Episode 8 — Aptitude and Reasoning / 8.8 — Average
8.8.c Solved Examples
Basic Level
Problem 1: Simple Average
Find the average of 15, 20, 25, 30, and 35.
Solution:
Sum = 15 + 20 + 25 + 30 + 35 = 125
Count = 5
Average = 125 / 5 = 25
Shortcut: These are consecutive multiples of 5 (an AP). Average = middle term = 25.
Answer: 25
Problem 2: Finding Sum from Average
The average weight of 25 students in a class is 42 kg. What is the total weight of all students?
Solution:
Sum = Average x Count
= 42 x 25
= 1,050 kg
Answer: 1,050 kg
Problem 3: Finding a Missing Number
The average of five numbers is 28. Four of the numbers are 20, 25, 30, and 35. Find the fifth number.
Solution:
Total sum = 5 x 28 = 140
Sum of four known numbers = 20 + 25 + 30 + 35 = 110
Fifth number = 140 - 110 = 30
Answer: 30
Problem 4: Average of First n Natural Numbers
Find the average of the first 80 natural numbers.
Solution:
Average of first n natural numbers = (n + 1) / 2
Average = (80 + 1) / 2 = 81 / 2 = 40.5
Answer: 40.5
Problem 5: Average of Consecutive Numbers
Find the average of all integers from 21 to 59.
Solution:
Average = (First + Last) / 2 = (21 + 59) / 2 = 80 / 2 = 40
Count = (59 - 21) + 1 = 39
Answer: 40
Problem 6: Each Element Increased by a Constant
The average of 10 numbers is 23. If each number is increased by 7, what is the new average?
Solution:
When every element is increased by k, new average = old average + k.
New average = 23 + 7 = 30
Answer: 30
Problem 7: Adding a New Element
The average of 6 numbers is 18. A seventh number, 39, is added. Find the new average.
Solution:
Old sum = 6 x 18 = 108
New sum = 108 + 39 = 147
New average = 147 / 7 = 21
Answer: 21
Medium Level
Problem 8: Weighted Average
In a company, 60 workers earn an average of Rs 15,000 per month and 40 managers earn an average of Rs 25,000 per month. Find the average salary of all employees.
Solution:
Combined average = (60 x 15000 + 40 x 25000) / (60 + 40)
= (900000 + 1000000) / 100
= 1900000 / 100
= Rs 19,000
Answer: Rs 19,000
Problem 9: Finding the Replaced Number
The average of 12 numbers is 45. If one number is replaced by 63 and the new average becomes 47, find the replaced number.
Solution:
Change in average = 47 - 45 = 2
Change in sum = 12 x 2 = 24
New number - Old number = 24
63 - Old number = 24
Old number = 63 - 24 = 39
Answer: 39
Problem 10: Removing an Element to Get a New Average
The average of 9 numbers is 40. If one number is removed, the average of the remaining 8 numbers becomes 38. Find the removed number.
Solution:
Old sum = 9 x 40 = 360
New sum = 8 x 38 = 304
Removed number = 360 - 304 = 56
Answer: 56
Problem 11: Average Speed (Equal Distances)
Ravi drives from home to office at 30 km/h and returns by the same route at 50 km/h. Find his average speed for the round trip.
Solution:
Average Speed = 2 x S1 x S2 / (S1 + S2)
= 2 x 30 x 50 / (30 + 50)
= 3000 / 80
= 37.5 km/h
Note: The arithmetic mean would be 40, but that is wrong for equal distances.
Answer: 37.5 km/h
Problem 12: Wrong Entry Correction
A teacher calculated the average marks of 30 students as 68. It was later discovered that one student's marks were recorded as 85 instead of 58. Find the correct average.
Solution:
Wrong sum = 30 x 68 = 2040
Correct sum = 2040 - 85 + 58 = 2013
Correct average = 2013 / 30 = 67.1
Answer: 67.1
Problem 13: Age Problem -- New Member Joins
The average age of 8 members of a team is 32 years. A new member aged 23 years joins. What is the new average age?
Solution:
Old sum = 8 x 32 = 256
New sum = 256 + 23 = 279
New average = 279 / 9 = 31 years
Answer: 31 years
Problem 14: Age Problem -- After Some Years
The average age of 5 friends is 24 years now. What will their average age be after 6 years?
Solution:
After t years, every person is t years older.
New average = 24 + 6 = 30 years
Answer: 30 years
Problem 15: Combined Average of Three Groups
Three sections A, B, and C of a class have 30, 25, and 20 students with average marks 60, 70, and 75 respectively. Find the overall average.
Solution:
Combined average = (30 x 60 + 25 x 70 + 20 x 75) / (30 + 25 + 20)
= (1800 + 1750 + 1500) / 75
= 5050 / 75
= 67.33
Answer: 67.33
Problem 16: Batting Average Increase
A cricketer has a batting average of 36 after 24 innings. How many runs must he score in the 25th innings to raise his average to 38?
Solution:
Old total = 24 x 36 = 864
Required new total = 25 x 38 = 950
Runs needed in 25th innings = 950 - 864 = 86
Answer: 86 runs
Problem 17: Average of First n Even Numbers
Find the average of the first 30 even natural numbers.
