ACT Math: Averages
Averages is a Statistics topic that appears constantly on the ACT, both as its own question type and buried inside word problems about grades, sports statistics, and mixtures. This free, complete lesson covers mean, median, mode, and range, finding a missing value given the average, weighted averages, and how outliers and new data points change an average. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, in the same four-answer-choice, calculator-allowed format as the current ACT. Everything here is free, and you can keep practicing afterward with the full ACT Math Question Bank linked below.
1. Mean, Median, Mode, and Range
The mean (average) is the sum of all values divided by how many values there are. The median is the middle value when the data is sorted, averaging the two middle values if there's an even count. The mode is the value that appears most often. The range is the difference between the largest and smallest value.
Worked Example: Find the mean, median, mode, and range of {4, 7, 7, 9, 13}.
Mean = (4 + 7 + 7 + 9 + 13) ÷ 5 = 40 ÷ 5 = 8
Median (already sorted, middle value): 7. Mode (appears twice): 7. Range: 13 − 4 = 9.
1. Find the mean of the data set: 12, 15, 18, 20, 25
Explanation: Sum: 12 + 15 + 18 + 20 + 25 = 90. Mean: 90 ÷ 5 = 18.
2. Find the median of the data set: 3, 8, 5, 12, 9
Explanation: Sorted: 3, 5, 8, 9, 12. With 5 values, the middle one is 8.
3. Find the median of the data set: 4, 9, 6, 11
Explanation: Sorted: 4, 6, 9, 11. With 4 values (even count), average the two middle values: (6 + 9) ÷ 2 = 7.5.
4. Find the mode of the data set: 2, 5, 5, 7, 9, 5, 3
Explanation: 5 appears three times, more than any other value, so it's the mode.
5. Find the range of the data set: 14, 22, 8, 31, 19
Explanation: Range = largest − smallest = 31 − 8 = 23.
2. Finding a Missing Value Given the Average
If you know the average and the number of values, you can find their sum: sum = average × count. This is the key to finding a missing value, find the required total sum, then subtract the known values.
Worked Example: Tom has taken 5 of 8 equally weighted tests and has an average score of exactly 78.0. What score does he need on the 6th test to bring his average after 6 tests up to exactly 80.0?
Sum of the first 5 tests: 5 × 78 = 390. Required sum after 6 tests: 6 × 80 = 480.
6th score = 480 − 390 = 90
1. The mean of four numbers is 15. If three of the numbers are 10, 18, and 20, what is the fourth number?
Explanation: Required sum: 4 × 15 = 60. Sum of the three known numbers: 10 + 18 + 20 = 48. Fourth number: 60 − 48 = 12.
2. A student has scores of 82, 90, and 76 on three tests. What score is needed on a fourth test to have an average of 85?
Explanation: Required sum: 4 × 85 = 340. Sum of the three known scores: 82 + 90 + 76 = 248. Fourth score: 340 − 248 = 92.
3. The average of five numbers is 20. If four of the numbers are 15, 22, 18, and 25, what is the fifth number?
Explanation: Required sum: 5 × 20 = 100. Sum of the four known numbers: 15 + 22 + 18 + 25 = 80. Fifth number: 100 − 80 = 20.
4. A basketball player has scored an average of 18 points per game over her first 6 games. How many total points has she scored?
Explanation: Total = average × count = 18 × 6 = 108.
3. Weighted Averages
A weighted average is used when different groups or categories don't contribute equally. Multiply each value by its weight (how many items it represents, or its percentage), add the results, and divide by the total weight.
Worked Example: If 20 students scored an average of 80 on a test and 30 students scored an average of 90, what is the average score for all 50 students?
(20 × 80) + (30 × 90) = 1,600 + 2,700 = 4,300
4,300 ÷ 50 = 86
1. A class of 25 students has an average test score of 78. Another class of 15 students has an average of 88. What is the combined average for all 40 students?
Explanation: (25 × 78) + (15 × 88) = 1,950 + 1,320 = 3,270. Divide by the total count: 3,270 ÷ 40 = 81.75.
2. A student's final grade is a weighted average: homework counts 20%, tests count 50%, and the final exam counts 30%. If the student earns 90 on homework, 80 on tests, and 85 on the final exam, what is the final grade?
