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Intern  Joined: 27 Mar 2013
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If ab ≠ 0, is ab > a/b ?  [#permalink]

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If ab ≠ 0, is ab > a/b ?

(1) |b| > 1

(2) ab + a/b > 0
Math Expert V
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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If ab ≠ 0, is ab > a/b ?

Is $$ab > \frac{a}{b}$$? --> is $$ab - \frac{a}{b}>0$$? --> $$\frac{a(b^2-1)}{b}>0$$? --> $$\frac{a}{b}*(b^2-1)>0$$?

(1) |b| > 1 --> both sides are non-negative, thus we can square: $$b^2>1$$ --> $$b^2-1>0$$. We need to know the sign of a/b. Not sufficient.

(2) ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0. We need to know the sign of b^2-1. Not sufficient.

(1)+(2) $$b^2-1>0$$ and $$\frac{a}{b} > 0$$, thus their product $$\frac{a}{b}*(b^2-1)>0$$. Sufficient.

Hope it's clear.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Bunuel wrote:
If ab ≠ 0, is ab > a/b ?

Is $$ab > \frac{a}{b}$$? --> is $$ab - \frac{a}{b}>0$$? --> $$\frac{a(b^2-1)}{b}>0$$? --> $$\frac{a}{b}*(b^2-1)>0$$?

(1) |b| > 1 --> both sides are non-negative, thus we can square: $$b^2>1$$ --> $$b^2-1>0$$. We need to know the sign of a/b. Not sufficient.

(2) ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0. We need to know the sign of b^2-1. Not sufficient.

(1)+(2) $$b^2-1>0$$ and $$\frac{a}{b} > 0$$, thus their product $$\frac{a}{b}*(b^2-1)>0$$. Sufficient.

Hope it's clear.

Hi Bunell,

Can you please explain this question through plugging in. For (2) , ab>-(a/b) .. So for what values this will give yes and no and for which values after combining 1 and 2 we will get the req answer

Also wanted to add. From (2) a/b and ab has to be of the same signs . i tried plugging in values in 2. a=8,b=4 so ab=32 and a/b=2 . Similarly a=-4,b=-2 , ab=8, a/b=2 and also with a=1/8,b=8 so ab=1, a/b=1/72. So for all values ab>a/b.

Please let me knw what i am doin wrong
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Manik12345 wrote:
Bunuel wrote:
If ab ≠ 0, is ab > a/b ?

Is $$ab > \frac{a}{b}$$? --> is $$ab - \frac{a}{b}>0$$? --> $$\frac{a(b^2-1)}{b}>0$$? --> $$\frac{a}{b}*(b^2-1)>0$$?

(1) |b| > 1 --> both sides are non-negative, thus we can square: $$b^2>1$$ --> $$b^2-1>0$$. We need to know the sign of a/b. Not sufficient.

(2) ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0. We need to know the sign of b^2-1. Not sufficient.

(1)+(2) $$b^2-1>0$$ and $$\frac{a}{b} > 0$$, thus their product $$\frac{a}{b}*(b^2-1)>0$$. Sufficient.

Hope it's clear.

Hi Bunell,

Can you please explain this question through plugging in. For (2) , ab>-(a/b) .. So for what values this will give yes and no and for which values after combining 1 and 2 we will get the req answer

Also wanted to add. From (2) a/b and ab has to be of the same signs . i tried plugging in values in 2. a=8,b=4 so ab=32 and a/b=2 . Similarly a=-4,b=-2 , ab=8, a/b=2 and also with a=1/8,b=8 so ab=1, a/b=1/72. So for all values ab>a/b.

Please let me knw what i am doin wrong

For (2):

If a=b=1 (ab + a/b > 0), then ab = a/b. Hence in this case the answer to the question whether ab > a/b is NO.

If a=b=2 (ab + a/b > 0), then ab > a/b. Hence in this case the answer to the question whether ab > a/b is YES.

Hope it's clear.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Wow! Great approach Buneul. It never occurred to me to simplify the equation in the question.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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If ab ≠ 0, is ab > a/b ?

Is ab > \frac{a}{b}? --> is ab - \frac{a}{b}>0? --> \frac{a(b^2-1)}{b}>0? --> \frac{a}{b}*(b^2-1)>0?

