An otherwise identical controlled pair sells at $285,000 with 2 garage bays and $275,000 with 1. What contribution is isolated for that specific one-bay increment, without extrapolating to other garage sizes?
Correct Answer
B) $10,000
(285000−275000)÷(2−1)=10000. Common errors: dividing the difference by the two bays gives $5,000; dividing the two-bay price by its bay count gives $142,500; averaging the two prices gives $280,000.
Why This Is the Correct Answer
Option B is correct because the price difference is $285,000 - $275,000 = $10,000, and the bay difference is 2 - 1 = 1. Dividing $10,000 by 1 isolates a $10,000 contribution for that one-bay increment.
Why the Other Options Are Wrong
Option A: $5,000
Option A, $5,000, divides the difference by two bays: ($285,000 - $275,000) / 2 = $5,000. The pair differs by only one bay, so the contribution is $10,000 / 1 = $10,000.
Option C: $142,500
Option C, $142,500, divides the two-bay sale price by its bay count: $285,000 / 2 = $142,500. That treats the whole price as garage value; the paired difference is $285,000 - $275,000 = $10,000.
Option D: $280,000
Option D, $280,000, averages the two prices: ($285,000 + $275,000) / 2 = $280,000. The contribution comes from the difference between the prices, which is $10,000.
Subtract Prices, Divide Difference
Contribution equals the difference in price divided by the difference in feature count, not the total count.
How to use: Use this whenever a paired-sale question asks for a per-feature contribution.
Exam Tip
Divide by the difference in the feature (here 2 - 1 = 1), not by the feature count of one sale.
Common Mistakes to Avoid
- -Dividing the price difference by the total number of bays
- -Averaging prices instead of subtracting them
- -Applying a one-bay increment to other garage sizes without support
Concept Deep Dive
Analysis
This question tests extracting the contribution of a single feature increment from a controlled pair. The two sales differ only by one garage bay, so the price difference divided by the one-bay difference measures that increment. The result applies to that specific increment and should not be extrapolated to other garage sizes without support.
Background Knowledge
Paired sales analysis compares nearly identical properties to isolate one characteristic's effect on value. The contribution of a feature increment is the price difference divided by the size of the feature difference, and the contribution of a second bay may differ from that of a first or third.
Real-World Application
Two matched resales in a subdivision differ only by garage size: one has two bays and sold for $285,000, the other one bay at $275,000. The appraiser uses about $10,000 as support for a one-bay garage difference among similar homes.
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