一道蓄水池注水的数学题,不知道怎么求解

2025-03-16 08:48:16
推荐回答(4个)
回答1:

甲乙丙丁各开1小时进水(1/2)+(1/5)-(1/4)-(1/6)=7/60
原来有1/6的水,还剩下1-(1/6)=5/6=50/60
经过7个循环之后,还剩下(50/60)-7*(7/60)=1/60
再就是甲开始注水,需要的时间是(1/60)÷(1/3)=1/20小时
所以,总共需要的时间是4×7+(1/20)=28.05小时

回答2:

水池总量为1,则甲单位时间进水1/3,乙1/5,丙-1/4,丁-1/6,;甲乙丙丁四个为一组,一组进水7/60。总要求进水量为5/6(50/60)。考虑第一次注满,那必须是几个完整组加甲乙一次,这样进水最多。故为3组进水为21/60,再加甲20/60,加乙12/60,大于50/60.。(考虑2组时,不够。)

回答3:

如果是4次,则最初的1/6+28/60=38/60,
因为1/3>1/4>1/5>1/6
所以经过一次+1/3-1/4+1/5-1/6之内的任意步骤,总和都比单独+1/3的和小
所以中间轮流完整的次数为5次,这样得到1/6+35/60=45/60
当第6次开水管时,45/60+1/3=65/60>1
所以第六次开水管,单开到甲管时,水已满

望采纳

回答4:

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