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Re: Is 4m  3n > 3m  n + m  2n ?
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18 Jul 2014, 11:59
vietmoi999 wrote: this is not og questions and should not be studied If Bunuel is presenting a question , I would definitely study it Sent from my iPhone using Tapatalk



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Re: Is 4m  3n > 3m  n + m  2n ?
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23 Nov 2016, 15:39
Statement 2 says 2n<m which translates to m2n>0. I thought, by filling in numbers, this would mean that 3mn certainly is bigger than 0. However, in the explanation above m=4 and n=3 are given as an example to fill in. But if m =4 and n=3 then 2n is not smaller than m. This would mean that these numbers could not be used with this premise right? m has to be bigger than 2n so m has to be >0. I'd say statement 2 is sufficient to answer the question. Could someone help me please?



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Re: Is 4m  3n > 3m  n + m  2n ?
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24 Nov 2016, 00:47
esmeeaugustijn wrote: Statement 2 says 2n<m which translates to m2n>0. I thought, by filling in numbers, this would mean that 3mn certainly is bigger than 0. However, in the explanation above m=4 and n=3 are given as an example to fill in. But if m =4 and n=3 then 2n is not smaller than m. This would mean that these numbers could not be used with this premise right? m has to be bigger than 2n so m has to be >0. I'd say statement 2 is sufficient to answer the question. Could someone help me please? If m =4 and n=3, then \((2n = 6) < (m = 4)\)
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Re: Is 4m  3n > 3m  n + m  2n ?
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12 Mar 2017, 20:41
Bunuel wrote: SOLUTIONIs \(3m  n + m  2n > 4m  3n\)?One of the properties of absolute values says that \(x+y\geqx+y\). Note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign). So, the strict inequality (>) holds when \(xy<0\). ( Check here: http://gmatclub.com/forum/tipsandhint ... l#p1381430) Notice that if we denote \(x=3m  n\) and \(y=m  2n\), then \(x+y=4m3n\). So, the question becomes: is \(x+y>x+y\)? Thus, the qeustion basically asks whether \(x\) and \(y\), or which is the same \(3m  n\) and \(m  2n\), have the opposite signs. (1) \(m > 0\). Clearly insufficient as no info about \(n\). Not sufficient. (2) \(2n < m\). This implies that \(m2n>0\). If \(m=3\) and \(n=1\), then \(3m  n>0\) (so in this case \(3m  n\) and \(m  2n\) will have the same sign) but if \(m=4\) and \(n=3\), then \(3m  n<0\) (so in this case \(3m  n\) and \(m  2n\) will have different signs sign). Not sufficient. (1)+(2) We have that \(m > 0\), or which is the same \(5m>0\) and \(m>2n\). Add them: \(6m>2n\). Reduce by 2 and rearrange: \(3mn>0\). Thus, both \(m2n\) and \(3mn\) are positive, so we have a NO answer to the question. Sufficient. Answer: C. Kudos points given to correct solutions above.Try NEW Absolute Value PS question. Could someone please help to explain \(x=3m  n\) and \(y=m  2n\), then \(x+y=4m3n\). So, the question becomes: is \(x+y>x+y\)? Thus, the qeustion basically asks whether \(x\) and \(y\), or which is the same \(3m  n\) and \(m  2n\), have the opposite signs? I thought it needs to be the same sign.
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Re: Is 4m  3n > 3m  n + m  2n ?
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13 Mar 2017, 00:21
ziyuen wrote: Bunuel wrote: SOLUTIONIs \(3m  n + m  2n > 4m  3n\)?One of the properties of absolute values says that \(x+y\geqx+y\). Note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign). So, the strict inequality (>) holds when \(xy<0\). ( Check here: http://gmatclub.com/forum/tipsandhint ... l#p1381430) Notice that if we denote \(x=3m  n\) and \(y=m  2n\), then \(x+y=4m3n\). So, the question becomes: is \(x+y>x+y\)? Thus, the qeustion basically asks whether \(x\) and \(y\), or which is the same \(3m  n\) and \(m  2n\), have the opposite signs. (1) \(m > 0\). Clearly insufficient as no info about \(n\). Not sufficient. (2) \(2n < m\). This implies that \(m2n>0\). If \(m=3\) and \(n=1\), then \(3m  n>0\) (so in this case \(3m  n\) and \(m  2n\) will have the same sign) but if \(m=4\) and \(n=3\), then \(3m  n<0\) (so in this case \(3m  n\) and \(m  2n\) will have different signs sign). Not sufficient. (1)+(2) We have that \(m > 0\), or which is the same \(5m>0\) and \(m>2n\). Add them: \(6m>2n\). Reduce by 2 and rearrange: \(3mn>0\). Thus, both \(m2n\) and \(3mn\) are positive, so we have a NO answer to the question. Sufficient. Answer: C. Kudos points given to correct solutions above.Try NEW Absolute Value PS question. Could someone please help to explain \(x=3m  n\) and \(y=m  2n\), then \(x+y=4m3n\). So, the question becomes: is \(x+y>x+y\)? Thus, the qeustion basically asks whether \(x\) and \(y\), or which is the same \(3m  n\) and \(m  2n\), have the opposite signs? I thought it needs to be the same sign. Plug the numbers and check. Try x=1 and y=2 AND x=1 and y=2.
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Re: Is 4m  3n > 3m  n + m  2n ?
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11 Mar 2018, 05:04
Bunuel wrote: SOLUTION
Is \(3m  n + m  2n > 4m  3n\)?
Notice that if we denote \(x=3m  n\) and \(y=m  2n\), then \(x+y=4m3n\). So, the question becomes: is \(x+y>x+y\)? Thus, the qeustion basically asks whether \(x\) and \(y\), or which is the same \(3m  n\) and \(m  2n\), have the opposite signs.
(1) \(m > 0\). Clearly insufficient as no info about \(n\). Not sufficient.
(2) \(2n < m\). This implies that \(m2n>0\). If \(m=3\) and \(n=1\), then \(3m  n>0\) (so in this case \(3m  n\) and \(m  2n\) will have the same sign) but if \(m=4\) and \(n=3\), then \(3m  n<0\) (so in this case \(3m  n\) and \(m  2n\) will have different signs sign). Not sufficient.
Dear BunuelCan you prove that statement 2 is insufficient through algebraic approach? I failed to do so Thanks



