Mixture Problem 7 Solution

This is the solution to a new algebra mixture problem as asked by Califax who asked, "A 50-lb solution of acid and water is 20% (by weight). How much pure acid must be added to this solution to make it 30% acid?"

1/5 = 20 %  .....  3/10 = 30 %       x = the unknown amount in lbs we want to add of  100% or 1/1 or 1 or just x of pure acid.

50(1/5) + x = 3/10 (50 + x)  multiply by 10...

100 + 10x = 3(50 + x)

100 + 10x = 150 + 3x

7x = 50  =   50/7  = 7 and 1/7th lbs of pure acid needs to be added to make the solution 30% acid.

Mixture Problem # 6 Solution

This is the solution to the algebra mixture word problem # 7 which read:

"If a merchant has two types of tea, one worth $2.70 per kilogram and the other worth $3.00 per kilogram, how many kilograms of each type should the merchant use in order to produce 30 kilograms of a blend that is worth $2.95 per kilogram?"

Even though this involves money we treat it like a mixture problem.

You got to unknowns but if you look at it as having x = 1 amount and y = 30 -x then you just have x, and 30 -x so...

Its the x amount * the 270 cent worth tea plus the 30 - x amount * the 300 cent worth tea is going to equal 30 * 295 cent blend

270 * x + 300(30-x) = 30(295)

270x + 9000 - 300x = 8850

-30x = -150

30x = 150

x = 150/30 = 5

so 5 kilograms of the 270 cent or $2.70 tea and 30 -5 or 25 kilograms of the 300 cent or $3.00 tea would need to be mixed together to make 30 kilograms of the $2.95 cent tea.

check 270 * 5 + 300 * 25 = 30*295

1350 + 7500 = 8850
8850 = 8850

Tags: Mixture Problem Solutions, algebra word problem solutions

Mixture Problem 5 Solution

This is sort of like the opposite of the last problem. Instead of diluting it your giving 100% pure alcohol to increase its level. As a fraction 100% is always just 1. anything else is always over a 100 so 30/100 = 3/10ths



So in this problem you would have:

(3/10)20 + 1(x) = (20 + x)(1/2)

6 + x = 10 + 1x/2 *multiply by 2

12 + 2x = 20 + x

x = 8

So 8 quarts of pure alcohol would need to be added to increase the % to 50% from 30%.

just to check:

(3/10)20 + 8 = (20 + 8) 1/2

6 + 8 = 14 is true

Mixture Problem 4 Solution

75% is the same thing as 3/4 ths, which makes an easier setup than using .75 in the problem.

(3/4) * 8 gallons = 24/4 = 6 gallons--- so there is 6 gallons of pure alcohol in the 8 gallons.

The problem asks how much water needs to be added to dilute it to only 50% or so only half of the total quarts is alcohol.

A non-algebra way to think of it is to treat it like an average.

((3/4)*8 + 0(8) ) / (8 +x) = 1/2 which is basically saying if you have 8 total cups which is 75% alcohol how many extra cups (x) would you need to add to 8 to make 1/2 or 50%

Keep in mind that water is the same thing as saying 0% alcohol and then treat it like the other mixture problems.



So the problem would read:

(3/4)8 + 0(x) = (1/2)(8+x)

6 + 0 = 4 + 1/2(x) * multiply by 2
12 = 8 + x

x = 4

So 4 quarts of water would need to be added to dilute the alcohol solution to 50%.

Mixture Problem 3 Solution

This mixture problem deals with dollars instead of percentages. But, works the same way as before.

She has 10 pounds of $3 coffee for a total worth of $30 and x pounds of $5 coffee. The question asks of how much of the x $5 coffee is needed to mix with the 10 pounds of $3 coffee to make (10 + x) pounds of $4.50 coffee with a total value of $4.50 * (10 + x)

to set this up we would then have:

3(10) + 5x = (10 + x) * 4.50

30 + 5x = 45 + 4.50 x *multiply by 100 to get rid of decimal

3000 + 500x = 4500 + 450x

50x = 1500

x = 1500/50

x = 30

so 30 pounds of the $5 grade must be added to the 10 pounds of $3 grade to make $4.50 grade-- If you think about it that makes sense. If you had exactly the same amount of $5 coffee (10 pounds) as the $3 coffee-- then that would make an average grade of $4 coffee--- so you know it would have to be a lot more than just 10 pounds to increase the value close to $5-- which 30 pounds does.

But, just to check..

Does

30 + 5(30) = (10 + 30) * 4.50 ?

30 + 150 = 40(4.50)

180 = 180 correct

Mixture Problem 2 Solution

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.

Mixture Problem 1 Solution

A easy way to set up percentages is to always put that percentage as a number over 100. So 15% percent is the same thing as 15/100 ...

So if a mixture of 40 quarts has 15% alcohol then there is 40 X (15/100) = 4 X (15/10) = 60/10 =

6 quarts of alcohol in the mixture.