#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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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.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.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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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.)
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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to find whether 1/a > a/(b^4 +3) or 1/a - a/(b^4 +3) >0If a does not equal to zero, is 1/a > a/(b^4 +3)?
(1) a^2=b^2
(2) a^2=b^4
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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Is 1/a>a/(b^4+3)?-----1)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.
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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The answer choices seem to contradict each other.(1) a^2=b^2
(2) a^2=b^4
(1) a=b (2) a=b^2
Should you be on the lookout for such instances? Or is this irrelevant?
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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
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: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.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.
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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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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)
(i haven't even had the chance to look at that book yet)
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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.
If you spot errors in any of our materials, please send an email to [email protected] and they will make sure the appropriate person is alerted. Thanks!
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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...
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...