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How do we know that only 44, 43, and 40 are not possible? When we solve it case by case for different numbers of correct, incorrect, and unanswered questions, we can arrive at this conclusion. But what is the shortest approach? What is the underlying thought behind it?
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In a test consisting of 15 questions, 3 marks are awarded for a correct answer, 1 mark is deducted for an incorrect answer and no mark is awarded for a non-attempted question.

If a student attempts at least one question in the paper, what is the number of distinct scores that he can get?

Questions attempted = n; Correct = c ; Incorrect = n - c; Score = 3c - (n-c) = 4c - n
Scores = {3n, 3n - 4, 3n - 8,.... , -n}

Impossible scores = {44,43,40} : 3 scores

All Scores = number of all scores from -15 to 45 - number of Impossible scores = 61 - 3 = 58

IMO B
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Hi Csanamika,

You've verified it case by case; what you're missing is the one clean idea that makes 44, 43, and 40 fall out instantly. It's already hiding inside chetan2u's and desertEagle's posts - let me make it explicit.

Start from the top and ask "how do I step down?"

The maximum is 45 (all 15 correct). Now ask: from a fully-correct paper, what are the only ways to lower your score?

- Turn one correct answer into an unattempted one - you lose the +3, so the score drops by 3.
- Turn one correct answer into a wrong one - you lose the +3 and pick up a -1, so the score drops by 4.

There is no move that lowers your score by 1 or by 2. So every score below the max must be:

45 - (some combination of 3's and 4's).

Which drops are impossible?

Now it's a tiny sub-question: which whole numbers can't be written as a sum of 3's and 4's?

- 1 - impossible - so 44 is impossible.
- 2 - impossible - so 43 is impossible.
- 5 - impossible (3+? and 4+? both overshoot) - so 40 is impossible.
- 3, 4, 6, 7, 8, ... - all possible (6=3+3, 7=3+4, 8=4+4, and everything after keeps working) - so 42, 41, 39, 38, ... are all reachable.

That's the whole thing: the only unreachable amounts are 1, 2, and 5, which knock out exactly 44, 43, and 40. Everything from -15 up to 45 except those three is fair game - 61 - 3 = 58.

Lock the idea in

Try the same rules with just 5 questions (max = 15). Below 15 you can again only subtract 3's and 4's, so you can't get 15-1 = 14, 15-2 = 13, or 15-5 = 10 - the same three gaps at the top. Same principle, smaller numbers.

Answer: B

Csanamika
How do we know that only 44, 43, and 40 are not possible? When we solve it case by case for different numbers of correct, incorrect, and unanswered questions, we can arrive at this conclusion. But what is the shortest approach? What is the underlying thought behind it?
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