Back

Averages

ALL MODULES

CAT 2025 Lesson : Averages - Common Types

6. Common Types of Average Questions

6.1 Age based questions

With no changes to a group of n people,
1) Average age of the group will increase by

\bm{1}

year for every year that passes by.
2) Total age of the group will increase by

\bm{n}

years for every year that passes by.

Example 13

The average age of a family of 6 members, that include Mary, John and Oliver, was

45

2010

. Mary and John are married to each other and had a baby in

2014

. In

2016

, Oliver died at the age of

62

. What was the average age of the family in

2018

, if nobody else had a baby or died during this period?
(

1

)

39

(

2

)

41

(

3

)

43

(

4

)

45

Solution

Average age of the

6

members in

2014 = 45 + 4 = 49

Total age of the

6

members in

2014 = 49 \times 6 = 294

With a just-born baby, total age of

\bm{7}

members in

\bm{2014}

remains the same at

\bm{294}

.

The total age would have increased to

294 + 7 \times 2 = 308

= 308 in

2016

.

As a person aged

62

died, average in

2016 = \dfrac{308 - 62}{6} = 41

Average age of the members in

2018 = 41 + 2 = 43

Answer:

43

6.2 Consecutive Integers

For

n

consecutive integers

x_{1}, x_{2}, ... , x_{n}

1

) As they are in AP, Average of these integers,

\overline{x} = \dfrac{x_{1} + x_{n}}{2}

∴ Averages of consecutive integers will be in multiples of

\bm{0.5}

2

) If the smallest term

x_{1}

is removed, new average

= \overline{x} + 0.5

3

) If the largest term

x_{n}

is removed, new average

= \overline{x} - 0.5

∴ New average when an integer is removed is between

\overline{x} - 0.5

and

\overline{x} + 0.5

Example 14

A set of consecutive positive integers beginning with

1

is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is

35 \dfrac{7}{17}

. What was the number erased?
[CAT 2001]
(

1

)

7

(

2

)

8

(

3

)

9

(

4

) None of these

Solution

Average of a set of consecutive natural number is always a multiple of

0.5

. When one number is removed from this set, the average increases or decreases by a maximum of

0.5

as explained above.

When a number is removed, the average becomes

35 \dfrac{7}{17}

, which is between

35

and

35.5

. Therefore, the initial average should have been either

\bm{35}

\bm{35.5}

.

Denominator for the new average has

17

. Therefore, the new number of terms is a multiple of

\bm{17}

.

Let n be the initial number of terms on the board.
Average of

n

consecutive natural numbers

= \dfrac{n + 1}{2}

If the initial average

= 35

, then

\bm{n = 69}

. And, if initial average

35.5, \bm{n = 70}

. When the number of terms is reduced by

1

, we get the number of terms, which should be a multiple of

17

. This is possible only when the initial number of terms

\bm{n = 69}

and the new number of terms is

68 (17 \times 4)

.

∴ Initially, there were

69

integers on the board and the average was

35.

Sum of

69

integers

= 35 \times 69 = \bm{2415}

Sum of

68

integers

= 35 \dfrac{7}{17} \times 68 = \dfrac{602}{17} \times 68 = \bm{2408}

The number erased

= 2415 - 2408 = 7

Answer: (1)

7

6.3 Items joining or leaving a group

When

1

or more items join or leave a group,

New Average

= \dfrac{\text{New Sum of Values}}{\text{New Number of Items}}

Alternatively, we can assume that the items joining or leaving the group have the same value as the initial average and add or subtract the relative impact. This is shown in the following example's alternative solution.

Example 15

The average weight of

63

students in a class is

77

kg. When a new student joined the class, this average increased to

78

kg. What is the weight of the new student?

Solution

Let the weight of the new student be $x$ .

Initial Weight

= 63 \times 77 = 4851

New average

= \dfrac{4851 + x}{63 + 1} = 78

⇒

4851 + x

4992

⇒

x

141

Alternatively (Recommended Method)

Average of

63

students in the class

= 77

kg

Average remains unchanged, if the new student's weight was 77 kg.

However, the average increases by

\bm{1}

kg for the new total of

\bm{64}

students. This means

1

) The new student brought in additional weight

2

) Additional weight

= 1 \times 64 =

\bm{64}

kg

New student's weight

=

Assumed Weight

+

Additional Weight

= 77 + 64 = 141

Answer:

141

kg

Note: Calculation in the second method is minimal.

Example 16

The average weight of

56

students in a class is

43

kg. When

2

students left the class, the new average increased to

43.5

kg. What is the average weight of the

2

students who left the group?

Solution

Average of

56

students in the class

= 43

kg

Average remains unchanged if the

2

students had weighed

43

kg each, a total of

\bm{86}

kg.

However, when they left, the average increases by

\bm{0.5}

kg for the new total of

\bm{54}

students. This means

1

) The students who left took away less weight (or weighed less than the average)

2

) Reduced weight

= 0.5 \times 54 =

\bm{27}

kg

Weight of students who left

=

Assumed Weight

-

Reduced Weight

= 86 - 27 = 59

Average weight of the

2

students

= \dfrac{59}{2} = 29.5

kg

Answer:

29.5

kg

Note: The solution includes the thought process, which might make it long. But the calculation is minimal.

Example 17

Average runs scored by a player is the total runs divided by the number of matches played by him/her. Smriti had average of

30

runs in the first

24

matches. After the next

4

matches, her average reduced to

28.5

runs. What is the average of the runs she scored in her last

4

matches?

Solution

Average of

24

matches

= 30

Average remains unchanged if she scored

30

runs in each of the next

4

matches, totalling to

\bm{120}

runs

However, average decreases by

\bm{1.5}

runs for the new number of

\bm{28}

matches.

1

) She scored less runs in the additional matches

2

) Reduced Runs

= 1.5 \times 28 = \bm{42}

runs

Run in last

4

matches

= 120 - 42 = 78

Average in last

4

matches

= \dfrac{78}{4} = 19.5

Answer:

19.5

runs

6.4 Replacement

When one or more items leave a group and the same number of items join the group, the number of items remain unchanged. In these questions, we can only find the difference between the values of items leaving the group and those joining the group. We, however, need additional information to find their exact values.

Example 18

The average height of 11 players in a team is 178 cm. When Joel leaves the team and Jeff joins, the average height reduces by 3 cm.Which of the following is true?
(

1

) Joel is shorter than Jeff by

30

cm

(

2

) Joel is shorter than Jeff by

33

cm

(

3

) Joel is taller than Jeff by

30

cm

(

4

) Joel is taller than Jeff by

33

Solution

Total height of

10

players and Joel

= 178 \times 11 = 1958

Total height of

10

players and Jeff

= (178 - 3) \times 11 = 1925

∴ Joel is taller than Jeff by

1958 - 1925 = 33

cm

Alternatively (Recommended Method)

With Joel leaving and Jeff joining, the number of players is unchanged at

11

players.

As the average height reduces, Jeff's height is lower than Joel.

Impact on total height is

3

cm for

11

players.

∴ Joel is taller than Jeff by

\bm{3 \times 11 = 33}

cm

Answer: (

4

) Joel is taller than Jeff by

33

Loading Video....