A wholesaler mixed coffee beans worth $6/kg with another worth $8.80/kg. The 16 kg mixture was worth $6.70/kg. How many kg of each type were used?Really urgent! Please help.

2 Answers

embizze's profile pic

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

Let x represent the number of kg of the $6 coffee, and y the number of kg of the $8.80 coffee.

Since the number of kg is 16 we have x+y=16

Since the mixture is worth $6.70 per kg, and the number of kg is 16, we have a total value of $107.20. Then 6x+8.8y=107.2

Solve the system simultaneously using any method. I will use linear combinations:



Multiply the first equation by 6 and subtract from the second equation to get:

2.8y=11.2 ==>y=4 ==>x=12

So there were 12kg of $6 coffee and 4kg of $8.80 coffee.


lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

Let x be the number of kilograms of the coffee beans worth $6/kg.
And y be the number of kilograms of the coffee beans worth $8.80/kg.

Then, set up the equation using the given conditions.

Since the total kilogram of the resulting mixture is 16kg, then:

`x + y =16` (Let this be EQ1.)

And, since the cost of the mixture is $6.70/kg, the other equation is:

`6x + 8.8y = 6.7(16)`

`6x + 8.8y =107.2`

To remove the decimal point, multiply both sides by 10.

`10*(6x+8.8y) =107.2*10`

`60x+88y=1072` (Let this be EQ2.)

Then, let's use elimination method of system of equations.
So, multiply EQ1 by 60.

`60(x + y = 16)`


Then, subtract it from EQ2.

`- 60x+60y=960`
            `28y = 112`

Divide both sides by 28.

`(28y)/28 = 112/28`


To solve for x, substitute this value of y to EQ1.

`x + y = 16`

`x+ 4 = 16 `


Hence, 12 kgs of coffee bean worth $6/kg and 4 kgs of coffee beans worth $8.80 are mixed to have a 16kg mixture worth $6.70/kg.