GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 19 Feb 2020, 20:50

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 61302
If x, y, and z are integers, with x < y< x, what is the average  [#permalink]

### Show Tags

18 Nov 2014, 07:26
11
00:00

Difficulty:

95% (hard)

Question Stats:

46% (02:31) correct 54% (02:41) wrong based on 201 sessions

### HideShow timer Statistics

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.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 61302
Re: If x, y, and z are integers, with x < y< x, what is the average  [#permalink]

### Show Tags

19 Nov 2014, 07:04
3
4
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.

_________________
##### General Discussion
Manager
Joined: 10 Sep 2014
Posts: 96
Re: If x, y, and z are integers, with x < y< x, what is the average  [#permalink]

### Show Tags

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]

### Show Tags

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.
Senior Manager
Joined: 27 Aug 2014
Posts: 371
Location: Netherlands
Concentration: Finance, Strategy
Schools: LBS '22, ISB '21
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]

### Show Tags

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.
Senior Manager
Joined: 27 Aug 2014
Posts: 371
Location: Netherlands
Concentration: Finance, Strategy
Schools: LBS '22, ISB '21
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]

### Show Tags

19 Nov 2014, 09:47
aah missed to look deeper into the cases from statement A, thanks for the explanation Bunuel
VP
Joined: 24 Nov 2016
Posts: 1218
Location: United States
Re: If x, y, and z are integers, with x < y< x, what is the average  [#permalink]

### Show Tags

21 Jan 2020, 06:19
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) $$x + z \lt 3$$

x,y,z = integers

(1) (x+y)z=5 insufic

x,y,z: -3,-2,-1…-5*-1=5
x,y,z: 0,1,5…-1*5=5
x,y,z: -3,4,5…1*5=5

(2) x+z<3 insufic

(1&2) insufic

x+z<3: -3+-1=-4<3
x+z<3: -3+5=2<3

Ans (E)
Re: If x, y, and z are integers, with x < y< x, what is the average   [#permalink] 21 Jan 2020, 06:19
Display posts from previous: Sort by