SOLUTION: How many milligrams of uranium with 72% purity and 84.8% must be mixed to obtain 9 milligrams with 80% purity?

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
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

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.

Solved.

---------------

It is a standard and typical problem on mixtures. See the lessons
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    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for dry substances like coffee beans, nuts, cashew and peanuts
    - Word problems on mixtures for dry substances like candies, dried fruits
    - Word problems on mixtures for dry substances like soil and sand
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
    - Advanced mixture problems
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Also, you have this free of charge online textbook in ALGEBRA-I in this site
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The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".

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Free of charge online textbook in ALGEBRA-I
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Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

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%.

CHECK:

0.0848(5.625)+.072(3.375) = .477+.243 = .72

.08(9) = .72

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