Quantitative Aptitude · Measures of Central Tendency and Dispersion

Arithmetic Mean for CA Foundation: Formulas, Methods and Solved MCQs

Updated 1 October 2026 · Fact-checked

Arithmetic mean is the sum of all observations divided by their number. For grouped data, use Σfx ÷ Σf. Use weights for weighted mean, total sums for combined mean, and the step-deviation method to shrink large values. For a corrected mean, fix the total first, then divide again.

Understand Arithmetic Mean

The arithmetic mean (AM) is the value you get when you share the total equally among all observations. If five students score 10, 12, 14, 16 and 18, the total is 70 and the mean is 14. It is the most common average and the one CA Foundation tests most.

The key idea is that the mean is tied to the total. Mean × number of observations = sum of observations. Almost every tricky question (combined mean, corrected mean, a changed or removed item) is solved by moving between mean and total.

For a frequency distribution, each value x appears f times, so the total is Σfx and the count is Σf. For a class-interval table, use the mid-point of each class as x. The result is an estimate, not an exact value.

A weighted mean is used when items are not equally important, such as marks in subjects with different credits. Each value is multiplied by its weight. A combined mean joins two or more groups into one, and it is a weighted mean where the weights are the group sizes.

The step-deviation method is a calculation shortcut. You shift the origin to an assumed mean A and divide by a common factor h, so you work with small numbers. The answer is identical to the direct method if done correctly.

Key formulas to remember

Simple AM (ungrouped): x̄ = Σx ÷ n; Equivalent to Σx = n × x̄. Use this to move between mean and total.
AM of a frequency distribution: x̄ = Σfx ÷ Σf; For class intervals, x is the mid-point of the class.
Weighted AM: x̄w = Σwx ÷ Σw; w is the weight of each value x.
Combined AM of two groups: x̄ = (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂); Extends to more groups by adding terms in numerator and denominator.
Step-deviation method: d = (x − A) ÷ h; x̄ = A + h × (Σfd ÷ Σf); A is an assumed mean and h is a common factor of the class width or differences.
Corrected mean: Correct Σx = Wrong Σx − wrong value + correct value; Wrong Σx = n × wrong mean. Divide the correct total by n (or the new n if items are added or removed).
Properties of AM: Σ(x − x̄) = 0; if y = a + bx then ȳ = a + b·x̄; Adding, subtracting, multiplying or dividing every item changes the mean in the same way. The mean is affected by extreme values.

How to solve Arithmetic Mean questions

Use this order for any AM question. It works for ungrouped, grouped, combined and corrected-mean problems.

1Identify the type: simple, grouped, weighted, combined, or a correction to a given mean.
2Convert the mean into a total using total = mean × number of items. Do this for every group or the wrong data.
3For class intervals, find mid-points. For weighted problems, list each value with its weight.
4Apply the matching formula. For large values, subtract a convenient A and divide by a common h to get d.
5Adjust the total if needed (remove a wrong item, add the correct one, add or remove observations).
6Divide the final total by the correct count. Check that you have not used the old n.
7Sanity check: the mean must lie between the smallest and largest value, and a combined mean must lie between the group means, nearer the larger group.

Quickest way: Total-and-balance shortcut with option elimination

When to use it: Use in the MCQ paper for combined mean, corrected mean and any grouped data with big numbers.

Think in totals, not means. Write each mean as n × mean at once.
For two groups, use the balance idea: the combined mean sits closer to the group with more items, and the distances are in inverse ratio of the sizes.
For a changed mean, find the change in total, then divide by n. Example: a value 10 too low in 50 items raises the mean by 10 ÷ 50 = 0.2.
For grouped data, pick A as a mid-point near the middle and compute only Σfd. Then x̄ = A + h × Σfd ÷ Σf.
Use range elimination: discard any option outside the smallest and largest value, or outside the two group means.
If a grouped table needs more than three minutes, mark it and move on. A wrong answer costs 0.25 marks.

