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19 Jul 2019, 01:45
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
56% (01:30) correct
44%
(01:23)
wrong
based on 64
sessions
History
Date
Time
Result
Not Attempted Yet
Here's how I think it can be solved, feel free to point out any wrongdoing. I want to improve myself, too..
(ab)^2 > ab ?
Or (ab)^2- ab > 0
ab(ab-1) > 0
Or ab> 0 and ab > 1.. well, this can't be true,
If ab > 1 then we don't need ab > 0
We must reverse the sign ab < 0 ( I saw it on a question, and it helps in such dangling conditions) or ab > 1
We need to know either of this or both.
Statement 1) A+3 = 0
A = -3
B+6= 0
B= -6
That's sufficient to prove one of our condition
Ab > 1
Therefore it will be sufficient for all positive values beyond it.
Statement 2) 0>a>b
A and B, if Integers, will give a yes, as long as they're on the same side of the number line, I.e a and b have same sign..
ab >0
But be careful,
We need to check for fractions, too, as nowhere it's mentioned that A and B are integers.
1/64 > 1/8
I took A= -1/4 and b= -1/2
This gives us a NO..
Therefore Statement 2) isn't sufficient..
Hence A) is our answer
Posted from my mobile device
(ab)^2 > ab ?
Or (ab)^2- ab > 0
ab(ab-1) > 0
Or ab> 0 and ab > 1.. well, this can't be true,
If ab > 1 then we don't need ab > 0
We must reverse the sign ab < 0 ( I saw it on a question, and it helps in such dangling conditions) or ab > 1
We need to know either of this or both.
Statement 1) A+3 = 0
A = -3
B+6= 0
B= -6
That's sufficient to prove one of our condition
Ab > 1
Therefore it will be sufficient for all positive values beyond it.
Statement 2) 0>a>b
A and B, if Integers, will give a yes, as long as they're on the same side of the number line, I.e a and b have same sign..
ab >0
But be careful,
We need to check for fractions, too, as nowhere it's mentioned that A and B are integers.
1/64 > 1/8
I took A= -1/4 and b= -1/2
This gives us a NO..
Therefore Statement 2) isn't sufficient..
Hence A) is our answer
Posted from my mobile device
firas92
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19 Jul 2019, 02:15
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(1) a + 3 = b + 6 = 0
a=-3, b=-6
ab=18, (ab)^2=18^2
So (ab)^2 > ab
1 is sufficient
(2) 0 > a > b
If a=-3, b=-6,
(ab)^2 > ab
If a=-0.1, b=-0.5
ab=0.05 and ab^2=0.0025
(ab)^2 < ab
2 is insufficient
Answer is (A)
Posted from my mobile device
a=-3, b=-6
ab=18, (ab)^2=18^2
So (ab)^2 > ab
1 is sufficient
(2) 0 > a > b
If a=-3, b=-6,
(ab)^2 > ab
If a=-0.1, b=-0.5
ab=0.05 and ab^2=0.0025
(ab)^2 < ab
2 is insufficient
Answer is (A)
Posted from my mobile device
