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If x, y, and z are integers, with x < y< x, what is the average [#permalink]
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18 Nov 2014, 08:26
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Re: If x, y, and z are integers, with x < y< x, what is the average [#permalink]
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18 Nov 2014, 08:46
1. I can only think of 1 possibility that work for statement 1. x=0,y=1,z=5 in which case the avg would be 3. I'm going to lean towards this statement being true.
2. Knowing that both equations must be correct and also agree with each other, I now know that there is another possibility for statement 1 since what I came up with does not fit this equation. Regardless, this equation is not sufficient by itself because there could be infinite values for x + z to be less than 3. Statement 2 is insufficient.
3. Now that I know there is another combo of #s (besides the ones I came up with in step 1) that must work for statement 1, I know that statement 1 is also insufficient. Down to answer choices C and E. After a little more thought I came up with If x=3, y=2, z=1, the avg is 2. This combo of #s also fits statement 1 and we now know it's insufficient.
4. Both equations combined tell us that the only combo of #s we can use is x=3,y=2,z=1 for an avg of 2.
Answer C!



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Re: If x, y, and z are integers, with x < y< x, what is the average [#permalink]
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18 Nov 2014, 12:45
The way I see it, statement 1 implies that the sum of x,y,z has to be 6 (x+y=1 and z=5, or x+y=5 and z=1).
I cant see any value in statement two.
So I guess the answer is A.



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Re: If x, y, and z are integers, with x < y< x, what is the average [#permalink]
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18 Nov 2014, 13:40
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I tried the problem in the following way:
from statement 1: (x+y)z = 5, as the product is positive, (x+y) and z should be both positive or negative, if positive, then, (x+y+z) = 6, we get AM as 2. but if (x+y) = 5 and z = 1, we get (x+y+z) = 6, or AM = 2. two different answers, so NSF.
from statement 2, NSF as no info on y.
statement 1 + 2, x+y and z should be negative as the sum of x+z is less than 3, as from statement 1, if x+y = 1 and z = 5, then the sum will be greater than 3, the other possibility z = 1 and x+y = 5 is not possible as z>y>x. so, AM = 2, sufficient. my answer [C]



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Re: If x, y, and z are integers, with x < y< x, what is the average [#permalink]
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19 Nov 2014, 08:04
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Bunuel wrote: Tough and Tricky questions: Algebra. If \(x\), \(y\), and \(z\) are integers, with \(x \lt y \lt z\), what is the average (arithmetic mean) of \(x\), \(y\), and \(z\)? (1) \((x + y)z = 5\) (2) \(x + z \lt 3\) Kudos for a correct solution. Official Solution:If \(x\), \(y\), and \(z\) are integers, with \(x \lt y \lt z\), what is the average (arithmetic mean) of \(x\), \(y\), and \(z\)? Since the average of \(x\), \(y\), and \(z\) is the sum of the three variables, divided by 3, we can rephrase the question as "what is the value of \(x + y + z\)?" We also note the restrictions on the possible values of \(x\), \(y\), and \(z\)  the variables must be integers in ascending order from \(x\) to \(z\) (not necessarily consecutive). Moreover, they must be different integers, since the inequality \(x \lt y \lt z\) indicates no equality among any of the variables. We note these conditions, but at this stage there is no simple way to apply them in a further rephrasing of the question. Statement (2): INSUFFICIENT. We start with statement (2), which is the easier statement (it only contains 2 of the 3 variables). We can quickly come up with sets of values that satisfy this statement and the given conditions but that have different sums (or averages). For instance, the set \(x = 0\), \(y = 1\), and \(z = 2\) meets all conditions \((x + z = 2 \lt 3\), all variables are integers and in ascending order), with \(x + y + z = 3\). Another set (\(x = 1\), \(y = 0\), and \(z = 1\)) also meets all conditions but sums to 0. Thus, there is no single value determined by this statement. Statement (1): INSUFFICIENT. The equation states that \(x + y\) (which must be an integer) multiplied by \(z\) (another integer) equals 5. Since 5 is a prime number, there are only 2 pairs of integers that multiply together to 5: 1 and 5, and 1 and 5. (Don't forget about the negative possibilities). Keeping the conditions that \(x \lt y \lt z\), we can construct the only sets that work: \(x + y = 1\) and \(z = 5\) (There's no way to assign \(z = 1\) and \(x + y = 5\) while preserving \(x \lt y \lt z\).) \(x = 0\), \(y = 1\), \(z = 5\), \(sum = 6\) \(x = 1\), \(y = 2\), \(z = 5\), \(sum = 6\) \(x = 2\), \(y = 3\), \(z = 5\), \(sum = 6\) \(x = 3\), \(y = 4\), \(z = 5\), \(sum = 6\) \(x + y = 5\) and \(z = 1\) \(x = 3\), \(y = 2\), \(z = 1\), \(sum = 6\) Since there are 2 possible sums, this statement is insufficient. Statements (1) and (2) together: INSUFFICIENT. Using the sets determined with Statement (1), we check the value of \(x + z\) for each case, keeping only the cases in which \(x + z\) is less than 3. Two cases remain. Case 1: \(x = 3\), \(y = 4\), \(z = 5\), \(x + z = 2 \lt 3\), \(x + y + z = 6\) Case 2: \(x = 3\), \(y = 2\), \(z = 1\), \(x + z = 4 \lt 3\), \(x + y + z = 6\) Since the two cases yield different sums, we cannot determine a single value for that sum. Answer: E.
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Re: If x, y, and z are integers, with x < y< x, what is the average [#permalink]
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19 Nov 2014, 10:47
aah missed to look deeper into the cases from statement A, thanks for the explanation Bunuel



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Re: If x, y, and z are integers, with x < y< x, what is the average [#permalink]
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