If x, y, and z are positive integers, is xz even?

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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Since we have 3 variables (x, y and z) and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

If x = 2, y = 1, z = 1, then xz is an even number and the answer is 'yes'.
If x = 1, y = 2, z = 1, then xz is an odd number and the answer is 'no'.

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

Combining 1 and 2,
xy=even and yz=even
If y is even, then x and z can be odd or even, then xz can be either even or odd
If y is odd, then x and z has to be both even, then xz = even
Hence, insufficient.
IMO E

E....x, y, and z could all equal 1 or any other odd positive integer, thus xy and xz would both be odd.

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