### Important Aptitude Formulas on Average:

**1.**Average or Arithmetic mean of n quantities =Sum of n quantities/n

Here, n = Number of quantities.

**2.**Average of n consecutive natural numbers = (n + 1)/2

{Total of n consecutive natural number = n (n + 1)/2

For e.g. mathematically, 1 + 2 + 3 +…………. + 20 = 20 (20 + 1)/2 = (20 21)/2 = 210

Using formula; Here, n = 20, Therefore, Total = 20 (20 + 1)/2 = 210}.

**3.**Average of squares of n consecutive natural numbers = (n + 1) (2n + 1)/6

**4.**Average cubes of n consecutive natural numbers = n (n + 1)²/4

**5.**Average of n consecutive even numbers = (n + 1)

{ For e.g. mathematically, 2 + 4 + 6 + 8 + 10 + 12 = 42/6 = 7

Using formula; Here, n = 6, Therefore, average = (n + 1) = (6 + 1) = 7 }

**6.**Average of squares of n consecutive even numbers = 2 (n + 1) (2n + 1)/3

**7.**Average of n consecutive odd numbers = n

{ For e.g. mathematically, 1 + 3 + 5 + 7 + 9 + 11 = 36/6 = 6

Using formula; Here, n = 6, Therefore, average is n = 6 }

**8.**Average speed = distance traveled/Time taken

**9.**Average of two different speeds ‘

*x’*and ‘

*y’*to travel same distance = 2

*xy*/(

*x*+

*y)*

{
Suppose a man covers a certain distance at x kmph and an equal distance
at y kmph, then the average speed during the whole journey is 2

*xy*/(*x*+*y*) kmph. }**10.**Average of three different speeds

*x, y, z*to travel equal distance = 3

*xyz*/(

*xy*+

*yz*+

*zx)*

**11. If a person is replace by another person by which;**

(I) Average is increased, then

Age of new comer = Age of person left + no. of persons × increase in average age.

(II) Average is decreased, then

Age of new comer = Age of person left - no. of persons × decrease in average age

**12. If a person joins a group without replacing any person by which;**

(I) Average increased, then

Age of new comer = Previous average age + no. of persons (including new comer) × increase in average age.

Or, Age of new comer = Increased average age + no. of persons originally in the group × increase in average age.

(II) Average decreased, then

Age of new comer = Previous average age - no. of persons (including new comer) × decrease in average age.

**13. If a person leaves the group but nobody joins the group by which;**

(I) Average increased, then

Age of man left = Previous average age + no. of present persons × increase in average age.

(II) Average decreased, then

Age of man left = Previous average age + no. of present persons × decrease in average age.

**14. If the average marks obtained by**

*‘x’*candidates in an examination is*‘n’*. If the average mark of passed candidates is*‘p’*and that of failed candidates is*‘q’*. Then
(I) No. of passed candidates: = [Total candidates (Total average. - Failed average.)]/(Passed average - Failed Average)

(II) No. of failed candidates: = [Total candidates (Passed average. - Total average.)]/(Passed average - Failed Average)

**15.**The geometric mean of numbers,

*x*

*₁*

*, x*

*₂*

*, x*

*₃*

*… x*is given by,

_{n}

__Example: 01__**The average of 5 numbers is 7. When 3 new numbers are added, the average of the 8 numbers is 8.5. The average of the 3 new numbers is:**

__Solution:__
Sum of 3 new numbers = (8 ×8.5) -
(5 × 7) = 33

∴ Their average = 33/3 =
11

__Example: 02__**A cricketer has a certain average for 9 innings. In the tenth inning, he scores 100 runs, thereby increasing his average by 8 runs. His new average is:**

__Solution:__
Let, average of 9 innings be x

Then, (9x + 100)/10 = x + 8

or,
10x + 80 = 9x + 100

or,
x = 20

∴ The new average is = (x
+ 8) = 28 runs.

__Example: 03__**In a cricket team, the average of 11 players is 28 years. Out of these, the average ages of 3 groups of 3 players each are 25 years, 28 years and 30 years respectively. If in these groups, the captain and the youngest player are not included and the captain is 11 years older than the youngest player. What is the age of the captain?**

__Solution:__
Let the age of youngest player be x

Then, (3 × 25) + (3 × 28) + (3 × 30) + x + x
+ 11 = 11 × 28

or, 75 + 84 + 90 + 2x + 11 = 308

or, 2x = 48

or, x = 24

∴ Age
of the captain = (x + 11) = (24 + 11) = 35 years.

