Question

A coffee manufacturer uses Colombian and Brazilian coffee beans to produce two blends, robust and mild. A pound of the robust blend requires 12 ounces of Colombian beans and 4 ounces of Brazilian beans. A pound of the mild blend requires 6 ounces of Colombian beans and 10 ounces of Brazilian beans. If the company has 1800 ounces of Colombian beans and 760 ounces of Brazilian beans, how many pounds of each type of coffee should the manufacturer make to use all their beans exactly?

Answer 1

1800ounces of robust blend requires 1350 ounces of Colombian beans and 450 ounces of Brazilian beans

760ounces of Mild beans requires 285ounces of Colombian beans and 475ounces of Brazilian beans

the total of Colombian beans needed in both is 1350+285 =1635 ounce

the total of Brazilian beans needed is 450+475 =925 ounce

Step-by-step explanation:

A pound of the robust blend requires 12 ounces of Colombian beans and 4 ounces of Brazilian beans

Let robust blend be represented as Rb

Let Colombian beans be represented as Cb

Let Brazilian beans be represented as Bb

• The ounce is abbreviated oz

• 1 pound is 16oz

1 pound of Rb = 12oz of Cb + 4oz of Bb

16oz of Rb = 12oz of Cb + 4oz of Bb( it is in ratio of 12:4 which is 3:1, meaning Cb is three parts of four while Bb takes one parts of four)

To make 1800 ounces of Rb

Divide by 4 = 1800/4 = 450

Brazilian beans is one part which is 450oz

Colombian beans is three parts = 3 x 450 = 1350oz

1800oz of Rb = 1350oz of Cb + 450oz of Bb

A pound of the mild blend requires 6 ounces of Colombian beans and 10 ounces of Brazilian beans.

Let mild blend be represented as Mb

Let Colombian beans be represented as Cb

Let Brazilian beans be represented as Bb

• The ounce is abbreviated oz

• 1 pound is 16oz

1 pound of Rb = 6oz of Cb + 10oz of Bb

16oz of Rb = 6oz of Cb + 10oz of Bb( it is in ratio of 6:10 which is 3:5, adding 3 and 5 we get 8; meaning Cb is three parts of eight while Bb takes five parts of eight)

To make 760 ounces of Mb

Divide by 8 = 760/8 = 95

Colombian beans is three parts of eight which is 3 x 95 =285oz

Brazilian beans is five parts of eight which is 5 x 95 = 475oz

760oz of Mb = 285oz of Cb + 475oz of Bb

1800ounces of robust blend requires 1350 ounces of Colombian beans and 450 ounces of Brazilian beans

760ounces of Mild beans requires 285ounces of Colombian beans and 475ounces of Brazilian beans

the total of Colombian beans needed in both is 1350+285 =1635 ounce

the total of Brazilian beans needed is 450+475 =925 ounce

Answer 2

•8.74 pounds of robust blend

• 1.25 pounds of mild blend

Step-by-step explanation:

We are given:

•12 ounces, 6 ounces of mild blend Colombian beans

•4 ounces, 10 ounces of mild blend Brazilian beans.

From the question, we get the following equations:

1) Colombian: 12x + 6y = 1800

2) Brazilian: 4x + 10y = 760

Let's multiply the Brazilian equation by 3, we now have:

Colombian: 12x + 6y = 1800

Brazilian : 12x + 30y = 2280

Solving simultaneously, we have:

-24y = - 480

Therefore,

y = -480/-24

y= 20 oz

y = 20 oz ( divide by 16 to convert to pounds) = 1.25 pounds

Let's solve for x from Brazilian equation:

4x + 10y = 760

Therefore, since y = 20, we have:

4x + 10(20) = 760

= 4x + 200 = 760

= 4x = 760 - 200

= 4x = 560

Therefore

x = 560/4

x = 140

140 oz = 8.74 pounds

Therefore, they should make;

8.74 pounds of robust blend and 1.25 pounds of mild blend.