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.
Similar Questions
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Important Questions from Linear Equation in 2 Variable
What is the solution of the following equations ?
2x + 3y = 12 and 3x − 2y = 5
Two positive numbers differ by 1280. When the greater number is divided by the smaller number, the quotient is 7 and the remainder is 50. The greater number is:
When 5 children from class A join class B, the number of children in both classes is the same. If 25 children from B, join A, then the number of children in A becomes double the number of children in B. The ratio of the number of children in A to those in B is:
If (x + 6y) = 8, and xy = 2, where x > 0, what is the value of (x 3+ 216y 3)?
If 8k 6+ 15k 3– 2 = 0, then the positive value of \(\left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right)\) is :
