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RS Aggarwal Solutions for Class 9 Chapter 18: Mean, Median and Mode of Ungrouped Data Download PDF
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Exercise 18(A)
1. Find the mean of:
(i) the first eight natural numbers
(ii) the first ten odd numbers
(iii) the first seven multiples of 5
(iv) all the factors of 20
(v) all prime numbers between 50 and 80.
Solution:
(i) We know that
First eight natural numbers = 1, 2, 3, 4, 5, 6, 7 and 8
So we get
Mean = sum of numbers/ total numbers
By substituting the values
Mean = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)/8
On further calculation
Mean = 36/8
By division
Mean = 4.5
Therefore, the mean of the first eight natural numbers is 4.5.
(ii) We know that
First ten odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19
So we get
Mean = sum of numbers/ total numbers
By substituting the values
Mean = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/10
On further calculation
Mean = 100/10
By division
Mean = 10
Therefore, the mean of first ten odd numbers is 10.
(iii) We know that
First seven multiples of five = 5, 10, 15, 20, 25, 30 and 35
So we get
Mean = sum of numbers/ total numbers
By substituting the values
Mean = (5 + 10 + 15 + 20 + 25 + 30 + 35)/7
On further calculation
Mean = 140/7
By division
Mean = 20
Therefore, the mean of first seven multiples of five is 20.
(iv) We know that
All the factors of 20 = 1, 2, 4, 5, 10 and 20
So we get
Mean = sum of numbers/ total numbers
By substituting the values
Mean = (1 + 2 + 4 + 5 + 10 + 20)/6
On further calculation
Mean = 42/6
By division
Mean = 7
Therefore, the mean of all the factors of 20 is 7.
(v) We know that
All prime numbers between 50 and 80 = 53, 59, 61, 67, 71, 73 and 79
So we get
Mean = sum of numbers/ total numbers
By substituting the values
Mean = (53 + 59 + 61 + 67 + 71 + 73 + 79)/7
On further calculation
Mean = 463/7
So we get
Mean = 66 1/7
Therefore, the mean of all prime numbers between 50 and 80 is 66 1/7.
2. The number of children in 10 families of a locality are 2, 4, 3, 4, 2, 0, 3, 5, 1 and 6.
Find the mean number of children per family.
Solution:
It is given that number of children in 10 families of a locality are 2, 4, 3, 4, 2, 0, 3, 5, 1 and 6.
We know that
Mean = sum of numbers/ total numbers
By substituting the values
Mean = (2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6)/ 10
On further calculation
Mean = 30/10
By division
Mean = 3
Therefore, the mean number of children per family is 3.
3. The following are the numbers of books issued in a school library during a week:
105, 216, 322, 167, 273, 405 and 346.
Find the average number of books issued per day.
Solution:
It is given that the number of books issued are 105, 216, 322, 167, 273, 405 and 346.
We know that
Mean = sum of numbers/ total numbers
By substituting the values
Mean = (105 + 216 + 322 + 167 + 273 + 405 + 346)/ 7
On further calculation
Mean = 1834/7
By division
Mean = 262
Therefore, the average number of books issued per day is 262.
4. The daily minimum temperature recorded (in ^{o}F) at a place during six days of a week was as under:
Monday 
35.5 
Tuesday 
30.8 
Wednesday 
27.3 
Thursday 
32.1 
Friday 
23.8 
Saturday 
29.9 
Find the mean temperature.
Solution:
We know that
Mean temperature = Sum of temperatures/ Number of days
By substituting the values
Mean temperature = (35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9)/6
On further calculation
Mean = 179.4/6
By division
Mean = 29.9^{ o}F
Therefore, the mean temperature is 29.9^{ o}F.
5. If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 13, find the value of x and hence find the mean of the last three observations.
Solution:
We know that
Number of observations = 5
It is given that mean = 13
We can write it as
[x + (x + 2) + (x + 4) + (x + 6) + (x + 8)]/5 = 13On further calculation
5x + 20 = 13 (5)
So we get
5x + 20 = 65
By subtraction
5x = 45
By division
x = 9
By substituting the value of x
We know that the last three observations are
9 + 4 = 13
9 + 6 = 15
9 + 8 = 17
We know that
Mean of last three observations = (13 + 15 + 17)/3
On further calculation
Mean of last three observations = 45/3
By division
Mean of last three observations = 15
Therefore, the mean of last three observations is 15.
6. The mean weight of 6 boys in a group is 48kg. The individual weights of five of them are 51kg, 45kg, 49kg, 46kg and 44kg. Find the weight of the sixth boy.
Solution:
It is given that
Mean weight of 6 boys = 48kg
So we get
Mean weight = sum of the weight of 6 boys/6 = 48
We know that
Sum of weight of six boys = 48 (6) = 288kg
So the sum of weight of 5 boys = (51 + 45 + 49 + 46 + 44) = 235kg
We get
Weight of sixth boy = sum of weight of six boys – sum of weight of 5 boys
By substituting the values
Weight of sixth boy = 288 – 235
By subtraction
Weight of sixth boy = 53kg
Therefore, the weight of sixth boy is 53 kg.
7. The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. Find the correct mean.
Solution:
It is given that
Mean of the marks of 50 students = 39
So we get
Sum of these marks = 39 (50) = 1950
We can write it as
Corrected sum of the marks = 1950 – wrong number + correct number
By substituting the values
Corrected sum of the marks = 1950 – 23 + 43
So we get
Corrected sum of the marks = 1970
So the correct mean = 1970/50
By division
Correct mean = 39.4
Therefore, the correct mean is 39.4.
