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Is mn > 0?
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01 Feb 2020, 18:00
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74% (01:09) correct 26% (01:00) wrong based on 41 sessions
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GMATBusters’ Quant Quiz Question 1 For questions from previous quizzes click hereIs mn > 0? 1) m^2 > n^2 2) n < 0
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Re: Is mn > 0?
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01 Feb 2020, 18:09
The official solution is as follows: Attachment:
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Re: Is mn > 0?
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01 Feb 2020, 18:13
1) m^2>n^2 if m=4 and n=3, then m^2=16>n^2=9. and mn=7>0 However, if m=4 and n=3, then m^2=16>n^2=9. and mn=1<0 Clearly, this statement is insufficient. 2) In the case above, n<0 in both cases. Therefore, it follows that this statement is also insufficient. 1)+2) Because of the explanation in 2), both statements together are insufficient as well Answer: E



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Re: Is mn > 0?
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01 Feb 2020, 18:17
mn > 0 ? 1) \(M^2 > n^2\) possible values are 1. Both positive  m=3 , n=2 , then \(m^2 > n^2\) and this gives mn > 0 ? as Yes 2. Both negative  m=3 , n = 2 , this gives \(m^2 > n^2\) and mn < 0? as No Two different answers, so Option 1 not sufficient. 2) N < 0 This option does not give any information about M. Hence insufficient. Lets try (1) + (2) Lets consider both n <0 and \(m^2 > n^2\) Lets plug in below given values 1) m=3 , n = 2 , this gives \(m^2 > n^2\). MN < 0 ? = No 2) m=1 , n = 2 , this gives \(m^2 > n^2\). MN < 0 ? = Yes Two different solution. Hence C is insufficient. Ans: E
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Re: Is mn > 0?
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01 Feb 2020, 18:54
the given inequality m  n> 0 can be modified into m>n ?
st1:m^2>n^2 this can be written as m > n which means that m & n can take multiple values (positive or negative) which will satisfy the given condition. this not sufficient to determine which variable is greater.
st 2: n<0 there is no information about m. hence not sufficient.
combining the two statement, no concrete information can be made. the following cases are possible, m > n & n<0 if n= 1 and m= 2 then m > n becomes 2 > 1 but m<n if n=1 and m =6 then m > n becomes 6 > 1 but m>n so different case are possible hence it is not possible to determine the unique case. hence insufficient. The answer would be E.



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Re: Is mn > 0?
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01 Feb 2020, 19:10
Question : m>n? Type : Y/N Statement 2 : n<0 Not Sufficient. We are left with A,C and E. Statement 1: m^2>n^2 m=2 n=1 gives us a Yes. m=2 n=1 gives us a No. Not Sufficient. We are left with C and E. Combined: m^2>n^2 n<0 n=1 m=2 gives us Yes. n=2 m=3 gives us No. Combined not sufficient. Answer E.
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Re: Is mn > 0?
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01 Feb 2020, 19:25
Is mn > 0? 1) m^2 > n^2 2) n < 0
1) This tells us that m is greater that n. So, if m>0, mn will >0. However, that is not provided, so Not Sufficient 2) This tell us that at least one of the two numbers is less than 0, however, it says nothing about m, so Not Sufficient 1&2) statement one told us that if m>0, then mn>0. Statement 2 told us that n<0. However, it doesn't tell us about m. so Not sufficient.
Answer: E



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Re: Is mn > 0?
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01 Feb 2020, 19:59
Is mn > 0? 1) m^2 > n^2 2) n < 0
Answer  E
From 1) m > n > m > n i.e. mn > 0 when m>0; m < n i.e. m+n < 0 when m<0 > Not sufficient
From 2) n < 0 > Not sufficient
Combining 1) and 2) > we know mn will be possible to know only when we know sign of m, sign of n doesn't affect (mn) or (m+n)..
we can prove by numbers m = 6, n = 5 > n < 0, m > n, m n = 11 > 0; m= 6, n = 5 > m > n, mn = 1 < 0
Hence not sufficient.



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Re: Is mn > 0?
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01 Feb 2020, 20:22
Is mn > 0? 1) m^2 > n^2 2) n < 0 (1), m^2 > n^2 => m^2n^2 >0 => (m+n) (mn)>0 Lets put some values of m & n m=4, n=2, 6*2>0 m=4, n=2, 2*6>0 m=4,n=2, 6*2>0 m=4, n=2, 2*6>0....So we really kind decide whether really mn>0. INSUFFICIENT (2) n<0 that means n is a negative number (may be integer or noninteger). Dont know anything else regarding m. INSUFFICIENT. (1)+(2) in the equation, (m+n)(mn)>0 => (mn)(m+n)>0 (where n=n), we dont know about m. INSUFFICIENT. Answer is E
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Re: Is mn > 0?
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01 Feb 2020, 21:45
Statement 1: m^2 > n^2 (m+n)(mn)>0
for mn>0 the m+n needs to be +ve, here we know nothing about m and n. Insufficient
Statement 2: n<0, insufficient, since nothing is mentioned about m, could be positive or negative.
Now, combining both the statements: if m is +ve, n is ve, and m>n, number wise without sign then mn is > 0, but if n is a greater number in negative then mn< 0. if m is also ve, mn will be <0,
So without any information about m, we cannot say for sure, if mn > 0 So answer is E.



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Re: Is mn > 0?
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02 Feb 2020, 00:28
#1 m^2 > n^2 test with m=+/2 and n = +/1 we get yes and no insufficient #2 n<0 m not know insufficient from 1 &2 for n being ve , if m is +ve then yes and if m is ve then no IMO E; sufficient
Is mn > 0? 1) m^2 > n^2 2) n < 0



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Re: Is mn > 0?
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02 Feb 2020, 01:23
Is mn > 0? rephrasing Is m>n? 1) m^2 > n^2 taking square root on both the sides m > n Sign of m and n is unknown Insufficient 2) n < 0 m can be negative or positive Insufficient (1)+(2); Lets take cases case1: n = 1, m =2 case2: n = 1, m =2 Insufficient E is correct
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Re: Is mn > 0?
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02 Feb 2020, 05:32
Is mn > 0? 1) m^2 > n^2 2) n < 0
Rewriting the question as Is m>n?
St 1  m^2>n^2
m=2 n=1m^2>n^2 > m>n  NO m=2 n=1m^2>n^2 > m>n  YES
ST 1 not sufficient
St 2 n<0 no information about m, mcould be +ve, in that case m>n  YES m could be much negative than n, in that case m>n  NO
combiningly m^2>n^2 & n<0 m=2 n=1 > m>n  NO m=1 n=0.5 m>n  YES
together not sufficient answer is E








