Is there any formula for finding the smallest no. of chapters needed to be learnt, based on the number of questions? I am a junior.
I understand that this is a highly unconventional and specific question, so bear with me. Also, this is my first time using the site, so be a little lenient with the downvotes.
I want to know if there is a formula to find the amount of chapters you will need to study for an exam, based on the total no. of chapters, questions asked and questions needed to be solved
Such that no matter which chapters I study, as long as I study the required amount, I will be able to solve at least the amount of questions needed to be solved.
So in a history exam, there are 10 chapters, and of those chapters, they can ask one question on each chapter. But in the paper, only 7 questions will be asked. Out of which we have to do any 5. What is the minimum number of chapters I have to learn, irrespective of what those chapters are, so that I can always answer 5 out of the 7 (assuming they are impartial for asking each question).
I'm just doing this for fun, as an emergency, in the event that I have to. It is very undesirable. And I wanna make that clear. Obviously I should study all chapters that I have, that's common sense. But JUST IN CASE I need to, I'd like to have a formula handy.
In this supposed formula, you input the total number of questions / chapters, out of the total the questions that they will ask, and questions needed to be solved out of the asked questions.
I am just using this method as an emergency IN CASE I need it.
Please excuse my informality and the strange nature of my question.
 A: Well, let's see. First let's solve your particular case, then generalize it for any arbitrary number of questions/chapters.
Suppose you have learned $x$ chapters. Then there are $10-x$ chapters which you did not learn.
As a malicious examiner, to prevent you from answering 5 questions, I should include all of these $10-x$ chapters into the test. And, because there are 7 questions in total, I must include $7-(10-x)=x-3$ more chapters.
But you have learned all the remaining chapters. My desire is that you cannot answer any five of them, and to achieve it, I should not include 5 or more. So I win if $x-3 < 5$ (the number of remaining questions which I must include, all of which you have learned, must be less than number of questions you wish to answer). And I lose if $x-3 \geqslant 5$, because in this case you will definitely answer some 5 questions. This is equivalent to $x \geqslant 5+3=8$.
So, in your particular case, the answer is 8.
Let's now assume that there are $n$ chapters in total, and there are $k$ questions in the test, and to pass the test you must answer at least $t$ questions. Obviously $t \leqslant k \leqslant n$.
Suppose again you have learned $x$ chapters. So, repeating the reasoning from above, being an angry examiner I should include all of $n-x$ chapters that you didn't learn, and after that I need to include $k-(n-x)=x+k-n$ more questions. But you learned all of the remaining questions.
You wish to answer $t$ questions, so, I win if $x+k-n<t$. And I lose if $x+k-n \geqslant t$, which is equivalent to $x \geqslant t-k+n$. Thus, the least $x$ for you to win is $t-k+n$.
In your particular case it's $5-7+10=8$, as we have seen above.
