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Question 1112776: The coffee factory would like to sell 3 types of coffee. Coffee A consists of 300 gram of Arabica beans and 200 gram of Robusta beans. Coffee B consists of 200 gram of Arabica beans, 200 gram of Kenya beans and 100 gram of Robusta beans. Coffee C consists of 100 gram of Arabica beans, 200 gram of Kenya beans and 200 gram of Robusta beans. The coffee factory have 30 kilogram of Arabica beans, 15 kilogram of Kenya beans and 25 kilogram of Robusta beans. How many pack of Coffee A, B and C will be packed if the factory would like to use up all of the beans?
Answer by ikleyn(52786)   (Show Source):
You can put this solution on YOUR website! .
Let a = # of packs of Coffee A;
b = # of packs of Coffee B;
c = # of packs of Coffee C.
Then, counting the balances for each type of coffee beans, you have this system of 3 equations in 3 unknowns a, b and c:
0.3a + 0.2b + 0.1c = 30 (kilograms of Arabica) (1) (<<<---=== counting Arabica)
0.2a + 0.1b + 0.2c = 25 (kilogram of Robusta) (2) (<<<---=== counting Robusta)
0.2b + 0.2c = 15 (kilogram of Kenya) (3) (<<<---=== counting Kenya)
There are many ways to solve it.
I chose the simplest way, using the online solver of this site
https://www.algebra.com/algebra/homework/Matrices-and-determiminant/cramers-rule-3x3.solver
giving me the answer a= 65, b= 30 and c= 45.
The last step that I did was checking:
0.3*65 + 0.2*30 + 0.1*45 = 30 ! Correct !
0.2*65 + 0.1*30 + 0.2*45 = 25 ! Correct !
0.2*30 + 0.2*45 = 15 ! Correct !
Solved.
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