OG数学题求解

geroro

Awesome!

YOYOHuang

直接最极端化吧算出范围即可。E-S的范围即为E的范围，设那10个偶数和为X，20个奇数和为Y，则E的最大值就是X最大减去Y最小，E最小就是X取最小-Y取最大。X-Y最大值：当X里的每个小数取整时都比原来数大了将近1时，Y比原来每个数仅仅只小了0.1，即X-Y=10*1-0.1*20=8；X-Y最小值就是X最小值-Y的最大值：当X里的每个小数取整都比原来的数大了仅仅0.1，Y将近小了1，即X-Y=10*（0.1）-20*0.1=-19，所以只有-16和6在范围内。

cyy043

链接真是贼好啊！！！！

carrieli0805

链接赞！实际差值就是在 -16到6范围之间，不是之间所有的数，但一定不会超过这个范围

budinvv

連結網址講得不錯，但考試真要碰到，應該會用猜的.....

Rayquelin

神链接！讲得很到位

姜姜0525

Hi, I have found a simpler solution that is worth checking out.
Max 0.8 -(0.9)
Min 0.2 -(0.1)

For 10 even Max - 8, Min - 2
For 20 odd Max - (-18), Min - (-2)

Possibilities for (E-S)
= 8+ (-2) = 6 (II - Yes)
= 2 + (-18) = -16 (I - Yes）
这是那个网站上一个人发表的
我的做法差不多 首先是找极端值 10个偶数:全是0.2 E比S多8 全是0.8 E比S多2。20个奇数：全是0.1 E比S少2 全是0.9 E比S少18
得到 8，2，-2，-18 四个数
可能的结果就是 -10 -16 6 或者0！所以选答案B

xinxuer

Solution:

When reading through this question, notice that we are asked which of the following is a POSSIBLE value of E – S. This tells us that we will not be looking for a definite answer here.

We are given that list T has 30 decimals, and that the sum of this list is S.

Next we are given that each decimal whose tenths digit is even is rounded up to the nearest integer and each decimal whose tenths digit is odd is rounded down to the nearest integer.

We are next given that E is the sum of these resulting integers.

Finally, we are given that 1/3 of the decimals in list T have a tenths digit that is even. This this means that 2/3 have a tenths digit that is odd. This means we have 10 decimals with an even tenths digit and 20 decimals with an odd tenths digit.

This is very helpful because we are going to use all this information to create a RANGE of values. We will calculate both the maximum value of E – S and the minimum value of E – S.

Another way to say this is that we want the maximum value of the sum of our estimated value in list T minus the sum of the actual values in list T and also the minimum value of the sum of the estimate values in list T minus the sum of the actual values in list T. Thus, we need to determine the largest estimated values and the smallest estimated values for the decimals in list T.

To do this, let’s go back to some given information:

Each decimal whose tenths digit is even is rounded up to the nearest integer.

Each decimal whose tenths digit is odd is rounded down to the nearest integer.

Let’s first compute the maximum estimated values for the decimals in list T. To get this, we want our 10 decimals with an even tenths digit to be rounded UP as MUCH as possible and we want our 20 decimals with an odds tenths digit to be rounded DOWN as LITTLE as possible. Thus, we can use decimals of 1.2 (for the even tenths place) and 1.1 (for the odd tenths place). Let’s first calculate S, or the sum of the 30 decimals. We know we have 10 decimals of 1.2 and 20 decimals of 1.1, so we can say:

S = 10(1.2) + 20(1.1) = 12 + 22 = 34

Now we can round up 1.2 to the nearest integer and round down 1.1. We see that 1.2 rounded up to the nearest integer is 2, and 1.1 rounded down to the nearest integer is 1. So now we can calculate E, or the sum of the resulting integers. We can say:

E = 10(2) + 20(1) = 20 + 20 = 40

Thus, the maximum value of E – S, in this case, is 40 – 34 = 6.

Let’s next compute the minimum estimated values for the decimals in list T. To get this we want our 10 decimals with an even tenths digit to be rounded UP as LITTLE as possible and we want out 20 decimals with an odds tenths digit to be rounded DOWN as MUCH as possible. Thus, we can use decimals of 1.8 (for the even tenths place) and 1.9 (for the odd tenths place). Let’s first calculate S, or the sum of the 30 decimals. We know we have 10 decimals of 1.2 and 20 decimals of 1.1, so we can say:

S = 10(1.8) + 20(1.9) = 18 + 38 = 56

Now we can round up 1.8 to the nearest integer and round down 1.9 to the nearest integer. 1.8 rounded up to the nearest integer is 2, and 1.9 rounded down to the nearest integer is 1. So now we can calculate E, or the sum of the resulting integers, so we can say:

E = 10(2) + 20(1) = 20 + 20 = 40

Thus, the minimum value of E – S, in this case, is 40 – 56 = -16.

Thus, the possible range of E – S is between -16 and 6, inclusive. We see that I and II fall within this range.

Note that you can use any combination of decimals here as long as they fit the criteria of positive and non-integer.

Answer: B

changfengpx

更简单的理解就是都取极端值，比如：
S假设有：    0.1   0.2   0.8   0.9
E进位为：     0      1     1       0
E-S对应：   -0.1   0.8   0.2    -0.9
E-S求和： 就是20个奇小数的误差跟10个偶小数的误差加起来，只要挑E-S最大的和最小的分别加起来就好
E-S最大值： -0.1*20+0.8*10=6
E-S最小值：-0.9*20+0.2*10=-16
因此在区间-16和6之间，选B
  

Raycyn

其实很好理解，我一开始卡在把rounded理解为四舍五入，
你看odd的时候下取整，那么这个时候e-s是个负数：区间为20*[-0.9,-0.1]为x
同理，当为even的时候，e－s是个正数，区间10*［0.2，0.8］为y
然后求x＋y的区间