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Is n < 0? (1) m < n (2) -n < m

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Bunuel
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Is n < 0? (1) m < n (2) -n < m [#permalink] New post   26 Sep 2019, 04:44
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A
B
C
D
E

Difficulty:

45% (medium)

Question Stats:

62% (01:31) correct 38% (01:14) wrong based on 141 sessions
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Is n < 0?

(1) m < n
(2) -n < m
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C
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Harsh9676
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Is n < 0? (1) m < n (2) -n < m [#permalink] New post   Updated on: 26 Sep 2019, 06:59
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IMO E

From 1

m-n<0, which means without knowing m's sign we cannot find whether n<0.

From 2, we get -n-m<0, here as well we need to know m's sign.

Combined, we can add both the inequalities as the signs are the same and get -2n<0.

From this we cannot know whether n<0.

Posted from my mobile device

Originally posted by Harsh9676 on 26 Sep 2019, 04:53.
Last edited by Harsh9676 on 26 Sep 2019, 06:59, edited 1 time in total.
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Re: Is n < 0? (1) m < n (2) -n < m [#permalink] New post   26 Sep 2019, 05:40
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Answer is E
a) Nothing known about m
b) -n<m
=> n>m
Again same as a
Combining both we dont get any answer.So answer must be E
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Is n < 0? (1) m < n (2) -n < m [#permalink] New post   Updated on: 09 Oct 2019, 05:54
Bunuel wrote:
Is n < 0?

(1) m < n
(2) -n < m


1. m<n
Not sufficient( m=-6 and n =-4 YES but m=2 and n= 3 NO)
2. -n<m
Not sufficient( n=10 and m =10 NO but n=-10 and m=11 YES)
Add 1 and 2
m<n
-n<m
m-n<m+n
-n<n
case 1. n>0 . -equation a
This will give
-2n<0
2n>0
n>0 equation b
=from equation a and b
=n>0
case2. n<0 (eq-a)
first multiply by - sign
n>-n
as n is negative -n is positive so we don't have to change the sign now
2n>0
n>0
From 1) and 2)
n>0
C:)

Originally posted by satya2029 on 26 Sep 2019, 06:37.
Last edited by satya2029 on 09 Oct 2019, 05:54, edited 1 time in total.
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Is n < 0? (1) m < n (2) -n < m [#permalink] New post   26 Sep 2019, 08:40
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Bunuel wrote:
Is n < 0?

(1) m < n
(2) -n < m


(1) m < n
--> n > m
'm' can be +ve or -ve --> Insufficient

(2) -n < m
--> n > -m
'm' can be +ve or -ve --> Insufficient

Combining (1) & (2),
n > m & n > -m
--> n is both greater than m & -m on the number line
So, the common solution is "n" will ALWAYS lie to the right of origin --> Sufficient

Eg: Case 1: Let m = 2 --> n > 2 & n > -2 --> n >2 is the solution --> n is ALWAYS greater than 0
Case 2: Let m = -2 --> n > -2 & n > -(-2) --> n > -2 & n > 2 --> n >2 is the solution --> n is ALWAYS greater than 0

IMO Option C

Pls Hit kudos if you like the solution
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Is n < 0? (1) m < n (2) -n < m [#permalink] New post   26 Sep 2019, 08:58
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n < 0?

St 1: m < n ==> m - n < 0 ==> Insufficient since we do not know the signs of m and n

St 2: - n < m ==> m + n > 0 ==> Insufficient since we do not know the signs of m and n

Sts 1 and 2 combined:
Multiplying both sides of st 1 by -1 ==> - (m - n > 0) ==> -m + n > 0
Add -m + n > 0 and m + n > 0 (St 2) ==> 2n > 0 ==> n > 0 ==> Sufficient (C is the answer)
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Re: Is n < 0? (1) m < n (2) -n < m [#permalink] New post   26 Sep 2019, 11:22
Each statement individually is insufficient.

Combining both:

-n<m<n.
Means n has to be positive.
C is sufficient.
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Re: Is n < 0? (1) m < n (2) -n < m [#permalink] New post   31 Oct 2020, 22:57
Expert Reply
Bunuel wrote:
Is n < 0?

(1) m < n
(2) -n < m



It can be answered in 30 seconds, if you take correct steps.

(1) m < n
(2) -n < m

There is nothing known about signs of m or n when you look at the individual statements.

Add both the inequalities given, m<n and -n<m => \(m+(-n)<n+m\)
\(m-n<n+m......2n>0....n>0\)
Answer is always NO

C
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Re: Is n < 0? (1) m < n (2) -n < m [#permalink] New post   01 Nov 2020, 06:47
Is n < 0?

(1) m < n

Insufficient.

(2) -n < m

Insufficient.

(1&2) m - n < n + m
0 < 2n
0 < n

N is positive. SUFFICIENT.

Answer is C.
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Re: Is n < 0? (1) m < n (2) -n < m [#permalink] New post   26 Feb 2022, 17:47
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Re: Is n < 0? (1) m < n (2) -n < m [#permalink]
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