How many students advanced to the next round?

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In summary, there were 100 students participating in a math contest with four questions. 90 students got the first question correct, 85 got the second question correct, 60 got the third question correct, and 40 got the fourth question correct. At least 38 students advanced to the next round by getting three or more questions right. The largest number of students who could miss at least two questions is 62. There is no general method for finding the optimal solution, but it is helpful to find upper and lower bounds and consider both passing and failing scenarios. In this case, there are 125 wrong answers, so the optimal solution is for 38 students to advance by getting 4, 4, 2, and 2
  • #1
jb235711
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Can someone give me general guidance on how to solve this problem (it's not homework):
There were 100 students taking a math contest consisting of four questions. 90 of them got problem 1 correct. 85 of them got problem 2 correct. 60 of them got problem 3 correct. 40 of them got problem 4 correct. Anyone who got 3 or more problems correct advanced to the next round. At least how many students advanced to the next round?
 
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  • #2
What's the largest number who can miss at least two?
 
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  • #3
There is no general method for finding the optimal solution to problems of this type, however a good start is often to find an upper and/or lower bound. It is also usually a good idea to look at both sides of the data - the question is phrased in terms of passing by getting at least 3 questions right; try looking at the number that fail by getting 2 questions wrong. How many questions are answered wrong overall?
 
  • #4
Thanks for your helps. I got it. There are 275 right answers. Each student gets two right answers. The remaining 75 are distributed so that the number that pass is minimized, 2 question to 37 students and the remaining question to the 38th student.
 
  • #5
jb235711 said:
Thanks for your helps. I got it. There are 275 right answers. Each student gets two right answers. The remaining 75 are distributed so that the number that pass is minimized, 2 question to 37 students and the remaining question to the 38th student.

Is this a solution, have you checked that it satisfies the constraints?

I was definitely thinking there wasn't an easy way to solve this.
 
  • #6
I totally thought there was no easy solution and started doing inclusion-exclusion and stuff, but apparently it is easy. And I double checked that there exists a distribution satisfying 38 students advancing i.e 37 students get 4 right, 1 student gets 3 right and 62 students get 2 right. You can also do 38 students get 4 right, 61 students get 2 right, and 1 student gets 1 right which doesn't change the number of students that advance. From these arrangements you can verify case by case e.g. taking a right answer from a student that has four correct and giving it to a student that has 2 correct will increase the number advancing, and so forth.
 
  • #7
To prove that this is an optimal solution you need to answer my question - how many questions are answered incorrectly?
 
  • #8
... from which you can get the answer to Bystander's question
 
  • #9
Deleted
 
  • #10
I'm bowing out of this question. When I look at it, I see a problem with 16 variables and 5 constraints, and as much as I try to convince myself that it is actually is easy, something niggles at me saying, nah, it's not.

I agree with the analysis that says 38 is a lower bound. I still struggle to show that it is a solution. I'm good at puzzles but this has me stumped. So best of luck to those who can solve it.
 
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  • #11
MrAnchovy said:
To prove that this is an optimal solution you need to answer my question - how many questions are answered incorrectly?
Yes, there are 125 wrong answers, so the most you can fail is 62, so the least that can advance is 38.
 
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FAQ: How many students advanced to the next round?

What is the inclusion-exclusion problem?

The inclusion-exclusion problem is a mathematical counting technique used to find the number of elements that belong to at least one of several sets. It is often used in combinatorics and probability to solve counting problems.

How do you solve an inclusion-exclusion problem?

To solve an inclusion-exclusion problem, you first need to identify all the sets involved. Then, you use the principle of inclusion-exclusion, which states that the total number of elements in all sets is equal to the sum of the number of elements in each individual set, minus the number of elements in the intersection of every pair of sets, plus the number of elements in the intersection of every trio of sets, and so on. This process continues until all possible intersections have been considered.

What is the principle of inclusion-exclusion?

The principle of inclusion-exclusion is a counting technique that allows us to find the number of elements in the union of several sets. It states that the total number of elements in all sets is equal to the sum of the number of elements in each individual set, minus the number of elements in the intersection of every pair of sets, plus the number of elements in the intersection of every trio of sets, and so on.

When is the inclusion-exclusion principle used?

The inclusion-exclusion principle is used when we need to find the number of elements in the union of several sets. It is often applied in combinatorics and probability to solve counting problems such as finding the number of ways to arrange objects or the probability of certain events occurring.

Can the inclusion-exclusion principle be applied to more than two sets?

Yes, the inclusion-exclusion principle can be extended to any number of sets. The formula remains the same, with the number of terms increasing as the number of sets increases. For example, for three sets A, B, and C, the formula would be A+B+C-(A∩B)-(A∩C)-(B∩C)+(A∩B∩C).

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