8. The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean?
Solution:
Consider x_{1}, x_{2}, …….. x_{24} as the given numbers
So we get
Mean = (x_{1} + x_{2}, …….. + x_{24})/24
It is given that mean = 35
(x_{1} + x_{2}, …….. + x_{24})/24 = 35
By cross multiplication
x_{1} + x_{2}, …….. + x_{24} = 840 ……. (1)
Take (x_{1} + 3), (x_{2} + 3), ……… (x_{24} + 3) as new numbers
So the mean of new numbers = [(x_{1} + 3) + (x_{2} + 3) + ……… + (x_{24} + 3)]/24
From equation (1) we get
[(x_{1} + 3) + (x_{2} + 3) + ……… + (x_{24} + 3)]/24 = (840 + 72)/24On further calculation
[(x_{1} + 3) + (x_{2} + 3) + ……… + (x_{24} + 3)]/24 = 912/24By division
Mean of new numbers = 38
Therefore, the mean of the new numbers is 38.
9. The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will be the new mean?
Solution:
Consider x_{1}, x_{2}, ……. x_{20} as the given numbers
So we get
Mean = (x_{1} + x_{2} + ……. + x_{20})/20
It is given that mean = 43
(x_{1} + x_{2} + ……. + x_{20})/20 = 43
By cross multiplication
x_{1} + x_{2} + ……. + x_{20} = 860 …… (1)
Take (x_{1} – 6), (x_{2} – 6) …… (x_{20} – 6) as the new numbers
So the mean of new numbers = [(x_{1} – 6) + (x_{2} – 6) + …… + (x_{20} – 6)]/20
From equation (1) we get
[(x_{1} – 6) + (x_{2} – 6) + …… + (x_{20} – 6)]/20 = (860 – 120)/20On further calculation
[(x_{1} – 6) + (x_{2} – 6) + …… + (x_{20} – 6)]/20 = 740/20By division
Mean of new numbers = 37
Therefore, the new mean of numbers is 37.
10. The mean of 15 numbers is 27. If each number is multiplied by 4, what will be the mean of the new numbers?
Solution:
Consider x_{1}, x_{2}, ……. x_{15} as the given numbers
So we get
Mean = (x_{1} + x_{2} + ……. + x_{15})/15
It is given that mean = 27
(x_{1} + x_{2} + ……. + x_{15})/15 = 27
By cross multiplication
x_{1} + x_{2} + ……. + x_{15} = 405 …… (1)
Take (x_{1} × 4), (x_{2} × 4), ……… (x_{15} × 4) as new numbers
So the mean of new numbers = [(x_{1} × 4) + (x_{2} × 4) + ……… + (x_{15} × 4)]/15
From equation (1) we get
[(x_{1} × 4) + (x_{2} × 4) + ……… + (x_{15} × 4)]/15 = (405 × 4)/ 15On further calculation
Mean of new numbers = 1620/15 = 108
Therefore, the mean of new numbers is 108.
11. The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?
Solution:
Consider x_{1}, x_{2}, ……. x_{12} as the given numbers
So we get
Mean = [x_{1} + x_{2} + ……. + x_{12}]/12
It is given that mean = 40
(x_{1} + x_{2} + ……. + x_{12})/12 = 40
By cross multiplication
x_{1} + x_{2} + ……. + x_{12} = 480 …… (1)
Take (x_{1} ÷ 8), (x_{2} ÷ 8), ……… (x_{12} ÷ 8) as new numbers
So the mean of new numbers = [(x_{1} ÷ 8) + (x_{2} ÷ 8) + ……… + (x_{12} ÷ 8)]/12
From equation (1) we get
[(x_{1} ÷ 8) + (x_{2} ÷ 8) + ……… + (x_{12} ÷ 8)]/12 = (480 ÷ 8)/12On further calculation
Mean of new numbers = 60/12 = 5
Therefore, the mean of the new numbers is 5.
12. The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of the new set of 20 numbers.
Solution:
Consider x_{1}, x_{2}, ……. x_{20} as the given numbers
So we get
Mean = (x_{1} + x_{2} + ……. + x_{20})/20
It is given that mean = 18
(x_{1} + x_{2}, ……. + x_{20})/20 = 18
By cross multiplication
x_{1} + x_{2} + ……. + x_{20} = 360 …… (1)
Take (x_{1} + 3), (x_{2} + 3), ……… (x_{10} + 3) as first ten new numbers
So the mean of new set of 20 numbers = [(x_{1} + 3) + (x_{2} + 3) + ……… + (x_{10} + 3) + x_{11} + …….. + x_{20}]/ 20
We get
Mean of new set of 20 numbers = [(x_{1} + x_{2} + ……. + x_{20}) + 3 × 10]/ 20
From equation (1) we get
Mean of new set of 20 numbers = (360 + 30)/ 20 = 19.5
Therefore, the mean of new set of 20 numbers is 19.5.
13. The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.
Solution:
It is given that
Mean of six numbers = 23
So we get the sum of six numbers = 23 (6) = 138
It is given that
Mean of five numbers = 20
So we get the sum of five numbers = 20 (5) = 100
We know that
Excluded number = sum of six numbers – sum of five numbers
By substituting the values
Excluded number = 138 – 100 = 38
Therefore, the excluded number is 38.
14. The average height of 30 boys was calculated to be 150cm. It was detected later that one value of 165cm was wrongly copied as 135cm for the computation of the mean. Find the correct mean.
Solution:
It is given that
Mean height of 30 boys = 150cm
So the total height = 150 (30) = 4500cm
We know that
Correct sum = 4500 – incorrect value + correct value
By substituting the values
Correct sum = 4500 – 135 + 165 = 4350
So we get
Correct mean = Correct sum/30
By substituting the values
Correct mean = 4530/30 = 151 cm
Therefore, the correct mean is 151 cm.
15. The mean weight of a class of 34 students is 46.5kg. If the weight of the teacher is included, the mean rises by 500g. Find the weight of the teacher.
Solution:
It is given that
Mean weight of 34 students = 46.5kg
So the total weight = 34 (46.5) = 1581kg
We know that
Mean weight of 34 students and teacher = 46.5 + 0.5 = 47kg
So the total weight of 34 students and teacher = 47 (35) = 1645kg
The weight of teacher = Weight of 34 students and teacher – total weight
By substituting the values
Weight of the teacher = 1645 – 1581 = 64kg
Therefore, the weight of the teacher is 64kg.
