need 70 percent alcohol solution. how many ounces of 30% must be mixed with 40oz of 80% to obtain 70% solution?

Question

i need to see the work as well as the solution please

Answer 1

 

Let x = amount of 30% solution added, in ounces

Let 40 + x = amount of 70% solution created, in ounces

 

30% = .3, 80% = .8, 70% = .7

 

From given info, we can construct the following 1-variable equation:

 

(.8)40 + .3x = .7(40 + x) 

 

32 + .3x = 28 + .7x

 

-28 + 32 = .7x - .3x

 

4 = .4x

 

x = 10

 

We need to add 10 ounces.

Answer 2

An easy way to solve this easily is to look at the amount first without the concentration:

 

Let x = amount of 30% solution

 

Amount of 30% + Amount of 80% = Amount of 70%

 

x + 40 = (x+40) //yes they are the same, but the emphasis of the parentheses denotes that it's a whole//

 

From there, just multiply each term by their corresponding concentration:

 

x(.30) + 40(.80) = (x+40)(.70)

 

.3x + 32 = .7x + 28

 

.4x = 4

 

x = 10 oz