**SOLVED EXAMPLE**

**PART 1**

__Question No. 01__
The average
age of 30 boys of a class is equal to 14 years. When the age of the class teacher
is included the average becomes 15 years. Find the age of the class teacher.

(A) 30 years

(B) 35 years

(C) 45 years

(D) 52 years

Answer:
Option C

__Explanation:__
Total age of
30 boys = 14 × 30 = 420 years

Total age
when the teacher is included = 15 × 31 = 465 years

∴ Age
of the class teacher = 465 - 420 = 45 years

**Alternate Method: Direct Formula**

Age of new
entrant = New average + (No. of old members × change in average)

= 15 + 30 (15
- 14) = 45 years.

__Question No. 02__
The captain
of a cricket team of 11 members is 26 years old and the wicket keeper is 3
years older. If the ages of these two are excluded, the average age of the
remaining players is one year less than the average age of the whole team. What
is the average age of the team?

(A) 23 years

(B) 24 years

(C) 25 years

(D) None of
these

Answer:
Option A

__Explanation:__
Let the
average age of the whole team by

*x*years.
∴ 11

*x*- (26 + 29) = 9(*x*-1)
=> 11

*x*- 9*x*= 46
=> 2

*x*= 46
=>

*x*= 23.
So, average
age of the team is 23 years.

__Question No. 03__
The average
age of boys in the class is twice the number of girls in the class. If the
ratio of boys and girls in the class of 36 be 5 : 1, what is the total of the
ages (in years) of the boys in the class?

(A) 380

(B) 342

(C) 372

(D) 360

Answer:
Option D

__Explanation:__
Number of
boys = (36 ×

^{5}/_{6}) = 30 years
Numbers of
girls = 6

Average age
of boys = (2 × 6) = 12 years

Total age of
boys = (30 × 12) years = 360 years.

__Question No. 04__
A car owner
buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive
years. What approximately is the average cost per litre of petrol if he spends
Rs. 4000 each year?

(A) Rs. 7.98

(B) Rs. 8

(C) Rs. 8.50

(D) Rs. 9

Answer:
Option A

__Explanation:__
Total
quantity of petrol consumed in 3 years

= {(4000/7.50)
+ (4000/8) + (4000/8.50)} litres

= 4000 {(2/15)
+ (1/8) + (2/17)} litres

= (76,700/51)
litres

Total amount
spent = Rs. (3 × 4000) = Rs. 12,000

∴ Average
cost = Rs. (12,000 × 51)/ 76,700

= Rs. 6120/767

= Rs. 7.98

__Question No. 05__
In Ankit's
opinion, his weight is greater than 65 kg but less than 72 kg. His sister does
not agree with Ankit and she thinks that Ankit's weight is greater than 60 kg
but less than 70 kg. His mother's view is that his weight cannot be greater
than 68 kg. If all are them are correct in their estimation, what is the
average of different probable weights of Ankit?

(A) 67 kg.

(B) 68 kg.

(C) 69 kg.

(D) None of
these

Answer:
Option A

__Explanation:__
Let Ankit's
weight be

*x*kg.
According to
Ankit, 65 <

*x*< 72
According to
Ankit's sister, 60 <

*x*< 70
According to
Ankit's mother,

*x*≤ 68
The values
satisfying all the above conditions are 66, 67 and 68.

∴ Required
average = (66 + 67 + 68)/3 = 201/3 = 67 kg.

__Question No. 06__
Indian
shooting team of 8 persons joins in a shooting competition. One of them scored
85 points. If he had scored 92 points, the average score for the team would
have been 84. The numbers of points, the team scored was

(A) 672

(B) 665

(C) 645

(D) 588

Answer:
Option B

__Explanation:__
Let the
total score be x.

Therefore, (

*x*+ 92 - 85)/8 = 84
=> (

*x*+ 7) = 672
=>

*x*= 665

__Question No. 07__
If the
average marks of three batches of 55, 60 and 45 students respectively is 50,
55, 60, then the average marks of all the students is:

