SOLUTION: Four gallons of a 45% acid solution is obtained by mixing a 90% solution with a 30% solution. How many gallons of each solution must be used to obtain the desired mixture?

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons  -> Linear Equations Lesson -> SOLUTION: Four gallons of a 45% acid solution is obtained by mixing a 90% solution with a 30% solution. How many gallons of each solution must be used to obtain the desired mixture?      Log On


   

Question 282711: Four gallons of a 45% acid solution is obtained by mixing a 90% solution with a 30% solution. How many gallons of each solution must be used to obtain the desired mixture?
Found 3 solutions by richwmiller, oberobic, mananth:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
.9*x+.3*y=.45*4
x+y=4
x=1
y=3

Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
When solving solution problems, you need to focus on how much 'pure' stuff you need.
In this case, you need to produce 4 gal. of a desired 45% solution.
That means you will need .45*4 = 1.8 gal. of the 'pure' stuff.
The simplest way would be to take 1.8 of 'pure' stuff and add 2.2 gal. of water, but you do not have that option.
Instead you have to use 90% and 30% solutions.
.
Defining the amounts of is critical.
x = amount of 90% solution
y = amount of 30% solution
4 = total amount of 45% solution
.
So y = 4-x, which will make the job of solving easier.
.
.9x + .3(4-x) = .45*4
.9x + 1.2 -.3x = 1.8
.6x + 1.2 = 1.8
.6x = .6
x = 1
.
That means you would propose to use 1 gal. of 90% solution + 3 gal. of 30% solution.
.
Check by determining if that is enough 'pure' stuff.
1*.9 = .9
3*.3 = .9
.9 + .9 = 1.8, which is what you needed.
.
Answer:
Mix 1 gal. of 90% solution with 3 gal. of 30% solution to obtain 4 gal. of 45% solution.
.
Done.

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Four gallons of a 45% acid solution is obtained by mixing a 90% solution with a 30% solution. How many gallons of each solution must be used to obtain the desired mixture?
let the quantity of 90% solution added be x
the total solution obtained is 4 gallons
So the remaing solution will be 30% solution
=4-x gallons
The sum of the acid content of mixed solutions is equal to acid content in the mixture
0.9x+0.3*(90-x)= 0.45 *4
0.9x+0.27 - 0.3x= 1.80
0.6x= 1.80-0.27
0.6x= 1.53
x=1.53 / 0.6
x= 2.55 gallons which is 90% acid solution
Balance will be 30% acid solution.
4-2.55 = 1.45 gallons