The weighted average is a calculation that takes into account the varying degrees of importance of the numerical values in a data set. In calculating a weighted average, each number in the data set is multiplied by a predetermined weight before the final calculation is made.
A weighted average can be more accurate than a simple average in which all numbers in a data set are assigned an identical weight.
The average weight of class A = 30 kg
The average weight of class B = 40 kg
Average weight of both the classes = 30+40/2 = 70/2= 35 kg
Can this be the right answer? No, this is a very common wrong answer for this question. When the average of two groups of items is to be determined, we cannot just take a simple average of the averages of the two groups. Instead, we should take a weighted average based on the number of items in the two groups. Weighted averages are not particularly tricky, but they can be very time-consuming. You can expect to see this type of question at least once in any competitive exam.
From the above example,
If there were 40 students in class A and 60 students in class B,
the combined average weight is (40 x 30+60 x 40)/(40 + 60) = 1200 + 2400/100= 3600/100= 36 kg
There is a great shortcut for weighted average problems, one that very few students seem to have seen before. It is the See-saw method. This method is one of the most difficult methods in algebra. If you go with the theoretical explanation this method is going to be a nightmare. So let us start with a practical understanding.
Assume that there are three students sitting on one side of a see-saw and one student on the other side. And also assume their individual weights are all equal.
How can we balance them?
We can balance the see-saw by either adding two students (on the side where there is one student) or by adjusting the pivot point towards the heavier side, we can make it in equilibrium. It can be observed that the distance between the place where they are seated and the pivot is the inverse of the weights on either pan.
Let us apply this concept to the above example:
The ratio of number of students in each class is 40: 60 or 2: 3.
The average of the two classes put together should be in between their individual averages of 30 and 40.
This difference of 10 kg is the equivalent of the distances in the see-saw and it has to be split in the inverse ratio that is, 3:2. Therefore, the average is 30 + (3/5) × 10 = 36 kg.
Solved Examples
Q.1.The average goal 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 the maximum number of players who scored at least 18 goals?
Solution:
To maximize the number of players who scored 18 and above goals, we will assume that only one person has scored 20. To counter him, we will have one person who will score 12 goals.
15-2 =13 players left
Now to maximize the 18 and above goals, for every two players who are scoring 18, we will have one player scoring 12. This is done, to arrive at an 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 or more goals = 9.
Q.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. Finds the weight of P.
Solution:
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
Q.3.The average age of 29 students is 18. If the age of the teacher is also included the average age of the class becomes 18.2. Find the age of the teacher.
Solution:
Let the average age of the teacher =x
(29 × 18 + x × 1)/30
Solving for x, we get x = 24
Using the shortcut, based on the same method we used previously
Calculate the change in average = 18.2-18 = 0.2
This change in 0.2 is reflected over a sample size of 30
New age is increased by 30 * 0.2 = 6 years above the average 18+6 = 24.
Q.4.The average age of Mr Mark’s 3 children is 8 years. A new baby is born. Find the average age of all his children.
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
Q.5. In Class I there are 12 students with an average age of 20 years and in Class II there are 16 students with an average age of 23 years. What is the average age of both classes?
Solution:
((12 ×20)+ (16 ×23))/((12+16)) = (240+368)/28 = 608/28 = 20.7 years
Q.6.The average of a batsman in 16 innings is 36. In the next innings, he scored 70 runs. What will be his new average?
Solution:
New average = (old sum+ new score)/(total number of innings) = ((16 ×36)+70)/((16+1)) = 38
Q.7.The numbers 1, 2, 3, 4, 5 and 6 have weights 0.5, 0.1, 0.1, 0.1, 0.1 and 0.1 respectively. What is the weighted mean?
Solution:
Weighted Mean
= 0.5 × 1 + 0.1 × 2 + 0.1 × 3 + 0.1 × 4 + 0.1 × 5 + 0.1 × 6
= 0.5 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6
= 2.5
Q.8.In a gymnastics competition, the overall score is calculated from points for execution and difficulty, calculated with weights of 80% and 20% respectively. Grace scored 9.8 for execution and had an overall score of 9.5. What was her score for difficulty?
Solution:
Let her score for difficulty be d
Therefore her overall score is her weighted average
= 0.8 × 9.8 + 0.2 × d
= 7.84 + 0.2d
But her overall score should be 9.5
Therefore 7.84 + 0.2d = 9.5
⇒ 0.2d = 1.66
⇒ d = 1.66 ÷ 0.2 = 8.3
So, her score for difficulty was 8.3
Q.9.In Bobby’s school, math grades for the year are calculated from assignments, tests and a final exam. Assignments count 30%, tests 20%, and the final exam 50%. If Bobby has an assignment grade of 85, a test grade of 72, and an exam of 61, what is Bobby’s overall grade?
Solution:
Bobby’s overall grade is the weighted mean
= 0.3 × 85 + 0.2 × 72 + 0.5 × 61
= 25.5 + 14.4 + 30.5
= 70.4