I'm not sure of how to solve word problems. The question is,"One lawn fertilizer is 24% nitrogen and another is 12% nitrogen. How much of fertilizer should be mixed to obtain 100 kg of fertilizer that is 21% nitrogen?"

Answer 1

see the other post.

Answer 2

Let m1 and m2 be the masses of fertilizer at 24% and 21% respectively. Then write:

Eq. 1: m1 + m2 = 100
Eq : (.24m1 + .21 m2)/(m1 + m2)= .21

The second equation is called a weighted average. Solving simultaneously yields:

m1 = 75 kg
m2 = 25 kg

Checking:

(.24×75 + .21× 25)/100 = .18 + .03 = .21
QED

Answer 3

Let x be mass of 24% mixture. Let y be mass of 12% mixture.
Equation would be, x+y=100
0.24x+0.12y= (0.21)(100)