Word problem

Question

A chemist mixed some 15% peroxide solution with some 6% peroxide solution to make 300 ml of a 9% peroxide solution. How many milliliters of the 15% solution were used?

Answer 1

Hi Jackie,

 

This is a typical algebra word problem.

 

Let's see what we have and what we need to find/calculate.

 

F = ml of 15% peroxide solution

S = ml of 6% peroxide solution

 

 

F + S = 300

.15F + .06S = 300(.09)

 

Let's multiply the second equation by 100 to get rid of the decimals

 

F + S = 300
15F + 6S = 300(9)

 

With the elimination method, we need to eliminate one of the variables.  Let's eliminate the S.  To do so, multiply the first equation by -6

 

-6F - 6S = 300(-6)
15F + 6S = 300(9)

 

Add the two equations together

 

9F = 300(-6) + 300(9)

9F = -1800 + 2700 

9F = 900

F = 100

 

Using the very first equation,

F + S = 300

Evaluating this at F = 100 and solving for S

S = 300-100 = 200

 

So we need 100 ml of 15% peroxide solution and 200 ml of 6% peroxide solution.