数学机经238，问题

fantasy1989

238、DS：banquet offer 3 kinds of dinners, respectively dinners of chiken, beef and fish.Their number of the dinners  满足7:5:2. the number of the dinners of fish is more than 5。what's the total number of the diners in the banquet?(狗主选B)
（1）fish beef的总数<30
(2) chiken总数<25
Answer：B
由已知得，fish要more than 5，即至少C:B:F=21:15:6
（1）fish+beef<30且为5+2=7的倍数，存在情况21和28
（2）chicken<25且为21倍数，即唯一情况21

不是说fish 大于5吗，（1）不是只能是21吗

kinghades

题目错误了吧~