**Dear Readers,**

We are providing you **Important Short Tricks on Average Questions** which are usually asked in **Bank Exams**. Use these below given short cuts to solve questions within minimum time. These shortcuts will be very helpful for your upcoming All **Bank Exam 2016.**

To make the **chapter easy for you all,** we are providing you all some **Important Short Tricks to Average Questions** which will surely make the chapter easy for you all.

**How to solve average questions?**

**Average of numbers= {sum of numbers/total No. of numbers}**

**Or you can use the below formula**

**Average= (sum of 1st term+sum of last term)/2**

Suppose you are given a question as given below:

**Find the average of 5,7,9,11,13?**

**Solution-**

**Here is a very simple trick for this type of questions.**

In this type of questions, first look for the **difference of numbers**. Since the difference is constant throughout i.e. 2. And given numbers are odd in numbers i.e. total 5 types of numbers.

Then in this type of questions, the answer is always the middle term i.e. 9 (in above question).

But if the numbers were given even in numbers means there are six numbers like 5, 7, 9 , 11, 13,15. Then the answer is always sum of middle terms or sum of opposite terms divided by 2 i.e. (5+15)/2 or (7+13)/2 or (9+11)/2. The answer is always same. In the above case the average will be 10.

Let us look at a simple question now.

**Example 1:**

The average mark of 70 students in a class is 80. Out of these 70 students, if the average mark of 40 students is 75, what is the average mark of the remaining 30 students?

Average of 70 students = 80.

Hence total marks of all the 70 students = 70×80 = 5600

Out of these 70, average of 40 students = 75

Hence total marks of these 40 students = 40×75 = 3000.

So the total marks of the remaining 30 students = 5600-3000 = 2600

Hence the average of the remaining 30 students = 2600/30 = 86.66

Now there will be a serious difference in the time taken if the numbers given here are not multiples of 10. Just look at example 2 below:

**Example 2:**

“Average of 75 students is 82, out of which average of 42 students is 79. What is the average of the remaining 33 students?

**Let’s go back to Example 1**

Given that average of 70 students is 80. This class is split into two groups of 40 and 30. I am sure you will agree with me that if the average of each of these groups is 80, then the average of the total group i.e. 70 students will also be 80.

But the first group of 40 students, instead of 80 they had an average of only 75. So definitely the average of the remaining 40 students must be greater than 80.

The loss incurred because of the first group must be compensated by the second group.

Let us look at the loss incurred because of the first group

We want them to have an average of 80, but they managed only 75. So we lost an average of 5 upon 40 students. So the loss in the sum = 40×5.

Now this loss of 40×5 must be compensated by our second group i.e. 30 students.

So their average must be not only the initial 80, but also the average meant to compensate the loss incurred because of the first group.

Hence the average of the remaining 30 students = 80 + (40×5)/30 = 80 + 20/3 = 86.66

Now let us look at Example 2 where the numbers are not comfortable.

**Example 3:**

Average of 75 students is 82, out of which average of 42 students is 79.

What is the average of the remaining 33 students?

Average of 75 members = 82.

**Two groups of 42 and 33 and we want each group to have an average of 82.**

But the first group i.e. 42 students they had an average of 79.

We fell short by 3 in the average.

So the loss in the sum= 3×42.

So the average of the second group i.e. 33 students must be 82 + (3×42)/33 = 82+ 42/11 = 85.82

So folks it’s time to get into some more different situations that arise when we calculate averages.

**Problem 1:**

Let’s look at one more sum similar to the one discussed above.

The **average height** of 74 students in a class is 168 cm out of which 42 students had an** average height** of 170 cm. Find the average height of the remaining 32 students.

**Solution:**

Average of 74 students = 168.

Here we can observe that, in the case of the first group we had a gain.

Since the average of 42 students is 170 cm, the gain is 2 cm in the average upon 42 students.

So the gain in the sum = 42×2

Hence average of the remaining 32 students = 168 – (42×2)/32= 168-(21/8) = 165.37

**Problem 2:**

The average height of 40 students in a class is 172 cm. 30 students whose average height is 172.5 cm left the class and 40 students whose average height is 170.5 cm joined the class. Find the average height of the present class.

**Solution:**

Going back to the ideal situation, we want the average of the students leaving the class as well as joining the class to be 172 so that the average remains the same.

But it is given that the average of the 40 students leaving the class is 172.5 (more than 172).

So we will incur a loss of 0.5 cm in the average upon 30 students.

Hence the loss in the sum = 0.5x 30 = 15 cm

Also, since the average of the 40 students joining the class is 170.5 (less than 172) we will incur a loss in this case as well. The loss in the average is 1.5 cm upon 40 students.

Hence the loss in the sum = 1.5×40 = 60 cm

Thus the total loss in the sum = 75 cm. This loss will be shared by 50 students which is the present strength of the class.

Hence the average of the present class = 172 – 75/50 = 170.5 cm.

**Thanks **

Examorient Team