a craftsperson has 2 alloys of silver. the 1st one is 70%pure silver amd the second one is 50%. how many kilograms of each must be mixed to have 12kg of an alloy which is 65% silver?
let the amount of 70% silver be x kg
then the amount of 50% silver must be 12-x kg
.7x + .5(12-x) = .65(12)
70x + 50(12-x) = 65(12)
I am sure you can take it from there.
(I got x=9)
To determine how many kilograms of each alloy should be mixed, we can use a system of equations. Let's assign variables to the unknown quantities:
Let x be the amount (in kilograms) of the 70% pure silver alloy to be used.
Let y be the amount (in kilograms) of the 50% pure silver alloy to be used.
We have two equations based on the silver content and total weight of the mixture:
Equation 1: (Amount of pure silver in alloy 1) + (Amount of pure silver in alloy 2) = Total pure silver in the mixture
Equation 2: (Amount of alloy 1) + (Amount of alloy 2) = Total weight of the mixture
Now, let's write these equations:
Equation 1: (0.70x) + (0.50y) = 0.65(12)
Equation 2: x + y = 12
The first equation represents the silver content, where 0.70 represents 70% and 0.50 represents 50% pure silver. The percentage of silver in the mixture is given as 65% of 12kg.
Now we solve the system of equations.
Multiply Equation 2 by -0.70:
-0.70x - 0.70y = -8.4
Add Equation 1 and the modified Equation 2:
(0.70x + 0.50y) + (-0.70x - 0.70y) = 0.65(12) - 8.4
0.20y = 1.8
y = 1.8 / 0.20
y = 9
Substitute the value of y into Equation 2 to find x:
x + 9 = 12
x = 12 - 9
x = 3
So, 3kg of the 70% pure silver alloy and 9kg of the 50% pure silver alloy should be mixed to obtain 12kg of an alloy that is 65% silver.