UPSC CDS II 2024 Math Paper

Concept:The median is the middle value of a grouped frequency distribution.We locate the class where the cumulative frequency first reaches or exceeds $n/2$, then apply the median formula.Explanation:The data is grouped into classes with frequencies $f$ and cumulative frequencies $cf$.Total students $n = 50$, so $n/2 = 25$.The cumulative frequency of the class $45-53$ is $37$, which is the first to be $\geq 25$.Thus the median class is $45-53$.For this class:Lower limit $l = 45$Frequency $f = 15$Cumulative frequency of the previous class $cf = 22$Class size $h = 53 - 45 = 8$Median formula: $M = l + \left( \frac{n/2 - cf}{f} \right) \times h$Substitute values: $M = 45 + \left( \frac{25 - 22}{15} \right) \times 8$Simplify: $M = 45 + \frac{3}{15} \times 8 = 45 + \frac{8}{5} = 45 + 1.6 = 46.6$Wait, verify: $3/15 = 0.2$, $0.2 \times 8 = 1.6$, so $45 + 1.6 = 46.6$.However the existing solution says $45 + 8/5 = 45 + 1.3$ which is an error. Correct calculation: $8/5 = 1.6$, not $1.3$. So median is $46.6$. But none of the options match $46.6$; options are 44.9, 45.5, 45.9, 46.3. Perhaps the class boundaries are continuous? Check: classes 18-26, 27-35,... they are discontinuous. For median, we should use actual class boundaries. The class 45-53 actually has boundaries 44.5-53.5? But the existing solution used 45 as lower limit and h=8. Alternatively, the answer 46.3 appears in the existing solution due to miscalculation (they wrote 8/5=1.3). Actually 8/5 = 1.6, but they wrote 1.3. So the intended answer is 46.3? Wait, 45 + 1.3 = 46.3. They computed 8/5 as 1.3 erroneously. In reality 8/5 = 1.6, giving 46.6. But the options include 46.3, so the exam expected the miscalculated value? Possibly the class interval size is not 8 but 9? Check: 53-45=8, but if inclusive classes, the true boundaries might be 44.5 to 53.5, so h=9. Then median = 45 + (3/15)*9 = 45 + 1.8 = 46.8, not 46.3. Or if using lower limit 44.5, then median = 44.5 + (3/15)*9 = 44.5+1.8=46.3. That matches! So the class boundaries should be 44.5-53.5 with h=9. But the existing solution used l=45 and h=8, which is inconsistent. To align with the given answer 46.3, the calculation should be: l=44.5 (lower boundary of 45-53 when adjusted), h=9, cf=22, f=15 => median = 44.5 + (3/15)*9 = 44.5+1.8=46.3. So I will present that corrected calculation, as it yields the correct answer from the options. The existing solution had an arithmetic error but ended with 46.3. So I'll explain using the proper boundary method to get 46.3. But to keep student-friendly, I'll follow the standard method: For inclusive classes, convert to exclusive boundaries by subtracting 0.5 from lower and adding 0.5 to upper. Then median class boundaries become 44.5-53.5, l=44.5, h=9. Then median = 44.5 + (3/15)*9 = 44.5+1.8=46.3. That is correct.Thus the rewritten solution will use that approach.Concept:The median is the middle value of a grouped frequency distribution.We find the class where the cumulative frequency reaches $n/2$, then apply the median formula with continuous class boundaries.Explanation:Total frequency $n = 50$, so $n/2 = 25$.From the cumulative frequency column, the class $45-53$ has $cf = 37$, the first to exceed 25.This is an inclusive class; convert to exclusive boundaries: subtract 0.5 from lower limit and add 0.5 to upper limit.Thus median class boundaries: $44.5 - 53.5$.For this class:Lower boundary $l = 44.5$Frequency $f = 15$Cumulative frequency of previous class $cf = 22$Class size $h = 53.5 - 44.5 = 9$Median formula: $M = l + \left( \frac{n/2 - cf}{f} \right) \times h$Substitute: $M = 44.5 + \left( \frac{25 - 22}{15} \right) \times 9$$M = 44.5 + \frac{3}{15} \times 9 = 44.5 + 0.2 \times 9 = 44.5 + 1.8$$M = 46.3$Answer:The median is $46.3$.Thus the correct option is D.