Can someone help explain the steps to this problem?
An alloy of tin is 15% tin and weighs 20 pounds. A second alloy is 10% tin. How much (to the nearest tenth lb.) of the second alloy must be added to the first alloy to get a 12% mixture.
? pounds
An alloy of tin is 15% tin and weighs 20 pounds. A second alloy is 10% tin. How much (to the nearest tenth lb.) of the second alloy must be added to the first alloy to get a 12% mixture.
? pounds
cw
Favorite Answer
Setup you equation like this, where B is the weight of the second alloy. Your total amount of tin stays constant so you setup the left side of the equation as the amount of tin before the alloys are combined, the right side is the amount of tin afterwards:
0.15(20)+0.10(B)=0.12(20+B)
That was the hard part, now it's just a standard 1 variable equation to solve:
reduce the equation:
3 +.1B = 2.4 +.12B
combine like terms:
0.6 = .02B
solve for B:
B = 30.0 lb
Colin S
.15 * 20 + .10 *x = .12 * (20 + x)
basically, you have to set it up such that the total tin content is the same from what you're mixing (20 lbs of .15 tin and x pounds of .10 tin) to what you're making (20 plus x pounds of .12 tin)
3 + .1 x = .12 x + 2.4
.6 = .02 x
x = 30 lbs
Jerome J
Let x = No. of lbs of 2nd alloy to add
Total weight after mixture = x + 20
Form equation of tin
(0.12)(x+20) = (0.15)(20) + (0.10)x
0.12x + 2.4 = 3 + 0.10x
0.12x - 0.10x = 3.0 - 2.4
0.02x = 0.6
x = 0.6 / 0.02 = 30 lbs of 2nd alloy
Pakyuol
working equation: 0.15x + 0.10y = 0.12(x + y)
You have an infinite # of solutions.
.