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标题: [求助]天山 4-29 [打印本页]

作者: knight9514    时间: 2005-12-7 07:34     标题: [求助]天山 4-29

Q29

The toll for crossing a certain bridge is $0.75 each crossing. Drivers who frequently use the bridge may instead purchase a sticker each month for $13 and then pay only $0.3 each crossing during that month. If a particular driver will cross the bridge twice on each of x days next month and will not cross the bridge on any other day, what is the least value of x for which this driver can save money by using the sticker?

A 14

B 15

The answer is B but I think the answer is A

0.75×2 ×N13+0.3×2 ×N


作者: asasasas1    时间: 2005-12-7 23:22

我的解法是:

(31/x)(2)(0.3)+13 < (31/x)(2)(0.75)

求X的最小值....

可是怎麼算不出來啊... 邏輯哪裡錯了呢?


作者: himba    时间: 2005-12-8 00:15

X如果是相隔的天数的话,最小值应该是1的,我想它应该问的是x的最大值,就是15没错.
作者: 小精灵    时间: 2005-12-8 20:20

发表一下愚见

cross the bridge twice on each of x days next month : 在下月,X天中的每一天过桥两次,即为只有X天过桥,因此过桥总次数为 2X,

使用sticker省钱,则其成本必然小于等于没有使用sticker的成本

13+2X * 0.3<=2X * 0.75 解之得:X>=14.4 故最小值为15 答案B






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