If y is the average of x odd consecutive integers and |z - 6/4| = 1/2,

If x is even, then the average of even number of odd consecutive integers (y), will be even. For example, the average of {1, 3} is 2.
If x is odd, then the average of odd number of odd consecutive integers (y), will be odd. For example, the average of {1, 3, 5} is 3.

Thus:
If x = even, then y = even;
If x = odd, then y = odd.

Next, |z - 6/4| = 1/2:
z - 6/4 = 1/2 --> z = 2;
-(z - 6/4) = 1/2 --> z = 1.
Since, z can be even as well as odd, then it takes no part in deciding whether the options are even or odd. So, we can ignore it.


I. xy(z + 1) is even: xy can be odd if both are odd and even if both are even. So, this option is not necessarily even. Discard.

II. x(z + y) is even: if x and y are odd, and z is even, then x(z + y) = odd(odd + even) = odd but if if x and y are odd, and z is odd too, then x(z + y) = odd(odd + odd) = even. So, this option is not necessarily even. Discard.

III. (x^2 - x)yz is even: (x^2 - x)yz = x(x - 1)yz. Since either x or x - 1 is even, then the whole expression must be even irrespective of the values of x, y, and z.

Answer: B.

Hope it's clear.
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Notice that the average of x no. of first odd consecutive integers = x

e.g.
average of 1,3,5 is 3 ; x=3 ( first three odd numbers)

average of 1,3,5,7 is 4; x =4 ( first four odd numbers )

average of 1, 3, 5, 7,9 is 5 ; x=5 ( first five odd numbers )



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Sir,

Just out of curiosity, in how many minutes can we solve this problem?

As we need to first find out the values of z and then check it with the z equation.

Then we take all the values and check all the options. so total 6 cases.

As you can see in the stats, average time of correct answer is 03:36 minutes, which means that it's possible to solve in about 2-3 minutes.
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y is the average of x odd consecutive integers
lets consider (1+3+5)/ 3 =3 and (1+3+5+7)/4 = 4.
Hence when x is even,y is even and when x is odd,y is odd as well.

now |z - 6/4| = 1/2
case 1) z - 6/4=1/2
z=2
case 2) z - 6/4= -1/2
z=1
that means z can take even value as well as odd value.

I. xy(z + 1) is even.
case 1) even*even (even+1)
even*even*odd= even.
case 2) odd*odd*(even+1)
odd*odd*odd= odd
hence I is false. Note: I have not considered each and every possible case

II. x(z + y) is even.
odd* (odd + even)= odd
hence II is false.

III. (x^2 - x)yz is even.

x can be even as well as odd
case 1) (even^2 - even) = even
case 2) (odd^2 - odd)= even
In both cases X turns out to be even.

as even * anything is even, given statement is true.

Hence answer is III only, option B

Kudos if it helps

there are only 2 possible cases of z, same to y and x.
Now, just check each I, II, III
=> B is the only answer

Let consider the odd consecutive set with the first term is 1.
Is x is even assume x = 2n, => the last term is 4n-1 => y = 2n, even
If x is odd assume x = 2n+1 => last term is 4n+1 => y = 2n+1, odd
From that, we can come to conclude that y even(odd) if x eve (odd)
z could be calculated easily, so odd or even what ever.
=> Answer B

The answer choices shortened my time usage.

If neither Option I nor Option II is correct, from the answer choices, only (B) is left.

I did not even test III.

With a lot of moving pieces, I almost always plug in.

1) Find z

\(|z - \frac{6}{4}| = \frac{1}{2}\)

\((z - \frac{3}{2} = \frac{1}{2})\) OR \((z -\frac{3}{2} = -\frac{1}{2}\))

\(z = 2\) OR
\(z = 1\)

2) Assign x, derive y
Let x = 3 numbers: (1,3,5)
y = average = 3

Let x = 4 numbers: (1,3,5,7)
y = 4

3) Possibilities
\(x = 3, y = 3\)
\(x = 4, y = 4\)
\(z = 1\) OR
\(z\) = 2

4) Assess

I. xy(z + 1) is even

To disprove, need all odds: O*O(*O) = O

We must pick x = 3, y = 3.

Make (z + 1) odd. Choose z = 2

xy(z + 1)

x = 3, y = 3, z = 2:
(3*3)(2+1) = 9. Not even. NO

II. x(z + y) is even

We need 2 odds.
Must pick x = 3, y = 3

Make (z + 3) odd; choose z = 2

x(z + y)
3(2 + 3) = 15. Not even. NO

From the answer choices, only III is possible.

All the others have I and/or II, neither of which is correct. I did not test Option III, but it must be the answer.

Answer B
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I'm not sure if this is the correct method or not. However, I used a simple trick.

Z will be even (negative or positive doesn't matter). Now the options are talking about which of the following MUST be true not COULD be true. So I simply picked whichever option had z multiplied with it and it was only one option and I got B straight away. I solved this in 57s this way.

*even x odd/even will always give even.

either i am right or extremely lucky