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Bunuel
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If abc ≠ 0, is abc > 0 ? 11 Aug 2017, 01:19
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95% (hard)

Question Stats:

36% (02:04) correct 64% (01:40) wrong based on 107 sessions

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If \(abc ≠ 0\), is \(abc > 0\) ?


(1) \(|a –b | = |a| - |b|\)

(2) \(|b + c| = |b| + |c|\)
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E
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If abc ≠ 0, is abc > 0 ? 11 Aug 2017, 17:06
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Bunuel
If \(abc ≠ 0\), is \(abc > 0\) ?


(1) \(|a –b | = |a| - |b|\)

(2) \(|b + c| = |b| + |c|\)
A brilliant question!

Thanks,
Mike :-)
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If abc ≠ 0, is abc > 0 ? 11 Aug 2017, 17:55
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from 2 you get that b,c should have the same sign therefore the question collapses to a>?0 NS.
from 1 you get that a,b should have same sign (try different signs and get contradiction) the question collapses to c>?0 NS.
Combine together, NS as if all tree positive then its true if all negative you get false.
WDYT?



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If abc ≠ 0, is abc > 0 ? 11 Aug 2017, 20:16
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Bunuel
If \(abc ≠ 0\), is \(abc > 0\) ?


(1) \(|a –b | = |a| - |b|\)

(2) \(|b + c| = |b| + |c|\)

Hi..

Clearly each sentence talks of two variables, hence insufficient individually..

Combined..
(1) \(|a –b | = |a| - |b|\)
I tells us that a>b or a=b and also both a and B are of same sign.
You can square also to find it..
\(a^2+b^2-2ab=a^2+b^2-2|a||b|........ 2|a||b|-2ab=0...\)
So a*b is POSITIVE and both are of same SIGN.

(2) [m]|b + c| = |b| + |c|
Here B and C are of same sign.

So combined all three are of same SIGN..
But if all three are positive, and is YES
If all three are NEGATIVE, and is NO
Insuff

C
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If abc ≠ 0, is abc > 0 ? 12 Aug 2017, 00:15
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#STATEMENT 1
- Plugin value, we can get a&b both positive or a&b both negative.
- Hence, insufficient.

#STATEMENT 2
- Plugin value, we can get b&c both positive or b&c both negative.
- Hence, insufficient.

Anyway statement #1 and #2 only talk about 2 out of 3 variable - INSUFFICIENT from the begining.

#BOTH STATEMENT TOGETHER
We have two combinations :
1. a,b and c all POSITIVE = YES for the question.
2. a,b and c all NEGATIVE = NO for the question.
INSUFFICIENT

Therefore, the answer is E.

Is this correct? :oops: :oops: :oops:
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If abc ≠ 0, is abc > 0 ? 12 Aug 2017, 00:19
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chetan2u
Bunuel
If \(abc ≠ 0\), is \(abc > 0\) ?


(1) \(|a –b | = |a| - |b|\)

(2) \(|b + c| = |b| + |c|\)

Hi..

Clearly each sentence talks of two variables, hence insufficient individually..

Combined..
(1) \(|a –b | = |a| - |b|\)
I tells us that a>b or a=b and also both a and B are of same sign.
You can square also to find it..
\(a^2+b^2-2ab=a^2+b^2-2|a||b|........ 2|a||b|-2ab=0...\)
So a*b is POSITIVE and both are of same SIGN.

(2) [m]|b + c| = |b| + |c|
Here B and C are of same sign.

So combined all three are of same SIGN..
But if all three are positive, and is YES
If all three are NEGATIVE, and is NO
Insuff

C

chetan2u

Why you choose C although both statement insufficient?
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If abc ≠ 0, is abc > 0 ? 12 Aug 2017, 11:20
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Bunuel
If \(abc ≠ 0\), is \(abc > 0\) ?


(1) \(|a –b | = |a| - |b|\)

(2) \(|b + c| = |b| + |c|\)

Let's take 2 & -2 and plug them across the statements (or choose any 2 values). I choose a number and its apposite to make calculations easy.

(1) \(|a –b | = |a| - |b|\)

(a,b) = (2, 2)
\(|2 –2 | = |2| - |2|\)= 0

(a,b) = (-2, -2)
\(|-2 +2 | = |2| - |2|\)= 0

(a,b) = (-2, 2) and (2,- 2) are invalid as RHS does not equal LHS

Conclusion is a & b must have same sign but c can be any number with any sign

Insufficient

(2) \(|b + c| = |b| + |c|\)

(b,c) = (2, 2)
\(|2 + 2| = |2| + |2|\)= 4

(b,c) = (-2,-2)
\(|-2 -2| = |-2| + |-2|\)= 4

(b,c) = (-2, 2) and (2,- 2) are invalid as RHS does not equal LHS

Conclusion is b & c must have same sign but a can be any number with any sign

Insufficient

Combining 1 & 2

Case I: 2, 2 , 2.................Answer is Yes

Case II: -2, -2,-2..............Answer is NO

Insufficient

Answer: E
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