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# If ab<ac , which is greater b or c ? i. a<0 2. c<0

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Intern
Joined: 27 Dec 2010
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If ab<ac , which is greater b or c ? i. a<0 2. c<0 [#permalink]

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01 Jan 2011, 12:48
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If ab<ac , which is greater b or c ?

i. a<0
2. c<0

This question if from Kaplan Premier 2009. Practice quiz #5

As per the answer its is A. The explanation if a is -ve the we can derive that b>c..

But 'a' can be positive also , let assume that a=.5
in this case B<c

So we are nor pretty sure about the value of 'a'. So how the answer is A.
[Reveal] Spoiler: OA
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Joined: 02 Sep 2009
Posts: 39575
Re: DS doubt - Algebra - Equation [#permalink]

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01 Jan 2011, 13:02
Riju1191 wrote:
If ab<ac , which is greater b or c ?

i. a<0
2. c<0

This question if from Kaplan Premier 2009. Practice quiz #5

As per the answer its is A. The explanation if a is -ve the we can derive that b>c..

But 'a' can be positive also , let assume that a=.5
in this case B<c

So we are nor pretty sure about the value of 'a'. So how the answer is A.

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

Given that $$ab<ac$$ --> $$ac-ab>0$$ --> $$a(c-b)>0$$, so $$a$$ and $$c-b$$ are either both positive or both negative.

(1) a<0 --> as $$a$$ is negative then $$c-b$$ is negative too --> $$c-b<0$$ --> $$c<b$$. Sufficient.
(2) c<0. Clearly insufficient.

As for your doubt: statement (1) says that $$a<0$$, so $$a$$ is negative and can not equal to 0.5 (or any other non-negative value).

Hope it's clear.
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Re: DS doubt - Algebra - Equation [#permalink]

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01 Jan 2011, 13:29
If ab<ac , which is greater b or c ?

i. a<0
2. c<0

As you said
Both a or c-b should be +ve or both should be -ve.

Let assume
a= .1 its is a positive
b = .1
c= .2

AB = .1*.1 = 0.01
AC = .1*.2 = 0.02

so ab< ac

In this case
if a < 0 and a is -ve then b>c
but if a<0 and +ve then C>b

So we are not pretty sure.. Again in the question it is not mentioned that a,b,c are integer
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Joined: 02 Sep 2009
Posts: 39575
Re: DS doubt - Algebra - Equation [#permalink]

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01 Jan 2011, 13:36
Riju1191 wrote:
If ab<ac , which is greater b or c ?

i. a<0
2. c<0

As you said
Both a or c-b should be +ve or both should be -ve.

Let assume
a= .1 its is a positive
b = .1
c= .2

AB = .1*.1 = 0.01
AC = .1*.2 = 0.02

so ab< ac

In this case
if a < 0 and a is -ve then b>c
but if a<0 and +ve then C>b

So we are not pretty sure.. Again in the question it is not mentioned that a,b,c are integer

I think you are confused with notation: statement (1) says that $$a<0$$, which means that $$a$$ is negative, so it can not equal to 0.1 or any other non-negative value.
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Re: DS doubt - Algebra - Equation [#permalink]

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01 Jan 2011, 23:27
hi guys .... i am a beginner in prep for GMAT. I left mathematics a very very long ago ... i have a question regarding the problem...

the question itself says

ab < ac ............can't we subtract a from either side straight away which will yield ...

b < c directly from the stem itself. We have answer without any further condition. a could be anything but if ab<ac it has to as per my understanding b<c.

anyone can throw light on it ...?
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Joined: 02 Sep 2009
Posts: 39575
Re: DS doubt - Algebra - Equation [#permalink]

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02 Jan 2011, 02:25
hi guys .... i am a beginner in prep for GMAT. I left mathematics a very very long ago ... i have a question regarding the problem...

the question itself says

ab < ac ............can't we subtract a from either side straight away which will yield ...

b < c directly from the stem itself. We have answer without any further condition. a could be anything but if ab<ac it has to as per my understanding b<c.

anyone can throw light on it ...?

I think you mean divide instead of subtract.

Never multiply or reduce (divide) inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

So you can not divided (reduce) $$ab<ac$$ by $$a$$ and write $$b<c$$ because you don't know whether $$a$$ is positive or negative: if it's positive then you should write $$b<c$$ but if its negative then you should flip the sign and write $$b>c$$. Statement (1) says $$a<0$$, so when reducing by $$a$$ we should flip the sign and write $$b>c$$.

