Rephrasing the Q

\frac{1}{a} > \frac{a}{(b^4 + 3)}

Cross multiplying

Is \frac{1}{a^2} > \frac{1}{(b^4 + 3)}

(1) a^2 = b^2

\frac{1}{b^2} > \frac{1}{(b^4 + 3)}

When denominator increases the value of the fraction decreases. We dont need to find it but we know that value of the first term will be greater than the second term.

(2) a^2 = b^4

\frac{1}{b^4} > \frac{1}{(b^4 + 3)}

When denominator increases the value of the fraction decreases. We dont need to find it but we know that value of the first term will be greater than the second term.

Hence D.

yes I get the answer e. But book has A as answer.......

from statement 1 we get |a| = |b|

put in a =2, b^4 =16...we get 1/2 > 2/19 so yes,

but put in a=-2 and b^4 =16 we get -1/2 < -2/19 so no.

Insufficient. similarly statement 2 too depends upon sign of a. Statements together to can't resolve the ambiguity of sign. Am i solving it wrong or is statement 1 a misprint in the book, could it be a =b^2 in statement 1 ?

I got E too.

1) a^2 = b^2 will hold true for a,b = 1 or -1

When a,b = 1 then the inequality holds true but when a.b = -1 then inequality doesn't hold. Insufficient.

2) a^2 = b^4. Same as above. Insufficient.

1 and 2 together) The same case can be applied again. Insufficient again.

Not sure how the answer is A.

Why did you cross multiply?
It's an inequality; you don't know whether a is positive or negative. Depending on the sign of a the inequality may or may not be reversed.

eg. 1/-3>-3/2
by cross multiplying
2>9
Now, that's incorrect.

melguy,
to my understanding, if you multiply or divide by a negative value then the inequality sign should be reversed.
2 < 3
multiply both sides by - 1
Is -2 < -3 ???? its a Big NO.
so we cannot multiply by a unknown variable.
Correct me if i am wrong

guys, the problem is written incorrectly
it should be (1) a = b^2 instead of (1) a^2 = b^2
check the Errata
http://www.manhattangmat.com/errata-adv-quant-4th.cfm

after this, it is easy.
(1) A can be only 1 since, b^2 is non-negative. B can be 1 or -1. this is Sufficient 1>1/[(1)^2+3] or 1>1/[(-1)^2+3]
(2) A can be 1 or -1. B can be 1 or -1. Insufficient
1>1/[(1)^2+3] or 1>1/[(-1)^2+3]
BUT
-1 not > 1/[(1)^2+3] or -1 not >1/[(-1)^2+3]

I've prepared a clear solution. I hope it's correct, I think this problem is quite easy:

In the penultimate row what I ment is obviously Let's root and not Let's divide by a^2
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Hi Imdreamer, I think I know what’s your issue here.
Please have a look at this three terms and describe the difference to yourself:

a^2

-a^2

(-a)^2

When executed, these three terms return the following results, respectively:

a^2

-a^2

a^2

Of course, the murky part here is the the middle row: Why negative squared stays negative? Because you don’t square the negative sign, just the variable! Yes, all the difference lies in the parentheses.



Now, let’s look at the 2nd Statement again:

2)a^2=b^4

Plug in Smart Numbers to imagine better:

4^2=2^4

Root on both sides:

4=2^2

Is this really TRUE? Nope. Well, at most in half, because BOTH:

4=2^2 AND 4=(-2)^2 ARE TRUE.

And this is our translated Statement 1) which is Sufficient, because any squared element will always be positive.


But now move back to the original equation from the Statement 2):

a^2=b^4 means:

4^2=2^4 OR 4^2 = (-2)^4 OR (-4)^2=2^4 OR (-4)^2=(-2)^4

All four equations are valid.



AND THE SAME ROOTED:

4=2^2 OR 4 = (-2)^2 OR (-4)=2^2 OR (-4)=(-2)^2


Of course the two last ones are NOT valid. But we can see that only after rooting, and this is the answer that the Statement 1) gives us.


The insufficiency of the Statement 2) lies in the squared term a, because any squared element will be positive. So unsquared a could be greater or less than zero: a>0 OR a<0. In our example a=4 OR a=(-4), because the square of both 4 and (-4) will equal 16.


Back to the Statement 2):

So a^2=b^4 rooted is:

|a|=b^2, because:

a=b^2 (for a>0) OR a=-b^2 (for a<0).

So, when 2) a^2=b^4, then b could be a positive or a negative number (2 or (-2) in the example), just like a (4 and (-4)). According to the rules for Absolute Value, we write it as follows:

AND |a|=a OR -a

AND |b|=b OR -b

(Knowing that, we’ve made a damp squib of this Absolute Value GMAT trap.)

Thus, we can’t decide which inequality from the stem is true basing on the Statement 2) only, because even when the sign of b is here irrelevant (because it is squared in the stem inequality), we can’t ignore the sign of a (because we use it not squared to multiply the inequality). We can’t determine which of the two inequalities is valid.
The answer is A.


P.S. The basics of the Absolute Value problem are covered in Manhattan GMAT Guide 1 - Number Properties, Chapter 3: Positives & Negatives.
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Hey alexpavlos, Statement 1) says a=b^2. There's an error in the book which's been denounced in this thread.
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Hi bunuel
can you help with this question.....the book has given statement 1 is sufficient...it says that since b^2 is always positive so "a" must also be positive......but i think a can be negative also......

regards
Archit