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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.
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harshavmrg
E * O = O
O * E = O

This is wrong.
Going by examples when you multiply the even(4)with the odd(3) the answer is 12 which is even.

hence whenever a Even is multiplied to a even or odd number it is always an even number. I hope you take note of this point.

Bunuel
Notice that we are not told that unknowns can only be integers.
This is one of the first points to try in such questions when nothing is mentioned about the type of numbers.
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gmihir
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?

Hi,

Whenever you are trying to solve such type of question, do not direct assume all variables as integers (if it is not mentioned).

x,y,z & t can be any real number for which the above relationships are true.

Regards,
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true that cyberjadugar!
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gmihir
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?
The question does not say that x,y,z,t are integers. They can be fraction also.
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gmihir
Is \(x + y – z + t\) even?

(1) \(x + y + t\) is even
(2) \(t*z\) is 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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that's
very good insight
vibhav
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.
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