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# If a does not = 0, is 1/a > a/(b^4 + 3)

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Senior Manager
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If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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12 Sep 2011, 17:42
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If a does not equal to zero, is 1/a > a/(b^4 +3)?

(1) a = b^2
(2) a^2 = b^4

This comes from the MGMAT advanced quant guide #4-9.
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Jul 2013, 06:05, edited 1 time in total.
Edited the question.
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Re: DS question from MGMAT advanced quant [#permalink]

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11 Mar 2012, 11:16
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pavanpuneet wrote:
From the last post from Manager, he says: from (1)
a = b^2 ==> b= Mod(a)
from here, you can say that (b^4+3) > a^2.
cz b^2 = a or, b^4 = a^2 (squaring both side)
==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain...
from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

THE QUESTION SHOULD READ:
If a does not equal to zero, is 1/a > a/(b^4 +3)?

Notice that we cannot cross-multiply $$\frac{1}{a} > \frac{a}{b^4 +3}$$, since we don't know whether $$a>0$$ or $$a<0$$, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with $$b^4 +3$$, since this expression is always positive: $$b^4 +3=nonnegative+positive=positive$$. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is $$\frac{1}{b^2} > \frac{b^2}{b^4 +3}$$? Now, since $$b^2>0$$ then here we can safely cross-multiply and the question becomes: is $$b^4+3>b^4$$? --> is $$3>0$$? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is $$\frac{1}{a} > \frac{a}{a^2 +3}$$? We have the same problem: we cannot cross-multiply since we don't know the sign of $$a$$. If $$a=1$$ then the answer to the question would be YES but if $$a=-1$$ then the answer to the question would be NO. Not sufficient.

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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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12 Mar 2012, 02:42
Its clear now! Thanks again for the prompt response Bunuel
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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22 May 2012, 19:14
Bunuel never ceases to amaze me. His reasoning is so sharp and his methods so efficient, I bet he conquers every problem in under 30 seconds. I have been following his posts extensively hoping to learn to think like he does. Lifelong fan right here.
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Re: DS question from MGMAT advanced quant [#permalink]

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07 Jan 2014, 05:26
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Bunuel wrote:
pavanpuneet wrote:
From the last post from Manager, he says: from (1)
a = b^2 ==> b= Mod(a)
from here, you can say that (b^4+3) > a^2.
cz b^2 = a or, b^4 = a^2 (squaring both side)
==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain...
from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

THE QUESTION SHOULD READ:
If a does not equal to zero, is 1/a > a/(b^4 +3)?

Notice that we cannot cross-multiply $$\frac{1}{a} > \frac{a}{b^4 +3}$$, since we don't know whether $$a>0$$ or $$a<0$$, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with $$b^4 +3$$, since this expression is always positive: $$b^4 +3=nonnegative+positive=positive$$. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is $$\frac{1}{b^2} > \frac{b^2}{b^4 +3}$$? Now, since $$b^2>0$$ then here we can safely cross-multiply and the question becomes: is $$b^4+3>b^4$$? --> is $$3>0$$? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is $$\frac{1}{a} > \frac{a}{a^2 +3}$$? We have the same problem: we cannot cross-multiply since we don't know the sign of $$a$$. If $$a=1$$ then the answer to the question would be YES but if $$a=-1$$ then the answer to the question would be NO. Not sufficient.

Hi Bunuel,
Great explanation. seeing a^2 in case 2 i quickly jumped and marked D as my answer. But here's a doubt that i have:
in case 2: what is given to us ... is a^2=b^4.
now, since i cant cross multiply my question's equation to get a^2 and then substitute with the value for a^2 n get the answer. so instead, cant i just take a square-root on both sides of my case2 equation.??

which gives me......
a^2 = b^4 ................ now taking root on both sides, i get ((p.s: i can take square-root on both sides as i know that it wont affect the sign(-/+ve) of either of a or b.))
a = b^2 .................. now that i know a is +ve .. i substitute just like I did in the first case n get the answer as D...wouldn't this be correct???

Cheers!
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Re: DS question from MGMAT advanced quant [#permalink]

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07 Jan 2014, 05:40
rawjetraw wrote:
Bunuel wrote:
pavanpuneet wrote:
From the last post from Manager, he says: from (1)
a = b^2 ==> b= Mod(a)
from here, you can say that (b^4+3) > a^2.
cz b^2 = a or, b^4 = a^2 (squaring both side)
==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain...
from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

THE QUESTION SHOULD READ:
If a does not equal to zero, is 1/a > a/(b^4 +3)?

