Average – All Important Concepts and Solved Problems

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What is average?

In simple words, average is collecting things from persons and giving it back to everyone equally. In this process it may happen that no one actually had the value he/she has got after average.

How to find average?

There could be two type of data. One is ungrouped data and another is grouped data (where frequency will be given).

\text{Average} = \displaystyle\frac{\text{Sum \ of \ item}}{\text{No \ of \ item}}, for ungrouped data

\text{Average} = \displaystyle\frac{\sum fx}{\sum f}, for grouped data

Sum = Avg × n where, n= \sum f OR no of item

In average problems, the most important term is Sum.

Formula Based Average Problems

As we have stated earlier that sum is the main term, just remember the sum. Avg can be calculated just by dividing the sum by number n.

Description	Sum	Avg = \displaystyle\frac{\text{Sum}}{\text{n}}
First n natural number	\displaystyle\frac{n(n+1)}{2}	\displaystyle\frac{n+1}{2}
First n even number	n(n+1)	n+1
First n odd number	n2	n
First n perfect square (12, 22, 32, …, n2 )	\displaystyle\frac{n(n+1)(2n+1)}{6}	\displaystyle\frac{(n+1)(2n+1)}{6}
First n perfect cube (13, 23, 33, …, n3 )	\displaystyle\left[\frac{n(n+1)}{2}\right]^2	\displaystyle\frac{n(n+1)^2}{4}

Question: Find the average of first 18 natural numbers?
(a) 9
(b) 9.5
(c) 18
(d) 19

Ans: (b) 9.5

Explanation:

Sum = \displaystyle\frac{n(n+1)}{2}

Avg = \displaystyle\frac{n+1}{2} = \displaystyle\frac{18+1}{2} = 9.5

Question: Find the average of first 50 even numbers.
(a) 50
(b) 51
(c) 25
(d) 27

Ans: (b) 51

Explanation:

Sum = n(n + 1)

Avg = n + 1 = 50 + 1 = 51

Question: Find the average of squares of first 11 natural numbers.
(a) 46
(b) 66
(c) 110
(d) 115

Ans: (a) 46

Explanation:

Sum = \displaystyle\frac{n(n+1)(2n+1)}{6}

Avg = \displaystyle\frac{(n+1)(2n+1)}{6} = \displaystyle\frac{12\times 23}{6} = 46

Question: Calculate the average of squares of natural numbers from 1 to 10.
(a) 38.5
(b) 385
(c) 30.25
(d) 302.5

Ans: (a) 38.5

Explanation:

Sum = \displaystyle\frac{n(n+1)(2n+1)}{6}

Here, n = 10

Avg = \displaystyle\frac{(n+1)(2n+1)}{6} = \displaystyle\frac{11\times 21}{6} = 38.5

Question: Find the average of cubes of first 10 natural numbers?
(a) 300
(b) 302.5
(c) 315
(d) 240

Ans: (b) 302.5

Explanation:

Sum = \displaystyle\left[\frac{n(n+1)}{2}\right]^2 = \displaystyle\frac{n^2{(n+1)}^2}{4}

Avg = \displaystyle\frac{n(n+1)^2}{4} = \displaystyle\frac{10\times 121}{4} = 302.5

Question: Find the average of first 20 multiples of 3.
(a) 31.5
(b) 10.5
(c) 30
(d) 60

Ans: (a) 31.5

Explanation:

3 ⟶ 3, 6, 9, 12, 15
    = 3×1, 3×2, 3×4, … , 3×20

Sum = \sum 3n = 3\sum n = 3\times \displaystyle\frac{n(n+1)}{2}

Avg = 3\times \displaystyle\frac{n+1}{2} = 3 × 10.5 = 31.5, This is nothing but avg of first 20 natural numbers times 3

Question: The average of first five multiples of 4 is
(a) 9
(b) 12
(c) 15
(d) 10

Ans: (b) 12

Explanation:

Here, n = 5

Avg of first 5 natural numbers = 3

Avg = 4 × 3 = 12

Question: If the average of first n consecutive odd number is 60, then find the sum of the numbers.
(a) 1830
(b) 3600
(c) 3660
(d) 7381

Ans: (b) 3600

first n consecutive odd number is another way to say first n odd numbers

For first n odd numbers sum = n² and avg = n

It is given that avg = 60

So, n = 60

Hence, sum = 60² = 3600

Change in each element is the change in average

Average is a assumed to be equal value amongst the each member of a group. Whatever change is made with each member, equal change will be reflected in average. See the following questions for clarity.

