Average
Key Concepts
Solved Examples: 50
Practice Questions: 5
Solved Examples
Question 1: The average of 12 numbers is 48. If the average of the first 7 numbers is 44, what is the average of the remaining 5 numbers?
Total of all 12 numbers = 12 × 48 = 576
Total of first 7 numbers = 7 × 44 = 308
Total of remaining 5 numbers = 576 − 308 = 268
Average of remaining 5 numbers = 268 ÷ 5 = 53.6
Total of first 7 numbers = 7 × 44 = 308
Total of remaining 5 numbers = 576 − 308 = 268
Average of remaining 5 numbers = 268 ÷ 5 = 53.6
Question 2: The average of 20 numbers is 56. The average of the first 8 numbers is 49 and the average of the last 7 numbers is 64. Find the average of the remaining 5 numbers.
Total of 15 numbers 15 × 72 = 1080
Total of 9 numbers 9 × 68 = 612
Total of 4 numbers 4 × 79 = 316
So, the sum of the remaining 2 numbers is
1080 − 612 − 316 = 152
∴ Average of the remaining 2 numbers
152 ÷ 2 = 76
Total of 9 numbers 9 × 68 = 612
Total of 4 numbers 4 × 79 = 316
So, the sum of the remaining 2 numbers is
1080 − 612 − 316 = 152
∴ Average of the remaining 2 numbers
152 ÷ 2 = 76
Question 3: The average of 15 numbers is 72. If the average of 9 of these numbers is 68 and the average of 4 others is 79, find the average of the remaining numbers.
Total of remaining 17 numbers = 17 × 63 = 1071
Original total = 1071 + 92 = 1163
Original average = 1163 ÷ 18 = 64.61
Original total = 1071 + 92 = 1163
Original average = 1163 ÷ 18 = 64.61
Question 4: The average of 9 numbers is 42. If eight of the numbers are 35, 38, 40, 41, 43, 45, 46 and 48, find the missing number.
Total of 9 numbers = 9 × 42 = 378
Sum of the given numbers = 35 + 38 + 40 + 41 + 43 + 45 + 46 + 48 = 336
Missing number = 378 − 336 = 42
Sum of the given numbers = 35 + 38 + 40 + 41 + 43 + 45 + 46 + 48 = 336
Missing number = 378 − 336 = 42
Question 5: The average of 24 numbers is 46. A new number is added, and the average becomes 48. Find the new number.
Original total = 24 × 46 = 1104
New total = 25 × 48 = 1200
New number = 1200 − 1104 = 96
New total = 25 × 48 = 1200
New number = 1200 − 1104 = 96
Question 6: The average of 16 numbers is 75. Two numbers, 84 and 96, are added to the group. What is the new average?
Original total = 16 × 75 = 1200
New total = 1200 + 84 + 96 = 1380
Total numbers = 16 + 2 = 18
New average = 1380 ÷ 18 = 76⅔
New total = 1200 + 84 + 96 = 1380
Total numbers = 16 + 2 = 18
New average = 1380 ÷ 18 = 76⅔
Question 7: The average of seven consecutive integers is 48. Find the largest integer.
For an odd number of consecutive integers, the average is the middle integer.
Middle integer = 48
The seven integers are:
45, 46, 47, 48, 49, 50, 51
Largest integer = 51
Middle integer = 48
The seven integers are:
45, 46, 47, 48, 49, 50, 51
Largest integer = 51
Question 8: The average of eight consecutive integers is 36.5. Find the smallest integer.
For an even number of consecutive integers, the average lies halfway between the two middle integers.
The eight integers are:
33, 34, 35, 36, 37, 38, 39, 40
Smallest integer = 33
The eight integers are:
33, 34, 35, 36, 37, 38, 39, 40
Smallest integer = 33
Question 9: The average of six consecutive odd integers is 50. Find the sum of the largest and the smallest integers.
The six consecutive odd integers are:
45, 47, 49, 51, 53, 55
Smallest integer = 45
Largest integer = 55
Required sum = 45 + 55 = 100
45, 47, 49, 51, 53, 55
Smallest integer = 45
Largest integer = 55
Required sum = 45 + 55 = 100
Question 10: The average of nine consecutive multiples of 4 is 68. Find the smallest multiple.
The average is the middle multiple.
Middle multiple = 68
There are four multiples on either side.
Smallest multiple = 68 − (4 × 4)
= 68 − 16
= 52
The nine multiples are:
52, 56, 60, 64, 68, 72, 76, 80, 84
Middle multiple = 68
There are four multiples on either side.
Smallest multiple = 68 − (4 × 4)
= 68 − 16
= 52
The nine multiples are:
52, 56, 60, 64, 68, 72, 76, 80, 84
Question 11: The average score of 25 students in Class A is 68, while the average score of 35 students in Class B is 74. Find the average score of all the students together.
Total score of Class A = 25 × 68 = 1700
Total score of Class B = 35 × 74 = 2590
Combined average
= (1700 + 2590) ÷ (25 + 35)
= 4290 ÷ 60
= 71.5
Total score of Class B = 35 × 74 = 2590
Combined average
= (1700 + 2590) ÷ (25 + 35)
= 4290 ÷ 60
= 71.5
Question 12: A company has 45 employees with an average age of 32 years and 30 employees with an average age of 41 years. Find the average age of all employees.
Total age of first group = 45 × 32 = 1440
Total age of second group = 30 × 41 = 1230
Combined average
= (1440 + 1230) ÷ (45 + 30)
= 2670 ÷ 75
= 35.6 years
Total age of second group = 30 × 41 = 1230
Combined average
= (1440 + 1230) ÷ (45 + 30)
= 2670 ÷ 75
= 35.6 years
Question 13: A warehouse stores 120 kg of rice costing ¤48 per kg, 80 kg costing ¤54 per kg, and 50 kg costing ¤60 per kg. Find the average cost per kilogram.
Total cost
= (120 × 48) + (80 × 54) + (50 × 60)
= 5760 + 4320 + 3000
= 13080
Total quantity = 120 + 80 + 50 = 250 kg
Average cost per kg
= 13080 ÷ 250
= ¤52.32 per kg
= (120 × 48) + (80 × 54) + (50 × 60)
= 5760 + 4320 + 3000
= 13080
Total quantity = 120 + 80 + 50 = 250 kg
Average cost per kg
= 13080 ÷ 250
= ¤52.32 per kg
Question 14: A sports club consists of 4 coaches, 12 adult players and 8 junior players. The average age of the coaches is 46 years, that of the adult players is 29 years and that of the junior players is 14 years. What is the average age of all the members of the club?