Solution:
First 30 even numbers: 2, 4, 6, ..., 60
Average of first n even natural numbers = n + 1 = 30 + 1 = 31
Verification: (2 + 60) / 2 = 31. Correct.
Answer: 31
Problem 18: Group Data Average
The following table shows the distribution of daily wages (in Rs) of 50 workers:
Wages (Rs) 200-300 300-400 400-500 500-600 600-700 Workers 8 12 15 10 5
Find the average daily wage.
Solution:
Mid-values: 250, 350, 450, 550, 650
Sum = 8x250 + 12x350 + 15x450 + 10x550 + 5x650
= 2000 + 4200 + 6750 + 5500 + 3250
= 21,700
Average = 21700 / 50 = Rs 434
Answer: Rs 434
Advanced Level
Problem 19: Finding Ratio from Combined Average (Alligation)
The average salary in department A is Rs 8,000 and in department B is Rs 12,000. If the average salary of both departments combined is Rs 9,200, find the ratio of employees in A to B.
Solution (Alligation method):
8000 12000
\ /
\ /
9200
/ \
/ \
12000 - 9200 9200 - 8000
= 2800 = 1200
n_A : n_B = 2800 : 1200 = 7 : 3
Verification:
Combined = (7 x 8000 + 3 x 12000) / (7 + 3)
= (56000 + 36000) / 10
= 92000 / 10
= 9200. Correct.
Answer: 7 : 3
Problem 20: Two Wrong Entries
The average of 50 numbers was calculated as 38.5. It was later found that two numbers 45 and 55 were misread as 35 and 40 respectively. Find the correct average.
Solution:
Wrong sum = 50 x 38.5 = 1925
Error = (45 - 35) + (55 - 40) = 10 + 15 = 25
Correct sum = 1925 + 25 = 1950
Correct average = 1950 / 50 = 39
Answer: 39
Problem 21: Average Speed with Three Legs
A man travels from A to B at 20 km/h, B to C at 30 km/h, and C to D at 60 km/h. If AB = BC = CD, find his average speed for the entire journey.
Solution:
Let each distance = d km.
Total distance = 3d
Time for AB = d/20
Time for BC = d/30
Time for CD = d/60
Total time = d/20 + d/30 + d/60
Finding LCM of 20, 30, 60 = 60:
= 3d/60 + 2d/60 + d/60
= 6d/60
= d/10
Average speed = Total distance / Total time
= 3d / (d/10)
= 30 km/h
Using the formula:
Avg Speed = 3 x S1 x S2 x S3 / (S1*S2 + S2*S3 + S1*S3)
= 3 x 20 x 30 x 60 / (20x30 + 30x60 + 20x60)
= 108000 / (600 + 1800 + 1200)
= 108000 / 3600
= 30 km/h
Answer: 30 km/h
Problem 22: Complex Age Problem
The average age of a husband, wife, and their child 3 years ago was 27 years. The average age of the wife and child 5 years ago was 20 years. What is the present age of the husband?
Solution:
3 years ago, sum of ages = 3 x 27 = 81
Present sum of all three = 81 + 3 x 3 = 81 + 9 = 90
5 years ago, sum of wife and child = 2 x 20 = 40
Present sum of wife and child = 40 + 2 x 5 = 40 + 10 = 50
Present age of husband = 90 - 50 = 40 years
Answer: 40 years
Problem 23: Successive Innings Problem
A batsman's average after 50 innings was 46. After the 51st innings, his average increased by 1. After the 52nd innings, his average further increased by 1. Find his score in the 51st and 52nd innings.
Solution:
After 50 innings: Average = 46, Sum = 50 x 46 = 2300
After 51st innings: Average = 47, Sum = 51 x 47 = 2397
Score in 51st innings = 2397 - 2300 = 97
After 52nd innings: Average = 48, Sum = 52 x 48 = 2496
Score in 52nd innings = 2496 - 2397 = 99
Answer: 97 and 99 runs
Problem 24: Overlapping Groups
The average marks of 50 students in a class are 72. The average marks of the top 20 students is 86 and the average marks of the bottom 25 students is 58. Find the average marks of the remaining 5 students.
Solution:
Total sum = 50 x 72 = 3600
Sum of top 20 = 20 x 86 = 1720
Sum of bottom 25 = 25 x 58 = 1450
Sum of remaining 5 = 3600 - 1720 - 1450 = 430
Average of remaining 5 = 430 / 5 = 86
Answer: 86
Problem 25: Performance Analysis
The average runs scored by a team in the first 10 matches is 180. In the next 5 matches, the average is 220. What should be the average in the remaining 5 matches so that the overall average for all 20 matches is 200?
Solution:
Target total = 20 x 200 = 4000
Sum from first 10 = 10 x 180 = 1800
Sum from next 5 = 5 x 220 = 1100
Required sum from last 5 = 4000 - 1800 - 1100 = 1100
Required average = 1100 / 5 = 220
Answer: 220
Problem 26: Median and Mean Together
The average (mean) of 7 numbers is 15. The median is 12. If each number is increased by 5, find the new mean and new median.
