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Re: DS question from MGMAT advanced quant [#permalink]
11 Mar 2012, 10:16

3

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Expert's post

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This post was BOOKMARKED

pavanpuneet wrote:

From the last post from Manager, he says: from (1) a = b^2 ==> b= Mod(a) from here, you can say that (b^4+3) > a^2. cz b^2 = a or, b^4 = a^2 (squaring both side) ==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain... from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

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

Notice that we cannot cross-multiply \frac{1}{a} > \frac{a}{b^4 +3}, since we don't know whether a>0 or a<0, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with b^4 +3, since this expression is always positive: b^4 +3=nonnegative+positive=positive. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is \frac{1}{b^2} > \frac{b^2}{b^4 +3}? Now, since b^2>0 then here we can safely cross-multiply and the question becomes: is b^4+3>b^4? --> is 3>0? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is \frac{1}{a} > \frac{a}{a^2 +3}? We have the same problem: we cannot cross-multiply since we don't know the sign of a. If a=1 then the answer to the question would be YES but if a=-1 then the answer to the question would be NO. Not sufficient.

Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]
22 May 2012, 18:14

Bunuel never ceases to amaze me. His reasoning is so sharp and his methods so efficient, I bet he conquers every problem in under 30 seconds. I have been following his posts extensively hoping to learn to think like he does. Lifelong fan right here.

Re: DS question from MGMAT advanced quant [#permalink]
07 Jan 2014, 04:26

Bunuel wrote:

pavanpuneet wrote:

From the last post from Manager, he says: from (1) a = b^2 ==> b= Mod(a) from here, you can say that (b^4+3) > a^2. cz b^2 = a or, b^4 = a^2 (squaring both side) ==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain... from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

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

Notice that we cannot cross-multiply \frac{1}{a} > \frac{a}{b^4 +3}, since we don't know whether a>0 or a<0, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with b^4 +3, since this expression is always positive: b^4 +3=nonnegative+positive=positive. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is \frac{1}{b^2} > \frac{b^2}{b^4 +3}? Now, since b^2>0 then here we can safely cross-multiply and the question becomes: is b^4+3>b^4? --> is 3>0? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is \frac{1}{a} > \frac{a}{a^2 +3}? We have the same problem: we cannot cross-multiply since we don't know the sign of a. If a=1 then the answer to the question would be YES but if a=-1 then the answer to the question would be NO. Not sufficient.

Answer: A.

Hi Bunuel, Great explanation. seeing a^2 in case 2 i quickly jumped and marked D as my answer. But here's a doubt that i have: in case 2: what is given to us ... is a^2=b^4. now, since i cant cross multiply my question's equation to get a^2 and then substitute with the value for a^2 n get the answer. so instead, cant i just take a square-root on both sides of my case2 equation.??

which gives me...... a^2 = b^4 ................ now taking root on both sides, i get ((p.s: i can take square-root on both sides as i know that it wont affect the sign(-/+ve) of either of a or b.)) a = b^2 .................. now that i know a is +ve .. i substitute just like I did in the first case n get the answer as D...wouldn't this be correct???

Re: DS question from MGMAT advanced quant [#permalink]
07 Jan 2014, 04:40

Expert's post

rawjetraw wrote:

Bunuel wrote:

pavanpuneet wrote:

From the last post from Manager, he says: from (1) a = b^2 ==> b= Mod(a) from here, you can say that (b^4+3) > a^2. cz b^2 = a or, b^4 = a^2 (squaring both side) ==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain... from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

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

Notice that we cannot cross-multiply \frac{1}{a} > \frac{a}{b^4 +3}, since we don't know whether a>0 or a<0, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with b^4 +3, since this expression is always positive: b^4 +3=nonnegative+positive=positive. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is \frac{1}{b^2} > \frac{b^2}{b^4 +3}? Now, since b^2>0 then here we can safely cross-multiply and the question becomes: is b^4+3>b^4? --> is 3>0? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is \frac{1}{a} > \frac{a}{a^2 +3}? We have the same problem: we cannot cross-multiply since we don't know the sign of a. If a=1 then the answer to the question would be YES but if a=-1 then the answer to the question would be NO. Not sufficient.

Answer: A.

