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20 Jan 2021, 06:38
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
68% (01:07) correct
32%
(01:10)
wrong
based on 130
sessions
History
Date
Time
Result
Not Attempted Yet
Updated on: 20 Jan 2021, 07:21
Originally posted by Nipungupta9081 on 20 Jan 2021, 06:48.
Last edited by Nipungupta9081 on 20 Jan 2021, 07:21, edited 1 time in total.
Last edited by Nipungupta9081 on 20 Jan 2021, 07:21, edited 1 time in total.
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IMO E
(1) x^3 > 0
This tells us X is positive = X could be 0 or 1.
X^2 = X^3 ------- INSUFFICIENT
(2) x≠1
X could be 0 or -1. --------- INSUFFICIENT
Joining both Statements
X^3 > 0 and X ≠ 1
Still cant get a definite value ------ INSFUFFIENT
Joining 1 + 2
Still INSUFFICIENT
(1) x^3 > 0
This tells us X is positive = X could be 0 or 1.
X^2 = X^3 ------- INSUFFICIENT
(2) x≠1
X could be 0 or -1. --------- INSUFFICIENT
Joining both Statements
X^3 > 0 and X ≠ 1
Still cant get a definite value ------ INSFUFFIENT
Joining 1 + 2
Still INSUFFICIENT
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20 Jan 2021, 07:05
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(1) \(x^3\) > 0
means that x is +ve.
If x is greater than 1 than we get no.
If x is less than 1 than we get yes; x=\(\frac{1}{2}\), \(x^2\)=\(\frac{1}{4}\) and \(x^3\)=\(\frac{1}{8}\); \(x^2\)>\(x^3\)
Insufficient.
(2) x≠1
Clearly insufficient as we do not know the polarity of x.
1+2
Statement 1 still stands
Insufficient.
Option E
means that x is +ve.
If x is greater than 1 than we get no.
If x is less than 1 than we get yes; x=\(\frac{1}{2}\), \(x^2\)=\(\frac{1}{4}\) and \(x^3\)=\(\frac{1}{8}\); \(x^2\)>\(x^3\)
Insufficient.
(2) x≠1
Clearly insufficient as we do not know the polarity of x.
1+2
Statement 1 still stands
Insufficient.
Option E
