SOLUTION: Gerry mixes different solutions with concentrations of 25%,40% and 50% to get 200 liters of a 32% solution. Also, it takes twice the liters of the 40% solution to equal the same li
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: Gerry mixes different solutions with concentrations of 25%,40% and 50% to get 200 liters of a 32% solution. Also, it takes twice the liters of the 40% solution to equal the same li
Log On
Question 188471: Gerry mixes different solutions with concentrations of 25%,40% and 50% to get 200 liters of a 32% solution. Also, it takes twice the liters of the 40% solution to equal the same liters of the 25%,find how many liters of each kind he uses. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Gerry mixes different solutions with concentrations of 25%,40% and 50% to
get 200 liters of a 32% solution.
Also, it takes twice the liters of the 40% solution to equal the same liters of the 25%,
find how many liters of each kind he uses.
:
Let x = amt of 40% solution
then
2x = amt of 25% solution
and
(200-3x) = 50% solution
:
.25(2x) + .40x + .50(200-3x) = .32(200)
:
.50x + .40x + 100 - 1.5x = 64
:
.50x + .40x - 1.5x = 64 - 100
:
-.60x = -36
x =
x = +60 liters of 40% solution
and
2(60) = 120 liters of 25% solution
and
200 - 60 - 120 = 20 liters of the 50% solution
;
:
Check:
.25(120) + .4(60) + .5(20) = .32(200)
30 + 24 + 10 = 64
