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:

(1) a+b>0 --> a-ax>0 --> a(1-x)>0 --> either a>0 and 1-x>0, so x<1ORa<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>-1ORa<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.

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.
_________________

(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.

Answer: E.

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.

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)
_________________

(1) a+b>0 --> a-ax>0 --> a(1-x)>0 --> either a>0 and 1-x>0, so x<1ORa<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>-1ORa<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.

Answer: E.

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

(1) a+b>0 --> a-ax>0 --> a(1-x)>0 --> either a>0 and 1-x>0, so x<1ORa<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>-1ORa<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.

Answer: E.

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?
_________________

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]
12 May 2014, 08:22

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 < . Can Somebody please explain this

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]
13 May 2014, 00:16

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 < . Can Somebody please explain this

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.”

gmatclubot

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0
[#permalink]
13 May 2014, 00:16