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.



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Let the number of red, blue and green chairs be x, y and z respectively.
total number of chairs = 100
The cost of a red chair is Rs.240
The cost of a blue chair is Rs.260
the cost of a green chair is Rs.300
The total cost of chair is Rs.25,000
divide (2) by 10
divide again by 2
12x+13y+15z=1250 --->(3)
from equation (1)
x+y =100-z
from equation (3)
12x+13y = 1250-15z
take z=k where k is an arbitrary constant and is a real number 
so the above equations become 
| 1 1     |
| 12 13 | = 13-12 = 1 which is not equal to zero.
by cramers rule the equations have a unique solution.
replace the first column in the determinant by 100-k and 1250-15k 
1300-13k -1250+15k = 2k+50/1= 2k+50 = x
replace the second column in the determinant by 100-k and 1250-15k
1250-15k -(12)(100-k) = 1250-15k -1200+12k = 50 -3k = y 
z = k
k should lie between 0 and 16 since 50-3k should not acquire a negative value.
the solution is ( 2k+50, 50-3k, k ) k lies between 0 and 16
substitute k as 0 you get ( 50,50,0)
when k is 1 you get ( 52,47,1)
when k is 2 you get ( 54,44,2)
the above mentioned solutions are takin in the order of (x,y,z) respectively .

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