Two-equation pricing (chairs and tables): Ten chairs and six tables together cost ₹6,200. Three chairs and two tables together cost ₹1,900. What is the cost of 4 chairs and 5 tables?

Introduction / Context:
We are given two linear price equations with two unknowns: chair price (C) and table price (T). Solve the system to determine C and T, and then compute the requested bundle price for 4 chairs and 5 tables.

Given Data / Assumptions:

• 10C + 6T = 6200
• 3C + 2T = 1900
• Uniform prices; no bulk discounts implied.

Concept / Approach:
Use elimination. Match the table coefficients by multiplying the second equation by 3 and subtract from the first to isolate the chair price. Solve for C, then back-substitute to find T. Finally compute 4C + 5T.

Step-by-Step Solution:

Verification / Alternative check:
Check both original equations with C = 500, T = 200: 10*500 + 6*200 = 6200 and 3*500 + 2*200 = 1900. Both satisfy perfectly.

Why Other Options Are Wrong:
₹3,300, ₹3,500, ₹3,800, ₹2,900 are not equal to 4*500 + 5*200 = ₹3,000.

Common Pitfalls:
Miscalculating multipliers during elimination or mixing coefficients leads to wrong unit prices; always verify by substitution before finalizing the bundle cost.

Final Answer:
₹ 3,000