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In the previous articles, we discussed how to calculate the average of a given series. But what if we eliminate, add, or replace some terms from that series. Will the average be affected? If yes, then how?

To find the answer, let us go through the concepts given below.

Concept 1

Let the average of n quantitiesa1, a2, a3……anbeA1.  Now a new number N is added to the series which increases the average to A2, then thevalue of the new number added will be:

N=n x (A2 – A1)+ A2

Or N= n × (increase in value of average) + A2

Derivation:

The average of n quantitiesa1, a2, a3……an is:

= A1    …….    (1)

After the addition of new number N, the average becomes,

= A2   ……   (2)

Subtracting 1 from 2, we get,

N = ((n+1) x A2) –(n x A1)

N=n x (A2– A1)+ A2

Example: The average age of 12 students is 40. If the age of the teacher is also included, then the average becomes 42. Then what will be the age of the teacher?

Solution: Here average age of students = 40

Number of students = 12

The new average age after including teacher = 42

So here n = 12, A1 = 40, A2 = 42

Using the formula N = n x (A2 – A1) + A2 we have,

N = 12 × 2 + 42 = 24 + 42 = 66

Therefore, the age of the teacher is 66 years.

Example: The average age of 30 students is 15 years. If the age of class teacher is included, the average increases by 1. What is the age of the class teacher?

Solution: We have N= n× (increase in value of average) +A2

N= 30 × 1 +16 = 46 years

Concept 2

Let average of n quantitiesa1, a2, a3……anbeA1. If a number is removed from the series and this decreases the average to A2, then thevalue of the number removed will be:

N = n x (A1– A2) + A2

Or N= n × (decrease in value of average) + A2

Derivation:

The average of n quantitiesa1, a2, a3……anis

…….    (1)

After the removal of number N, the average becomes,

……   (2)

Subtracting2 from 1, we get

N = n x (A1– A2) + A2

Example: The average age of 30 students and a teacher is 15 years. If the age of teacher is excluded, the average decreases by 1.What is the age of the class teacher?

Solution:Here n = 31, A1 = 15, A2 = 14

We have N= n × (decrease in value of average) + A2

N= 31 × 1+14=45years

Concept 3

Let average of n quantitiesa1, a2, a3……anbe A1. Now a number from the series say a1is replaced with a new numbersay M, and becauseof this,the average increased to A2, then,

M = a1 + n x (A2 – A1)

Or New term = Replaced Term + (increased in average × number of terms).

Derivation:

The average of n quantitiesa1, a2, a3……an is,

…….  (1)

Let us say a1 is replaced with M, then,

……   (2)

Subtracting(1) from (2), we get:

Ma1= n x (A2 – A1)

M= a1+ n x (A2 – A1)

Example: The average age of 6 students is increased by 2 years when one student whose age was 13 years replaced by a new boy.Find the age of the new boy.

Solution: The age of the boy will be = Age of the replaced boy +(increase in average × number of terms) i.e., the age of the newly added boy = 13 + 2 x 6 = 25

To get a tight hold on the above concepts let us solve the exercise on calculating averages below.

EXERCISE

Question 1:  Average weight of 5 men increased by 4 kg when one of them whose weight is 96kg is replaced by another man. What is the weight of the new man?

(1)108

(2) 116

(3) 126

(4) 130

Solution: Option 2

Let the initial average bez.So, the total weight will be 5z.

Let the weight of the new man be y kg and the new average is z+ 4.

We have 5z – 96 + y = 5 x (z + 4)

Þ 5z – 96 + y = 5z + 20

Þ y = 96 + 20 = 116 kg

Or we can use theformula:

Weight of new person added = weight of the person removed + (no of person ×increase in average)

Weight of new person added = 96+5 × 4=116 kg

Question 2: The average of a batsman after 30 innings was 50 runs per innings. If after the 31 innings his average increased by 4 runs,then what was his score in 31stinning?

(1) 174

(2) 175

(3) 176

(4) 177

Solution: Option 1

Here, Thenew average = 50+4=54 runs

N = n × (increase in value of average) + A2

N= 30 × 4 +54=174 runs

Or we have

Runs in 31st inning = runs in 31 innings – runs in 30 innings

Runs in 31st inning= 31 × 54- 30 × 50=1674 – 1500 = 174 runs

Question 3: The average age of 10 students is 18 years. If the age of class teacher is included, the average increases by 1. What is the age of the class teacher?

(1) 30

(2) 40

(3) 29

(4) 28

Solution: Option 3

N = n × (increase in value of average) + A2

N= (10 × 1) + 19 = 10 + 19 = 29 years.

Question 4: The batting average of a cricket player for 64 innings is 62 runs. His highest score exceeds his low­est score by 180 runs. Exclud­ing these two innings, the aver­age of remaining innings becomes 60 runs. His highest score was:

(1) 180 runs

(2) 209 runs

(3) 212 runs

(4) 214 runs

Solution: Option 4

Let the cricketer’s highest score bey runs. Then the lowest will be y – 180.

Therefore, the sum will be 60 × 62 + y + y – 180 = 64 × 62

Question 5: A cricketer has a mean score of 60 runs in 10 innings. Find out how many runs are to be scored in the eleventh innings to raise the mean score to 62?

(1) 83

(2) 82

(3) 80

(4) 81

Solution: Option 2

Runs in 10 innings = 60 × 10=600

Runs in 11 innings = 62 × 11=682

So, runs scored in 11th innings =682-600=82

### Averages Questions: Problems on averages you should solve for competitive examination preparation

Welcome to this exercise on Problems on Averages. In this exercise, we build on the basic concepts for finding the Average. As you explore this topic, you will come across questions where you will be needing to find averages where either a new term is added or subtracted, or a term is replaced. Such questions need optimized tackling and can be solved with ease by using the formulas and understanding the relationships highlighted in this Averages Questions article. The Averages Questions exercise comes into

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