(A) 53.33

(B) 54.68

(C) 55

(D) None of
these

Answer:
Option B

__Explanation:__
Required
average = {(55 × 50) + (60 × 55) + (45 × 60)} / (55 + 60 + 45)

= (2750 + 3300
+ 2700)/160

= 8750/160

= 54.68

__Question No. 08__
Three years
ago, the average age of Raju, Ranjan and Ravi was 27 years and that of Ranjan
and Ravi, 5 years ago was 20 years. Raju’s present age is

(A) 30 years

(B) 35 years

(C) 40 years

(D) 48 years

Answer:
Option C

__Explanation:__
Present age
of (Raju + Ranjan + Ravi) = {(27 × 3) + (3 × 3)} = 90 years.

Present age
of (Ranjan + Ravi) = {(20 × 2) + (2 × 5)} = 50 years.

Therefore,
Raju’s present age = (90 - 50) = 40 years.

__Question No. 09__
The average
monthly income of ‘P’ and ‘Q’ is Rs. 5050. The average monthly income of ‘Q’
and ‘R’ is Rs. 6250 and the average monthly income of ‘P’ and ‘R’ is Rs. 5200.
The monthly income of ‘P’ is:

(A) 3500

(B) 4000

(C) 4050

(D) 5000

Answer:
Option B

__Explanation:__
Let

*P*,*Q*and*R*represent their respective monthly incomes. Then, we have:*P*+

*Q*= (5050 × 2) = 10,100 …..... (i)

*Q*+

*R*= (6250 × 2) = 12,500 …….. (ii)

*P*+

*R*= (5200 × 2) = 10,400 …….. (iii)

Adding (i),
(ii) and (iii), we get: 2(

*P*+*Q*+*R*) = 33,000 or*P*+*Q*+*R*= 16,500 …..... (iv)
Subtracting
(ii) from (iv), we get

*P*= 4000.
∴

*P*'s monthly income = Rs. 4000

__Question No. 10__
The average
of 5 consecutive numbers is ‘

*n*’. If the next two numbers are also included, the average will
(A) Increase
by 1

(B) Remain
same

(C) Increase
by 1.4

(D) Increase
by 2

Answer:
Option A

__Explanation:__
Let, 5
consecutive numbers be, x, x+1, x+2, x+3 and x+4

Their
average is = (5x + 10)/5 = (x + 2)

The next two
numbers are = (x + 5) and (x + 6)

Therefore,
Average of the numbers = [(5x + 10) + (x + 5) + (x + 6)]/7 = (7x + 21)/7 = (x
+3)

So, the
average increased by 1.

**PART 2**

__Question No. 01__
A cricketer,
whose bowling average is 12.4, takes 5 wickets for 26 runs and thereby
decreases his average by 0.4. The number of wickets taken by him before his
last match is

(A) 85

(B) 82

(C) 84

(D) 83

Answer:
Option A

__Explanation:__
Let the
number of wickets taken before the last match be

*x*.
Then, (12.4

*x*+ 26)/(*x*+ 5) = 12
=> 12.4

*x*+ 26 = 12*x*+ 60
=> 0.4

*x*= 34
=>

*x*= 34/0.4 = 340/4 = 85

__Question No. 02__
The average
weight of men ‘A’, ‘B’ and ‘C’ is 84 kg. Another man ‘D’ joints the group and
the average now becomes 80 kg. If another man ‘E’, whose weight is 3 kg more
than that of ‘D’, replaces ‘A’, then the average weight of ‘B’ ‘C’ ‘D’ and ‘E’
becomes 79 kg. The weight of ‘A’ is

(A) 70 kg

(B) 72 kg

(C) 75 kg

(D) 80 kg

Answer:
Option C

__Explanation:__
A + b + C =
(84 × 3) = 252 kg

A + b + C +
D = (80 × 4) = 320 kg.

Therefore, D
= (320 - 252) kg = 68 kg.

Given, E =
(68 + 3) kg = 71 kg.

Therefore, B
+ C + D + E = (79 × 4) = 316 kg.

Now, (A + b
+ C + D) - (B + C + D + E) = (320 - 316) kg = 4 kg.

So, A - E =
4 kg.

=> A = 4
+ E = 75 kg.

__Question No. 03__
The average
weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is
45.15 kg. Find the average weights of all the boys in the class.

