CONCEPT 1
An average is the most likely middle value of a set of data. It is the mean value of a data set around which the numbers are clustered and hence it is a representative of the set.
Average = Sum of all the values of the set of numbers / Total number of values
Example (JMET 2009)
A student in a business school calculates his cumulative average.
QT and QB were his last two tests.
83 marks in QT increases average by +2 marks
75 marks in QB increases average by +1
Next test is reasoning. If he gets 51 in reasoning, his average will be ---?
Soln:
Average -- a number of tests -- n (Before QT)
After QT
a+2 = (an + 83)/ (n+1) => a+2n = 81 ---- (1)
After OB
a+3= (an + 83 +75)/ (n+2) => 2a+3n = 152 ---- (2)
Solving 1 and 2 we get, n= 10, a = 61
After Reasoning,
Average = (an+83+75+51)/ (n+3) Sub values of a and n,
we get average = 63
NOTE:
1. If the value of each item of the group ----> increases or decreases by the same value x
The average of the group ----> increases or decreases by x
2. This is useful in dealing with ages. If average age of a group of people is x years
After n years , the average will be x+n
Why? With each passing year, the age of each person increases by 1
3. If the value of each item of the group ---> multiplied/ divided by the same value x (x not equal to 0)
average of the group ----> multiplied/ divided by x
4. Net deficit due to the numbers below average = Net surplus of the numbers above average.
CONCEPT 2
ALTERNATE METHOD TO CALCULATE AVERAGES:
When it is used? If we have large group of numbers that may lead to tedious calculations.
Another way to call the method is assumed average method.
A simple example: Calculate the average of 102, 105, 92, 103, 96 and 98.
Assume any arbitrary average between 92 and 105 (between smallest value and largest value)
Let us take arbitrary average = 100. Now, find the difference of each number
102- 100 = 2, 105-100= 5. 92-100= -8, 103-100 = 3, 96-100= -4, 98-100= -2.
2. 5. -8, 3, -4, -2 are all called as deviations. Sum of all deviations = -4
Actual average = 100-(4 /6) = 99.33
CONCEPT 3
CALCULATION OF AVERAGES WHEN NEW VALUES ARE ADDED:
When all the values of a set is replaced by the average, the sum will not change.
Consider set of 5 numbers; 43, 51, 74, 60 and 22.
Average = (22+43+51+60+74)/5 = (250/5) = 50
The concept is, if we replace all the 5 values of the set to 50, the sum will still remain 250. (5*50 = 250)
If a new item 68 gets added, then the new average will be (250+68)/(5+1) = 318/6 = 53
A more simple way: If a new number 68 is added, it is 18 more than the original average.
So, the number should distribute 18 equally to all the 6 numbers including itself ==> 18/6 = 3 to all the values
The new average is 50 +3 = 53
Thus the formula is,
New average after addition = >
Original avg + (Value of newly added item (V) - Original average (O)) / The number of items after addition
If V-O is negative, the the average will reduce.
A converse approach:
Given that the average of 5 numbers is 50 and when another number is added, the average becomes 53.
Find the newly added number.
Ans: (53*6) - (50*5) = 68
or Easy way --> Original average (5 nos) = 50, New average (6 nos) = 53
So, 3 is distributed equally to all the 6 numbers by the new number, (3*6)= 18 --> New no. = 50+18= 68
Example 1:
The average of 4 numbers is 12, 13, 17, and 18 is (12+13+17+!8)/4 = 60/4 = 15
Suppose if a 5th number is added, the average becomes 16. Find the newly added number.
To solve this, visualise addition of a fifth number, which increases average by 1.
15+ 1 = 16
15+ 1 = 16
15+ 1 = 16
15+ 1 = 16
Thus +1 appearing 4 times is due to the fifth number, which is able to maintain the average as 16 first and then give 'one' to each of the first 4. So, the fifth number is 16+4 = 20.
Example 2:
Average wt of players = 60 kg
If a player of 68 kg joins the group, average weight = 61 kg
No. of players at initial stage = ?
Soln:
Assume, each player weight = 60 kg.
Another player joins, the average = 61 kg (This means he brings extra 8 kg which has to be distrubuted equally.) Since, the average increases by 1, 8 kg would be distributed to 8 players
So, initially there will be 7 players.
Example 3:
Average weight of a school football team ( 22 members including the goal keeper) decreases by half a kilogram when the goalie is not included . Average age initially = 60 kg Goal keeper's weight?
Soln:
After goalie removal, the average will be 60 - 0.5 = 59.5 kg
So, the goalie age will be 0.5*21 = 10.5 kg in excess of the initial average => 10.5+ 60 = 70.5 kg.
Example 4:
The average age of class of 30 students and a teacher reduces by 0.5 years, if we exclude the teacher. If the initial average is 14 years, find the age of the class teacher.
Soln:
The teacher after fulfilling the average of 14 years is also able to give 0.5 years to each of 30 students
(30*0.5 =15) ==> he has 15 years to give over and maintain his own average of 14 years.
Age of teacher = 14 + (30 * 0.5) = 29 years.
(Note: The problem should be viewed as change in average from 13.5 to 14 when teacher is included)
Example 5:
The average marks of a group of 20 students on a test is reduced by 4 when the topper who scored 90 marks is replaced by a new student. How many marks did the new student have?
Soln:
The new student should have scored (20 *4) 80 marks less than the topper.So, the mark of new student = 90-80 = 10 marks.
Example 6:
The average age of 3 students A, B and C is 48 marks.
Another student D joins the group => the new average = 44
Another student E who has 3 more marks than D joins the group => the average = x
Average of 4 students B,C,D and E becomes 43 marks. How much did A get in the exam?
Soln:
Solve while reading
By first sentence, Total marks of 3 students(A+B+C) = 48*3 = 144 marks.
If D, joins the total marks of 4 students(A+B+C+D) = 44*4 = 176 marks. ---- (1)
176-144= 32 marks is what D has got.
E's mark will be 32+3 = 35 marks.
Total marks of 4 students (B+C+D+E) = 43*4 = 172 marks ----1(2)
Subtracting 1 and 2 we get, A-E = 4 marks. E's mark is 35 => A's mark is 35+4 = 39 marks.
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