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Is 4m  3n > 3m  n + m  2n ?
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11 Mar 2018, 05:16
Mo2men wrote: Dear BunuelCan you prove that statement 2 is insufficient through algebraic approach? I failed to do so Thanks Hey Mo2men , Here I go: We are given 2n < m This means m  2n > 0 and RHS could be 4m  3n or 3n  4m Now, I will take the 2nd and 3rd modulus positive while I am not sure of the 1st modulus, so I will evaluate both + and  scenarios for it. We know that our RHS = 4m  3n because 4m > 3n. When 3m  n > 0 We have LHS = 3m  n + m  2n = 4m  3n , which is also equal to RHS if it is 4m  3n. But not otherwise. When 3m  n < 0 We have LHS =  3m + n + m  2n = 2m  n , which is not equal to RHS. Hence, 2nd statement is giving us both scenarios. Hence, insufficient. Does that make sense?
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Re: Is 4m  3n > 3m  n + m  2n ?
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11 Mar 2018, 05:25
abhimahna wrote: Mo2men wrote: Dear BunuelCan you prove that statement 2 is insufficient through algebraic approach? I failed to do so Thanks Hey Mo2men , Here I go: We are given 2n < m This means m  2n > 0 and if m  2n > 0 => m > 2n or 2m > 4n or 4m > 8n. => 4m > 3n as wellNow, I will take the 2nd and 3rd modulus positive while I am not sure of the 1st modulus, so I will evaluate both + and  scenarios for it. We know that our RHS = 4m  3n because 4m > 3n. When 3m  n > 0 We have LHS = 3m  n + m  2n = 4m  3n , which is also equal to RHS When 3m  n < 0 We have LHS =  3m + n + m  2n = 2m  n , which is not equal to RHS. Hence, 2nd statement is giving us both scenarios. Hence, insufficient. Does that make sense? Hi abhimahnaThe highlighted dedcution is not always true Take Bunuel example above m = 4 & n =3 Statement 2 says : m > 2n ......... 4 > 6...true In your conclusion that 4m > 3n, you can consider same same example above 16 > 9..........which is not true.



