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数学机经238,问题
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 |
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