Is x odd?
(1) x/2 is not an integer
(2) 2x + 3 is odd.
Can some experts show me the correct solution to figure the sufficient statement?
OA C
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Is x odd
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Statement 1:lheiannie07 wrote:Is x odd?
(1) x/2 is not an integer
(2) 2x + 3 is odd
Since x/2 is not an integer, x cannot be EVEN.
Be sure to test NON-INTEGER values.
Case 1: x=1, with the result that x/2 = 1/2
In this case, the answer to the question stem is YES.
Case 2: x=1/2, with the result that x/2 = 1/4
In this case, the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.
Statement 2:
2x + 3 = 1, 3, 5, 7, 9, 11...
Subtracting 3 from every value, we get:
2x = -2, 0, 2, 4, 6, 8...
Dividing every value by 2, we get:
x = -1, 0, 1, 2, 3, 4...
The resulting blue list indicates that x can be ANY INTEGER VALUE.
Since x can be odd or even, INSUFFICIENT.
Statements combined:
Statement 2 --> x must be an integer.
Statement 1 --> x cannot be even.
Since x must be an integer that is not even, x is ODD.
SUFFICIENT.
The correct answer is C.
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Dear Mitch,GMATGuruNY wrote:Statement 1:lheiannie07 wrote:Is x odd?
(1) x/2 is not an integer
(2) 2x + 3 is odd
Since x/2 is not an integer, x cannot be EVEN.
Be sure to test NON-INTEGER values.
Case 1: x=1, with the result that x/2 = 1/2
In this case, the answer to the question stem is YES.
Case 2: x=1/2, with the result that x/2 = 1/4
In this case, the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.
Statement 2:
2x + 3 = 1, 3, 5, 7, 9, 11...
Subtracting 3 from every value, we get:
2x = -2, 0, 2, 4, 6, 8...
Dividing every value by 2, we get:
x = -1, 0, 1, 2, 3, 4...
The resulting blue list indicates that x can be ANY INTEGER VALUE.
Since x can be odd or even, INSUFFICIENT.
Statements combined:
Statement 2 --> x must be an integer.
Statement 1 --> x cannot be even.
Since x must be an integer that is not even, x is ODD.
SUFFICIENT.
The correct answer is C.
If the question stem says: is the integer x odd?, then is statement 1 sufficient on its own?
Thanks
