请教PP数学

laihuiyuan

1.At a certain supplier,a machine of type A costs 20,000 and a machine of type B costs 50,000.Each can be purchased by making a 20 percent down payment and repaying the remainder of the cost and the financial charges over a period of time.If the finance charges are equal to 40 percent of the remainder of the cost,how much less would 2 machines of type A cost than 1 machine of type B under this arrangement.
答案是13200
但是我怎么算都是10800=2*（20000+20000*0.2*0.4）-（50000+50000*0.2*0.4）
2.n除以3余2，t除以5余3，问nt除以15余几？
（1）n-2为5的倍数
（2）t为3的倍数
3.Amy's grade was the 90th percentile of the 80 grades for her class. Of the 100 grades from another class,19 were higher than Amy's and the rest were lower.If no other grade was the same as Amy's grade,the Amy's grade was what percentile of the grades of the two classes combined?
这道题题意不是很懂，大家能翻译下吗？
大概觉得应该是(80*0.9+80)/(80+100)约等于85%，但是题目没有问大概等于多少啊。。。

laihuiyuan

谢谢~~不好意思忘写答案了，是C哈

sunny9653

第二题是不是C呀

initusur

n=3x+2  
t=5y+3，
nt=15z+p,求p
化简一下，也就是(3x+2)(5y+3)=15z+p，15xy+9x+10y+6=15z+p，求p

根据条件1，n=5u+2.  那3x+2=5u+2
3x=5u   
3x是5的倍数。那9x肯定也是15的倍数
根据条件2，t=3v. 那3v= 5y+3
5y是3的倍数。5y也是5的倍数。所以5y也是15的倍数，所以10y也是15的倍数

单独条件1或者2都不能说明什么
1，2都满足的话，那么9x+10y就是15的倍数。那么15xy+9x+10y+6=15z+p就可以化简为15q=p-6，由于P Q都是整数所以P=6
所以1，2都满足就可以推导出余数为6   所以选C

laihuiyuan

谢啦~~2道都是没理解题意，等第二题的解释。。。

maseryou

1.Machine A:  down payment:  remainder:40000-8000=32000 finance charge:  total:40000+12800=52800Machine B:50000 down payment: remainder:50000-10000=40000 finance charge:  total:50000+16000=66000
difference between two arrangement:66000-52800=13200

2,不太确定

3，说amy的分数是80个分数中的90%，她的分数在另一个有 100个分数中有19个分数比她的高，剩下的比她的低， 如果没有任何分数是跟amy的分数相同， 那么她的分数在这两个班的总分数中占百分之几？
80*(1-90%) =8  There are 8 people ahead of Amy in her class
Combined with another class, there are 19+8=27 people’s grades ahead of Amy’s.
# of people whose grades below Amy’s / total people of two class  (100+80-27)/(100+80)=85%