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Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Last edited by Bunuel on 17 Mar 2013, 01:37, edited 1 time in total.

Statement (1) : Tells the sign of (C) means sufficient.

Statement (2) : a+b < 0

=> a < -b

a b a<-b Is a>b? + + Not Poss N/A - - Yes No - + Yes No + - Yes No

Why the answer is not (D).

Please tell what I am missing above

Rgds, TGC
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Before cross multiplying or multiplying numerator and denominator by c you should keep in mind that c should not be equal to 0 So which statement proves c won't be 0 (1) c < 0 This statement implies that c is not equal to 0 (2) a + b < 0 You aren't sure whether c would be 0 or not. Hence, the answer should be A

(1) c < 0. Multiply (a-b)/c<0 by negative c and flip the sign a-b>0 --> a>b. Sufficient.

(2) a + b < 0. The sum of two numbers is less than zero. Can we tell which of them is greater? (Can we tell whether a>b or a<b?) No, consider a=1, b=-2 and c=-1 AND a=-2, b=1 and c=1. Not sufficient.

(1) c < 0. Multiply (a-b)/c<0 by negative c and flip the sign a-b>0 --> a>b. Sufficient.

(2) a + b < 0. The sum of two numbers is less than zero. Can we tell which of them is greater? (Can we tell whether a>b or a<b?) No, consider a=1, b=-2 and c=-1 AND a=-2, b=1 and c=1. Not sufficient.

Answer: A.

What is wrong in the solution that I gave?

Rgds, TGC
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Statement (1) : Tells the sign of (C) means sufficient.

Statement (2) : a+b < 0

=> a < -b

a b a<-b Is a>b? + + Not Poss N/A - - Yes No - + Yes No + - Yes No

Why the answer is not (D).

Please tell what I am missing above

Rgds, TGC

Consider one of the cases in your approach above: if both a and b are negative, it's possible that a>b as well as a<b. For example, a=-1, b=-2, and c=-1 AND a=-2, b=-1, and c=1.

Statement (1) : Tells the sign of (C) means sufficient.

Statement (2) : a+b < 0

=> a < -b

a b a<-b Is a>b? + + Not Poss N/A - - Yes No - + Yes No + - Yes No

Why the answer is not (D).

Please tell what I am missing above

Rgds, TGC

Consider one of the cases in your approach above: if both a and b are negative, it's possible that a>b as well as a<b. For example, a=-1, b=-2, and c=-1 AND a=-2, b=-1, and c=1.

Hope it helps.

Why to take c variable in picture .

Given that a<-b I think appropriate would be a = -1 b = -2 (here a>b still holding a<-b)

a = -2 b = -1 (here a <b still holding a<-b)

Hence,wrong
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Statement (1) : Tells the sign of (C) means sufficient.

Statement (2) : a+b < 0

=> a < -b

a b a<-b Is a>b? + + Not Poss N/A - - Yes No - + Yes No + - Yes No

Why the answer is not (D).

Please tell what I am missing above

Rgds, TGC

Consider one of the cases in your approach above: if both a and b are negative, it's possible that a>b as well as a<b. For example, a=-1, b=-2, and c=-1 AND a=-2, b=-1, and c=1.

Hope it helps.

Why to take c variable in picture .

Given that a<-b I think appropriate would be a = -1 b = -2 (here a>b still holding a<-b)

a = -2 b = -1 (here a <b still holding a<-b)

Hence,wrong

Hi TGC good to see you back on the forums..........people gather courage from life's storms.

I think the process you suggested is correct and has got no flaws in it....its concise...i do not find any use of third variable here.

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