In aptitude (Linear Equations and Word Problems), 2 tables and 3 chairs together cost ₹3,500, while 3 tables and 2 chairs together cost ₹4,000. Assuming each table has the same price and each chair has the same price, what is the cost of one table (in ₹)?

Introduction / Context:
This question is a classic linear equations word problem where item prices must be deduced from combined totals. It tests your ability to translate a real world purchasing scenario into algebraic equations and then solve simultaneous equations to find the unknown prices. Such reasoning is important for both quantitative aptitude and everyday financial decision making.

Given Data / Assumptions:

• 2 tables plus 3 chairs cost ₹3,500.
• 3 tables plus 2 chairs cost ₹4,000.
• Each table has a fixed price, say T rupees.
• Each chair has a fixed price, say C rupees.
• We must find T, the cost of one table.

Concept / Approach:
We form two linear equations based on the given total costs: 2T + 3C = 3,500 3T + 2C = 4,000 These two equations in two unknowns can be solved by elimination or substitution. The elimination method is usually faster for exam settings. Once we solve for T and C, we pick the value of T that matches one of the answer options.

Step-by-Step Solution:
Step 1: Let the price of one table be T and the price of one chair be C. Step 2: From the first condition, write 2T + 3C = 3,500. Step 3: From the second condition, write 3T + 2C = 4,000. Step 4: Multiply the first equation by 3 to get 6T + 9C = 10,500. Step 5: Multiply the second equation by 2 to get 6T + 4C = 8,000. Step 6: Subtract the second new equation from the first new equation: (6T + 9C) - (6T + 4C) = 10,500 - 8,000. Step 7: This simplifies to 5C = 2,500. Step 8: Solve for C: C = 2,500 / 5 = 500. Step 9: Substitute C = 500 into 2T + 3C = 3,500. Step 10: So 2T + 3 * 500 = 3,500, which gives 2T + 1,500 = 3,500. Step 11: Therefore, 2T = 2,000 and T = 1,000. Step 12: Thus the cost of one table is ₹1,000.

Verification / Alternative check:
Check with the second original equation. With T = 1,000 and C = 500: 3T + 2C = 3 * 1,000 + 2 * 500 = 3,000 + 1,000 = 4,000. This matches the given total for the second combination, confirming that the values are consistent. The first combination also works, since 2 * 1,000 + 3 * 500 = 2,000 + 1,500 = 3,500.

Why Other Options Are Wrong:
₹500, ₹1,500, ₹2,000, and ₹750 do not satisfy both equations when matched with any nonnegative chair price. Only T = ₹1,000, C = ₹500 makes both combined totals correct, so the other options are invalid.

Common Pitfalls:
Some students guess values from the options without checking both equations, or they make arithmetic errors while eliminating variables. Misaligning signs during subtraction is another frequent issue. A good practice is to rewrite the equations carefully, align like terms, and check the final values in both original equations to confirm accuracy.

Final Answer:
The cost of one table is ₹1,000.