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20 Jun 2023, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (01:27) correct
26%
(01:30)
wrong
based on 27
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If x, y, z and w are positive integers, is x odd?
(1) 7x + 8y + 4z + 5w is odd.
(2) 3x + 2y + 8z + 2w is even.
(1) 7x + 8y + 4z + 5w is odd.
(2) 3x + 2y + 8z + 2w is even.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
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20 Jun 2023, 02:08
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Let's examine each statement separately:
(1) 7x + 8y + 4z + 5w is odd.
The sum of an even number of even integers is even, and the sum of an odd number of even integers is odd. The sum of any number of odd integers is always odd.
If 8y and 4z are even, then 7x + 5w must be odd. This is true if and only if x and w have opposite parity (i.e. one is odd and the other is even). Therefore, we cannot determine if x is odd from this statement alone.
(1) is insufficient.
(2) 3x + 2y + 8z + 2w is even.
The sum of an even number of even integers is even, and the sum of an odd number of even integers is odd. The sum of any number of odd integers is always odd.
If 2y, 8z, and 2w are even, then 3x must be even. This is true if and only if x is even. Therefore, we can determine that x is even from this statement alone.
(2) is sufficient.
The answer is (B), statement (2) alone is sufficient to determine that x is even.
(1) 7x + 8y + 4z + 5w is odd.
The sum of an even number of even integers is even, and the sum of an odd number of even integers is odd. The sum of any number of odd integers is always odd.
If 8y and 4z are even, then 7x + 5w must be odd. This is true if and only if x and w have opposite parity (i.e. one is odd and the other is even). Therefore, we cannot determine if x is odd from this statement alone.
(1) is insufficient.
(2) 3x + 2y + 8z + 2w is even.
The sum of an even number of even integers is even, and the sum of an odd number of even integers is odd. The sum of any number of odd integers is always odd.
If 2y, 8z, and 2w are even, then 3x must be even. This is true if and only if x is even. Therefore, we can determine that x is even from this statement alone.
(2) is sufficient.
The answer is (B), statement (2) alone is sufficient to determine that x is even.
20 Jun 2023, 08:42
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target is x odd
#1
7x + 8y + 4z + 5w is odd.
7x+e+5w
case 1 ; if w is odd then x has to be even
case 2 : if w is even then x has to be odd
insufficient
#2
3x + 2y + 8z + 2w is even.
3x+e
x has to be even
sufficient option B
#1
7x + 8y + 4z + 5w is odd.
7x+e+5w
case 1 ; if w is odd then x has to be even
case 2 : if w is even then x has to be odd
insufficient
#2
3x + 2y + 8z + 2w is even.
3x+e
x has to be even
sufficient option B
Bunuel
