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.
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.
`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.
`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.
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.