16. The mean weight of a class of 36 students is 41kg. If one of the students leaves the class then the mean is decreased by 200g. Find the weight of the student who left.
Solution:
It is given that
Mean weight of 36 students = 41kg
So the total weight = 41 (36) = 1476kg
It is given that if one of the students leaves the class then the mean is decreased by 200g
So the new mean = 41 – 0.2 = 40.8kg
We get the total weight of 35 students = 40.8 (35) = 1428kg
So the weight of student who left = total weight of 36 students – weight of 35 students
By substituting the values
Weight of the student who left the class = 1476 – 1428 = 48kg
Therefore, the weight of the student who left is 48kg.
17. The average weight of a class of 39 students is 40kg. When a new student is admitted to the class, the average decreases by 200g. Find the weight of the new student.
Solution:
It is given that
Mean weight of 39 students = 40kg
So the total weight = 40 (39) = 1560kg
It is given that when a new student is admitted to the class, the average decreases by 200g
So the new mean = 40 – 0.2 = 39.8kg
We know that
Total weight of 40 students = 39.8 (40) = 1592kg
So the weight of new student = total weight of 40 students – total weight of 39 students
By substituting the values
Weight of new student = 1592 – 1560 = 32kg
Therefore, the weight of the new student is 32kg.
18. The average weight of 10 oarsmen in a boat is increased by 1.5kg when one of the crew who weighs 58kg is replaced by a new man. Find the weight of the new man.
Solution:
It is given that
Weight of 10 oarsmen is increased by 1.5kg
So the total weight increased = 1.5 (10) = 15kg
It is given that one of the crew who weights 58kg is replaced by a new man
So we get the weight of new man = weight of man replaced + total weight increased
By substituting the values
Weight of the new man = 58 + 15 = 73kg
Therefore, the weight of a new man is 73kg.
19. The mean of 8 numbers is 35. If a number is excluded then the mean is reduced by 3. Find the excluded number.
Solution:
It is given that
Mean of 8 numbers = 35
So the total sum = 35 (8) = 280
It is given that if a number is excluded then the mean is reduced by 3
So the new mean = 35 – 3 = 32
We know that total sum of 7 numbers = 32 (7) = 224
So we get
Excluded number = sum of 8 numbers – sum of 7 numbers
By substituting the values
Excluded number = 280 – 224 = 56
Therefore, the excluded number is 56.
20. The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread are 52 and 8 instead of 152 and 88 respectively. Find the correct mean.
Solution:
It is given that
Mean of 150 items = 60
So the total sum = 150 (60) = 9000
We know that
Correct sum of items = sum of 150 items – sum of wrong items + sum of right items
By substituting the values
Correct sum of items = 90000 – (52 + 8) + (152 + 88)
On further calculation
Correct sum of items = 9000 – 60 + 240
So we get
Correct sum of items = 9180
So the correct mean = correct sum of items/ total number of items
By substituting the values
Correct mean = 9180/150 = 61.2
Therefore, the correct mean is 61.2.
21. The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16^{th} result.
Solution:
It is given that
Mean of 31 results = 60
So the total sum = 31 (60) = 1860
We know that
Mean of first 16 results = 58
So the total sum = 16 (58) = 928
Mean of last 16 results = 62
So the total sum = 16 (62) = 992
We get the 16^{th} result = total sum of first 16 results + total sum of last 16 results – total sum of 31 results
By substituting the values
16^{th} result = 928 + 992 – 1860
On further calculation
16^{th} result = 1920 – 1860 = 60
Therefore, the 16^{th} result is 60.
22. The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46, find the 6^{th} number.
Solution:
It is given that
Mean of 11 numbers = 42
So the total sum = 42 (11) = 462
Mean of first 6 numbers = 37
So the total sum = 37 (6) = 222
Mean of last 6 numbers = 46
So the total sum = 6 (46) = 276
We know that
6^{th} number = total sum of last 6 numbers + total sum of first 6 numbers – total sum of 11 numbers
By substituting the values
6^{th} number = 276 + 222 – 462
So we get
6^{th} number = 498 – 462 = 36
Therefore, the 6^{th} number is 36.
23. The mean weight of 25 students of a class is 52kg. If the mean weight of the first 13 students of the class is 48kg and that of the last 13 students is 55kg, find the weight of the 13^{th} student.
Solution:
It is given that
Mean weight of 25 students = 52kg
So the total weight = 25 (52) = 1300kg
Mean weight of first 13 students = 48kg
So the total weight = 13 (48) = 624kg
Mean weight of last 13 students = 55kg
So the total weight = 13 (55) = 715kg
We know that
Weight of 13^{th} student = total weight of the first 13 students + total weight of last 13 students – total weight of 25 students
By substituting the values
Weight of 13^{th} student = 624 + 715 – 1300
On further calculation
Weight of 13^{th} student = 39kg
Therefore, the weight of the 13^{th} student is 39kg.
24. The mean score of 25 observations is 80 and the mean score of another 55 observations is 60. Determine the mean score of the whole set of observations.
Solution:
It is given that
Mean score of 25 observations = 80
So the total score = 25 (80) = 2000
Mean score of 55 observations = 60
So the total score = 55 (60) = 3300
We know that
Total number of observations = 25 + 55 = 80
So the total score = 2000 + 3300 = 5300
Mean score = total score/total number of observations
By substituting the values
Mean score = 5300/80
So we get
Mean score = 66.25
Therefore, the mean score of the whole set of observations is 66.25.
25. Arun scored 36 marks in English, 44 marks in Hindi, 74 marks in mathematics and x marks in science. If he has secured an average of 50 marks, find the value of x.
Solution:
It is given that
Average marks in 4 subjects = 50
So the total marks = 50 (4) = 200
We know that
36 + 44 + 75 + x = 200
On further calculation
155 + x = 200
So we get
x = 200 – 155
By subtraction
x = 45
Therefore, the value of x is 45.
26. A ship sails out to an island at the rate of 15 km/hr and sails back to the starting point at 10 km/hr. Find the average sailing speed for the whole journey.