Notice that we cannot cross-multiply $$\frac{1}{a} > \frac{a}{b^4 +3}$$, since we don't know whether $$a>0$$ or $$a<0$$, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with $$b^4 +3$$, since this expression is always positive: $$b^4 +3=nonnegative+positive=positive$$. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is $$\frac{1}{b^2} > \frac{b^2}{b^4 +3}$$? Now, since $$b^2>0$$ then here we can safely cross-multiply and the question becomes: is $$b^4+3>b^4$$? --> is $$3>0$$? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is $$\frac{1}{a} > \frac{a}{a^2 +3}$$? We have the same problem: we cannot cross-multiply since we don't know the sign of $$a$$. If $$a=1$$ then the answer to the question would be YES but if $$a=-1$$ then the answer to the question would be NO. Not sufficient.

Hi Bunuel,
Great explanation. seeing a^2 in case 2 i quickly jumped and marked D as my answer. But here's a doubt that i have:
in case 2: what is given to us ... is a^2=b^4.
now, since i cant cross multiply my question's equation to get a^2 and then substitute with the value for a^2 n get the answer. so instead, cant i just take a square-root on both sides of my case2 equation.??

which gives me......
a^2 = b^4 ................ now taking root on both sides, i get ((p.s: i can take square-root on both sides as i know that it wont affect the sign(-/+ve) of either of a or b.))
a = b^2 .................. now that i know a is +ve .. i substitute just like I did in the first case n get the answer as D...wouldn't this be correct???

Cheers!

The point is that $$\sqrt{a^2}=|a|$$, not a. For example if a=-1, then $$\sqrt{a^2}=\sqrt{1}=1=|a|=-a$$, not a.

Hope it's clear.
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Re: DS question from MGMAT advanced quant [#permalink]

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08 Jan 2014, 01:29
whoops ! thanks a ton bunuel. Got it.
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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08 Jan 2014, 02:05
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rjkaufman21 wrote:
If a does not equal to zero, is 1/a > a/(b^4 +3)?

(1) a = b^2
(2) a^2 = b^4

This comes from the MGMAT advanced quant guide #4-9.

Statement I is sufficient:
(1/b^2) > a/(b^4 + 3)

We can cross multiply since both the denominators are positive for sure
b^4 + 3 > b^2(b^2)
B^4 + 3 > B^4 - Well that is true every time

Statement II is insufficient
a^2 = b^4
If a is and b are positive then the equation will turn to: B^4 + 3 > B^4 In that case YES
If a is negative and b is positive then a = -b^2, substituting it we will get:
b^4 + 3 < -1 (b^2)(b^2)
b^4 + 3 < -(b^4) - That is not true

So we have a YES and a NO hence statement II is insufficient

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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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01 Aug 2014, 02:58
rjkaufman21 wrote:
If a does not equal to zero, is 1/a > a/(b^4 +3)?

(1) a = b^2
(2) a^2 = b^4

This comes from the MGMAT advanced quant guide #4-9.

a = b^2 -> 1/a = 1/b^2 = b^2/ b^4
b^4 < b^4 + 3 -> 1/a = b^2/b^4> b^2/b^4 +3
So (1) is sufficient

a^2 = b^4 -> a = b^2 or a=-b^2
Since there are two possible answers for a, 1/a can be either larger or smaller than a/(b^4+3)

So A is the answer
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If a ≠ 0, is (1/a) > a/(b^4+3) ? [#permalink]

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04 Aug 2014, 05:06
$$If A \neq 0, is$$ $$\frac{1}{A}$$ $$>$$ $$\frac{A}{B^4+3} ?$$

(1) $$A = B^2$$
(2) $$A^2 = B^4$$
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Re: If a ≠ 0, is (1/a) > a/(b^4+3) ? [#permalink]

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04 Aug 2014, 05:41
Hi ,

I have to find out whether the inequality is true or false and whatever the result it is enough to answer the question.

I solved it this way . Since nothing about A and B is mentioned . I have an option even to pick up irrational numbers also.

Choice A : A=B^2 , 1/B^2 > B^2/(B^4+3) => 3>0 .. which is true . So i can answer the question.

Choice B: A^2 = B^4 => A= +B^2 or A=-B^2 . When A = +B^2 then the result is same as choice A and it is true.

When A = -B^2 then -1/B^2 > -B^2/(B^4 +3) => -(B^4 +3 ) > -B^4 , now this equation becomes (B^4 + 3) < B^4 as "-" reverses the equality .

3 < 0 which is false , hence this statement alone is not sufficient .

Both together : It is not even worth to try both together because Choice A is a subset of Choice B.

