SOLUTION: How many liters of a 20% acid solution must be mixed with a 60% solution to obtain 40 liters of a 35% solution? Also please include a chart if u could.

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Question 1054919: How many liters of a 20% acid solution must be mixed with a 60% solution to obtain 40 liters of a 35% solution? Also please include a chart if u could.
Found 2 solutions by jorel555, Theo:
Answer by jorel555(1290) (Show Source):
You can put this solution on YOUR website!
Let n be the amount of 60% solution. Then:
.6n+.2(40-n)=.35(40)
.4n=6
n=15
You need 15 liters of 60% solution and 25 liters of 20% solution for your desired mixture. ☺☺☺☺

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
let x = the number of liters of the 20% solution.

let y = the number of liters of the 60% solution.

you want x + y to be equal to 40 liters.

x is the number of liters total in the first solution.
y is the number of liters total in the second solution.

you want .2 * x + .6 * y to be equal to .35 * 40

.2 * x is the number of liters of acid in the first solution.
.6 * y is the number of liters of acid in the second solution.
.35 * 40 is the number of liters of acid in the final solution.

you have two equations that need to be solved simultaneously.

they are:

x + y = 40
.2x + .6y = .35*40

simplify these equations to get:

x + y = 40
.2x + .6y = 14

you can solve by substitution or by elimination or by graphing.
i will solve this one by graphing.
this means to graph both equations and find the intersection.

the graph looks like this:

the graph says the intersection is at the coordinate point of (25,15).

this means that x = 25 and y = 15.

x is the number of liters of the 20% solution.
y is the number of liters of the 60% solution.

the formula of .2x + .6y = 14 becomes .2 * 25 + .6 * 15 = 14.
simplify this equation to get 14 = 14.
this confirms the solution is good.