Multiple choice

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. 16

  2. 20

  3. 22

  4. 26
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total SP = 264, Profit = 10%, so Total CP = 264 / 1.1 = 240. Let x be cost of B, then A is x+8. Let q be quantity of B, then 10-q is quantity of A. q <= 10-q, so 2q <= 10, q <= 5. Total CP = q(x) + (10-q)(x+8) = 240. qx + 10x + 80 - qx - 8q = 240. 10x - 8q = 160. x = (160 + 8q) / 10 = 16 + 0.8q. To maximize x, maximize q. Max q = 5. x = 16 + 0.8(5) = 16 + 4 = 20.

AI explanation

The total cost price of the 10-litre mixture is 264 divided by 1.1, which equals Rs. 240, making the average cost per litre Rs. 24. Let paint B cost Rs. x per litre, so paint A costs Rs. (x + 8) per litre. By the mixture and alligation method, the ratio of B to A is (24 - (x + 8)) : (x - 24) = 1 : 1, which means the quantities of B and A are equal at 5 litres each. Solving (16 - x) = (x - 24) gives 2x = 40, so the highest possible cost of paint B is Rs. 20 per litre.

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