[联谊] 国内小学五年级的数学题！

oldwolf · 发表于 2004-11-18 16:39

How come, I alreay provide you a distribution showing that there is a possibility that 30 students may fail the example. How can you say at most 28 students may fail. If you look at the distribution of wrong answers among students, you will get the conclusion:

Look at this situation: 100 students, 70 answered all 5 questions correctly. The other 30 failed 3 questions each one. Here is how they failed:

5 failed Question 2, 4, 5
6 failed Question 3, 4, 5
10 failed Question 1,4,5
4 failed Quesiton 2, 3, 1
5 failed Question 3, 1, 5

Total failed students 5+6+10+4+5 = 30

Total students failed Question 1: 10+4+5 = 19
Total students failed Question 2: 5+4 = 9
Total students failed Question 3: 6+4+5 = 15
Total students failed Question 4: 5+6+10 = 21
Total students failed Question 5: 5+6+10+5 = 26

For the above situlation, please count how many students failed the exam. It is 30?

9 = 5 0 0 4 0          Question 2
15= 0 6 0 4 5          Question 3
19= 0 0 10 4 5       Question 1
21= 5 6 10 0 0       Question 4
26= 5 6 10 0 5       Question 5

total
failed
students 5 + 6 + 10 + 4 + 5 = 30

Post by 天秤座
I would like to say that: Oldwolf didn't provide a clear algorithm, in stead, GiganticBalls has provided a good analysis. Only the result needs some correction. The following is the answer:

at most: 9 students failed all 5 questions ( GiganticBalls is right at this stage).

===============Pay attention here================

AT MOST: 19 students failed 4 questions. calculation is:
9 failed: Q1+Q2+Q4+Q5;
10 failed: Q1+Q3+Q4+Q5;
0 failed all 5 questions.

AT MOST: 28 students failed 3 questions. calculation is:
19 faild: Q1+Q4+Q5;
7 failed: Q2+Q3+Q5;
2 failed: Q2+Q3+Q4;
0 failed 4 or more questions.

Therefore:

不及格的最多有19+7+2=28人，
则及格的最少有100-28=72人。
===============================

天秤座 · 发表于 2004-11-18 16:45

最多9人错5道。
最多19人错4道 (9人错1，2，4，5题； 10人错1，3，4，5题)。
最多28人错3道(19人错1，4，5题； 7人错2，3，5题； 2人错2，3，4题)。

所以:
不及格的最多有19+7+2=28人，
则及格的最少有100-28=72人。

Post by 天秤座
I would like to say that: Oldwolf didn't provide a clear algorithm, in stead, GiganticBalls has provided a good analysis. Only the result needs some correction. The following is the answer:

   at most: 9 students failed all 5 questions ( GiganticBalls is right at this stage).

   ===============Pay attention here================

   AT MOST: 19 students failed 4 questions. calculation is:
   9 failed: Q1+Q2+Q4+Q5;
   10 failed: Q1+Q3+Q4+Q5;
   0 failed all 5 questions.

   AT MOST: 28 students failed 3 questions. calculation is:
   19 faild: Q1+Q4+Q5;
   7 failed: Q2+Q3+Q5;
   2 failed: Q2+Q3+Q4;
   0 failed 4 or more questions.

   Therefore:

   不及格的最多有19+7+2=28人，
   则及格的最少有100-28=72人。
   ===============================

oldwolf · 发表于 2004-11-18 16:48

天秤座, Just do me a favour, count the number of my previous message. Tell me the truth.:rolleyes:

Post by 天秤座
最多9人错5道。
最多19人错4道 (9人错1，2，4，5题； 10人错1，3，4，5题)。
最多28人错3道(19人错1，4，5题； 7人错2，3，5题； 2人错2，3，4题)。

所以:
不及格的最多有19+7+2=28人，
则及格的最少有100-28=72人。

eihpos · 发表于 2004-11-18 16:51

干吗不列方程?有AK还用弓箭吗?:confused: :confused: :confused: :confused:

oldwolf · 发表于 2004-11-18 16:52

Before post your intelligent method, please look at the following situation. Tell me how many students failed the exam. :confused:

Look at this situation: 100 students, 70 answered all 5 questions correctly. The other 30 failed 3 questions each one. Here is how they failed:

5 failed Question 2, 4, 5
6 failed Question 3, 4, 5
10 failed Question 1,4,5
4 failed Quesiton 2, 3, 1
5 failed Question 3, 1, 5

Total failed students 5+6+10+4+5 = 30

Total students failed Question 1: 10+4+5 = 19
Total students failed Question 2: 5+4 = 9
Total students failed Question 3: 6+4+5 = 15
Total students failed Question 4: 5+6+10 = 21
Total students failed Question 5: 5+6+10+5 = 26

For the above situlation, please count how many students failed the exam. It is 30?

oldwolf · 发表于 2004-11-18 16:57

I do have a equation, list the numbers just for someone to understand. To prove my answer is right.

Equation is a little bit difficult to understand.

Post by eihpos
干吗不列方程?有AK还用弓箭吗?:confused: :confused: :confused: :confused:

天秤座 · 发表于 2004-11-18 17:37

Thanks, you are right. It's 70 rather than 72.

It is almost not possible to get this result without solving a set of equations. That is why I did a quick estimate, it seems not a correct one.

Post by oldwolf

I do have a equation, list the numbers just for someone to understand. To prove my answer is right.

Equation is a little bit difficult to understand.

木又子 · 发表于 2004-11-18 20:26

起用了多台高性能计算机:

http://my.cnd.org/modules/newbb/viewtopic.php?topic_id=32280&forum=1

berriuqam · 发表于 2004-11-18 20:56

呵呵,喜欢做思维体操的人还不少.再来一题,还是老规矩,不准列方程,不准编程,不准.......,用小学算术水平解题.

1000米长的距离，甲在一端，乙、丙在另一端相向而行，甲、乙、丙的速度分别是4公里/小时、6公里/小时、10公里/小时，丙在行进中，遇到甲就折返，遇到乙也折返，直到甲乙相遇。问丙和甲共相遇几次？和乙相遇几次？

eihpos · 发表于 2004-11-18 21:43

干吗不让列方程不让编程?有AK还用弓箭吗?你以为你是只儿豁阿歹呀!干吗不让列方程不让编程?有AK还用弓箭吗?你以为你是只儿豁阿歹呀!干吗不让列方程不让编程?有AK还用弓箭吗?你以为你是只儿豁阿歹呀!干吗不让列方程不让编程?有AK还用弓箭吗?你以为你是只儿豁阿歹呀!:confused: :confused: :confused: :confused: :confused: :confused: :confused:

[联谊] 国内小学五年级的数学题！

~, how can I explain it.

现用中文post结果: 72

~~~~~~

Please look at my example, and count it.

You are right. It's 70 rather than 72.

参考答案在此.

哲别,原名只儿豁阿歹,蒙古帝国神箭手,军事家,军事统帅