Mean, Median, Mode, In Reverse

When we are first taught about mean, median, and mode in algebra our first homework problems typically go something like this: Find the mean median and mode of the following set of numbers, 2, 5, 6, 6, 8, 10, 12. Everything is a whole number, they are already in order, and there are minimal calculations needed.

(For those of you keeping score, and since I know you took the time to figure it out 😉 the mean of that set was 7, median = 6, and mode = 6. Good job!)

But then things get a little harder. Fast forward to the last section of your homework and guess what?


You’re given the mean, median, and mode but not the numbers. You have to create the set of numbers. You were probably never shown how to do this because it’s not as easy as giving someone a formula. Most of the time people can fumble around on a problem like this and accidentally get an answer. The good news is there are many many ways to do this, because as long as your numbers have the specified values you are ok. But that’s not good enough for us. Here’s what to do.

When Things Are Reversed, Median Comes First

The best plan of action is to start with the median. The reason is simple. There can only be one median, and it MUST be in the middle of the data set. Thus we start with the median value and then strategically add an equal amount of numbers greater than and less than it. Each time we add on one side, we must add on the other.

In an ideal situation we would add the median and move to step #2. But if you are told in the homework problem that the total count of numbers in the set has to be EVEN then we must take another step. Since an even number of data points means there are two numbers “in the middle” we know that the median is the average value of them. In this case we still want to begin with the median, but we will choose two numbers that have the median as their average. Although we can do this many ways, we should keep it simple by choosing the closest numbers.

For example, if we are told our data set must have 14 total numbers and a median of 9, we know we must choose two middle numbers. I would suggest we start with 8 and 10. They average out to be 9, so it is still our median. NOTE: we could also simply choose 9 and 9 as our middle numbers, and that would work. Just be careful because if you do it this way you’ve created a median and a mode, since the number is repeated. Either way,

When Things Are Reversed, Mode Comes Second

After we have chosen one number (or two if we are forced) as our median, the next thing we want to address is the mode. A key thing to remember about mode is that it has to occur most often, but there is no rule as to how many times. So our second step is to add the mode number to our data set a few times. I usually suggest 3 times, because when you calculate mean there must be some room to add numbers to reach the exact calculation, so we should give ourselves some room for error in case a number needs to show up twice. Some times we are told that we can only have a few numbers in our set so space is limited. Regardless, it is best to start with median then add mode.

Remember, when you add the mode, you MUST add other numbers on the other side of the median. So if the mode is less than the median and you added it three times, you must add three numbers to your data set that are greater than the median. I usually just try to add three numbers in a row on the other side of the median.

In some rare cases you will be told there is no mode. This is great actually, because you can skip this step altogether and begin building for the mean.

Now Calculate The Mean In Reverse

Now we have to get very creative. There are many ways to do this next step. This is why some problems can be hard and others can be easy. Remember the mean is the sum of all the numbers divided by how many numbers are in the set. At this point you will have several numbers. But chances are they do not yet have the mean that you desire. So the best way is to put variables in the data set and solve with algebra. This will be a trial and error process, and may involve adding many numbers to your set.

First we add two variables (call them A and B, or whatever you wish) that are less than the median. Then add two variables (C and D) that are greater than the mean. We will take all the numbers and add them. Then algebraically isolate the A + B + C + D and it will equal a specific number. Then try to come up with a successful combination of numbers that add to solve the equation AND obey their place in the set.

This can be complicated, so let’s take a look at some examples.

Mean, Median, Mode in Reverse Examples:

Mean = 20, Median = 23, Mode = 17

First start with the Median number and add it to your set. Then we add the mode. In this case we add 20 three individual times. Our list so far is 20, 20, 20, 23. Since we added three numbers less than 23, we must add 3 numbers that are greater than 23. As you can watch in the video above in a perfect world the three numbers we added here would also have solved the Mean. This would have solved the whole problem.

Instead, since it didn’t work I just decided to add the numbers 24, 25, and 26 since they were greater than 23 but were not too big. With median and mode completed, we needed to move on to the last step. We added a number greater than 23 (variable D) and another less than 23 (variable E).

We substitute all this into the mean formula and get 20 (the mean) = (17 + 17 + 17 + E +23 + 24 + 25 + 26 + D) all divided by 9. We did inverse operations to remove the divided by 9. This means we multiplied both sides by 9. Then we added all the numbers, giving us 149. Our simplified step was 180 = 149 + E + D. We did a second inverse operation to get E + D by itself. This means we subtracted 149 from each side. 180 – 149 gives us 31. Our final equation that we needed to solve was 31 = E + D.

Remember, there are many ways to solve this, but we must choose numbers where one is less than 23 and the other is greater than 23. In this case I just chose E = 4. When this is substituted in we have 31 = 4 + D. This simplifies to D = 27. Our final set of numbers is 4, 17, 17, 17, 23, 24, 25, 26, 27. Remember this is not the ONLY possible answer.

