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Is c > 2 if a + b + c > 0 ? (1) c > a + b + 2 (2) a + b + 2 < 0

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Is c > 2 if a + b + c > 0 ? (1) c > a + b + 2 (2) a + b + 2 < 0  [#permalink]

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New post 01 Sep 2015, 22:36
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A
B
C
D
E

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Question Stats:

59% (02:18) correct 41% (01:58) wrong based on 108 sessions

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Re: Is c > 2 if a + b + c > 0 ? (1) c > a + b + 2 (2) a + b + 2 < 0  [#permalink]

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New post 02 Sep 2015, 02:42
2
Ans:B.
1. a + b + c > 0 .... Eq 1 (Given)

a) - a - b + c > 2 .... Eq 2,
Add Eq 1 & Eq 2 => C > 1
Not sufficient.

b) - a - b > 2 .... Eq 3

Add eq 1 & eq 3... => C > 2.

Sufficient.
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Re: Is c > 2 if a + b + c > 0 ? (1) c > a + b + 2 (2) a + b + 2 < 0  [#permalink]

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New post 02 Sep 2015, 03:38
1
Is c > 2 if a + b + c > 0 ?

(1) c > a + b + 2

(2) a + b + 2 < 0


statement 1
a + b + c > 0 ----1

c > a + b + 2
c-a-b >2----2


adding 1 and 2
we get c>1 not sufficient

statement 2
a + b + 2 < 0

and a + b + c > 0 --- we subtract both of the inequality
to get c>2

B sufficient

Ans B
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Is c > 2 if a + b + c > 0 ? (1) c > a + b + 2 (2) a + b + 2 < 0  [#permalink]

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New post Updated on: 06 Sep 2015, 20:43
1
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


Is c > 2 if a + b + c > 0 ?

(1) c > a + b + 2

(2) a + b + 2 < 0


Transforming the original condition and the question, we have c>-a-b from c>2? and thus we need to know the relation between -a-b and 2. From 2), we find that a+b<-2 and a+b+c>0, therefore -c<a+b<-2, -c<-2, c>2 thus it is sufficient, and B is the answer.
In case of 1), if c = 5 then a=b=1 and the answer is yes, but if c =2 then a=b=-0.5 and the answer is no. Thus it is not sufficient.
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Originally posted by MathRevolution on 03 Sep 2015, 04:38.
Last edited by MathRevolution on 06 Sep 2015, 20:43, edited 1 time in total.
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Re: Is c > 2 if a + b + c > 0 ? (1) c > a + b + 2 (2) a + b + 2 < 0  [#permalink]

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New post 05 Sep 2015, 03:51
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In statement (1) and (2) we have a+b common and Question also have a+b
To simplify, assume a+b=x

So,the Question becomes Is c>2 if x+c>0?

Statement (1)c>x+2,Sum of x+c>0 and c>x+2 (we can sum inequalities if their signs are in the same direction):

x+c+c>0+x+2

or,c>1

Not Sufficient

Statement (2) x+2<0,Subtract x+2<0 from x+c>0 (we can subtract inequalities if their signs are in the opposite directions):

x+c-x-2>0

or, c>2


Sufficient

So the Correct Answer is B
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Re: Is c > 2 if a + b + c > 0 ? (1) c > a + b + 2 (2) a + b + 2 < 0  [#permalink]

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New post 06 Sep 2015, 06:51
Solution:

a+b+c > 0 ==> c > -(a+b) -->(1)

Statement1 : c > a + b + 2 --> (2)
Add (1) and (2), then 2c > 2 --> c > 1
Therefore, insufficent.

Statement2 : a + b + 2 < 0 ==> 2 < -(a+b) ==> -2 > a+b -->(3)
add (1) and (3),then c-2 > 0 ==> c > 2
Sufficient.

So, Option B
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Re: Is c > 2 if a + b + c > 0 ? (1) c > a + b + 2 (2) a + b + 2 < 0  [#permalink]

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