Average — also called mean — is one of the most frequently tested topics in every major competitive exam. SSC CGL, RRB NTPC, IBPS PO, and school-level examinations all include average problems in various forms: finding the mean of a data set, finding a missing value when the average is known, calculating weighted averages, and solving word problems involving changing averages.
The conventional method — add all values and divide by count — is reliable but slow. When a question asks for the average of 8 numbers with three-digit values, that addition alone takes 15–20 seconds before you even begin dividing. Over an entire exam, this adds up to minutes of unnecessary time expenditure.
The shortcuts in this guide reduce average calculations to mental operations — using assumed mean, pattern recognition, and algebraic relationships to reach answers in two to three steps rather than ten.
Understanding Average — The Foundation
Before shortcuts, the core formula:
Mean = Sum of all values ÷ Number of values
Rearranged forms used constantly in exams:
- Sum = Mean × Count
- Missing value = (New mean × New count) − (Old mean × Old count)
- Count = Sum ÷ Mean
Memorizing these three forms eliminates the need to re-derive relationships during the exam.
Shortcut 1: The Assumed Mean Method
This is the most powerful average shortcut and the one that produces the greatest time saving. Instead of adding all values, you pick a convenient reference point and work only with the deviations from it.
The Method:
- Choose an assumed mean A (pick a round number close to the middle of the data)
- Find deviation of each value from A: d = value − A
- Find mean of deviations: d̄ = sum of deviations ÷ count
- Actual mean = A + d̄
Worked Example 1: Average of 47, 53, 51, 49, 55
- Assumed mean A = 50
- Deviations: −3, +3, +1, −1, +5
- Sum of deviations = −3+3+1−1+5 = 5
- d̄ = 5/5 = 1
- Actual mean = 50 + 1 = 51
Worked Example 2: Average of 248, 253, 246, 251, 257, 245
- Assumed mean A = 250
- Deviations: −2, +3, −4, +1, +7, −5
- Sum = −2+3−4+1+7−5 = 0
- d̄ = 0/6 = 0
- Actual mean = 250 + 0 = 250
Worked Example 3: Average of 318, 325, 312, 330, 320
- Assumed mean A = 320
- Deviations: −2, +5, −8, +10, 0
- Sum = −2+5−8+10+0 = 5
- d̄ = 5/5 = 1
- Actual mean = 320 + 1 = 321
Worked Example 4: Average of 1,248 / 1,256 / 1,244 / 1,260 / 1,252
- Assumed mean A = 1,252
- Deviations: −4, +4, −8, +8, 0
- Sum = 0 → d̄ = 0
- Actual mean = 1,252
Why this is faster: The deviations are always small numbers — single or double digits — regardless of how large the original values are. Adding small deviations is orders of magnitude faster than adding large values.
Shortcut 2: The Middle Value Shortcut for Evenly Spaced Numbers
When numbers form an arithmetic progression (each consecutive pair has the same difference), the mean equals the middle value — no calculation required.
The Rule:
For an odd-count arithmetic sequence: Mean = Middle term
For an even-count arithmetic sequence: Mean = Average of the two middle terms
Worked Examples:
Average of 13, 17, 21, 25, 29:
- Common difference = 4, count = 5 (odd)
- Middle term = 3rd term = 21
Average of 24, 30, 36, 42, 48, 54:
- Count = 6 (even)
- Two middle terms = 36 and 42
- Mean = (36+42)/2 = 39
Average of consecutive numbers 15 to 35:
- First = 15, Last = 35
- Mean = (15+35)/2 = 25
- (Always true for consecutive or evenly spaced sequences: Mean = (First + Last)/2)
Average of first n natural numbers:
- Mean = (n+1)/2
- Average of 1 to 50 = 51/2 = 25.5
- Average of 1 to 100 = 50.5
Average of first n even numbers:
- Mean = n+1
- Average of first 10 even numbers (2,4,...,20) = 11
Average of first n odd numbers:
- Mean = n
- Average of first 8 odd numbers (1,3,5,...,15) = 8
Shortcut 3: The Effect of Adding or Removing a Value
Exam questions frequently ask: "If a new member joins a group, what is the new average?" or "If one value is removed, how does the average change?" These can be solved in one step.
Formula for adding a new value:
New mean = Old mean + (New value − Old mean) / New count
Formula for removing a value:
New mean = (Old mean × Old count − Removed value) / New count
Worked Example 1:
Average of 6 numbers is 42. A 7th number, 63, is added. New average?
- New mean = 42 + (63−42)/7 = 42 + 21/7 = 42 + 3 = 45
Worked Example 2:
Average of 8 students is 35. One student scoring 67 leaves. New average?
- New mean = (35×8 − 67)/7 = (280−67)/7 = 213/7 = 30.43
Worked Example 3:
Average of 10 numbers is 25. If one number is replaced by 55 (original was 15), new average?
- Increase in sum = 55−15 = 40
- Increase in mean = 40/10 = 4
- New mean = 25+4 = 29
Key insight: Every time a value is replaced, the change in mean = (new value − old value) / count. This single formula handles replacement questions in one line.
Shortcut 4: Finding a Missing Value When Average Is Known
This is the direct application of the rearranged formula: Missing value = (Mean × Count) − Sum of known values
Worked Example 1:
Average of 5 numbers is 48. Four of them are 42, 51, 39, 55. Find the fifth.
