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Averages, Mean, and Medians
The Arithmetic Mean
The arithmetic mean is the most commonly used measure of central tendency. It takes all values into account and redistributes them equally — if every person in a group gave the same amount, that amount would be the mean.
$$\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{\sum x_i}{n}$$
The arithmetic mean is the sum of all observations divided by the number of observations.
$\bar{x}$=Arithmetic mean(same as data)
$x_1, x_2, \ldots, x_n$=Individual observations(same as data)
$n$=Number of observations(count)
$\sum x_i$=Sum of all observations(same as data)
All values are equal
→The mean equals every individual value.
$n = 2$
→The mean is the midpoint between the two numbers.
Sum ↔ Mean Link: This is the most powerful relationship in average problems: $\sum x_i = \bar{x} \times n$. If you know any two of sum, mean, and count, you can find the third.
Affected by Outliers: A single very large or very small value can pull the mean far from the centre — a billionaire walking into a room of average earners spikes the mean dramatically.
Always a Value the Data Can Take: Unlike the median, the mean doesn't have to equal any actual observation, but it always lies between the minimum and maximum.
Two properties of the arithmetic mean solve many exam-style problems quickly without recalculating from scratch.
$$\sum x_i = n \bar{x} \quad | \quad \bar{x} + a = \text{new mean when each value increases by } a$$
The total sum equals count times mean, and shifting every observation by a constant shifts the mean by the same constant.
$a$=A constant added to (or subtracted from) every observation(same as data)
Each value is multiplied by $k$
→The new mean is $k \times \bar{x}$.
Each value is multiplied by $k$ then $a$ is added
→The new mean is $k\bar{x} + a$.
Deviation Sum is Zero: The sum of deviations from the mean is always zero: $\sum(x_i - \bar{x}) = 0$. Positive and negative deviations cancel perfectly.
Combined Mean Shortcut: If group 1 has mean $\bar{x}_1$ with $n_1$ items and group 2 has mean $\bar{x}_2$ with $n_2$ items, the combined mean is $\bar{x}_c = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$.
Weighted Average
A weighted average assigns different importance (weights) to different values. Not all observations count equally — a final exam worth 50% matters more than a quiz worth 10%, even though both produce a single score.
$$\bar{x}_w = \frac{w_1 x_1 + w_2 x_2 + \cdots + w_n x_n}{w_1 + w_2 + \cdots + w_n} = \frac{\sum w_i x_i}{\sum w_i}$$
Each value is multiplied by its weight before summing, then divided by the total weight.
$\bar{x}_w$=Weighted average(same as data)
$x_i$=Individual values(same as data)
$w_i$=Weight assigned to each value(varies (count, percentage, etc.))
All weights are equal
→The weighted average reduces to the ordinary arithmetic mean.
Blend / Mixture Problems: When two groups are mixed (e.g., blending tea priced at Rs 400/kg and Rs 600/kg), the price of the mixture is the weighted average of the two prices.
Alligation Method: A quick visual technique for mixture problems — the ratio of quantities equals the ratio of differences between the mixture value and each group's value.
Percentage Weights: If weights are given as percentages (e.g., 40% midterm, 60% final), you can multiply directly: $\bar{x}_w = 0.4 \times \text{midterm} + 0.6 \times \text{final}$.
The Median
The median is the middle value when all observations are arranged in order. It splits the data into two equal halves — 50% of values lie below it and 50% above it. Unlike the mean, it is unaffected by extreme values.
$$\text{Median} = \begin{cases} x_{\frac{n+1}{2}} & \text{if } n \text{ is odd} \\ \frac{x_{\frac{n}{2}} + x_{\frac{n}{2}+1}}{2} & \text{if } n \text{ is even} \end{cases}$$
For odd $n$, the median is the single middle value. For even $n$, it is the average of the two middle values.
$n$=Number of observations(count)
$x_k$=The $k$-th value in sorted order(same as data)
$n = 1$
→The median equals the single value (same as the mean).
All values are identical
→The median equals that value (same as the mean).
Sort First, Always: The median requires ordered data. For unsorted data like {7, 1, 4}, first sort to {1, 4, 7}, then pick the middle.
Resistant to Outliers: For salaries of {30K, 35K, 40K, 2,000K}, the median is Rs 37,500 but the mean is Rs 526,250. The median tells a more honest story.
Odd vs Even Matters: With 7 data points, the median is the 4th value ($\frac{7+1}{2} = 4$). With 8 data points, average the 4th and 5th values.