Total age of coaches = 4 × 46 = 184
Total age of adult players = 12 × 29 = 348
Total age of junior players = 8 × 14 = 112
Total age of all members = 184 + 348 + 112 = 644
Total number of members = 4 + 12 + 8 = 24
Average age = 644 ÷ 24 = 26.83 years
Total age of adult players = 12 × 29 = 348
Total age of junior players = 8 × 14 = 112
Total age of all members = 184 + 348 + 112 = 644
Total number of members = 4 + 12 + 8 = 24
Average age = 644 ÷ 24 = 26.83 years
Question 15: A grocer has sales of ¤5,850, ¤6,420, ¤6,780, ¤6,150 and ¤5,900 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of ¤6,000?
Total sales required for 6 months 6 × ¤6,000 = ¤36,000
Sales for the first 5 months ¤5,850 + ¤6,420 + ¤6,780 + ¤6,150 + ¤5,900 = ¤31,100
Required sales in the sixth month ¤36,000 − ¤31,100 = ¤4,900
∴ The required sale in the sixth month is ¤4,900.
Sales for the first 5 months ¤5,850 + ¤6,420 + ¤6,780 + ¤6,150 + ¤5,900 = ¤31,100
Required sales in the sixth month ¤36,000 − ¤31,100 = ¤4,900
∴ The required sale in the sixth month is ¤4,900.
Question 16: The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
At most, 19 numbers can be greater than zero.
The Sum Rule: For the average of 20 numbers to be zero, their total sum must equal zero (20 x 0 = 0).
The Balance: You can have 19 positive numbers, as long as the 20th number is a negative number large enough to cancel them all out.
The Sum Rule: For the average of 20 numbers to be zero, their total sum must equal zero (20 x 0 = 0).
The Balance: You can have 19 positive numbers, as long as the 20th number is a negative number large enough to cancel them all out.
Question 17: The average monthly income of A and B is ¤4,800. The average monthly income of B and C is ¤6,200 and the average monthly income of A and C is ¤5,400. What is the monthly income of A?
From the given information
A + B = 2 × ¤4,800 = ¤9,600
B + C = 2 × ¤6,200 = ¤12,400
A + C = 2 × ¤5,400 = ¤10,800
Adding the first and third equations
A + B + A + C = ¤9,600 + ¤10,800
2A + B + C = ¤20,400
But B + C = ¤12,400
∴ 2A + ¤12,400 = ¤20,400
or, 2A = ¤8,000
or, A = ¤4,000
∴ The monthly income of A is ¤4,000.
A + B = 2 × ¤4,800 = ¤9,600
B + C = 2 × ¤6,200 = ¤12,400
A + C = 2 × ¤5,400 = ¤10,800
Adding the first and third equations
A + B + A + C = ¤9,600 + ¤10,800
2A + B + C = ¤20,400
But B + C = ¤12,400
∴ 2A + ¤12,400 = ¤20,400
or, 2A = ¤8,000
or, A = ¤4,000
∴ The monthly income of A is ¤4,000.
Question 18: The average age of a husband, wife and their child 4 years ago was 26 years, and the average age of the wife and the child 6 years ago was 18 years. What is the present age of the husband?
Let the present ages of husband, wife and child be H, W and C respectively.
4 years ago:
(H − 4 + W − 4 + C − 4) ÷ 3 = 26
(H + W + C − 12) ÷ 3 = 26
H + W + C − 12 = 78
H + W + C = 90 …(1)
6 years ago:
(W − 6 + C − 6) ÷ 2 = 18
(W + C − 12) ÷ 2 = 18
W + C − 12 = 36
W + C = 48 …(2)
Subtract equation (2) from equation (1):
H + W + C − (W + C) = 90 − 48
H = 42
4 years ago:
(H − 4 + W − 4 + C − 4) ÷ 3 = 26
(H + W + C − 12) ÷ 3 = 26
H + W + C − 12 = 78
H + W + C = 90 …(1)
6 years ago:
(W − 6 + C − 6) ÷ 2 = 18
(W + C − 12) ÷ 2 = 18
W + C − 12 = 36
W + C = 48 …(2)
Subtract equation (2) from equation (1):
H + W + C − (W + C) = 90 − 48
H = 42
Question 19: A car owner buys petrol at ¤60, ¤72 and ¤90 per litre for three successive years. If he spends ¤3,600 each year on petrol, what is the average cost per litre of petrol over the three years?
Petrol purchased in the first year ¤3,600 ÷ ¤60 = 60 litres
Petrol purchased in the second year ¤3,600 ÷ ¤72 = 50 litres
Petrol purchased in the third year ¤3,600 ÷ ¤90 = 40 litres
Total petrol purchased 60 + 50 + 40 = 150 litres
Total amount spent 3 × ¤3,600 = ¤10,800
So the average cost per litre ¤10,800 ÷ 150 = ¤72
∴ The average cost of petrol is ¤72 per litre.
Petrol purchased in the second year ¤3,600 ÷ ¤72 = 50 litres
Petrol purchased in the third year ¤3,600 ÷ ¤90 = 40 litres
Total petrol purchased 60 + 50 + 40 = 150 litres
Total amount spent 3 × ¤3,600 = ¤10,800
So the average cost per litre ¤10,800 ÷ 150 = ¤72
∴ The average cost of petrol is ¤72 per litre.
Question 20: In Arun’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Arun and he thinks that Arun’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
Let Arun’s weight be X kg.
According to Arun:
65 < X < 72
According to Arun’s brother:
60 < X < 70
According to Arun’s mother:
X ≤ 68
The values satisfying all the above conditions are:
66, 67 and 68
Required average:
= (66 + 67 + 68) ÷ 3
= 201 ÷ 3
= 67 kg
According to Arun:
65 < X < 72
According to Arun’s brother:
60 < X < 70
According to Arun’s mother:
X ≤ 68
The values satisfying all the above conditions are:
66, 67 and 68
Required average:
= (66 + 67 + 68) ÷ 3
= 201 ÷ 3
= 67 kg
Question 21: The average weight of A, B and C is 50 kg. If the average weight of A and B is 44 kg and that of B and C is 46 kg, then what is the weight of B?
Let the weights of A, B and C be A kg, B kg and C kg respectively.
Average weight of A, B and C:
(A + B + C) ÷ 3 = 50
A + B + C = 150 …(1)
Average weight of A and B:
(A + B) ÷ 2 = 44
A + B = 88 …(2)
Average weight of B and C:
(B + C) ÷ 2 = 46
B + C = 92 …(3)
Adding equations (2) and (3):
(A + B) + (B + C) = 88 + 92
A + 2B + C = 180
From equation (1):
A + B + C = 150
Subtracting:
(A + 2B + C) − (A + B + C) = 180 − 150
B = 30
Average weight of A, B and C:
(A + B + C) ÷ 3 = 50
A + B + C = 150 …(1)
Average weight of A and B:
(A + B) ÷ 2 = 44
A + B = 88 …(2)
Average weight of B and C:
(B + C) ÷ 2 = 46
B + C = 92 …(3)
Adding equations (2) and (3):
(A + B) + (B + C) = 88 + 92
A + 2B + C = 180
From equation (1):
A + B + C = 150
Subtracting:
(A + 2B + C) − (A + B + C) = 180 − 150
B = 30
Question 22: A library has an average of 480 visitors on Sundays and 240 visitors on other days. What is the average number of visitors per day in a month of 30 days that begins with a Sunday?
