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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

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.

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

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17 May 2017, 22:54

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.

Answer: B.

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.

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.

Answer: B.

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.
_________________

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, 12:29

2

This post was BOOKMARKED

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.

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

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10 Oct 2017, 00:36

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. => Answer B