- Required sum = 48 × 5 = 240
- Known sum = 42+51+39+55 = 187
- Missing value = 240−187 = 53
Worked Example 2:
Average marks of 6 students is 72. Five students scored 68, 74, 80, 65, 70. Find the sixth student's marks.
- Required sum = 72×6 = 432
- Known sum = 68+74+80+65+70 = 357
- Missing marks = 432−357 = 75
Worked Example 3:
Average of 7 observations is 63. The average of the first 3 is 57 and the last 3 is 68. Find the 4th observation.
- Total sum = 63×7 = 441
- First 3 sum = 57×3 = 171
- Last 3 sum = 68×3 = 204
- 4th observation = 441−171−204 = 66
Shortcut 5: Weighted Average
Weighted average appears in problems involving mixtures, class averages combined across sections, and combined group averages. The formula differs from simple mean.
Formula:
Weighted Mean = (w₁x₁ + w₂x₂ + ... + wₙxₙ) / (w₁ + w₂ + ... + wₙ)
Where w = weight (frequency, count, or proportion) and x = value.
Worked Example 1:
Class A has 30 students with average marks 75. Class B has 20 students with average 85. Combined average?
- Weighted mean = (30×75 + 20×85) / (30+20)
- = (2250 + 1700) / 50
- = 3950/50 = 79
Worked Example 2:
A shopkeeper mixes 20 kg of rice at ₹40/kg with 30 kg at ₹50/kg. Average price per kg?
- = (20×40 + 30×50) / (20+30)
- = (800+1500) / 50
- = 2300/50 = ₹46/kg
Worked Example 3:
A car travels 120 km at 60 km/h and 180 km at 90 km/h. Average speed?
- Total distance = 300 km
- Time 1 = 120/60 = 2 hours
- Time 2 = 180/90 = 2 hours
- Average speed = 300/4 = 75 km/h
Important note: Average speed is always weighted by time, not distance. Never take the simple average of speeds unless equal times are involved.
Shortcut 6: The Alligation Method for Averages
When two groups are combined and you know their individual averages and the combined average, alligation finds the ratio of group sizes in one step — without algebra.
The Method:
Draw a cross:
- Place the two individual averages on the left (top and bottom)
- Place the combined average in the middle
- Subtract diagonally: each difference gives the ratio of the other group
Worked Example 1:
Group A average = 60, Group B average = 80, combined average = 68. Find ratio of A:B.
60 80
\ /
68
/ \
80-68 68-60
= 12 = 8
Ratio A:B = 12:8 = 3:2
Worked Example 2:
A mixture of two types of rice costs ₹48/kg. Type 1 costs ₹40/kg, Type 2 costs ₹60/kg. Ratio?
40 60
48
60-48 48-40
= 12 = 8
Ratio Type1:Type2 = 12:8 = 3:2
Worked Example 3:
Average age of men = 35, women = 28, combined = 32. Find ratio of men:women.
35 28
32
28-32 32-35
= -4 = -3
Take absolute values: Ratio men:women = 4:3
Shortcut 7: Age-Based Average Problems
Age problems with averages are a staple of SSC and RRB exams. They follow predictable patterns.
Pattern 1: Average age increases with time
If average age of a group is A today, after T years the average age will be A+T.
- Average age of 5 people = 30. Average age after 3 years = 33
Pattern 2: New member joins
New member's age = New average × New count − Old average × Old count
Worked Example:
Average age of 4 people = 32. A 5th person joins and average becomes 30. Age of 5th person?
- 5th person's age = 30×5 − 32×4 = 150 − 128 = 22
Pattern 3: One member leaves, one joins
Change in sum = (Joining age − Leaving age)
Change in mean = Change in sum / Count
Worked Example:
Average age of 6 people = 40. One aged 52 leaves and one aged 28 joins. New average?
- Change in sum = 28−52 = −24
- Change in mean = −24/6 = −4
- New average = 40−4 = 36
Common Average Problems in Competitive Exams
Problem Type 1: Error Correction
A student calculates average of 8 numbers as 42. Later finds one number was read as 56 instead of 65. Correct average?
- Error = 65−56 = +9
- Increase in sum = 9
- Increase in mean = 9/8 = 1.125
- Correct average = 42+1.125 = 43.125
Problem Type 2: Consecutive Numbers Average
Average of 5 consecutive even numbers is 34. Find the largest.
- Middle (3rd) number = 34
- Sequence: 30, 32, 34, 36, 38
- Largest = 38
Problem Type 3: Runs in Cricket
A batsman's average after 20 innings is 45. He scores 87 in the 21st innings. New average?
- New average = (45×20 + 87)/21 = (900+87)/21 = 987/21 = 47
Problem Type 4: Dismissal Average
A batsman averages 40 in 10 innings. In the 11th he scores 0 and his average drops by 4. Find his score. Wait — if average drops by 4 to 36:
- New sum = 36×11 = 396
- Old sum = 40×10 = 400
- 11th innings score = 396−400 = −4?
- Recalibrate: average drops by 4 means new average = 40−4 = 36
- New sum should be = 36 × 11 = 396
- Old sum = 400
- Score = 396−400 → problem implies a specific score; always set up equation from sum formula
Quick Reference Summary
| Situation | Shortcut |
|---|---|
| Values close to a round number | Assumed mean method |
| Evenly spaced / consecutive values | Middle term or (First+Last)/2 |
| Adding/removing one value | Mean ± change/new count |
| Finding missing value | (Mean × Count) − Known sum |
| Two groups combined | Weighted average formula |
| Finding group size ratio | Alligation cross method |
| Age after T years | Add T to current average |