A mixture of 6% disinfectant solution is to be made from 8% and 5% disinfectant solutions...How much of each

Question

A mixture of 6% disinfectant solution is to be made from 8% and 5% disinfectant solutions...How much of each should be used if 33 gallons of 6% solution are needed?

____ gallons of 8% solution.

_____ gallons of 5% solution.

How to solve this?

Answer 1

 

 

 

Let x =  gallon of 8% solution,

let y =gallon  of 5% solution.

 

For mixture problems, we will use table

 

 

 

 

                                              gallon sol'n             percent acid       total gallon acid

                            8% sol'n     x                           0.08                          0.08x

                            5% sol'n     y                            0.05                          0.05y

                            mixture        x + y =  33             0.06                  (0.06)(33) = 1.98

 

 

 

 

 

Since x + y = 33,

then x = 33 – y. Using this, we can substitute for x in our grid, and eliminate one of the variables

                                                      gallon sol'n          percent acid         gallon acid

                                 8% sol'n       33 – y                      0.08                   0.08(33 – y)

                                 5% sol'n         y                            0.05                     0.05y

                                  mixture          x + y = 33              0.06             (0.06)(33) =1.98

 

 

 

 

When the problem is set up like this, you can use the last column to write your equation: The gallon of acid from the 8% solution, plus the gallon of acid in the 5% solution, add up to the gallon of acid in the 6% solution. Then:

0.08(33 – y) + 0.05y = 1.98

2.64-0.08y+0.05y=1.98
2.64-0.03y=1.98
-0.03y = 1.98-2.64
-0.03y = -0.66

y=-0.66/-0.03

 

y=22

 

 

so we need 22 gallon of 5% sol

 

x= 33-y=33-22=11

 

11 gallon of 8% sol

 

 

Answer 2

Hi Lelia, the way to solve mixture problems is to create a table with the information given.

 

Let x represent the amount of 6% solution needed.

 

Then 33 - x is the amount of 8% solution needed.

 

Solution        Amount              Decimal               Total

-------------------------------------------------------------------

5%                   x                      .05              .05 * x

 

8%              33 - x                    .08              .08 * (33 - x)

 

6%                 33                      .06              .06 * 33

 

The total amount of 5% solution plus the total amount of 8% solution must equal the total amount of 6% solution.

 

Equation:

 

.05 * x + .08 * (33 - x) = .06 * 33

 

Multiply all terms by 100 to clear the decimal:

 

5 * x + 8 * (33 - x) = 6 * 33

 

Clear the parenthesis:

 

5x + 264 - 8x = 198

 

Combine terms:

 

264 - 3x = 198

 

Subtract 198 from both sides:

 

66 - 3x = 0

 

Add 3x to both sides:

 

3x = 66

 

Divide both sides by 3:

 

x = 22

 

You need 22 gallons of 5% solution and 11 gallons of 8% solution to make 33 gallons of 6% solution.

 

Questions?