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Arithmetic II

Averages

ALL MODULES

CAT 2025 Lesson : Averages - Median, Mode & Past Questions

7. Median and Mode

7.1 Median

Median is a middle value that lies between the higher half and the lower half of a finite list of numbers. The median, unlike mean, is not influenced by outliers.

For

n

terms arranged in ascending or descending order,

Median

= {\left(\dfrac{n + 1}{2} \right)}^{th}

term where

\bm{n}

is odd

Median

=

Mean of

{\left( \dfrac{n}{2} \right)}^{th}

term and

{\left( \dfrac{n}{2} + 1 \right)}^{th}

term where

\bm{n}

is even

Example 19

What is the median of

87, 54, 32, 47, 9006, 38

Solution

The

6

terms in ascending order are

32, 38, 47, 54, 87, 9006.

Median

=

Mean of

{\left( \dfrac{6}{2} \right)}^{th}

term and

{\left( \dfrac{6}{2} + 1 \right)}^{th} =

Mean of

3^{\text{rd}}

and

4^{\text{t}}

terms

= \dfrac{47 + 54}{2} = 50.5

Answer:

50.5

Note: In this example,

9006

will be considered an outlier as it is more than

100

times the largest value.

This might affect the mean but not the median. For instance, increasing the last term from

9006

to any bigger number does not affect the median, which will remain unchanged at

50.5

Example 20

If the mean and median of

5

natural numbers

1, 4, 6, 11, x

are equal, then how many different values can

x

take?
(

1

)

0

(

2

)

1

(

3

)

2

(

4

)

3

Solution

As there are

5

terms, the median is the

{\left( \dfrac{5 + 1}{2} \right)}^{th} = 3^{\text{rd}}

term. This

3^{\text{rd}}

term can be

4, 6

x

.

Mean

= \dfrac{1 + 4 + 6 + 11 + x}{5} = \dfrac{22 + x}{5}

Scenario 1: Median

= 4

\dfrac{22 + x}{5} = 4

⇒

x = -2

(Rejected, as

-2

is not a natural number)

Scenario 2: Median

= 6

\dfrac{22 + x}{5} = 6

⇒

\bm{x = 8}

Data set is {

1, 4, 6, 8, 11

}, where

6

is the median. This is accepted, i.e.,

x

can take only

1

value.

Scenario 3: Median

= x

\dfrac{22 + x}{5} = x

⇒

\bm{x = 5.5}

(Rejected, as

5.5

is not a natural number)

∴ Scenario

2

is the correct answer.

Answer: (

2

)

1

7.2 Mode

For n terms, mode is the term(s) with the maximum number of occurrences.

A set of numbers can have only

1

mean and

1

median. However, the set can have one or more modes.

Example 21

What is the mode for

2, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 8, 8, 8

?
(

1

)

4

(

2

)

8

(

3

)

4

8

(

4

)

4

and

8

Solution

4

and

8

occur the most number of times (

3

times).

∴ This set has

2

modes, i.e.

4

and

8

.

Answer: (

4

)

4

and

8

Note: Option (

3

) states "

4

8

". This is incorrect as both

4

and

8

are modes and not either one of them.

8. Past Questions

Question 1

Consider the set S

=

{

2, 3, 4

, ...,

2n + 1

}, where

n

is a positive integer larger than

2007

. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X

-

Y?
[CAT 2007]

		$0$
		$1$
		$\dfrac{n}{2}$
		$\dfrac{n + 1}{2n}$
		$2008$

Observation/Strategy

1

) The first even and odd terms are

2

and

3

respectively.

2

) The last even and odd terms are

2n

and

2n + 1

respectively.

3

) As the even or odd terms are in Arithmetic Progression, their average is the average of the first and last terms.

Average of odd integers in S

=

= \dfrac{3 + 2n + 1}{2} = n + 2

Average of even integers in S

=

= \dfrac{2 + 2n}{2} = n + 1

-

= n + 2 - (n + 1) = 1

Answer: (

2

)

1

Question 2

Rajiv is a student in a business school. After every test he calculates his cumulative average. QT and OB were his last two tests.

83

marks in QT increased his average by

2

75

marks in OB further increased his average by

1

. Reasoning is the next test, if he gets

51

in Reasoning, his average will be _____?
[XAT 2008]

		$63$
		$62$
		$61$
		$60$
		$59$

Observation/Strategy

1

) As the impact on average by each of the

2

tests are given, we can form

2

linear equations. So, we can solve

2

variables with these.

2

) Number of tests taken before QT can be one variable and the average marks before QT can be the other variable

Let the number of tests before QT be

n

and the average marks before QT be

a

. ∴ Total marks before QT

= an

After scoring

83

in QT, average increased by

2

.
New Average

= \dfrac{an + 83}{n + 1} = a + 2

⇒

an + 83 = an + a + 2n + 2

⇒

a + 81 - 2n \longrightarrow

(A)

After scoring

75

in OB, average increased by

1

.

New Average

= \dfrac{an + 83 + 75}{n + 2} = a + 3

⇒

an + 158 = an + 2a + 3n + 6

⇒

\bm{a = \dfrac{152 - 3n}{2}}\longrightarrow

(B)

Equating (A) and (B), we get

81 - 2n = \dfrac{152 - 3n}{2}

⇒

\bm{n = 10}

Substituting this in (A) or (B), we get

\bm{a = 61}

∴ If he gets

51

in his next test,
New Average

= \dfrac{an + 83 + 75 + 51}{n + 3} = \dfrac{610 + 83 + 75 + 51}{13} = \dfrac{819}{13} = 63

Answer: (

1

)

63

Question 3

Ramesh analysed the monthly salary figures of five vice presidents of his company. All the salary figures (in lakhs) are integers. The mean and the median salary figures are

5

lakh, and the only mode is

8

lakh. Which of the options below is the sum (in lakh) of the highest and the lowest salaries?
[XAT 2012]

		$9$
		$10$
		$11$
		$12$
		None of these

Observation/Strategy
Observation/Strategy

1

) As there are

5

terms, the median of

5

is the

3^{\text{rd}}

term.

2

) As the only mode is

8

, the

4^{\text{th}}

and

5^{\text{th}}

terms should be

8

, and none of the other terms should be equal.

3

) As the mean is

5

for

5

terms, sum of the terms

= 5 \times 5 = 25

4

) The salary figures (in lakhs) are integers.

Let the other

2

terms be

x

and

y

, such that

x < y

Sum of terms

= x + y + 5 + 8 + 8 = 25

⇒ x + y = 4

<br><br> As

x \ne y

, the only way this is possible is if

x = 1

and

y = 3

<br><br> Sum of highest and lowest salaries

= 8 + 1 = \bm{9}

<br><br> Answer: (

)

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