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Q. 1: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
1. 26
2. 16
3. 20
4. 22
Correct Answer: 3. 20
Solution: let cost of paint A = x / liter
So cost of paint B = (x-8)/liter
Cost of 10 liter of the mixture = 264/1.1 = 240
Cost per liter of mixture = 240/10 = 24
For highest possible cost of paint B, both A and B should be in equal ratio , so
Cost of mixture = (x+ x -8)/2
24 = (2x -8)/2
x = 28
thus cost of paint B = 28 -8 = 20 per liter
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