Let x be the number of chicken in farm 1 and y be the number of chicken in farm 2.

statement 1 : avg. wt of chicken in farm 1 =13 pounds. (insufficient)
statement 2: insufficient in itself.

combine both : 12x-13y=6. By hit and trial x=7 and y=6 satisfies. (went a little further, couldn't come up with any more values of x and y which would satisfy this equation) hence, farm 2 has 6 chicken.

You were close, But made mistake here.

The statement 2 says that the TOTAL weight of Chicken in farm 2 is 6 less than the TOTAL weight of Chicken in farm 1. So the above method you used is wrong. As 12 and 13 are the AVERAGE weight of the chickens in the respective farms, not total weight.

Official Explanation :

We need to find the number of chicken in farm 2(N2)

(1) Average weight is 13. So as per the average formula we know that

\(13 = \frac{Total (T2)}{N2}\)

No enough info, so in sufficient.

(2) As per the statement T2= T1-6

Since we dont know the total weight of chicken in farm 1- Insufficient.

Combining (1) & (2),

\(Average = \frac{T2}{N2}\)

We know T2= T1-6, when we substitute.

\(13= \frac{(T1-6)}{N2}\)

Since we still don't know the total weight(T1) of the chickens in the farm 1. So both statement together are not sufficient.

Answer is E.
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The average weight of chicken in farm 1 is given as 12 pounds and i have considered x chicken in total. SO total weight should be 12x. what is the flaw in this?

Oh ok. My mistake

In DS we just need to find whether the two given statement are sufficient to find the answer. We don't have to plug in. If at all we do that, we may have different values for X and Y.

In this problem, we still don't know the total value of chicken in farm 1. Hence we are not able to find the number of chicken in farm 2. So E.
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Nothing wrong. You should find x. How are you going to do that?
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There are other solutions. For example, x=20 and y=18.
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Yes I tried for other values also. And x=20, y=18 is also a solution. Hence, E. ( I would go with E because of this reason)

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