Common mistakes in Arithmetic Mean

Averaging the group means directly in a combined mean problem.
It looks easy and works only when the groups are equal in size.
Fix: Always weight by group size: (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂).
Using class limits or class widths instead of mid-points in grouped data.
Students rush and read the class column as the value column.
Fix: Write a separate x column with mid-point = (lower + upper) ÷ 2 before doing anything else.
Dividing the corrected total by the old n after adding or removing items.
The habit of dividing by n carries over from the original question.
Fix: Write the new count explicitly. Remove or add items to both the total and the count.
Forgetting to multiply back by h and add A in the step-deviation method.
Students stop after finding Σfd ÷ Σf.
Fix: Finish with x̄ = A + h × (Σfd ÷ Σf). Compare it to the range of data to catch errors.
Forgetting that a change applied to every item changes the mean the same way.
Students treat each item separately or confuse mean with spread.
Fix: If each item gets +a, the mean gets +a. If each is multiplied by b, the mean is multiplied by b.
Mixing up weight and value in weighted mean.
Word problems list quantities and prices in different orders.
Fix: Decide first what is being averaged (for example, price). The other quantity is the weight.

Worked examples

Example 1

The mean of 20 observations is 15. It was later found that one value 25 was wrongly read as 52. The correct mean is: (a) 13.65 (b) 14.35 (c) 15.00 (d) 16.35

Show the solution

Wrong total = 20 × 15 = 300.
Correct total = 300 − 52 + 25 = 273.
Correct mean = 273 ÷ 20 = 13.65.

Answer: (a) 13.65

Example 2

A group of 30 students has a mean mark of 60. Another group of 20 students has a mean mark of 45. The combined mean of all 50 students is: (a) 52.5 (b) 54 (c) 55 (d) 57

Show the solution

Total of group 1 = 30 × 60 = 1,800.
Total of group 2 = 20 × 45 = 900.
Combined total = 2,700 for 50 students.
Combined mean = 2,700 ÷ 50 = 54.
Check: 54 lies between 45 and 60 and is nearer 60, the larger group. Option (a) 52.5 is the simple average of 60 and 45, so it is a trap.

Answer: (b) 54

Example 3

For the following data, the mean is: Class 0–10, 10–20, 20–30, 30–40 with frequencies 4, 6, 7, 3. Options: (a) 19.5 (b) 20.5 (c) 21.5 (d) 23

Show the solution

Mid-points: 5, 15, 25, 35. Take A = 25 and h = 10.
d = (x − 25) ÷ 10: −2, −1, 0, 1.
Σf = 4 + 6 + 7 + 3 = 20.
Σfd = 4(−2) + 6(−1) + 7(0) + 3(1) = −8 − 6 + 0 + 3 = −11.
Mean = 25 + 10 × (−11 ÷ 20) = 25 − 5.5 = 19.5.
Recheck by the direct method: Σfx = 20 + 90 + 175 + 105 = 390; 390 ÷ 20 = 19.5.

Answer: (a) 19.5

Exam tips

Convert every mean into a total first. Most questions then become one line of arithmetic.
In combined mean questions, check that your answer lies between the group means and nearer the larger group before marking it.
Expect corrected-mean questions with two errors, or with an item added or removed. Update both the total and the count.
For grouped data, a smart choice of A and h can cut your calculation time. Always recheck the sign of d for classes below A.
Avoid guessing when you cannot eliminate at least two options, because each wrong answer costs 0.25 marks.

Practice questions from Measures of Central Tendency and Dispersion

Arithmetic Mean: frequently asked questions

What is the difference between simple and weighted arithmetic mean?

Simple mean treats every item as equally important and divides the total by the count. Weighted mean multiplies each item by its importance (weight) and divides by the sum of weights. Use weighted mean when items differ in importance, such as credits or quantities.

How do I find the combined mean of two groups?

Multiply each group mean by its size to get its total, add the totals, and divide by the total number of items. The formula is (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂). Do not simply average the two means unless the groups are equal in size.

When should I use the step-deviation method?

Use it when values or mid-points are large or the class width is common, such as 10 or 20. It reduces the numbers you multiply. The final answer must be the same as in the direct method.

How do I solve corrected mean questions?

Find the wrong total as n × wrong mean. Subtract the wrong value, add the correct one, and divide by the correct count. If items are added or removed, change the count as well.