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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. 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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