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16 Sep 2014, 01:52
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E
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55%
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Question Stats:
61% (02:06) correct
39%
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wrong
based on 54
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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) \(x + z \lt 3\)
(1) \((x + y)z = 5\)
(2) \(x + z \lt 3\)
16 Sep 2014, 01:52
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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.
Answer: E
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
