Directions: Study the table carefully and answer the given question: | Publishing Houses | Number of Books Published | Ratio of Academic and Non-academic Books | Percentage of Books distributed | Number of distributors in publishing house | | :--- | :--- | :--- | :--- | :--- | | M | 28200 | 7 : 3 | 81 | 17 | | N | 32200 | 5 : 9 | 74 | 23 | | O | 29700 | 6 : 5 | 92 | 18 | | P | 31200 | 8 : 5 | 86 | 24 | | Q | 33800 | 7 : 6 | 79 | 25 | | R | 35700 | 11 : 6 | 82 | 21 | | S | 37800 | 5 : 13 | 89 | 24 | What is the average number of non-academic books published by publishers R and S?
Aptitude
Statistics
Difficulty: Medium
Choose an option
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A18750
-
B18850
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C19950
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D18950
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E19990
Answer
Correct Answer: 19950
Explanation
### Concept & Weighted Averages
Calculate the specific volume of non-academic books for both publishers using their respective ratios, then compute their arithmetic mean.
$$ \text{Average} = \frac{\text{Value 1} + \text{Value 2}}{2} $$
### Step-by-Step Solution
1. **Calculate Non-academic Books for R:**
- Total Books = $35700$
- Ratio (Academic : Non-academic) = $11 : 6$. Total parts = $17$.
- Non-academic = $35700 \times \left(\frac{6}{17}\right) = 2100 \times 6 = 12600$.
2. **Calculate Non-academic Books for S:**
- Total Books = $37800$
- Ratio (Academic : Non-academic) = $5 : 13$. Total parts = $18$.
- Non-academic = $37800 \times \left(\frac{13}{18}\right) = 2100 \times 13 = 27300$.
3. **Calculate the Average:**
- $\text{Average} = \frac{12600 + 27300}{2} = \frac{39900}{2} = 19950$.
### Exam Strategy & Shortcut
Notice the common factor when dividing by the ratio parts: $35700 / 17 = 2100$ and $37800 / 18 = 2100$. This makes calculation rapid. You can just sum the active ratio parts multiplied by this common factor: $\frac{(2100 \times 6) + (2100 \times 13)}{2} = \frac{2100 \times 19}{2} = 1050 \times 19 = 19950$.
### Common Pitfall
Using the academic portion of the ratio instead of the non-academic portion. Always map the requested category to its correct ratio position (the second number in this case).
### Final Answer
Therefore, the correct answer is **19950**.