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

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Bunuel

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01 Apr 2017, 10:05

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jamescath

Bunuel

uzzy12

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.

Bunuel I have a doubt in equality:
a>0 and 1−x>0, so x<1 OR a<0 and 1−x<0, so x>1. not able to get how did a>0 turned to a<0 and x<1 to 1-x<0

We have \(a(1-x)>0\). For a product of two numbers to be positive they must have the same sign. Two scenarios:

1. Both are positive:
\(a>0\) and \(1-x>0\).
\(1-x>0\) is the same as \(x<1\).

2. Both are negative:
\(a<0\) and \(1-x<0\).
\(1-x<0\) is the same as \(x>1\).

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Bunuel

uzzy12

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.

Bunuel Please help me in understanding this by (1)+(2) we came up to \(a>0\), now from question we have \(-b=ax\) which means product of ax is negative and by that logic would not x be <0 as we have got \(a>0\).... Please help me in understanding flaw in my logic

The highlighted part is not correct.

From \(b=-ax\) knowing that \(a>0\), does not necessarily mean that x is negative. Consider x = -1/2 and x = 1/2. Notice that we know nothing about b, we don't know whether it's positive or negative.

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Hi Bunuel

Not very sure how you got x>1 region on the numberline.

From stmt 1, I got a-ax>0 or a(1-x)>0 or a(x-1)<0 and on the number line ---(-)----1-----(+)-------- ie x<1.

Pls clarify.

Bunuel

uzzy12

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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Hi Bunuel

Not very sure how you got x>1 region on the numberline.

From stmt 1, I got a-ax>0 or a(1-x)>0 or a(x-1)<0 and on the number line ---(-)----1-----(+)-------- ie x<1.

Pls clarify.

Bunuel

uzzy12

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.

\(a-ax>0\) --> \(a(1-x)>0\)

So, a and 1-x have the same sign:

Both are positive: \(a>0\) and \(1-x>0\), so \(x<1\)

Both are negative: \(a<0\) and \(1-x<0\), so \(x>1\).

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Bunuel

uzzy12

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.

Hi Bunuel

Why are we doing OR in ur statements?

b=-ax. Hence, a-ax>0 then a(1-x)>0. therefore, a>0 or 1-x >0 so 1>x. Thats all..How can we say x<1?

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zanaik89

Bunuel

uzzy12

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.

Hi Bunuel

Why are we doing OR in ur statements?

b=-ax. Hence, a-ax>0 then a(1-x)>0. therefore, a>0 or 1-x >0 so 1>x. Thats all..How can we say x<1?

We have that \(a(1-x)>0\), so the product of two multiples a and 1 - x is positive. This to be true either both must be positive OR both must be negative.

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uzzy12

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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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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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

Regards,
Abheek

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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!

Bunuel

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

For more check Ultimate GMAT Quantitative Megathread

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uzzy12

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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uzzy12

If ax + b = 0, is x > 0

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

Solution:

The easiest method to solve a DS question depends on the question. For certain questions (especially for statements which are not sufficient to provide an answer), the question can be easily solved by plugging in numbers. Other questions require you to use algebraic manipulations to arrive at the correct answer. In either case, you can use a combination of both methods; i.e. use algebraic manipulations to get some information about the statements and plug in numbers only after that so that you don’t waste time trying unnecessary cases.

Let me show what I mean on one of the examples you provided.

In the first example, assuming that a is non-zero, we can write x = -b/a. This expression tells us that x is -1 times the ratio of b to a; i.e. if b and a have the same sign then x is negative and if b and a have opposite signs, then x is positive. When plugging in numbers for a and b, this information tells me to only look for examples where a and b have the same and opposite signs, so that if I tried a = 1 and b = -1, I won’t waste time trying a = -1 and b = 1 since I know both will produce a positive x.

Statement One Alone:

For an example where a and b have the same sign, we can simply take a = b = 1. In this case, x + 1 = 0 and hence, x = -1 i.e. x > 0 is not true.

For an example where a and b have opposite signs, we can take a = 2 and b = -1. In this case, 2x - 1 = 0 and hence, x = 1/2 i.e. x > 0 is true.

Statement one alone is not sufficient.

Statement Two Alone:

For an example where a and b have the same sign, we can take a = 2 and b = 1. In this case, 2x + 1 = 0 and hence, x = -1/2 i.e. x > 0 is not true.

For an example where a and b have opposite signs, we can take a = 1 and b = -1. In this case, x - 1 = 0 and hence, x = 1 i.e. x > 0 is true.

Statement two alone is not sufficient.

Statements One and Two Together:

Notice that if a + b > 0 and a - b > 0 are true at the same time, then adding the inequalities side by side, we determine that 2a > 0; i.e. a > 0.

