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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n| [#permalink]
1
Kudos
Expert Reply
Bunuel wrote:
Is \(m > n\) ?


(1) \(m^2 + n^2 > 2mn\)

(2) \(|m|n = m|n|\)


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Solution


Step 1: Analyse Question Stem


We need to find whether m > n or not.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: \(m^2 +n^2 >2mn\)
    • \(m^2 + n^2 -2mn > 0\)
    • \((m-n)^2 > 0\)
      o Here, we can’t say where m > n or not, because the power of m-n is 2, and even power of any number always gives us positive number.
Hence, statement is not sufficient, we can eliminate answer options A and D.
Statement 2: \(|m|n = m|n|\)
    • EIther m and n both will have same sign or at least one of them is zero.
    • Case I: If m = 1 and n = 2,
      o Then, |m|n = m|n| = 2, m < n
    • Case II: If m = -1 and n = -2,
      o Then |m|n = m|n| = -1, m > n
    • We are getting different results
    • We can’t answer the question with this information.
Hence, statement 2 is also not sufficient.

Step 3: Analyse Statements by combining.


From statement 1: \((m-n)^2 > 0\)
From statement 2: \(|m|n = m|n|\)
    • Case I: If m = 0 and n = 2, then |m|n = m|n| = 0
      o 0 < 2, and m < n
    • Case II: If m = 2 and n = 0, then |m|n = m|n| = 0
      o 2 > 0 and m < n
    • We are getting different results.
Hence, the correct answer is Option E.
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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n| [#permalink]
#1
\(m^2 + n^2 > 2mn\)
can be written as
(m-n)^2>0
in that case either of m or n can be -ve or both can be +ve and we get possibilities of m<n and m>n insufficient

#2
\(|m|n = m|n|\)

possible when m=n or both m & n are +ve values
insufficient
from 1 &2
value of m & n +ve but we cannot determine conclusively that m>n or n>m
insufficient
OPTION E

Bunuel wrote:
Is \(m > n\) ?


(1) \(m^2 + n^2 > 2mn\)

(2) \(|m|n = m|n|\)


Project PS Butler


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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n| [#permalink]
2
Kudos
Is \(m > n\) ?


(1) \(m^2 + n^2 > 2mn\) --> insuff: (m-n)^2>0 => |m-n| > 0, so m >n or m<n; we can prove the same by number picking also, let's say m=3 & n=2, then the answer is yes, if m=-3 & n=-2, then the answer is no

(2) \(|m|n = m|n|\) --> insuff: let's say m=3 & n=2, then the answer is yes, if m=-3 & n=-2, then the answer is no

combining (1) & (2) => let's say m=3 & n=2, then the answer is yes, if m=-3 & n=-2, then the answer is no
Answer: E
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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n| [#permalink]
1
Kudos
Is \(m > n\) ?


(1) \(m^2 + n^2 > 2mn\)

(2) \(|m|n = m|n|\)

1) \((m-n)^2 > 0\). so, 0 < m-n < 0. Not sufficient
2) m and n are of the same sign. Not sufficient.
Together, when m =2, n=3, both conditions met. Again, when m =-4, n=-2, Again, both hold true. Not sufficient.
E is the answer.
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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n| [#permalink]
1
Kudos
Is m>n

Statement 1: mod(m-n)>0
Either m>n or n>m
Insufficient
Statement 2: mod m n= m mod n
Either m,n have same sign or at least 1 is 0
m=0, n=-1, m>n
m=1,n=3, n>m
Insufficient
Both statements together:
(m-n)^2>0 and mod m n= m mod n
Still Insufficient
The above examples work here as well
E
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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n| [#permalink]
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