For grouped data presented in a frequency table, the median is estimated using the cumulative frequency to locate the median class, then applying an interpolation formula.
$$\text{Median} = L + \left(\frac{\frac{n}{2} - F}{f}\right) \times h$$
Linearly interpolates within the median class to estimate where the middle value falls.
$L$=Lower boundary of the median class(same as data)
$n$=Total frequency (sum of all frequencies)(count)
$F$=Cumulative frequency of the class before the median class(count)
$f$=Frequency of the median class(count)
$h$=Class width (upper boundary − lower boundary)(same as data)
Find the Median Class: Compute cumulative frequencies and identify the first class whose cumulative frequency reaches or exceeds $\frac{n}{2}$.
Assumption of Uniformity: This formula assumes data is evenly spread within each class — it's an estimate, not exact.
Mode and Range
The mode is the value that appears most frequently in a dataset. A dataset can have one mode (unimodal), two modes (bimodal), or many modes (multimodal). If every value appears exactly once, there is no mode.
$$\text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$
For grouped data, this formula estimates the mode within the modal class (the class with the highest frequency).
$L$=Lower boundary of the modal class(same as data)
$f_1$=Frequency of the modal class(count)
$f_0$=Frequency of the class just before the modal class(count)
$f_2$=Frequency of the class just after the modal class(count)
$h$=Class width(same as data)
Ungrouped Data: Simply count occurrences — the value with the highest count is the mode. For {3, 5, 5, 7, 8, 8, 8}, the mode is 8.
Useful for Categorical Data: The mean and median require numerical data, but the mode works for categories too — e.g., the most popular car colour.
No Unique Mode Possible: If two or more values tie for highest frequency, the data is bimodal or multimodal.
The range is the simplest measure of dispersion — the difference between the largest and smallest values. It tells you how spread out the data is, but only uses two values and ignores everything in between.
$$\text{Range} = x_{\max} - x_{\min}$$
Subtract the smallest observation from the largest to get the total spread.
$x_{\max}$=Largest value in the dataset(same as data)
$x_{\min}$=Smallest value in the dataset(same as data)
Quick Check: The range is fast to compute and gives an immediate sense of variability — perfect for a first glance.
Sensitive to Outliers: One extreme value can make the range misleadingly large. The interquartile range (IQR) is a more robust alternative.
Coefficient of Range: For comparing spreads of different datasets: $\frac{x_{\max} - x_{\min}}{x_{\max} + x_{\min}} \times 100$.
Average Speed Problems
When two journeys cover the same distance at different speeds, the average speed is NOT the simple mean of the two speeds. Because you spend more time at the slower speed, it must pull the average down — you need the harmonic mean.
$$\bar{v} = \frac{2v_1 v_2}{v_1 + v_2}$$
For equal distances, the average speed is the harmonic mean of the two speeds — total distance divided by total time.
$\bar{v}$=Average speed over the entire journey(distance/time)
$v_1$=Speed during the first leg(distance/time)
$v_2$=Speed during the second leg(distance/time)
$v_1 = v_2$
→Average speed equals $v_1$ (no difference between arithmetic and harmonic mean).
Derivation: For distance $d$ each way, total distance $= 2d$, total time $= \frac{d}{v_1} + \frac{d}{v_2} = d\left(\frac{1}{v_1} + \frac{1}{v_2}\right)$. Average speed $= \frac{2d}{d\left(\frac{v_1 + v_2}{v_1 v_2}\right)} = \frac{2v_1 v_2}{v_1 + v_2}$.
Always Less Than Arithmetic Mean: For $v_1 \neq v_2$, the harmonic mean is always strictly less than the arithmetic mean $\frac{v_1 + v_2}{2}$.
Three Equal Parts: For three equal distances at speeds $v_1, v_2, v_3$: $\bar{v} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}}$.
When two (or more) time periods have different speeds, the average speed is simply the arithmetic mean of the speeds, because each speed contributes equally to the total time.
$$\bar{v} = \frac{v_1 + v_2}{2} \quad \text{(for equal time intervals)}$$
With equal time spent at each speed, total distance = $(v_1 t + v_2 t)$ and total time = $2t$, so average speed = $(v_1 + v_2)/2$.
$t$=Duration of each time interval(time)
Same Time vs Same Distance: Same time → arithmetic mean. Same distance → harmonic mean. Memorise this distinction — it's a frequently tested difference.
General Case: For time periods $t_1, t_2, \ldots$ at speeds $v_1, v_2, \ldots$: $\bar{v} = \frac{v_1 t_1 + v_2 t_2 + \cdots}{t_1 + t_2 + \cdots}$ (a weighted average of speeds).