A month of 30 days beginning with Sunday will have:
Number of Sundays = 5
Number of other days = 25
Total visitors on Sundays:
= 480 × 5
= 2400
Total visitors on other days:
= 240 × 25
= 6000
Total visitors in the month:
= 2400 + 6000
= 8400
Average number of visitors per day:
= 8400 ÷ 30
= 280
Number of Sundays = 5
Number of other days = 25
Total visitors on Sundays:
= 480 × 5
= 2400
Total visitors on other days:
= 240 × 25
= 6000
Total visitors in the month:
= 2400 + 6000
= 8400
Average number of visitors per day:
= 8400 ÷ 30
= 280
Question 23: A student’s marks were wrongly entered as 96 instead of 76. Due to this error, the average marks of the class increased by 1/2 mark. What is the total number of students in the class?
Difference between the wrongly entered marks and the actual marks:
= 96 − 76
= 20 marks
Increase in total marks = 20
Increase in average marks = 1/2 mark
Let the number of students in the class be x.
Increase in average:
= Increase in total marks ÷ Number of students
1/2 = 20 ÷ x
x = 20 × 2
x = 40
Number of students in the class = 40
= 96 − 76
= 20 marks
Increase in total marks = 20
Increase in average marks = 1/2 mark
Let the number of students in the class be x.
Increase in average:
= Increase in total marks ÷ Number of students
1/2 = 20 ÷ x
x = 20 × 2
x = 40
Number of students in the class = 40
Question 24: The captain of a football team of 11 members is 26 years old and the goalkeeper is 3 years older than the captain. If the ages of these two players are excluded, the average age of the remaining players is 1 year less than the average age of the whole team. What is the average age of the team?
Let the average age of the whole football team be X years.
Total age of 11 players = 11X
Age of captain = 26 years
Age of goalkeeper = 26 + 3 = 29 years
Total age of captain and goalkeeper:
= 26 + 29
= 55 years
Age of remaining 9 players = 11X − 55
According to the question:
Average age of remaining players = X − 1
Therefore, (11X − 55) ÷ 9 = X − 1
11X − 55 = 9X − 9
11X − 9X = 55 − 9
2X = 46
X = 23
Total age of 11 players = 11X
Age of captain = 26 years
Age of goalkeeper = 26 + 3 = 29 years
Total age of captain and goalkeeper:
= 26 + 29
= 55 years
Age of remaining 9 players = 11X − 55
According to the question:
Average age of remaining players = X − 1
Therefore, (11X − 55) ÷ 9 = X − 1
11X − 55 = 9X − 9
11X − 9X = 55 − 9
2X = 46
X = 23
Question 25: The average weight of 24 items is 35 kg. If one item weighing 22 kg is removed and replaced by a new item, the average weight of the group drops by exactly 250 grams. Find the weight of the new item.
Original total weight 24 × 35 = 840 kg
The average drops by 250 grams = 0.25 kg.
New average 35 − 0.25 = 34.75 kg
New total weight 24 × 34.75 = 834 kg
So, the total weight decreases by 840 − 834 = 6 kg
The 22 kg item is removed and replaced by the new item.
∴ Weight of the new item 22 − 6 = 16 kg
The average drops by 250 grams = 0.25 kg.
New average 35 − 0.25 = 34.75 kg
New total weight 24 × 34.75 = 834 kg
So, the total weight decreases by 840 − 834 = 6 kg
The 22 kg item is removed and replaced by the new item.
∴ Weight of the new item 22 − 6 = 16 kg
Question 26: The average score of 15 students in an aptitude test is 64. When one student’s score is removed and replaced by a new student’s score, the average score of the class increases to 66. If the score of the new student is 92, find the score of the student who was removed.
Original total score 15 × 64 = 960
New total score 15 × 66 = 990
So increase in the total score 990 − 960 = 30
The new student’s score is 92, so the removed student’s score must be
92 − 30 = 62
∴ The score of the student who was removed is 62.
New total score 15 × 66 = 990
So increase in the total score 990 − 960 = 30
The new student’s score is 92, so the removed student’s score must be
92 − 30 = 62
∴ The score of the student who was removed is 62.
Question 27: The average age of 8 family members is 40 years. A senior member leaves the family and is replaced by a new member who is 24 years old. If the new average age of the family drops to 36 years, find the age of the senior member who left.
Original total age 8 × 40 = 320 years
New total age 8 × 36 = 288 years
So decrease in the total age 320 − 288 = 32 years
The new member is 24 years old. Hence, the senior member who left was
24 + 32 = 56 years
∴ The age of the senior member who left is 56 years.
New total age 8 × 36 = 288 years
So decrease in the total age 320 − 288 = 32 years
The new member is 24 years old. Hence, the senior member who left was
24 + 32 = 56 years
∴ The age of the senior member who left is 56 years.
Question 28: A delivery truck carries 10 packages with an average weight of 18 kg. Two packages weighing 15 kg and 23 kg are removed and replaced by two new identical packages. If the new average weight of all the packages on the truck increases to 20 kg, find the individual weight of each new package.
Original total weight 10 × 18 = 180 kg
New total weight 10 × 20 = 200 kg
Increase in total weight 200 − 180 = 20 kg
The two removed packages weighed 15 + 23 = 38 kg
Let the weight of each new package be x kg.
The two new packages weigh 2x kg
Since the total weight increased by 20 kg
2x − 38 = 20
or, 2x = 58
or, x = 29 kg
∴ The individual weight of each new package is 29 kg.
New total weight 10 × 20 = 200 kg
Increase in total weight 200 − 180 = 20 kg
The two removed packages weighed 15 + 23 = 38 kg
Let the weight of each new package be x kg.
The two new packages weigh 2x kg
Since the total weight increased by 20 kg
2x − 38 = 20
or, 2x = 58
or, x = 29 kg
∴ The individual weight of each new package is 29 kg.
Question 29: The average of 11 results is 60. If the average of the first 6 results is 58 and the average of the last 6 results is 63, find the value of the 6th result.
Total of all 11 results 11 × 60 = 660
Total of the first 6 results 6 × 58 = 348
Total of the last 6 results 6 × 63 = 378
When the totals of the first 6 and last 6 are added, the 6th result is counted twice, while all other results are counted once.
6th result 348 + 378 − 660 = 66
The value of the 6th result is 66.
Total of the first 6 results 6 × 58 = 348
Total of the last 6 results 6 × 63 = 378
When the totals of the first 6 and last 6 are added, the 6th result is counted twice, while all other results are counted once.
6th result 348 + 378 − 660 = 66
The value of the 6th result is 66.
