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one canned juice drink is 255 orange juice; another is 10% orange juice. How many of each should be mixed together in order to get 15L that is 18% orange juice

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4 Answers


Hi, David.

The best way to set these problems up so that they make sense is to put them in a table format.

Use the variable x to represent the amount of 25% juice and y to represent the amount of 10% juice.

                         liters juice                percent oj               total liters oj

25% juice                  x                          .25                            .25x

10% juice                  y                          .10                            .10y

mixture                x + y = 15                  .18                        (15)(.18) = 2.7


You can use the last column to write your equation, that x liters of the 25% juice and y liters of the 10%juice give you 15 liters of 18% juice, or:

.25x + .1y = 2.7

Since x + y = 15, then y = 15 - x.  We can substitute 15-x  in our equation for  y  to eliminate one of the variables:

.25x + .1(15 - x) = 2.7

Use the distributive property to multiply:

.25x + 1.5  - .1x = 2.7        

then combine like terms:

.15x + 1.5 = 2.7                  

subtract 1.5 from both sides:

.15x = 1.2

and divide by .15:


X represented the amount of 25% juice, and we needed 15 liters all together, so we need 8 liters of 25% juice and 7 liters of 10% juice.


Hope this helps!



This question is a mixture question. These are often given in terms of liquids, but other mixtures are possible. There have been good answers for your specific problem. I present here a general approach to mixture problems.

Step 1: Analyze the problem

As with any text problem it is important to first realize what information you are given. Mixture problems contain information about mixture concentrations (how much P (part) is in the mixture), mixture amounts (can be measured in mass (e.g. kilogram), but most often in volume (e.g. liter).

  1. Question: Which part of the mixture does the question ask about (what is P)?
  2. Question: What is the total mixture?
  3. Question: What is unit used for the amount of the mixture.
  4. Question: What is unit used for the concentration of the mixture (e.g. percent, kg/liter, mol/liter).
  5. Determine the unit for the part from the answers to questions 3 and four.

In your example

  1. The part is pure orange juice (OJ)
  2. The whole mixture is the juice drink (Mix).
  3. The Mix amounts are measured in liter.
  4. The Mix concentrations are measured in percent.
  5. IA mixture concentration given in percent always means that the part and the mixture are measured in the same units. So here this means that OJ is also measured in liter.

Step 2: Set up the problem

Always set up mixture problems in a table. (What I suggest is very similar to Kathye P.'s table, but the rows and columns are switched. The advantage is that you can read off the equations needed in step 3more easily.

The general table for mixture problems:

Mxiture 1                      Mixture 2              Resulting mixture

Amount of Mix         <fill in this row in the units determined in question 3 above.
                                  if the value is unknown, use a variable.>

Concentration          <fill in this row in the units determined in question 4 above.
of Mix                         if the value is unknown, use a variable.>

Amount of P             <fill in this row in the units determined under 5 above.
                                 This is done by multiplying in each column the values of the two rows above>

In your example

Mxiture 1 (25%)               Mixture 2 (10%)                Resulting mixture

Amount of Mix              x                                      y                                15 L

Concentration             0.25                                 0.1                               0.18
(convert % to
a number)          

Amount of OJ             0.25x                               0.1y                              0.18 * 15


Step 3: Set up your calculation

The two rows with the amount of the part (P) and the amount of the mix give you your equations. Just add a plus between the first two values and an equal between the second and the third value. You will get two equations with two unknowns. Multiply out any numeric values, and then proceed to solve the two equations.  (If you don't know this, you need to spend some serious time reviewing it.)

In your example

from the first table row             x + y = 15

from the last table row             0.25x + 0.1y  = 0.18 * 15
   simplified                             0.25x + 0.1y = 2.7



x + y = 15
0.25x + 0.1y = 2.7


This way of setting up mixture problems will work no matter what is asked for. The most common question asks for volume of the mixtures necessary to produce a certain mixture, but other questions are possible.



This approach works for a variety of problems in amounts are not given directly but via ratios.

Common categories are

  • Two people/groups or machines working at different speeds (speeds are given in some unit per hour, such as room per hour, lawn per hour, items manufactured per hour, etc.
  • A vehicle traveling at two different speeds.

If you are interested in learning the general set-up so you don't have to learn different approaches for these similar problems, send me an email.





Hi David,

I think you mean 25%, not 255. If that is NOT what you mean, then ignore my answer. Assuming I’m right, here is how I would go about solving this one.

This is a mixing problem. There are volumes and concentrations to consider. What we need to find is the volume of each of the different orange juices that we will mix to make the final 15 L.

Let V be the volume of the 25% orange juice. We know that the total volume is 15 L, so it makes sense that the volume of the 18% orange juice is 15 – V.

Why? Because V + (15 – V) = 15, the final volume.

What else do we know? We know the concentrations of the 2 juices and the concentration of the final juice. I like to convert the percents to decimals. The formula to keep in mind is C1V1 = C2V2, or in the case of a mixing problem, C1V1 + C2V2+...+CnVn = CTVT, where CT is the final concentration and VT is the final volume.

In this case:

.25V + .1(15-V) = .18(15)

Solve for V:

Start by distributing the .1 and doing the multiplications.

.25V + 1.5 -.1V = 2.7

Subtract 1.5 from both sides

.25V + 1.5 -.1V – 1.5 = 2.7 – 1.5

.25V -.1V = 1.2

Combine like terms

.15V = 1.2

Divide by .15

V = 1.2/.15

V = 8

Now relate this to what we said V was, the volume of the 25% juice.

Volume of 25%: 8 L

Volume of 10%: 15 L – 8 L = 7 L

I am not sure what you mean by 255 orange juice.  I am assuming that it is a percentage, but it can't be 255%.

Assuming that you mean 25.5% orange juice and 10% orange juice, I am going to try to explain how to solve the problem, but not give the answer (as you request).

You need to start by designating variables for the volume of each concentration of orange juice.  For example, you can say that x is the volume of 25.5% orange juice and y is the volume of 10% orange juice.

Using the chosen variables, set up one equation where the volumes add up to volume of the final solution (x + y = 15).

Continuing with the same variables, set up an equation that involves the percentages multiplied by their volumes (.255x + .10y = .18 X 15).

Solve the system of equations using either substitution or elimination method to get the final solution.