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Hoozan
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The solution to the question as it currently stands is incorrect. The answer should be E.

Question: If a ≠ b, what is the value of a + b?

(1) a^2 - b^2 = 6
This can further be broken down into:
(a+b)(a-b) = 6
We cannot get a numerical answer for this.

(2) ab = 12
Here we can many solutions including fractions and negatives:
2*6
36*1/3

Taking the two together once again we don't have an answer.
Please check the question or let me know if I am missing something.


I have made the changes to the question. the (1) statement says a^2 - b^2 = 0 and not "6"
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(1) a^2 - b^2 = 0

(a-b)(a+b)=0

so either a-b=0 or a+b=0 but a-b can't be zero ( Given a ≠ b)
A = Suff

(2) ab = 12
a x b = 3 x 4 or 12 x 1 or 6x2 , multiple answers are possible so B = Insufficient
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Hoozan
If a ≠ b, what is the value of a + b?

(1) a^2 - b^2 = 0
(2) ab = 12

Hoozan
I think both the statements are inconsistent with each other. As per statement 1, we know a+b=0 (also sufficient statement)
Now for statement 2 analysis, if -
(a+b)^2= a^2 + b^2 +2ab = 0 (from 1)
so, a^2 + b ^2 = -24 ----> Not possible as the sum of 2 squares can't be negative.

As far as I know even if one of the statements is insufficient, they shouldn't be inconsistent with each other.
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Hoozan
If a ≠ b, what is the value of a + b?

(1) a^2 - b^2 = 0
(2) ab = 12

Hoozan
I think both the statements are inconsistent with each other. As per statement 1, we know a+b=0 (also sufficient statement)
Now for statement 2 analysis, if -
(a+b)^2= a^2 + b^2 +2ab = 0 (from 1)
so, a^2 + b ^2 = -24 ----> Not possible as the sum of 2 squares can't be negative.

As far as I know even if one of the statements is insufficient, they shouldn't be inconsistent with each other.

a^2 - b^2 = 0 and ab = 12 has two sets of (a, b) satisfying it: \(a = -2\sqrt{3}, \ b = -2\sqrt{3}\) and \(a = 2\sqrt{3}, \ b =2\sqrt{3}\). However, for both of these sets, a = b, which contradicts the info given in the stem.

So, yes, you are right, the question is flawed. Tagging it as Poor Quality and locking.

Thank you!

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