Question 951263: By weight, one fertilizer is 10% potassium, 40% nitrogen, and 50% phosphorus. A second fertilizer has percents of 10, 30, 60, respectively, and a third has percents of 0, 20, 80, respectively. How much of each must be mixed to get 200 pounds with 8, 31, 61 percent?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! By weight, one fertilizer is 10% potassium, 40% nitrogen, and 50% phosphorus.
A second fertilizer has percents of 10, 30, 60, respectively,
and a third has percents of 0, 20, 80, respectively.
How much of each must be mixed to get 200 pounds with 8, 31, 61 percent?
:
let a = amt of 1st fertilizer
let b = amt of 2nd fertilizer
let c = amt of third fertilizer
we know
a + b + c = 200; the total amt equation
and
Write an equation for each substance using decimals for percent
.10a + .10b + 0c = .08(200); potassium
.40a + .30b + .20c= .31(200); nitrogen
.50a + .60b + .80c = .61(200); phosphorus
:
.10a + .10b + 0c = 16
.40a + .30b + .20c = 62
.50a + .60b + .80c = 122
:
Multiply the 1st substance equation by 10, subtract from the total amt equation
a + b + c = 200
a + b + 0 = 160
-----------------Subtraction eliminate a and b, find c
0 + 0 + c = 40 lb of phosphorus
:
using the 2nd and 3rs substance equations, replace c with 40
.40a + .30b + .20(40) = 62
.50a + .60b + .80(40) = 122
:
.40a + .30b + 8 = 62
.50a + .60b + 32 = 122
:
.40a + .30b = 54
.50a + .60b = 90
:
multiply the 1st "two unknown" equation by 2 subtract the 2nd
.80a + .60b = 108
.50a + .60b = 90
--------------------subtraction eliminates b find a
.30a + 0 = 18
a = 18/.3
a = 60 lb of potassium
:
Find b (nitrogen)
200 - 40 - 60 = 100 lb of nitrogen
:
"How much of each must be mixed to get 200 pounds with 8, 31, 61 percent?
60, 100, 40
:
:
You can check this for yourself in one of the substance equations
.50(60) + .60(100) + .80(40) = .61(200); phosphorus
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