How do I solve this ??

Question

a mixture of 6% disinfectant solution is to be made from 8% and 5% disinfectant solutions. How much of each solution should be used if 24 gallons of 6% solution are needed?

Answer 1

We have a strong solution (8%) and at weak solution (5%) to make a 6% solution. Let x = gallons of 8% solution to use, and y = gallons of 5% solution to use. So, we can make 2 equations from these facts. (1) x + y = 24 (total number of gallons to be made). (2) 0.08x + 0.05y = 0.06(24). Then solve equation 1 for x: x = 24 - y. Then substitute this value of x into equation (2): 0.08(24 - y) + 0.05 y = 1.44; 1.92 - 0.08y + 0.05y = 1.44; 0.48 = 0.03y; y = 16 gallons. If total is 24 gallons and we use 16 gallons of 5%, we must have 24 - 16 = 8 gallons of 8%.

 

Answer 2

I like to set up a chart for these kind of problems.  

 

So I have 

8% solution + 5% solution = 6% solution

 

So how much of each solution will I need to make 24 gallons of the 6% solution.  So if I let the 8% solution be x gallons, then the 5% solution will be what is left over or 24-x

 

8% solution + 5% solution = 6% solution

       x                  24-x             24

 

Then I need to multiply the top by the bottom first changing the percents to decimals.  

.08 x + .05(24-x) = .06(24)

Then solve

.08x + 1.2 - .05x = 1.44

.03x + 1.2 = 1.44

.03x + 1.2-1.2 = 1.44 - 1.2

.03x = .24

  x = 8 gallons.  

So I need 8 gallons of the 8 % solution, 24-8 = 16 gallons of the 5% solution to equal 24 gallons of a 6% solution.