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I am creating scoreboards for a test that has 3 sections (40, 50 and 60 questions).

Correct answers count for 1 point, meaning the highest scores are 40, 50 and 60 respectively.

Skipped answers count for 0 points.

Wrong answers have a "penalty" of -0.25.

I would like to create a list of all the possible scores that someone could get on each section. For the 40 question section, 40/40 is possible, while 39.75/40 is not (since if you got 39 correct and 1 wrong, you'd actually be at 38.75).

I am not sure whether to be thinking of skipped questions as "plus zero" or as "minus one", and similarly I am not sure whether to think of wrong answers as "minus 0.25" or "minus 1.25". I believe both ideas work but am I confusing myself with which direction to think as I try to make a list of all possible scores.

I tried to make a matrix where x = the number of correct answers, y = the number of wrong answers, z = the number of skipped answers and S = your final score.

This is what I came up with:

x - 0.25y = S x + y + z = 40 (or 50, or 60, depending on the section) x, y and z are integers between 0 and 40 (or 50, or 60, depending on the section)

Solving this matrix yields

solved matrix

I don't know where to go from here. I tried to solve the list of all possible scores manually but am getting quickly confused and would like a method to how to be sure I'm getting all the possible scores and eliminated all the impossible ones.

Thanks a lot!

Lorncat
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  • There's probably a generating function approach to this but I don't know how to set it up. This setup here has a free variable. – Sean Roberson Aug 16 '22 at 00:06
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    I advise starting with perfection, and then regarding skipped questions as $(-1)$ and missed questions as $(-1.25)$. The challenge is that $(5)$ skipped questions equals $(4)$ missed ones. – user2661923 Aug 16 '22 at 00:21
  • I could be mistaken, but it seems like the only impossible scores are ${39.75, 39.5, 39.25, 38.5, 38.25, 37.25}.$ It seems like all other scores, between $(-10)$ and $(40)$ inclusive, in increments of $(0.25)$ are possible. So, the computation is $$(50 \times 4) + 1 - (3 + 2 + 1).$$ – user2661923 Aug 16 '22 at 00:55
  • Re my last comment, the idea is that you immediately know that all of the following scores are possible ${0,1,2\cdots,40}$. Below $(0)$, you deduct $(-.25)$ for each question that changes from skipped to missed. Further, you know that each of the scores $A = {36, 36.25, 36.5,36.75}$ are possible. Then, I could be mistaken, but it seems that if $x\in A$ is a possible score, then so is $(x-1), (x-2), \cdots.$ Again, I could be mistaken about this. – user2661923 Aug 16 '22 at 01:04

2 Answers2

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Let $C$ be the number of correct answers, $W$ the number of wrong answers, $Q$ the total number of questions in the section, and $S$ the final score multiplied by 4 (just to keep everything in integers, it doesn't really affect anything). Then:

  1. $C, W \geq 0$
  2. $C + W \leq Q$
  3. $S = 4C - W$

Applying some fairly loose limits, we can quickly see that $-Q \leq S \leq 4Q$ is the only range where we might be able to achieve solutions.

Then we can note that with $C = 0$ you get $S = -W$ and $0 \leq W \leq Q$, and so you can achieve every possible score $S = -Q, \ldots, 0$. For positive scores, let $W = 4A + B$, where $0 \leq B \leq 3$. Then we get that $S = 4C - W = 4C - (4A + B) = 4(C - A) - B \leq 4C - B$. Notice that then if $A \neq 0$, we can replace $W$ with $B$ and $C$ with $C - A$ to achieve the same score, and so if we can't achieve a particular score by setting $A = 0$ then we can't achieve it at all. (There's a little bit of algebra hidden in all of that that I leave as an exercise.)

So then the valid positive scores are the ones that we can achieve when $W = 0, 1, 2, 3$, and $C = 1, 2, \ldots, Q - W$. Running through each of those gives:

$W$ $C$ $S$
$0$ $1, 2, \ldots, Q$ $4, 8, \ldots, 4Q$
$1$ $1, 2, \ldots, Q - 1$ $3, 7, \ldots, 4Q - 5$
$2$ $1, 2, \ldots, Q - 2$ $2, 6, \ldots, 4Q - 10$
$3$ $1, 2, \ldots, Q - 3$ $1, 5, \ldots, 4Q - 15$

In other words, in addition to every non-positive score from $-Q$ to $0$, we can achieve every positive score from $1$ to $4Q$ with 6 exceptions: $4Q - 11$, $4Q - 7$, $4Q - 6$, $4Q - 3$, $4Q - 2$, and $4Q - 1$.

ConMan
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Hint： $$40-((1-0)x+(1-(-0.25))y)$$ where $x, y$ are nonnegative integers and $$x+y \le 40$$

miracle173
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Li Jiarui
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