<PRE><B>Question 62080</B>
A soil analysis of a lawn determines 50 pounds of fertilizer containing 20 % nitrogen is needed. How can this mixture be made from two different fertilizers one containing 15 % nitrogen and the other containing 24% nitrogen? 
:
Let x = amt of 15% fertilizer
It tells us that 50 pound are required so:
Let (50-x) = amt of 24% fertilizer: (the two ferilizers have to add up to 50)
:
The percent fertilizer equation in decimal form:
.15x + .24(50-x) = .20(50)
.15x + 12 - .24x = 10
.15x - .24x = 10 - 12
-.09x = -2
x = -2/-.09
x = +22.22 pounds of the 15% required
:
50 - 22.22 = 27.78 pounds of the 24%
:
Check .15(.22.22) + .24(27.78) = 10.00
:
Did this make sense to you?? 

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