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Intern  Joined: 07 Sep 2014
Posts: 12
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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Difficulty:   55% (hard)

Question Stats: 70% (03:13) correct 30% (03:07) wrong based on 655 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

Originally posted by vinyasgupta on 08 Mar 2015, 10:01.
Last edited by Bunuel on 08 Mar 2015, 10:54, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: If y is the average of x odd consecutive integers and |z - 6/4| = 1/2,  [#permalink]

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5
4
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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1
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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I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.-Mohammad Ali
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If y is the average of x odd consecutive integers and |z - 6/4| = 1/2,  [#permalink]

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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.
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Joined: 02 Sep 2009
Posts: 58453
Re: If y is the average of x odd consecutive integers and |z - 6/4| = 1/2,  [#permalink]

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

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

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

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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

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

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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

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

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_________________ Re: If y is the average of x odd consecutive integers and |z - 6/4| = 1/2,   [#permalink] 14 Oct 2018, 02:06
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