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MGMAT Problem (Advanced Quant)

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MGMAT Problem (Advanced Quant)

by OneTwoThreeFour » Thu Jun 09, 2011 2:13 pm
#4-9

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

(1) a^2=b^2

(2) a^2=b^4

Thanks!

Spoiler:
The correct answer is A, but I think the answer should be E. Based on statement 1, a and b can have the same numbers but with different signs. Therefore, I think E should be the solution.
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by Ian Stewart » Thu Jun 09, 2011 2:46 pm
OneTwoThreeFour wrote:#4-9

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

(1) a^2=b^2

(2) a^2=b^4

Thanks!

Spoiler:
The correct answer is A, but I think the answer should be E. Based on statement 1, a and b can have the same numbers but with different signs. Therefore, I think E should be the solution.
Using either statement, we can have a=b=1, in which case 1/a is bigger than a/(b^4 +3), since 1 is bigger than 1/4, or we can have a=b=-1, in which case 1/a is smaller than a/(b^4 + 3), since -1 is smaller than -1/4. So the answer is E. If the OA is A, it's wrong.
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by OneTwoThreeFour » Thu Jun 09, 2011 3:07 pm
Thanks Ian!

Are there any MGMAT tutors that want to chime in? (I think I also found another error in the book; For the MGMAT tutors, is there an email address or link where I can send in potential typos? The book has been excellent so far, but I understand first editions are always a bit rougher around the edges than future editions.)
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by cans » Fri Jun 10, 2011 8:23 am
If a does not equal to zero, is 1/a > a/(b^4 +3)?

(1) a^2=b^2

(2) a^2=b^4
to find whether 1/a > a/(b^4 +3) or 1/a - a/(b^4 +3) >0
or (b^4 + 3 - a^2)/a >0
or a(b^4 - a^2 + 3) >0
a) a^2 = b^2
a(a^4 - a^2 + 3) >0
(a^4 - a^2 + 3) this is always greater than 0 as D>0
thus a>0 ?? insufficient
b)a^2 = b^4 -> a>0???
insufficient
a&b) a^2 = b^2 = b^4 -> b^2=1 (b^2 is not equal to 0, as a is not equal to 0)
a^2=1 or a=+-1
insufficient
IMO E
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by manpsingh87 » Fri Jun 10, 2011 10:10 am
OneTwoThreeFour wrote:#4-9

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

(1) a^2=b^2

(2) a^2=b^4

Thanks!

Spoiler:
The correct answer is A, but I think the answer should be E. Based on statement 1, a and b can have the same numbers but with different signs. Therefore, I think E should be the solution.
Is 1/a>a/(b^4+3)?-----1)
1)a^2=b^2;
a^2-b^2=0;
a=b; or a=-b;
put a=b we have;
b/b^4+3;
let b=1; 1/1+3=1/4 and 1/b=1 hence 1/b>b/b^4+3;
now let b=1/2; we have b/b^4+3 8/49; and 1/b=2; therefore 1/b>b/b^4+3;

for a=-b;
let b=-1; -1/1+3=-1/4 therefore 1/b<b/b^4+3;

therefore 1 alone is not sufficient to answer the question.

2)

a^2=b^4;
b^2(b^2-a^2)=0;either b^2=0 or (b^2-a^2)=0
for b=0; we have 1/a>0;
when a>0; then 1/a>0 will hold true and for a<0; 1/a will not hold true, for example for a=-(1/2);we have 1/a<0;

now for (b^2-a^2)=0; as seen from 1) different results are possible, hence 2 alone is also not sufficient to answer the question.

combining 1 and 2 we have;
(b^2-a^2)=0; as a common solution, as seen in 1) which clearly is insufficient..!!

hence answer should be E
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by Sanjay2706 » Sat Jun 11, 2011 5:46 am
IMO A seems right.
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by OneTwoThreeFour » Mon Jun 13, 2011 5:00 pm
Why do you think A is right Sanjay? I would love to hear your thoughts on the problem.
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by nafiul9090 » Tue Jun 14, 2011 8:31 am
i think E fits the bill
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by Gurpinder » Tue Jun 14, 2011 9:12 am
(1) a^2=b^2

(2) a^2=b^4
The answer choices seem to contradict each other.

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

Should you be on the lookout for such instances? Or is this irrelevant?
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by nubu » Tue Jun 14, 2011 10:40 pm
Hi Ian

In statement 1, why could you point out only 1 case a = b = +/- 1. IMO, from a^2 = b^2 we may have a = b = +/-1 or 2 or 3 or ....? Am I misunderstanding something?

Thanks a lot
Ian Stewart wrote:
OneTwoThreeFour wrote:#4-9

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

(1) a^2=b^2

(2) a^2=b^4

Thanks!

Spoiler:
The correct answer is A, but I think the answer should be E. Based on statement 1, a and b can have the same numbers but with different signs. Therefore, I think E should be the solution.
Using either statement, we can have a=b=1, in which case 1/a is bigger than a/(b^4 +3), since 1 is bigger than 1/4, or we can have a=b=-1, in which case 1/a is smaller than a/(b^4 + 3), since -1 is smaller than -1/4. So the answer is E. If the OA is A, it's wrong.
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by Ian Stewart » Tue Jun 14, 2011 11:03 pm
nubu wrote:

Using either statement, we can have a=b=1, in which case 1/a is bigger than a/(b^4 +3), since 1 is bigger than 1/4, or we can have a=b=-1, in which case 1/a is smaller than a/(b^4 + 3), since -1 is smaller than -1/4. So the answer is E. If the OA is A, it's wrong.


I wasn't giving every solution to the equation; I was only proving the statement is not sufficient. To do that, we just need one example which gives a 'yes' answer to the question, and one which gives a 'no' answer. There are lots of other examples I could have chosen, but 1 and -1 seemed the simplest.
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by lunarpower » Thu Jun 16, 2011 3:00 am
the advanced quant book was just released a couple of weeks ago, so it will probably contain errors -- thanks for pointing this out. i'll bring it to the writers' attention.
(i haven't even had the chance to look at that book yet)
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by Stacey Koprince » Thu Jun 16, 2011 5:09 am
Yes, we definitely keep track so that we can fix errors in future printings, and we also post errata lists online so that you can check to see whether something is already a known error. I don't think we have an errata list up for this book yet, but we will soon.

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by sumgb » Sun Aug 07, 2011 3:00 pm
my take on this problem is ...

stmnt 1, a=b=1, true; a=b=-1 false so insuff.
stmnt 2, since a^2 = b^4 this implies a = b^2 (taking square root) which means a is +ve since b^2 cant be negative. since a is not equal to 0. stmnt 2 is sufficient.

to test this, if a=b=1, then true; if a=4 b=2 then true; if a = 9 b =3 in any case 1/a is bigger than a/(b^4 +3). (sign of variable b does not matter as the expression contains b^4 which is always +ve) Hence stmnt 2 sufficient.

IMO B

Am I not allowed to take sq roots of the given statements?
Can experts please comment on my approach and correct it if I am missing something? please...
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by sumgb » Sun Aug 07, 2011 3:03 pm
What's the takeaway here please?
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