CAT 2025 Lesson : Averages - Basics & Assumed Mean

1. Introduction

There will be about $1$ to $3$ questions on Averages in the Quantitative Ability section in entrance tests. Additionally, Data Interpretation questions also typically involve Percentages, Ratios and/or Averages. Therefore, it is imperative for you to have a strong understanding of Averages.

As a student, you might have calculated your own average marks across subjects, or the average marks of your class in a particular subject. Other common examples of averages in real life include characteristics such as height or weight. Averages are used in statistics to showcase the approximate values of any variable, such as the average income of a state, the average rainfall in a year, the average amount spent on food by families etc.

Data points tend to cluster around a central value. This is called central tendency. The commonly used measures of such central tendency are Mean, Median and Mode.

By Mean, we typically refer to Arithmetic Mean or average. The other kinds of Means - Geometric Mean and Harmonic Mean - are covered later in this chapter.

Other statistical concepts such as variance and standard deviation have not come up in entrance tests, and hence, we are not covering these concepts.

2. Arithmetic Mean or Average

Arithmetic Mean (AM) or Average is the sum of each item of a data set divided by the total number of items in the data set.

The formula for arithmetic mean, denoted as $\overline{x}$ for $n$ items, namely $x_{1}, x_{2}, x_{3}, ..., x_{n}$ , is

$\overline{x} = \dfrac{x_{1} + x_{2} + x_{3} + ... + x_{n}}{n}$ ⇒ $\overline{x} = \dfrac{\displaystyle\sum_{i=1}^n x_{i}}{n}$

Note: $\displaystyle\sum_{i=1}^n x_{i}$ is a representation of “sum of all $x_{i}$ values, where $i$ takes all integer values from $1$ till $n$ (both inclusive)”. Therefore, $\displaystyle\sum_{i=1}^n x_{i} = x_{1} + x_{2} + x_{3} + ... + x_{n}$

Example 1

Mr. And Mrs. Roy have $5$ children whose weights are $50$, $56$, $48$, $32$ and $40$ kg. What is the average weight of their children?

Solution

Average weight $= \dfrac{50 + 56 + 48 + 32 + 40}{5}$

$= \dfrac{226}{5} = 45.2$ kg

Answer: $45.2$ kg

2.1 Properties of Average

$1$) The average of any group of items always lies between the smallest and largest values of that group.

$2$) If the value of every item in a group is increased by $k$, then the average of the group increases by $k$.

$3$) If the value of every item in a group is decreased by $k$, then the average of the group decreases by $k$.

$4$) If the value of every item in a group is multiplied by $k$, then the average of the group becomes $k$ times itself.

$5$) If the value of every item in a group is divided by $k$, then the average of the group becomes ($\dfrac{1}{k}$) times itself.

Example 2

In a test, P, Q, R and S scored $40, 80, 60$ and $20$ marks respectively. If upon revaluation, they each got $6$ marks more, what is the average marks secured by them in the test?

Solution

Initial Average marks $= \dfrac{40 + 80 + 60 + 20}{4} = \dfrac{200}{4} = 50$

As each of their marks is increased by $6$,
New average marks $= 50 + 6 = 56$

Answer: $56$

2.2 Average of numbers in Arithmetic Progression

Arithmetic Progression (AP) is covered in Sequences & Progressions lesson. If a sequence is in AP, then the difference between any $2$ consecutive terms remains constant. Therefore, AP is of the form $a, a + d, a + 2d, ... , a + (n - 1)d$.

Average of numbers in AP $= \dfrac{(\text{first term} + \text{last term})}{2}$

Consecutive natural numbers are also in AP. For the first $n$ natural numbers, $1$ and $n$ are the first and last terms respectively.

∴ Average of first n natural numbers $= \dfrac{n + 1}{2}$

The following are additional averages derived from the above.
Average of first n odd natural numbers $=$ n
Average of first n even natural numbers $=$ n $+ 1$

Example 3

What is the average of the $15$ consecutive natural numbers starting from $45$?

Solution

We have learnt in Number theory that the number of consecutive integers between $x$ and $y$ (both inclusive) is $y - x + 1$.

Here, $x = 45$ and the number of terms is $15$.

∴ $y - 45 + 1 = 15$
⇒ $y = 59$

$15$ consecutive numbers starting from $45$ ends at $59$ (both inclusive) are in AP.

∴ Average $= \dfrac{45 + 59}{2} = \dfrac{104}{2} = 52$

Answer: $52$

Example 4

What is the average of the first $12$ multiples of $13$?

Solution

$\dfrac{(13 \times 1) + (13 \times 2) + ... + (13 \times 12)}{12} = 13 \times \dfrac{(1 + 2 + ... + 12)}{12}$

This is nothing but $13$ times the average of the first $12$ natural numbers.

The average of the first $12$ natural numbers $= \dfrac{1 + 12}{2} = 6.5$

∴ Average of the first $12$ multiples of $13 = 13 \times 6.5 = 84.5$

Answer: $84.5$

2.3 Assumed Mean or Assumed Average

So far, we have been calculating averages by the most common method - adding all the values and dividing the result by the number of values. You will find this difficult if there are a lot of large values. In such cases, another easier technique called Assumed Mean or Assumed Average method can be used.

In this method, any number can be taken as the Assumed Mean. Where the numbers are large and close to each other, taking a middle term as the Assumed Mean makes calculations easier.

Where $a$ is the Assumed Mean and $n$ is the Number of Terms,

Actual Average $= a + \dfrac{\displaystyle\sum_{i=1}^n (x_{i} - a)}{n}$

Taking $d_{i} = x_{i} - a$, the above can also be expressed as

Actual Average $= a + \dfrac{\displaystyle\sum_{i=1}^n d_{i}}{n}$

Let us use an example to calculate the average using this method.

Example 5

Eight students scored $465, 453, 485, 470, 479, 474, 463$ and $459$ marks in a certain test. What is the average of these marks?

Solution

The numbers are huge and close to one another. In the above set of numbers, $485$ is the highest and $453$ is the lowest. $470$ is somewhere between these and makes subtraction easy.

∴ The assumed mean method can be used here. Let the assumed mean be $470$.

Marks($x_{i}$)	Assumed Mean(a)	$\bm{d_{i} = (x_{i} - a)}$
$465$	$470$	$-5$
$453$	$470$	$-17$
$485$	$470$	$15$
$470$	$470$	$0$
$479$	$470$	$9$
$474$	$470$	$4$
$463$	$470$	$-7$
$459$	$470$	$-11$
Total		$\bm{-12}$

Average $= a + \dfrac{\displaystyle\sum_{i=1}^n d_{i}}{n} = 470 + \dfrac{(-12)}{8} = 470 - 1.5 = 468.5$

Answer: $468.5$