# A small seminar hall can hold 100 chairs.Three different colours(red,blue and green) of chairs are available. The cost of a red chair is Rs.240, cost of a blue chair is Rs.260, and the cost of a green chair is Rs.300. The total cost of chair is Rs.25,000. Find atleast 3 different solution of the number of chairs in each colour to be purchased.

Toolbox:
• Consistency of a system of 2 ( or 3 ) linear equations in 2 ( or 3 ) unknowns. Case 1 : If $\Delta \neq 0$ thesystem is consistent and has a unique solution. Cramer's rule can be used. Case 2 : If $\Delta = 0$, there are two / ( four) possibilities:
• 2a: If $\Delta = 0$, and atleast one of the values $\Delta_ x, \Delta_ y ( or \: \Delta_ z )$ is nonzero, the system is inconsistent and has no solution.
• 2b: If $\Delta = 0\: and \: \Delta_x=\Delta_y ( = \Delta_z)=0$ and at least one element ( second order minor ) of $\Delta$ is nonzero, the system is consistent with infinitely many solution. The system reduces to one equation ( two equations )
• 2c: If $\Delta = 0$ and $\Delta_x=\Delta_y = \Delta_z=0$ and all their 2 x 2 minors are 0, and $\Delta$ has atleast one nonzero element, then the system is consistent with infinitely many solutions. The system reduces to one equation.
• 2d: If $\Delta = 0, \Delta_x=\Delta_y = \Delta_z=0$, all 2 x 2 minors of $\Delta$ are zero but there is at least one 2 x 2 minor of $\Delta_x, \Delta_y \: or \: \Delta_z$ that is nonzero, then the system is inconsistent and has no solution.
Step 1
Let $x , y, z$ be the number of red, blue and green chairs, costing Rs. 240, Rs. 260 and Rs. 300 each respectively.
The total number of chairs = 100
$x +y+z=100$
The total cost of the chairs = Rs.25000
$240x+260y+300z = 25000$
Step 2
These are two equations in three unknowns and have infinitely many solutions. Let $z = k,\: k \in R$ ( since the numbers have to be whole numbers). The equations reduce to
$x+y=100-k$
$240x+260y=25000-300k$
$\Delta = \begin{vmatrix} 1 & 1 \\ 240 & 260 \end{vmatrix} = 260-240=20 \neq 0.$ Cramer's rule can be used.
Step 3
$\Delta_x = \begin{vmatrix} 100-k & 1 \\ 25000-300k & 260 \end{vmatrix} = 26000-260k-25000+300k$
$= 1000+40k$
$\Delta_y = \begin{vmatrix} 1 & 100-k \\ 240 & 25000-300k \end{vmatrix} = 25000-300k-24000+240k$
$= 1000-60k$
$x = \large\frac{\Delta_x}{\Delta} = \large\frac{1000+40k}{20} = 50+2k$
$y = \large\frac{\Delta_y}{\Delta} = \large\frac{1000-60k}{20} = 50-3k$
$(x, y, z) = (50+2k, \: 50-3k, \: k)$ For whole number values of all three.
Step 4
Three possible solutions
when $k = 0\: \: \: \: (x,y,z) = (50, 50, 0)$
$\: \: \: \: \: \: \: \: \: k = 5\: \: \: \: (x,y,z) = (60, 35, 5)$
$\: \: \: \: \: \: \: \: \: k = 10\: \: \: \: (x,y,z) = (70, 20, 10)$

edited Jun 3, 2013
Can I put k=1,2,3 ?
How can I solve this using Cramer's Rule?