威尔逊 50分钟 |
设水速为1,四人的船速为a,b,c,d,则他们离目的距离为60a,60b,60c,60d。有: 60a/(a-1)=75 => a=5 60b/(b-1)=70 => b=7 60c/(c+1)=50 => c=5 60d/(d+1)=45 => d=3 回程时间分别为: 60a/(a+1)=50 60b/(b+1)=52.5 60c/(c-1)=75 60d/(d-1)=90 |
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