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

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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 24 Feb 2013, 19:18
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Archit143 wrote:
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


Responding to a pm:

A response above clarifies that this is an errata.
The correct statement 1 is a = b^2 instead of a^2 = b^2.

In that case

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

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

Is\frac{1}{a} > \frac{a}{(b^4 + 3)} ?
Is \frac{(b^4 + 3)}{a} > a ?
Is \frac{(b^4 + 3)}{a} - a > 0 ?
Is \frac{(b^4 + 3 - a^2)}{a} > 0 ?

(1) a = b^2
This tells us that 'a' must be positive. Further, squaring, we get a^2 = b^4
Hence, the question becomes:
Is 3/a > 0.
It must be since a is positive. Sufficient

(2) a^2 = b^4
Doesn't tell us anything about the sign of a.
The question becomes:
Is 3/a > 0?
We cannot say. Not Sufficient

Answer (A)
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 25 Feb 2013, 03:46
VeritasPrepKarishma wrote:
Archit143 wrote:
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


Responding to a pm:

A response above clarifies that this is an errata.
The correct statement 1 is a = b^2 instead of a^2 = b^2.

In that case

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

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

Is\frac{1}{a} > \frac{a}{(b^4 + 3)} ?
Is \frac{(b^4 + 3)}{a} > a ?
Is \frac{(b^4 + 3)}{a} - a > 0 ?
Is \frac{(b^4 + 3 - a^2)}{a} > 0 ?

(1) a = b^2
This tells us that 'a' must be positive. Further, squaring, we get a^2 = b^4
Hence, the question becomes:
Is 3/a > 0.
It must be since a is positive. Sufficient

(2) a^2 = b^4
Doesn't tell us anything about the sign of a.
The question becomes:
Is 3/a > 0?
We cannot say. Not Sufficient

Answer (A)

Hi karishma
Statement 1 is a^2 = b^2 and not a = b^2

Can you clear something that i am missing here...
by the way thanks for explaining and statement is insufficient that clear....


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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 25 Feb 2013, 04:05
Expert's post
Archit143 wrote:
VeritasPrepKarishma wrote:
Archit143 wrote:
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


Responding to a pm:

A response above clarifies that this is an errata.
The correct statement 1 is a = b^2 instead of a^2 = b^2.

In that case

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

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

Is\frac{1}{a} > \frac{a}{(b^4 + 3)} ?
Is \frac{(b^4 + 3)}{a} > a ?
Is \frac{(b^4 + 3)}{a} - a > 0 ?
Is \frac{(b^4 + 3 - a^2)}{a} > 0 ?

(1) a = b^2
This tells us that 'a' must be positive. Further, squaring, we get a^2 = b^4
Hence, the question becomes:
Is 3/a > 0.
It must be since a is positive. Sufficient

(2) a^2 = b^4
Doesn't tell us anything about the sign of a.
The question becomes:
Is 3/a > 0?
We cannot say. Not Sufficient

Answer (A)

Hi karishma
Statement 1 is a^2 = b^2 and not a = b^2

Can you clear something that i am missing here...
by the way thanks for explaining and statement is insufficient that clear....


Archit


It's an error in the MGMAT book. They have given it in their errata. The link of their errata is given in this post (on the previous page):
if-a-is-not-equal-to-zero-is-1-a-a-b-122266.html#p989860
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 25 Feb 2013, 04:34
Thanks Karishma for clearing the doubts...Can you pls explain what does IaI = IbI mean....in simpler terms....
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 13 Mar 2013, 17:01
If b^4 + 3 > a^2
=> a^2 - b^4 < 3

Now

2. a^2 = b^4
=> a^2 -b^4 = 0
It tells a^2 - b^4 < 3 is sufficient

Wondering what wrong assumption I am making with above.
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 13 Mar 2013, 20:56
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Archit143 wrote:
Thanks Karishma for clearing the doubts...Can you pls explain what does IaI = IbI mean....in simpler terms....


|a| = |b| means that the distance of a from 0 is equal to the distance of b from 0.
This means, if a = 5, b = 5 or -5
Similarly, if a = -5, b = 5 or -5

So, imagine the number line. There are two points at a distance of 5 from 0. a and b could lie on any one of these points.
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 13 Mar 2013, 21:01
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kaushalsp wrote:
If b^4 + 3 > a^2
=> a^2 - b^4 < 3

Now

2. a^2 = b^4
=> a^2 -b^4 = 0
It tells a^2 - b^4 < 3 is sufficient

Wondering what wrong assumption I am making with above.


'Is \frac{1}{a} > \frac{a}{(b^4 + 3)}' is NOT the same as 'Isb^4 + 3 > a^2?'

