A student took five papers in an examination, where full marks were same on each paper. Her marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all these papers together, the candidate obtained 60% of the total marks. Then the number of papers in which she got more than 50% marks is

### Concept & Proportional Thresholds To compare proportional marks to a percentage threshold, we must establish a relationship between the "ratio parts" of the marks obtained and the "maximum marks" per paper. $$\text{Threshold} = \text{Percentage} \times \text{Maximum Marks per Paper}$$ ### Step-by-Step Solution 1. **Define Obtained Marks:** Let the marks obtained in the five papers be $6x, 7x, 8x, 9x,$ and $10x$. 2. **Calculate Total Obtained Marks:** Total marks obtained = $6x + 7x + 8x + 9x + 10x = 40x$. 3. **Define Maximum Marks:** Let the maximum marks for each single paper be $M$. Total maximum marks for all 5 papers = $5M$. 4. **Equate Using Given Percentage:** The total marks obtained ($40x$) is $60\%$ of the total maximum marks ($5M$). $40x = 0.60 \times 5M$ $40x = 3M$ $M = \frac{40x}{3} \approx 13.33x$. 5. **Determine the 50% Threshold:** We need to find papers where marks are greater than $50\%$ of $M$. $50\%$ of $M = 0.5 \times \frac{40x}{3} = \frac{20x}{3} \approx 6.67x$. 6. **Compare Marks to Threshold:** We check which papers have marks $> 6.67x$: - Paper 1: $6x$ (Less) - Paper 2: $7x$ (More) - Paper 3: $8x$ (More) - Paper 4: $9x$ (More) - Paper 5: $10x$ (More) She scored more than $50\%$ in 4 papers. ### Exam Strategy & Shortcut Ignore variables. Assume the sum of the ratio parts (40) is exactly the total marks obtained. Since 40 represents $60\%$ of the grand total, the grand total is $\frac{40}{0.6} = 66.67$. Since there are 5 identical papers, the maximum marks per paper is $\frac{66.67}{5} = 13.33$. $50\%$ passing threshold per paper = $\frac{13.33}{2} = 6.67$. Looking at the ratio numbers ($6, 7, 8, 9, 10$), the numbers $7, 8, 9, 10$ are greater than $6.67$. Thus, 4 papers. ### Common Pitfall Attempting to calculate the average of the ratio numbers directly without tying them to the maximum possible marks, leading to arbitrary and incorrect threshold comparisons. ### Final Answer Therefore, the correct answer is **4**.