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Is x + y – z + t even?

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Is x + y – z + t even? [#permalink]

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New post Updated on: 29 Dec 2016, 15:50
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Is \(x + y – z + t\) even?

(1) \(x + y + t\) is even
(2) \(t*z\) is odd
[Reveal] Spoiler: OA

Originally posted by gmihir on 22 May 2012, 09:12.
Last edited by stonecold on 29 Dec 2016, 15:50, edited 2 times in total.
Edited the question.
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Re: Is x + y – z + t even? [#permalink]

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New post 22 May 2012, 09:35
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gmihir wrote:
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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Re: Is x + y – z + t even? [#permalink]

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New post 22 May 2012, 10:51
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gmihir wrote:
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?


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.

Two different answers. Not sufficient.

Answer: E.

Hope it's clear.
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Re: Is x + y – z + t even? [#permalink]

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New post 22 May 2012, 20:26
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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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Re: Is x + y – z + t even? [#permalink]

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New post 24 May 2012, 07:51
harshavmrg wrote:
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 wrote:
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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Re: Is x + y – z + t even? [#permalink]

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New post 24 May 2012, 20:24
gmihir wrote:
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.

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Re: Is x + y – z + t even? [#permalink]

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New post 24 May 2012, 21:18
true that cyberjadugar!
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Re: Is x + y – z + t even? [#permalink]

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New post 29 Jul 2014, 15:54
gmihir wrote:
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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Re: Is x + y – z + t even? [#permalink]

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New post 21 Dec 2017, 11:23
Expert's post
gmihir wrote:
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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Re: Is x + y – z + t even?   [#permalink] 21 Dec 2017, 11:23
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