Solution:
Consider x km as the distance of mark from the starting point
So the time taken by the ship to reach the mark = x/15 hours
We know that
The time taken by the ship to reach the starting point from the mark = x/10 hours
So we get
Total time taken = x/15 + x/10
By taking LCM
Total time taken = x/6 hours
We know that total distance covered = x + x = 2x km
So the average sailing speed for the whole journey = 2x ÷ x/6
It can be written as
Average sailing speed for the whole journey = (2x × 6)/ x
So we get
Average sailing speed for the whole journey = 12 km/hr
Therefore, the average sailing speed for the whole journey is 12 km/hr.
27. There are 50 students in a class, of which 40 are boys. The average weight of the class is 44kg and that of the girls is 40kg. Find the average weight of the boys.
Solution:
It is given that
Total number of students = 50
So total number of girls = 50 – 40 = 10
Average weight of the class = 44kg
So the total weight = 44 (50) = 2200kg
Average weight of 10 girls = 40kg
So total weight = 40 (10) = 400kg
We know that total weight of 40 boys = total weight of 50 students – total weight of 10 girls
By substituting the values
Total weight of 40 boys = 2200 – 400 = 1800kg
So we get
Average weight of boys = total weight of 40 boys/ number of boys
By substituting the values
Average weight of boys = 1800/40 = 45kg
Therefore, the average weight of boys = 45kg.
28. The aggregate monthly expenditure of a family was ₹ 18720 during the first 3 months, ₹ 20340 during the next 4 months and ₹ 21708 during the last 5 months of a year. If the total savings during the year be ₹ 35340 find the average monthly income of the family.
Solution:
We know that
Total earnings of the year = ₹ (3 × 18720 + 4 × 20340 + 5 × 21708 + 35340)
On further calculation
Total earnings of the year = ₹ (56160 + 81360 + 108540 + 35340)
So we get
Total earnings of the year = ₹ 281400
We know that
Average monthly income of the family = Total earning of the year/ Number of months
By substituting the values
Average monthly income of the family = 281400/ 12
So we get
Average monthly income of the family = ₹ 23450
Therefore, the average monthly income of the family is ₹ 23450.
29. The average weekly payment to 75 workers in a factory is ₹ 5680. The mean weekly payment to 25 of them is ₹ 5400 and that of 30 others is ₹ 5700. Find the mean weekly payment of the remaining workers.
Solution:
It is given that
Average weekly payment of 75 workers = ₹ 5680
So the total payment = ₹ (75 × 5680) = ₹ 426000
Mean weekly payment of 25 workers = ₹ 5400
So the total payment = ₹ (25 × 5400) = ₹ 135000
Mean weekly payment of 30 workers = ₹ 5700
So the total payment = ₹ (30 × 5700) = ₹ 171000
We know that number of remaining workers = 75 – 25 – 30 = 20
So the total weekly payment of remaining 20 workers = Total weekly payment of 75 workers – total weekly payment of 25 workers – total weekly payment of 30 workers
By substituting the values
Total weekly payment of remaining 2 workers = 426000 – 135000 – 171000 = ₹ 120000
So the mean weekly payment of remaining 20 workers = 120000/20 = ₹ 6000
Therefore, the mean weekly payment of the remaining workers is ₹ 6000.
30. The mean marks (out of 100) of boys and girls in an examination are 70 and 73 respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.
Solution:
Consider x: 1 as the ratio of number of boys to the number of girls
So we get
Sum of marks of boys = 70x
Sum of marks of girls = 73 (1) = 73
We know that
Sum of marks of boys and girls = 71 (x + 1)
It can be written as
70x + 73 = 71 (x + 1)
On further calculation
7x + 73 = 71x + 71
So we get
x = 2
Therefore, the ratio of the number of boys to the number of girls is 2: 1.
31. The average monthly salary of 20 workers in an office is ₹ 45900. If the manager’s salary is added, the average salary becomes ₹ 49200 per month. What’s manager’s monthly salary?
Solution:
It is given that
Mean monthly salary of 20 workers = ₹ 45900
So the total monthly salary = ₹ (20 × 49500) = ₹ 918000
Mean monthly salary of 20 workers and manager = ₹ 49200
So the total monthly salary = ₹ (21 × 49200) = ₹ 1033200
We know that
Manager’s monthly salary = Total monthly salary of 20 workers and manager – Total monthly salary of 20 workers
By substituting the values
Manager’s monthly salary = 1033200 – 918000 = ₹ 115200
Therefore, the manager’s monthly salary is ₹ 115200.
Exercise 18(B)
1. Obtain the mean of the following distribution:
Variable (x_{i}) 
4 
6 
8 
10 
12 
Frequency (f_{i}) 
4 
8 
14 
11 
3 
Solution:
Variable (x_{i}) 
Frequency (f_{i}) 
f_{i} x_{i} 
4 
4 
16 
6 
8 
48 
8 
14 
112 
10 
11 
110 
12 
3 
36 

Σ f_{i} = 40 
Σ f_{i} x_{i} = 322 
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
By substituting the values
Mean = 322/40 = 8.05
2. The following table shows the weights of 12 workers in a factory:
Weight (in kg) 
60 
63 
66 
69 
72 
No. of workers 
4 
3 
2 
2 
1 
Find the mean weight of the workers.
Solution:
Weight (in kg) (x_{i}) 
No. of workers (f_{i}) 
f_{i} x_{i} 
60 
4 
240 
63 
3 
189 
66 
2 
132 
69 
2 
138 
72 
1 
72 

Σ f_{i} = 12 
Σ f_{i} x_{i} =771 
We know that
Mean weight of the workers = Σ f_{i} x_{i}/ Σ f_{i}
By substituting the values
Mean weight of the workers = 771/12 = 64.250kg
Therefore, the mean weight of the workers is 64.250kg.
3. The measurements (in mm) of the diameters of the heads of 50 screws are given below:
Diameter (in mm) (x_{i}) 
34 
37 
40 
43 
46 
Number of screws (f_{i}) 
5 
10 
17 
12 
6 
Calculate the mean diameter of the heads of the screws.
