xy >0---->As x and y has same sign if both are negative or positive xy will always be positive. Thus, this has to be true

D. yz<0----> z=negative and y could be negative or positive and yz will change according to the value y takes

E. \(\frac{x^2}{z}\)<0----> \(x^2\) is always positive and z is negative thus \(\frac{positive}{negative}\)=negative. Thus, this has to be true.

F. xyz<0----> If x &y are positive and z is negative thus, xyz is negative & If x& y are negative still xyz is negative. Thus, this is true

IMO B, C, E, F

Hope this helps!

\(x^3\)\(y^5\)>0

As both the power of x & y are odd, for \(x^3\)\(y^5\)>0, both x &y should have same sign i.e. either positive or negative.

\(x^2\)\(z^3\)<0

Here, \(x^2 \)is positive but \(z^3\) is negative and thus, z is negative as well.

So, we know x & y both have same sign and z is negative.

A. x>0---> x can be positive or negative not necessarily true.

B. z<0---->Yes z is negative thus this has to true

C. xy >0---->As x and y has same sign if both are negative or positive xy will always be positive. Thus, this has to be true

D. yz<0----> z=negative and y could be negative or positive and yz will change according to the value y takes

E. \(\frac{x^2}{z}\)<0----> \(x^2\) is always positive and z is negative thus \(\frac{positive}{negative}\)=negative. Thus, this has to be true.

F. xyz<0----> If x &y are positive and z is negative thus, xyz is negative & If x& y are negative still xyz is negative. Thus, this is true

IMO B, C, E, F

Hope this helps!

Re: If x^3y^5 > 0, and x^2z^3 <0, which of the following must be
[#permalink]
26 Apr 2021, 18:51

I grouped into two scenarios to satisfy each inequality:

Scenario 1:

+x and +y with +x and -z

Scenario 2:

-x and -y with -x and -z

Evaluate each choice making sure it is satisfied in both Scenarios.

Scenario 1:

+x and +y with +x and -z

Scenario 2:

-x and -y with -x and -z

Evaluate each choice making sure it is satisfied in both Scenarios.

Re: If x^3y^5 > 0, and x^2z^3 <0, which of the following must be
[#permalink]
18 May 2022, 20:46

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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