Which 5 Types Of Average Problems You Can Expect In Bank Exams (And How To Solve Them)

Dear Reader, do you want to know important types of average problems you can expect in bank exams? You will find those below…

Type 1: Number Series Summation Based Averages: What Are They?

In number system topic you will see questions like, “find sum of first 25 natural numbers“, “find sum of squares of first 30 natural numbers“, etc. If these types of questions are combined with averages, you will get type 1 of average problems.

For example, below question falls under Type 1:

Example Question 1: Find the average of first 30 natural numbers:

To solve this, you have to know the formula for first n natural numbers, which is n(n+1)/2

Therefore, sum of first 30 natural numbers = 30 x (31)/2 = 465

Also, you know that average = sum of observations/ number of observations

Therefore, average of first 30 natural numbers = Sum of first 30 natural numbers / 30
= 465/30
= 15.5

View more number series formulas given below:

Type 2: Consecutive Even/Odd Type Problems:

You can expect problems based on average of consecutive odd or even numbers. To be clear, let us see an example.

Example Question 2: Average of 4 consecutive odd numbers is 16. Find the second odd number in the series.

Let us assume four odd numbers to be x,x+2,x+4,x+6

Note: Here we are assuming the numbers to be x, x+2, x+4 and x+6 because, difference between any two consecutive odd numbers is 2. If the question is on consecutive even numbers, again you can assume in similar way because, difference between two consecutive even numbers is also 2. Therefore, assumption is same for both odd and even number series.

Average of the above four odd numbers = Sum of the 4 odd numbers / 4 = 16 …(16 is given in question)

Therefore, (x)+(x+2)+(x+4)+(x+6)/4 = 16
(4x+12)/4 = 16
4x+12=64
4x=52
x=13

Second odd number = x+2 = 13+2 = 15

Type 3 : Change In Average Based Problems

You will find problems based on change in average when a new person joins or leaves an existing group.
Below example will help you to understand this type fully.

Example Question 3: In a classroom at a cbse school in Jaipur, average weight of 5 students is 70 Kgs. The average increased to 73, after Rohit joined as a new student. What is the weight of Rohit?

Case I: Before Rohit Joined
Average weight of 5 students = Sum of weights of the 5 students / 5
70=Sum of weights of the 5 students/5
Or Sum of weights of the 5 students = 70 x 5 = 350 Kg

Case 2: After Rohit Joined
If you assume x to be Rohit’s weight, new average = Sum of 6 weights of students / 6
= (Sum of weights of old 5 students + Weight of Rohit) / 6
= (350 + x)/6 = 73 (In question, you will find 73 is the new average)
Or, 350 + x = 438
x = 88 Kgs = Rohit’s weight

Type 4: Multiple Groups Based Average Problems

You will get problems involving 2 or more groups, their individual averages and combined average of all the groups.

You will understand type 4 after reading the below example…

Example Question 4: A total of 35 soccer players is divided into two teams of 15 and 20 players. The average weight of the first team is 60 Kg and that of the second team is 70 Kg. The team manager wants to know the average weight of the whole team. Help him to find the answer.

You know the formula for average = Sum of observations / Number of observations
Therefore, Sum of observations = Average x Number of observations
You will be using the above formula for this example.

Now consider group 1:
Number of members in group 1 = 15
Average weight for group 1 = 60
Sum of weights of all 15 members in group 1 = 15 x 60 = 900

Now consider group 2:
Number of members in group 2 = 20
Average weight for group 2 = 70
Sum of weights of all 20 members in group 2 = 20 x 70 = 1400

Now, average of the entire group = Sum of weights of all 35 members / 35
= Sum of weights of all members of group 1 + Sum of weights of all members of group 2 / 35
= 900 + 1400 / 35
= 2300/35
= 65.71

Type 5: Distance And Speed Based Averages

In bank exams, you may also get average questions based on distance and speed.

Example Question 5: A man travels by motorcycle from his home to office. He covers his first half of journey at 40 km/hr and realizes he is late. He then increases his speed by 50% for his remaining journey. Find his average speed for the whole distance (from home to office).

The man covers first half of his journey at 40 km/hr.
He increases his speed by 50% for remaining journey. Therefore, speed for his remaining journey = (50% of 40 + 40)
= 60km/hr.

You can solve this problem using two methods: Direct method and Shortcut method.

Here is your direct method:

Average speed = Total distance / Total time
Let the total distance for the whole journey is 2d
Then, first half distance = d and Second half distance = d

Time taken to travel first half = First half distance / Speed for first half journey
= d/40

Time taken to travel second half = Second half distance / Speed for second half journey
= d/60

Total time taken = d/40 + d/60

Now you can apply the above values in the below formula:
Average speed = Total distance / Total time
= 2d / (d/40 + d/60)
= 2 / (1/40 + 1/60)
= 2 / (10/240)
= 480/10
= 48 Kmph

Now, here is your shortcut method:

For such problems, you can use the below shortcut formula:
If x is the speed for one half of the journey and y is the speed for remaining half
The average speed for the whole journey = 2xy/x+y

In our case, x = 40 kmph and y = 60 kmph
Therefore, average speed = 2xy/x+y
= 2 x 40 x 60 / (40 + 60)
= 4800/100 = 48 kmph.