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The question is asking whether |a|= b-c. Hence if the answer is no then also it gives a solution for the DS.

a + b

a=-1, c= 0 means -1+1 = b = 0
hence |-1| is not equal to b-c = 0-0

a=-1,c=1 means -1+1 = b = 0
hence |-1| is not equal to b-c = 0-1

hence sufficient. Thus C.
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The question is asking whether |a|= b-c. Hence if the answer is no then also it gives a solution for the DS.

a + b

a=-1, c= 0 means -1+1 = b = 0
hence |-1| is not equal to b-c = 0-0

a=-1,c=1 means -1+1 = b = 0
hence |-1| is not equal to b-c = 0-1

hence sufficient. Thus C.

I go with C as well.
If you can say |a| # b-c . this is sufficient for DS problem
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fanatico
Is |a| = b - c ?
(1) a + c != b
(2) a < 0

a=-1, b=100, c=0; a+c=-1; a+c != b; |a|=1, b-c=100, |a| != b-c
a=-1, b=2, c=1; a+c=0; a+c!=b; |a|=1, b-c=1, |a|=b-c

1.
Not Sufficient.

2.
Not Sufficient.

Combining both;
Not Sufficient.

Ans: "E"
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fluke
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Is |a| = b - c ?
(1) a + c != b
(2) a < 0

a=-1, b=100, c=0; a+c=-1; a+c != b; |a|=1, b-c=100, |a| != b-c
a=-1, b=2, c=1; a+c=0; a+c!=b; |a|=1, b-c=1, |a|=b-c


Ans: "E"

How can we fix the B's value.
B gets a derived value from the equation in 1.
If we can fix up value for b then its obvious that the answer will be E.

I have used the derived values.
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fluke
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Is |a| = b - c ?
(1) a + c != b
(2) a < 0

a=-1, b=100, c=0; a+c=-1; a+c != b; |a|=1, b-c=100, |a| != b-c
a=-1, b=2, c=1; a+c=0; a+c!=b; |a|=1, b-c=1, |a|=b-c


Ans: "E"

How can we fix the B's value.
B gets a derived value from the equation in 1.
If we can fix up value for b then its obvious that the answer will be E.

I have used the derived values.

I have showed that the expression in the stem may or may not hold true without violating either of the given statements. If you see a problem, please point it out.
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I second fluke, please check my solution also. However, to answer your question, b's value can be fixed because if you move around a and c on the RHS of !=, then the variable on LHS gets a derived value. In other words, a,b and c are all variables, and the equation can have several values of a, b and c for which it can be true, fluke has tried to show the veracity of that using some sample values of variables - "number plugging" :) .
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Clearly 1 & 2 are not sufficient on their own.

taken together:

1. a ! = b - c or -a != c - b
2. |a| = -a
Q now is "is -a = b - c?"

Nothing can tell us this

eg 1. a = -4, b = 7, c = 3 meets all condition and answers as yes.
2. a = -4, b = 7, c = 5 meets all condition and answers as no.

I agree with "fluke"...
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fanatico
Is |a| = b - c ?
(1) a + c != b
(2) a < 0

Apparently, the statements alone are not sufficient.

if a = -1 , b = 5 and c = 3 the answer is NO
if a = -10 , b = 14 and c = 4, the answer is Yes.

Hence E

Reasoning:
since a + c != b => a = b - c! since a <0
we can say that |a| = c! - b which is also equal to b -c
now the question becomes is 2b= c+c!

write down numbers from 1 to 6 and their FACTORIALS.
if b = 2 and c = 2, the a becomes 0. neglect this.
if b = 14 and c = 4, a becomes -10. take this value.

the reason why I have taken even numbers of C because the factorial above 1 will always yield an even number.
=> 2b= c+c! can not hold true if c is not even.

nice question.
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this is a question from Grockit
Ans is C
and they have given an explanation
Bunuel can you check whether my process below is right or wrong

Since Mod a is always +ve so cant we write Is b-c>0 i.e. b>c
Hence C should be the answer
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Archit143
this is a question from Grockit
Ans is C
and they have given an explanation
Bunuel can you explain what is the correct answer and procedure

Discussed here: is-a-b-c-1-a-c-is-not-equal-to-b-2-a-105351.html

Hope it helps.
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