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If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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Updated on: 08 Mar 2015, 10:54
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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
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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.




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Re: If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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08 Mar 2015, 11:07
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.
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Re: If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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08 Mar 2015, 11:46
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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If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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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. 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.



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Re: If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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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. 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 23 minutes.
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Re: If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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02 Sep 2017, 13:29
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
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Re: If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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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



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If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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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 4n1 => 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



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If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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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. Answer B
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Re: If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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14 Oct 2018, 02:06
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Re: If y is the average of x odd consecutive integers and z  6/4 = 1/2,
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