Data Sufficiency

G_Rug wrote:Is the integer x odd?

1) 2(y+x) is an odd integer
2) 2y is an odd integer

Target question: Is the integer x odd?

Statement 1: 2(y+x) is an odd integer
There are several pairs of values that meet this condition. Here are two:
Case a: x = 3 and y = 0.5 in which case integer x is odd
Case b: x = 2 and y = 0.5 in which case integer x is even
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 2y is an odd integer
No information about x so statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
Statement 1 says that 2(y+x) is an odd integer
In other words, 2y + 2x is odd
Statement 2 says that 2y is some odd #.
When we combine this with statement 1, we get: some odd # + 2x is odd
If x is an integer, then 2x is always even, which means some odd # + 2x will ALWAYS be odd, regardless of whether or not x is odd.
So, the combined statements are NOT SUFFICIENT

To demonstrate this, consider these two conflicting cases:
Case a: x = 3 and y = 0.5 in which case integer x is odd
Case b: x = 2 and y = 0.5 in which case integer x is even

Answer = E

Cheers,
Brent

G_Rug wrote:Is the integer x odd?

1) 2(y+x) is an odd integer
2) 2y is an odd integer


I don't agree with the book's answer. Here's my approach:

1) SUFF. Looking at statement 1, it is clear that y and/or x are fractions, and they add to a fraction with an odd number in the numerator (i.e. 1/2 + 2/2 = 3/2, multiplied by 2 will yield 3). Given that the question already suggests that x is an integer ("is the integer x odd?"), y must be the fraction. This means x is an even number (2,4,6,8), so that the sum of y and x yield an odd number in the numerator. This is sufficient to answer the question.

The problem with your solution is above, in blue. We cannot conclude that x is even based on statement 1.
Consider these cases:
Case a: x = 3 and y = 0.5 (here, 2(y+x) is an odd integer and x is odd)
Case b: x = 2 and y = 0.5 (here, 2(y+x) is an odd integer and x is even)

IMPORTANT: In your example, you show that 1/2 + 2/2 = 3/2, multiplied by 2 will yield 3
Here, y = 0.5 and x = 1, so in your example, x is not even (as you suggest)

Cheers,
Brent

ella willius wrote:If X And Y Are Integers Greater Than 1 Is X A Multiple Of Y (1) 3y^2 + 7 Y = X. sufficient or not ?

Asking whether x is a multiple of y is the same as asking whether x = (y)(some integer)
For example, 12 is a multiple of 3 because 12 = (3)(4)

Given: x = 3y^2 + 7y
Factor to get x = y(3y + 7)
In other words: x = y(some integer)
So, it MUST be the case that x is a multiple of y. (The statement is SUFFICIENT)

Cheers,
Brent