An otherwise identical controlled pair sells at $350,000 with 2 garage bays and $335,000 with 1. What contribution is isolated for that specific one-bay increment? This does not establish a constant rate for all garage sizes.
Correct Answer
B) $15,000
(350000−335000)÷(2−1)=15000. Common errors: dividing the difference by the two bays gives $7,500; dividing the two-bay price by its bay count gives $175,000; averaging the two prices gives $342,500.
Why This Is the Correct Answer
Option B is correct because the price difference is 350,000 − 335,000 = $15,000, and the bay difference is 2 − 1 = 1. Dividing gives $15,000 for that one-bay increment. The stem notes this does not establish a constant rate for all garage sizes.
Why the Other Options Are Wrong
Option A: $7,500
Option A, $7,500, is wrong because it divides the $15,000 difference by the two bays of the larger property: 15,000 ÷ 2 = 7,500. The pair differs by only one bay, so the contribution is 15,000 ÷ 1 = $15,000.
Option C: $175,000
Option C, $175,000, is wrong because it divides the two-bay sale price by its bay count: 350,000 ÷ 2 = 175,000. That spreads the whole property price over the bays rather than isolating the price difference; the correct contribution is $15,000.
Option D: $342,500
Option D, $342,500, is wrong because it averages the two sale prices: (350,000 + 335,000) ÷ 2 = 342,500. An average price says nothing about the contribution of the extra bay, which is 350,000 − 335,000 = $15,000.
Price gap over unit gap
Divide the difference in price by the difference in units, not by the total units.
How to use: Use this for any paired-sale question that asks for a per-unit contribution.
Exam Tip
Watch for distractors that divide by total units or use a whole sale price instead of the price difference.
Common Mistakes to Avoid
- -Dividing the price difference by total bays instead of the bay difference
- -Assuming the per-bay contribution is the same for every garage size
- -Using total price instead of the price difference
Concept Deep Dive
Analysis
This question tests isolating the contribution of a one-unit increment from a controlled pair. The price difference is divided by the difference in units, which is one bay. The result applies to that specific increment, not as a constant rate for every garage size.
Background Knowledge
Paired sales analysis isolates the value contribution of a difference by comparing otherwise similar sales. The contribution of a feature may not be linear; the second bay may contribute differently from the third. Appraisers apply a derived figure only where the market evidence supports it.
Real-World Application
Two matched sales differ only in garage size, one with two bays and one with one. The appraiser derives $15,000 for the second bay and applies it to comparables differing from the subject by that same increment. For a three-bay comparable, the appraiser looks for separate support.
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