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【OG20-P168-139题】
If X is the hundredths digit in the decimal 0.1X and if Y is the thousandths digit in the decimal 0.02Y, where X and Y are nonzero digits, which of the following is closest to the greatest possible value of 0.1X0.02Y
分析A选项
分析B选项
分析C选项
分析D选项
分析E选项
答案 C
截图附上解题思路:
950267voay
太坑了吧,我还以为是0.1*X
Crystal8
1297678eu
谁能帮我翻译一下这个题
林老师啦
真尼玛坑爹啊 190/21≈9
emily666
题目都不难,但是时间压力下容易想不到思路
baoyuecao
这个好坑啊,看成乘法就真的没法做了...
锤子的爱情回复 baoyuecao
确实
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2021-01-21 20:32:50
VirDeer777
Y是1,X是9的时候分数最大。190/21=9.047.... 所以跟9最接近
乔治8198
这题有点陷阱的,因为小数点后面位数不是很多,所以不能简单地把0.19近似为0.2,把0.021近似为0.02,而是按照原值计算。
Xxxffll
为什么不是和10更近?
Yang1116回复Xxxffll
因为,离9更近...啊。
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2018-10-18 01:14:26
441420ztyf回复Xxxffll
9+1/21,和10差20/21,和9只差1/21,哪个更近
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2020-08-07 23:56:33
艾默
题目什么意思?๑•́₃•̀2伐开心
Piaowei1111
X是小数0.1*的百位数,Y是0.02*的千位数,取x的最大值9, y的最小值1. 0.19/0.021约等于9
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