Question: Average age of 7 members of a family is 33 years. What will be the average age of them after 4 years?
(a) 34
(b) 37
(c) 37.5
(d) 52

Ans: (b) 37

Explanation:

Current avg = 33 years

After 4 years age of everyone will increase by 4.

Hence new avg = 33 + 4 = 37

Question: Average of 10 observations is 36. If every result is divided by 2, what will be the new average of 10 observations?
(a) 34
(b) 22
(c) 18
(d) 17

Ans: (c) 18

Explanation:

New avg = \displaystyle\frac{\text{old avg}}{2} = \displaystyle\frac{36}{2} = 18

Finding Single Missing Number

Question: Average of 11 observations is 50. If average of first 5 observations is 47 and that of last 5 observations is 52, then find the 6th number.
(a) 55
(b) 45
(c) 50
(d) 64

Ans: (a) 55

Explanation:

By Traditional Method

Sum (11 numbers) = 50 × 11 = 550

Sum (first 5 numbers) = 47 × 5 = 235

Sum (last 5 numbers) = 52 × 5 = 260

6th Number = 550 – (235 + 260) = 550 – 495 = 55

This is how we are taught to find answer in schools. But believe me there are many advanced method which can lead us to the answer orally or with very less work. See the following method which we will call as Deviation Method.

By Deviation Method

You are requested to watch this 1 min video for explanation written below

Group avg of 11 observations is 50. It is assumed that every observation is 50.

But it is given that avg of first 5 observations is 47. So, each observation in this group is 3 less. So, total deviation in this group is -3 × 5 = -15

Again avg of last 5 observations is 52. Each observation is 2 more. So, here deviation is +2 × 5 = +10

Net deviation = -15 + 10 = -5

We know, total deviation in avg = 0 always

This -5 will be balance by the only remaining 6th number.

Hence 6th number = 50 + 5 = 55.

For rest of the problems we’ll not write the story part.

Question: Average of 67 numbers is 87. If average of first 32 numbers is 83 and that of last 34 numbers is 89, then find the 33rd number.
(a) 147
(b) 226
(c) 261
(d) 218

Ans: (a) 147

Explanation:

33rd number = 87 + 60 = 147

Question: Average of 25 observations is 18. If average of first 13 observations is 16 and that of last 11 observations is 20, then find the 14th number.
(a) 14
(b) 22
(c) 26
(d) 28

Ans: (b) 22

Explanation:

14th No = 18 + 4 = 22

Question: In a series of 11 matches of cricket, average runs of Rohit Sharma is 85. In first 4 matches his average runs was 72 and in next 6 matches his average runs was 90. How many runs scored by Rohit Sharma in last match?
(a) 115
(b) 120
(c) 97
(d) 107

Ans: (d) 107

Explanation:

Runs in last match = 85 + 22 = 107

Finding Unknown from Multiple Missing Numbers

Question: Average of 25 numbers is 67. Average of first 10 numbers is 64 and that of last 13 numbers is 69. If 11th number is 25% of 12th number, then find the 12th number.
(a) 27.6
(b) 135
(c) 108.8
(d) 110.4

Ans: (d) 110.4

Explanation:

11th : 12th = 1 : 4

12th no = \displaystyle\frac{4}{5}\times 138 = 110.4

Question: A batsman played 40 innings with average score of 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average score of his remaining innings is 48 runs. What is the highest score of this batsman?
(a) 132
(b) 156
(c) 174
(d) 182