Question 30: The average temperature of a city for the first 4 days of a week is 38 degrees, and the average temperature for the last 4 days of the same week is 41 degrees. If the average temperature for the entire 7-day week is 39 degrees, find the temperature recorded on the 4th day.
Total temperature for the 7 days 7 × 39 = 273°
Total temperature for the first 4 days 4 × 38 = 152°
Total temperature for the last 4 days 4 × 41 = 164°
When these two 4-day totals are added, the temperature of the 4th day is counted twice.
Temperature on the 4th day 152 + 164 − 273 = 43°
∴ The temperature recorded on the 4th day was 43°.
Total temperature for the first 4 days 4 × 38 = 152°
Total temperature for the last 4 days 4 × 41 = 164°
When these two 4-day totals are added, the temperature of the 4th day is counted twice.
Temperature on the 4th day 152 + 164 − 273 = 43°
∴ The temperature recorded on the 4th day was 43°.
Question 31: The average weight of a shipment of 50 machinery components is 120 kg. If the factory decides to reduce the weight of every component by exactly 8% to save on raw materials, find the new average weight of the shipment.
Original average weight = 120 kg
Each component’s weight is reduced by 8%.
Reduction in average weight 8% of 120 = 120 × 8/100 = 9.6 kg
New average weight 120 − 9.6 = 110.4 kg
∴ The new average weight of the shipment is 110.4 kg.
Each component’s weight is reduced by 8%.
Reduction in average weight 8% of 120 = 120 × 8/100 = 9.6 kg
New average weight 120 − 9.6 = 110.4 kg
∴ The new average weight of the shipment is 110.4 kg.
Question 32: A student logs the scores of 10 exams. The scores are all distinct positive integers arranged in strictly ascending order. The average score of all 10 exams is 80. If the average of the first 6 exams is 75, find the absolute maximum possible score that the student could have achieved in the final 10th exam.
Total score of all 10 exams 10 × 80 = 800
Total score of the first 6 exams 6 × 75 = 450
∴ Total score of the final 4 exams 800 − 450 = 350
Let the scores be x₁ < x₂ < x₃ < x₄ < x₅ < x₆ < x₇ < x₈ < x₉ < x₁₀
To maximize x₁₀, we must make x₇, x₈ and x₉ as small as possible.
Since the first 6 scores are distinct positive integers and all are less than x₇, the largest possible sum of the first 6 scores for a given x₇ is
(x₇ − 1) + (x₇ − 2) + (x₇ − 3) + (x₇ − 4) + (x₇ − 5) + (x₇ − 6)
= 6x₇ − 21
This sum must be at least 450
6x₇ − 21 ≥ 450
6x₇ ≥ 471
∴ x₇ ≥ 78.5
Since the scores are integers x₇ ≥ 79
To maximize x₁₀, take the smallest possible values
x₇ = 79, x₈ = 80, x₉ = 81
Therefore x₁₀ = 350 − (79 + 80 + 81)
= 350 − 240
= 110
This maximum is achievable. For example, the first six scores can be
70, 73, 74, 75, 76, 78
Their sum is 450, and the complete sequence can be 70, 73, 74, 75, 76, 78, 79, 80, 81, 110
∴ The absolute maximum possible score in the 10th exam is 110.
Total score of the first 6 exams 6 × 75 = 450
∴ Total score of the final 4 exams 800 − 450 = 350
Let the scores be x₁ < x₂ < x₃ < x₄ < x₅ < x₆ < x₇ < x₈ < x₉ < x₁₀
To maximize x₁₀, we must make x₇, x₈ and x₉ as small as possible.
Since the first 6 scores are distinct positive integers and all are less than x₇, the largest possible sum of the first 6 scores for a given x₇ is
(x₇ − 1) + (x₇ − 2) + (x₇ − 3) + (x₇ − 4) + (x₇ − 5) + (x₇ − 6)
= 6x₇ − 21
This sum must be at least 450
6x₇ − 21 ≥ 450
6x₇ ≥ 471
∴ x₇ ≥ 78.5
Since the scores are integers x₇ ≥ 79
To maximize x₁₀, take the smallest possible values
x₇ = 79, x₈ = 80, x₉ = 81
Therefore x₁₀ = 350 − (79 + 80 + 81)
= 350 − 240
= 110
This maximum is achievable. For example, the first six scores can be
70, 73, 74, 75, 76, 78
Their sum is 450, and the complete sequence can be 70, 73, 74, 75, 76, 78, 79, 80, 81, 110
∴ The absolute maximum possible score in the 10th exam is 110.
Question 33: A researcher measures the lifespan of 20 specimens. The calculated average lifespan is exactly 45 days. It is later discovered that the lifespans of three specific specimens were completely omitted from the calculation due to a database error. The actual lifespans of these three missing specimens are locked in a strict ratio of 2 : 3 : 4. If including these three specimens increases the overall true average lifespan of all 20 specimens to 48 days, find the actual lifespan of the longest-living omitted specimen.
The original average of the 20 specimens was 45 days, so the calculated total lifespan was
20 × 45 = 900 days
After including the three omitted specimens, the true average becomes 48 days
20 × 48 = 960 days
∴ Total lifespan of the three omitted specimens
960 − 900 = 60 days
Their lifespans are in the ratio 2 : 3 : 4
Total ratio parts 2 + 3 + 4 = 9
Value of one ratio part 60 ÷ 9 = 20/3 days
The longest-living specimen corresponds to 4 parts
4 × 20/3 = 80/3 days
= 26⅔ days
∴ The actual lifespan of the longest-living omitted specimen is 26⅔ days.
20 × 45 = 900 days
After including the three omitted specimens, the true average becomes 48 days
20 × 48 = 960 days
∴ Total lifespan of the three omitted specimens
960 − 900 = 60 days
Their lifespans are in the ratio 2 : 3 : 4
Total ratio parts 2 + 3 + 4 = 9
Value of one ratio part 60 ÷ 9 = 20/3 days
The longest-living specimen corresponds to 4 parts
4 × 20/3 = 80/3 days
= 26⅔ days
∴ The actual lifespan of the longest-living omitted specimen is 26⅔ days.
Question 34: The average of 15 numbers is 72. It is discovered that three numbers were completely omitted from this calculation. The three missing numbers are in the ratio 2 : 3 : 5. If including these three numbers increases the overall average of all 18 numbers to 75, find the largest of the three missing numbers.
Original total of the 15 numbers 15 × 72 = 1080
Total of all 18 numbers after including the missing numbers
18 × 75 = 1350
∴ Sum of the three missing numbers 1350 − 1080 = 270
The missing numbers are in the ratio 2 : 3 : 5
Total ratio parts 2 + 3 + 5 = 10
Value of one ratio part 270 ÷ 10 = 27
The largest missing number corresponds to 5 parts 5 × 27 = 135
∴ The largest of the three missing numbers is 135.
