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Question 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

- 20
- 16
- 22
- 26

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Given, The amount of B in mixture ≤ Amount of A in mixture

Selling Price = Rs. 264 and Profit = 10 %

264 = 1.1 × CP

CP = \\frac{264}{1.1}) = Rs. 240 for 10 litres

CP per litre = Rs. 24

We also know that, CP per litre of A = CP per litre of B + 8
Let CP per litre of B = x So, CP per litre of A = x + 8

Given Amount of B ≤ Amount of A,

Maximum possible cost of B occurs when B = A

We need to choose values for A and B in such a way that the quantities remain same.But A should be 8 more than B

Therefore, A should have a CP per litre of Rs. 28 and B should have a CP per litre of Rs 20

B cannot be assigned any more than this as the amount of B would become more than A

Example: Let us assign B to be Rs. 21 and A to be Rs. 29 which would result in a ratio of 3:5 where B amounts more than A, which doesn’t satisfy the condition

Max possible cost of Paint B = **Rs. 20 **per litre

The question is **"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" **

Choice A is the correct answer.

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