Solution:
Diameter (in mm) (x_{i}) 
Number of screws (f_{i}) 
f_{i} x_{i} 
34 
5 
170 
37 
10 
370 
40 
17 
680 
43 
12 
516 
46 
6 
276 

Σ f_{i} = 50 
Σ f_{i} x_{i} = 2012 
We know that
Mean diameter of the heads of the screws = Σ f_{i} x_{i}/ Σ f_{i}
By substituting the values
Mean diameter of the heads of the screws = 2012/50 = 40.24mm
Therefore, the mean diameter of the heads of the screws is 40.24mm.
4. The following data give the number of boys of a particular age in a class of 40 students.
Age (in years) 
15 
16 
17 
18 
19 
20 
Frequency (f_{i}) 
3 
8 
9 
11 
6 
3 
Calculate the mean age of the students.
Solution:
Age (in years) (x_{i}) 
Frequency (f_{i}) 
f_{i} x_{i} 
15 
3 
45 
16 
8 
128 
17 
9 
153 
18 
11 
198 
19 
6 
114 
20 
3 
60 

Σ f_{i} = 40 
Σ f_{i} x_{i} = 698 
We know that
Mean age of the students = Σ f_{i} x_{i}/ Σ f_{i}
By substituting the values
Mean age of the students = 698/40 = 17.45 years
Therefore, the mean age of the students is 17.45 years.
5. Find the mean of the following frequency distribution:
Variables (x_{i}) 
10 
30 
50 
70 
89 
Frequency (f_{i}) 
7 
8 
10 
15 
10 
Solution:
Variables (x_{i}) 
Frequency (f_{i}) 
f_{i} x_{i} 
10 
7 
70 
30 
8 
240 
50 
10 
500 
70 
15 
1050 
89 
10 
890 

Σ f_{i} = 50 
Σ f_{i} x_{i} = 2750 
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
By substituting the values
Mean = 2750/50 = 55
6. Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages (in ₹ ) (x_{i}) 
250 
300 
350 
400 
450 
Number of workers (f_{i}) 
8 
11 
6 
10 
5 
Solution:
Daily wages (in ₹ ) (x_{i}) 
Number of workers (f_{i}) 
f_{i} x_{i} 
250 
8 
2000 
300 
11 
3300 
350 
6 
2100 
400 
10 
4000 
450 
5 
2250 

Σ f_{i} = 40 
Σ f_{i} x_{i} = 13650 
We know that
Mean of daily wages of 40 workers in a factory = Σ f_{i} x_{i}/ Σ f_{i}
By substituting the values
Mean of daily wages of 40 workers in a factory = 13650/40 = ₹ 341.25
Therefore, the mean of daily wages of 40 workers in a factory is ₹ 341.25.
7. If the mean of the following data is 20.2, find the value of p.
Variable (x_{i}) 
10 
15 
20 
25 
30 
Frequency (f_{i}) 
6 
8 
p 
10 
6 
Solution:
Variable (x_{i}) 
Frequency (f_{i}) 
f_{i} x_{i} 
10 
6 
60 
15 
8 
120 
20 
p 
20p 
25 
10 
250 
30 
6 
180 

Σ f_{i} = 30 + p 
Σ f_{i} x_{i} = 610 + 20p 
It is given that
Mean = 20.2
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
So we get
(610 + 20p)/ (30 + p) = 20.2
On further calculation
610 + 20p = 606 + 20.2p
So we get
0.2p = 4
By division
p = 20
Therefore, the value of p is 20.
8. If the mean of the following data is 8, find the value of p.
x 
3 
5 
7 
9 
11 
13 
f 
6 
8 
15 
p 
8 
4 
Solution:
x 
f 
f_{i} x_{i} 
3 
6 
18 
5 
8 
40 
7 
15 
105 
9 
p 
9p 
11 
8 
88 
13 
4 
52 

Σ f_{i} = 41 + p 
Σ f_{i} x_{i} = 303 + 9p 
It is given that
Mean = 8
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
So we get
(303 + 9p)/ (41 + p) = 8
On further calculation
303 + 9p = 8 (41 + p)
So we get
303 + 9p = 328 + 8p
It can be written as
9p – 8p = 328 – 303
We get
p = 25
Therefore, the value of p is 25.
9. Find the missing frequency p for the following frequency distribution whose mean is 28.25.
x 
15 
20 
25 
30 
35 
40 
f 
8 
7 
p 
14 
15 
6 
Solution:
x 
f 
f_{i} x_{i} 
15 
8 
120 
20 
7 
140 
25 
p 
25p 
30 
14 
420 
35 
15 
525 
40 
6 
240 

Σ f_{i} = 50 + p 
Σ f_{i} x_{i} = 1445 + 25p 
It is given that
Mean = 28.25
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
So we get
(1445 + 25p)/ (50 + p) = 28.25
On further calculation
1445 + 25p = 1412.50 + 28.25p
So we get
28.25p + 25p = 1445 + 1412.50
It can be written as
3.25p = 32.5
By division
p = 10
Therefore, the value of p is 10.
10. Find the value of p for the following frequency distribution whose mean is 16.6.
x 
8 
12 
15 
p 
20 
25 
30 
f 
12 
16 
20 
24 
16 
8 
4 
Solution:
x 
f 
f_{i} x_{i} 
8 
12 
96 
12 
16 
192 
15 
20 
300 
p 
24 
24p 
20 
16 
320 
25 
8 
200 
30 
4 
120 

Σ f_{i} = 100 
Σ f_{i} x_{i} = 1228 + 24p 
It is given that
Mean = 16.6
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
So we get
(1228 + 24p)/ 100 = 16.6
On further calculation
1228 + 24p = 1660
So we get
24p = 1660 – 1228
By subtraction
24p = 432
We get
p = 18
Therefore, the value of p is 18.