Total of all 18 numbers after including the missing numbers
18 × 75 = 1350
∴ Sum of the three missing numbers 1350 − 1080 = 270
The missing numbers are in the ratio 2 : 3 : 5
Total ratio parts 2 + 3 + 5 = 10
Value of one ratio part 270 ÷ 10 = 27
The largest missing number corresponds to 5 parts 5 × 27 = 135
∴ The largest of the three missing numbers is 135.
Question 35: A group of 10 people has an average age of 25 years. If two new members join the group, the overall average age of all the members increases by exactly 2 years. If the age of one of the new members is exactly 12 years greater than the other, find the individual age of the younger new member.
Original total age of 10 people 10 × 25 = 250 years
The new average is 25 + 2 = 27 years
Total age of 12 people 12 × 27 = 324 years
∴ Total age of the two new members 324 − 250 = 74 years
Let the age of the younger member be x years.
The older member is 12 years older x + 12
Therefore x + (x + 12) = 74
or, 2x + 12 = 74
or, 2x = 62
or, x = 31
∴ The younger new member is 31 years old.
The new average is 25 + 2 = 27 years
Total age of 12 people 12 × 27 = 324 years
∴ Total age of the two new members 324 − 250 = 74 years
Let the age of the younger member be x years.
The older member is 12 years older x + 12
Therefore x + (x + 12) = 74
or, 2x + 12 = 74
or, 2x = 62
or, x = 31
∴ The younger new member is 31 years old.
Question 36: The average number of copies of a book sold per day by a shopkeeper is 54 in the initial 7 days and 58 in the initial 8 days after the book launch. On the 9th day, she sells 14 copies fewer than on the 8th day, and the average number of copies sold per day from the 2nd day to the 9th day becomes 62. Find the number of copies sold on the first day of the book launch.
Average sales in the first 7 days = 54
Total copies sold in the first 7 days 7 × 54 = 378
Average sales in the first 8 days = 58
Total copies sold in the first 8 days 8 × 58 = 464
Therefore, copies sold on the 8th day 464 − 378 = 86
On the 9th day, she sells 14 fewer copies 86 − 14 = 72
Average sales from the 2nd to the 9th day = 62
There are 8 days from the 2nd to the 9th day.
∴ Total copies sold from the 2nd to the 9th day 8 × 62 = 496
Total copies sold from the 1st to the 9th day 464 + 72 = 536
Copies sold on the 1st day is 536 − 496 = 40
∴ The number of copies sold on the first day was 40.
Total copies sold in the first 7 days 7 × 54 = 378
Average sales in the first 8 days = 58
Total copies sold in the first 8 days 8 × 58 = 464
Therefore, copies sold on the 8th day 464 − 378 = 86
On the 9th day, she sells 14 fewer copies 86 − 14 = 72
Average sales from the 2nd to the 9th day = 62
There are 8 days from the 2nd to the 9th day.
∴ Total copies sold from the 2nd to the 9th day 8 × 62 = 496
Total copies sold from the 1st to the 9th day 464 + 72 = 536
Copies sold on the 1st day is 536 − 496 = 40
∴ The number of copies sold on the first day was 40.
Question 37: The average salary of 5 managers and 25 engineers in a company is ¤72,000. If each manager receives a 20% salary increase while the salaries of the engineers remain unchanged, the average salary of all 30 employees would increase by 5%. What is the average salary, in ¤, of the engineers?
Original total salary of all 30 employees 30 × ¤72,000 = ¤2,160,000
The new average salary would be ¤72,000 × 1.05 = ¤75,600
New total salary 30 × ¤75,600 = ¤2,268,000
So increase in the total salary ¤2,268,000 − ¤2,160,000 = ¤108,000
This increase comes entirely from the managers’ 20% salary increase.
Therefore, the original total salary of the 5 managers was
¤108,000 ÷ 20% = ¤540,000
So total salary of the 25 engineers is ¤2,160,000 − ¤540,000 = ¤1,620,000
Average salary of the engineers ¤1,620,000 ÷ 25 = ¤64,800
∴ The average salary of the engineers is ¤64,800.
The new average salary would be ¤72,000 × 1.05 = ¤75,600
New total salary 30 × ¤75,600 = ¤2,268,000
So increase in the total salary ¤2,268,000 − ¤2,160,000 = ¤108,000
This increase comes entirely from the managers’ 20% salary increase.
Therefore, the original total salary of the 5 managers was
¤108,000 ÷ 20% = ¤540,000
So total salary of the 25 engineers is ¤2,160,000 − ¤540,000 = ¤1,620,000
Average salary of the engineers ¤1,620,000 ÷ 25 = ¤64,800
∴ The average salary of the engineers is ¤64,800.
Question 38: A class has 36 students whose average weight is 42 kg. m new students join the class, and their average weight is n kg. If it is known that m + n = 54, what is the maximum possible average weight of the class now?
Original average weight = 42 kg.
For the overall average to increase, the new students must have an average weight greater than 42 kg.
∴ n > 42
Given m + n = 54
or, n = 54 − m
The new students add an extra weight of m(n − 42)
This extra weight is distributed among all 36 + m students.
New average 42 + m(n − 42) ÷ (36 + m)
Substituting n = 54 − m we get
42 + m(12 − m) ÷ (36 + m)
Since m is a positive integer, check the values around the maximum
For m = 5:
Increase = 5 × 7 ÷ 41 = 35 ÷ 41 ≈ 0.854
Average ≈ 42.854 kg
For m = 6
Increase = 6 × 6 ÷ 42 = 36 ÷ 42 = 0.857
Average ≈ 42.857 kg
For m = 7
Increase = 7 × 5 ÷ 43 = 35 ÷ 43 ≈ 0.814
Average ≈ 42.814 kg
∴ The maximum occurs when m = 6 and n = 48
New average 42 + 36 ÷ 42
= 42 + 6/7
= 42.86 kg approximately.
∴ The maximum possible average weight of the class is 42.86 kg.
For the overall average to increase, the new students must have an average weight greater than 42 kg.
∴ n > 42
Given m + n = 54
or, n = 54 − m
The new students add an extra weight of m(n − 42)
This extra weight is distributed among all 36 + m students.
New average 42 + m(n − 42) ÷ (36 + m)
Substituting n = 54 − m we get
42 + m(12 − m) ÷ (36 + m)
Since m is a positive integer, check the values around the maximum
For m = 5:
Increase = 5 × 7 ÷ 41 = 35 ÷ 41 ≈ 0.854
Average ≈ 42.854 kg
For m = 6
Increase = 6 × 6 ÷ 42 = 36 ÷ 42 = 0.857
Average ≈ 42.857 kg
For m = 7
Increase = 7 × 5 ÷ 43 = 35 ÷ 43 ≈ 0.814
Average ≈ 42.814 kg
∴ The maximum occurs when m = 6 and n = 48
New average 42 + 36 ÷ 42
= 42 + 6/7
= 42.86 kg approximately.
