Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 23 Aug 2016, 22:25

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0

Author Message
TAGS:

### Hide Tags

Intern
Joined: 14 Sep 2009
Posts: 29
Followers: 0

Kudos [?]: 35 [2] , given: 22

If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

24 Aug 2010, 17:41
2
KUDOS
29
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

56% (02:25) correct 44% (01:18) wrong based on 592 sessions

### HideShow timer Statistics

If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 34393
Followers: 6245

Kudos [?]: 79323 [25] , given: 10016

### Show Tags

24 Aug 2010, 18:00
25
KUDOS
Expert's post
27
This post was
BOOKMARKED
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0

Given: $$b=-ax$$. Question: is $$x>0$$

(1) $$a+b>0$$ --> $$a-ax>0$$ --> $$a(1-x)>0$$ --> either $$a>0$$ and $$1-x>0$$, so $$x<1$$ OR $$a<0$$ and $$1-x<0$$, so $$x>1$$. Not sufficient.

(2) $$a-b>0$$ --> $$a+ax>0$$ --> $$a(1+x)>0$$ --> either $$a>0$$ and $$1+x>0$$, so $$x>-1$$ OR $$a<0$$ and $$1+x<0$$, so $$x<-1$$. Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> $$(a+b)+(a-b)>0$$ --> $$a>0$$, so we have first range from (1): $$x<1$$ and first case from (2): $$x>-1$$ --> $$-1<x<1$$, so $$x$$ may or may not be negative. Not sufficient.

_________________
Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 690
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Followers: 15

Kudos [?]: 136 [10] , given: 15

### Show Tags

24 Aug 2010, 20:57
10
KUDOS
As usual Bunuel's approach is elegant.

Here is how I did it:

x=-b/a; x>0 means a and b have to be different sign. I am usually not good at picking numbers, but in this case if you see a+b>0 and a-b>0 it is clear that 5+2>0 5-2>0 and 5+(-2)>0 5-(-2)>0 hence we cannot say that a and b will have the same sign. Hence E.
_________________

Consider kudos, they are good for health

Manager
Status: what we want to do, do it as soon as possible
Joined: 24 May 2010
Posts: 114
Location: Vietnam
WE 1: 5.0
Followers: 2

Kudos [?]: 55 [0], given: 315

### Show Tags

15 Mar 2011, 23:59
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0

Given: $$b=-ax$$. Question: is $$x>0$$

(1) $$a+b>0$$ --> $$a-ax>0$$ --> $$a(1-x)>0$$ --> either $$a>0$$ and $$1-x>0$$, so $$x<1$$ OR $$a<0$$ and $$1-x<0$$, so $$x>1$$. Not sufficient.

(2) $$a-b>0$$ --> $$a+ax>0$$ --> $$a(1+x)>0$$ --> either $$a>0$$ and $$1+x>0$$, so $$x>-1$$ OR $$a<0$$ and $$1+x<0$$, so $$x<-1$$. Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> $$(a+b)+(a-b)>0$$ --> $$a>0$$, so we have first range from (1): $$x<1$$ and first case from (2): $$x>-1$$ --> $$-1<x<1$$, so $$x$$ may or may not be negative. Not sufficient.

you've made things easier bunuel. +1
_________________

Consider giving me kudos if you find my explanations helpful so i can learn how to express ideas to people more understandable.

SVP
Joined: 16 Nov 2010
Posts: 1673
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 34

Kudos [?]: 470 [0], given: 36

### Show Tags

17 Mar 2011, 00:09
I've also tried this using numbers:

From (1) we can see that:

let a = 4 b = 6

So a * - 3/2 + b = 0 and x > 0

But if a = -4 and b = 6

then a * 3/2 + 6 = 0 and x < 0

So (1) is not sufficient

From (2) :

If a = 2, b = -2

then a *1 + b = 0 and x > 0

if a = 4 and b = 2 then

a* -1/2 + 2 = 0 whereby x < 0

Combining (1) and (2):

a > 0 , but b may or may not be > 0, so the answer is E
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Manhattan GMAT Instructor
Joined: 29 Apr 2010
Posts: 126
Followers: 55

Kudos [?]: 321 [0], given: 1

### Show Tags

18 Mar 2011, 14:03
awesome explanations both bunuel and mainhoon --these two approaches would probably serve two different types of students really well. nice work!

there's also a potential blend of the two-- mainhoon's rephrase (are the signs of a and b different) and the stacking/adding of inequalities when evaluating choice c for that same rephrased question (if your head gets turned around when too many numbers are involved)
_________________

JP Park | Manhattan GMAT Instructor | Los Angeles

Manhattan GMAT Discount | Manhattan GMAT Reviews

Director
Joined: 29 Nov 2012
Posts: 900
Followers: 14

Kudos [?]: 872 [4] , given: 543

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

30 Jun 2013, 07:48
4
KUDOS
3
This post was
BOOKMARKED
given $$ax=- b$$ then x = $$\frac{-b}{a}$$ question becomes is $$\frac{-b}{a}>0$$

