1. Smart College plans to purchase a number of new computer sets for the coming school year. Two computer campanies, A and B, are soliciting sales. They both charge $8000 for a computer set. however, Company A offers a 30% discount on purchases of 11th and over, while company B offer a 15% discount regardless of quantity of purchase.
Smart College finally decides to purchase from company A for they are cheaper, what is the minimum number of computer sets ordered by Smart College?
MATHS 不等式! help!
2007-04-09 5:22 am
回答 (3)
2007-04-09 5:46 am
✔ 最佳答案
Assume Q = qty of purchasePrice purchase form A company < Price purchase form B company
1st 10pcs prices + 11th-Qth pcs onwards prices in Co. A < Qpcs price in Co.B
(8000 x 10) + (8000 x (Q-10) x 70%) < 8000 x Q x 85%
80000 + 5600Q - 56000 < 6800Q
24000 < 1200Q
20
2007-04-08 21:55:08 補充：
補充中文版: 假設Q=電腦數量要選A公司的話, 價錢必需少於向B公司購買由於A公司的頭十台沒有折扣, 所以不等式是:在A公司使的錢
2007-04-08 21:57:34 補充：
在A公司使的錢 少於 在B公司使的錢 頭十台的價錢 + 第十一至Q台的價錢再7折 少於 Q台的價錢再85折
2007-04-08 22:00:15 補充：
Q必需大於20
2011-01-12 12:08 am
where did u get this question from??
2007-04-09 6:01 am
Let the number of computer sets ordered by Smart College is X
8000*X*(1-15%)= 6800X ----(Co. B)
8000*(X-10)*(1-30%)+80000=5600X-56000+80000=5600X+24000 ----(Co. A)
5600X+24000<6800X
24000<1200X
X>20
IF X =20
8000*20*0.85=136,000
5600*20+24000=136,000
IF X>21,
for example X=21
8000*21-.85=142,800
5600*21+24000=141,600
A company is the cheaper supplier for Smart College if they purchase more than 20 sets of computer.
8000*X*(1-15%)= 6800X ----(Co. B)
8000*(X-10)*(1-30%)+80000=5600X-56000+80000=5600X+24000 ----(Co. A)
5600X+24000<6800X
24000<1200X
X>20
IF X =20
8000*20*0.85=136,000
5600*20+24000=136,000
IF X>21,
for example X=21
8000*21-.85=142,800
5600*21+24000=141,600
A company is the cheaper supplier for Smart College if they purchase more than 20 sets of computer.
參考: Myself
收錄日期: 2021-04-25 20:32:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070408000051KK04553