∴ The maximum possible average weight of the class is 42.86 kg.
Question 39: The difference in weight between the heaviest student and the lightest student in a class of 28 students is 18 kg. If the lightest student is excluded, the new average weight of the class increases by 1% of the original average weight. If the heaviest student is excluded, the new average weight of the class drops to 98% of the original average weight. Find the original average weight of the class.
Let the original average weight be A kg.
Total weight of the 28 students 28A
When the lightest student is excluded, the number of students becomes 27 and the new average is 101% of the original average.
Lightest weight
= 28A − 27 × 1.01A
= 28A − 27.27A
= 0.73A
When the heaviest student is excluded, the new average is 98% of the original average.
Heaviest weight
= 28A − 27 × 0.98A
= 28A − 26.46A
= 1.54A
The difference between the heaviest and lightest weights is 18 kg.
1.54A − 0.73A = 18
0.81A = 18
A = 18 ÷ 0.81
A = 22 ²⁄₉
∴ The original average weight is 22 ²⁄₉ kg.
Total weight of the 28 students 28A
When the lightest student is excluded, the number of students becomes 27 and the new average is 101% of the original average.
Lightest weight
= 28A − 27 × 1.01A
= 28A − 27.27A
= 0.73A
When the heaviest student is excluded, the new average is 98% of the original average.
Heaviest weight
= 28A − 27 × 0.98A
= 28A − 26.46A
= 1.54A
The difference between the heaviest and lightest weights is 18 kg.
1.54A − 0.73A = 18
0.81A = 18
A = 18 ÷ 0.81
A = 22 ²⁄₉
∴ The original average weight is 22 ²⁄₉ kg.
Question 40: A family consists of a mother, a father, and some children. The average age of the members of the family is 22 years, the father is 50 years old, and the average age of the mother and children is 18 years. How many children are in the family?
Let the number of children be n.
Total number of family members = n + 2
Total age of the family 22(n + 2)
The total age of the mother and children 18(n + 1)
Since the father’s age is 50 years
22(n + 2) = 50 + 18(n + 1)
or, 22n + 44 = 50 + 18n + 18
or, 22n + 44 = 18n + 68
or, 4n = 24
or, n = 6
∴ There are 6 children in the family.
Total number of family members = n + 2
Total age of the family 22(n + 2)
The total age of the mother and children 18(n + 1)
Since the father’s age is 50 years
22(n + 2) = 50 + 18(n + 1)
or, 22n + 44 = 50 + 18n + 18
or, 22n + 44 = 18n + 68
or, 4n = 24
or, n = 6
∴ There are 6 children in the family.
Question 41: A college has raised 72% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people to be solicited?
Let the total number of people to be solicited be N.
People already solicited = 60% of N
= 0.6N
Amount already raised = 0.6N × Rs. 600
= Rs. 360N
This represents 72% of the total amount required.
∴ Total amount required
= Rs. 360N ÷ 0.72
= Rs. 500N
Amount still required = Rs. 500N − Rs. 360N
= Rs. 140N
Remaining people to be solicited
= 40% of N
= 0.4N
∴ Average donation required from each remaining person
= Rs. 140N ÷ 0.4N
= Rs. 350
∴ The average donation from the remaining people should be Rs. 350.
People already solicited = 60% of N
= 0.6N
Amount already raised = 0.6N × Rs. 600
= Rs. 360N
This represents 72% of the total amount required.
∴ Total amount required
= Rs. 360N ÷ 0.72
= Rs. 500N
Amount still required = Rs. 500N − Rs. 360N
= Rs. 140N
Remaining people to be solicited
= 40% of N
= 0.4N
∴ Average donation required from each remaining person
= Rs. 140N ÷ 0.4N
= Rs. 350
∴ The average donation from the remaining people should be Rs. 350.
Question 42: A student appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then what is the total number of tests?
Let the total number of tests be n and the original average score be A.
Total score = nA
When the first 10 tests are excluded
nA − 200 = (n − 10)(A + 1)
Expanding nA − 200 = nA + n − 10A − 10
Cancelling nA
10A − n = 190 …①
When the last 10 tests are excluded:
nA − 300 = (n − 10)(A − 1)
Expanding nA − 300 = nA − n − 10A + 10
Cancelling nA
10A + n = 310 …②
Adding ① and ②
20A = 500
A = 25
Substituting in ②
250 + n = 310
or, n = 60
∴ The total number of tests is 60.
Total score = nA
When the first 10 tests are excluded
nA − 200 = (n − 10)(A + 1)
Expanding nA − 200 = nA + n − 10A − 10
Cancelling nA
10A − n = 190 …①
When the last 10 tests are excluded:
nA − 300 = (n − 10)(A − 1)
Expanding nA − 300 = nA − n − 10A + 10
Cancelling nA
10A + n = 310 …②
Adding ① and ②
20A = 500
A = 25
Substituting in ②
250 + n = 310
or, n = 60
∴ The total number of tests is 60.
Question 43: The average of x, y and z is 52. x is as much more than the average as y is less than the average. Find the value of z.
Average of x, y and z = 52
So, x + y + z = 52 × 3
or, x + y + z = 156
Since x is as much more than the average as y is less than the average
x − 52 = 52 − y
or, x + y = 104
Substituting
104 + z = 156
or, z = 52
∴ z = 52
So, x + y + z = 52 × 3
or, x + y + z = 156
Since x is as much more than the average as y is less than the average
x − 52 = 52 − y
or, x + y = 104
Substituting
104 + z = 156
or, z = 52
∴ z = 52
Question 44: Onions are sold for 5 consecutive months at the rates of ¤10, ¤20, ¤40, ¤40, and ¤80 per kg, respectively. A family spends a fixed amount of money on onions during each of the first three months, and then spends half that amount during each of the next two months. What is the average expense on onions, in ¤ per kg, over these 5 months?
Let the amount spent during each of the first three months be ¤A.
∴ Amount spent during each of the next two months = ¤A⁄2
The quantities of onions purchased are
Month 1 = A⁄10 kg
Month 2 = A⁄20 kg
Month 3 = A⁄40 kg
Month 4 = A⁄80 kg
Month 5 = A⁄160 kg
Total expenditure A + A + A + A⁄2 + A⁄2 = 4A
Total quantity purchased A⁄10 + A⁄20 + A⁄40 + A⁄80 + A⁄160
= (16A + 8A + 4A + 2A + A)⁄160
= 31A⁄160 kg
∴ Average expense per kg = 4A ÷ 31A⁄160
= 640⁄31
≈ ¤20.65 per kg
∴ Amount spent during each of the next two months = ¤A⁄2
The quantities of onions purchased are
Month 1 = A⁄10 kg
Month 2 = A⁄20 kg
Month 3 = A⁄40 kg
Month 4 = A⁄80 kg
Month 5 = A⁄160 kg
Total expenditure A + A + A + A⁄2 + A⁄2 = 4A
Total quantity purchased A⁄10 + A⁄20 + A⁄40 + A⁄80 + A⁄160
= (16A + 8A + 4A + 2A + A)⁄160
= 31A⁄160 kg
∴ Average expense per kg = 4A ÷ 31A⁄160
= 640⁄31
≈ ¤20.65 per kg
Question 45: The arithmetic mean of the scores of 30 students in an examination is 48. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 25, then what is the maximum possible score of the toppers?
