Algebraic Solution for Chair and Table Costs
Let C be the cost of one chair and T be the cost of one table.
From the problem statement, we can set up two linear equations:
- Equation 1: $2C + 1T = 880$
- Equation 2: $1C + 2T = 980$
Solving the System of Equations
We can solve this system using the elimination method.
- Multiply Equation 1 by 2 to make the coefficients of T match: $2 \times (2C + T) = 2 \times 880$ $4C + 2T = 1760$ (Equation 3)
- Subtract Equation 2 from Equation 3: $(4C + 2T) - (C + 2T) = 1760 - 980$ $3C = 780$
- Solve for C: $C = \frac{780}{3}$ $C = 260$
- Substitute the value of C back into Equation 1: $2(260) + T = 880$ $520 + T = 880$
- Solve for T: $T = 880 - 520$ $T = 360$
Therefore, the cost of one chair is ₹260 and the cost of one table is ₹360.
Verification
Check the costs using the second condition (1 chair and 2 tables cost ₹980):
$1C + 2T = 1(260) + 2(360) = 260 + 720 = 980$
The costs satisfy both conditions.
The cost of each chair is ₹260 and the cost of each table is ₹360.
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