Mean = 20, Median = 20, Mode = 20

In example problems it is common to be given parameters where two or more of them equal the same number. In this case if you’re clever, I know what you are thinking. If we used a set of one number, and that number was “20” we would be finished. Mean, median, and mode all equal 20!

This might work, but some homework problems make you have at least 5 numbers in the set, and they can’t all be the same. If that’s the case, let’s look at how to solve this.

We start with the median, which is 20. Then we move to mode, which is also 20. We will add 20 two more times to our set. Keep it balanced by adding a 20 to the left and one to the right. To get the mean, we know our target needs to be 20. Let’s add two variables, one less than 20 and one greater than 20. Since the first step to finding the mean is adding everything we get A + 20 + 20 + 20 + B. This divided by 5 (since we have 20 three times and an A and a B). All this equals 5.

Do inverse operations to eliminate the divided by 5. Do this by multiplying each side by 5. This leaves us with 100 = A + 60 + B. We do another inverse operation by subtracting each side by 60. This leaves us with 40 = A + B.

Now we are able to be creative. We want A + B to equal 40, but must have A less than 20 and B greater than 20. I just chose A = 15 to proceed. We substitute that into our last formula, which gives us 40 = 15 + B. We subtract 15 from each side and are left with 25 = B. We know our final set of numbers is 15, 20, 20, 20, 25.

Two More Examples of Mean Median Mode in Reverse:

Mean = 8, Median = 7, Mode = 12, #’s in set = 5

In this case our data set can only have 5 numbers, so we must choose wisely. We begin with the median as always. We add 7 to our set. Then we move to the mode. Since the mode is 12 and it must show up more than the others, we just add it twice, since we don’t have much room to spare. Since we added two numbers greater than the median, in order to keep it balanced we must add two numbers less than 7. But be careful because adding these final two numbers is all we have left. We need them to not only be less than 7, but we need them to also solve our mean.

Since as of now we don’t know what the numbers are we will call them variables A and B. We must add all the numbers as the first step to find the mean, so we add A + B + 7 + 12 + 12. We divide this by 5, and the whole thing is equal to 8. We do inverse operations on the 5, which means we multiply each side by 5. 8 times 5 gives us 40, and this equals A + B + 7 + 12 + 12. This simplifies to 40 = A + B + 31. We subtract 31 from each side, giving us 9 equals A + B. A and B both need to be less than 7, so I chose B equals 6. When we substitute this back in we have 9 = A + 6. Subtract 6 from both sides and we have 3 = A. Since both A and B are less than 7 we did not change the median, and since they are different we didn’t change the mode.

Our final answer is 3, 6, 7, 12, 12.

Mean = 15, Median = 7, Modes = 20 and 2, #’s in set > 12

Begin with the median of 7. Next we will add the modes. We are lucky that one mode is less than the median and one is greater than the median, so the set stays balanced when we add them. I chose to add the modes three times each. Right now our set is 2, 2, 2, 7, 20, 20, 20. We must add 6 more numbers to achieve our total set size. As we add numbers we must always keep the mean in mind.

Let’s add variables A, B, and C in the set below 7 and D, E, and F above 7. They don’t yet have to be in order (we can do that later) but it is important to remember that 3 will be less than the median and 3 will be greater than the median. We add everything together and divide it by how many numbers we have. This equals the mean of 15.

So our first algebra step is: 15 = (2 + 2 + 2 + A + B + C + 7 + D + E + F + 20 + 20 + 20) divided by 13. We do inverse operations by multiplying each side by 13. 15 times 13 gives us 195. We add the numbers on the other side which gives us 73 + A + B + C + D + E + F. Subtract 73 from each side, which gives us a simplified version of the mean. 122 equals A + B + C + D + E + F. Here is where we get creative. Since there aren’t too many numbers to choose from that are less than 7, I want to go ahead and just try a few numbers for A, B, and C to see if we could make them work.

Remember, this can be a trial and error process, so beware this step might have to be repeated with other guesses. I chose A = 4, B = 5, and C = 6. We need to substitute these in to see what we have left. 122 = 4 + 5 + 6 + D + E + F. This simplifies to 122 = 15 + D + E + F. Subtract 15 from both sides to get 107 = D + E + F. This means our last step is to pick three numbers than are greater than 7 than all add up to 107. I think we can do this. We don’t want to mess up the modes so I chose D =  10 and E = 30. You can choose what you’d like if you’ve made it this far. We are left trying to find F so that it satisfies our equations. We can substitute in to 107 = 10 + 30 + F to solve for F. Add the 10 and 30 to get 40, and subtract it from both sides. 107 – 40 gives us our answer, 67 = F.

Our final answer to this last example problem is 2, 2, 2, 4, 5, 6, 7, 10, 20, 20 , 20, 30, 67.

Whew, good job. It can be tough on problems that have many possible answers because there is an overwhelming feeling of possibly messing it up. I would love to know what you thought of these example problems. Did they help? Leave a comment below or send me and email at

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