Total score of the 30 students 30 × 48 = 1440
Let the score of each topper be T.
There are 25 other students. To maximize the toppers’ score, the scores of these 25 students must be as small as possible.
Since their scores are distinct integers and the lowest score is 25, their scores can be
25, 26, 27, …, 49
Sum of these 25 scores
25 + 49) × 25 ÷ 2 = 925
∴ Total score of the five toppers 1440 − 925 = 515
Since all five toppers have the same score
T = 515 ÷ 5
= 103
∴ The maximum possible score of each topper is 103.
Let the score of each topper be T.
There are 25 other students. To maximize the toppers’ score, the scores of these 25 students must be as small as possible.
Since their scores are distinct integers and the lowest score is 25, their scores can be
25, 26, 27, …, 49
Sum of these 25 scores
25 + 49) × 25 ÷ 2 = 925
∴ Total score of the five toppers 1440 − 925 = 515
Since all five toppers have the same score
T = 515 ÷ 5
= 103
∴ The maximum possible score of each topper is 103.
Question 46: Dan is four times as old as Ben, and Carl is twice as old as Dan. If Dan’s age is 2 years less than the average age of all three, then what is Carl’s age, in years?
Let Ben’s age be x years.
Dan’s age = 4x years
Carl’s age = 8x years
Average age of the three
= (x + 4x + 8x) ÷ 3
= 13x ÷ 3
According to the question
4x = 13x ÷ 3 − 2
12x = 13x − 6
x = 6
∴ Carl’s age = 8 × 6
= 48 years
Dan’s age = 4x years
Carl’s age = 8x years
Average age of the three
= (x + 4x + 8x) ÷ 3
= 13x ÷ 3
According to the question
4x = 13x ÷ 3 − 2
12x = 13x − 6
x = 6
∴ Carl’s age = 8 × 6
= 48 years
Question 47: A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs, and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and his lowest score was x runs. What is the smallest possible value of x?
Total score in the first n innings = 30n
Total score in all n + 2 innings = 29(n + 2)
The last two innings contributed
= 38 + 15
= 53
∴ 30n + 53 = 29(n + 2)
or, 30n + 53 = 29n + 58
or, n = 5
So, the batsman played 5 innings in the first group.
Total score in these 5 innings = 5 × 30
= 150
Each of these scores is less than 38. To make the lowest score as small as possible, the other four scores should be as large as possible.
Since the scores are integers and each is less than 38, each of the other four scores can be at most 37.
∴ Maximum total of the other four scores = 4 × 37
= 148
So, the smallest possible score is x = 150 − 148 = 2
∴ The smallest possible value of x is 2 runs.
Total score in all n + 2 innings = 29(n + 2)
The last two innings contributed
= 38 + 15
= 53
∴ 30n + 53 = 29(n + 2)
or, 30n + 53 = 29n + 58
or, n = 5
So, the batsman played 5 innings in the first group.
Total score in these 5 innings = 5 × 30
= 150
Each of these scores is less than 38. To make the lowest score as small as possible, the other four scores should be as large as possible.
Since the scores are integers and each is less than 38, each of the other four scores can be at most 37.
∴ Maximum total of the other four scores = 4 × 37
= 148
So, the smallest possible score is x = 150 − 148 = 2
∴ The smallest possible value of x is 2 runs.
Question 48: A research institute enrolled 18 participants fewer in Program A than in Program B, and all participants completed the program. The total number of days taken by participants in Program A was 270 days, while the total for Program B was 252 days. If the average completion time for participants in Program A was 2 days more than the average completion time for participants in Program B, how many participants were enrolled in Program A?
Let the number of participants enrolled in Program A be n.
∴ Number of participants enrolled in Program B = n + 18
Average completion time in Program A = 270 ÷ n
Average completion time in Program B = 252 ÷ (n + 18)
According to the question
270 ÷ n = 252 ÷ (n + 18) + 2
Multiplying by n(n + 18)
270(n + 18) = 252n + 2n(n + 18)
or, 270n + 4860 = 252n + 2n² + 36n
or, 2n² + 18n − 4860 = 0
Dividing by 2
n² + 9n − 2430 = 0
or, (n + 54)(n − 45) = 0
Since the number of participants cannot be negative:
n = 45
Check:
Program A average = 270 ÷ 45 = 6 days
Program B has 63 participants 52 ÷ 63 = 4 days
Difference = 6 − 4 = 2 days
∴ Program A enrolled 45 participants.
∴ Number of participants enrolled in Program B = n + 18
Average completion time in Program A = 270 ÷ n
Average completion time in Program B = 252 ÷ (n + 18)
According to the question
270 ÷ n = 252 ÷ (n + 18) + 2
Multiplying by n(n + 18)
270(n + 18) = 252n + 2n(n + 18)
or, 270n + 4860 = 252n + 2n² + 36n
or, 2n² + 18n − 4860 = 0
Dividing by 2
n² + 9n − 2430 = 0
or, (n + 54)(n − 45) = 0
Since the number of participants cannot be negative:
n = 45
Check:
Program A average = 270 ÷ 45 = 6 days
Program B has 63 participants 52 ÷ 63 = 4 days
Difference = 6 − 4 = 2 days
∴ Program A enrolled 45 participants.
Question 49: The average weight of 24 athletes increases by 3 kg when two of them leave the group. If the average weight of these two athletes is one-fourth of the average weight of the original 24 athletes, what is the average weight, in kg, of the remaining 22 athletes?
Let the original average weight of the 24 athletes be A kg.
Total weight of the 24 athletes = 24A
Average weight of the two athletes who leave = A ÷ 4
Total weight of these two athletes = 2A ÷ 4
= A ÷ 2
The new average weight increases by 3 kg = A + 3
Total weight of the remaining 22 athletes = 22(A + 3)
The total weight can also be expressed as 24A − A ÷ 2
∴ 22(A + 3) = 24A − A ÷ 2
or, 22A + 66 = 47A ÷ 2
or, 44A + 132 = 47A
or, 3A = 132
or, A = 44
∴ Average weight of the remaining 22 athletes
= A + 3
= 47 kg.
Total weight of the 24 athletes = 24A
Average weight of the two athletes who leave = A ÷ 4
Total weight of these two athletes = 2A ÷ 4
= A ÷ 2
The new average weight increases by 3 kg = A + 3
Total weight of the remaining 22 athletes = 22(A + 3)
The total weight can also be expressed as 24A − A ÷ 2
∴ 22(A + 3) = 24A − A ÷ 2
or, 22A + 66 = 47A ÷ 2
or, 44A + 132 = 47A
or, 3A = 132
or, A = 44
∴ Average weight of the remaining 22 athletes
= A + 3
= 47 kg.
