Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

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.