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Bunuel
If a and b are integers, is a + 8b even?

(1) 2a + b is even

(2) a^b is even


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Ans is B.
Question basically asks if A is Even
Statement 1: here we can say B is even but nuthing can be said about A.
Statement 2: This clearly says A is even because if A was odd a^b can never be even. So Sufficient.
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Answer is B as any integer irrespective of its exponent (apart from 0) remains odd or even as the initial value.
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Bunuel
If a and b are integers, is a + 8b even?

(1) 2a + b is even

(2) a^b is even


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is a + 8b even? in other words is a even, because 8b is always even ?
(1) 2a+b=even means b is even, but this gives us no info about a Not sufficient
(2)a^b=even means a is even and b is not equal to zero Sufficient
Answer (B)
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Bunuel
If a and b are integers, is a + 8b even?

(1) 2a + b is even

(2) a^b is even


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It's a Yes/No question.

a+8b is even? --> We need to know whether 'a' is even or odd as 8b is always even.

Statement 1 : 2a+b

Don't provide any idea about a, it can be odd or even

Insufficient

Statement 2: a^b is even

Hence a is even [even^(odd or even)=even]

Sufficient

Answer : B
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a and b are integers, is a + 8b even?
(1) 2a + b is even
(2) a^b is even

If we modify the original condition according to the Variable Approach Method, from a+8b, 8b is always even, so we ultimately want to know whether a is an even number.
From condition 2, in order for a^b to be even, the product of even numbers are always even, so a=even, and this condition is sufficient.
But for condition 1, 2a is always even regardless of the value of a, so we cannot determine the value of a, so this is a insufficient condition, and the answer is therefore (B).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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I think the answer is "E" and here is my logic:
you have 4 possible combos: (o for odd, e for even)
1) AeBe:
2) AeBo: out because condition 1
3) AoBe:
4) AoBo: out because condition 1 ,out because condition 2
Thus, two surviving options still can say anything about A.
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