Hence A is the answer .

Key turning point : to realize that A = +B^2 and A = - B^2.
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If a ≠ 0, is (1/a) > a/(b^4+3) ? [#permalink]

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04 Aug 2014, 06:12
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pratikshr wrote:
$$If A \neq 0, is$$ $$\frac{1}{A}$$ $$>$$ $$\frac{A}{B^4+3} ?$$

(1) $$A = B^2$$
(2) $$A^2 = B^4$$

It's a Yes/No question. From question itself we can deduce that $$(B^4+3)$$ will always be positive.
Rephrasing question -> $$\frac{B^4+3}{A}$$ $$>$$ $$\frac{A}{1} ?$$
Note that ,we can not multiply A in this inequality since A might be negative too.

Stmt 1 -
$$A = B^2$$ => A is square of some number (B in this case) so A will always be positive. So multiply with A both side ->
squaring statement 1 at both sides gives us $$A^2 = B^4$$ so we now can multiply with A at both sides as A will always be positive => Sufficient

Stmt 2 -
$$A^2 = B^4$$ => Here A can be negative also as shown below -
$$A = B^2$$ or $$A = -B^2$$
So this statement is not Sufficient.

Hence A.
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If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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08 Aug 2014, 06:42
I would like to know what's wrong with my thinking.

(1) Since a^2 = b^2 are squares, a could be positive or negative. Hence we don't know if the inequality holds.

(2) Since a^2 = b^4 a must be positive because b^2=a --> b>0. Inequality holds.

You see I'm confused.

Btw: Do I have a different question than you guys? In my book statement 1 states a^2=b^2?
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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12 Aug 2014, 03:24
pratikshr wrote:
$$If A \neq 0, is$$ $$\frac{1}{A}$$ $$>$$ $$\frac{A}{B^4+3} ?$$

(1) $$A = B^2$$
(2) $$A^2 = B^4$$

Merging similar topics. Please refer to the discussion above.
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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14 Aug 2014, 01:48
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Can you explain this part?

a^2=b^4 --> is $$\frac{1}{a} > \frac{a}{a^2 +3}$$?

Bunuel wrote:
pavanpuneet wrote:
From the last post from Manager, he says: from (1)
a = b^2 ==> b= Mod(a)
from here, you can say that (b^4+3) > a^2.
cz b^2 = a or, b^4 = a^2 (squaring both side)
==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain...
from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

THE QUESTION SHOULD READ:
If a does not equal to zero, is 1/a > a/(b^4 +3)?

Notice that we cannot cross-multiply $$\frac{1}{a} > \frac{a}{b^4 +3}$$, since we don't know whether $$a>0$$ or $$a<0$$, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with $$b^4 +3$$, since this expression is always positive: $$b^4 +3=nonnegative+positive=positive$$. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is $$\frac{1}{b^2} > \frac{b^2}{b^4 +3}$$? Now, since $$b^2>0$$ then here we can safely cross-multiply and the question becomes: is $$b^4+3>b^4$$? --> is $$3>0$$? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is $$\frac{1}{a} > \frac{a}{a^2 +3}$$? We have the same problem: we cannot cross-multiply since we don't know the sign of $$a$$. If $$a=1$$ then the answer to the question would be YES but if $$a=-1$$ then the answer to the question would be NO. Not sufficient.

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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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06 Mar 2015, 21:59
pipe19 wrote:
I would like to know what's wrong with my thinking.

(1) Since a^2 = b^2 are squares, a could be positive or negative. Hence we don't know if the inequality holds.

(2) Since a^2 = b^4 a must be positive because b^2=a --> b>0. Inequality holds.

You see I'm confused.

Btw: Do I have a different question than you guys? In my book statement 1 states a^2=b^2?

(1) doesn't say a^2=b^2, it says a=b^2, and because of this we know that a is positive and we can cross multiply without having to switch the inequality

(2) doesn't tell us anything by itself, and b^2=a comes from the (1) and so we could technically say that they both together work, but because (1) works on its own then this is irrelevant.
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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25 Aug 2015, 03:34
Bunuel

Thanks for the explanation !
I just wanted to ask , if first statement was like

a) a^2=b^2

Will this statement still be sufficient to answer the above question ?
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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12 Dec 2015, 00:49
Bunuel wrote:
pavanpuneet wrote:
From the last post from Manager, he says: from (1)
a = b^2 ==> b= Mod(a)
from here, you can say that (b^4+3) > a^2.
cz b^2 = a or, b^4 = a^2 (squaring both side)
==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain...
from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

THE QUESTION SHOULD READ:
If a does not equal to zero, is 1/a > a/(b^4 +3)?

