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

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If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink] New post 24 Aug 2010, 16:41
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If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0
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Re: GMAT PREP DS [#permalink] New post 24 Aug 2010, 17:00
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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.

Answer: E.
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Re: GMAT PREP DS [#permalink] New post 24 Aug 2010, 19:57
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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.
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Re: GMAT PREP DS [#permalink] New post 15 Mar 2011, 22:59
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.

Answer: E.

you've made things easier bunuel. +1
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Re: GMAT PREP DS [#permalink] New post 16 Mar 2011, 23:09
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
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Re: GMAT PREP DS [#permalink] New post 18 Mar 2011, 13:03
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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)
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink] New post 30 Jun 2013, 06:48
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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?
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink] New post 27 Jul 2013, 09:24
Awesomely explained by @frozzy
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink] New post 27 Jul 2013, 10:17
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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.
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Re: GMAT PREP DS [#permalink] New post 25 Oct 2013, 06:42
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.

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
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Re: GMAT PREP DS [#permalink] New post 25 Oct 2013, 07:26
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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.

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?
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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