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

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

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New post 20 Nov 2017, 10:19
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Difficulty:

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Question Stats:

60% (01:00) correct 40% (01:10) wrong based on 124 sessions

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Is a + b > 0?

(1) |a| > |b|
(2) a < b

Self made - tricky

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

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New post 20 Nov 2017, 10:51
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chetan2u wrote:
Is a+b>0?
(1) |a|>|b|
(2) a<b


From Stmnt 1 (a+b)(a-b)>0. So insufficient.
From Stmn 2 (a-b) < 0. Insufficient.

Combining 1 & 2...

a+b<0.

Please let me know if anything wrong with my approach.
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Re: Is a + b > 0? (1) |a| > |b| (2) a < b  [#permalink]

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New post 20 Nov 2017, 10:57
Is a + b > 0?

(1) |a| > |b|
Case 1: a=-7, b=-2 (a+b < 0)
Case 2: a=4, b=3 (a+b > 0) (Insufficient)

(2) a < b
Case 1: a=-7, b=-2 (a+b < 0)
Case 2: a=3, b=4 (a+b > 0) (Insufficient)

Combining the information on both the statements, the expression
a+b can be negative only (Sufficient)(Option C)
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Re: Is a + b > 0? (1) |a| > |b| (2) a < b  [#permalink]

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New post 26 Nov 2017, 22:46
(1) |a| > |b| means that the distance of 'a' from zero is greater than the distance of 'b' from zero, on the number line. But both could be positive (in which case a > b and a+b>0), both could be negative (in which case a < b and a+b<0) or a positive, b negative (in which case a>b and a+b>0) or a negative, b positive (in which case a<b and a+b<0).
So Insufficient.

(2) a < b
Again this doesnt tell us whether a+b is positive or negative. Insufficient.

Combining the two statements, if a<b and |a|>|b|, it could happen in two cases:
I) either both are negative but a is more away from zero than b
II) or a is negative, b is positive but distance of a from zero is more than distance of b from zero

In either of the two cases, sum will be negative. Or a+b<0. Sufficient.

Hence C answer
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Re: Is a + b > 0? (1) |a| > |b| (2) a < b  [#permalink]

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New post 07 Dec 2017, 11:08
chetan2u wrote:
Is a + b > 0?

(1) |a| > |b|
(2) a < b


Awesome question! seriously Chetan this is easy but tricky
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Re: Is a + b > 0? (1) |a| > |b| (2) a < b &nbs [#permalink] 07 Dec 2017, 11:08
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