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Re: Is 4m  3n > 3m  n + m  2n ?
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11 Mar 2018, 05:41
Mo2men wrote: Hi abhimahnaThe highlighted dedcution is not always true Take Bunuel example above m = 4 & n =3 Statement 2 says : m > 2n ......... 4 > 6...true In your conclusion that 4m > 3n, you can consider same same example above 16 > 9..........which is not true. Hey Mo2men , I agree to your point. I missed the nature of variables point. Anyways, to conclude we can still have two different RHS values and some of the scenarios would work while others won't. Hence, B is insufficient.
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Re: Is 4m  3n > 3m  n + m  2n ?
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11 Mar 2018, 16:30
abhimahna wrote: Hey Mo2men , I agree to your point. I missed the nature of variables point. Hi abhimahnaIt seems you forgot math and focus on other school issues now I was challenged by statement 2 and reached algebraic proof and I hope Bunuel could feedback. The xlation might be long for details. Question: 3m−n+m−2n>4m−3n Quote Bunuel solution: One of the properties of absolute values says that x+y≥x+y. Note that "=" sign holds for xy≥0 (or simply when x and y have the same sign). So, the strict inequality (>) holds when xy<0. Notice that if we denote x=3m−n and y=m−2n, then x+y=4m−3n. So, the question becomes: is x+y>x+y? Thus, the question basically asks whether x and y, or which is the same 3m−n and m−2n, have the opposite signs. Statement 2: \(2n < m\) Case 1:m > 0 & n>= 0 m > 2n & 2n >n....hence m>n and consequently 4m > 3n & 3m > n First term: 3m−n = +++ second term: m−2n = +++ x & y have same sign.............Answer to question is NOCase 2:m > 0 & n < 0 Same as above excalty x & y have same sign.............Answer to question is NOThe trick starts here when 'm' is negative and 'n' is negative because m could be larger or less than n. (let m=4 & n =2) or ( m=2 & n =3). Both examples hold statement 2 true. let's see the algebraic approach By plugging numbers, I proved that '3mn' is negative and hence x & y have different sign and answer is Yes. But I could not reach it algebraically. I hope to get some help. Thanks



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Is 4m  3n > 3m  n + m  2n ?
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12 Mar 2018, 00:43
Mo2men wrote: Hi abhimahnaIt seems you forgot math and focus on other school issues now I was challenged by statement 2 and reached algebraic proof and I hope Bunuel could feedback. The xlation might be long for details. Question: 3m−n+m−2n>4m−3n Quote Bunuel solution: One of the properties of absolute values says that x+y≥x+y. Note that "=" sign holds for xy≥0 (or simply when x and y have the same sign). So, the strict inequality (>) holds when xy<0. Notice that if we denote x=3m−n and y=m−2n, then x+y=4m−3n. So, the question becomes: is x+y>x+y? Thus, the question basically asks whether x and y, or which is the same 3m−n and m−2n, have the opposite signs. Statement 2: \(2n < m\) Case 1:m > 0 & n>= 0 m > 2n & 2n >n....hence m>n and consequently 4m > 3n & 3m > n First term: 3m−n = +++ second term: m−2n = +++ x & y have same sign.............Answer to question is NOCase 2:m > 0 & n < 0 Same as above excalty x & y have same sign.............Answer to question is NOThe trick starts here when 'm' is negative and 'n' is negative because m could be larger or less than n. (let m=4 & n =2) or ( m=2 & n =3). Both examples hold statement 2 true. let's see the algebraic approach By plugging numbers, I proved that '3mn' is negative and hence x & y have different sign and answer is Yes. But I could not reach it algebraically. I hope to get some help. Thanks Hey Mo2men , Lol, I have to politely deny your statement "It seems you forgot math and focus on other school issues now". I am pretty much sure that if I give the exam at this moment also, I will end up getting Q50. Anyways, you need to understand the fact that there are n number of ways to solve any question. The way I approached above can also be used to break the 2nd statement and call it insufficient. And the way I solved is also an algebraic way to solve any modulus questions. You need to understand that if we have any complex modulus, we can always end up making multiple possibilities and see which case will actually work. That's what I did. Hence. I got 2nd statement insufficient. I hope that makes sense.
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Is 4m  3n > 3m  n + m  2n ?
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12 Mar 2018, 04:06
abhimahna wrote: Hey Mo2men , Lol, I have to politely deny your statement "It seems you forgot math and focus on other school issues now". I am pretty much sure that if I give the exam at this moment also, I will end up getting Q50. Anyways, you need to understand the fact that there are n number of ways to solve any question. The way I approached above can also be used to break the 2nd statement and call it insufficient. And the way I solved is also an algebraic way to solve any modulus questions. You need to understand that if we have any complex modulus, we can always end up making multiple possibilities and see which case will actually work. That's what I did. Hence. I got 2nd statement insufficient. I hope that makes sense. I'm pretty sure you get Q50 my friend I wish you to continue success in your mba journey.



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