GRE Quant: Why doesn't plugging in work in this problem?

[PhdApplicant311]

Plugging in Numbers (Test it!) had been a life-saver.... but why doesn't it work here? 

If a – b > a + b, where a and b are integers, which of the following must be true?  Indicate all that apply

a < 0 
 
b < 0

ab < 0 

I tried a = 3 and b = -2 

3 - ( -2 ) > 3 + (-2) 
3 + 2 > 3 - 2
5 > 1

a < 0 or 3 < 0 -  False, or - 2 or b is less than 0 - True, 3 x -2 = -6 or ab < 0- True

However, the answer is only choice B or b < 0 is true. 

Another option I tried was a = - 3 and b = - 2, which satisfies the equation but gives me another set of answers. 

Sorry if I am being daft here! Appreciate your help.  

 

 

 

 

[gughok]

Plugging in Numbers (Test it!) had been a life-saver.... but why doesn't it work here? 

If a – b > a + b, where a and b are integers, which of the following must be true?  Indicate all that apply

a < 0 
 
b < 0

ab < 0 

I tried a = 3 and b = -2 

3 - ( -2 ) > 3 + (-2) 
3 + 2 > 3 - 2
5 > 1

a < 0 or 3 < 0 -  False, or - 2 or b is less than 0 - True, 3 x -2 = -6 or ab < 0- True

However, the answer is only choice B or b < 0 is true. 

Another option I tried was a = - 3 and b = - 2, which satisfies the equation but gives me another set of answers. 

Sorry if I am being daft here! Appreciate your help.  

 

 

 

 

Note the question asks which must be true - it isn't enough to plug one example in! You have to be sure that the answers you get will be true for any pair of a and b satisfying the inequality given. I would advise using straightforward algebra to find that (b < 0) must be true:

a - b > a + b (subtract a from both sides)

- b > b (add b to both sides)

0 > b (which is the same as...)

b < 0

So, whenever the inequality holds, it must be the case that b < 0 is true. However, your a = 3 and b = -2 example shows that (a < 0) need not be true, and if you try a = -1 and b = -2, you will find that (ab < 0) is also not necessarily true.

Therefore the only inequality that must be true is the one that follows algebraically from the original inequality: b < 0

[PhdApplicant311]

Thank you gughok! 

[Brent@GreenlightGRE]

gughok has nicely demonstrated the advantages of using algebra over plugging in numbers. In general, plugging in numbers is useful for determining statements that are not true. 
For example, the values a = -1 and b = -1 show that the statement ab < 0 is NOT true, but it doesn't prove that the statements a < 0 and b < 0 are true. 

Cheers,

Brent - Greenlight Test Prep

 

[PhdApplicant311]

Thank you Greenlight! 

UPDATE: I've just been revising inequalities from the Manhattan 5 LB book and I feel so silly for asking this now!
This is a super simple inequality problem as it turns out. 

 

[GREMasterEMPOWERRichC]

Hi PhdApplicant311,

GRE questions often require a bit of 'flexible' thinking on the part of the solver (the GRE is a critical thinking Test, after all). Sometimes questions come with 'twists' that have to be accounted for. Here, we're asked for what 'MUST be true' which really means "which of the following is ALWAYS TRUE, no matter how many different examples you come up with?" This is similar to a QC in that coming up with one TEST is probably not going to be enough to get you to the correct answer. Here, the 'twist' really measures the 'thoroughness' of your thinking.

After you come up with your first TEST (A is positive, B is negative), you have to think about the OTHER possibilities (your next TEST was A is negative, B is negative). Etc. For Roman Numeral questions, it's also really helpful to pay attention to the 5 answer choices - they're often designed to help you avoid having to do some of the math.

GRE Masters aren't born, they're made,

Rich

[PhdApplicant311]

Thank you Rich!