SOLUTION: This mixture problem has me stuck. The problem: A hospital needs a 50% dextrose solution. Only a 75% and 40% solution are in stock. How much of the 75% solution should be mixed wit

Algebra.Com
Question 993766: This mixture problem has me stuck. The problem: A hospital needs a 50% dextrose solution. Only a 75% and 40% solution are in stock. How much of the 75% solution should be mixed with 500 oz of the 40% solution to get a 50% dextrose solution. My teacher showed us how to set up the chart with the percentages and how to multiply across and add down but the column where the variables is confusing me.

Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
sometimes those charts are helpful.

i'm not sure how helpful they are here.

the basic problem is this.

you have two solutions.
we'll call them A and B.

A is a 40% solution
B is a 75% solution

C is the new solution that you want that is a 50% solution.

you have 500 ounces of the 40% solution.
you have x ounces of the 75% solution.
you will have 500 + x ounces of the 50% solution.

your equation will be:

.40 * 500 + .75 * x = .50 * (500 + x)

you need to solve for x.

simplify the equation to get:

200 + .75 * x = 250 + .50 * x

subtract .50 * x from both sides of the equation and subtract 200 from both sides of the equation to get:

.75 * x - .50 * x = 250 - 200

combine like terms to get:

.25 * x = 50

divide both sides of this equation by .25 to get:

x = 50 / .25 = 200

go back to your original equation of .40 * 500 + .75 * x = .50 * (500 + x) and replace x with 200 to get:

.40 * 500 + .75 * 200 = .50 * (500 + 200)

simplify this equation to get:

200 + 150 = 250 + 100

combine like terms on each side of this equation to get:

350 = 350.

this confirms the value of x is equal to 200 to be good.

i had difficulty fitting a chart to this problem.

best i could come up with it this:

mix                      A             B            C
total ounces             500           x            x + 500
persent solution         40%           75%          50%
ounces solution          .4 * 500      .75 * x      .50 * (x + 500)






Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
This mixture problem has me stuck. The problem: A hospital needs a 50% dextrose solution. Only a 75% and 40% solution are in stock. How much of the 75% solution should be mixed with 500 oz of the 40% solution to get a 50% dextrose solution. My teacher showed us how to set up the chart with the percentages and how to multiply across and add down but the column where the variables is confusing me.
Yes, the chart is good, but I’ll do the problem without it

Let the amount of 75% solution to be mixed, be S
Then amount of 75% solution in the resulting mixture = .75S
Amount of 40% solution to be mixed: 500 oz
Then amount of 40% solution in the resulting mixture = .4(500)
Equation for the resulting mixture is: .75S + .4(500) = .5(S + 500)
.75S + 200 = .5S + 250
.75S - .5S = 250 – 200
.25S = 50
S, or amount of 75% solution to mix = , or oz
RELATED QUESTIONS

A hospital laboratory needs 12L of a 30% solution. The hospital only has 25% and 40% of... (answered by Boreal,solver91311)
According to new hospital standards a certain disinfectant solution needs to be 20%... (answered by josmiceli)
Jamal needs 1000 ml of 60 % - ethanol solution for his chemistry experiment.the school... (answered by josgarithmetic)
Stuck need help with this problem: A chemist needs a 110 mL of 32% solution, but only... (answered by josgarithmetic,MathTherapy)
A chemist needs 240ml of a 50% Solution but only has 32% and 76% solutions to mix. How... (answered by richwmiller)
If I have a 70% dextrose solution. How many grams of dextrose are in... (answered by )
My friend sent me a mixture problem, asking for help, and this one stumps me. "The... (answered by josgarithmetic,MathTherapy)
I've been on this problem for a very long time. I've been trying to translate the problem (answered by josgarithmetic,ikleyn,MathTherapy)
Please help me solve this problem. This is the last problem on my homework. Its hard for... (answered by rothauserc)