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just took a practice test in a Kaplan book and had a few [#permalink]
06 Sep 2004, 07:28

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just took a practice test in a Kaplan book and had a few issues with it, so I thought I would post it here...you all may have addressed it prior, if so I would appreciate a link or a re-do:

If ab<ac, which is greater, b or c?

1. a<0
2. c<0

I didn't agree with the answer they give...thanks for the input in advance...

As per the questions ab<ac means depending on the sign of a this inequality will change

if a is + ive then, divide the inequality with a on both sides, it becomes b<c

if a is - ive then, divide the inequality with a on both sides, it becomes

b>c

1) says a is negative - sufficient 2) insufficient. we don't know the value of b.

i agree with your analysis...BUT, how do you explain plugging in some numbers that do not work into #1? For instance: 1) a<0, so let a= -5...then you have -5b<-5c, if you'll notice, the opening statement asks if b<c. I can plug numbers into -5b<-5c so that b<c AND b>c. For instance, b could be 1 and c could be 10 (b<c) OR b could be -10 and c could be -20 (b>c)....so I would say if there are numbers that could make #1 both true and false, then it has to be insufficient, doesn't it?

The question asks whether b is greater or c. When you pick numbers for b and c, you are making an assumption. That is not the way to approach DS problems.

As per the questions ab<ac means depending on the sign of a this inequality will change

if a is + ive then, divide the inequality with a on both sides, it becomes b<c

if a is - ive then, divide the inequality with a on both sides, it becomes

b>c

1) says a is negative - sufficient 2) insufficient. we don't know the value of b.

i agree with your analysis...BUT, how do you explain plugging in some numbers that do not work into #1? For instance: 1) a<0, so let a= -5...then you have -5b<-5c, if you'll notice, the opening statement asks if b<c. I can plug numbers into -5b<-5c so that b<c AND b>c. For instance, b could be 1 and c could be 10 (b<c) OR b could be -10 and c could be -20 (b>c)....so I would say if there are numbers that could make #1 both true and false, then it has to be insufficient, doesn't it?

When you choose b = 1 and c = 10 in your scenario above, your original relation ab<ac for a = -5 ceases to be true. Therefore, these values of b and c are invalid for the original condition to hold with a negative a.