Aptitude Made Easy | Average – codewindow.in

Various competitive examinations ask questions regularly based on Average. Average
is one of the the important aspect which we see in each of our observation.

Why average?

When you plan to take bike/car you will see how much average that car will be giving,
even you go for interview or placement you will be asked for average percentage of
graduation degree.
When we know average, we get high level idea. Basically to make us capable of getting
overall idea looking at data, average questions are asked in exams.

Important formulae:

1) Average =

 Sum of all numbers/ Number Count

2) Average speed when we travel from Source place A to B and come back
from B to A [Assuming distance will be equal as source and destination A, B]
2xy/x + y

Where:

x speed from Source(A) to Destination (B)
y speed from Destination(B) to Source (A)

Problem 1:
David obtained 76, 65, 82, 67, 85 marks out of 100 in English, Maths, Physics,
Chemistry and Biology. What is average of his marks?

Solution :

Average = Total of all marks / Number CountNumber of Subject

              = (76+65+82+67+85) / 5

              = 375/5

              = 75

Answer is 75 marks

Problem 2

A Student was asked to find arithmetic mean of numbers 3, 11, 7, 9, 15, 13, 8, 19, 7, 21,
14, x. He find mean to be 12. What should be number in place of x.

Solution:

Average is 12 and numbers count is also 12.

Average = Total of all marks / Number CountNumber of Subject

12          = (13+11+7+9+15+13+8+19+7+21+14+x) / 12

12          = (137+x) / 12

137 + x  = 144

x = 144 – 137 = 7

Answer is 7

Problem 3

What is Average of 50 natural numbers ?

Solution :

Sum of 1 to 50 (n natural numbers)
Sum of 1st n natural numbers = n*(n+1)/2

                                                 = 50*(50+1)/2

                                                 = 25 * 51

                                                 = 1275

Average = Sum of All Numbers / Number Count 

              = 1275 / 50

              = 255 / 10

              = 25.5

Simple Tricks

If you observe properly when you are identifying average of consecutive number:

= (1st Number + Last Number) / 2

= (1+50) / 2

= 25.5 

Answer is 25.5

Problem 4

Average of weight if A, B, C is 45 kg. If average weight of A and B is 40 kg. Average
weight of B and C is 43 kg. What is weight of B?

Solution :

Average weight of A, B and C is 45

Average = Sum of All Weights / Person Count

45 = Sum of All Weights / 3

Total of weights of A +B + C = 45 * 3 = 135
A and B average weight is 40
So A + B = 40 * 2 = 80
B and C average weight is 43
So B + C = 43 * 2 = 86
To find value of B we should first get value of either C or A
A + B = (A+B+C) – C
80 = 135 – C
C = 55 kg
To get value of B we have to solve
B + C = 86
B + 55 = 86
B= 31 kg

Answer is 31 kg

Problem 5:

Average of 50 numbers is 30. If 35 and 50 are removed from list what will be average of
remaining numbers?

Solution :

As average of 50 numbers is 30
So Total of all numbers would be
50 * 30 = 1500
As 35 and 40 discarded total would we reduced by
35 + 40=75

New Total would be: 1500 – 75 = 1425
As 2 numbers removed, numbers count will be 28

New Average = New Total / New number Count

                      = 1424 / 28

                      = 29.68

Answer is 29.68

Problem 6

Average of 11 numbers is 10.9. If average of 1st 6 numberd is 10.5 and that of last 6
numbers is 11.4. Find out middle number in series.
Solution :

Average = Sum of all numbers / Number count

10.9 = Sum of all numbers / 11

Total of all numbers = 10.9 * 11 = 119.9
1st 6 numbers average is 10.5
Total of 1st 6 numbers = 10.5 * 6 = 63
Last 6 numbers average is 11.4
Total of last 6 numbers = 11.4 * 6 = 68.4

Total of 12 numbers = Total of 1st 6 numbers + Total of last 6 numbers – Middle number
This is because as you can see middle number is common in both places and added
twice.

119.9 = 63 + 68.4 – Middle number
Middle number = 131.4 – 119.9 = 11.5

Answer is 11.5

Problem 7

Average of 36 students in group 14 years. When teachers age is added to it average
increases by 1. What is teacher’s age?

