All too many weaklings are also cowards, and few cowards fail to be fools.
Evidence
All too many weaklings Many A's are B are also cowards
Most cowards are fools Most B's are C
Conclusion
There must be someone Some A's are C who is both weakling and fool
Evaluate
Hey, I just feel the need to stick up for us weaklings, But I have a feeling that's not what LSAT is testing. :)
We can tell this argument is working off of quantified statements and the conclusion is trying go for some Trait Overlap inference.
However, this is the only time we've seen the quantifier "all too many".
Consider this statement: all too many celebrities have their own clothing line
Am I saying every celebrity has their own clothing line? Am I saying most celebrities (more than 51%)? Am I saying some celebrities (at least one)?
It's a very unspecific quantity, but it's definitely not All or Most. The author is expressing his opinion that too many celebrities have clothing lines. He might think that "one celebrity is already too much". But since that's such an extreme position, it's better to interpret this as "many," which means "at least a handful".
We can be flexible with how the answer presents that quantity since it's so unspecific, but it should probably show up as Some/Many.
Few A's fail to be B = Most A's are B, since "few = less than 50%". If less than half fail to be B, then more than half succeed at being at B.
The argument was flawed because there doesn't have to be any overlap between weaklings and fools. Consider this parallel: All too many celebrities have their own fashion lines. Most fashion lines are created by people with at least a decade of experience making clothes. Thus, some celebrities have at least a decade of experience making clothes.
That doesn't need to be true. 51% (or more) of fashion lines come from people with experience, but it's highly possible that all the celebrity fashion lines are found in that 49% of fashion lines coming from people without 10+ yrs of experience.
Goal
We want two quantified premises that go A to B, B to C, and then a conclusion that thinks there's some overlap between A and C.
P1: Many A's are B P2: Most B's are C C: Some A's are C