Guessing JEE score

I first checked if any of the scores are outside the possible range. Since maximum score is $90 * 4 = 360$ , none did.

I then wrote down problem statement as a system of two equations: if $a$ is number of correct answers, $b$ is number of wrong, and $c$ not answered, then $\begin{cases} 4*a - b = score \\ a + b + c = 90 \end{cases}$

The problem is in finding $a$ , $b$ , $c$ that do not satisfy this system.

Most exam scores can be constructed by taking out 1 to 3 wrong answers out of $4*a$ and padding the rest with not answered, e.g.

$67 = 4 * 17 - 1$ where $a = 17, b =1$ and $c = 90 - 17 - 1 = 72$

So if any numbers do not satisfy the system, they must be for exam scores near maximum, where there is not enough not answered questions to pad with.

Taking 345: $\begin{cases} 4*a - b = 345 \\ a + b + c = 90 \end{cases} <=>$ $\begin{cases} 4*a - 90 + a + c = 345 \\ b = 90 - a - c \end{cases} <=>$ $\begin{cases} 5*a + c = 435 \\ b = 90 - a - c \end{cases}$

Since 435 is divisible by 5, we assume $c=0$ . Then $a=435/5 = 87, b = 3$ . This means score 345 has $a, b, c$ , satisfying the system.

Taking 349, we can reuse an earlier result: $a=87, b=3, c=0, score = 345$ If 349 is valid, it's $a, b, c$ must be close to the ones above.

With a little experimentation, varying a, b, and c shows that there are none to satisfy the system for $score = 349$ .

The answer is $\boxed{349 (C)}$ .

Since there is one mark deducted for every wrong answer all the lower scores are possible.(Just take the score to the nearest multiple of 4 and cut 1 mark for every wrong answer) So v will see the scores which are close to the maximum marks A-345 is possible if one answers 87 correctly and 3 incorrectly D- this is possible in more than one way if one gets 312 and loses 3 or gets 316 and loses 7 C- this is not possible because if one gets 349 he must have answered 88 questions or more which is not possible.