oldwolf

Here is a simple algorithm to solve it.


The persions who didn't answer correctly the
first question: 9
second : 15
third : 19
fourth: 21
fifth: 26


sort the above numbers:

9
15
19
21
26

Take the last 3 numbers: 19 21 26. The minimum of the these 3 numbers is the answer: 19


This algorithm can be applied to other questions like 10000 students answer 1000 questions and can be easily solved in a computer.

Post by berriuqam
100个人回答五道试题，有81人答对第一题，91人答对第二题，85人答对第三题，79人答对第四题，74人答对第五题，答对三道题或三道题以上的人算及格，那么，在这100人中，至少有（ ）人及格。

请说出算法！不准列方程,要用小学算术的水平解这道题.

oldwolf

Here is a simple algorithm to solve it.


The persions who didn't answer correctly the first question: 19
                   second        : 15
                   third           

Post by berriuqam
100个人回答五道试题，有81人答对第一题，91人答对第二题，85人答对第三题，79人答对第四题，74人答对第五题，答对三道题或三道题以上的人算及格，那么，在这100人中，至少有（ ）人及格。

请说出算法！不准列方程,要用小学算术的水平解这道题.

cliff2002

The kind of question remind me 高中语文的文章 <<孔以及>> 中 ”回香豆的回有几种写法”。
except for very small number of extradinary students, most students just memorize the steps instructed by teachers,ths kind of education will sacrifice all other kinds of capacity such as imagination, creativity, humanity...


Remember what Chinese lack most is "innovation" , even chinese students gained 100 in test, so what?

oldwolf

A quick answer to other quesitons:

The maximum number of students who failed 4 questions is: 15


The maximum number of students who failed 5 questions is: 9

Post by oldwolf
Here is a simple algorithm to solve it.


The persions who didn't answer correctly the first question: 19
second : 9
third : 15
fourth: 21
fifth: 26


Sort the above numbers:

9
15
19
21
26

Take the last 3 numbers: 19 21 26.

The minimum of the last 3 numbers is the answer: 19.


This algorithm can be applied to other questions like 1000 students answer 50 questions and can be quickly and easily solved in a computer.


This question is a good one to test students studying computer algorithms.


Oldwolf

berriuqam

发此帖纯属娱乐,和打麻将没什么区别,您认为腰酸背疼的打一晚上麻将有何意义?请大家把精力集中在解法上,题外话不要多谈了.

huck

No, this is definitely not a question for computer algorithm. It is such a simple question that a smart brain needs only one flash to solve. I dont understand why an algorithm is needed while any kids knowing division could give the answer.
BTW, I could figure out the answer in a few seconds but dont understand the basis of sorting in your answer? It is funny!

oldwolf

Here is why it should be 70.


The total wrong answers is

9+15+19+21+26 = 90

The maximum number of studnets who failed 3 questions could be 90 / 3 = 30

In this case, it is possible that 30 students failed the exam. The following is how the distribution of wrong answers are.


wrong
answers

9 =   5    0    0       4     0
15= 0    6    0       4     5
19= 0    0    10     4     5
21= 5    6    10     0     0
26= 5    6    10     0     5

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


Therefore, the maxium number of failed students is 30. So at least 100 - 30 = 70 students pass the exam.


But just in this case, the maximum number of failed students ncould be 30. If the distribution of failed question changes, it is possible the maximum number clouldn't reach 30.

Post by huck
No, this is definitely not a question for computer algorithm. It is such a simple question that a smart brain needs only one flash to solve. I dont understand why an algorithm is needed while any kids knowing division could give the answer.
BTW, I could figure out the answer in a few seconds but dont understand the basis of sorting in your answer? It is funny!

oldwolf

Here is why it should be 70.


The total wrong answers is

9+15+19+21+26 = 90

The maximum number of studnets who failed 3 questions could be 90 / 3 = 30

In this case, it is possible that 30 students failed the exam. The following is how the distribution of wrong answers are.


wrong
answers

9 =    5    0    0       4     0
15= 0    6    0       4     5
19= 0    0    10     4     5
21= 5    6    10     0     0
26= 5    6    10     0     5

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


Therefore, the maxium number of failed students is 30. So at least 100 - 30 = 70 students pass the exam.


But just in this case, the maximum number of failed students ncould be 30. If the distribution of failed question changes, it is possible the maximum number clouldn't reach 30.

Post by huck
No, this is definitely not a question for computer algorithm. It is such a simple question that a smart brain needs only one flash to solve. I dont understand why an algorithm is needed while any kids knowing division could give the answer.
BTW, I could figure out the answer in a few seconds but dont understand the basis of sorting in your answer? It is funny!

oldwolf

Here is why it should be 70.


The total wrong answers is

9+15+19+21+26 = 90

The maximum number of studnets who failed 3 questions could be 90 / 3 = 30

In this case, it is possible that 30 students failed the exam. The following is how the distribution of wrong answers are.


wrong
answers

9 =     5    0    0       4     0
15= 0    6    0       4     5
19= 0    0    10     4     5
21= 5    6    10     0     0
26= 5    6    10     0     5

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


Therefore, the maxium number of failed students is 30. So at least 100 - 30 = 70 students pass the exam.


But just in this case, the maximum number of failed students ncould be 30. If the distribution of failed question changes, it is possible the maximum number clouldn't reach 30.

Post by huck
No, this is definitely not a question for computer algorithm. It is such a simple question that a smart brain needs only one flash to solve. I dont understand why an algorithm is needed while any kids knowing division could give the answer.
BTW, I could figure out the answer in a few seconds but dont understand the basis of sorting in your answer? It is funny!

oldwolf

Here is why it should be 70.


The total wrong answers is

9+15+19+21+26 = 90

The maximum number of studnets who failed 3 questions could be 90 / 3 = 30

In this case, it is possible that 30 students failed the exam. The following is how the distribution of wrong answers are.


wrong
answers

9 =      5    0    0       4     0
15= 0    6    0       4     5
19= 0    0    10     4     5
21= 5    6    10     0     0
26= 5    6    10     0     5

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


Therefore, the maxium number of failed students is 30. So at least 100 - 30 = 70 students pass the exam.


But just in this case, the maximum number of failed students ncould be 30. If the distribution of failed question changes, it is possible the maximum number clouldn't reach 30.

Post by huck
No, this is definitely not a question for computer algorithm. It is such a simple question that a smart brain needs only one flash to solve. I dont understand why an algorithm is needed while any kids knowing division could give the answer.
BTW, I could figure out the answer in a few seconds but dont understand the basis of sorting in your answer? It is funny!