In this article, you are going to learn the basic concepts of average or arithmetic mean. Average is a very important area as far as the competitive exams are concerned. You will see questions based on arithmetic mean in almost all the competitive exams.
- Average is basically the Arithmetic mean of given items.
Arithmetic mean formula
- Mathematically, Arithmetic Mean= average = Sum of terms/ No. of terms.
Properties of average
- When the difference between all the items is same (and the number of terms is odd), then the average is equal to the middle term. The average of the first and last term would also be the average of all the terms of the sequence.
- If x is added to all the items, then the average increases by x.
- If x is subtracted from all the items, then the average decreases by x.
- If every item is multiplied by x, then the average also gets multiplied by x.
- If every item is divided by x, then the average also gets divided by x.
Example 1: Find the arithmetic mean of first five prime numbers.
Solution: First five prime numbers are 2, 3, 5, 7 and 11. The mean or the average formula is = Sum/no. = (2 + 3 + 5 + 7 +11)/5 = 5.6.
Example 2: Find the average of 56, 41, 59, 52, 42 and 44.
Solution: If the given items are very close to each other, a shortcut may be used instead of applying the average formula. Assume that the average is 50, as the terms are around 50 only. Now calculating the difference of all items with 50, you are getting + 6, - 9, +9, + 2, - 8, - 6. Now, if you add these numbers, the result comes out to be - 6. That implies this - 6 will be equally distributed among all the terms and every term will be reduced by - 6/6 = - 1 i.e. average will be reduced by 1. Thus the real average will be 50 - 1 = 49.
From this example, you can learn that it is not necessary to apply the arithmetic mean formula in every question where you have to find the average, seeing the values of numbers, sometimes doing it logically would be of great help. This technique can reduce your time taken for a question by a huge margin. Needless to say time is the most important factor in a competitive paper.
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Example 3: The average salary of 20 officers in a company is 300. The average salary of non-officers in the same company is 120. Find the number of the non-officers if the average of the whole organization is 150.
Solution: There are two methods to solve this. One is the conventional method. Let the number of Non-officers be N. Now the equation can be made, knowing that the sum of the salaries of all the officers and non-officers would give the total salary of the organization. The equation will be 20 × 300 + 120 × N = 150 (20 + N). Solving this, N = 100.
Besides this, you can think logically that every officer is getting (300 - 150 = 150) Rs. 150 more than the average. Total of the extra money taken by all the officers together is 150 × 20 = 3000. Now every non-officer is getting (150 - 120) = Rs. 30 less than the average. How many officers will get Rs. 3000 less i.e. 3000/30 = 100. Hence there are 100 non-officers.
From this example, you can learn that it is not necessary to apply the mean formula in every case. Sometimes applying the extra item given/ taken can also help to find the average without much effort and calculations.
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Averages: Key Learning
- This article tells us the approach to solve average problems in various forms. The techniques explained in this article, would make to solve the average questions in quick span of time and with much lesser effort and better accuracy. As the topic is very important, do remember to solve sufficient number of questions till you become an expert in the area.
If you have any queries related to the examples or any concept mentioned in the article, please post them in the comments section below.