Word problem, yeah a PROBLEM that's an understatement! Help?

Question

One canned juice drink is 20% orange juice; another is 10% orange juice. How many liters of each should be mixed together in order to get 10L that is 11% orange juice?? WOW I can't even begin to comprehend how to do this...

How many liters of 20% OJ should be in the mixture?

How many liters of 10% OJ should be in the mixture?

Answer 1

9 liters of 10% and 1 liter of 20% should do it. ~
Essentially 10% of the 20% solution is spread over all 10 liters, which adds 1% per liter.
See? Piece of cake. Did that in my head.

Answer 2

Hi,

I wrote an answer for this and was almost done when my keyboard froze and I lost everything. Worse yet, I had to uninstall my keyboard and re-install it so that it finally works! Anyhow, here is what I wanted to give you:


First, the traditional way:

Let x = number of gallons of 10% OJ needed
Then 10 - x = number of gallons of 20% OJ needed

If we multiply the percent of orange juice times the liters of each solution, it gives us how many liters of pure orange juice we have. So our equation is %OJ x liters of 10% OJ + %OJ x liters of 20% OJ = 11%(10 total liters) I will omit decimals and just use 10,11,and 29 as the per cents. This equation is:

10x + 20(10 - x) = 11 (10)

10x + 200 - 20x = 110

-10x + 200 = 110

-10x = -90

x = 9 This means 9 liters of 10% OJ will be needed to mix with 10 - 9 or 1 liter of 20% OJ.


Here's the logical way to look at a mixture problem like this:

--+-------+---------+
10........11.........20
<-1/10-><-9/10->

You are mixing 10% and 20% solutions - these are 10 numbers apart. The mixture you are trying to make is at 11%, which is 1/10 of the entire distance from the 10% solution while it is 9/10 of the entire distance from the 20% solution.


Now switch these fractions and you need 9/10 of the 10 liters or 9 liters to be made from 10% OJ while you need to have 1/10 of 10 liters or 1 liter of 20% OJ for the mixture.



Another example of this:

You are mixing a 30% solution with an 80% solution to make 40 liters of a 60% solution. how much of each will you need?

-+---+---+---+---+---+
.30.40.50.60.70..80

30 and 80 are 50 apart. 60 is 30 away from the 30% solution and 20 away from the 80% end.

30 out of 50 total is 3/5 on 30% end and 20 out of 50 is 2/5 of total difference on the 80% end. NOW, REVERSE the fractions -

You will need 2/5 of 40 liters to be from the 30% solution.
This is 2/5 x 40 or 16 liters of 30% solution.

You will need 3/5 of 40 liters to be from the 80% solution.
This is 3/5 x 40 or 24 liters of 80% solution.


I hope that gives you a "logical" way to work out these problems, as well as by using equations.

I hope that helps!! :-)

Answer 3

x=liters of 20%
y=liters of 10%

x+y=10 (or x=10-y)
.20x+.10y=.11(x+y)
.20(10-y)+.10y=.11(10-y+y)
2-.2y+.10y=1.1
.9=.1y
y=9 if we use 9 liters of 10%, then we must use 1 liter of 20% since the total needs to be 10 liters.

Answer 4

You want 10 Litres of 0.11 concentration.

You need x Litres of 0.1 plus (10 - x) Litres of 0.2

0.1*x + 0.2*(10 - x) = 10 * 0.11

0.1 x + 2 - 0.2 x = 1.1
-0.1 x = 1.1 - 2 = - 0.9
x = 9
(therefore 10-x = 1)

9 of 10% plus 1 of 20%