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19 Jul 2019, 03:37
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
58% (01:45) correct
42%
(01:51)
wrong
based on 93
sessions
History
Date
Time
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Not Attempted Yet
19 Jul 2019, 03:54
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If \(x^5\) + yz >0, Is x>0 ?
1. \(x^2\)y\(z^3\)<0
=> For x : x <0 or x >0
=> If y >0 then z <0
=> If y<0 then z>0
For the equality to holds those are all possible consideratons.
But We cannot say anything about x.
So, Insufficient.
2. \(z^3\)<0<\(y^5\)
=> This means z is definitively negative. z<0
=> Also y is definitively positive. y>0
As the odd powers of a negative number will always result in a negative number. And odd powers of a positive number will always result in a positive number.
Now, if y> 0, z<0:
And If \(x^5\) + yz >0,
then \(x^5\) + (+y)*(-z) >0
=> \(x^5\) -(yz) >0
=> for this equality to holds true, x should be positive, means x>0. As, only (positive number) - (negative number) can result in greater than 0.
Sufficient.
IMO the answer is B.
Please hit kudos if you like the solution.
1. \(x^2\)y\(z^3\)<0
=> For x : x <0 or x >0
=> If y >0 then z <0
=> If y<0 then z>0
For the equality to holds those are all possible consideratons.
But We cannot say anything about x.
So, Insufficient.
2. \(z^3\)<0<\(y^5\)
=> This means z is definitively negative. z<0
=> Also y is definitively positive. y>0
As the odd powers of a negative number will always result in a negative number. And odd powers of a positive number will always result in a positive number.
Now, if y> 0, z<0:
And If \(x^5\) + yz >0,
then \(x^5\) + (+y)*(-z) >0
=> \(x^5\) -(yz) >0
=> for this equality to holds true, x should be positive, means x>0. As, only (positive number) - (negative number) can result in greater than 0.
Sufficient.
IMO the answer is B.
Please hit kudos if you like the solution.
19 Jul 2019, 04:26
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RajatVerma1392
If \(x^5\) + yz >0, Is x>0 ?
1. \(x^2\)y\(z^3\)<0
=> For x : x <0 or x >0
=> If y >0 then z <0
=> If y<0 then z>0
For the equality to holds those are all possible consideratons.
But We cannot say anything about x.
So, Insufficient.
2. \(z^3\)<0<\(y^5\)
=> This means z is definitively negative. z<0
=> Also y is definitively positive. y>0
As the odd powers of a negative number will always result in a negative number. And odd powers of a positive number will always result in a positive number.
Now, if y> 0, z<0:
And If \(x^5\) + yz >0,
then \(x^5\) + (+y)*(-z) >0
=> \(x^5\) -(yz) >0
=> for this equality to holds true, x should be positive, means x>0. As, only (positive number) - (negative number) can result in greater than 0.
Sufficient.
IMO the answer is B.
Please hit kudos if you like the solution.
1. \(x^2\)y\(z^3\)<0
=> For x : x <0 or x >0
=> If y >0 then z <0
=> If y<0 then z>0
For the equality to holds those are all possible consideratons.
But We cannot say anything about x.
So, Insufficient.
2. \(z^3\)<0<\(y^5\)
=> This means z is definitively negative. z<0
=> Also y is definitively positive. y>0
As the odd powers of a negative number will always result in a negative number. And odd powers of a positive number will always result in a positive number.
Now, if y> 0, z<0:
And If \(x^5\) + yz >0,
then \(x^5\) + (+y)*(-z) >0
=> \(x^5\) -(yz) >0
=> for this equality to holds true, x should be positive, means x>0. As, only (positive number) - (negative number) can result in greater than 0.
Sufficient.
IMO the answer is B.
Please hit kudos if you like the solution.
Statement1:
=> If y >0 then z<0
=> If y<0 then z>0
In both conditions, y*z will be negative.
x^5>-yz-->(-yz) is postive number
if x^5 is greater than positive number, x will be greater than 0
Right?
Stament1 is Sufficient.
P.S.(let me know If it is wrong)
