word problem

Question

You need 425 mL of a 45% alcohol solution. On hand, you have a 15% alcohol mixture. How much of the 15% alcohol mixture and pure alcohol will you need to obtain the desired solution?

You will need ____mL of the 15% solution
and _____mL of pure alcohol.

Answer 1

This problem is similar to the previous problem.

Only difference is the numbers. It is done the same way.

 

 Let X represent the amount of 15% alcohol solution.

 Let Y represent the amount of pure alcohol.

 

  X + Y = 425

 

 .15X + 1*Y = .45*425 = 191.25

 

Solving the first equation for Y and multiplying the 

second equation by 100 to clear the decimals, the

system is now:

 

 Y = 425 - X

15X + 100Y = 19125

 

Substituting the first equation into the second:

 

15X + 100 ( 425 - X) = 19125

 

15X + 42500 - 100X = 19125

 

42500 - 85X = 19125

 

42500 - 19125 = 85X

 

23375 = 85x

 

x=275

 

275 mL of the 15% alcohol solution is needed

as well as 425 - 275 = 150 mL of pure alcohol.

 

As a check, the total alcohol in the mixture is

  .15*275 + 150 = 191.25

 

That's 191.25/425 = .45 = 45% alcohol in the mixture.

 

The answer is in bold above, is checked, and proven to be correct.