Average Methods Example 1

Shortcut tricks on average are one of the most important topics in exams. Time takes a huge part in competitive exams. If you know time management then everything will be easier for you. Most of us miss that part. Here in this page we give few examples on Average shortcut tricks. These shortcut tricks cover all sorts of tricks on Average. Visitors are requested to carefully read all shortcut examples. You can understand shortcut tricks on Average by these examples.

First of all do a practice set on math of any exam. Choose any twenty math problems and write it down on a page. Do first ten maths using basic formula of this math topic. You also need to keep track of timing. Write down the time taken by you to solve those questions. Now read our examples on average shortcut tricks and practice few questions. After doing this go back to the remaining ten questions and solve those using shortcut methods. Again keep track of Timing. You will surely see the improvement in your timing this time. But this is not all you want. You need more practice to improve your timing more.

Math section in a competitive exam is the most important part of the exam. That doesn’t mean that other sections are not so important. But if you need a good score in exam then you have to score good in maths. A good score comes with practice and practice. You should do your math problems within time with correctness, and this can be achieved only by using shortcut tricks. But it doesn’t mean that you can’t do math problems without using any shortcut tricks. You may do math problems within time without using any shortcut tricks. You may have that potential. But other peoples may not do the same. For those we prepared this average shortcut tricks. We try our level best to put together all types of shortcut methods here. But we may miss few of them. If you know anything else rather than this please do share with us. Your little help will help so many needy.

Average Methods Example 1
In this average methods Example 1 we discuss some examples which help your better understanding in average chapter which came in most competitive exams.This is the basic theory of average which applied in question to obtain answers here is Average Methods of example 1 in different form of examples.

This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below  given some more example for your better practice.

 

 

Example 1 :
In a School 8 student average weight is increased by 3.5 kg, when new student comes in place of one of them than weight become 55 kg, what would be the weight of the new student ?
Answer :
Step 1: At first we find the total weight increased So, ( 8 X 3.5 )kg = 28 kg.
Step 2: Now the weight of new Student is = ( 55 + 28 )kg = 83 kg.
So, the weight of the new student is 83 kg.

 

 

Example 2:
The Average weight 3 girls A, B, and C is 55 kg, While the average weight of three boys B, D, and E is 57 kg. What is the average weight of A, B, C, D, E ?
Answer :
Step 1: At First we find 3 girls total weight ( A + B + C ) = ( 55 X 3 ) = 165 kg. and Total 3 boys weight of ( B + D + E ) = ( 57 x 3 ) = 171.
Step 2: Adding both weight of ( A + 2 B + C + D + E ) = ( 165 + 171 ) = 336 kg.
Note: So, to find the average weight of A B C D and E, we duty to know B’s weight, which is not given. So the data is inadequate.

 

 

Example 3:
There are two section X and Y of a College, consisting of 36 and 44 students respectively. if the average weight of section X is 40 Kg and that of section Y is 35 Kg. Find the average weight of the whole College (in Kg) ?
Answer:
Step 1: At first the total ( X + Y ) college students are ( 36 + 44 ) = 80.
Step 2: college X total student weight and college Y total student weight is ( 36 x 40 + 44 x 35 ) = 2980
Average weight of the whole college is = 2980 / 80 = 37.25.

 

 

Example 4 :
Find the average of all the numbers between 6 and 34 which are divisible by 5.
Answer:
Step 1: 10 + 15 + 20 + 25 + 30 which are divisible by 5 in between 6 and 34
So average of number is 10 + 15 + 20 + 25 + 30 / 5 = 100 / 5 = 20.

 

 

Example 5:
Divya obtain 56, 75, 78, 86 and 88 marks out of 100 in Physics, Life Science, Mathematics, Biology, and Computer. What are her average marks ?
Answer :
Step 1: Average of all subject marks 56 + 75 + 78 + 86 + 88 / 5
=383 / 5 = 76.6.

 

 

Example 6:
In an examination it is required to get 34% of the aggregate marks to pass. A student gets 270 marks and is declared failed by 7% marks. What are the maximum aggregate marks a student can get ?
Answer :
Step 1: 34% = 270 + 7%
Step 2: 27% = 270
Step 3: 100% = 270 x 100 / 27 = 1000 marks.

 

 

 

 

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