(1) |b| > 1 --> both sides are non-negative, thus we can square: b^2>1 --> b^2-1>0. We need to know the sign of a/b. Not sufficient.

(2) ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0. We need to know the sign of b^2-1. Not sufficient.

(1)+(2) b^2-1>0 and \frac{a}{b} > 0, thus their product \frac{a}{b}*(b^2-1)>0. Sufficient.

Hi bunuel,

I did not get the highlighted part of the solution.
Can you please explain why both a and b cannot be negative. I think if we consider values of a and b = -2 and -3 respectively. ab + a/b will remain positive overall.

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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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GmatDestroyer2013 wrote:
If ab ≠ 0, is ab > a/b ?

Is ab > \frac{a}{b}? --> is ab - \frac{a}{b}>0? --> \frac{a(b^2-1)}{b}>0? --> \frac{a}{b}*(b^2-1)>0?

(1) |b| > 1 --> both sides are non-negative, thus we can square: b^2>1 --> b^2-1>0. We need to know the sign of a/b. Not sufficient.

(2) ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0. We need to know the sign of b^2-1. Not sufficient.

(1)+(2) b^2-1>0 and \frac{a}{b} > 0, thus their product \frac{a}{b}*(b^2-1)>0. Sufficient.

Hi bunuel,

I did not get the highlighted part of the solution.
Can you please explain why both a and b cannot be negative. I think if we consider values of a and b = -2 and -3 respectively. ab + a/b will remain positive overall.

Both ab and a/b cannot be negative, not a and b.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Yes, I agree. Thank you for the quick response.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Why cant I solve the equation as ab2 > a hence b2>1 hence (a) the ans ???
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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ashutoshbarawkar wrote:
Why cant I solve the equation as ab2 > a hence b2>1 hence (a) the ans ???

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

So, we can neither cross-multiply ab > a/b by b nor reduce it by a.

If b is positive, then yes we'll get ab^2 > a but if b is negative, then when multiplying by negative value we should flip the sign of the inequality and in this case we would get ab^2 < a.

The same way if a is positive, then b > 1/b but if it's negative, then b < 1/b.

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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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1
ab > a/b ?

We can simplify the stem to:

$$ab-\frac{a}{b} > 0$$ or $$\frac{a}{b}* (b^2 - 1)$$> 0 ?

(1) $$|b| > 1$$ => $$b^2 > 1$$. Thus, $$(b^2 - 1) > 0$$. However, we know nothing about $$\frac{a}{b}$$. Insufficient

(2) $$ab+\frac{a}{b} > 0$$ => $$\frac{a}{b}*(b^2 + 1) > 0$$. Since, $$b^2 > 0$$, $$(b^2 + 1) > 0$$. Thus, $$\frac{a}{b} > 0$$. However, we know nothing about $$(b^2-1)$$. Insufficient

Together, $$\frac{a}{b} > 0$$ & $$(b^2-1)> 0$$ => $$\frac{a}{b}*(b^2-1) > 0$$. Sufficient.

Therefore, answer is C.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Can someone please explain why C is the answer. I still dont understand the explanation. From statement (2) ab + a/b > 0 we know that both a and b are either positive or negative. but without information on the relationship between a and b, we can't determine the answer. Statement (1) just tells us that either b<-1 or b>1. But we dont know anything about a. For example, if a=b, then the answer to the question is no. But if a=2 and b=3, then the answer to the question is yes. So both statement combined are insufficient. Please help!
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GMAT 1: 710 Q49 V38 GMAT 2: 760 Q48 V47 If ab ≠ 0, is ab > a/b ?  [#permalink]

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kritiu wrote:
Can someone please explain why C is the answer. I still dont understand the explanation. From statement (2) ab + a/b > 0 we know that both a and b are either positive or negative. but without information on the relationship between a and b, we can't determine the answer. Statement (1) just tells us that either b<-1 or b>1. But we dont know anything about a. For example, if a=b, then the answer to the question is no. But if a=2 and b=3, then the answer to the question is yes. So both statement combined are insufficient. Please help!

Form statement you get ab>-a/b

Check the excel sheet, I have put some numbers on it. I couldn't add a table here. I am still learning the ropes of this trade.
Attachments Workbook1.xlsx [35.33 KiB]

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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Bunuel wrote:
ashutoshbarawkar wrote:
Why cant I solve the equation as ab2 > a hence b2>1 hence (a) the ans ???

