SOLUTION: By weight (A) fertilizer is 20%K, 30%N and 50%P, (B) fertilizer is 10, 30, 60 and (C) fertilizer is 0, 40, 60. How much of each must be mixed to obtain 200lbs of 10%K, 32.5%N, and

Question 831449: By weight (A) fertilizer is 20%K, 30%N and 50%P, (B) fertilizer is 10, 30, 60 and (C) fertilizer is 0, 40, 60. How much of each must be mixed to obtain 200lbs of 10%K, 32.5%N, and 57.5%P.
.20k + .30n + .50p (A)
.10k + .30n + .60p (B)
.00k + .40n + .60p (C)
.10k + 32.5n + .60p = 200 (D)
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Ak + Bk + Ck = Dk ?????
An + Bn + Cn = Dn ?????
Ap + Bp + Cp = Dp ?????

Answer by Elomeht(22) (Show Source): You can put this solution on YOUR website!
The important point to note is that the total weight of the mixture is 200. Start with that.
Since the desired ratio of K/N/P will be 10%/32.5%/57.5%, we know that the total amounts are:
K amount = 0.1 times 200 = 20
N amount = 0.325 times 200 = 65
P amount = 0.575 times 200 = 115
Let the amount of each fertilizer be A, B and C respectively.
Then
1. .2A + .1B = 20
2. .3A + .3B + .4C = 65
3. .5A +.6B + .6C = 115

From 1, A = 100 - .5B
From 2, .3(100 - .5B) + .3B + .4C = 65; 30 -.15B + .3B + .4C = 65;
B = (35 - .4C)/.15
So in 3, .5[100 - .5(35 - .4C)/.15] + .6[(35 -.4C)/.15] + .6C = 115;
50 - .25(35 - .4C)/.15 + 4(35 - .4C) + .6C = 115;
50 - (8.75 - .1C)/.15 + 140 - 1.6C + .6C = 115;
7.5 -8.75 + .1C + 21 - .24C + .09C = 17.25
2.5 = 0.05C
C = 2.5/0.05 = 50;
B = [35 - (.4 times 50)]/.15 = 15/.15 = 100
A = 100 - (.5 times 100) = 100 - 50 = 50

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