If a = 2 and b = 1, then 2x + 1 = 0 and hence, x = -1/2. Notice that a = 2 and b = 1 satisfy both a + b > 0 and a - b > 0. In this case, x < 0.

If a = 2 and b = -1, then 2x - 1 = 0 and hence, x = 1/2. Notice that a = 2 and b = - 1 satisfy both a + b > 0 and a - b > 0. In this case, x > 0.

Statements one and two together is not sufficient.

Answer: E

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One glance at the statements is enough to tell that this question is more about a and b and less about x. The equation given as data just furthers the point.

ax + b = 0. Since the question stem asks if x>0, we keep x on one side of the equation.
x = - \(\frac{b }{ a}\).

Therefore, the question stem can now be rephrased as “Is –\(\frac{b }{ a}\) > 0?”.

-(\(\frac{b}{a}\)) will be positive if (\(\frac{b}{a}\)) is negative; If (\(\frac{b}{a}\)) has to be negative, b and a should be of opposite signs.

The question can be rephrased to “Are a and b of opposite signs?”. You will see that the statements now make a lot more sense because the question is also about the signs of a and b.

From statement I alone, a+b > 0.

Note that a sum or two variables is never enough to tell you a lot about the signs, compared to a product or a division of terms.
At the most, a+b>0 can tell you that both numbers cannot be ZERO and both cannot be negative at the same time. That’s about it!

Statement I alone is insufficient to say if a and b, are of opposite signs. Answer option A and D can be eliminated. Possible answer options are B, C and E.

From statement II alone, a-b>0.

A similar thing can be said about a difference of terms.
At the most, a-b>0 tells us that both numbers are not ZERO. Nothing more, nothing less!

Statement II alone is insufficient to answer the question. Answer option B can be eliminated. Possible answer options are C and E.

Combining statements I and II, we have the following:

From statement I, a + b > 0; from statement II, a – b > 0.

Since both inequalities have the same sign, they can be added. In fact, GMAT expects you to know this property of inequalities and use it to combine statements.

Adding the two inequalities, we have
2a > 0 or a > 0.

The combination of statements tells us that a is positive but provides us with no information about b.
The combination of statements is insufficient. Answer option C can be eliminated.

The correct answer option is E.

Hope that helps!
Aravind B T

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I have a question here.

from statement 1 : a+b >0
from statement 2 : -a+b < 0 derived by multiplying the inequality by -1

Now subtracting (1) from (2). Since I know that on subtracting the inequalities, the sign of inequality is taken from the inequality I subtract from, we get a<0.

Is my approach correct? If yes, how do I get both a<0 and a>0 ?

Please help.

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When you meet a question requiring to find out whether a variable is positive or negative, don't forget it could be ZERO.

In two statements: a + b > 0; a – b > 0
If a > 0 and b = 0, then x = 0 will satisfy the equation ax + b = 0. And we know ZERO is neither positive nor negative, so neither statement is sufficient.

Jatin108

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28 Oct 2022, 07:56

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can we just mark E because none of the statements talk about "X" ?

and also we can't be sure of Sign of A and B from the statements given

chetan2u

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28 Oct 2022, 10:05

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Jatin108

can we just mark E because none of the statements talk about "X" ?

and also we can't be sure of Sign of A and B from the statements given

No, it may not be correct to mark E just because choices don’t not talk of x because x itself is related to a and b.
Of course, ‘not being sure of sign of a and b’ helps us in answering the question.

But, ‘whether knowing signs of a and b is sufficient to answer the question’ would come only after analysing the equation ax+b=0. So, merely looking at choices may not be sufficient on its own.

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30 Jan 2023, 11:44

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Hi Bunuel chetan2u KarishmaB - Two quick questions

Wanted to focus on the equation in the question stem : ax + b = 0

Q1) What is "b" in the above blue equation ? - is "b" the constant ?

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All equations have the basic template

Quote:

The standard form for linear equations in two variables is Ax+By=C.

Q2) Where is the "Y variable" (i.e. 2nd variable) in the above blue equation ?

Is the "Y variable" missing - thus I can infer, Y = 0 ?

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01 Feb 2023, 21:14

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jabhatta2

Quote:

The standard form for linear equations in two variables is Ax+By=C.

Q2) Where is the "Y variable" (i.e. 2nd variable) in the above blue equation ?

Is the "Y variable" missing - thus I can infer, Y = 0 ?

Yes, a and b are constants.

ax + b = 0 means x = -b/a
If I were to represent this on the XY plane, this would be a line parallel to the Y axis passing through x = -b/a. Hence y can take any value. There are no constraints on the value of y.