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Manhattan GMAT problem - Inequalities- possibly wrong?

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by njj023 » Sun Jul 10, 2011 4:37 pm
From the Advanced Math textbook

If a is not equal to 0, is (1/a) > a/(b^4 + 3)?

1. a^2 = b^2

2. a^2 = b^2

[spoiler]The OA is A but I dont get why it says the following for Statement 1 - "b^2 must be positive so a is positive"

Why must a have to be positive? B and A could both be -3 for example and Statement 1 would still hold[/spoiler]

Thanks for the help guys
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Source: — Data Sufficiency |
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by vineeshp » Sun Jul 10, 2011 6:37 pm
Can you post the full text? Also, option B seems to be same as option A. Am I missing something?
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by njj023 » Sun Jul 10, 2011 6:51 pm
Sorry here are the options again. Had a small typo in option 2

1. a^2 = b^2

2. a^2 = b^4

The full text says the following:

"1. SUFFICIENT: b^2 must be positive so a is positive. Therefore we only have to evaluate the top branch of the flow chart:

Is b^4 + 3 > a^2?
=> Is b^4 + 3 > (b^2)^2
=> Is b^4 + 3 > b^4?
=> Is 3 > 0"

I am pretty sure this is a typo on their part. I think Statement 1 is supposed to say "a = b^2"

Otherwise even their computation wouldn't make sense because they are evaluating a^2 to be b^4 above

Thanks
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by vineeshp » Sun Jul 10, 2011 8:12 pm
Yes I agree. It is a typo on their part. :)
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by navami » Tue Jul 12, 2011 9:24 am
option A and B looks same
Let me try to conclude with the option A alone.
the question is .
1/a > a/(B^4 + 3) ????
or if b^4 + 3 > a^2

now option 1 says a^2 = b^2
so we can rewrite our equation as
b^4 + 3 = a^2 x a^2 + 3 = a^2 ( a^2 + 3/ a^2)

now a^2 is always +ve
so do ( a^2 + 3/ a^2)
and a^2 ( a^2 + {some positive vaule}) is always greater thamn a^2
hence we have the ans
This time no looking back!!!
Navami
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by amit2k9 » Tue Jul 12, 2011 10:47 pm
a a=b = 1| -1 gives different results. hence POE.

b a=b= 1|-1 or +|- 4|2 give different results. not sufficient.

a+b 1|-1 satisfies both hence C not sufficient too.
E it is.

for a= b^2 means a=1,4,9 and so on. LHS > RHS always.

Hence A it is.
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