2023 CSAT, all questions
UPSC Prelims 2023 · CSAT
Question 65: Basic Numeracy
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 ?
a16
b17
c23
d35
AnswerUPSC’s answer is (d).
99% of the total 4 x 100 = 400 marks is 396. So 4 marks are lost across the four papers, and no paper can lose more than 100. The number of ways to spread 4 lost marks over 4 papers is the number of non-negative solutions of x1 + x2 + x3 + x4 = 4, which is 7C3 = 35.
Why each option is right or wrong
- (a) Wrong: far too few; the count is 7C3 = 35.
- (b) Wrong: far too few; the count is 7C3 = 35.
- (c) Wrong: the count is 7C3 = 35.
- (d) Right: 4 lost marks over 4 papers gives 7C3 = 35.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2023, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.