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If the average (arithmetic mean) of the assessed values of \(\mathrm{x}\) houses is \($212,000\) and the average of the assessed values of \(\mathrm{y}\) other houses is \($194,000\), what is the average of the assessed values of the \(\mathrm{x + y}\) houses?
We need to find the average assessed value of all houses combined, given:
To calculate this combined average, we must determine the weighted average. This requires knowing either the exact values of x and y, or their ratio. Without this information, we cannot determine how much weight each group contributes to the overall average.
Key insight: For sufficiency, we need information that allows us to calculate one unique average value for all \(\mathrm{x + y}\) houses.
Statement 1: \(\mathrm{x + y = 36}\)
This tells us there are 36 houses total, but doesn't specify how many belong to each group. Let's test whether different distributions yield different averages:
Test Case 1: Equal distribution (\(\mathrm{x = 18, y = 18}\))
Test Case 2: Unequal distribution (\(\mathrm{x = 24, y = 12}\))
Since different distributions of the 36 houses produce different average values ($203,000 vs $206,000), we cannot determine a unique answer.
Statement 1 is NOT sufficient.
This eliminates choices A and D.
Now we forget Statement 1 completely and analyze Statement 2 independently.
Statement 2: \(\mathrm{x = 2y}\)
This establishes a fixed 2:1 ratio between the two groups. Let's see if this ratio alone determines the average:
Using \(\mathrm{x = 2y}\):
Notice that y cancels out completely! This means regardless of the actual number of houses, as long as the ratio is 2:1, the average will always be $206,000.
[STOP - Sufficient!]
Statement 2 is sufficient.
This eliminates choices C and E.
Statement 2 alone provides the ratio needed to calculate a unique weighted average, while Statement 1 alone does not.
Answer Choice B: "Statement 2 alone is sufficient, but Statement 1 alone is not sufficient."