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

1

This post received KUDOS

Expert's post

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

Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink]
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....

Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink]
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....

Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink]
13 Mar 2013, 20:56

Expert's post

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. _________________

Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink]
13 Mar 2013, 21:01

Expert's post

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 'Is\(b^4 + 3 > a^2\)?'

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) _________________

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

Re: If a NOT=0, is 1/a>a/(b^4+3) ? [#permalink]
31 Jul 2013, 09:53

Expert's post

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 by\(a^2 \to a(b^4+3)>a^3 \to\) 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 that\(b^2 = b^4 \to b^2(b^2-1) = 0 \to b^2=a^2 = 1 [b^2 \neq{0}\) as\(a\neq{0}]\) 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. _________________

Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink]
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?

Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink]
15 Aug 2013, 22:14

1

This post received KUDOS

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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