is 1/a-b < b-a ???
1. a<b
2.1> abs a-b
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inequality
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parallel_chase
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I think there is something wrong with the second statement. While posting a question kindly check if you have posted the question correctly or not.
Statement I is sufficient.
Therefore the answer will be either A or D depending on the second statement.
Statement I is sufficient.
Therefore the answer will be either A or D depending on the second statement.
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Vignesh.4384
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Hi parallel_chase,
About option A.
I simplified the question as follows
Question 1/(b-a) < (a-b)
1 < (b-a) (a-b)
1 > (b-a) (b-a)
1 > (b-a)^2
Is this correct ?
If this is correect then we cant conclude from option A right ?
About option A.
I simplified the question as follows
Question 1/(b-a) < (a-b)
1 < (b-a) (a-b)
1 > (b-a) (b-a)
1 > (b-a)^2
Is this correct ?
If this is correect then we cant conclude from option A right ?
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parallel_chase
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Vignesh.4384 wrote:Hi parallel_chase,
About option A.
I simplified the question as follows
Question 1/(b-a) < (a-b)
1 < (b-a) (a-b)
1 > (b-a) (b-a)
1 > (b-a)^2
Is this correct ?
If this is correect then we cant conclude from option A right ?
I think everything is correct except for the last step.
-1 > (b-a)^2.
If you do this the statement I is sufficient because for any value of a or b it will always be greater than -1, the answer would be no and statement will be sufficient.
Let me know what you think.
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Vignesh.4384
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parallel_chase
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Dont thank me yet.Vignesh.4384 wrote:I make stupid silly mistakes al the time ..
Thanks parallel_chase. i got it
If you notice you cannot simplify the equation, because you dont know the values of a or b weather they are positive or negative, therefore you cannot simplify the equation because you are dividing and multiplying the variables.
If you could do that, this can never be a data sufficiency question.
Hence you have to take the original inequality as is.
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I think the above solutions are overcomplicating matters. From 1:arorag wrote:is 1/a-b < b-a ???
1. a<b
2.1> abs a-b
a < b, so b - a > 0 and a - b < 0.
If a-b is negative, so is 1/(a-b). So of course it's less than b-a, which is positive. Sufficient.
From 2 (which I've understood to say 1 > |a-b|), a-b could be positive or negative. Insufficient. A.
That said, if you did want to do this algebraically, from 1) you know that a-b<0. So if you multiply both sides of the inequality by (a-b), you need to reverse the inequality:
1/(a-b) < (b-a)
1 > (b-a)(a-b)
Then multiplying by -1, we again need to reverse the inequality:
-1 < (a-b)(a-b)
-1 < (a-b)^2
which must be true; (a-b)^2 can't be less than zero.
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