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

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Intern
Joined: 07 Sep 2014
Posts: 16
Location: United States
GMAT Date: 03-14-2015
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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08 Mar 2015, 10:01
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Question Stats:

70% (02:32) correct 30% (02:39) wrong based on 552 sessions

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If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, which of the following MUST be true?

I. xy(z + 1) is even.

II. x(z + y) is even.

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

(A) I only
(B) III only
(C) I and II
(D) II and III
(E) I, II, and III
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Mar 2015, 10:54, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 44319
Re: If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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08 Mar 2015, 11:07
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vinyasgupta wrote:
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, which of the following MUST be true?

I. xy(z + 1) is even.

II. x(z + y) is even.

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

(A) I only
(B) III only
(C) I and II
(D) II and III
(E) I, II, and III

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.

Hope it's clear.
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Re: If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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08 Mar 2015, 11:46
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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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Joined: 29 Jul 2015
Posts: 13
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Concentration: Marketing, Strategy
WE: Information Technology (Retail Banking)
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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17 May 2017, 23:54
Bunuel wrote:
vinyasgupta wrote:
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, which of the following MUST be true?

I. xy(z + 1) is even.

II. x(z + y) is even.

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

(A) I only
(B) III only
(C) I and II
(D) II and III
(E) I, II, and III

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.

Hope it's clear.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 44319
Re: If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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18 May 2017, 00:04
Animatzer wrote:
Bunuel wrote:
vinyasgupta wrote:
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, which of the following MUST be true?

I. xy(z + 1) is even.

II. x(z + y) is even.

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

(A) I only
(B) III only
(C) I and II
(D) II and III
(E) I, II, and III

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.

Hope it's clear.

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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Posts: 71
Re: If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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02 Sep 2017, 13:29
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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
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Joined: 12 Dec 2016
Posts: 1947
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
Re: If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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16 Sep 2017, 09:07
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
Intern
Joined: 23 Sep 2017
Posts: 14
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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10 Oct 2017, 01:36
vinyasgupta wrote:
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, which of the following MUST be true?

I. xy(z + 1) is even.

II. x(z + y) is even.

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

(A) I only
(B) III only
(C) I and II
(D) II and III
(E) I, II, and III

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.
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Joined: 22 May 2016
Posts: 1420
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, [#permalink]

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10 Oct 2017, 15:16
vinyasgupta wrote:
If y is the average of x odd consecutive integers and |z - 6/4| = 1/2, which of the following MUST be true?

I. xy(z + 1) is even.

II. x(z + y) is even.

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

(A) I only
(B) III only
(C) I and II
(D) II and III
(E) I, II, and III

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.

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If y is the average of x odd consecutive integers and |z - 6/4| = 1/2,   [#permalink] 10 Oct 2017, 15:16
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