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

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

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New post 15 Jun 2018, 08:48
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uzzy12 wrote:
If ax + b = 0, is x > 0

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



Given ax + b = 0, hence x = -b/a, & is x > 0?

Statement 1:

a + b > 0

a = 1, b = 0, x = 0
a = -1, b = 2, x = 2 > 0

Statement 1 alone is Not Sufficient.

Statement 2:

a - b > 0

a = 1, b = 0, x = 0
a =1, b = -1, x = 1 > 0

Statement 2 alone is Not Sufficient.

Combining both statements,
a = 1, b = 0, x = 0
a = 2, b = -1, x = 1/2

Combing both statements is Not Sufficient.

Answer E.


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

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New post 14 Oct 2018, 00:01
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

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


(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.


Sum (1) and (2), we can get a>0. Subtract (1) from (2), we can get b>0. In this setting, ax+b=0 means x is negative. Thus, I thought C is correct. What am I wrong?
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If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

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New post 24 Nov 2018, 16:35
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A video explanation can be found here:
https://www.youtube.com/watch?v=cYPgNTx-XK8&t=5s

For DS questions, always ask "What would I need to know in order to answer this question?" In this case, isolate x:

ax + b = 0

ax = -b

x = -b/a

Whether x is positive depends upon the signs of a and b -- The real question is, "What are the signs of a and b?"

(1) either both numbers are positive, or one is positive and one is negative. Insifficient.

(2) same problem. Insufficient.

Together, if you add the two inequalities you'll have 2a > 0, which means a > 0, but that's only half of what we need to know...

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

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New post 22 Feb 2019, 11:09
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I love this question. Probably, because one can solve it under 1 minute if you are clear on the rules of positive & negative (in inequality).

The question is saying the following: ax + b =0 => b=-ax or => -b/a = x

So the rephrased question is not to check x > 0 but is -b/a > 0. Now this has two implications as follows:-

1. Case I: If a > 0, b < 0 [Note: ',' indicates and]
2. Case II: If a <0, b > 0

So another deep down will give us the following phrasing:- Is Case I or Case II true (If either one of them is true, then you can see we will get our answer to the question)

Now, Statement A: a + b > 0 i.e. we don't the 'sign' of either a or b: both could be positive; one could be large positive, one could be small negative. So clearly, Not Sufficient

Moving on to statement B: a - b > 0 i.e. same as the above statement only with a minor tweak, hence not sufficient

Combining both the statement: Now inequalities can be added in the same direction; therefore, we get the following:-
2a > 0 => a > 0
Do you think this information is enough to answer the question?. The answer is a resounding 'No'. Why is that?
Well, the statement only gives us the sign of a; however, we don't know whether b is either > 0 or less < 0 or even equal to 0.

Kudos, if you liked the solution

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

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New post 08 Mar 2019, 08:26
There are four possibilities for the signs of a and b: ++, --, +- or -+
Picking numbers:

(1) a + b > 0

a=1,b=1 --> 1x + 1 = 0, x is -1
a=1, b=-1 --> 1x - 1 = 0, x is 1
(can stop here, insufficient)
a=-1, b=1 --> -1x + 1 = 0, x is 1
a=-1, b=-1 --> -1x + 1 = 0, x is 1

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

a=2, b=1 --> 2x + 1 = 0, x is -1/2
a=2, b=-1 --> 2x - 1 = 0, x is is 1/2
Also insufficient

1&2 together don't give us anything new, E
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0  [#permalink]

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New post 17 Apr 2019, 23:23
Bunuel
On your solution of (1)+(2) in which you sum (1) and (2) as they have same sign>> (a+b)+(a−b)>0 and get a>0.
Question is whenever I encountered same sign, can I also subtract LHS terms >> (a+b)-(a−b)>0 and get b>0.
I was mistakenly did so with this Q and from (1)+(3); I got both a>0 (combine LHS) and b>0 (Subtract LHS), which results in X<0, if ax + b = 0, corresponding to ans. C

Please correct me on this point so that next time I can apply only the combination of LHS and not subtraction in context of equation with same sign.

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

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New post 17 Apr 2019, 23:31
abc555 wrote:
Bunuel
On your solution of (1)+(2) in which you sum (1) and (2) as they have same sign>> (a+b)+(a−b)>0 and get a>0.
Question is whenever I encountered same sign, can I also subtract LHS terms >> (a+b)-(a−b)>0 and get b>0.
I was mistakenly did so with this Q and from (1)+(3); I got both a>0 (combine LHS) and b>0 (Subtract LHS), which results in X<0, if ax + b = 0, corresponding to ans. C

Please correct me on this point so that next time I can apply only the combination of LHS and not subtraction in context of equation with same sign.

Thank you!


Check : Manipulating Inequalities (adding, subtracting, squaring etc.).

For more check Ultimate GMAT Quantitative Megathread


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

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New post 18 May 2019, 07:31
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uzzy12 wrote:
If ax + b = 0, is x > 0

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


Target question: Is x > 0

Given: ax + b = 0

Statement 1: a + b > 0
At this point, we have 1 equation, 1 inequality, and THREE variables.
Even if we had 2 EQUATIONS and 3 variables, we probably wouldn't be able to make any conclusions about whether x is positive or negative.
Given this, let's TEST some values
There are several values of a, b and x that satisfy statement 1 (and the given equation ax + b = 0). Here are two:
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: a - b > 0
This is the same scenario as statement 1, so let's TEST some values
There are several values of a, b and x that satisfy statement 1 (and the given equation ax + b = 0). Here are two:
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0   [#permalink] 18 May 2019, 07:31

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