Ans: (c) 174

Explanation:

Lowest + Highest = 2 × 50 + 76 = 176

176 – 172 = 4, this will be equally distributed

So, the parts are 2 and 2 + 172 = 174

[Similar Question] : A batsman played 30 innings with average score of 40 runs. His highest score exceeds his lowest score by 100 runs. If these two innings are excluded, the average score of his remaining innings is 38 runs. What is the highest score of this batsman?
(a) 118 (b) 100 (c) 18 (d) 36 {Ans: (a) 118}

Questions on similar concept for practice
2018 (Q1); 2016 (Q8); 2014(Q2, Q6)

Kindly note the question numbers and practice from here.

Finding Unknown When Number Overlaps

Question: Average of 15 numbers is 40. If average of first 8 numbers is 36 and that of last 8 numbers is 42, then find the 8th number.
(a) 24
(b) 30
(c) 96
(d) 68

Ans: (a) 24

Explanation:

Let, the 8th number = x

Sum(15) = 15 × 40 = 600
Sum(first 8) = 8 × 36 = 288
Sum(last 8) = 8 × 42 = 336

Now, 288 + 336 – x = 600
⇒ 624 – x = 600
⇒ x = 24

Aliter (By Deviation Method)

8th number = 40 – 16 = 24

Question: Average of 15 numbers is 40. If average of first 8 numbers is 30 and that of last 8 numbers is 50, then find the 8th number.
(a) 30
(b) 45
(c) 40
(d) 35

Ans: (c) 40

Explanation:

Questions on similar concept for practice
2020 (Q5); 2019 (Q3); 2018 (Q3); 2013 (Q5)

Kindly note the question numbers and practice from here.

Average Due to Change in Existing Group

The following some problems are involves average when new member(s) included, or member(s) excluded or replacement is done.

Question: Average of a group of 24 students is 32 years. When age of teacher is also included average age is increased by 0.8 years. Calculate the age of the teacher.
(a) 26
(b) 22
(c) 36
(d) None of these

Ans: (d) None of these

Explanation:

By Traditional Method

Let, age of teacher = x years

Age of students = 32 × 24 = 768

Age of students plus teacher = 32.8 × 25 = 820

∴ 768 + x = 820
⇒ x = 52

Aliter (By Deviation Method)

If the teacher comes with same avg age, there would be no change in new avg. But he comes with such an age that he increases avg age by 0.8 years, i.e. he add age of 0.8 years to everyone including himself. So his age will be existing avg plus what extra he comes with.

His age = 32 + 0.8 × 25 = 32 + 20 = 52

Question: The average weight of 8 persons increased by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
(a) 80
(b) 85
(c) 75
(d) 90

Ans: (b) 85

Explanation:

This is a replacement case. So, no of member will remain same = 8.

By Traditional Method

Let, avg age = x, and age of new person = y

we need to find out y = ?

8x – 65 + y = 8 (x + 2.5)
⇒ 8x – 65 + y = 8x + 8 × 2.5
⇒ y = 65 + 8 × 2.5
       = 65 + 20
       = 85

Aliter (By Deviation Method)

If the new person was also 65 kg, there was no change in avg. But he increases weight of everyone by 2.5 kg including himself.

∴ Weight of new person = 65 + 8 × 2.5 = 65 + 20 = 85

Question: The average weight of 10 members of a family increases by 2.5 kg when a new member comes in place of one of them weighing 70 kg. Find the weight of new member.
(a) 45 kg
(b) 75 kg
(c) 72.5 kg
(d) 95 kg

Ans: (d) 95 kg

Explanation:

Wt. of new member = 70 + 10 × 2.5 = 95

Question: Average marks of a group of 19 students is 76. When marks of another student is included, average decreased by 2. Calculate the marks of another student.
(a) 48
(b) 46
(c) 36
(d) 40

Ans: (c) 36

Explanation:

