Question 1162264: How many milligrams of uranium with 72% purity and 84.8% must be mixed to obtain 9 milligrams with 80% purity? Found 3 solutions by ankor@dixie-net.com, ikleyn, greenestamps:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! How many milligrams of uranium with 72% purity and 84.8% must be mixed to obtain 9 milligrams with 80% purity?
:
let x = amt of 84.8% uranium
Total is to be 9 mg, therefore
(9-x) = amt of 72%
:
Mixture equation in decimal form
.848x + .72(9-x) = .80(9)
.848x + 6.48 - .72x = 7.2
.848x - .72x = 7.2 - 6.48
.128x = .72
x = .72/.128
x = 5.625 mg of 84.8% uranium
then
9 - 5.625 = 3.375 mg of 72%
:
:
Check on your calc
.848(5.625) + .72(3.375) = .8(9)
4.77 + 2.43 = 7.2
x milligrams of the 84.8% purity
and (9-x) milligrams of the 72% purity.
The amount of the pure uranium in ingredients is 0.848x + 0.72*(9-x) milligrams.
The amount of the pure uranium in the mixture (alloy ?) is 0.8*9 milligrams.
The mass balance equation for pure uranium is
0.848x + 0.72*(9-x) = 0.8*9 milligrams.
From the equation
x = = 5.625.
ANSWER. 5.625 milligrams of the 84.8% purity and the rest, 9-5.625 = 3.375 milligrams of the 72% purity.
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.
Read them and become an expert in solution mixture word problems.
If a formal algebraic solution is not required, here is a quick and easy informal path to solving a "mixture" problem like this.
(1) The 80% target percentage is 5/8 of the way from 72% to 84.8%:
72 to 84.8 is a difference of 12.8
72 to 80 is a difference of 8
8/12.8 = 80/128 = 5/8
That means 5/8 of the mixture needs to be the higher percentage ingredient.
ANSWER: 5/8 of 9mg = 45/8 = 5.625mg of 84.8% purity; the other 3.375mg of 72%.
