## PrepTest 49, Section 4, LR 24

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s1m4

Posts: 214
Joined: Tue Jan 12, 2010 10:04 pm

### PrepTest 49, Section 4, LR 24

#24

Can somebody please explain why A is correct, and how other choices could be eliminated? (B is easy to eliminate, but the others not so much.)

Very tough parallel flaw question..

Cambridge LSAT

Posts: 257
Joined: Mon Aug 24, 2009 3:26 pm

### Re: PrepTest 49, Section 4, LR 24

A simple example with numbers will help to illustrate the flaw:
# of employees: 100
# of employee computer programmers: 51
# of computer programmers (total): 20,000
# of computer programmers with excellent salaries: 10,001
# of computer programmers without excellent salaries: 9,999

All the computer programmers at this company could belong to the set of computer programmers who don't have excellent salaries.

Similarly in choice A, none of the numbers are given, so there is no way to know for certain that the two sets overlap.
# of Molly's classmates: 25
# of Molly's classmates who are gardeners: 13
# of gardeners (total): 20,000
# of gardeners with a great deal of patience: 10,001
# of gardeners without a great deal of patience: 9,999

chewdak

Posts: 106
Joined: Fri Apr 03, 2009 5:54 pm

### Re: PrepTest 49, Section 4, LR 24

Another "most" question.
We know "most" just means "majority" or over 50%.
Most A are B.
Most B have C.
Therefore at least one A has C.
This is faulty as explained above by cambridgelsat.
(A) parallels the structure above
(B) changes structure by using "some" instead of "at least one"
(C) changes conclusion so that it becomes 'Therefore at least one A, who is B, has C'.
(D) Changes premise from 'Most A are B' to "Most B in A are D"
(E) This one is actually true, not what we are looking for.
Last edited by chewdak on Fri Jan 22, 2010 5:50 pm, edited 1 time in total.

s1m4

Posts: 214
Joined: Tue Jan 12, 2010 10:04 pm

thx guys !!