It is currently 24 May 2020, 18:29

### GMAT Club Daily Prep

#### 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

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

# ab2c3 > 0 a2b2c3 < 0

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Intern
Joined: 22 Aug 2018
Posts: 29
Followers: 0

Kudos [?]: 5 [0], given: 4

ab2c3 > 0 a2b2c3 < 0 [#permalink]  18 Sep 2018, 04:33
00:00

Question Stats:

62% (00:33) correct 37% (00:47) wrong based on 40 sessions
$$ab^2c^3 > 0$$
$$a^2b^2c^3 < 0$$

 Quantity A Quantity B $$ab^2$$ $$ac$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

How to approach this question?
[Reveal] Spoiler: OA
Intern
Joined: 10 Sep 2018
Posts: 20
Followers: 0

Kudos [?]: 23 [0], given: 9

Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]  18 Sep 2018, 04:46
the first inequality is true if a<0 & c<0 or a>0 & c>0.

second inequality is true if c<0.

that's why it must be that a<0.

so we have a<0, b^2>0, and c<0.

thus the answer is B.
Intern
Joined: 22 Aug 2018
Posts: 29
Followers: 0

Kudos [?]: 5 [0], given: 4

Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]  18 Sep 2018, 04:54
saifee wrote:
the first inequality is true if a<0 & c<0 or a>0 & c>0.

second inequality is true if c<0.

that's why it must be that a<0.

so we have a<0, b^2>0, and c<0.

thus the answer is B.

I did not understand this - > "that's why it must be that a<0"
Intern
Joined: 10 Sep 2018
Posts: 20
Followers: 0

Kudos [?]: 23 [0], given: 9

Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]  18 Sep 2018, 05:00
read the first line. because we've got c<0 from the second inequality, we must conclude a<0, too.
Retired Moderator
Joined: 07 Jun 2014
Posts: 4806
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 162

Kudos [?]: 2654 [1] , given: 394

Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]  18 Sep 2018, 15:53
1
KUDOS
Expert's post
Start from the second inequality

$$a^2b^2c<0$$. Since $$]a^2b^2$$ cannot be less than zero, as they are squares. c must be less than 0. Now with that information look at the first inequality.

$$ab^2c^3>0$$ since c<0 a must be less than 0.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Joined: 22 Aug 2018
Posts: 29
Followers: 0

Kudos [?]: 5 [0], given: 4

Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]  19 Sep 2018, 09:05
sandy wrote:
Start from the second inequality

$$a^2b^2c<0$$. Since $$]a^2b^2$$ cannot be less than zero, as they are squares. c must be less than 0. Now with that information look at the first inequality.

$$ab^2c^3>0$$ since c<0 a must be less than 0.

If we in the beginning itself, divide both the sides by "a", then we will have $$b^2$$ and $$c$$ left, in that case, wouldn't A become bigger?
Retired Moderator
Joined: 07 Jun 2014
Posts: 4806
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 162

Kudos [?]: 2654 [1] , given: 394

Re: ab2c3 > 0 a2b2c3 < 0 [#permalink]  19 Sep 2018, 17:25
1
KUDOS
Expert's post
rajlal wrote:

If we in the beginning itself, divide both the sides by "a", then we will have $$b^2$$ and $$c$$ left, in that case, wouldn't A become bigger?

You cant divide with a if you dont know its sign.

Division with a negative number would flip the inequality.

For example

-1 < 5 divide both sides with -1 you get 1> -5.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Re: ab2c3 > 0 a2b2c3 < 0   [#permalink] 19 Sep 2018, 17:25
Display posts from previous: Sort by

# ab2c3 > 0 a2b2c3 < 0

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.