Missing Value and Algebraic Average Problems
Many average problems give you the mean and all but one observation, asking you to find the missing value. The key is always the sum–mean relationship: the total sum must equal the mean times the count.
$$\text{Missing value} = n \bar{x} - \sum \text{(known values)}$$
Subtract the sum of known values from the required total sum to isolate the missing value.
$n$=Total number of observations (including the missing one)(count)
$\bar{x}$=Given average (mean)(same as data)
New Person Joining: If a person with value $x$ joins a group of $n$ people with average $\bar{x}$, the new average is $\frac{n\bar{x} + x}{n + 1}$.
Person Leaving: If a person with value $x$ leaves a group of $n$ people with average $\bar{x}$, the new average is $\frac{n\bar{x} - x}{n - 1}$.
Replacement: If a person with value $a$ is replaced by a person with value $b$ in a group of $n$, the average changes by $\frac{b - a}{n}$.
Sometimes the average of a sequence or set of consecutive numbers is needed. For consecutive integers (or any arithmetic sequence), the mean equals the midpoint of the first and last terms.
$$\bar{x} = \frac{a + l}{2} \quad \text{(for equally spaced values)}$$
In any evenly spaced list, the mean equals the average of the first and last values.
$a$=First term(same as data)
$l$=Last term(same as data)
$n$ is odd
→The mean is the actual middle term.
Consecutive integers from $p$ to $q$
→Mean = $\frac{p + q}{2}$, count = $q - p + 1$.
Consecutive Integers Shortcut: The average of $1, 2, 3, \ldots, n$ is $\frac{n+1}{2}$. The average of $m, m+1, \ldots, n$ is $\frac{m+n}{2}$.
Sum from Average: Sum of first $n$ natural numbers $= \frac{n(n+1)}{2}$. Sum of first $n$ even numbers $= n(n+1)$. Sum of first $n$ odd numbers $= n^2$.
Integer Check: If the mean is not an integer, the individual values cannot all be integers (unless the count is even and the mean is a half-integer).
Application Strategy
Worked Examples
Problem Statement
A student's scores in 5 subjects are 72, 85, 90, 68, and 78. After a sixth subject, her average rises to 82. What did she score in the sixth subject?
arithmetic meanmissing valuesum-mean relationship
Strategic Toolkit
$$\text{Missing value} = n\bar{x} - \sum x_i$$
Step 1
Calculate the sum of the first 5 scores
Sum = 72 + 85 + 90 + 68 + 78 = 393
$$72 + 85 + 90 + 68 + 78 = 393$$
Step 2
Calculate the required total for 6 subjects
Required sum = 6 × 82 = 492
$$6 \times 82 = 492$$
Step 3
Find the sixth score
Sixth score = 492 − 393 = 99
$$\text{Sixth score} = 492 - 393 = 99$$
Problem Statement
A course has three components: assignments (weight 20%), midterm (weight 30%), and final exam (weight 50%). A student scores 75 in assignments, 60 in the midterm, and 88 on the final. What is the overall weighted average?
weighted averagepercentage weights
Strategic Toolkit
$$\bar{x}_w = 0.20 \times 75 + 0.30 \times 60 + 0.50 \times 88$$
Step 1
Multiply each score by its weight
Assignments: 0.20 × 75 = 15. Midterm: 0.30 × 60 = 18. Final: 0.50 × 88 = 44
$$15 + 18 + 44$$
Step 2
Sum the weighted scores
Weighted sum = 15 + 18 + 44 = 77
$$\bar{x}_w = 15 + 18 + 44 = 77$$
Step 3
Interpret
The student's overall weighted average is 77/100. Notice the final exam pulled the average up significantly despite the weak midterm.
Problem Statement
A delivery driver travels from Warehouse A to Warehouse B at 60 km/h and returns along the same route at 40 km/h. What is the average speed for the entire round trip?
average speedharmonic meanequal distances
Strategic Toolkit
$$\bar{v} = \frac{2v_1 v_2}{v_1 + v_2}$$
Step 1
Identify the formula
Same distance each way → harmonic mean, not arithmetic mean.
