<PRE><B>Question 1162264</B>
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</PRE>