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
Explanation:
Let the 10 litres of mixture has ‘Y’ litres of A and (10 – Y) litres of B. Let cost of paint B be Rs. X and that of A be Rs. (X + 8).
We know that, Y ≥ (10 – Y) ⇒ Y ≥ 5
The trader makes 10% profit by selling this mixture at Rs. 264.
∴ Cost price of the mixture = 2641.1 = Rs. 240
∴ (X + 8) × Y + (10 – Y) × X = 240
∴ 10X + 8Y = 240
∴ X = 24 – 0.8Y
For maximum value of X, we need to consider minimum value of Y.
∴ X = 24 – (0.8 × 5) = Rs. 20
Hence, option (a).
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