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If x^3y^5 > 0, and x^2z^3 <0, which of the following must be
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13 Jul 2020, 09:59
13 Jul 2020, 09:59
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53% (01:10) correct
46% (01:07) wrong based on 127 sessions
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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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Official Answer
B,C,E,F
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If x^3y^5 > 0, and x^2z^3 <0, which of the following must be
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15 Jul 2020, 08:18
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
General Discussion
Re: If x^3y^5 > 0, and x^2z^3 <0, which of the following must be
[#permalink]
20 Feb 2021, 07:40
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!
