Bunuel wrote:
In what ratio must a person mix three kinds of tea costing $60/kg, $75/kg and $100 /kg so that the resultant mixture when sold at $96/kg yields a profit of 20%?
A. 1 : 2 : 4
B. 3 : 7 : 6
C. 1 : 4 : 2
D. 1 : 4 : 1
E. 1 : 5 : 2
For a selling price of $96 per kilogram to yield a profit of 20%, the average cost per kilogram of the mixture must be $80.
From cheapest to most expensive, let the three teas be x, y and z.
Since the three costs for x, y and z -- $60 per kilogram, $75 per kilogram, and $100 per kilogram -- must be combined to yield a MIXTURE of x+y+z with an average cost of $80 per kilogram, the following equation is implied:
60x + 75y + 100z = 80(x+y+z)
60x + 75y + 100z = 80x + 80y + 80z
20z = 20x + 5y
4z = 4x + yThe resulting blue equation has an INFINITE number of solutions.
Option C is one solution for 4z=4x+y --> x=1, y=4, z=2
If 1 kilogram of x, 4 kilograms of y, and 2 kilograms of z are combined to form a 7-kilogram mixture, the average cost per kilogram \(= \frac{1*60 + 4*75 + 2*100}{7} = \frac{560}{7} = 80\)
Another solution for 4z=4x+y --> x=2, y=4, z=3
If 2 kilograms of x, 4 kilograms of y, and 3 kilograms of z are combined to form a 9-kilogram mixture, the average cost per kilogram \(= \frac{2*60 + 4*75 + 3*100}{9} = \frac{720}{9} = 80\)
Another solution for 4z=4x+y --> x=3, y=4, z=4
If 3 kilograms of x, 4 kilograms of y, and 4 kilograms of z are combined to form an 11-kilogram mixture, the average cost per kilogram \(= \frac{3*60 + 4*75 + 4*100}{11} = \frac{880}{11} = 80\)
Another solution for 4z=4x+y --> x=4, y=20, z=9
If 4 kilograms of x, 20 kilograms of y, and 9 kilograms of z are combined to form a 33-kilogram mixture, the average cost per kilogram \(= \frac{4*60 + 20*75 + 9*100}{33} = \frac{2640}{33} = 80\)
As the examples above illustrate, there are an infinite number of ways to combine the three teas to yield a mixture with an average cost of $80 per kilogram.
Since there is no unique ratio for the three teas, the problem is flawed.