Averages seem simple — and the formula is — but the questions can get sneaky. The most common traps involve average speed (it’s NOT the simple average of speeds!) and problems where elements are added or removed. Let’s master the patterns.
What Is an Average?
Average (arithmetic mean) = Sum of all values / Number of values
In simple language, the average tells us “if all values were equal, what would each value be?”
Critical warning: Average speed for equal distances is 2ab/(a+b), which is the harmonic mean, NOT the arithmetic mean (a+b)/2. This is the #1 trap in average speed problems.
The “Sum” Approach — The Master Key
Most average problems become easy if we think in terms of sums, not averages.
The core idea: If the average of n numbers is A, then their sum is n×A. When things change (elements added, removed, replaced), we track the change in sum and recalculate.
Average Speed — The Big Trap
Equal Distances at Different Speeds
Example: A person travels from A to B at 40 km/h and returns at 60 km/h. What’s the average speed?
WRONG answer: (40+60)/2 = 50 km/h. This is wrong because they spend MORE time at the slower speed.
RIGHT answer: Average speed = 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/h
Let’s verify: Say distance = 120 km each way.
- Time going: 120/40 = 3 hours
- Time returning: 120/60 = 2 hours
- Average speed = 240/5 = 48 km/h ✓
Equal Times at Different Speeds
If a person drives at 40 km/h for 2 hours and 60 km/h for 2 hours, THEN we can use simple average: (40+60)/2 = 50 km/h. Because the time is the same for both speeds.
Three Different Speeds for Equal Distances
Average speed = 3abc / (ab + bc + ca)
Effect of Adding / Removing Elements
Adding a New Element
Example: The average of 10 numbers is 25. When an 11th number is added, the average becomes 27. What is the 11th number?
- Old sum = 10 × 25 = 250
- New sum = 11 × 27 = 297
- 11th number = 297 - 250 = 47
Removing an Element
Example: The average of 20 numbers is 35. If one number (56) is removed, what’s the new average?
- Old sum = 20 × 35 = 700
- New sum = 700 - 56 = 644
- New average = 644/19 = 33.89 (approximately)
Replacing an Element
Example: The average of 8 numbers is 14. If one number 24 is replaced by 8, what’s the new average?
- Change in sum = 8 - 24 = -16
- Change in average = -16/8 = -2
- New average = 14 - 2 = 12
Shortcut: When replacing one element in a group of n: Change in average = (New value - Old value) / n
Weighted Average
When groups of different sizes have different averages, we can’t just average the averages. We need to weight them.
Example: Section A has 30 students with average marks 70. Section B has 20 students with average marks 80. What’s the overall average?
Weighted average = (30×70 + 20×80) / (30+20) = (2100 + 1600)/50 = 3700/50 = 74
Notice it’s closer to 70 than 80, because Section A has more students. Makes sense!
Age-Based Average Problems
These show up ALL the time. The key insight is: when time passes, everyone’s age increases by the same amount.
Example: The average age of a family of 5 members is 30 years. If a baby is born, what’s the new average after 2 years?
- Current sum = 5 × 30 = 150
- After 2 years: each of the 5 existing members is 2 years older → sum increases by 10
- Baby is 2 years old
- New sum = 150 + 10 + 2 = 162
- New average = 162/6 = 27 years
Cricket Run Rate Problems
Example: In a 50-over match, the target is 282. After 40 overs, a team has scored 200. What run rate is needed in the remaining overs?
- Runs needed = 282 - 200 = 82
- Overs remaining = 10
- Required run rate = 82/10 = 8.2 runs per over
Consecutive Numbers
Useful shortcut: The average of consecutive numbers equals the middle number.
- Average of 3, 4, 5, 6, 7 → middle number = 5
- Average of first n natural numbers = (n+1)/2
- Average of first n even numbers = (n+1)
- Average of first n odd numbers = n
Worked Examples
Example 1: The average marks of 30 students is 60. The average of the top 10 is 75, and the average of the bottom 10 is 40. What’s the average of the remaining 10?
- Total sum = 30 × 60 = 1800
- Top 10 sum = 10 × 75 = 750
- Bottom 10 sum = 10 × 40 = 400
- Remaining sum = 1800 - 750 - 400 = 650
- Remaining average = 650/10 = 65
Example 2: A batsman has an average of 45 after 20 innings. How many runs must he score in the 21st innings to increase his average to 48?
- Current sum = 20 × 45 = 900
- Required sum after 21 innings = 21 × 48 = 1008
- Runs in 21st innings = 1008 - 900 = 108
Example 3: The average weight of a class of 24 students is 35 kg. If the teacher’s weight is included, the average increases by 400g. What is the teacher’s weight?
- Students’ total weight = 24 × 35 = 840 kg
- New average = 35.4 kg for 25 people
- New total = 25 × 35.4 = 885 kg
- Teacher’s weight = 885 - 840 = 45 kg
Shortcut: Teacher adds (25 × 0.4) = 10 kg more than the old average. So teacher weighs 35 + 10 = 45 kg.
Example 4: A car travels from city A to B at 60 km/h, B to C at 40 km/h, and C to D at 80 km/h. If AB = BC = CD, find the average speed.
Using the three-equal-distance formula: Average speed = 3 × 60 × 40 × 80 / (60×40 + 40×80 + 80×60) = 3 × 192000 / (2400 + 3200 + 4800) = 576000 / 10400 = 55.38 km/h (approximately)
Example 5: The average of 5 consecutive odd numbers is 41. Find the largest number.
For consecutive odd numbers, the average = middle number = 41. The 5 numbers: 37, 39, 41, 43, 45. Largest = 45
Common Exam Patterns
- “Average speed for equal distances” → Use 2ab/(a+b), NOT (a+b)/2
- “Average changes when element added/removed” → Track sums, find the unknown
- “Average of n consecutive numbers” → It’s always the middle number
- “Weighted average of two groups” → (n₁×avg₁ + n₂×avg₂) / (n₁ + n₂)
- “How much to score to change average to X?” → Required sum - Current sum
- “Average age after n years” → Add n to every person’s age (sum increases by n × count)
Practice Problems
Q1: The average of 11 numbers is 36. If the average of the first 6 is 32 and the average of the last 6 is 37, find the 6th number.
Q2: A person goes to office at 30 km/h and returns at 20 km/h. If the total travel time is 5 hours, find the distance to the office.
Q3: The average weight of 8 people increases by 2.5 kg when a new person replaces one weighing 65 kg. What is the weight of the new person?
Answers:
A1: Total sum = 11 × 36 = 396. First 6 sum = 6 × 32 = 192. Last 6 sum = 6 × 37 = 222. The 6th number is counted in both groups. So: 192 + 222 - 396 = 18. The 6th number is 18.
A2: Let distance = d. Time = d/30 + d/20 = 5. → (2d + 3d)/60 = 5 → 5d = 300 → d = 60 km.
A3: Change in total weight = 8 × 2.5 = 20 kg increase. New person’s weight = 65 + 20 = 85 kg.