A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is
Let the quantities of the paints A and B in the mixture sold be a litres and b litres respectively.
Value at which the entire mixture is sold = Rs. 264
Profit percent made = 10%
Value at which the entire mixture is bought = 264 × 100/110 = Rs. 240
Price at which the entire mixture is bought = Rs. 24 per litre
Let the cost of B be Rs. x per litre.
Cost of A = Rs. (x + 8) per litre
[(x + 8)a + xb]/10 = 24
Maximum cost of B will occur when a is minimum.
Since, b <= a. So, minimum value of a is 5.
Accordingly b = 5.
(x + 8)(5) + x(5) = 240
10x + 40 = 240
x = 20
The correct option is C.