Question 403138: Bob has to mix a solution of .67 herbicide and .46 herbicide to form 42 liters of a .55 solution. How much .67 solution does Bob use? Found 3 solutions by stanbon, mananth, rvquartz:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Bob has to mix a solution of .67 herbicide and .46 herbicide to form 42 liters of a .55 solution. How much .67 solution does Bob use?
---------------
Equation:
cide + cide = cide
0.67x + 0.46(42-x) = 0.55(42)
----
Multiply thru by 100 to get:
67x + 46*42 - 46x = 55*42
----
21x = 9*42
---
x = 18 liters (amt. of 0.67 solution needed)
=====================
cheers,
stan H.
You can put this solution on YOUR website! Bob has to mix a solution of .67 herbicide and .46 herbicide to form 42 liters of a .55 solution. How much .67 solution does Bob use?
percent ---------------- quantity
67.00% ---------------- x
46.00% ---------------- 42-x
55.00% ---------------- 42 (Mixture)
...
67x+46(42-x)=55*42
You can put this solution on YOUR website! based on the problem statement, we can create a mathematical model that says
x + y = 42 liters
where
x = number of liters of the 67% strength solution
y = number of liters of the 46% strength solution
also based on the problem statement, we can create another mathematical model that says
0.67x + 0.46y = 0.55(42) = 23.1 liters (note: that is the amount of 'pure' 100% acid)
now we have two separate independent equations and two unknowns
we can substitute and solve
y = (42 - x) so
0.67x + 0.46(42 - x) = 23.1
0.67x - 0.46x + 19.32 = 23.1
0.21x + 19.32 = 23.1
0.21x = 3.78
x = 3.78/0.21
x = 18 liters of the 67% strength solution
y = 24 liters of the 46% strength solution
(
