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If ab ≠ 0, is ab > a/b ?
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25 May 2014, 12:32
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If ab ≠ 0, is ab > a/b ? (1) b > 1 (2) ab + a/b > 0
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Re: If ab ≠ 0, is ab > a/b ?
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31 May 2014, 04:41
Bunuel wrote: If ab ≠ 0, is ab > a/b ?
Is \(ab > \frac{a}{b}\)? > is \(ab  \frac{a}{b}>0\)? > \(\frac{a(b^21)}{b}>0\)? > \(\frac{a}{b}*(b^21)>0\)?
(1) b > 1 > both sides are nonnegative, thus we can square: \(b^2>1\) > \(b^21>0\). 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^21. Not sufficient.
(1)+(2) \(b^21>0\) and \(\frac{a}{b} > 0\), thus their product \(\frac{a}{b}*(b^21)>0\). 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 ?
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31 May 2014, 05:09
Manik12345 wrote: Bunuel wrote: If ab ≠ 0, is ab > a/b ?
Is \(ab > \frac{a}{b}\)? > is \(ab  \frac{a}{b}>0\)? > \(\frac{a(b^21)}{b}>0\)? > \(\frac{a}{b}*(b^21)>0\)?
(1) b > 1 > both sides are nonnegative, thus we can square: \(b^2>1\) > \(b^21>0\). 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^21. Not sufficient.
(1)+(2) \(b^21>0\) and \(\frac{a}{b} > 0\), thus their product \(\frac{a}{b}*(b^21)>0\). 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 ?
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03 Jun 2014, 05:33
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 ?
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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^21)}{b}>0? > \frac{a}{b}*(b^21)>0?
(1) b > 1 > both sides are nonnegative, thus we can square: b^2>1 > b^21>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^21. Not sufficient.
(1)+(2) b^21>0 and \frac{a}{b} > 0, thus their product \frac{a}{b}*(b^21)>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 ?
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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^21)}{b}>0? > \frac{a}{b}*(b^21)>0?
(1) b > 1 > both sides are nonnegative, thus we can square: b^2>1 > b^21>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^21. Not sufficient.
(1)+(2) b^21>0 and \frac{a}{b} > 0, thus their product \frac{a}{b}*(b^21)>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 ?
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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 ?
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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 ?
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Re: If ab ≠ 0, is ab > a/b ?
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08 Nov 2014, 20:43
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 \((b^21)\). Insufficient
Together, \(\frac{a}{b} > 0\) & \((b^21)> 0\) => \(\frac{a}{b}*(b^21) > 0\). Sufficient.
Therefore, answer is C.



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Re: If ab ≠ 0, is ab > a/b ?
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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 ?
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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 ?
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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 crossmultiply 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 ?
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11 May 2015, 01:22
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 crossmultiply 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 ?
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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 Regards, Aj



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Re: If ab ≠ 0, is ab > a/b ?
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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 ?
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09 Sep 2015, 06:37
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 ? > aba/b >0 > a(b1/b) > 0 > \(\frac{a*(b^21)}{b} > 0\) > \(\frac{a*(b+1)(b1)}{b} > 0\), this is ONLY dependent on the sign of (b+1)(b1) 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)(b1) 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 ?
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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 ? > aba/b >0 > a(b1/b) > 0 > \(\frac{a*(b^21)}{b} > 0\) > \(\frac{a*(b+1)(b1)}{b} > 0\), this is ONLY dependent on the sign of (b+1)(b1) 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)(b1) 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 ?
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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 eyeopener. Number picking method just killed me by such problems... Here's my try for this problem. Question:\(\frac{a*(b^21)}{b}>0\) (1) b > 1 and b < 1 this tells us that this part is always \(b^21 > 0\), 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 \(b^21\) (1)+(2) From (1) we know that \(b^21 > 0\) 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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