A puzzle regarding marks!

Question

In a exam there are 50 questions. +1 for correct, -1/3 for incorrect and -1/6 for unattempting. Sarah scored 32 in that exam. Find the minimum number of questions she must have done wrong. How can I approach this question?

Answer 1

We have $$c-\tfrac13i-\tfrac16u=32\ .$$ Multiply by $6$ to simplify; also note that there are $50$ questions in the exam; so $$c+i+u=50\ ,\quad 6c-2i-u=192\ .$$ Solving, $$i=7c-242\ ,\quad u=292-8c\ .$$ Now clearly $c,i,u$ must be integers; also all must be non-negative; so $$\tfrac{242}7\le c\le\tfrac{292}8$$ which gives $35\le c\le36$. Therefore we have solutions $$(c,i,u)=(35,3,12),\ (36,10,4)\,,$$ possible values of $i$ are $3,10$, and the minimum value is $3$.

Answer 2

With this type of problem, it's easy to go through the space of possible answers. We'll do this by maximizing the number of unanswered questions (since this means that she could achieve her score with fewer wrong answers).

You know that Sarah got 32 points. So she got at least 32 questions right.

That leaves 18 questions. Suppose she didn't answer any of them. That subtracts $\frac{18}{6} = 3$ points. That leaves her with $29$ points.

Let's take the next multiple of $6$: suppose she left $12$ questions unanswered, which removes $\frac{12}{6} = 2$ points. This leaves her with $30$ points, but also $6$ more questions to be accounted for. So suppose she gets 3 questions right, giving her $33$ points, and $3$ questions wrong, so she ends up with $32$ points.

So the answer is $3$.

Answer 3

Let $c$ be the number of correct answers, $i$ of the incorrect answers and $u$ of the unattempted answers. We have the two equations $c - 1/3 i - 1/6 u = 32$ and $c + i + u = 50$. Plugging the second one into the first we get $$c - 1/3 i - 1/6 \cdot (50 - c - i) = 32.$$ Solving this equation for $i$, we get $$i = 7c - 242$$

Modulo $7$, we get $$i \equiv 3 \pmod 7$$

So $i\geq 3$.

For $i=3$, it follows that $c = (i+242)/7 = 35$ and $u = 50-i-c = 12$. Since $(i,c,u) = (3,35,12)$ in fact gives the score $32$, the minimum possible number of incorrect answers is $3$.

Answer 4

For these types of questions it's best to start by setting what our variable is and how we can put the given information into some kind of equation(s).

Let's say that $x$ is the number of incorrect answers on the test and $y$ is the number of unattempted answers. Then the number of correct answers would be $50 - x - y$. (Because all the questions are either correctly answered, incorrectly answered, or unanswered).

We have also that the score is 32. To get the score, we multiply the number of each type of answer by its value/penalty and add them all up, giving us:

$$1 \times (50 - x - y) - \frac{1}{3}x - \frac{1}{6}y = 32$$
Simplifying this gives us:
$$8x + 7y = 108$$

This is an equation with 2 unknowns to which we need only integer solutions (since the number of questions can only be an integer), otherwise known as Diophantine equations. These either have infinite solutions or none, and in this case the solutions are given by: $$x = 7n + 3, y = -8n + 12, n \in \mathbb{Z}$$

Of course, we need a combination which gives values within the range of the question, and we see $n = 0$ is the smallest one which does the trick: $x = 3, y = 12,$ and $50 - (3) - (12) = 35$, so:

3 incorrect questions
12 unanswered questions, and
35 correct questions.

Answer 5

With each wrong answer she lost ${4\over3}$ points, and with each unattempted answer ${7\over 6}$ points. This leads to ${4\over3}w+{7\over6}u=18$, or $$8w+7u=108\ .$$ Write this in the form $$w=3+7(15-w-u)$$ and realize that the smallest natural $w$ making this true modulo $7$ is $w=3$. This then enforces $u=12\geq0$, and as $3+12\leq50$ we are done.