Hi Bunuel, Great explanation. seeing a^2 in case 2 i quickly jumped and marked D as my answer. But here's a doubt that i have: in case 2: what is given to us ... is a^2=b^4. now, since i cant cross multiply my question's equation to get a^2 and then substitute with the value for a^2 n get the answer. so instead, cant i just take a square-root on both sides of my case2 equation.??

which gives me...... a^2 = b^4 ................ now taking root on both sides, i get ((p.s: i can take square-root on both sides as i know that it wont affect the sign(-/+ve) of either of a or b.)) a = b^2 .................. now that i know a is +ve .. i substitute just like I did in the first case n get the answer as D...wouldn't this be correct???

Please do share your thooughts. Cheers!

The point is that \sqrt{a^2}=|a|, not a. For example if a=-1, then \sqrt{a^2}=\sqrt{1}=1=|a|=-a, not a.

We can cross multiply since both the denominators are positive for sure b^4 + 3 > b^2(b^2) B^4 + 3 > B^4 - Well that is true every time

Statement II is insufficient a^2 = b^4 If a is and b are positive then the equation will turn to: B^4 + 3 > B^4 In that case YES If a is negative and b is positive then a = -b^2, substituting it we will get: b^4 + 3 < -1 (b^2)(b^2) b^4 + 3 < -(b^4) - That is not true

So we have a YES and a NO hence statement II is insufficient

Answer is A _________________

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If a ≠ 0, is (1/a) > a/(b^4+3) ? [#permalink]
04 Aug 2014, 05:12

1

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pratikshr wrote:

If A \neq 0, is\frac{1}{A}>\frac{A}{B^4+3} ?

(1) A = B^2 (2) A^2 = B^4

It's a Yes/No question. From question itself we can deduce that (B^4+3) will always be positive. Rephrasing question -> \frac{B^4+3}{A}>\frac{A}{1} ? Note that ,we can not multiply A in this inequality since A might be negative too.

Stmt 1 - A = B^2 => A is square of some number (B in this case) so A will always be positive. So multiply with A both side -> squaring statement 1 at both sides gives us A^2 = B^4 so we now can multiply with A at both sides as A will always be positive => Sufficient

Stmt 2 - A^2 = B^4 => Here A can be negative also as shown below - A = B^2 or A = -B^2 So this statement is not Sufficient.

Re: If a does not = 0, is 1/a > a/(b^4 + 3) [#permalink]
14 Aug 2014, 00:48

1

This post received KUDOS

Can you explain this part?

a^2=b^4 --> is \frac{1}{a} > \frac{a}{a^2 +3}?

Bunuel wrote:

pavanpuneet wrote:

From the last post from Manager, he says: from (1) a = b^2 ==> b= Mod(a) from here, you can say that (b^4+3) > a^2. cz b^2 = a or, b^4 = a^2 (squaring both side) ==> (b^4+X) > a^2 (where x is +ve no)...

Then why cant we say that option B alone is also sufficient and hence answer should be D..

To further explain... from stem: b^4+3>a^2; thus, from 1, a=b^2; squaring results a^2=b^4; thus clearly stem is satisfied. Further this condition is nothing but condition 2. Please explain where am I wrong?

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

Notice that we cannot cross-multiply \frac{1}{a} > \frac{a}{b^4 +3}, since we don't know whether a>0 or a<0, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity). Also notice that we don't have the same issue with b^4 +3, since this expression is always positive: b^4 +3=nonnegative+positive=positive. So, 321kumarsushant's first step is not correct.

(1) a=b^2 --> is \frac{1}{b^2} > \frac{b^2}{b^4 +3}? Now, since b^2>0 then here we can safely cross-multiply and the question becomes: is b^4+3>b^4? --> is 3>0? Answer ot this question is YES.Sufficient.

(2) a^2=b^4 --> is \frac{1}{a} > \frac{a}{a^2 +3}? We have the same problem: we cannot cross-multiply since we don't know the sign of a. If a=1 then the answer to the question would be YES but if a=-1 then the answer to the question would be NO. Not sufficient.

Answer: A.

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Re: If a does not = 0, is 1/a > a/(b^4 + 3)
[#permalink]
14 Aug 2014, 00:48

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