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标题: 求解 [打印本页]

作者: mustard12 时间: 2012-4-4 17:32 标题: 求解

gwd第一套中的，if 10的50次方-74 is written as an integer in base 10 notation，what is the sum of the digits in that interger。答案有 424 433 440 449 467. 大家能帮忙解释一下题干吗？

还有一道题，Running at their respective constant rates，machine X takes 2 days longer to produce w widgets than machine Y。At these rates ，if the two machines together produce 5/4w widgets in 3 days，how many days would it take machine X alone to produce 2w widgets。答案有4 6 8 10 12.
感觉好绕，没弄清楚意思，希望大家帮帮忙

作者: james2201 时间: 2012-4-5 08:58

100-74=26
1000-74=926
10000-74=9926
100000-74=99926
...
....
10^52-74=......26

"what is the sum of the digits in that interger"
9+9+9+...+2+6

謝謝!

作者: DayDreamin 时间: 2012-4-6 08:01

gwd第一套中的，if 10的50次方-74 is written as an integer in base 10 notation，what is the sum of the ...
mustard12 发表于 2012-4-4 17:32

第二條
time=quantity/rate
ratex=x, ratey=y

let quantity = 1
timex-timey=2

together=ratex+ratey
quantity=(5/4)*w
time=3

還沒有驗算
謝謝!

作者: mustard12 时间: 2012-4-7 06:31

可是，木有3这个答案啊……求再看看

作者: DayDreamin 时间: 2012-4-8 06:33

我不是說答案是 3....

題目說 3 days

作者: DayDreamin 时间: 2012-4-8 19:53

問題: 2W/Rx=?

(1) W/Rx - W/Ry = 2
(2) (5/4*W)/(Rx+Ry) = 3

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