A non-algebraic approach would see that in order to have x amount of quarts of boric acid solution you could use averages to figure out how many quarts at 30% boric acid solution you would need to mix with 2 quarts of 10% boric acid solution.

When I say averages I mean if you scored 10% on 2 tests in school how many 30% tests would you need to get you a 20% average?

Well you might have seen that 20% is exactly half way between 10% and 30%. You already know that if you had just 1 test at 10% you would need 1 test at 30% to get an average of 20%---- But, you have 2 tests at 10% - so you need 2 test at 30% to get the 20.

This is analogous to the boric acid solution. You would need 2 quarts of 30% and mix it with the 2 quarts of 10% in order to get 20% boric acid solution (4 quarts total).

Algebraically you can set it up like you would an average..

From earlier you had (10% + 10%+ 30% + 30%)/4 = 20%

It should be pointed out theres a couple ways to work with percents. You can either write it in decimal: 10% becoming .10 , 50% becoming .5

or you can just write your percentage (minus the percent sign) over 100:

10% becomes 10/100 , 50% becomes 50/100 for example

You can enter 10/100 and 50/100 into a calculator and you'll get the .10 and .5 respectively, which is the decimal version.

Once again from earlier we had (10% + 10%+ 30% + 30%)/4 = 20% which is the same thing as writing (10/100 + 10/100 + 30/100 + 30/100)/4 = 20/100 this equals...

(80/100)/4 = 20/100 =

(80/100) X (1/4) = 20/100 =

20/100 = 20/100 , which is correct since both sides of the "equals" are equal.

So to set this up with an unknown you have:

(10/100 + 10/100 + 30x/100)/(2+x) = 20/100

(20/100 + 30x/100) = (20/100) X (2+x)

multiply by 100 and the 100's on the denominator cancel out

20 + 30x = 20 X (2+x) =

20 + 30x = 40 + 20x =

10x = 20

x = 20/10 = 2

So there you have an algebraic way of solving it...

As a shortcut you can skip right to the chase algebraically by setting up the problem as (# of quarts from Jar A) X ((Percentage of Solution) + (x amount of quarts of Jar B)) X (Percentage of Solution) = (# of quarts from Jar A + x amount of quarts from Jar B) X (Needed Percentage of Solution)

in this case we had : 2 X (10/100) + x X (30/100) = (2+x) X (20/100)

= 20/100 + 30x/100 = 40/100 + 20x/100 multiply by 100

* could have just gotten rid of the 100's from the beginning but you can make sure of no mistakes this way especially when doing other problems

For all other mixture problems you can see that the percentages when written over 100 allow you to cancel the 100's -- so for now on problems like this you can start wrting it as

2(10) + 30x = 20(2+x) =

20 + 30x = 40 + 20x =

10x = 20

x = 2

OK I might have been long winded on this one, but I just wanted to say more than I had to on the 1st real mixture problem. I'll be less long winded on the next problem of this type.

## Sunday, March 9, 2008

### Mixture Problem 2 Solution

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