About Soroban Expertise I have a systematic and orderly way to organize the facts in a word problem, which (usually) leads clearly to the necessary equation. I think I can help with all types of word problems.
Experience 38 years of teaching college-level math, mostly at a two-year college.
Education/Credentials BS and MS in mathematics, SUNY Albany
Expert: Soroban Date: 2/13/2008 Subject: Remedial College Math
Question I can't seem to figure out how to start.
A chemist mixed some 15% peroxide solution with some 6%peroxide solution to make 300 ml of a 9% peroxide solution. How many ml of the 15% solution was used.
Answer Hello, Stacy!
These "mixture problems" drove me crazy
until I learned to organize the information.
We will consider the amount of peroxide at each solution.
We have some with 15% peroxide, another with 6% peroxide.
We want 300 ml which is 9% peroxide.
Let x = amount of 15% solution.
This contains: (15%)(x) = 0.15X ml of peroxide
Then 300-x = amount of 9% solution.
This contains: (9%)(300-x) = 0.09(300-x) ml of peroxide.
The mixture will contain: 0.15x + 0.09(300-x) ml of peroxide.
We know that the final mixture is 300 ml which is 9% peroxide.
This contains: (9%)(300) = (0.09)(300) = 27 ml. of peroxide.
In the last two sentences, we described the amount of peroxide
in the final mixture IN TWO WAYS.
So they must be equal, right?
There is our equation! . . . 0.15x + 0.09(300 - x) = 27
