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13 Aug 2017, 07:40
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Hi,
I was wondering if you could point out the flaw in my logic for the question below. I think I'm approaching it correctly but am not sure if I'm misinterpreting what you can/cannot do in a DS question.
Is a^3 > a^2?
(1) 1/a > a
(2) a^5 > a^3
Stmt (1):
This stmt only holds when 0<a<1 and when a is -ve; in both cases we see that the result actually gives us a no because a^3 < a^2. So I think this stmt is sufficient because both ranges of numbers we test give us the same result, which is that a^3 > a^2 is not true.
Stmt (2):
This stmt only holds when a>+1 and when -1<a<0; doesn't hold for values between 0-1, and also doesn't hold for -ve values <-1. When testing with values a>+1 we will always get a^3 > a^2; when testing with values -1<a<0 we always get the opposite because squaring a -ve number is always positive: hence, for this case a^3 < a^2. So I think this stmt is insufficient because it gives us opposite results.
Answer: A
Is my interpretation of how to use stmt (1) & stmt (2) correct? The way I look at it is that these stmts give us certain constraints and only for values that make those constraints true can we apply those to the main question that we have to prove. Is there a better way to approach this problem? I seem to make more errors on DS Inequality problems...
Thanks!
I was wondering if you could point out the flaw in my logic for the question below. I think I'm approaching it correctly but am not sure if I'm misinterpreting what you can/cannot do in a DS question.
Is a^3 > a^2?
(1) 1/a > a
(2) a^5 > a^3
Stmt (1):
This stmt only holds when 0<a<1 and when a is -ve; in both cases we see that the result actually gives us a no because a^3 < a^2. So I think this stmt is sufficient because both ranges of numbers we test give us the same result, which is that a^3 > a^2 is not true.
Stmt (2):
This stmt only holds when a>+1 and when -1<a<0; doesn't hold for values between 0-1, and also doesn't hold for -ve values <-1. When testing with values a>+1 we will always get a^3 > a^2; when testing with values -1<a<0 we always get the opposite because squaring a -ve number is always positive: hence, for this case a^3 < a^2. So I think this stmt is insufficient because it gives us opposite results.
Answer: A
Is my interpretation of how to use stmt (1) & stmt (2) correct? The way I look at it is that these stmts give us certain constraints and only for values that make those constraints true can we apply those to the main question that we have to prove. Is there a better way to approach this problem? I seem to make more errors on DS Inequality problems...
Thanks!
13 Aug 2017, 14:40
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Hi vihaan21,
For future reference, you should post your GMAT questions to the appropriate sub-forum. The DS forum can be found here:
https://gmatclub.com/forum/data-sufficiency-ds-141/
Your approach to this particular question is fine, but not all DS question are 'best approached' in this way. DS questions 'test' a number of skills that are NOT "math skills", including organization, accuracy, attention-to-detail, the ability to 'prove' that you're correct, etc. and these prompts have their own built-in 'patterns' that you can learn to take advantage of. Thus, you would likely need to do more than just work through a bunch of random DS prompts to maximize your skills in this area.
GMAT assassins aren't born, they're made,
Rich
For future reference, you should post your GMAT questions to the appropriate sub-forum. The DS forum can be found here:
https://gmatclub.com/forum/data-sufficiency-ds-141/
Your approach to this particular question is fine, but not all DS question are 'best approached' in this way. DS questions 'test' a number of skills that are NOT "math skills", including organization, accuracy, attention-to-detail, the ability to 'prove' that you're correct, etc. and these prompts have their own built-in 'patterns' that you can learn to take advantage of. Thus, you would likely need to do more than just work through a bunch of random DS prompts to maximize your skills in this area.
GMAT assassins aren't born, they're made,
Rich
