SOLUTION: A chemist has 20% and 60% solutions of acid available. How many liters of each solution should be mixed to obtain 25 liters of 28% acid solution?

Question 1146352: A chemist has 20% and 60% solutions of acid available. How many liters of each solution should be mixed to obtain 25 liters of 28% acid solution?
Found 3 solutions by Theo, josgarithmetic, greenestamps:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
x = the number of liters of the 20% solution.
y = the number of liters of the 60% solution.

you want 25 liters of a 28% solution.
that would require .28 * 25 = 7 liters of acid.

you have two equatons that need to be solved simultaneously.
they are:

x + y = 25
.2x + .6y = 7

multiply both sides of the first equation by .2 and leave the second equation as is to get:

.2x + .2y = 5
.2x + .6y = 7

subtract the first equation from the second to get:

.4y = 2

solve for y to get y = 5

since x + y = 25, then x = 20

your solution is that you need 20 liters of the 20% solution plus 5 liters of the 60% solution to get 25 liters of the 28% solution.

.2 * 20 + .6 * 5 = 7

7/25 = .28 = 28%

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
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A chemist has 20% and 60% solutions of acid available. How many liters of each solution should be mixed to obtain 25 liters of 28% acid solution?
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v, unknown volume of 60%
M=25
M-v, unknown volume of 20%
L=20
H=60
T=28

-

-----------------substitute, evaluate, then use for evaluating M-v.
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Here is a faster and easier way to find the answer to "mixture" problems like this.

(1) The target 28% is 1/5 of the way from 20% to 60%. (20 to 60 is a difference of 40; 20 to 28 is a difference of 8; 8/40 = 1/5.)
(2) That means 1/5 of the mixture should be the higher (60%) ingredient.

ANSWER: 1/5 of 25L, or 5L, of 60% acid; the other 20L of 20% acid.
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