From statement 1

$$a+b > 0$$ so the values for $$(a,b)>> (+,+) , (-,+), (+,-)$$

$$a-b>0$$ so the values for $$(a,b) >> (+,+), (+,-)$$

Combination still 2 cases remain. Hence E

Is this method safe?
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Intern
Joined: 17 May 2013
Posts: 8
Followers: 0

Kudos [?]: 13 [0], given: 7

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

27 Jul 2013, 10:24
Awesomely explained by @frozzy
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 630
Followers: 76

Kudos [?]: 1005 [0], given: 136

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

27 Jul 2013, 11:17
fozzzy wrote:
given $$ax=- b$$ then x = $$\frac{-b}{a}$$ question becomes is $$\frac{-b}{a}>0$$

From statement 1

$$a+b > 0$$ so the values for $$(a,b)>> (+,+) , (-,+), (+,-)$$

$$a-b>0$$ so the values for (a,b) >> (+,+), (+,-)

Combination still 2 cases remain. Hence E

Is this method safe?

Actually, all the possible cases are : (+,+) = 7,5 ; (+,-) = 7,-5 ; (-,-) = -5,-7.

Otherwise, IMO it is all good.
_________________
Intern
Joined: 18 Jan 2013
Posts: 8
GPA: 3.71
Followers: 0

Kudos [?]: 0 [0], given: 246

### Show Tags

25 Oct 2013, 07:42
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0

Given: $$b=-ax$$. Question: is $$x>0$$

(1) $$a+b>0$$ --> $$a-ax>0$$ --> $$a(1-x)>0$$ --> either $$a>0$$ and $$1-x>0$$, so $$x<1$$ OR $$a<0$$ and $$1-x<0$$, so $$x>1$$. Not sufficient.

(2) $$a-b>0$$ --> $$a+ax>0$$ --> $$a(1+x)>0$$ --> either $$a>0$$ and $$1+x>0$$, so $$x>-1$$ OR $$a<0$$ and $$1+x<0$$, so $$x<-1$$. Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> $$(a+b)+(a-b)>0$$ --> $$a>0$$, so we have first range from (1): $$x<1$$ and first case from (2): $$x>-1$$ --> $$-1<x<1$$, so $$x$$ may or may not be negative. Not sufficient.

Great solution sir, but can you correct me?? I took option a) because to satisfy a+b>0. one of them ,a or b, must be negative. So, if a is positive b is negative. x should be positive. On the contrary, if b is positive, a must be negative. So the x should be positive to keep the negative sign of a alive. Thank you. Waiting for your assistance
Math Expert
Joined: 02 Sep 2009
Posts: 34393
Followers: 6245

Kudos [?]: 79323 [0], given: 10016

### Show Tags

25 Oct 2013, 08:26
Priya15081 wrote:
Bunuel wrote:
Given: $$b=-ax$$. Question: is $$x>0$$

(1) $$a+b>0$$ --> $$a-ax>0$$ --> $$a(1-x)>0$$ --> either $$a>0$$ and $$1-x>0$$, so $$x<1$$ OR $$a<0$$ and $$1-x<0$$, so $$x>1$$. Not sufficient.

(2) $$a-b>0$$ --> $$a+ax>0$$ --> $$a(1+x)>0$$ --> either $$a>0$$ and $$1+x>0$$, so $$x>-1$$ OR $$a<0$$ and $$1+x<0$$, so $$x<-1$$. Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> $$(a+b)+(a-b)>0$$ --> $$a>0$$, so we have first range from (1): $$x<1$$ and first case from (2): $$x>-1$$ --> $$-1<x<1$$, so $$x$$ may or may not be negative. Not sufficient.

Great solution sir, but can you correct me?? I took option a) because to satisfy a+b>0. one of them ,a or b, must be negative. So, if a is positive b is negative. x should be positive. On the contrary, if b is positive, a must be negative. So the x should be positive to keep the negative sign of a alive. Thank you. Waiting for your assistance

Why does $$a+b>0$$ imply that either a or b is negative? Why cannot both be positive?
_________________
Intern
Joined: 24 Dec 2013
Posts: 2
Followers: 0

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

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

12 May 2014, 09:22
I don't understand fossey or bunuel's solution.
Bunuel's soln- I'm kinda confused on how you are marking as both either > and > or < and < .
Moderator
Joined: 25 Apr 2012
Posts: 728
Location: India
GPA: 3.21
Followers: 42

Kudos [?]: 615 [0], given: 723

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

13 May 2014, 01:16
nivi123 wrote:
I don't understand fossey or bunuel's solution.
Bunuel's soln- I'm kinda confused on how you are marking as both either > and > or < and < .

