The cost of 21 tables and 35 chairs is ₹ 41825. What is the cost of 9 tables and 15 chairs?

### Concept & Logic\n\nThis problem can be efficiently solved using the concept of ratios and scaling. If the quantities of both items are reduced by the same common factor, their total cost is reduced by that same factor.\n\n### Step-by-Step Solution\n\n* **Given:** \n Cost of 21 tables + 35 chairs = ₹ 41825\n\n* **Deduction:** \n Let $T$ be the cost of one table and $C$ be the cost of one chair.\n $$21T + 35C = 41825$$\n\n Notice that 21 and 35 have a greatest common divisor (GCD) of 7. Let's divide the entire equation by 7:\n $$3T + 5C = \\frac{41825}{7} = 5975$$\n\n We need to find the cost of 9 tables and 15 chairs, which is equivalent to $9T + 15C$. \n Notice that $9T + 15C$ is exactly 3 times the simplified equation:\n $$9T + 15C = 3 \\times (3T + 5C)$$\n\n* **Calculation:**\n $$3 \\times 5975 = 17925$$\n\n### Exam Strategy & Shortcut\n\nRecognize the direct proportionality between the given quantities and the required quantities:\n$$\\text{Ratio} = \\frac{9}{21} = \\frac{15}{35} = \\frac{3}{7}$$\nSimply multiply the total initial cost by $\\frac{3}{7}$:\n$$\\text{Required Cost} = 41825 \\times \\frac{3}{7} = 5975 \\times 3 = 17925$$\nThis avoids trying to find individual values for $T$ and $C$, which is impossible with only one equation.\n\n### Common Pitfall\n\nStudents often waste time attempting to set up a system of linear equations to solve for individual prices of a table and a chair. Since there is only one equation with two variables, the individual values cannot be determined, leading to confusion or selecting "None of these". Always look for a common scalar multiple first.\n\n### Final Answer\n\n**Therefore, the correct answer is ₹ 17925.**