Step 2
Apply the harmonic mean formula
v̄ = (2 × 60 × 40) / (60 + 40) = 4800 / 100 = 48 km/h
$$\bar{v} = \frac{2 \times 60 \times 40}{60 + 40} = \frac{4800}{100} = 48 \text{ km/h}$$
Step 3
Verify with total distance and time
Let distance = 120 km each way. Time out = 120/60 = 2 h. Time back = 120/40 = 3 h. Average = 240/5 = 48 km/h. Confirmed.
$$\frac{240}{2+3} = 48 \text{ km/h}$$
Problem Statement
The weekly earnings (in Rs) of 9 workers are: 3200, 4500, 2800, 5100, 3900, 4200, 2900, 3800, 4600. Find the median weekly earning.
mediansorted order
Strategic Toolkit
$$\text{Median} = x_{\frac{n+1}{2}}$$
Step 1
Sort the data in ascending order
Sorted: 2800, 2900, 3200, 3800, 3900, 4200, 4500, 4600, 5100
Step 2
Identify the median position
n = 9 (odd), so median position = (9 + 1) / 2 = 5th value
$$\text{Position} = \frac{9+1}{2} = 5$$
Step 3
Read the median
The 5th value in the sorted list is Rs 3,900.
$$\text{Median} = 3{,}900$$
Problem Statement
The average weight of 8 players on a cricket team is 62 kg. If a 70 kg player is replaced by a 58 kg player, what is the new average weight?
arithmetic meanreplacementsum-mean relationship
Strategic Toolkit
$$\text{New average} = \bar{x} + \frac{b - a}{n}$$
Step 1
Find the change in total weight
The replacement removes 70 kg and adds 58 kg, a net decrease of 12 kg.
$$\text{Change} = 58 - 70 = -12 \text{ kg}$$
Step 2
Distribute the change across all players
Each player's share of the change = −12 / 8 = −1.5 kg
$$\frac{-12}{8} = -1.5 \text{ kg}$$
Step 3
Calculate the new average
New average = 62 − 1.5 = 60.5 kg
$$\text{New average} = 62 + \frac{58 - 70}{8} = 62 - 1.5 = 60.5 \text{ kg}$$
Memorize formulas
Arithmetic Mean
Sum divided by count
Formula
$\bar{x} = \frac{\sum x_i}{n}$
Weighted Average
Sum of weighted values divided by sum of weights
Formula
$\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}$
Harmonic Mean (2 speeds, same distance)
Average speed when equal distances are covered at different speeds
Formula
$\bar{v} = \frac{2v_1 v_2}{v_1 + v_2}$
Median (odd n)
The middle value of sorted data
Formula
$\text{Median} = x_{\frac{n+1}{2}}$
Mode (grouped data)
Estimated mode within the modal class
Formula
$L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) h$
Review
1
The arithmetic mean of the numbers 14, 22, 18, and 26 is:
Ans
20
Review
Tap to reveal
Sum all values
Add the four numbers together.
$$14 + 22 + 18 + 26 = 80$$
Divide by the count
Apply the arithmetic mean formula by dividing the sum by the number of values.
$$\bar{x} = \frac{80}{4} = 20$$
2
A student's final grade in a course is determined by two midterms (25% each) and a final exam (50%). If the student scores 72, 68, and 84 on these assessments respectively, what is the weighted average grade?
Ans
77
Review
Tap to reveal
Apply weights to each score
Multiply each score by its corresponding weight: midterm 1 (25%), midterm 2 (25%), final (50%).
$$0.25 \times 72 = 18,\quad 0.25 \times 68 = 17,\quad 0.50 \times 84 = 42$$
Sum the weighted contributions
Add the weighted scores and divide by the total weight (which equals 1 in this case).
$$\bar{x}_w = \frac{18 + 17 + 42}{1} = 77$$
3
A delivery driver covers a 120 km route at 60 km/h and returns over the same route at 80 km/h. What is the average speed for the entire round trip?
Ans
$\frac{480}{7}$ km/h
Review
Tap to reveal
Calculate time for each leg
Compute the time taken for the outbound and return trips using time = distance/speed.
$$t_1 = \frac{120}{60} = 2 \text{ h},\quad t_2 = \frac{120}{80} = 1.5 \text{ h}$$
Apply the average speed formula
Average speed = total distance / total time. Total distance = 240 km, total time = 3.5 h.
$$\bar{v} = \frac{240}{3.5} = \frac{480}{7} \approx 68.57 \text{ km/h}$$
4
Find the median of the data set: 13, 7, 21, 16, 9, 25, 11
Ans
13
Review
Tap to reveal
Sort the data
Arrange all seven values in ascending order to locate the middle position.
$$7,\ 9,\ 11,\ 13,\ 16,\ 21,\ 25$$
Identify the median position
For odd $n$, the median is at position $\frac{n+1}{2}$.
$$\text{Position} = \frac{7+1}{2} = 4 \implies \text{Median} = 13$$
5
The average of 5 numbers is 20. If the number 30 is removed, what is the average of the remaining 4 numbers?
