Negative Marking and the Attempt Decision
A candidate's C-CAT mock has three sections, A, B, and C, each with 50 questions. Marking is +3 for a correct answer, -1 for a wrong answer, and 0 for an unattempted question. Sectional qualifying marks are 60 out of 150 in each section, and the overall qualifying mark is 225 out of 450. In the mock the candidate attempted 40 questions in A with 32 correct, 35 in B with 21 correct, and 15 in C with 6 correct. The candidate has 90 minutes in the real test and currently averages 1.2 minutes per question in A, 1.5 in B, and 2.0 in C. A coach proposes that the candidate spend the remaining practice time attempting 20 more C questions at the current C accuracy. The candidate wants to know whether that plan can reach the C cutoff and what accuracy would actually be required.
1. Compute the candidate's net score in each section and overall. Also compute the percentage of the maximum marks.
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Net = 3C - W, where C is correct and W is wrong; unattempted questions add 0.
- A: attempted 40, correct 32, wrong 8. Net = 96 - 8 = 88 out of 150 = 58.7%.
- B: attempted 35, correct 21, wrong 14. Net = 63 - 14 = 49 out of 150 = 32.7%.
- C: attempted 15, correct 6, wrong 9. Net = 18 - 9 = 9 out of 150 = 6.0%.
- Overall: attempted 90, correct 59, wrong 31. Net = 177 - 31 = 146 out of 450 = 32.4%.
Only A clears the 60-mark sectional cutoff; B and C do not. The overall score is far below 225.
2. Evaluate the coach's plan: attempt 20 more C questions at 40% accuracy. Can it reach the 60-mark C cutoff? What accuracy on all 50 C questions would be needed?
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For one C question with accuracy p, expected net = 3p - 1(1-p) = 4p - 1. At p=0.40, expected net per question is 0.6 marks. Twenty more questions add about 12 marks, taking C from 9 to about 21, still 39 short of 60. Even attempting all 50 at 40% yields 50(4*0.40 - 1) = 50(1.6 - 1) = 30 marks, only half the cutoff. To reach 60 by attempting all 50: 50(4p - 1) >= 60, so 4p - 1 >= 1.2, giving p >= 0.55. Thus 55% accuracy on all 50 C questions is the minimum; the coach's plan is insufficient by a wide margin.
3. Suppose C accuracy rises to 55% and all 50 are attempted, while A and B remain exactly as in the mock. Compute the new overall net and percentage. Is the overall cutoff met?
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C net = 50(4*0.55 - 1) = 50(2.2 - 1) = 50(1.2) = 60. A=88 and B=49 remain unchanged. Overall = 88 + 49 + 60 = 197 out of 450 = 43.8%. The overall cutoff is 225 (50%), so 197 is 28 marks short. Meeting the C sectional cutoff does not guarantee overall qualification; A's 88 and B's 49 leave too much work. The candidate would need 28 more net marks elsewhere, for example about 9-10 more correct answers with no extra wrong answers, or a combination of higher accuracy and more attempts in B.