Overview: Averages

Table of Contents
1. Introduction to the Concept of Average
2. Basic Formulae and Common Results
3. Important Properties and Simple Tricks
4. Useful Shortcuts and Solved Examples
5. Strategy and Common Patterns in Competitive Questions
6. Additional Examples for Practice
7. Conclusion
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Introduction to the Concept of Average

Average is the sum of different terms (data) divided by the total number of terms.

Average = (Sum of given terms) / (Total number of terms)

For example: Find the average of given terms 12, 16, 17, 19, 21.

Sol:
Total number of terms = 5
Sum of terms = 12 + 16 + 17 + 19 + 21 = 85
Average = 85 / 5 = 17

Basic Formulae and Common Results

The following formulae are frequently used and should be remembered for quick calculations.

• Average of first n natural numbers = (n + 1) / 2
• Average of first n even numbers = n + 1
• Average of first n odd numbers = n
• Average of n consecutive natural numbers = (First number + Last number) / 2
• Average of squares of first n natural numbers = (n + 1)(2n + 1) / 6

Important Properties and Simple Tricks

These properties reduce computation and help in many short-answer questions.

• If each number is increased by the same value a, the average increases by a.
• If each number is decreased by the same value a, the average decreases by a.
• If each number is multiplied by the same non-zero value a, the average is multiplied by a.
• If each number is divided by the same non-zero value a, the average is divided by a.
• If the total of all numbers is known as T and there are n numbers, then average = T / n.
• To combine two groups: if group A has average A1 with n1 items and group B has average A2 with n2 items, combined average = (n1·A1 + n2·A2) / (n1 + n2).
• Removing or adding an item: if an item x is removed from a list of n items with average m, new average = (n·m - x) / (n - 1). If item x is added, new average = (n·m + x) / (n + 1).

Useful Shortcuts and Solved Examples

(1) Failed/Passed Students - Finding Counts from Averages

If the average marks of n students is m, the average marks of the passed students is p and the average marks of the failed students is f, then the number of failed students (x) is

x = n(p - m) / (p - f)

Example 1: The average marks obtained by 125 students in an exam is 29. If the average marks of passed students is 36 and that of failed students is 11. What is the number of failed students?

Sol:
Total students n = 125
Overall average m = 29
Average of passed students p = 36
Average of failed students f = 11
Number of failed students = 125 × (36 - 29) / (36 - 11)
Number of failed students = 125 × 7 / 25
Number of failed students = 35

(2) Using Change in Average to Find New Average

If a batsman in his nth innings scores s and thereby increases his average by t, then the average after the nth innings (the new average) is

New average = s - t(n - 1)

Example 2: A batsman in his 44^th innings makes a score of 86 and thereby increases his average by 1, Find the average after 44 innings?

Sol:
n = 44
s = 86
t = 1
Average after 44th innings = s - t(n - 1)
Average after 44th innings = 86 - 1 × (44 - 1)
Average after 44th innings = 86 - 43 = 43

(3) Weighted Average (General Concept)

When different items or groups have different counts (weights), the average is a weighted average. If values v1, v2, ..., vk have weights w1, w2, ..., wk respectively, then

Weighted average = (w1·v1 + w2·v2 + ... + wk·vk) / (w1 + w2 + ... + wk)

This concept is identical to combining groups and is used widely in problems where different sections, classes or categories contribute differently to the whole.

Example 3: Class A has 30 students with average 72. Class B has 20 students with average 68. Find the combined average of both classes.

Sol:
Sum of marks in Class A = 30 × 72 = 2160
Sum of marks in Class B = 20 × 68 = 1360
Total students = 30 + 20 = 50
Total sum = 2160 + 1360 = 3520
Combined average = 3520 / 50 = 70.4

Strategy and Common Patterns in Competitive Questions

Keep these strategies in mind when answering time-bound questions:

• Convert averages to totals when convenient: average × count = total.
• Use the effect of adding/removing known values to compute unknowns quickly.
• For changes in average by a small integer, the formula linking the new score and change in average is often quickest (as in the batsman example).
• When combining groups, compute group totals first then divide by total count rather than trying to average the averages without weights.
• For consecutive numbers or symmetric sets, use middle/median as the average; for an odd number of consecutive integers the middle term equals the average.

Additional Examples for Practice

Example: The average of 5 numbers is 20. One number is removed and the new average becomes 18. Find the removed number.

Sol:
Original total = 5 × 20 = 100
New total after removal = 4 × 18 = 72
Removed number = 100 - 72 = 28

Example: Three numbers have average 10. If two of them are 8 and 14, find the third number.

Sol:
Total of three numbers = 3 × 10 = 30
Sum of known two numbers = 8 + 14 = 22
Third number = 30 - 22 = 8

Conclusion

The concept of average is central to quantitative aptitude and is based on the fundamental relation: Average = Total ÷ Number of terms. 

• If a set has n numbers with average m, its total is n × m. 
• When a value x is added or removed, the new average becomes (n × m ± x) / (n ± 1).
•  When two groups with averages A₁ and A₂ and sizes n₁ and n₂ are combined, the overall average is (n₁ × A₁ + n₂ × A₂) / (n₁ + n₂). 
• For consecutive numbers, the average is (First term + Last term) / 2.