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Is x + y – z + t even? (1) x + y + t is even (2) tz is odd

why the answer is not C? as tz is odd, t and z both must be odd and and as x+y+t is even, even - odd = odd so C, can anyone please explain?

Though i did not understand why the OA is E

i would go through my understanding, and SME can help me out here

given that x + y + t - z is even so for this two terms (x + y + t) & z so to have a even result both the terms should be either even or both should be odd

E - E = E O - O = E

1) info about x + y + t is even .. good for us, but no information about z is given , hence INSUFFICIENT

2) tz is odd

O * O = O E * O = O O * E = O

hence we cannot make sure if z is even or ODD, hence INSUFFICIENT

so the answer choices A, B, D are knocked off...

now taking together 1 & 2

x + y + t = even and z is either even or odd hence the answer is E
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Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

why the answer is not C? as tz is odd, t and z both must be odd and and as x+y+t is even, even - odd = odd so C, can anyone please explain?

Notice that we are not told that unknowns can only be integers.

Even when we consider the statements together we can have the following cases: \(x+y=odd\), \(t=odd\) and \(z=odd\) then \((x+y)-z+t=odd-odd+odd=odd\) and the answer to the question is NO; \(x+y=\frac{3}{2}\), \(t=\frac{1}{2}\) and \(z=2\) then \((x+y)-z+t=\frac{3}{2}-2+\frac{1}{2}=0=even\) and the answer to the question is YES.

Picked E. because we are not told that all digits are integers. So in that case it the the ans may not be an integer. And if it is not an integer it is also neither even nor odd.

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.

Since we have 4 variables (x and y) and 0 equations, E is most likely to be the answer and so we should consider 1) & 2) first.

There is no restriction each of the variables \(x, y, z\) and \(t\) is an integer.

Condition 1) & 2):

\(x = 1, y = 2, z = 1, t = 1\) : The answer is no. \(x = 1/2, y = 1, z = 2, t = 1/2\) : The answer is yes.

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.
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