PrepTest 151, Section 2, Question 15
All the apartments on 20th Avenue are in(29 more words not shown)
The reasoning in the argument(12 more words not shown)
- Aoverlooks the possibility(12 more words not shown)
- Bdraws a conclusion(13 more words not shown)
- Cfails to consider(17 more words not shown)
- Dconfuses a condition(28 more words not shown)
- Efails to address(17 more words not shown)
Reading the argument
An average of two apartments per house doesn't mean most houses have two.
Try three old houses with six apartments altogether. Put one apartment in the first house, one in the second, and four in the third. Every apartment is in an old house, and there are exactly twice as many apartments as houses. Both premises are satisfied.
What we’re looking for
But two of the three houses contain only one apartment. The conclusion is false: most houses do not contain more than one. That is the flaw: the extra apartments can be concentrated in a single house rather than evenly distributed. The average alone doesn't establish what most houses contain.
Working the answers
- A
There could be other buildings on the avenue that aren't old houses. That wouldn't affect the comparison, as long as the apartments we're counting are in old houses, as the premise says. The existence of an office building or another non-apartment building doesn't explain the mistake about how these apartments are distributed.
- B
The premises give a total ratio and tell us where the apartments are. The conclusion adds a claim about most individual houses. Those aren't the same statements, as the one-one-four example shows: the premises can be true while the conclusion is false. The argument is defective, but not because it simply repeats its conclusion as a reason.
- C
Other rental accommodations aren't apartments, so they aren't part of the count used in the argument. A hotel or another kind of rental could exist without changing how many apartments each old house contains. That isn't the missing distribution information.
- D
This would mean treating something that can guarantee the conclusion as something the conclusion requires. But the two-to-one ratio doesn't guarantee that most houses contain multiple apartments, as the one-one-four example shows. Nor is that ratio required: counts of two, two, and one would give most of three houses multiple apartments, with only five apartments total. The error is taking an overall ratio as enough to settle the distribution, not treating a genuine guarantee as a requirement.
- ECorrect
Some houses could contain three or more apartments, accounting for enough of the total that most other houses contain only one. In our example, the four-apartment house supplies the extra units while the other two houses remain single-apartment houses. Without addressing this possible concentration, the total ratio doesn't establish the claim about most houses.
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