11. Find the missing frequencies in the following frequency distribution whose mean is 34.
x 
10 
20 
30 
40 
50 
6 
Total 
f 
4 
f_{1} 
8 
f_{2} 
3 
4 
35 
Solution:
x 
f 
f_{i} x_{i} 
10 
4 
40 
20 
f_{1} 
20f_{1} 
30 
8 
240 
40 
f_{2} 
40f_{2} 
50 
3 
150 
6 
4 
240 
Total 
35 = Σ f_{i} =19 + f_{1} + f_{2} 
Σ f_{i} x_{i} = 670 + 20f_{1} + 40f_{2} 
We know that
Σ f_{i} =19 + f_{1} + f_{2}
It can be written as
35 = 19 + f_{1} + f_{2}
So we get
f_{1} + f_{2} = 16 …… (1)
It is given that
Mean = 34
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
So we get
(670 + 20f_{1} + 40f_{2})/ 35 = 34
On further calculation
(335 + 10f_{1} + 20f_{2})/ 35 = 17
So we get
335 + 10f_{1} + 20f_{2} = 595
It can be written as
10f_{1} + 20f_{2} = 260
Dividing the equation by 10
f_{1} + 2f_{2} = 26 …….. (2)
By subtracting equation (1) from (2)
f_{2} = 10
Substituting in equation (1)
f_{1} + 10 = 16
So we get
f_{1} = 6
Therefore, the missing frequencies are 6 and 10.
12. Find the missing frequencies in the following frequency distribution whose mean is 50.
x 
10 
30 
50 
70 
90 
Total 
f 
17 
f_{1} 
32 
f_{2} 
19 
120 
Solution:
x 
f 
f_{i} x_{i} 
10 
17 
170 
30 
f_{1} 
30f_{1} 
50 
32 
1600 
70 
f_{2} 
70f_{2} 
90 
19 
1710 
Total 
120 = Σ f_{i} =68 + f_{1} + f_{2} 
Σ f_{i} x_{i} = 3480 + 30f_{1} + 70f_{2} 
We know that
Σ f_{i} =68 + f_{1} + f_{2}
It can be written as
120 = 68 + f_{1} + f_{2}
So we get
f_{1} + f_{2} = 52
f_{2} = 52 – f_{1} ……. (1)
It is given that
Mean = 50
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
So we get
(3480 + 30f_{1} + 70f_{2})/ 120 = 50
Substituting equation (1)
3480 + 30f_{1} + 70(52 – f_{1}) / 120 = 50
On further calculation
(3480 + 30f_{1} + 3640 – 70f_{1})/ 120 = 50
So we get
(7120 – 40f_{1})/ 120 = 50
It can be written as
6000 = 7120 – 40f_{1}
By subtraction
40f_{1} = 1120
By division
f_{1} = 28
By substituting it in equation (1)
f_{2} = 52 – 28 = 24
Therefore, the missing frequencies are 28 and 24.
13. Find the value of p, when the mean of the following distribution is 20.
x 
15 
17 
19 
20 + p 
23 
f 
2 
3 
4 
5p 
6 
Solution:
x 
f 
f_{i} x_{i} 
15 
2 
30 
17 
3 
51 
19 
4 
76 
20 + p 
5p 
100p + 5p^{2} 
23 
6 
138 

Σ f_{i} = 15 + 5p 
Σ f_{i} x_{i} = 5p^{2} + 100p + 295 
It is given that
Mean = 20
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
So we get
(5p^{2} + 100p + 295)/ (15 + 5p) = 20
On further calculation
5p^{2} + 100p + 295 = 300 + 100p
So we get
5p^{2} = 5
By division
p^{2} = 1
By taking square root
p = 1
Therefore, the value of p is 1.
14. The mean of the following distribution is 50.
x 
10 
30 
50 
70 
90 
f 
17 
5a + 3 
32 
7a – 11 
19 
Find the value of a and hence the frequencies of 30 and 70.
Solution:
x 
f 
f_{i} x_{i} 
10 
17 
170 
30 
5a + 3 
150a + 90 
50 
32 
1600 
70 
7a – 11 
490a – 770 
90 
19 
171 

Σ f_{i} = 60 + 12a 
Σ f_{i} x_{i} = 2800 + 640a 
It is given that
Mean = 50
We know that
Mean = Σ f_{i} x_{i}/ Σ f_{i}
So we get
(2800 + 640a)/ (60 + 12a) = 50
On further calculation
2800 + 640a = 3000 + 600a
So we get
40a = 200
By division
a = 5
We know that
Frequency of 30 = 5a + 3
By substituting the values
Frequency of 30 = 5 (5) + 3
So we get
Frequency of 30 = 28
We know that
Frequency of 70 = 7a – 11
By substituting the values
Frequency of 70 = 7 (5) – 11
So we get
Frequency of 70 = 24
Therefore, the value of a is 5 and the frequencies of 30 and 70 is 28 and 24.
