Averages can be defined as the central value in a set of data.
Example:
What is the average of the first five consecutive odd numbers?
Solution:
The first five consecutive odd numbers are: 1, 3, 5, 7, 9.
Here, the number of data or observations is 5 and the sum of these 5 numbers is 25.
So, average = 25 / 5 = 5.
Note:
Sum of 1st n consecutive natural numbers = [n(n+1)]/2
Average of 1st n consecutive natural numbers = (n+1)/2
Conventionally solving:
New average = (old sum+ new score)/(total number of innings) = ((16 ×36)+70)/((16+1)) = 38Shortcut technique:
- Step 1) Take the difference between the new score and the old average = 70 – 36= 34
- Step 2) This is 34 extra runs which are spread over 17 innings. So, the innings average will increase by 34/17 = 2
- Step 3) Hence, the average increases by => 36+2 = 38.Here are a few more average questions and their solutions using the same technique.
Example 2:
The average mark of 19 children in a particular school is 50. When a new student with marks 75 joins the class, what will be the new average of the class?
Step 1) Take the difference between the old average and the new marks = 75-50=25
Step 2) This score of 25 is distributed over 20 students => 25/20 = 1.25
Step 3) Hence, the average increases by 1.25=> 50+1.25 = 51.25.
Solution:
The new age will be 0 years. The difference between the old average and the new age = 0-8= -8
This age of 8 years is spread over 4 children => (-8/4= -2) Hence, the average reduces to 8-2= 6 years.
Conventionally solving:
Let the average age of the teacher = x(29 × 18 + x × 1)/30Solving for x, we get x = 24.
Shortcut Technique:
Using the shortcut, based on the same method used previously:
Step 1: Calculate the change in average = 18.2 – 18 = 0.2.This change in 0.2 is reflected over a sample size of 30.
The new age is increased by 30 × 0.2 = 6 years above the average i.e. 18 + 6 = 24; which is the age of the teacher.
The concept of an assumed mean is not new. It is widely used to reduce the calculation in finding the average in statistics where the data is huge.
Here, We will demonstrate the application of the assumed mean to solve some aptitude questions based on averages and weighted averages.
Let us take an example to understand the concept
Example: In a class of 30 students, the average age is 12 years. If the age of the class teacher is included, the new average age of the class becomes 13 years. Find the age of the class teacher.
Solution:
Standard Approach
Applying the standard approach, the total age of the 30 students = years. When the class teacher is included, the new total age of the class = years. Note that the increase in the total age is because of the class teacher only. Hence the age of the class teacher = 403 – 360 = 43 years
Deviation method
To understand the deviation method, let us simulate the problem. In the case of average age, assume that each student has 12 chocolates with them.
Therefore, he came with 30 + 13 = 43 chocolates. Or the age of the class teacher is 43 years.
Let us take one more example:
Example: If a person with age 55 joins a group of 5 persons with an average age of 49 years. What will be the new average age of the group?
Solution:
Standard approach:
Total age will be . And there will be 6 people now. So, the average will be 300/6 = 50.Deviation approach:
Since 55 is 6 more than 49, by joining of the new person, the total will increase by 6 and number of person increases to 6; the average will increase by 1.
So, the average is 49 + 1 = 50.
where x1, x2, x3, …, xn are averages and w1, w2, w3, .., wn are their respective weight-ages
Example: In a class of 25 boys and 15 girls, the average heights of the two groups of boys and girls are 150 cm and 140 cm respectively. Find the average height of the class.
Standard approach:
Total weight of the group of boys = and for the group of girls, the total weight =Therefore, the average height of the class =
Deviation Approach:
Assume that each boy and each girl is carrying 150 and 140 chocolates respectively.
- We have to distribute the total chocolates such that each boy and girl must carry an equal number of chocolates.
- Since each boy carries 10 chocolates more than each girl, let us take away 10 chocolates from each boy so that now each boy and girl carries 140 chocolates. The excess chocolates we have is equal to .
- These excess chocolates are to be equally distributed among the boys and the girls. so, everyone will be getting =6.25 more chocolates. Hence the average of the class = 140+6.25 = 146.25 cm.
Question 1:
The average goals scored by 15 selected players in EPL is 16. The maximum number of goals scored by a player is 20 and the minimum is 12. The goals scored by players are between 12 and 20. What can be a maximum number of players who scored at least 18 goals?
To maximize the number of players who scored 18 and above number of goals, one should assume that only one person has scored 20. To counter him, there will be one person who will score 12 goals.
i.e. 15 – 2 = 13 players left.
Now to maximize the 18 and above goals for every two players who are scoring 18, there will be one player scoring 12. This is done, to arrive at the average of 16. We will have 8 players with a score of 18 and 4 players with a score of 12. The last player will have a score of 16 Thus, the maximum number of people with 18 and more goals = 9.
Question 2:
The average weight of a group of 8 girls is 50 kg. If 2 girls R and S replace P and Q, the new average weight becomes 48 kg. The weight of P= Weight of Q and the weight of R = Weight of S.Another girl T is included in the group and the new average weight becomes 48 kg. Weight of T= Weight of R. Find the weight of P?
8 x 50 +R+S-P-Q= 48×8 R+S-P-Q=-16
P+Q-R-S= 16 R=S and P=Q
P-R=8
One more person is included and the weight = 48 kg
Let the weight be a = (48 × 8 + a)9/9 = 48
A = 48 kg= weight of R
=> Weight of P= 48+8= 56 kg.
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