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

So, we can neither cross-multiply ab > a/b by b nor reduce it by a.

If b is positive, then yes we'll get ab^2 > a but if b is negative, then when multiplying by negative value we should flip the sign of the inequality and in this case we would get ab^2 < a.

The same way if a is positive, then b > 1/b but if it's negative, then b < 1/b.

Hi Bunnel, Considering the above quote, is the below approach correct?

Given that ab>a/b

Case 1: if a > 0 then b > 1/b ; true if b>1
Case 2: if a < 0 then b< 1/b ; true is b <1

State 1 says lbl >1 cant conclude anything from this

State 2 says ab + a/b > 0 so from this we know ab & a/b are both positive which means ab is positive, meaning a & b have same sign.
Cant conclude anything further

From State 1 & 2;
lbl > 1 and ab have same sign so only case 1 valid.

Aj
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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AjChakravarthy wrote:
Bunuel wrote:
ashutoshbarawkar wrote:
Why cant I solve the equation as ab2 > a hence b2>1 hence (a) the ans ???

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

So, we can neither cross-multiply ab > a/b by b nor reduce it by a.

If b is positive, then yes we'll get ab^2 > a but if b is negative, then when multiplying by negative value we should flip the sign of the inequality and in this case we would get ab^2 < a.

The same way if a is positive, then b > 1/b but if it's negative, then b < 1/b.

Hi Bunnel, Considering the above quote, is the below approach correct?

Given that ab>a/b

Case 1: if a > 0 then b > 1/b ; true if b>1
Case 2: if a < 0 then b< 1/b ; true is b <1

State 1 says lbl >1 cant conclude anything from this

State 2 says ab + a/b > 0 so from this we know ab & a/b are both positive which means ab is positive, meaning a & b have same sign.
Cant conclude anything further

From State 1 & 2;
lbl > 1 and ab have same sign so only case 1 valid.

Aj

Hi AjChakravarthy,

Few suggestions on your analysis:

Case 1: if a > 0 then b > 1/b ; true if b>1--> b > 1/b is also true when -1 < b < 0. Consider b = -0.5, in this case -0.5 > -2. Hence we can't say that b > 1/b is true for only b > 1.

Case 2: if a < 0 then b< 1/b ; true is b <1 --> b < 1/b is not true for cases when -1 < b < 0. Hence we can't say b < 1/b is true for all values of b < 1.

From State 1 & 2;
lbl > 1 and ab have same sign so only case 1 valid
-- > |b| > 1 can be written as b > 1 if b is +ve or -b > 1 if b is -ve. This can be simplified to b > 1 or b < -1. Hence |b| > 1 can't be inferred as b being always positive.

Hope its clear!

Regards
Harsh
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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[/quote]

Hi AjChakravarthy,

Few suggestions on your analysis:

Case 1: if a > 0 then b > 1/b ; true if b>1--> b > 1/b is also true when -1 < b < 0. Consider b = -0.5, in this case -0.5 > -2. Hence we can't say that b > 1/b is true for only b > 1.

Case 2: if a < 0 then b< 1/b ; true is b <1 --> b < 1/b is not true for cases when -1 < b < 0. Hence we can't say b < 1/b is true for all values of b < 1.

From State 1 & 2;
lbl > 1 and ab have same sign so only case 1 valid
-- > |b| > 1 can be written as b > 1 if b is +ve or -b > 1 if b is -ve. This can be simplified to b > 1 or b < -1. Hence |b| > 1 can't be inferred as b being always positive.

Hope its clear!

Regards
Harsh[/quote]

Oh Damn!! Thanks a lot Harsh.

These things between -1 and +1 are dangerous things. They behave very differently and I always get these wrong Regards,
Aj
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GMAT 1: 640 Q48 V28 Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Despite reading the question and explanation again and again I still can't understand " ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0."

Bunuel could you please help? CEO  S
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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akirah wrote:
Despite reading the question and explanation again and again I still can't understand " ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0."

Bunuel could you please help? Let me try to explain.