Exercise 18(C)
1. Find the median of:
(i) 2, 10, 9, 9, 5, 2, 3, 7, 11
(ii) 15, 6, 16, 8, 22, 21, 9, 18, 25
(iii) 20, 13, 18, 25, 6, 15, 21, 9, 16, 8, 22
(iv) 7, 4, 2, 5, 1, 4, 0, 10, 3, 8, 5, 9, 2
Solution:
(i) By arranging the numbers in ascending order
We get
2, 2, 3, 5, 7, 9, 9, 10, 11
We know that n = 9 is odd
So we get
Median = ½ (n + 1) th term
By substituting the values
Median = ½ (9 + 1) th term
It can be written as
Median = value of the 5^{th} term = 7
We get
Median = 7
(ii) By arranging the numbers in ascending order
We get
6, 8, 9, 15, 16, 18, 21, 22, 25
We know that n = 9 is odd
So we get
Median = ½ (n + 1) th term
By substituting the values
Median = ½ (9 + 1) th term
It can be written as
Median = value of the 5^{th} term = 16
We get
Median = 16
(iii) By arranging the numbers in ascending order
We get
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
We know that n = 11 is odd
So we get
Median = ½ (n + 1) th term
By substituting the values
Median = ½ (11 + 1) th term
It can be written as
Median = value of the 6^{th} term = 16
We get
Median = 16
(iv) By arranging the numbers in ascending order
We get
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
We know that n = 13 is odd
So we get
Median = ½ (n + 1) th term
By substituting the values
Median = ½ (13 + 1) th term
It can be written as
Median = value of the 7^{th} term = 4
We get
Median = 4
2. Find the median of:
(i) 17, 19, 32, 10, 22, 21, 9, 35
(ii) 72, 63, 29, 51, 35, 60, 55, 91, 85, 82
(iii) 10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27
Solution:
(i) By arranging the numbers in ascending order
We get
9, 10, 17, 19, 21, 22, 32, 35
We know that n = 8 which is even
So we get
Median = ½ {(n/2)th term + (n/2 + 1)th term}
It can be written as
Median = ½ {4^{th} term + 5^{th} term)
By substituting the values
Median = ½ (19 + 21)
On further calculation
Median = ½ (40)
We get
Median = 20
(ii) By arranging the numbers in ascending order
We get
29, 35, 51, 55, 6, 63, 72, 82, 85, 91
We know that n = 10 which is even
So we get
Median = ½ {(n/2)th term + (n/2 + 1)th term}
It can be written as
Median = ½ {5^{th} term + 6^{th} term)
By substituting the values
Median = ½ (60 + 63)
On further calculation
Median = ½ (123)
We get
Median = 61.5
(iii) By arranging the numbers in ascending order
We get
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
We know that n = 12 which is even
So we get
Median = ½ {(n/2)th term + (n/2 + 1)th term}
It can be written as
Median = ½ {6^{th} term + 7^{th} term)
By substituting the values
Median = ½ (15 + 17)
On further calculation
Median = ½ (32)
We get
Median = 16
3. The marks of 15 students in an examination are
25, 19, 17, 24, 23, 29, 31, 4, 19, 20, 22, 26, 17, 25, 21.
Find the median score.
Solution:
By arranging the numbers in ascending order
We get
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
We know that n = 15 which is odd
So we get
Median = ½ (n + 1) th term
By substituting the values
Median = ½ (15 + 1) th term
It can be written as
Median = value of the 8^{th} term = 23
Therefore, the median score is 23.
4. The heights (in cm) of 9 students of a class are
148, 144, 152, 155, 160, 147, 150, 149, 145.
Find the median height.
Solution:
By arranging the numbers in ascending order
We get
144, 145, 147, 148, 149, 150, 152, 155, 160
We know that n = 9 which is odd
So we get
Median = [(n + 1)/2]th value
By substituting the values
Median = [(9 + 1)/2]th value
We get
Median = 5^{th} value = 149cm
Therefore, the median height is 149cm.
5. The weights (in kg) of 8 children are
13.4, 10.6, 12.7, 17.2, 14.3, 15, 16.5, 9.8.
Find the median weight.
Solution:
By arranging the numbers in ascending order
We get
9.8, 1.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
We know that n = 8 which is even
So we get
Median = ½ {(n/2)th term + (n/2 + 1)th term}
It can be written as
Median = ½ {4^{th} term + 5^{th} term)
By substituting the values
Median = ½ (13.4 + 14.3)
On further calculation
Median = ½ (27.7)
We get
Median = 13.85kg
Therefore, the median weight is 13.85kg.
6. The ages (in years) of 10 teachers in a school are
32, 44, 53, 47, 37, 54, 34, 36, 40, 50.
Find the median age.
Solution:
By arranging the numbers in ascending order
We get
32, 34, 36, 37, 4, 44, 47, 50, 53, 54
We know that n = 10 which is even
So we get
Median = ½ {(n/2)th term + (n/2 + 1)th term}
It can be written as
Median = ½ {5^{th} term + 6^{th} term)
By substituting the values
Median = ½ (40 + 44)
On further calculation
Median = ½ (84)
We get
Median = 42
Therefore, the median age is 42 years.
7. If 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41 are ten observations in an ascending order with median 24, find the value of x.
Solution:
By arranging the numbers in ascending order
We get
10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41
We know that n = 10 which is even
So we get
Median = ½ {(n/2)th term + (n/2 + 1)th term}
It can be written as
Median = ½ {5^{th} term + 6^{th} term)
By substituting the values
Median = ½ (x + 1 + x + 3)
On further calculation
Median = ½ (2x + 4)
We get
Median = x + 2
It is given that median = 24
So we get
x + 2 = 24
On further calculation
x = 24 – 2 = 22
Therefore, the value of x is 22.
8. The following observations are arranged in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93.
If the median is 65, find the value of x.
Solution:
We know that n = 10 which is even
It is given that median = 65
We can write it as
½ {(n/2)th term + (n/2 + 1)th term} = 65
By substituting the values
½ {(10/2)th term + (10/2 + 1)th term} = 65
So we get
½ {5^{th} value + 6^{th} value} = 65
It can be written as
(x + x + 2)/2 = 65
On further calculation
2x + 2 = 130
By subtraction
2x = 128
By division
x = 64
Therefore, the value of x is 64.
9. The numbers 50, 42, 35, (2x + 10), (2x – 8), 12, 11, 8 have been written in a descending order. If their median is 25, find the value of x.
Solution:
We know that n = 8 which is even
It is given that median = 25
We can write it as
½ {(n/2)th term + (n/2 + 1)th term} = 25
By substituting the values
½ {(8/2)th term + (8/2 + 1)th term} = 25
So we get
½ {4^{th} value + 5^{th} value} = 25
It can be written as
(2x + 10 + 2x – 8)/2 = 25
On further calculation
4x + 2 = 50
By subtraction
4x = 48
By division
x = 12
Therefore, the value of x is 12.
10. Find the median of the data
46, 41, 77, 58, 35, 64, 87, 92, 33, 55, 90.
In the above data, if 41 and 55 are replaced by 61 and 75 respectively, what will be the new median?
