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# If x, y, and z are integers, with x < y< x, what is the average

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Math Expert
Joined: 02 Sep 2009
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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, 07:26
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Difficulty:

95% (hard)

Question Stats:

40% (02:28) correct 60% (02:45) wrong based on 155 sessions

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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.

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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, 07:04
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3
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.

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##### General Discussion
Manager
Joined: 10 Sep 2014
Posts: 97
Re: If x, y, and z are integers, with x < y< x, what is the average  [#permalink]

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18 Nov 2014, 07:46
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.

Intern
Joined: 29 Oct 2014
Posts: 1
Re: If x, y, and z are integers, with x < y< x, what is the average  [#permalink]

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18 Nov 2014, 11:45
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.
Manager
Joined: 27 Aug 2014
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Concentration: Finance, Strategy
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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, 12:40
2
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.
Manager
Joined: 27 Aug 2014
Posts: 140
Concentration: Finance, Strategy
GPA: 3.9
WE: Analyst (Energy and Utilities)
Re: If x, y, and z are integers, with x < y< x, what is the average  [#permalink]

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19 Nov 2014, 09:47
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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07 Dec 2017, 05:31
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Re: If x, y, and z are integers, with x < y< x, what is the average &nbs [#permalink] 07 Dec 2017, 05:31
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