Solution :

Average weight = Sum of all weights / Number of students

Sum of all weights = 36 * 14 = 504

Assume teacher’s weight x.
Once teacher’s age is added average is increased by 1.
New average = 15
New Person count = 37
Student + Teacher weight = New average * New count
504 + x = 15 * 37
x = 555 – 504 = 51

Answer is Teacher’s age is 51 years

Problem 8

In 50 Over ODI match, in 1st 10 overs of cricket game run rate was only 3.2 What
should be run rate in remaining overs to reach target of 282?

Solution :

In 10 overs run rate was 3.2 runs
So total score after 10 overs = 10 * 3.2 = 32
Total target = 282 runs
Pending runs after 10 over = 282 – 32 = 250
Remaining overs = 50 – 10 = 40

Required run rate = Remaining Runs / Remaining Overs

= 250/40

= 6.25

Answer is Required run rate is 6.25

Problem 9

Grocer has sale Rs 6435, Rs 6927, Rs 6855, Rs 7230, Rs 6562 for 5 consecutive months. How much should be in 6th month to have average sale is 6500?

Solution :

In order to get average sale for 6 month 6500 Rs.
Total of sales of 5 months = 6500 * 6 = 39000 Rs

Total of initial 5 months = 6435 + 6927 + 6855 + 7230 + 6562
= 34009

6th Month sales should be = Total sales – 1st 5 month sales
= 39000 – 34009
= 4991

Answer is 6th month sale should be 4991 Rs

Problem 10

Motorist travel 150 km away at average of 30 km/hr and returns at speed of 50 km/hr
His average speed is?

Solution :

Please don’t confuse here with distance as it is not required at all.
Average speed when we travel from Source place A to B and come back from B to A
[Assuming distance will be equal as source and destination A, B] 

= 2xy / x + y

Where:

x speed from Source(A) to Destination (B)
y speed from Destination(B) to Source (A)

= (2*30*50)/(30+50)

= 3000/80

= 37.5

Answer is Average speed is 37.5 km/hr

Problem 11

A cricketer has certain average of 10 innings in 11th inning he scored 108 runs thereby
increasing average by 6 runs. His new average??

Solution :

We have last inning’s score as 108 and total number of innings 11.
Let us assume x is original average
So new average is x + 6
Total of 1st 10 innings + 108 = 11 * New average
10 * x + 108 = 11 (x + 6)
10x + 108 = 11x + 66
x = 108 – 66 = 42
Old average is 42
New average = x + 6 = 42 + 6 = 48

Answer is New average is 48

Problem 12
Average weight of 8 person is increases by 2.5 kg when new person comes in place of
one of them weighs 65 kg. What might be weight of new number?

Solution :

Average weight of 8 person increased by 2.5 kg.
So Original total weight would increase by 8 * 2.5 = 20 Kg
In order to get total increase by 20 Kg, new person’s weight should be more than 20
from old person

New person weight = old person weight + 20
New person weight = 65 + 20 = 85

Answer is new person’s weight is 85 Kg

Problem 13

Average of 10 number is 7 if each number is multiplied by 12 then average of new
number is

Solution :

Let us look at small example before we solve this:
Average of 1, 2, 3, 4, 5 will be 3
Let us multiply each number by 10
Average of 10, 20, 30, 40, 50 will be 30 [ 3 * 10 = 30 as all number multiplied by 10]
So you can observe if you multiply all number by common constant
Original average of 10 numbers is 7
If we multiply all numbers by 12, average would also will increase by 12
So new average = 7 * 12 = 84

Answer is 84

Problem 14

Average of 35 student is 16 years. Average of 21 students is 14 years
What is average age of remaining 14 students?

Solution :

35 student age average 16 years.
So total of 35 student ages = 35 * 16 = 460
21 students age average is 14 years
So total of 14 student ages = 21 * 14 = 294
We have to find average of pending 14 students
Total of ages of 35 students= Total of ages of 21 students +Total of ages of 14 students
460 = 294 + Total of Remaining 14 student ages
Total of remaining 14 student ages = 460 – 294 = 166

Average of 14 student ages = Total remaining 14 student ages / 14

= 166/14

= 19

Answer is average of remaining 14 students is 19 Years

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