Question

a mixture is desired that will contain 65%A, 20%B and 15%C, all by weight. The only raw materials from which this is to be made are: (a) 10%A, 60%B, 30%C (b) 90%A, 10%C (c) 98%A, 2%C. If we are to end up with 1000 Kg of the final mixture, how much of each raw material do we use?

Answer 1

The total weight required = 1000 Kg

out of this we need 650 grams of A, 200 grams of B , 150 grams of C

x gram of (a) and y gram of (b) and z gram of (c)

So the composition will be =

60x/100 = 200 grams

x = 333.33 grams ....(1)

Next will be

10/100 y + 2/100 z + 30/100x = 150

90y/100 + 98z/100 + 10x/100 = 650

we will put x = 333.33

0.1y + 0.02z + 0.3X 333.33 = 150 ...................(2)

0.9y + 0.98z + 0.1X 333.33 = 650 ...................(3)

Multiplying equation (2) with 9 and substracting it from (3)

0.18z + 899.9 - 0.98z - 299.9 = -700

-0.8z = -1300

z = 1625 grams ........(4)

Putting 4 in (2)

0.1y + 0.02 X 125 + 0.3X 333.33 = 150

0.1y + 2.5 + 99.9 = 150

0.1 y = 47.6

y = 476 grams