BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GMAT Prep - DS Inequalities

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Mon Oct 22, 2007 10:24 am
Thanked: 1 times

GMAT Prep - DS Inequalities

by oks » Fri Dec 26, 2008 4:13 pm
If ax + b = 0, is x > 0?

a. a+b > 0
b. a-b > 0

I thought the answer was B, but OA is E.

My approach: ax = -b, so either a > 0 and x < 0 or a <0 and x > 0.

If a-b> 0, a>b, which means that a is positive and x is negative.

Please explain. Thanks!
Top
Source: — Data Sufficiency |
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Mon Dec 08, 2008 6:53 am
Thanked: 1 times

by crazykrans » Fri Dec 26, 2008 4:38 pm
If ax + b = 0, is x > 0?

a. a+b > 0
b. a-b > 0

Oks, Ans is E

In your solution, make sure you consider the case a=2(positive) and b=-1 (negative)when evaluating a-b>0
Top
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Mon Oct 22, 2007 10:24 am
Thanked: 1 times

by oks » Sat Dec 27, 2008 3:33 pm
Crazykrans, I am not sure I follow.

If a=2(positive) and b=-1, then a is > 0, so the answer should be B?
Top
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 am
Thanked: 146 times
Followed by:2 members

Re: GMAT Prep - DS Inequalities

by parallel_chase » Sat Dec 27, 2008 3:45 pm
oks wrote:If ax + b = 0, is x > 0?

a. a+b > 0
b. a-b > 0
ax + b = 0
ax = -b

x > 0 ?? or is x = positive

Statement I

a+b > 0
a - ax > 0
a( 1-x) > 0

either a & (1-x) both positive or both negative.

x could be positive or negative

Insufficient.

Statement II
a-b > 0
a+ax > 0
a (1+x) > 0

either a & (1+x) both positive or both negative.

x could be either negative or positive

Insufficient.

Combining I & II

We only know that x could be positive or negative.

Hence E.

Hope this helps.
No rest for the Wicked....
Top
Legendary Member
Posts: 621
Joined: Wed Apr 09, 2008 7:13 pm
Thanked: 33 times
Followed by:4 members

by vittalgmat » Sat Dec 27, 2008 9:52 pm
Got E..
procedure is similar to parallel_chase. Let me add a little more detail.

from the stem
ax = -b OR b = -ax

stmt 1:
a+b > 0
=> a -ax > 0
=> a(1 -x) > 0

=> either a is +ve OR (1 -x) is +ve
=> ie 1 -x > 0 => x < 1. X could be +ve (values 0 and 1. maybe decimals) OR -ve (x can take any value < 1...)
So Not sufficient.


stmt2:
a -b >0
a - (ax) > 0
a(1 +x) > 0
a +ve OR 1 +x is +ve
1 +x > 0 => x > -1
Again X could be +ve or -ve.

Combining the two.
Here we have 2 possibilities.
1) a is +ve. Then we dont know anything about x. so not sufficient.

2) x > -1 and x < 1. There could be an infinite decimal numbers between these two values; some are +ve and some are -ve.
So insufficient as well.

From both cases we cant determine x +ve or -ve.
so E


HT Helps
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Sat Dec 27, 2008 9:58 pm
PC / Vittal,
Good explanations!

Regards,
Cramya
Top
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 am
Thanked: 82 times
Followed by:9 members
GMAT Score:720

by maihuna » Mon Dec 29, 2008 4:10 am
from stem x = -b/a

so one needs to determine whether a n b are of opposite sign, as that is the only way to make x positive...

from 1. a+b>0 a n b can be choosen not to satisfy req i.e a = 3 b =2 x-ve
a = 3 b = -1 x +ve

from 2. a-b >0 => a>b so still a = 3 b = 2 holds true x -ve
a=3 b=-1 x +ve
combine also doesnt help as for a = 3 b=2 x -ve
a = 3 b = -1 x +ve
Top
Post new topic Post Reply
• Page 1 of 1