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# X, Y and Z are positive integers, and X+2Y+2Z=13. Z=?

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X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]

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15 May 2012, 22:26
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X, Y and Z are positive integers, and X+2Y+2Z=13. Z=?

(1) X>Y>Z
(2) None of them is equal to 4
[Reveal] Spoiler: OA
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Joined: 02 Sep 2009
Posts: 39588
Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]

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15 May 2012, 23:36
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X, Y and Z are positive integers, and X+2Y+2Z=13. Z=?

(1) X>Y>Z. If z=2 then the least values of y and x are 3 and 4 respectively, in this case x+2y+2z=14>13, so z must be less than 2 and since given that it's a positive integer then it can only be 1. Sufficient.

(2) None of them is equal to 4. Not sufficient, consider: x=3, y=2, z=3 and x=3, y=3, z=2.

Hope it's clear.
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Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]

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27 May 2012, 20:43
(1) X>Y>Z. If z=2 then the least values of y and x are 3 and 4 respectively, in this case x+2y+2z=14>13, so z must be less than 2 and since given that it's a positive integer then it can only be 1. Sufficient.

Bunuel, any reason why you did not consider Z=1 so in this case case y=2 and z=3 and then the solution is insufficient.
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Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]

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28 May 2012, 03:35
rggoel9 wrote:
(1) X>Y>Z. If z=2 then the least values of y and x are 3 and 4 respectively, in this case x+2y+2z=14>13, so z must be less than 2 and since given that it's a positive integer then it can only be 1. Sufficient.

Bunuel, any reason why you did not consider Z=1 so in this case case y=2 and z=3 and then the solution is insufficient.

I guess you mean z=1, y=2 and x=3 (z<y<z). This values are not valid since they does not satisfy x+2y+2z=13.
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Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]

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30 May 2012, 06:01
another way to look at the question is that the sum is 13... and it is x+2y+2z, which means that z has to be odd as 2z+2y will be even and hence z has to be 1, bcoz any other odd and the condition x>y>z does not give 13.

Second condition is clearly insufficient.
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Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]

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30 May 2012, 09:40
pavanpuneet - Very good explanation. I liked it.
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Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]

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18 Sep 2015, 01:23
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Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]

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20 Sep 2015, 00:48
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

X, Y and Z are positive integers, and X+2Y+2Z=13. Z=?

(1) X>Y>Z
(2) None of them is equal to 4

Transforming the original condition and the question, we have 3 variables (x,y,z) and 1 equation (x+2y+2z=13) therefore we need 2 more equations to match the number of variables and equations. Since there is 1 each in 1) and 2), C has high probability of being the answer.

Using both 1) & 2) together, (x,y,z)=(7,2,1),(5,3,1) leads to z=1 and the answer is unique, therefore C being the answer. However, the key question is an integer problem, therefore we must apply common mistake type 4(A) and therefore try using the conditions separately. In case of 1), (x,y,z)=(7,2,1),(5,3,1) leads to z=1 and therefore is unique. Therefore the condition is sufficient.
In case of 2), (x,y,z)=(7,2,1),(3,3,2) gives us two answers z=1,2 and therefore the answer is not unique. Therefore the condition is not sufficient, and the answer is A. This problem is a typical 495051 level question, and there are roughly 4 of these questions in a test.

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=?   [#permalink] 20 Sep 2015, 00:48
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