Please Help Me Solve This!!!

Question

How many gallons of a 90% antifreeze solution must be mixed with 90 gallons of 25% antifreeze to get a mixture that is 80% antifreeze? Use the six-step method.

Answer 1

Hi Brooke, the first step to mixture problems is to set up a table with the information. Then it is a simple matter of solving an equation.

 

Table

 

Solution              Gallons      Percent         Totals

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

Antifreeze 90%        x             .90          .90 *   x            

Antifreeze 25%      90             .25          .25 * 90

 

Antifreeze 80%    90+x           .80          .80 * (90+x)

 

Equation:

 

.90 * x + .25 * 90 = .80 * (90+x)

 

I do not like to work with decimals, so multiply all terms by 100:

 

90 * x + 25 * 90 = 80 * (90+x)

 

90x + 2250 = 7200 + 80x

 

Subtract 80x from both sides:

 

10x + 2250 = 7200

 

Subtract 2250 from both sides:

 

10x = 4950

 

Divide both sides by 10:

 

x = 495

 

You need to mix 495 gallons of 90% antifreeze with 90 gallons of 25% antifreeze to get an 80% antifreeze solution.

 

Questions?

 

Answer 2

Let's say that the volume of 90% solution is v₁, So the total volume of the mixture is:

(V₁+90) gallons.

This volume will contain this amount of antifreeze:

 

(V₁+90)0.8=0.8V₁+72 (regardless of the unit)

 

This quantity will be the same as the following:

 

0.9V₁+90*0.25=0.9V₁+22.5

 

0.9v₁+22.5=0.8V₁+72

 

this will give the following answer:

V₁=495 gallons

 

I hope this will help

 

Regards,