UPSC Prelims 2023·CSAT·Quantitative Aptitude·Combinatorics and Probability

In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored by the students are in integers. A student can score 99% in n different ways. What is the value of n?

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Last updated 23 May 2026, 3:31 pm IST
  1. A16
  2. B17
  3. C23
  4. D35Correct

Explanation

The maximum marks for each of the four papers (P, Q, R, S) are 100. Total maximum marks for the examination = 4 papers * 100 marks/paper = 400 marks. To score 99% in the examination, the student needs to obtain: 99% of 400 marks = (99/100) * 400 = 396 marks. Let the marks scored in the four papers be p, q, r, and s, respectively. We know that 0 <= p, q, r, s <= 100 (marks are integers and cannot exceed the maximum). The sum of the marks must be 396: p + q + r + s = 396. Instead of directly solving this equation with upper bounds, it's simpler to consider the "marks lost" from the maximum possible score. The maximum possible score is 100 + 100 + 100 + 100 = 400. The student scored 396 marks. So, the total marks lost by the student across all four papers = 400 - 396 = 4 marks. Let x1 be the marks lost in paper P (i.e., 100 - p). Let x2 be the marks lost in paper Q (i.e., 100 - q). Let x3 be the marks lost in paper R (i.e., 100 - r). Let x4 be the marks lost in paper S (i.e., 100 - s). Since 0 <= p, q, r, s <= 100, it implies that 0 <= x1, x2, x3, x4 <= 100. The sum of the marks lost is: x1 + x2 + x3 + x4 = (100 - p) + (100 - q) + (100 - r) + (100 - s) x1 + x2 + x3 + x4 = 400 - (p + q + r + s) x1 + x2 + x3 + x4 = 400 - 396 x1 + x2 + x3 + x4 = 4 We need to find the number of non-negative integer solutions to this equation. The upper bound constraint (x_i <= 100) is automatically satisfied because the sum is only 4, so no individual x_i can exceed 4. This is a classic stars and bars problem. The number of non-negative integer solutions to x1 + x2 + ... + xk = n is given by the formula (n + k - 1) C (k - 1). Here, n = 4 (the sum of lost marks) and k = 4 (the number of papers/variables). Number of ways = (4 + 4 - 1) C (4 - 1) = 7 C 3 = (7 * 6 * 5) / (3 * 2 * 1) = 35 Therefore, a student can score 99% in 35 different ways. The final answer is D.
Quantitative Aptitude: In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored by the stude

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