Ans
17.5
Review
Tap to reveal
Find the total sum
Multiply the average by the count to get the total sum of all five numbers.
$$\text{Sum}_5 = 5 \times 20 = 100$$
Subtract the removed value and recalculate
Remove 30 from the sum and divide by the new count of 4.
$$\text{Sum}_4 = 100 - 30 = 70,\quad \bar{x} = \frac{70}{4} = 17.5$$
6
A factory produces items in two batches. Batch 1 has 300 units produced at a cost of Rs. 40 per unit. Batch 2 has 200 units produced at a cost of Rs. 60 per unit. What is the average cost per unit across both batches?
Ans
Rs. 48
Review
Tap to reveal
Calculate total cost for each batch
Multiply the unit count by the per-unit cost for each batch.
$$\text{Cost}_1 = 300 \times 40 = 12000,\quad \text{Cost}_2 = 200 \times 60 = 12000$$
Compute the weighted average
Divide the combined cost by the total number of units.
$$\bar{x}_w = \frac{12000 + 12000}{500} = \text{Rs. } 48 \text{ per unit}$$
7
In a survey, 12 people reported the following number of weekly online purchases: 3, 5, 3, 7, 2, 5, 3, 8, 5, 3, 6, 4. What is the mode and the range of this data?
Ans
Mode: 3, Range: 6
Review
Tap to reveal
Count the frequency of each value
Tally how many times each number appears in the data set to find the most frequent value.
$$3 \to 4,\quad 5 \to 3,\quad 2, 4, 6, 7, 8 \to 1 \text{ each}$$
Determine mode and range
The mode is the value with the highest frequency (3). The range is the difference between the maximum and minimum values.
$$\text{Mode} = 3,\quad \text{Range} = 8 - 2 = 6$$
8
A bus travels at 50 km/h for the first 3 hours and at 70 km/h for the next 2 hours. What is the average speed for the entire journey?
Ans
58 km/h
Review
Tap to reveal
Compute total distance and total time
Use $\text{distance} = \text{speed} \times \text{time}$ for each segment, then sum both distances and times.
$$d = 50 \times 3 + 70 \times 2 = 290 \text{ km},\quad t = 5 \text{ h}$$
Divide total distance by total time
Average speed is always total distance divided by total time, not the average of the speeds.
$$\bar{v} = \frac{290}{5} = 58 \text{ km/h}$$
9
The average salary of 5 employees in a small office is Rs. 40,000. When one employee earning Rs. 35,000 leaves and is replaced by a new employee, the average salary rises to Rs. 43,000. What is the new employee's salary?
Ans
Rs. 50,000
Review
Tap to reveal
Find the change in total salary
The average increased by Rs. 3,000 across 5 employees, so the total increase is $5 \times 3000$.
$$\Delta \text{Total} = 5 \times (43000 - 40000) = 15000$$
Add the increase to the departing employee's salary
The new employee must earn enough to cover both the departed salary and the total increase.
$$\text{New salary} = 35000 + 15000 = \text{Rs. } 50000$$
10
The average of $a$ and $b$ is 20, and the average of $b$ and $c$ is 28. If the average of $a$ and $c$ is 24, what is the value of $b$?
Ans
24
Review
Tap to reveal
Write the three sum equations
Convert each average statement into a sum of two variables.
$$a + b = 40,\quad b + c = 56,\quad a + c = 48$$
Solve for $a + b + c$
Add all three equations and divide by 2 to get the sum of all three variables.
$$2(a + b + c) = 144 \implies a + b + c = 72$$
Isolate $b$
Subtract the equation for $a + c$ from the total sum to find $b$.
$$b = 72 - 48 = 24$$
11
The average weight of 8 people increases by 2.5 kg when a person weighing 65 kg is replaced by a new person. What is the weight of the new person?
Ans
85 kg
Review
Tap to reveal
Calculate the total increase
Multiply the increase in average by the number of people to get the total weight change.
$$\Delta \text{Total} = 8 \times 2.5 = 20 \text{ kg}$$
Add to the replaced person's weight
The new person must account for both the departing weight and the extra 20 kg in total.
$$\text{New weight} = 65 + 20 = 85 \text{ kg}$$
Question 1 of 11
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1
The arithmetic mean of the numbers 14, 22, 18, and 26 is: