Word Problem?

Question

Gerry Gundersen mixes different soultions with concentratations of 25%, 40%, and 50% to get 200 liters of a 32% soultion. If he uses twice as much of the 25% soluion as of the 40% solution, find how many liters of each kind he uses.

Please no answers of "do your own homework." If I exactly understood the question, I wouldn't be asking for help. Any help is apperciated. Thanks!

Answer 1

Volume of 40% solution = X
Volume of 25% solution = 2X
Volume of 50% solution = Y

3X + Y = 200 litres.

Overall concentration of the 25%-40% mix will be:
(2 x 0.25 + 1 x 0.4)/3 = 0.3 = 30%

Because the fluids are in ratio of 2:1, that solution will be constant at 30%, no matter what volume is used.

Now, to get the actual amounts right:

(3X x 0.3 + Y x 0.5) = 0.32

By rearranging the first formula to get 3X = 200 - Y, we get:

([200 - Y] x 0.3 + Y x 0.5) = 0.32 x 200
60 - 0.3Y + 0.5Y = 64
0.2Y = 64 - 60
0.2Y = 4 litres
2Y = 40 litres
Y = 20 litres

3X + Y = 200
X = (200 - 20)/3
X = 60 litres

So:

25% solution = 2X = 120 litres
40% solution = X = 60 litres
50% solution = Y = 20 litres.

Will

Answer 2

look i'll explain it to u in simpler words----
25% of x + 40% of y + 50% of z = 32% of 200
this is on the principle that-
amt. of solute added = amt. of solute present in the solution

as, y=1/2 of x
replace in the above eq. u'll get a simultaneuos one..with x and z

as is already given---
total amt. of solution = total amt. of different solutions added.
hence---
x+y+z = 200
as y = 1/2 of x
thus,
x+1/2x +z=200
== 3/2 x + z = 200

now u have got 2 equations
solve them as simultaneous eq. u'll get the answer
for more hlp , mail me

Answer 3

x @.25, y@.4, z@.5

x+y+z=200
.25x+.4y+.5z=.32*200
2y =x
reduces to 2 equations with 2 unknowns
solve and get x=120, y=60, z=20

Answer 4

for the 25% solution,
let 'x' be total volume of the solution
so, the volume of the SOLUTE will be [25/100]x = 0.25x

for the 40% solution,
let 'y' be the total volume of the solution.
similarly, volume of the solute = 0.4y

for the 50% solution,
let 'z' be the total volume.
z will be 200-(x+y), since total volume of the final solution is 200 l.
so, volume of solute will be (0.5)*[200-(x+y)]

he uses twice as much 25% solution,
so, x must be equal to 2y.

the final equation:
[volume of solutes/volume of solution]*100=32
so,
[0.25x+0.4y+0.5{200-(x+y)} ]/200 litres=0.32

substituting x as 2y,
0.25(2y) + 0.4y + 0.5{200-(2y+y)} =0.32*200 litres

solve it if you really want to learn.really.
you will understand it much better if you actually solve it

Answer 5

.25(2x)+.40x+ .50(200-3x)=.32(200)

.5x + .4x +100-1.5x=64

-.6x=-36 -36/-0.6=60

x=60

Try to find "How to Solve It" by G. Polya
http://www.khake.com/page47

Answer 6

It seems that you have plenty of time to play on the computer, but no time to study. "Apperciate" that. You should find your own "soluion".

Answer 7

Yo, Surveyor! What are you doing, following this person around insulting them? Who's wasting more time, her or you?

Answer 8

huh?????? ask your teacher i dont understand it either
sorry