Question:
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 pro?t of 10%, then the highest possible cost of paint B, in Rs. per litre, is
- 26
- 16
- 20
- 22
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=264 Profit percent made=10%
Value at which the entire mixture is bought = $264\times \frac{100}{110}=240$
Price at which the entire mixture is bought=24 per litre Let the cost of B be x per litre.
Cost of A=(x+8)per litre
$\frac{(x+8) a+x b}{10}=24$
Maximum cost of B will occur when a is minimum. b<=a. So, minimum a is 5.
Corresponding b is 5. Then (x+8)(5)+x(5)=240 x=20
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