Solution:
When each element is increased by k:
New Mean = Old Mean + k = 15 + 5 = 20
New Median = Old Median + k = 12 + 5 = 17
Answer: New Mean = 20, New Median = 17
Problem 27: Combined Weighted Average with Percentages
In an exam, 30% of students scored an average of 40 marks, 50% scored an average of 60 marks, and the remaining 20% scored an average of 80 marks. Find the overall average.
Solution:
Let total students = 100 (for convenience)
Group 1: 30 students, average 40
Group 2: 50 students, average 60
Group 3: 20 students, average 80
Overall average = (30x40 + 50x60 + 20x80) / 100
= (1200 + 3000 + 1600) / 100
= 5800 / 100
= 58
Answer: 58 marks
Problem 28: Average with Replacement -- Find Original Average
When a number 53 is replaced by another number, the average of 15 numbers increases by 3. Find the new number.
Solution:
Change in sum = n x change in average = 15 x 3 = 45
New number = Old number + 45 = 53 + 45 = 98
Answer: 98
Problem 29: Family Age Problem with Birth of Child
The average age of a couple at the time of their marriage was 25 years. After 4 years, their first child was born. After another 4 years, their second child was born. What is the average age of the family when the second child is 5 years old?
Solution:
The couple married at ages summing to 2 x 25 = 50 years.
When second child is 5 years old:
Years since marriage = 4 + 4 + 5 = 13 years
Husband's + Wife's current ages = 50 + 2 x 13 = 50 + 26 = 76
First child's age = 4 + 5 = 9 years (born 4 years after marriage,
second child born 4 years later,
now 5 more years)
Second child's age = 5 years
Sum of all ages = 76 + 9 + 5 = 90
Average = 90 / 4 = 22.5 years
Answer: 22.5 years
Problem 30: Multi-Step Average Problem
The average weight of A, B, and C is 60 kg. The average weight of A and B is 55 kg. The average weight of B and C is 63 kg. Find the weight of B.
Solution:
A + B + C = 3 x 60 = 180 ... (1)
A + B = 2 x 55 = 110 ... (2)
B + C = 2 x 63 = 126 ... (3)
From (2): A = 110 - B
From (3): C = 126 - B
Substituting in (1):
(110 - B) + B + (126 - B) = 180
236 - B = 180
B = 56 kg
Verification:
A = 110 - 56 = 54 kg
C = 126 - 56 = 70 kg
Average = (54 + 56 + 70) / 3 = 180 / 3 = 60. Correct.
A + B = 54 + 56 = 110. Average = 55. Correct.
B + C = 56 + 70 = 126. Average = 63. Correct.
Answer: B = 56 kg
Problem 31: Bowling Average
A bowler has a bowling average of 24.6 after 25 matches (total wickets = 125). In the next match, he takes 5 wickets for 60 runs. What is his new bowling average?
Solution:
Bowling average = Total runs conceded / Total wickets
Old runs conceded = 24.6 x 125 = 3075
New runs conceded = 3075 + 60 = 3135
New total wickets = 125 + 5 = 130
New bowling average = 3135 / 130 = 24.115 (approx 24.12)
Answer: 24.12 (approximately)
Problem 32: Deviation Method Application
Find the average of: 497, 503, 492, 508, 495, 505, 498, 502.
Solution (Deviation method, assume A = 500):
Deviations from 500: -3, +3, -8, +8, -5, +5, -2, +2
Sum of deviations = (-3 + 3) + (-8 + 8) + (-5 + 5) + (-2 + 2) = 0
Average = 500 + 0/8 = 500
Answer: 500
Summary of Problem Types Covered
| # | Problem Type | Level |
|---|------------------------------------|----------|
| 1 | Simple average | Basic |
| 2 | Sum from average | Basic |
| 3 | Missing number | Basic |
| 4 | First n natural numbers | Basic |
| 5 | Consecutive range | Basic |
| 6 | Constant added to all | Basic |
| 7 | Adding a new element | Basic |
| 8 | Weighted average | Medium |
| 9 | Replacement | Medium |
| 10| Removing an element | Medium |
| 11| Average speed | Medium |
| 12| Wrong entry correction | Medium |
| 13| Age -- new member | Medium |
| 14| Age -- after years | Medium |
| 15| Combined groups | Medium |
| 16| Batting average | Medium |
| 17| First n even numbers | Medium |
| 18| Grouped data | Medium |
| 19| Alligation / ratio from average | Advanced |
| 20| Two wrong entries | Advanced |
| 21| Three-leg average speed | Advanced |
| 22| Complex age problem | Advanced |
| 23| Successive innings | Advanced |
| 24| Overlapping groups | Advanced |
| 25| Performance target | Advanced |
| 26| Mean and median combined | Advanced |
| 27| Weighted average with percentages | Advanced |
| 28| Replacement -- find new number | Advanced |
| 29| Family age with births | Advanced |
| 30| Multi-step (3 equations) | Advanced |
| 31| Bowling average | Advanced |
| 32| Deviation method application | Advanced |
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