76 – 2 × 20 = 76 – 40 = 36

Question: The average runs of a cricket player of 10 innings was 32. How many run must he make in his next inning so as to increase his average by 4?
(a) 76
(b) 46
(c) 36
(d) 32

Ans: (a) 76

Explanation:

32 + 4 × 11 = 32 + 44 = 76

Question: Average weight of m boys is 43. If the weight of teacher who weighs 63 kg is also included, the average becomes 45 kg. Find the value of m.
(a) 6
(b) 8
(c) 9
(d) 10

Ans: (c) 9

Explanation:

63 = 43 + 2(m + 1)
⇒ 2(m + 1) = 20
⇒ m + 1 = 10
⇒ m = 9

Aliter (Traditional Method)

Number = m, Avg = 43 ⇒ Sum = 43m

∴ 43m + 63 = 45(m + 1)
⇒ 45m + 45 = 43m + 63
⇒ 2m = 18
⇒ m = 9

Question: The average age of some students is 25. A teacher joins them of age 56.5, so that the new average of all students becomes 26.5. Find out how many students are there?
(a) 21
(b) 20
(c) 16
(d) 18

Ans: (b) 20

Explanation:

Let, number of students = x

∴ 56.5 = 25 + 1.5(x + 1)
⇒ 25 + 1.5x + 1.5 = 56.5
⇒ 1.5x + 26.5 = 56.5
⇒ 1.5x = 30
⇒ x = 20

Question: A batsman in his 12th innings makes score of 63 runs and thereby increases his average by 2. What is his average after 12th innings?
(a) 39
(b) 41
(c) 36
(d) 44

Ans: (b) 41

Explanation:

By Traditional Method

Up to 11th innings avg = x
After 12th innings avg = x + 2

We have to find out x + 2 = ?

Now, 11x + 63 = 12 (x + 2 )
⇒ 11x + 63 = 12x + 24
⇒ x = 39

∴ x + 2 = 41

By Deviation Method

In 12th innings, the batsman scores 63, for that avg increases by 2

To keep avg intact (as up to 11th innings) he should score = 63 – 2 × 12 = 63 – 24 = 39

This 39 is avg up to 11th innings

hence, avg after 12th innings = 39 + 2 = 41

Question: A batsman makes score of 87 runs in the 17th innings and thus increases his average by 3, Find his average after 17th innings?
(a) 39
(b) 46
(c) 42
(d) 40

Ans: (a) 39

Explanation:

Old avg = 87 – 3 × 17 = 87 – 51 = 36

New avg = 36 + 3 = 39

Question: 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. find the average age of the team.
(a) 23
(b) 56
(c) 76
(d) 18

Ans: (a) 23

Explanation:

C + WC = 26 + 29 = 55

To keep average intact they should have taken = 55 – 1 × 9 = 55 – 9 = 46

Hence avg age = \displaystyle\frac{46}{2} = 23

Aliter (by Traditional Method)

Age of captain = 26

Age f wicket keeper = 29

Let, avg age of team = x

Now, 11x – (26 + 29) = 9(x – 1)
⇒ 11x – 55 = 9x – 9
⇒ 2x = 46
⇒ x = 23

Question: The average age of 30 students of a class is 15 years. A student of 20 years left the class and two more students, whose age difference was 5 years, joined the class in place of him. If the average of the class was still 15 years, then find the age of the younger new student?
(a) 43
(b) 56
(c) 15
(d) 18

Ans: (c) 15

Explanation:

By Traditional Method

Let, age of new students are x and x + 5

Initial member = 30 and final member = 31

So, 15 × 30 – 20 + (2x + 5) = 15 × 31
⇒ 2x – 15 = 15
⇒ 2x = 30
⇒ x = 15

Aliter (by Deviation Method)

Group avg = 15

Member excluded: – 20 = -15 – 5, he created deficiency of 5 in sum
This will be balanced by new two members

Members added: 2 × 15 + 5 = 35

35 = 15 + 20

So, age of younger member is 15 years

Average with Alligation Concept

Though each problem of average can be solved using Alligation, a few problems will be solved more quickly in the cases like below

the above three averages will be given. In that case we could easily find out the ratio of numbers of group1 to group2.

