In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored…

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?
  1. 16
  2. 17
  3. 23
  4. 35

Solution

Total marks = 400, so 99% means 396 marks. We need integer scores $p, q, r, s$ with each between 0 and 100 and $p+q+r+s = 396$. Letting $p' = 100-p$ etc., $p'+q'+r'+s' = 4$ with each $0 \le p' \le 100$. The number of non-negative integer solutions is $\binom{4+3}{3} = \binom{7}{3} = 35$. So n = 35.

Asked in: CSAT 2023

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