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

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If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 22 Oct 2011, 11:26
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If a is not equal to zero, is 1/a > a / (b^4 + 3) ?

(1) a^2 = b^2
(2) a^2 = b^4
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Re: Inequality from manhattan advanced quant. [#permalink] New post 22 Oct 2011, 23:01
Aj85 wrote:
If a is not equal to zero, is 1/a > a / (b^4 + 3) ?

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


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.
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 08:12
Aj85 wrote:
If a is not equal to zero, is 1/a > a / (b^4 + 3) ?

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


Do you have a specific doubt?
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 08:52
blink005 wrote:
Aj85 wrote:
If a is not equal to zero, is 1/a > a / (b^4 + 3) ?

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


Do you have a specific doubt?



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 ?
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 09:56
I'd agree with you

Aj85 wrote:

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 ?


If their answer is A then I reckon the option should be a=b^2
as a is positive you can cross-multiply

b^4+3>a^2
a=b^2
so b^4+3>b^4 is true

A would be sufficient then.
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 10:50
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.
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 14:20
Quote:
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 ?


Not sure if I understand how you get negative. After we have rephrased the Q

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

the sign will always stay positive since it is square and raise to the power 4 of a number. You will always get a positive result regardless of the sign.

\frac{1}{(-2)^2}

\frac{1}{(2)^2}

\frac{1}{(-2)^4}

\frac{1}{(2)^4}

Also, in response to 'could it be a =b^2' - thats what you have mentioned in the statement.
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 18:33
melguy wrote:
Not sure if I understand how you get negative. After we have rephrased the Q

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

the sign will always stay positive since it is square and raise to the power 4 of a number. You will always get a positive result regardless of the sign.

\frac{1}{(-2)^2}

\frac{1}{(2)^2}

\frac{1}{(-2)^4}

\frac{1}{(2)^4}

Also, in response to 'could it be a =b^2' - thats what you have mentioned in the statement.


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.
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 19:31
Quote:
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.

Inequality and fractions work exactly the same way. May be cross multiply is not the right term but the answer will still be the same

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

Without touching anything else multiply both the sides by \frac{1}{(a)}.

The result will be

\frac{1}{a^2} > \frac{1}{(b^4 + 3)} (which is the rephrased Q)

Last edited by melguy on 23 Oct 2011, 21:26, edited 1 time in total.
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 20:47
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 :)
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Re: Inequality from manhattan advanced quant. [#permalink] New post 23 Oct 2011, 21:28
arjunbt wrote:
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 :)


My Apologies. I forgot one important thing - reverse the signs when multiplying or dividing by a negative number (when working with inequality). I will look at the problem again soon (when i reach my house). Hopefully i will get it right this time :) I am still having a hard time understanding how the solution is A.
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Re: Inequality from manhattan advanced quant. [#permalink] New post 24 Oct 2011, 02:38
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]
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Re: Inequality from manhattan advanced quant. [#permalink] New post 24 Oct 2011, 05:26
melguy wrote:
arjunbt wrote:
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 :)


My Apologies. I forgot one important thing - reverse the signs when multiplying or dividing by a negative number (when working with inequality). I will look at the problem again soon (when i reach my house). Hopefully i will get it right this time :) I am still having a hard time understanding how the solution is A.


That was my point. :|
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 13 Jan 2013, 09:39
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I've prepared a clear solution. I hope it's correct, I think this problem is quite easy:
Attachment:
File comment: The solution.
Is 1a.jpg
Is 1a.jpg [ 525.46 KiB | Viewed 2065 times ]

In the penultimate row what I ment is obviously Let's root and not Let's divide by a^2
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Last edited by HumptyDumpty on 13 Jan 2013, 23:21, edited 1 time in total.
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 13 Jan 2013, 16:38
HumptyDumpty wrote:
I've prepared a clear solution. I hope it's correct, I think this problem is quite easy:
Attachment:
Is 1a.jpg

In the penultimate row what I ment is obviously Let's square and not Let's divide by a^2



I understand the solution for A, but I have a doubt on B.. How can a square number be a negative? b^2 will be a positive number, therefore a will always be positive. This is the reason I marked D as an answer. Am I missing something?
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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 14 Jan 2013, 01:13
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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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 14 Jan 2013, 05:42
Aj85 wrote:
If a is not equal to zero, is 1/a > a / (b^4 + 3) ?

(1) a = b^2---->Correcting the statement as stated in later part of the thread
(2) a^2 = b^4



Now the given expression is

1/a> a/(b^4+3)Can be written as 1/a-a/(b^4+3) >0 (Note, We cannot cross multiply as we don't know the sign of the terms of the inequality)
Simplifying the expression, we get

(b^4+3- a^2)/ a (b^4+3) >0

From St1 we get a=b2, Substituting for a in the expression we get

(b^4+3- b^4)/ b^2(b^4+3)-------> 3/b^2(b^4+3) is always greater than zero for any value of b

from St 2, we get

a=b2 and a=-b2 which when applied to the expression will give us 2 ans i.e Expression>0 for a=b2 and expression<0 for a=-b2

Hence A alone is sufficient

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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 26 Jan 2013, 11:51
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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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1) [#permalink] New post 30 Jan 2013, 22:30
Y not take specific numbers and then try to solve..

means square of two numbers are equal when both the numbers are equal (barring their signs)

moreover, we can find such a number for which a^2=b^4

hope it helps!
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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, 08:08
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......

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Re: If a is not equal to zero, is 1/a > a / (b^4 + 3) ? (1)   [#permalink] 24 Feb 2013, 08:08
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