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Math Expert V
Joined: 02 Sep 2009
Posts: 53709

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Difficulty:   65% (hard)

Question Stats: 45% (01:45) correct 55% (01:22) wrong based on 38 sessions

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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) $$x + z \lt 3$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 53709

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Official Solution:

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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Joined: 26 Sep 2016
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Bunuel, hello!
If I am not mistaken we are looking for average. So if we have that sum everywhere is 6, then average is two, isnt it?

Bunuel wrote:
Official Solution:

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.

Math Expert V
Joined: 02 Sep 2009
Posts: 53709

### Show Tags

Jahfors wrote:
Bunuel, hello!
If I am not mistaken we are looking for average. So if we have that sum everywhere is 6, then average is two, isnt it?

Bunuel wrote:
Official Solution:

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.

In once case the sum is 6 and in another case the sum is -6:

Case 1: $$x = -3$$, $$y = 4$$, $$z = 5$$ --> $$x + y + z = 6$$ --> the average = 6/3 = 2.

Case 2: $$x = -3$$, $$y = -2$$, $$z = -1$$ --> $$x + y + z = -6$$ --> the average = -6/3 = -2.
_________________ Re: S98-03   [#permalink] 16 Apr 2018, 00:50
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# S98-03

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