(A) 47.55 kg

(B) 48 kg

(C) 48.55 kg

(D) 49.25 kg

Answer:
Option C

__Explanation:__
Required
average = {(50.25 × 16) + (45.15 × 8)} / (16 + 8)

= (804 +
361.20)/24

= 1165.20/24

= 48.55

__Question No. 04__
The average
age of a class is 15.8 years. The average age of the boys in the class is 16.4
years while that of girls is 15.4 years. What is the ratio of boys to girls in
the class?

(A) 1 : 2

(B) 3 : 4

(C) 3 : 5

(D) 2 : 3

Answer:
Option D

__Explanation:__
Let the
ratio be

*k*: 1
Then, (

*k*× 16.4) + (1 × 15.4) = (*k*+ 1) × 15.8
Or, (16.4 -
15.8)

*k*= (15.8 - 15.4)
Or,

*k*= 0.4/0.6 = 2/3
Therefore,
the required ratio =

^{2}/_{3}: 1 = 2 : 3

__Question No. 05__
The average
age of husband, wife and their child 3 years ago was 27 years and that of wife
and the child 5 years ago was 20 years. The present age of the husband is:

(A) 35 years

(B) 40 years

(C) 50 years

(D) None of
these

Answer:
Option B

__Explanation:__
Sum of the
present ages of husband, wife and child = {(27 × 3) + (3 × 3)} years = 90
years.

Sum of the
present ages of wife and child = {(20 × 2) + (5 × 2)} years = 50 years.

∴ Husband's
present age = (90 - 50) years = 40 years.

__Question No. 06__
The average
weight of

*A*,*B*and*C*is 45 kg. If the average weight of ‘*A*’ and ‘*B*’ be 40 kg and that of ‘*B*’ and ‘*C*’ be 43 kg, then the weight of ‘*B*’ is:
(A) 17 kg

(B) 20 kg

(C) 26 kg

(D) 31 kg

Answer:
Option D

__Explanation:__
Let A, B, C
represent their respective weights. Then, we have:

*A*+

*B*+

*C*= (45 × 3) = 135 …...... (i)

*A*+

*B*= (40 × 2) = 80 ......... (ii)

*B*+

*C*= (43 × 2) = 86 .........(iii)

Adding (ii)
and (iii), we get:

*A*+ 2*B*+*C*= 166 ........ (iv)
Subtracting
(i) from (iv), we get:

*B*= 31.
∴

*B*'s weight = 31 kg.

__Question No. 07__
Mr. '

*A*' travels to Mr. '*B*' 150 km away at an average speed of 55 km per hour and returns at 45 km per hour. His average speed for the whole journey in km per hour is
(A) 52.5

(B) 48.5

(C) 49.5

(D) 47.5

Answer:
Option C

__Explanation:__
Average
speed = 2

*xy*/(*x*+*y*)
= (2 × 55 × 45)/(55
+ 45) km/hr

= 49.5 km/hr.

__Question No. 08__
The average
of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

(A) 0

(B) 1

(C) 10

(D) 19

Answer:
Option D

__Explanation:__
Average of
20 numbers = 0.

∴ Sum
of 20 numbers (0 × 20) = 0.

It is quite
possible that 19 of these numbers may be positive and if their sum is

*a*then 20th number is (-*a*).

__Question No. 09__
A pupil's
marks were wrongly entered as 83 instead of 63. Due to that the average marks
for the class got increased by half (1/2). The number of pupils in the class
is:

(A) 10

(B) 20

(C) 40

(D) 73

Answer:
Option C

__Explanation:__
Let there
be

*x*pupils in the class
Total
increase in marks = (

*x*× ½) =*x*/2
∴

*x*/2= (83 - 63)
=>

*x*/2= 20
=>

*x*= 40

__Question No. 10__
A family
consists of two grandparents, two parents and three grandchildren. The average
age of the grandparents is 67 years, that of the parents is 35 years and that
of the grandchildren is 6 years. What is the average age of the family?

(A) 28

^{4}/_{7}
(B) 31

^{5}/_{7}
(C) 32

^{1}/_{7}
(D) None of
these

Answer:
Option B

__Explanation:__
Required
average = {(67 × 2) + (35 × 2) + (6 × 3)} / (2 + 2 + 3)

= (134 + 70 +
18)/7

= 222/7 = 31

^{5}/_{7}years.
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