Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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, \(a^2 = b^4 = (-9)^2 = 3^4\) _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...