A wholesaler mixed coffee beans worth $6/kg with another kind worth $8.80/kg. The 16kg mixture was worth $6.70/kg. How many kilograms of each type was used?
Let the amount of the $6 mixture used be x kg
let the amount of the $8.80 mixture be 16-x kg
then
6x + 8.8(16-x) = 6.7(16)
solve for x
thnx SO MUCH
6x+1408-88x=1072
you may move the decimal 1 place if you want
1408-82x=1072
1408-1072=82x
336=88x
x=336/88
To solve this problem, we can set up a system of equations based on the given information. Let's use two variables:
Let x be the number of kilograms of coffee beans worth $6/kg.
Let y be the number of kilograms of coffee beans worth $8.80/kg.
Based on the information given, we can write two equations:
1. The total weight equation: x + y = 16 (since the total mixture is 16kg)
2. The total value equation: (6x + 8.80y)/(x + y) = 6.70 (since the value of the mixture is $6.70/kg)
Now, we can solve this system of equations to find the values of x and y.
First, we'll simplify the second equation:
6x + 8.80y = 6.70(x + y)
6x + 8.80y = 6.70x + 6.70y
Next, rearrange the equation:
8.80y - 6.70y = 6.70x - 6x
2.10y = 0.70x
Now, divide both sides by 0.70:
y = (0.70/2.10)x
y = (1/3)x
Substitute this value of y into the first equation:
x + (1/3)x = 16
(4/3)x = 16
4x = 48
x = 12
Now, substitute this value of x back into the equation for y:
y = (1/3)(12)
y = 4
Therefore, the wholesaler used 12kg of coffee beans worth $6/kg and 4kg of coffee beans worth $8.80/kg.