Mind you, it is not given that 'a' is positive. You cannot cross multiply in an inequality if you do not know the sign of the varaible.

e.g.
a < b/c is not the same as ac < b

If we know that c is positive, then it is ok. Then a < b/c is same as ac < b
If instead, c is negative, then a < b/c is the same as ac > b (Note that the inequality sign has flipped)

Hence, statement (2) is not sufficient alone.
Statement (1) tells us the sign of a and we see that it is sufficient alone (check my solution above)
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If a NOT=0, is 1/a>a/(b^4+3) ? [#permalink] New post 31 Jul 2013, 07:19
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Last edited by Zarrolou on 31 Jul 2013, 07:29, edited 1 time in total.
Edited the question.
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Re: If a NOT=0, is 1/a>a/(b^4+3) ? [#permalink] New post 31 Jul 2013, 07:52
bagdbmba wrote:
1. If a NOT=0, is \frac{1}{a}>\frac{a}{b^4+3}

i. a^2=b^2
ii. a^2=b^4


IMO E

LET a = 1
according to statement 1==> b^4=1
now putting in equation
\frac{1}{1}>\frac{1}{1+3}==>satisfies
if a = -1 b^4 =1
\frac{1}{-1}>\frac{-1}{1+3}==>doesnt satisfies.

statement 2:
a^2 = b^2
let a=1 b^4 = 1
\frac{1}{1}>\frac{1}{1+3}==>satisfies

let a=-1 b^4 = 1
\frac{1}{-1}>\frac{-1}{1+3}===>doesnt satisfies.

combining both
a,b= 1 or -1

same cases will not satisfy

hence E
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Re: If a NOT=0, is 1/a>a/(b^4+3) ? [#permalink] New post 31 Jul 2013, 09:53
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bagdbmba wrote:
1. If a NOT=0, is \frac{1}{a}>\frac{a}{b^4+3}

i. a^2=b^2
ii. a^2=b^4


The question asks IS \frac{(b^4+3)}{a}>a \to Multiply both sides bya^2 \to a(b^4+3)>a^3 \to Is a(b^4+3-a^2)>0.

From F.S 1, the question becomes: Is a(a^4-a^2+3)>0.

Note that (a^4-a^2+3) will always be positive as because it can be represented as sum of 2 squares : (a^4-2a^2+1)+a^2+2 = (a^2-1)^2+(a^2+2)

Thus, the question now simply becomes : Is a>0

We clearly don't know that. Hence, Insufficient.

From F.S 2, the question becomes : Is a(b^4+3-a^2)>0 \to a(a^2+3-a^2)>0 \to Is 3a>0. Again, Insufficient.

Taking both together, we know thatb^2 = b^4 \to b^2(b^2-1) = 0 \to b^2=a^2 = 1 [b^2 \neq{0} asa\neq{0}]
Thus, a could be\pm1. Insufficient.

E.

Sidenote: The actual question is from Manhattan Gmat and they had published an errata for this question , which modifies the first fact statement to
a = b^2. Now, if that would have been the case,the answer would have been A. However, with the given condition, the answer is E.
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 31 Jul 2013, 09:56
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 14 Aug 2013, 10:11
Expert's post
VeritasPrepKarishma wrote:
Archit143 wrote:
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


Responding to a pm:

A response above clarifies that this is an errata.
The correct statement 1 is a = b^2 instead of a^2 = b^2.

In that case

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

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

Is\frac{1}{a} > \frac{a}{(b^4 + 3)} ?
Is \frac{(b^4 + 3)}{a} > a ?
Is \frac{(b^4 + 3)}{a} - a > 0 ?
Is \frac{(b^4 + 3 - a^2)}{a} > 0 ?

(1) a = b^2
This tells us that 'a' must be positive. Further, squaring, we get a^2 = b^4
Hence, the question becomes:
Is 3/a > 0.
It must be since a is positive. Sufficient

(2) a^2 = b^4
Doesn't tell us anything about the sign of a.
The question becomes:
Is 3/a > 0?
We cannot say. Not Sufficient

Answer (A)



Hi Karishma,
As per the above highlighted part, if a is '-ve' then a=-b^2, so \sqrt{a} will be \sqrt{-b^2}. Hence a becomes imaginary as b^2 is always positive.So here also a must be '+ve'.

But, we've considered a as '-ve' also! Could you please explain this?

Much appreciate your feedback.
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 15 Aug 2013, 22:14
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Expert's post
bagdbmba wrote:
VeritasPrepKarishma wrote:
Archit143 wrote:
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


Responding to a pm:

A response above clarifies that this is an errata.
The correct statement 1 is a = b^2 instead of a^2 = b^2.

In that case

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

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

Is\frac{1}{a} > \frac{a}{(b^4 + 3)} ?
Is \frac{(b^4 + 3)}{a} > a ?
Is \frac{(b^4 + 3)}{a} - a > 0 ?
Is \frac{(b^4 + 3 - a^2)}{a} > 0 ?

(1) a = b^2
This tells us that 'a' must be positive. Further, squaring, we get a^2 = b^4
Hence, the question becomes:
Is 3/a > 0.
It must be since a is positive. Sufficient

(2) a^2 = b^4
Doesn't tell us anything about the sign of a.
The question becomes:
Is 3/a > 0?
We cannot say. Not Sufficient

Answer (A)



Hi Karishma,
As per the above highlighted part, if a is '-ve' then a=-b^2, so \sqrt{a} will be \sqrt{-b^2}. Hence a becomes imaginary as b^2 is always positive.So here also a must be '+ve'.

But, we've considered a as '-ve' also! Could you please explain this?

Much appreciate your feedback.




a^2 = b^4

When you take the square root of both sides here, you get |a| = |b^2| = b^2
You do not get a = b^2.
Note that when you take square root of x^2 = y^2, you get |x| = |y|, not x = y

So a can still be negative. Say a = -9, b = 3
In this case, a^2 = b^4 = (-9)^2 = 3^4
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 09 Sep 2013, 03:27
Expert's post
Thanks Karishma for the explanation and apologies for late acknowledgement as I could figure it out later and hence this post slipped somehow :)

But, again much thanks. +1
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1)   [#permalink] 09 Sep 2013, 03:27
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