If x, y, and z represent consecutive integers, and x \textless y \textless z, which of the following equals y?
I. \(x+1\)
II. \(\frac{x+z}{2}\)
III. \(\frac{x+y+z}{3}\)
I. \(x+1\)
II. \(\frac{x+z}{2}\)
III. \(\frac{x+y+z}{3}\)
Show Hint
- I only
- I and II only
- I and III only
- II and III only
- I, II and III
The Correct Option is
Solution and Explanation
The problem deals with the properties of consecutive integers. Since the integers are consecutive and ordered (\(x \textless y \textless z\)), we can express \(y\) and \(z\) in terms of \(x\), or express all three in terms of \(y\). The latter is often simpler for this type of problem.
Step 2: Detailed Explanation:
Let's represent the integers in terms of the middle integer, \(y\).
Since they are consecutive, \(x\) is one less than \(y\), and \(z\) is one more than \(y\).
So, we have:
\(x = y - 1\)
\(y = y\)
\(z = y + 1\)
Now we test each of the three statements using these relationships.
Statement I: \(x+1\)
Substitute \(x = y - 1\) into the expression:
\[ (y - 1) + 1 = y \] Statement I is equal to \(y\). This is correct.
Statement II: \(\frac{x+z}{2}\)
This expression represents the average (mean) of the first and last integers. For any evenly spaced set, the mean is equal to the median (the middle value).
Let's verify by substitution:
\[ \frac{(y - 1) + (y + 1)}{2} = \frac{y - 1 + y + 1}{2} = \frac{2y}{2} = y \] Statement II is equal to \(y\). This is correct.
Statement III: \(\frac{x+y+z}{3}\)
This expression represents the arithmetic mean of all three integers. For consecutive integers, the mean is always the middle number.
Let's verify by substitution:
\[ \frac{(y - 1) + y + (y + 1)}{3} = \frac{y - 1 + y + y + 1}{3} = \frac{3y}{3} = y \] Statement III is equal to \(y\). This is correct.
Step 3: Final Answer:
All three statements, I, II, and III, are equivalent to \(y\).
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