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If ab ≠ 0, is ab > a/b ?

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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
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New post 25 May 2014, 12:45
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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.

Answer: C.

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

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New post 31 May 2014, 04:41
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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.

Answer: C.

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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New post 31 May 2014, 05:09
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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.

Answer: C.

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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New post 03 Jun 2014, 05:33
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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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New post 22 Jun 2014, 02:01
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.

Thanks in advance for your help.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 22 Jun 2014, 03:35
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.

Thanks in advance for your help.


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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New post 22 Jun 2014, 04:44
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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New post 30 Aug 2014, 23:34
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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New post 01 Sep 2014, 02:01
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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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New post 08 Nov 2014, 20:43
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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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New post 24 Dec 2014, 20:00
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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If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 09 Apr 2015, 22:21
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.
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 10 May 2015, 03:05
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.

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

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New post 11 May 2015, 01:22
1
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.

Thanks in advance,
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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New post 11 May 2015, 09:57
[/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 :stupid

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

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New post 09 Sep 2015, 05:49
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? :(
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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 09 Sep 2015, 06:37
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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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Re: If ab ≠ 0, is ab > a/b ?  [#permalink]

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New post 10 Sep 2015, 00:53
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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New post 04 Jan 2016, 14:44
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.

Answer C
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If ab ≠ 0, is ab > a/b ? &nbs [#permalink] 04 Jan 2016, 14:44

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