shailendra.sharma wrote:If a and x are integers, and a is even, is x even too?
(1) x^a is odd.
(2) a·x=12
OA -- C
Can experts help why OA shall be C. My answer is A.
If x^a is odd, it will be possible only when x is odd ==> A is sufficient. I am sure I am missing something here.
Your statement above (in green) is almost always true. However, if a = 0 (an even number) then x^a = 1, for values of x that are even or odd. For example 4^0 = 1 and 1 is odd, and 4 is even.
Here's my full solution:
Target question:
Is x even?
Given: a and x are integers, and a is even
Statement 1: x^a is odd.
There are several pairs of values that meet this condition. Here are two:
Case a: x = 3 and a = 2, in which case
x is odd
Case b: x = 4 and a = 0, in which case
x is even
Since we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: ax=12
There are several pairs of values that meet this condition. Here are two:
Case a: x = 3 and a = 4, in which case
x is odd
Case b: x = 6 and a = 2, in which case
x is even
Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined:
From statement 1, we see that the ONLY way that x can be even is if a = 0. Otherwise x must be odd.
From statement 2, we can conclude that a does not equal zero.
So, we can conclude that
x must be odd.
In other words,
x is not even.
Since we can answer the
target question with certainty, the combined statements are SUFFICIENT
Answer =
C
Cheers,
Brent
