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If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0
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24 Aug 2010, 17:41
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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
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24 Aug 2010, 18:00
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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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
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24 Aug 2010, 20:57
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 ab>0 it is clear that 5+2>0 52>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
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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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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
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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
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Re: GMAT PREP DS
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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)
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Re: GMATPREP  Please explain my flaw
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14 Apr 2011, 10:59
sonnco wrote: ax + b = 0, is x > 0? 1. a + b > 0 2. a  b > 0 How is the OA answer right?! Your input is greatly appreciated. \(ax+b=0\) \(ax=b\) \(x=\frac{b}{a}\) \(Is \hspace{3} x > 0\) \(Is \hspace{3} \frac{b}{a} > 0\) \(Is \hspace{3} \frac{b}{a} < 0\) OR Does a and b have opposite signs? 1. \(a + b > 0\) \(a>b\) It doesn't tell us anything about sign. a=5 b=2 a>b as 5>2 Same sign. No. a=5 b=2 a>b as 5>2 Opposite signs. Yes. Not Sufficient. 2. \(a  b > 0\) \(a>b\) It doesn't tell us anything about sign. a=5 b=2 a>b as 5>2 Same sign. No. a=5 b=2 a>b as 5>2 Opposite signs. Yes. Combining both; a > b Using same sample set: a=5, b=2, No. a=5, b=2, Yes. Not Sufficient. Ans: "E"
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Re: GMATPREP  Please explain my flaw
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14 Apr 2011, 11:27
sonnco wrote: ax + b = 0, is x > 0?
1. a + b > 0 2. a  b > 0
You can look at this geometrically  because the equation is that of a line, the question is set up so that you can look at it as a coordinate geometry question. The equation y = ax + b is the equation of a line with slope a and yintercept b. When we plug in y=0, we are finding the xintercept of the line. So the question is just asking if the xintercept of the line y=ax + b is positive. In a diagram, that would be true if our line crosses the xaxis to the right of the origin (0,0). Now, using both statements, if a > b and a > b, then a > b. So a is clearly positive, and our line has a positive slope and is climbing as you move to the right. We can now just draw two different scenarios on the coordinate plane, one where b is positive and one where b is negative. If you choose, say, b = 1 and a = 5, then if you quickly sketch your line you'll see that the xintercept of y = ax + b is negative, and if b=1 and a = 5 then if you quickly sketch your line you'll see that the xintercept of y=ax+b is positive, so not sufficient, and the answer is E.
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Re: GMATPREP  Please explain my flaw
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14 Apr 2011, 22:08
(1) Let a = 2, b = 6 2 * 3 + 6 = 0, x = 3 < 0 Let a = 2, b = 6 2 * 3 + 6 = 0, x = 3 > 0 (1) is insufficient (2) Let a = 6 b = 3 6 * 1/2  3 = 0, x = 1/2 > 0 Let a = 3 b = 4 3 * 4/3  4 = 0, x = 4/3 < 0 (2) is insufficient (1) and (2) 2a > 0 => a > 0 So b can be ve or +ve and a > b So x can be +ve or ve so that ax + b = 0 Let a = 6, b = 5 or 5 6 * 5/6 + 5 = 0, x < 0 6 * 5/6  5 = 0, x > 0 Answer  E
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Re: GMATPREP  Please explain my flaw
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14 Apr 2011, 22:46
sonnco wrote: ax + b = 0, is x > 0?
1. a + b > 0 2. a  b > 0
x = b/a. The problem boils down to.  Do we know sign of a and b??1) + 2). a + b > 0 a  b > 0 Adding we get a > 0. Also absolute value of a is greater than absolute value of b. However nothing can be concluded about the sign of b. Hence E.



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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0
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30 Jun 2013, 07:48
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 \((a,b)>> (+,+) , (,+), (+,)\) \(ab>0\) 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
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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 \((a,b)>> (+,+) , (,+), (+,)\)
\(ab>0\) 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
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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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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
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25 Oct 2013, 08:26
Priya15081 wrote: Bunuel wrote: Given: \(b=ax\). Question: is \(x>0\)
(1) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient.
(2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0
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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 < . 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
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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 < . 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*(1x) >0 now this is possible only if either a>0 and (1x)>0 or a <0 and (1x) <0 Hope it helps
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If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0
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29 Nov 2014, 16:42
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



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If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0
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30 Nov 2014, 21:24
we have a+b>0 ab>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? some please help... Bunuel



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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0
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01 Dec 2014, 03:54



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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0
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09 Apr 2015, 20:38
Would this approach be wrong?
Choose two numbers: a=3, b=2. a+b>0 and ab>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.




Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0 &nbs
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