Hi,

What is the product when 2 positive nos a,b are multiplied. The product is >0
When a positive number and a negative number is multipled the product is <0
When 2 negative nos are multiplied then the product is >0

Now in the given inequality arrived by simplifying the given expression we get that a*(1-x) >0 now this is possible only if either a>0 and (1-x)>0 or a <0 and (1-x) <0

Hope it helps
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Senior Manager
Joined: 10 Mar 2013
Posts: 290
GMAT 1: 620 Q44 V31
GMAT 2: 690 Q47 V37
GMAT 3: 610 Q47 V28
GMAT 4: 700 Q50 V34
GMAT 5: 700 Q49 V36
GMAT 6: 690 Q48 V35
GMAT 7: 750 Q49 V42
GMAT 8: 730 Q50 V39
Followers: 5

Kudos [?]: 85 [1] , given: 2404

If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

29 Nov 2014, 16:42
1
KUDOS
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0

Guessing numbers helped me.

x = -b/a
Is -b/a > 0?

(1) a > -b
a = 5, b =3 => N
a = 5, b=-3 => Y
NS
(2) a > b
Same numbers as in (1)
NS

(1) + (2)
a > |b|
Same numbers used in (1) and in (2)
NS
Intern
Status: Brushing up rusted Verbal....
Joined: 30 Oct 2013
Posts: 17
Location: India
Schools: AGSM '16
GMAT Date: 11-30-2014
GPA: 3.96
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 7 [0], given: 27

If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

30 Nov 2014, 21:24
we have a+b>0
a-b>0
if we add both the equations. a>0
if we subtract both the equations, b>0
then we know both have same signs....
hence can determine, sign of x=-b/a is >0 or not
What is wrong in my explanation?
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 34393
Followers: 6245

Kudos [?]: 79323 [1] , given: 10016

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

01 Dec 2014, 03:54
1
KUDOS
Expert's post
sudd1 wrote:
we have a+b>0
a-b>0
if we add both the equations. a>0
if we subtract both the equations, b>0
then we know both have same signs....
hence can determine, sign of x=-b/a is >0 or not
What is wrong in my explanation?
Bunuel

For two inequalities, you can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

So, we cannot subtract a+b>0 from a-b>0.

For more check: inequalities-tips-and-hints-175001.html
_________________
Intern
Joined: 09 Apr 2015
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 14

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

09 Apr 2015, 20:38
Would this approach be wrong?

Choose two numbers: a=3, b=2.
a+b>0 and a-b>0 are individually and together fulfilled.

ax+b= 3*x+2=0 => x=-2/3 < 0 => we have shown that both I and II are insufficient.
Math Expert
Joined: 02 Sep 2009
Posts: 34393
Followers: 6245

Kudos [?]: 79323 [0], given: 10016

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

10 Apr 2015, 04:19
meltedcheese wrote:
Would this approach be wrong?

Choose two numbers: a=3, b=2.
a+b>0 and a-b>0 are individually and together fulfilled.

ax+b= 3*x+2=0 => x=-2/3 < 0 => we have shown that both I and II are insufficient.

To get insufficiency you should get both a NO and an YES answers to the question. You have a NO answer, so to get that the statements are insufficient, you should choose numbers which give an YES answer to the question.
_________________
Manager
Joined: 04 Jan 2014
Posts: 103
Followers: 0

Kudos [?]: 29 [0], given: 20

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

12 Apr 2015, 06:15

ax + b = 0

So X=-b/a

1) a+b>0

So, a>-b--- for a>0

1>-b/a
1>x
or -a>-b--- for a<0
1<-x
x<-1

2) a-b>0

a>b---- for a>0
1<b/a
x<-1

-a>b--- for a<0
1<-b/a
1<x
VP
Joined: 09 Jun 2010
Posts: 1331
Followers: 3

Kudos [?]: 96 [0], given: 768

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

### Show Tags

27 Apr 2015, 21:42
this question is hard.

we should consider/study gmatprep questions carefully because those questions prepresent what gmat test and think.
Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0   [#permalink] 27 Apr 2015, 21:42

Go to page    1   2    Next  [ 28 posts ]

Similar topics Replies Last post
Similar
Topics:
1 If abc ≠ 0, is a > 0? (1) 3*a/b > 0 (2) b/c^2 < 0 3 16 Sep 2015, 03:51
4 If a^2 = b, is 1 > a > 0? 3 29 Jul 2015, 03:11
1 Is x=y ? a.x^2 - y =0 b.x-1>0 1 15 Jun 2011, 11:28
3 ax + b = 0, is x > 0? 1. a + b > 0 2. a - b > 0 How 6 14 Apr 2011, 10:41
11 Is a^2*b > 0? 19 17 Jun 2007, 21:58
Display posts from previous: Sort by