Explanation: 0.20(90) + 0.50(80) + 0.30(85) = 18 + 40 + 25.5 = 83.5.
3. A car travels 100 miles at 50 mph and then 100 miles at 25 mph. What is the average speed for the entire trip?
Explanation: Average speed is total distance divided by total time, not a simple average of the two speeds. Time for the first leg: 100 ÷ 50 = 2 hours. Time for the second leg: 100 ÷ 25 = 4 hours. Average speed: 200 miles ÷ 6 hours ≈ 33.3 mph.
4. A mixture is made from 3 liters of a solution that is 10% acid and 2 liters of a solution that is 25% acid. What is the acid concentration of the mixture?
Explanation: (3 × 0.10) + (2 × 0.25) = 0.3 + 0.5 = 0.8 liters of acid, out of 5 total liters: 0.8 ÷ 5 = 0.16 = 16%.
5. In a weighted average, a category worth 40% of the grade has a score of 70, and a category worth 60% has a score of 90. What is the weighted average?
Explanation: 0.40(70) + 0.60(90) = 28 + 54 = 82.
4. How Outliers and New Data Affect the Average
Adding, removing, or changing a data point shifts the mean, and the size of the shift depends on how far the new or removed value is from the rest of the set. The mean is sensitive to outliers, extreme values far from the rest of the data, while the median is much more resistant to them.
Worked Example: A data set of 5 values has a mean of 20. A sixth value of 50 is added. What is the new mean?
Original sum: 5 × 20 = 100. New sum: 100 + 50 = 150.
New mean = 150 ÷ 6 = 25
1. A data set of 4 numbers has a mean of 12. If a fifth number, 32, is added, what is the new mean?
Explanation: Original sum: 4 × 12 = 48. New sum: 48 + 32 = 80. New mean: 80 ÷ 5 = 16.
2. Which measure of central tendency is most affected by an extreme outlier?
Explanation: Since the mean incorporates every value's exact size into the sum, one extreme value can pull it substantially. The median only depends on which value is in the middle position, so it's far less sensitive to an outlier's exact size.
3. A data set is {5, 6, 7, 8, 100}. Which best explains why the median is a better measure of center than the mean for this set?
Explanation: The mean here is (5+6+7+8+100)/5 = 25.2, far above almost every value in the set, while the median, 7, sits right where most of the data actually is.
4. A set of 6 test scores has a mean of 84. If the lowest score, 60, is removed, what is the new mean of the remaining 5 scores?
Explanation: Original sum: 6 × 84 = 504. New sum: 504 − 60 = 444. New mean: 444 ÷ 5 = 88.8.
Common Mistakes to Avoid
- Forgetting to average the two middle values for an even-sized data set. With an odd number of values, the median is a single data point; with an even number, it's the average of the two middle values, a very common ACT trap.
- Averaging two speeds, rates, or percentages directly when a weighted average is required. A trip with two different speeds over equal distances (but unequal times) is not just the plain average of the speeds, use total distance divided by total time instead.
- Solving a missing-value problem by averaging the known values alone. You need the total required sum (average × total count) first, then subtract the known values, not just average what's already known.
- Weighting every category equally in a weighted average. If categories are worth different percentages or represent different group sizes, multiply each value by its own weight before adding, don't just take a plain average.
- Assuming the mean and median behave the same way. A question describing a skewed data set or an outlier is testing whether you know the mean shifts toward extreme values while the median does not.
Frequently Asked Questions
Is there a shortcut for missing-value average problems?
The key relationship, sum = average × count, turns every missing-value problem into simple subtraction once you calculate the required total sum. Write that relationship down first before touching the specific numbers in the problem.
How can I tell if a question wants a simple average or a weighted average?
If every value in the group carries equal importance (like unweighted test scores or a basic list of numbers), use a simple average. If the values represent different-sized groups, different percentages, or different amounts (like liters of a solution or class sizes), you need a weighted average.
Why does removing or adding one value change the mean so predictably?
Because the mean is built directly from the total sum, adding or removing a value changes that sum by exactly the value added or removed, and then the total is redivided by the new count. That's why working through the sum, rather than trying to average things directly, is the most reliable approach.
Where can I practice more problems like these?
The Averages quizzes in the ACT Math Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested on the ACT Math section.