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24 May 2022, 01:40
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
89% (00:47) correct
11%
(00:55)
wrong
based on 72
sessions
History
Date
Time
Result
Not Attempted Yet
24 May 2022, 02:13
Kudos
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(1) \(b^2–a>0\)
=> \(b^2 > a\)
We do not know value of 'b', so a can be +ve or -ve or 0. e.g., If b=2 then 'a' should be < 4 and 'a' can be -1, 0, 1, etc.
Statement 1 alone is not sufficient.
(2) \(ab^2<0\)
=> Square of a real number is always non-negative. Hence, \(b^2\) will always be >= 0
Product of \(a\) and \(b^2\) is -ve, implies 'b' is not equal to zero and 'a' is negative.
Statement 2 alone is sufficient.
Answer: B
=> \(b^2 > a\)
We do not know value of 'b', so a can be +ve or -ve or 0. e.g., If b=2 then 'a' should be < 4 and 'a' can be -1, 0, 1, etc.
Statement 1 alone is not sufficient.
(2) \(ab^2<0\)
=> Square of a real number is always non-negative. Hence, \(b^2\) will always be >= 0
Product of \(a\) and \(b^2\) is -ve, implies 'b' is not equal to zero and 'a' is negative.
Statement 2 alone is sufficient.
Answer: B
Moderators:
