If x, y and z are positive integers, is x - y odd?

(1) and (2) are clearly not enough on their own.

So from (1) and (2) together:

z^2 - z^2 + 2z - 1 = 2z-1 which is odd, so answer is C.

Yes, you can plug in numbers. Generally, in even odd questions, plugging numbers works. (mind you, generally, not always)

Break it down in the following way:

If x, y and z are positive integers, is x - y odd?
Question: Is one of x and y even and one odd? (because x - y will be odd only if one of them is even and one is odd)
(1) x=z^2
If z is even, x is even. If z is odd, x is odd. No info about y.
or if z = 2, x is 4. If z = 1, x is 1.

(2) y=(z-1)^2
If z is even, y is odd. If z is odd, y is even. No info about x.
or If z = 2, y = 1. If z = 1, y is 0 (even number).

Together:
If z is even, x is even and y is odd.
If z is odd, x is odd and y is even.
One is always odd, other is always even.
or
If z is 2, x = 4 and y = 1
If z = 1, x = 1 and y = 0

Answer (C)

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x, y and z are positive integers, is x - y odd?

(1) x=z^2
(2) y=(z-1)^2

There are 3 variables (x,y,z) but only 2 equations are given by the 2 conditions, so there is high chance (E) will become the answer.
Looking at the conditions together,
x=z^2, y=(z-1)^2, and as x=odd and y=even or x=even and y=odd, the answer to the question always becomes 'yes' and the answer seems like (C)
However, this is a question with commonly made mistakes, so just to make sure, if the conditions are examined separately,
we cannot know the value of y from condition 1, and condition 2 is not sufficient as well, so the answer becomes (C).

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

Here's an algebraic approach:

Target question: Is x-y odd?

Given: x, y, and z are positive integers

Statement 1: x = z²
There's no information about y, so there's no way to determine whether or not x-y is odd.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y = (z - 1)²
There's no information about x, so there's no way to determine whether or not x-y is odd.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: x = z²
Statement 2: y = (z-1)²
Subtract equations to get: x-y = z² - (z-1)²
Expand to get: x-y = z² - [z² - 2z + 1]
Simplify to get: x-y = 2z - 1
Since z is a positive integer, we know that 2z is EVEN, which means 2z-1 is ODD.
If 2z-1 is ODD, we can conclude that x-y is definitely ODD
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

Bunuel BrentGMATPrepNow from the two statements can we conclude that x and y are consecutive integers and hence x - y will be odd? rather then opening and solving (2)

x and y are not consecutive integers. If (2) were y = z^2 - 1 (instead of y = (z - 1)^2), then you'd be right.

Stem analysis:
x-y can only be odd if x is odd and y is even or vice versa. So we would need enough information to conclude that x and y are not both even or odd.

Statement one:
x = z^2 means that x is a perfect square and that means that x must be some perfect square >0. That information doesn't provide anything on if x is an even or odd integer.

Statement two:
y = (z-1)^2 is similar to statement two and along doesn't provide any information on whether y is an even or odd integer.

Both together:
Let's replace z with the sqrt(x), so y = (sqrt(x) - 1)^2. Since we know x is a perfect square then the sqrt(x) will be some even or odd integer MINUS 1. So that means that if X is even, then Y must be odd, and vice versa, which provides us suffecient information to conclude that one of x or y is even and odd, so x-y must be odd.

x - y will be odd only if x and y have opposite types i.e. {even, odd} or {odd, even}

(1) Says nothing about y
(2) Says nothing about x

Both (1) & (2) tells that, both numbers will be of opposite type as
z, z-1 will be two consecutive numbers having opposite types.
And, their squares will be also opposite types. Take any example, {1,2}, {2,3}, {4,5} ...
Either {even, odd} or {odd, even}.

Sufficient (C)

Hence, sufficient.

My question why I will subtract (2) and (1) why we can add or divide this two statements??

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