Question 50: A sports academy has 24 junior players and 36 senior players. In the first performance assessment, the average score of the senior players was 6 points higher than that of the junior players. In the second assessment, the average score of the senior players decreased by 3 points, while the average score of all players increased by 2 points. What was the average change in the score of the junior players?
The numbers of junior and senior players are in the ratio:
24 : 36 = 2 : 3
We can consider the groups as 2 parts and 3 parts.
The average score of all players increased by 2 points.
Total increase for the entire group = 5 × 2 = 10 parts
The average score of the senior players decreased by 3 points.
Total score change for the senior group = 3 × (−3) = −9 parts
The total score change for the entire group must equal the combined score changes of the two groups.
∴ 10 = −9 + Junior group change
Junior group change = 19 parts
Since the junior group represents 2 parts:
Average increase in the junior players’ scores:
= 19 ÷ 2
= 9.5 points
∴ The average score of the junior players increased by 9.5 points.
24 : 36 = 2 : 3
We can consider the groups as 2 parts and 3 parts.
The average score of all players increased by 2 points.
Total increase for the entire group = 5 × 2 = 10 parts
The average score of the senior players decreased by 3 points.
Total score change for the senior group = 3 × (−3) = −9 parts
The total score change for the entire group must equal the combined score changes of the two groups.
∴ 10 = −9 + Junior group change
Junior group change = 19 parts
Since the junior group represents 2 parts:
Average increase in the junior players’ scores:
= 19 ÷ 2
= 9.5 points
∴ The average score of the junior players increased by 9.5 points.
What are Averages?
Average, also known as the Arithmetic Mean, is a single value that represents the central or typical value of a group of numbers. It is calculated by adding all the given numbers and dividing the total by the number of observations.
Example: Find the average of 12, 18, 20, 25, and 30.
Sum of the numbers
= 12 + 18 + 20 + 25 + 30
= 105
Number of observations
= 5
Average
= 105 ÷ 5
= 21
Therefore, the average of 12, 18, 20, 25, and 30 is 21.
Example: Find the average of 12, 18, 20, 25, and 30.
Sum of the numbers
= 12 + 18 + 20 + 25 + 30
= 105
Number of observations
= 5
Average
= 105 ÷ 5
= 21
Therefore, the average of 12, 18, 20, 25, and 30 is 21.
Formula for Average
Arithmetic Mean – The sum of all observations divided by the number of observations.
Median – The middle value when the observations are arranged in ascending or descending order.
Mode – The value that occurs most frequently in a set of observations.
Median – The middle value when the observations are arranged in ascending or descending order.
Mode – The value that occurs most frequently in a set of observations.
Types of Average
Average = Sum of all observations ÷ Number of observations
Total = Average × Number of observations
Total = Average × Number of observations
Weighted Average
A Weighted Average is used when different groups contain different numbers of observations. In such cases, the overall average depends not only on the average of each group but also on the size of each group.
Effect of Adding or Removing an Observation
Adding or removing an observation changes the average depending on the value of that observation.
If a number greater than the current average is added, the average increases.
If a number less than the current average is added, the average decreases.
If a number equal to the current average is added, the average remains unchanged.
If a number greater than the current average is added, the average increases.
If a number less than the current average is added, the average decreases.
If a number equal to the current average is added, the average remains unchanged.
Average of Consecutive Numbers
The average of consecutive numbers follows a simple pattern.
For an odd number of consecutive numbers, the average is always the middle number.
For an even number of consecutive numbers, the average is the mean of the two middle numbers.
Examples:
Average of 11, 12, 13, 14, 15
= 13
Average of 21, 22, 23, 24
= (22 + 23) ÷ 2
= 22.5
For an odd number of consecutive numbers, the average is always the middle number.
For an even number of consecutive numbers, the average is the mean of the two middle numbers.
Examples:
Average of 11, 12, 13, 14, 15
= 13
Average of 21, 22, 23, 24
= (22 + 23) ÷ 2
= 22.5
Summary of Average
| Number | Divisibility Rule |
|---|---|
| 2 | Last digit is 0, 2, 4, 6 or 8. |
| 3 | Sum of the digits is divisible by 3. |
| 4 | The number formed by the last two digits is divisible by 4. |
| 5 | Last digit is 0 or 5. |
| 6 | The number is divisible by both 2 and 3. |
| 7 | Double the last digit and subtract it from the remaining number. Repeat until a two-digit number is obtained. |
| 8 | The number formed by the last three digits is divisible by 8. |
| 9 | Sum of the digits is divisible by 9. |
| 10 | Last digit is 0. |
| 11 | The difference between the sums of alternate digits is 0 or a multiple of 11. |
| 12 | The number is divisible by both 3 and 4. |
Common Mistakes
- Omitting Zero Values:
Students often exclude 0 from the count of items while calculating the average. Zero is also a valid observation and must be included in the denominator.
Formula:
Average = Total Sum ÷ Total Number of Items - Averaging the Speeds:
For a round trip with equal distances at speeds x and y, students often use (x + y) ÷ 2, which is incorrect.
Formula:
Average Speed = (2 × x × y) ÷ (x + y) - Incorrect Adjustments:
When an item is added, removed, or replaced, students often apply the change directly to the average instead of first finding the total sum.
Formula:
New Sum = (Old Average × Old Count) ± Change in Value - Ignoring Group Sizes:
Students often take the simple average of two group averages without considering the number of members in each group. When group sizes are different, a weighted average should be used.
Formula:
Weighted Average = Total Sum of All Groups ÷ Total Number of Items - Age and Time Shifts:
In age-based average problems, students often subtract or add years incorrectly. If the age of multiple people changes, the change must be applied to each person separately. For example, if the age of 5 people is considered 3 years ago, the total age decreases by 5 × 3 = 15 years.
Practice Questions
Question 1: The average age of 8 members of a committee is 42 years. When the age of the chairman is included, the average age increases by 3 years. If the chairman’s age is 12 years more than the oldest member among the original 8 members, what is the age of the chairman?
Question 2: The average marks of 30 students in a class is 68. If the marks of one student were incorrectly entered as 86 instead of 56, what would have been the actual average marks of the class?
Question 3: The average weight of 6 persons increases by 4 kg when a person weighing 72 kg is replaced by another person. What is the weight of the new person?
Question 4: The average monthly income of A, B and C is ₹18,000. The average income of B and C is ₹15,000 and the average income of A and B is ₹20,000. Find the monthly income of B.
Question 5: The average weight of 10 students is 45 kg. When the weight of the teacher is included, the average increases by 2 kg. If the teacher’s weight is 20 kg more than the heaviest student, find the teacher’s weight.
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