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OpenStudy (anonymous):
Solve this numerical syllogism for me: Eight of ten A's are B's; four of ten A's are C's. What is the minimum number of B's that are also C's?
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OpenStudy (osanseviero):
2 B´s are C´s
OpenStudy (anonymous):
2 Bs would be the minimum number of Bs that can be Cs.
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