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Bunuel
If (a-b)/c<0, is a>b?

(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?

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targetgmatchotu
(a-b)/c <0

=> c(a-b)/c^2 < 0
=> ac -bc <0
=> ac<bc

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,
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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.
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Bunuel
If \(\frac{(a - b)}{c} < 0\), is \(a > b\)?


(1) \(c < 0\)

(2) \(a + b < 0\)

Given: \(\frac{(a - b)}{c} < 0\)

Target question: Is \(a > b\)?

Statement 1: \(c < 0\)
In other words, c is negative.
So the given information becomes: \(\frac{(a - b)}{negative} < 0\)
From this, we can conclude that a - b is positive
In other words: a - b > 0
Add b to both sides of the inequality to get: a > b
The answer to the target question is YES, a is greater than b
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: \(a + b < 0\)
There are several values of a, b and c that satisfy statement 2 (and the given information). Here are two:
Case a: a = -3, b = 1, c = 1. In this case, the answer to the target question is NO, a is not greater than b
Case b: a = 0, b = -1, c = -1. In this case, the answer to the target question is YES, a is greater than b
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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We know that if x/y <1 in an inequality, both x and y are of different signs. So either x >0 and y<0 or x<0 and y>0

(a−b)/c<0 -So either (a-b) >0 and c<0 which means a>b with c<,0 or the other case is

(a-b)<0 and c>0 which would mean a<b with c>0

(1) c<0

So (a-b) >0 and hence a>b (Sufficient)

(2) a+b<0

We cannot judge here which of a or b is greater as we do not know their signs.

For example if a =5,b=2 then a+b >0 and since a>b ,the answer to question stem is a yes.

But if we have a=-5 and b=12 then a+b>0 but a<b ad answer to the question stem is a no.(Insufficient)
(option a)

Devmitra Sen
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What others have said to explain why statement 2 is insufficient are great, and I have another perspective to offer. Upon simplifying the question, it becomes "Is a-b >0 given (a-b) / c <0?" Now we know statement one alone is sufficient, so we only need to assess whether statement 2 is also sufficient alone and don't need to examine the two statements together (the usual AC/BCE rule for all DS questions). Looking at statement 2, there's no info on the value or sign of c which is crucial because if c is negative, a-b has to be greater than 0 to satisfy the given condition. The opposite for when c is positive. Thus with no info on c from statement 2, we don't have sufficient information to answer the question.
Therefore answer must be A.

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