IAS 2015 — CSAT Maths

Correct Option (B)

Let 'M' represent the marks allotted for each question. Let 'N' be the total number of questions in the test.

• The candidate attempted 8 questions and secured 50% marks in each.
• The total marks obtained by the candidate from these 8 questions can be calculated as: $8 \times (0.50 \times M) = \textbf{4M}$8×(0.50×M)=4M.
• The problem states that the candidate obtained a total of 40% in the test. The total marks for the entire test would be $N \times M$N×M.
• Therefore, the marks obtained ($4M$4M) constitute 40% of the total test marks ($N \times M$N×M).
• This relationship can be expressed as: $\textbf{4M = 0.40 \times (N \times M)}$4M = 0.40 \times(N \timesM).
• Since 'M' represents marks per question and cannot be zero, we can divide both sides of the equation by 'M': $\textbf{4 = 0.40 \times N}$4 = 0.40 \timesN.
• Solving for N: $N = \frac{4}{0.40} = \textbf{10}$N=0.404​=10.
• Hence, there were 10 questions in the test.

Incorrect Options:

• If the total number of questions was 8 (Option A), it would imply the candidate attempted all questions. In this scenario, scoring 50% in each question would result in a total score of 50% for the test, which contradicts the given condition of 40%.
• If the total number of questions was 15 (Option C), the total marks for the test would be $15M$15M. The percentage obtained by the candidate would be $\frac{4M}{15M} \times 100 = \frac{4}{15} \times 100 \approx \textbf{26.67\%}$15M4M​×100=154​×100≈26.67%, which is not 40%.
• If the total number of questions was 16 (Option D), the total marks for the test would be $16M$16M. The percentage obtained by the candidate would be $\frac{4M}{16M} \times 100 = \frac{1}{4} \times 100 = \textbf{25\%}$16M4M​×100=41​×100=25%, which is not 40%.