Solution:
We know that n = 11 which is odd
By arranging the numbers in ascending order
We get
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
So we get
Median = [(n + 1)/2]th value
By substituting the values
Median = [(11 + 1)/2]th value
We get
Median = 6^{th} value = 58
By replacing 41 and 55 by 61 and 75
By arranging the numbers in ascending order
We get
33, 35, 46, 58, 61, 64, 75, 77, 87, 90, 92
So we get
Median = [(n + 1)/2]th value
By substituting the values
Median = [(11 + 1)/2]th value
We get
Median = 6^{th} value = 64
Therefore, the new median is 64.
Exercise 18(D)
1. Find the mode of the following items.
0, 6, 5, 1, 6, 4, 3, 0, 2, 6, 5, 6
Solution:
By arranging the numbers in ascending order
We get
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
Observations (x) 
0 
1 
2 
3 
4 
5 
6 
Frequency 
2 
1 
1 
1 
1 
2 
4 
From the table we know that 6 occurs maximum number of times so the mode is 6.
2. Determine the mode of the following values of a variable.
23, 15, 25, 40, 27, 25, 22, 25, 20
Solution:
By arranging the numbers in ascending order
We get
15, 20, 22, 23, 25, 25, 25, 27, 40
Observations (x) 
15 
20 
22 
23 
25 
27 
40 
Frequency 
1 
1 
1 
1 
3 
1 
1 
From the table we know that 25 occurs maximum number of times so the mode is 25.
3. Calculate the mode of the following sizes of shoes sold by a shop on a particular day.
5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9
Solution:
By arranging the numbers in ascending order
We get
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Observations (x) 
1 
2 
3 
4 
5 
6 
7 
8 
9 
Frequency 
2 
1 
2 
1 
2 
2 
1 
1 
5 
From the table we know that 9 occurs maximum number of times so the mode is 9.
4. A cricket player scored the following runs in 12 oneday matches:
50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35.
Find his modal score.
Solution:
By arranging the numbers in ascending order
We get
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Observations (x) 
9 
19 
27 
28 
30 
32 
35 
50 
60 
Frequency 
1 
1 
1 
1 
1 
1 
1 
4 
1 
From the table we know that 50 occurs maximum number of times so the mode is 50.
5. If the mean of the data 3, 21, 25, 17, (x + 3), 19, (x – 4) is 18, find the value of x. Using this value of x, find the mode of the data.
Solution:
We know that
Number of observations = 7
It is given that mean = 18
It can be written as
(3 + 21 + 25 + 17 + x + 3 + 19 + x – 4)/7 = 18
On further calculation
2x + 84 = 126
By subtraction
2x = 42
By division
x = 21
By substituting the value of x
(x + 3) = 21 + 3 = 24
(x – 4) = 21 – 4 = 17
So we get
3, 21, 25, 17, 24, 19, 17
We know that 17 occurs maximum number of times so the mode is 17.
6. The numbers 52, 53, 54, 54, (2x + 1), 55, 55, 56, 57 have been arranged in an ascending order and their median is 55. Find the value of x and hence find the mode of the given data.
Solution:
We know that
Number of observations = 9
By arranging the numbers in ascending order
We get
52, 53, 54, 54, (2x + 1), 55, 55, 56, 57
So we get
Median = [(n + 1)/2]th value
By substituting the values
Median = [(9 + 1)/2]th value
We get
Median = 5^{th} value = 2x + 1
It is given that median = 55
We can write it as
2x + 1 = 55
On further calculation
2x = 54
By division
x = 27
By substituting the value of x
2x + 1 = 2(27) + 1 = 55
So we get
52, 53, 54, 54, 55, 55, 55, 56, 57
We know that 55 occurs the maximum number of times so the mode is 55.
7. For what value of x is the mode of the data 24, 15, 40, 23, 27, 26, 22, 25, 20, x + 3 found 25? Using this value of x, find the median.
Solution:
It is given that
Mode = 25
So we know that 25 occurs maximum number of times
Take x + 3 = 25
On further calculation
x = 22
We get
24, 15, 40, 23, 27, 26, 22, 25, 20, 25
By arranging the numbers in ascending order
We get
15, 20, 22, 23, 24, 25, 25, 26, 27, 40
So the number of observations = 10
We get
Median = ½ {(n/2)th term + (n/2 + 1)th term}
By substituting the values
Median = ½ {(10/2)th term + (10/2 + 1)th term}
So we get
Median = (5^{th} term + 6^{th} term)/2
On further calculation
Median = (24 + 25)/2
We get
Median = 49/2 = 24.5
8. The numbers 42, 43, 44, 44, (2x + 3), 45, 45, 46, 47 have been arranged in an ascending order and their median is 45. Find the value of x. Hence, find the mode of the above data.
Solution:
We know that
Number of observations = 9
It is given that median = 45
So we get
Median = [(n + 1)/2]th value
By substituting the values
Median = [(9 + 1)/2]th value
We get
Median = 5^{th} value = 2x + 3
It is given that median = 45
We get
2x + 3 = 45
On further calculation
2x = 42
By division
x = 21
By substituting the value of x
2x + 3 = 2(21) + 3 = 45
We get
42, 43, 44, 44, 45, 45, 46, 47
We know that 45 occurs maximum number of times so the mode is 45.
RS Aggarwal Solutions for Class 9 Maths Chapter 18: Mean, Median and Mode of Ungrouped Data
Chapter 18, Mean, Median and Mode of Ungrouped Data, has 4 exercises with problems based on finding arithmetic mean, frequency distribution, median of ungrouped data and mode. The major concepts which are explained in RS Aggarwal Solutions Chapter 18 are as follows:
 Arithmetic Mean
 Properties of Arithmetic Mean
 Mean for an ungrouped frequency distribution
 Median of ungrouped data
 Mode of ungrouped data
RS Aggarwal Solutions Class 9 Maths Chapter 18 – Exercise list
Exercise 18A Solutions 31 Questions
Exercise 18B Solutions 14 Questions
Exercise 18C Solutions 10 Questions
Exercise 18D Solutions 8 Questions
RS Aggarwal Solutions Class 9 Maths Chapter 18 – Mean, Median and Mode of Ungrouped Data
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