Interactive tool
Find the correct and wrong answer combinations that reach a marks target
In short
Enter the score you are aiming for and the marking scheme. The calculator returns the minimum correct answers needed with no wrong answers, the largest number of wrong answers that still works, and a table of feasible combinations.
Choose a preset or enter your own marks and penalty values. A target above the theoretical maximum is rejected.
Exam preset
JEE Main Paper 1: 75 questions, +4 correct, −1 incorrect.
Minimum correct answers with zero wrong answers
50correct
ceil(200 ÷ 4) = 50
Maximum wrong answers still feasible
20
Theoretical maximum score
300.00
Target
200.00
Questions available
75
These are arithmetic combinations only. Reaching a marks target is not a rank, percentile or admission outcome.
| Correct | Wrong | Unattempted | Score |
|---|---|---|---|
| 50 | 0 | 25 | 200.00 |
| 51 | 4 | 20 | 200.00 |
| 52 | 8 | 15 | 200.00 |
| 53 | 12 | 10 | 200.00 |
| 54 | 16 | 5 | 200.00 |
| 55 | 20 | 0 | 200.00 |
Each row is the minimum number of correct answers needed for that number of wrong answers.
Formula used
The largest feasible W is found by testing each value of W from 0 upwards, not by a shortcut assumption.
With no wrong answers, the minimum correct answers needed is ceil(T ÷ P), where T is the target and P is the marks per correct answer.
For any chosen number of wrong answers W, the minimum correct needed becomes ceil((T + (N × W)) ÷ P), where N is the penalty per wrong answer. A combination only counts if minimumCorrect + W fits inside the total question count Q.
1. Pick the exam preset so the question count and marking rule are filled in.
2. Enter your target score.
3. Read the zero-wrong minimum and the maximum feasible wrong answers.
4. Use the scenario table to see how the correct count rises as wrong answers rise.
With 75 questions at +4 and −1, zero wrong answers need ceil(200 ÷ 4) = 50 correct. With 4 wrong answers you need 51 correct, because 51 × 4 − 4 = 200.
With 20 wrong answers you need 55 correct, and 55 + 20 = 75 exactly fills the paper. With 21 wrong answers you would need 56 correct, and 56 + 21 = 77 does not fit, so the maximum feasible wrong count is 20.
Each row of the table is the least number of correct answers that reaches your target for that number of wrong answers. Doing better than a row is always possible; doing worse misses the target.
The maximum feasible wrong count is a arithmetic limit imposed by the paper size, not a recommendation about how much to guess.
It assumes one uniform marking rule across the paper, which is true for the JEE Main, NEET and NDA presets but not for JEE Advanced. It assumes whole-number question counts and rounds correct answers up, because part of a question cannot be answered.
Built Rank Sarthi pages that go with this calculation.
Preset question counts and marking rules come from official NTA and UPSC documents.
