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CAT 2017 paper had 100 questions. Each correct answer was awarded 3 marks and each wrong answer was penalized 1 negative mark. If certain number of aspirants attempted different number of questions but each one of them obtained the same overall score of 56, what is the largest possible size of this group of people?

(A) 21
(B) 16
(C) 19
(D) 14
(E) 25

The combinations can be 1.19 correct and 1 incorrect 2. 20 Correct and 4 incorrect so on 39 correct and 61 incorrect

No. of combinations = 39-19+1= 21

Option A

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How did you get the entire range? I had to check for it and see for pairs.

3x-y =56
x+y=100
Minimum value of x is 19 and maximum 39 as y can take 61 in that case. If x is greater than 40, 3x-y will always be greater than 56

x will take all integer values from 19 to 39. We have to count the numbers from 19 to 39
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Minimum is 19 correct, 1 incorrect, which is 20 questions for 56 points.

Each additional correct needs 3 incorrect for same score, which is 4 questions attempted.

How many sets of 4 between 20 and 100 ? 80/4=20.

Plus the first one =21

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