When you are given that ab+a/b > 0 ---> $$a* \frac{b^2+1}{b} >0$$ , now realise that $$b^2+1$$ is ALWAYS > 0 for all values of b and thus for making $$a* \frac{b^2+1}{b} >0$$ you only need to worry about a/b > 0

Once you see that the given statement 2 is reduced to a/b > 0 ---> this can only mean that both a and b have the SAME signs. Either both are <0 or both are >0. You can also see it this way:

when a/b >0--> consider 4 cases:

a>0 and b>0 ---> YES for a/b >0
a>0 and b<0 ---> NO for a/b >0
a<0 and b>0 ---> NO for a/b >0
a<0 and b<0 ---> YES for a/b >0

Thus, you get "YES" only when both a and b are of the SAME sign.

You do not have to worry about the actual sign of a or b as shown below:

Once you see that statements need to be combined, you see that the original equation is ab > a/b ? ---> ab-a/b >0 ---> a(b-1/b) > 0 ---> $$\frac{a*(b^2-1)}{b} > 0$$

---> $$\frac{a*(b+1)(b-1)}{b} > 0$$, this is ONLY dependent on the sign of (b+1)(b-1) as a/b will ALWAYS be positive for all values of a and b (as both a and b are of the SAME sign).

Statement 1 mentions that |b| >1 --> b>1 or b<-1 , for these ranges, (b+1)(b-1) is > 0 ALWAYS and hence you get a definite "YES", making C as the correct answer.

Hope this helps.
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GMAT 1: 640 Q48 V28 Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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Engr2012 wrote:
akirah wrote:
Despite reading the question and explanation again and again I still can't understand " ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0."

Bunuel could you please help? Let me try to explain.

When you are given that ab+a/b > 0 ---> $$a* \frac{b^2+1}{b} >0$$ , now realise that $$b^2+1$$ is ALWAYS > 0 for all values of b and thus for making $$a* \frac{b^2+1}{b} >0$$ you only need to worry about a/b > 0

Once you see that the given statement 2 is reduced to a/b > 0 ---> this can only mean that both a and b have the SAME signs. Either both are <0 or both are >0. You can also see it this way:

when a/b >0--> consider 4 cases:

a>0 and b>0 ---> YES for a/b >0
a>0 and b<0 ---> NO for a/b >0
a<0 and b>0 ---> NO for a/b >0
a<0 and b<0 ---> YES for a/b >0

Thus, you get "YES" only when both a and b are of the SAME sign.

You do not have to worry about the actual sign of a or b as shown below:

Once you see that statements need to be combined, you see that the original equation is ab > a/b ? ---> ab-a/b >0 ---> a(b-1/b) > 0 ---> $$\frac{a*(b^2-1)}{b} > 0$$

---> $$\frac{a*(b+1)(b-1)}{b} > 0$$, this is ONLY dependent on the sign of (b+1)(b-1) as a/b will ALWAYS be positive for all values of a and b (as both a and b are of the SAME sign).

Statement 1 mentions that |b| >1 --> b>1 or b<-1 , for these ranges, (b+1)(b-1) is > 0 ALWAYS and hence you get a definite "YES", making C as the correct answer.

Hope this helps.

So far, I was so confused with ab, a/b, a and b. This explanation is perfect. It cant get better than this. Thank you so much Engr2012
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If ab ≠ 0, is ab > a/b ?  [#permalink]

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gmatquant25 wrote:
If ab ≠ 0, is ab > a/b ?

(1) |b| > 1

(2) ab + a/b > 0

Solutions for this kind of problems made by Bunuel are just an eye-opener. Number picking method just killed me by such problems... Here's my try for this problem.

Question:$$\frac{a*(b^2-1)}{b}>0$$

(1) b > 1 and b < -1 this tells us that this part is always $$b^2-1 > 0$$, but we have no info about $$\frac{a}{b}$$. Not Sufficient
(2) $$\frac{a*(b^2+1)}{b}>0$$, $$b^2+1$$ is always positive, sothat a and b in the expression $$\frac{a}{b}$$ must have the same sign. But here we have no information about $$b^2-1$$

(1)+(2) From (1) we know that $$b^2-1 > 0$$ and from (2) that $$\frac{a}{b}$$ must have the same sign, meaning that it's always positive. Hence, the whole expression is positive.

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