Notice that we cannot cross-multiply $$\frac{1}{a} > \frac{a}{b^4 +3}$$, since we don't know whether $$a>0$$ or $$a<0$$, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with $$b^4 +3$$, since this expression is always positive: $$b^4 +3=nonnegative+positive=positive$$. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is $$\frac{1}{b^2} > \frac{b^2}{b^4 +3}$$? Now, since $$b^2>0$$ then here we can safely cross-multiply and the question becomes: is $$b^4+3>b^4$$? --> is $$3>0$$? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is $$\frac{1}{a} > \frac{a}{a^2 +3}$$? We have the same problem: we cannot cross-multiply since we don't know the sign of $$a$$. If $$a=1$$ then the answer to the question would be YES but if $$a=-1$$ then the answer to the question would be NO. Not sufficient.

Hi Bunuel,
Thanks for a great explanation.

I started out with something like this.
1) $$a = b^2$$.
so, on the RHS we have$$\frac{a}{(b^4 + 3)}$$. On substituing above value this gives -
$$\frac{a}{(a^2 + 3)}$$
Now if we take 1/a common from this fraction, we are left with
$$\frac{1}{a}* [\frac{(a^2}{(a^2 + 3)}]$$.

Clearly, the term $$\frac{(a^2}{(a^2 + 3)}$$ is less than 1. Lets just say this is T (T < 1)
When a = 2, then we have 1/2 > 1/2 * T

but when a = -2
this makes the RHS greater than the LHS since T < 1
-1/2 < -1/2 * T

Isn't it ? Did i go wrong somewhere ?
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Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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19 Feb 2016, 09:28
the explanations here really are as long as that in the manhattangmat book.
Why multiply when you already shouldnt multiply, since you dont know the value of a?
It then went ahead to almost miraculously say that "a" cant be negative in 1 but can be negative in 2.

I have a suggestion to those finding this question difficult.
WHEN YOU SEE THIS TYPE OF INEQUALITY, DONT CROSSMULTIPLY!!! DONT!! (it creates confusion)
just pick values that agrees with condition in 1 and do same for 2.
you will quickly realize that in 2 a^2 = b^2 satisfies even negative 4 as well as 4.
aNother point: dont try to manipulate statement 2. it will only lead to 1
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If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]

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19 Feb 2016, 10:28
Nez wrote:
the explanations here really are as long as that in the manhattangmat book.
Why multiply when you already shouldnt multiply, since you dont know the value of a?
It then went ahead to almost miraculously say that "a" cant be negative in 1 but can be negative in 2.

I have a suggestion to those finding this question difficult.
WHEN YOU SEE THIS TYPE OF INEQUALITY, DONT CROSSMULTIPLY!!! DONT!! (it creates confusion)
just pick values that agrees with condition in 1 and do same for 2.
you will quickly realize that in 2 a^2 = b^2 satisfies even negative 4 as well as 4.
aNother point: dont try to manipulate statement 2. it will only lead to 1

You do raise a couple of good points. As a matter of principle, in inequalities, NEVER multiply by a variable whose sign you do not know for sure. It is not recommended not because it causes "confusion" but it lets you eliminate one other case, making a statement seem sufficient while it actually was not sufficient.

You can modify the given question as , is$$\frac{1}{a} > \frac{a}{b^4+3}$$ ---> $$\frac{1}{a} - \frac{a}{b^4+3}>0$$ ---> $$\frac{b^4-a^2+3}{a*(b^4+3)} > 0$$

Per statement 1, $$a=b^2$$, substitute this and you will get $$\frac{b^4-a^2+3}{a*(b^4+3} = \frac{b^4-b^4+3}{b^2*(b^4+3)} = \frac{3}{b^2*(b^4+3)}$$, denominator is clearly > 0 making the expression $$\frac{b^4-a^2+3}{a*(b^4+3)} > 0$$. Hence sufficient.

Per statement 2, $$a^2=b^4$$ ---> $$a= \pm b^2$$. You will see that you end up getting a "yes" and a "no" to the question asked "is $$\frac{b^4-a^2+3}{a*(b^4+3)} > 0$$" when you alternate values of a in terms of b^2 and thus this statement is NOT sufficient.

A is thus the correct answer.

1 thing, manipulation of statement 2 is not just going to give you 1 , it will also give you $$-b^2$$ in addition.

Hope this helps.
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If a does not = 0, is 1/a > a/(b^4 + 3)   [#permalink] 19 Feb 2016, 10:28

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