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# Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n|

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Director
Joined: 02 Oct 2017
Posts: 701
Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n|  [#permalink]

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Updated on: 11 Jul 2018, 05:36
3
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:51) correct 42% (01:58) wrong based on 74 sessions

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Is m > n ?

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

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

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Give kudos if you like the post

Originally posted by push12345 on 10 Jul 2018, 21:14.
Last edited by Bunuel on 11 Jul 2018, 05:36, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n|  [#permalink]

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10 Jul 2018, 21:28
push12345 wrote:
IS m >n

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

2) |m|n =m|n|

Give kudos if you like question

(1) m^2 + n^2 > 2mn means that m^2 + n^2 - 2mn > 0
Or (m-n)^2 > 0. This will be true whenever m and n are unequal. So all this tells is that 'm' is not equal to 'n'. Not sufficient.

(2) |m|n =m|n|
This will be true either when both m/n are positive OR when both m/n are negative. So all this tells is that 'm' and 'n' have same sign. Not sufficient.

Combining the two statements, m and n have same signs and they are not equal. But still we cant say whether m > n or not.
Eg, m=3, n=2 satisfies both the statements. And m=2, n=3 also satisfies both the statements. Not sufficient.

(OA is B. Please check if theres some error)
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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n|  [#permalink]

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10 Jul 2018, 21:29
To find is m > n ?

Statement 1

$$m^2 + n^2 > 2mn$$

=> $$m^2 + n^2 - 2mn > 2mn - 2mn$$

=> $$(m-n)^2 > 0$$

=> m can be 3 and n can be 0 => m > n

=> m can be 0 and n can be 3 => m < n

Since we have two different answers, statement 1 is not sufficient

Statement 2

|m|n =m|n|

Again m can be 3 and n can be 0 => m > n

=> m can be 0 and n can be 3 => m < n

Since we have two different answers, statement 2 is not sufficient

Combining two statements together, we still can't find a unique value for m > n

Hence option 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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11 Jul 2018, 04:41
push12345 wrote:
Is m > n ?

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

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

M = 4 N = 2 M>N answer is yes
M=4 and N =8 M< N answer is no

Both these satisfy B , hence B cannot be the answer . I am also getting 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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11 Jul 2018, 05:30
i am also getting E.
Got a shock when the OA says its B.

Any OE?
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Joined: 12 Dec 2016
Posts: 12
Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n|  [#permalink]

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18 Jul 2018, 19:28
S1: m^2-2mn+n^2>0
So (m-n)^2>0 and here we need to apply a square root
Now the interesting part about applying a square root to cancel off the power of (m-n)^2 is that we have to consider both positive and negative outcomes
Meaning (m-n)>0 OR -(m-n)>0
This leads to m>n OR n>m and hence S1 is insufficient.

S2: |m|n = m|n|
So this basically says m & n have the same sign in order for S2 to satisfy itself, but we still do not know which is bigger. This leads to S2 being insufficient.

Trying to combine S1 & S2 still is insufficient. Answer is 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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18 Jul 2018, 19:40
Is m > n ?

10 seconds answer.... Look at m and n in both statements. Even if you interchange m and n, there is no change in equation
So we cannot say anything about m and n as both can take each others place

(1) m^2 + n^2 > 2mn
$$m^2+n^2>2mn...........m^2+n^2-2mn>0.........(m-n)^2>0$$
We cannot say if m-n>0 or n-m>0
Insufficient

(2) |m|n = m|n|
Nothing much but m and n have same sign..
Insufficient

Combined..
Nothing much
Insufficient

E
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: Is m > n ? (1) m^2 + n^2 > 2mn (2) |m|n = m|n| &nbs [#permalink] 18 Jul 2018, 19:40
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