Hope it's clear.
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Re: DS doubt - Algebra - Equation [#permalink]

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02 Jan 2011, 02:28
Thanks. Yes its clear.

Posted from my mobile device
Intern
Joined: 21 Jun 2012
Posts: 3
If ab < ac, which is greater, b or c? [#permalink]

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24 Aug 2012, 20:09
(Note: I found two older topics for this question, but they were each over a year old and I assumed there were bump rules here..)

This is from Kaplan CAT3:

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

(1) a < 0

(2) c < 0

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

I just have one issue with this problem: when I originally did it I tried picking numbers rather than just dividing by -a to get b > c. This is what I got:

(1) If a < 0, then perhaps a = -1. Then if b =3 and c=2, ab < ac and c > b. But if b = -2 and c = -3, ab < ac and b > c.
I'm not finding any rules I'm missing in the problem, so my answer was C.. or E... Why is the above reasoning wrong?

Thanks...
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Posts: 361
Location: US
Re: If ab < ac, which is greater, b or c? [#permalink]

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24 Aug 2012, 23:40
paldrich1287 wrote:
(Note: I found two older topics for this question, but they were each over a year old and I assumed there were bump rules here..)

This is from Kaplan CAT3:

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

(1) a < 0

(2) c < 0

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

I just have one issue with this problem: when I originally did it I tried picking numbers rather than just dividing by -a to get b > c. This is what I got:

(1) If a < 0, then perhaps a = -1. Then if b =3 and c=2, ab < ac and c > bBut if b = -2 and c = -3, ab < ac and b > c.
I'm not finding any rules I'm missing in the problem, so my answer was C.. or E... Why is the above reasoning wrong?

Thanks...

Above c>b in yellow is incorrect. It is c<b becasue b =3 and c=2. Hence answer is A.
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Re: If ab < ac, which is greater, b or c? [#permalink]

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25 Aug 2012, 00:00
paldrich1287 wrote:
(Note: I found two older topics for this question, but they were each over a year old and I assumed there were bump rules here..)

This is from Kaplan CAT3:

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

(1) a < 0

(2) c < 0

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

I just have one issue with this problem: when I originally did it I tried picking numbers rather than just dividing by -a to get b > c. This is what I got:

(1) If a < 0, then perhaps a = -1. Then if b =3 and c=2, ab < ac and c > b. But if b = -2 and c = -3, ab < ac and b > c.
I'm not finding any rules I'm missing in the problem, so my answer was C.. or E... Why is the above reasoning wrong?

Thanks...

$$ab<ac$$ can be rewritten as $$ab-ac<0$$ or $$a(b-c)<0$$, which means that $$a$$ and $$b-c$$ have opposite signs.

(1) $$a<0$$ then necessarily $$b-c>0.$$
Sufficient.

(2) $$c<0$$
Obviously not sufficient, as we know nothing about $$b.$$

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Posts: 39575
Re: If ab < ac, which is greater, b or c? [#permalink]

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25 Aug 2012, 02:19
paldrich1287 wrote:
(Note: I found two older topics for this question, but they were each over a year old and I assumed there were bump rules here..)

This is from Kaplan CAT3:

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

(1) a < 0

(2) c < 0

I just have one issue with this problem: when I originally did it I tried picking numbers rather than just dividing by -a to get b > c. This is what I got:

(1) If a < 0, then perhaps a = -1. Then if b =3 and c=2, ab < ac and c > b. But if b = -2 and c = -3, ab < ac and b > c.
I'm not finding any rules I'm missing in the problem, so my answer was C.. or E... Why is the above reasoning wrong?

Thanks...

Merging similar topics. Please refer to the solutions above.
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Joined: 21 Jun 2012
Posts: 3
Re: If ab < ac, which is greater, b or c? [#permalink]

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25 Aug 2012, 11:48
Capricorn369 wrote:
Above c>b in yellow is incorrect. It is c<b becasue b =3 and c=2. Hence answer is A.

D'oh! I figured it was something very obvious.. thanks..
Intern
Joined: 01 Aug 2010
Posts: 17
Re: If ab<ac , which is greater b or c ? i. a<0 2. c<0 [#permalink]

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02 Nov 2012, 16:01
the difficulty level tag assigned to this question is 600-700, but that is a wide range, can some one try to put it in a narrower band, i.e. is it a 610-630 or 670-700 type question.
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Re: If ab<ac , which is greater b or c ? i. a<0 2. c<0 [#permalink]

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23 Apr 2016, 10:47
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If ab<ac , which is greater b or c ? i. a<0 2. c<0   [#permalink] 23 Apr 2016, 10:47
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