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If w, x and y are integers such that w < x < y, are w, x and

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If w, x and y are integers such that w < x < y, are w, x and  [#permalink]

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New post Updated on: 12 Nov 2013, 14:05
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If w, x and y are integers such that w < x < y, are w, x and y consecutive integers?

(1) y - w = 2

(2) The average (arithmetic mean) of w, x and y is x

Originally posted by guerrero25 on 12 Nov 2013, 13:13.
Last edited by Bunuel on 12 Nov 2013, 14:05, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If w, x and y are integers such that w < x < y, are w, x and  [#permalink]

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New post 12 Nov 2013, 14:11
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If w, x and y are integers such that w < x < y, are w, x and y consecutive integers?

(1) y - w = 2 --> y = w + 2. Thus we have that w < x < (w + 2). There is only one integer between w and w + 2, which is w + 1. Thus the three integers are w, x = w + 1, and y = w + 2 --> w, x and y are constitutive integers. Sufficient.

(2) The average (arithmetic mean) of w, x and y is x. If w=0, x=5 and y=10, then the answer is NO but if w=0, x=1 and y=2, then the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.
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If w, x and y are integers such that w < x < y, are w, x and  [#permalink]

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New post 09 Apr 2018, 11:48
Bunuel wrote:
If w, x and y are integers such that w < x < y, are w, x and y consecutive integers?

(1) y - w = 2 --> y = w + 2. Thus we have that w < x < (w + 2). There is only one integer between w and w + 2, which is w + 1. Thus the three integers are w, x = w + 1, and y = w + 2 --> w, x and y are constitutive integers. Sufficient.

(2) The average (arithmetic mean) of w, x and y is x. If w=0, x=5 and y=10, then the answer is NO but if w=0, x=1 and y=2, then the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.


Hi Bunuel,

I understand the no. approach for St.2, but I don't understand what I am doing wrong.

I assumed that the nos. are consecutive. Let w=n, so x=n+1 and y=n+2
The avg of three no. would be [n+n+1+n+2]/[3]= n+1, which is equal to x, same as St.2, so the assumption is correct.

Kindly explain.
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Re: If w, x and y are integers such that w < x < y, are w, x and  [#permalink]

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New post 09 Apr 2018, 13:12
ShashwatPrakash wrote:
Bunuel wrote:
If w, x and y are integers such that w < x < y, are w, x and y consecutive integers?

(1) y - w = 2 --> y = w + 2. Thus we have that w < x < (w + 2). There is only one integer between w and w + 2, which is w + 1. Thus the three integers are w, x = w + 1, and y = w + 2 --> w, x and y are constitutive integers. Sufficient.

(2) The average (arithmetic mean) of w, x and y is x. If w=0, x=5 and y=10, then the answer is NO but if w=0, x=1 and y=2, then the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.


Hi Bunuel,

I understand the no. approach for St.2, but I don't understand what I am doing wrong.

I assumed that the nos. are consecutive. Let w=n, so x=n+1 and y=n+2
The avg of three no. would be [n+n+1+n+2]/[3]= n+1, which is equal to x, same as St.2, so the assumption is correct.

Kindly explain.


You sowed that (2) is true for consecutive integers but what if (2) is also true if the integers are not consecutive? In this case you'd get an YES and a NO answer to the question.
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Re: If w, x and y are integers such that w < x < y, are w, x and   [#permalink] 09 Apr 2018, 13:12
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