Question: The mean weight of 180 students in a school is 50 kg. The mean weight of boys is 60 kg and that of girls is 45 kg. Find the number of boys and girls in the school.
(a) 100 & 80
(b) 60 and 120
(c) 70 & 110
(d) 150 & 30

Ans: (b) 60 and 120

Explanation:

Boys = \displaystyle\frac{1}{3}\times 180 = 60

Girls = 120

Aliter (By Traditional Method)

Let, no of boys = x and that of girls = y

So, 60x + 45y = 50(x + y)
⇒ 60x + 45y = 50x + 50y
⇒ 10x = 5y
⇒ \displaystyle\frac{\text{x}}{\text{y}} = \displaystyle\frac{1}{2}

∴ x : y = 1 : 2

Hence, boys = 60 and girls = 120

Question: The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is :
(a) 14
(b) 28
(c) 21
(d) 30

Ans: (c) 21

Explanation:

1 part = 7
3 part = 21

Aliter (By Traditional Method)

Technician = 7, avg = 12000 ⇒ sum = 84000
Rest = x, avg = 6000 ⇒ sum = 6000x
Total = 7 + x, avg = 8000 ⇒ sum = 8000(x + 7)

So, 84000 + 6000x = 8000(x + 7)
⇒ 84 + 6x = 56 + 8x [By dividing both side by 1000]
⇒ 2x = 28
⇒ x = 14

∴ x + 7 = 21

Questions on similar concept for practice
2016 (Q4)

Kindly note the question numbers and practice from here.

Average with Ratio Concepts

Question: Of four numbers, whose average is 60, the first is one-fourth of the sum of the last three. The first number is :
(a) 12
(b) 48
(c) 36
(d) 32

Ans: (b) 48

Explanation:

If 1^st number is 1, sum of last 3 will be 4
Total sum = 1 + 4 = 5

5 part = 4 × 60
1 part = 48

Questions on similar concept for practice
2019 (Q5); 2017(Q11, Q20); 2013(Q1)

Kindly note the question numbers and practice from here.

Question: The average marks of 21 students in a class is 42. If 15 students with an average marks of 78 joins them, find the average marks of all students.
(a) 53
(b) 56
(c) 57
(d) 58

Ans: (c) 57

Explanation:

By Traditional Method

n = 21, avg = 42 ⇒ sum = 21 × 42 = 882

n = 15, avg = 78 ⇒ sum = 15 × 78 = 1170

Let, avg of group = x

∴ (21 + 15)x = 882 + 1170
⇒ 36x = 2052
⇒ x = 57

Aliter (By Deviation Method)

Here it is assumed that avg is 42 for all numbers.

For five numbers avg is 36 more. Hence more in sum = 5 × 36. This sum will be distributed among all students which is now 7 + 5 = 12, because we have taken the ratio.

∴ Avg of all students = 42 + \displaystyle\frac{36\times 5}{12} = 42 + 15 = 57

We could have have also assumed the bigger no as average

∴ Avg of all students = 78 – \displaystyle\frac{36\times 7}{12} = 78 – 36 = 57

But we would go with smaller no.

Question: In a class of 120 students, 45% are girls and the remaining are boys. If the average of the girls’ marks is 54 and that of boys’ is 46, what is the average of the whole class?
(a) 49.8
(b) 49.7
(c) 49.6
(d) 49.5

Ans: (c) 49.6

Explanation:

Avg of whole class = 46 + \displaystyle\frac{8\times 9}{20} = 46 + 3.6 = 49.6

Question: Average of n numbers is 58. If each of 65% of the number is increased by 16 and each of remaining number is decreased by 9, then the new average of the numbers :
(a) 67.125
(b) 64.75
(c) 65.25
(d) 66.5

Ans: (c) 65.25

Explanation:

65% : 35% = 13 : 7
Let, total no = 13 +7 = 20
Existing avg = 58

Now, let’s find out the net change in the average

Net change in the avg = +16 – \displaystyle\frac{25\times 7}{20} = 16 – 8.75 = 7.25

New avg = 58 + 7.25 = 65.25

Aliter (By traditional Method)

Change in avg = \displaystyle\frac{0.65n\times 16 - 0.35n\times 9}{n} = 10.40 – 3.15 = 7.25

Avg will increase by 7.25

New avg = 58 + 7.25 = 65.25

Question: Average weight of a class is 60 kg and average weight of boys in the class is 80 kg. Ratio of boys to girls in the class is 5 : 4. If there are 72 students in the class , then find the average weight of girls in the class.
(a) 54 kg
(b) 42 kg
(c) 35 kg
(d) 45 kg

Ans: (c) 35 kg

Explanation:

B : G = 5 : 4

For 5 boys contribution is 100 kg more in total. In another word we say that for 4 girls contribution is 100 kg less in total.

For each girl this is = \displaystyle\frac{-100}{4} = -25

So, avg weight of girls = 60 – 25 = 35

Aliter(By Traditional Method)

B : G = 5 : 4 and B + G = 72
∴ B = 40, G = 32

Let avg of girls = x

So, 40 × 80 + 32x = 72 × 60
⇒ 100 + x = 135
⇒ x = 35

Question: In a class of 50 students, 40% are girls. The average marks of the whole class is 64.4 and the average of the boys’ marks is 62. What is the average marks of the girls?
(a) 69.4
(b) 66.4
(c) 68
(d) 68.2

Ans: (c) 68

Explanation:

G : B = 40% : 60% = 2 : 3

Avg = 64.4 + 3.6 = 68

Wrong Average

In this type of problem we have to find out the total deviation in the wrong sum. And this deviation will be adjusted accordingly.

Deviation will be = value taken – actual value

Question: The average of 8 observations was 25.5. It was noticed later that two of those observations were wrongly taken. One observation was 14 more than the original value and the other observation was wrongly taken as 31 instead of 13. What will be the correct average of those 8 observations?
(a) 21
(b) 21.5
(c) 22.5
(d) 29.5

Ans: (b) 21.5

Explanation:

Wrong sum = 8 × 25.5 = 204

Deviations in sum:
        14 = 14
31 – 13 = 18
________________
Total     = 32

32 is extra taken in sum

So, actual sum = 204 – 32 = 172

Hence actual avg = \displaystyle\frac{172}{8} = 21.5

Aliter(By deviation Method)

Deviations in sum:
        14 = 14
31 – 13 = 18
________________
Total     = 32

32 is extra taken in sum

Actual avg = 25.5 – \displaystyle\frac{32}{8} = 25.5 – 4 = 21.5

Question: Average temperature of a place of 25 days is recorded as 15 ^oC. Later it was found that average temperature of 4 days which was 14 ^oC by mistake taken as 16.5 ^oC. Find the original average temperature of that place within that duration.
(a) 14.8 ^oC
(b) 14 ^oC
(c) 14.4 ^oC
(d) 14.6 ^oC

Ans: (d) 14.6 ^oC

Explanation:

Extra taken = (16.5 – 14) × 4 = 10

Actual avg temperature = 15 – \displaystyle\frac{10}{25} = 15 – 0.4 = 14.6

Question: The average marks of 14 students was calculated as 71. but later it was found that there was an error in noting but the marks of two students as 42 instead of 56 and 74 instead of 32. What will be the correct average of the students?
(a)53
(b) 66
(c) 69
(d) 59

Deviation in sum:
(42 – 56) + (74 – 32)
= – 46 + 74
= 28

28 is extra taken in sum

Actual avg = 71 – \displaystyle\frac{28}{14} = 71 – 2 = 69

Average of AP Series

We have a dedicated post on this topic. you may read that from here.

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