RS Aggarwal Solutions for Class 9 Maths Exercise 18A PDF
RS Aggarwal Solutions are created with the aim of strengthening skills and conceptual knowledge among students. Understanding the topics in Class 9 is important as some of the topics are repeated in higher grades as well. Accurate solutions are created by subject experts based on the RS Aggarwal textbook of current CBSE syllabus. The students can use PDF of solutions as reference material from exam point of view. RS Aggarwal Solutions for Class 9 Maths Chapter 18 Mean, Median and Mode of Ungrouped Data Exercise 18A are provided here.
RS Aggarwal Solutions for Class 9 Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18A Download PDF
Access RS Aggarwal Solutions for Class 9 Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18A
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 â€“ sum of weight of 5 boys
By substituting the values
Weight of sixth boy = 288 â€“ 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 â€“ wrong number + correct number
By substituting the values
Corrected sum of the marks = 1950 â€“ 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}, â€¦â€¦.. x_{24} as the given numbers
So we get
Mean = (x_{1} + x_{2}, â€¦â€¦.. + x_{24})/24
It is given that mean = 35
(x_{1} + x_{2}, â€¦â€¦.. + x_{24})/24 = 35
By cross multiplication
x_{1} + x_{2}, â€¦â€¦.. + x_{24} = 840 â€¦â€¦. (1)
Take (x_{1} + 3), (x_{2} + 3), â€¦â€¦â€¦ (x_{24} + 3) as new numbers
So the mean of new numbers = [(x_{1} + 3) + (x_{2} + 3) + â€¦â€¦â€¦ + (x_{24} + 3)]/24
From equation (1) we get
[(x_{1} + 3) + (x_{2} + 3) + â€¦â€¦â€¦ + (x_{24} + 3)]/24 = (840 + 72)/24On further calculation
[(x_{1} + 3) + (x_{2} + 3) + â€¦â€¦â€¦ + (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}, â€¦â€¦. x_{20} as the given numbers
So we get
Mean = (x_{1} + x_{2} + â€¦â€¦. + x_{20})/20
It is given that mean = 43
(x_{1} + x_{2} + â€¦â€¦. + x_{20})/20 = 43
By cross multiplication
x_{1} + x_{2} + â€¦â€¦. + x_{20} = 860 â€¦â€¦ (1)
Take (x_{1} â€“ 6), (x_{2} â€“ 6) â€¦â€¦ (x_{20} â€“ 6) as the new numbers
So the mean of new numbers = [(x_{1} â€“ 6) + (x_{2} â€“ 6) + â€¦â€¦ + (x_{20} â€“ 6)]/20
From equation (1) we get
[(x_{1} â€“ 6) + (x_{2} â€“ 6) + â€¦â€¦ + (x_{20} â€“ 6)]/20 = (860 â€“ 120)/20On further calculation
[(x_{1} â€“ 6) + (x_{2} â€“ 6) + â€¦â€¦ + (x_{20} â€“ 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}, â€¦â€¦. x_{15} as the given numbers
So we get
Mean = (x_{1} + x_{2} + â€¦â€¦. + x_{15})/15
It is given that mean = 27
(x_{1} + x_{2} + â€¦â€¦. + x_{15})/15 = 27
By cross multiplication
x_{1} + x_{2} + â€¦â€¦. + x_{15} = 405 â€¦â€¦ (1)
Take (x_{1} Ã— 4), (x_{2} Ã— 4), â€¦â€¦â€¦ (x_{15} Ã— 4) as new numbers
So the mean of new numbers = [(x_{1} Ã— 4) + (x_{2} Ã— 4) + â€¦â€¦â€¦ + (x_{15} Ã— 4)]/15
From equation (1) we get
[(x_{1} Ã— 4) + (x_{2} Ã— 4) + â€¦â€¦â€¦ + (x_{15} Ã— 4)]/15 = (405 Ã— 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}, â€¦â€¦. x_{12} as the given numbers
So we get
Mean = [x_{1} + x_{2} + â€¦â€¦. + x_{12}]/12
It is given that mean = 40
(x_{1} + x_{2} + â€¦â€¦. + x_{12})/12 = 40
By cross multiplication
x_{1} + x_{2} + â€¦â€¦. + x_{12} = 480 â€¦â€¦ (1)
Take (x_{1} Ã· 8), (x_{2} Ã· 8), â€¦â€¦â€¦ (x_{12} Ã· 8) as new numbers
So the mean of new numbers = [(x_{1} Ã· 8) + (x_{2} Ã· 8) + â€¦â€¦â€¦ + (x_{12} Ã· 8)]/12
From equation (1) we get
[(x_{1} Ã· 8) + (x_{2} Ã· 8) + â€¦â€¦â€¦ + (x_{12} Ã· 8)]/12 = (480 Ã· 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}, â€¦â€¦. x_{20} as the given numbers
So we get
Mean = (x_{1} + x_{2} + â€¦â€¦. + x_{20})/20
It is given that mean = 18
(x_{1} + x_{2}, â€¦â€¦. + x_{20})/20 = 18
By cross multiplication
x_{1} + x_{2} + â€¦â€¦. + x_{20} = 360 â€¦â€¦ (1)
Take (x_{1} + 3), (x_{2} + 3), â€¦â€¦â€¦ (x_{10} + 3) as first ten new numbers
So the mean of new set of 20 numbers = [(x_{1} + 3) + (x_{2} + 3) + â€¦â€¦â€¦ + (x_{10} + 3) + x_{11} + â€¦â€¦.. + x_{20}]/ 20
We get
Mean of new set of 20 numbers = [(x_{1} + x_{2} + â€¦â€¦. + x_{20}) + 3 Ã— 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 â€“ sum of five numbers
By substituting the values
Excluded number = 138 â€“ 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 â€“ incorrect value + correct value
By substituting the values
Correct sum = 4500 â€“ 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 â€“ total weight
By substituting the values
Weight of the teacher = 1645 â€“ 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 â€“ 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 â€“ weight of 35 students
By substituting the values
Weight of the student who left the class = 1476 â€“ 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 â€“ 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 â€“ total weight of 39 students
By substituting the values
Weight of new student = 1592 â€“ 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 â€“ 3 = 32
We know that total sum of 7 numbers = 32 (7) = 224
So we get
Excluded number = sum of 8 numbers â€“ sum of 7 numbers
By substituting the values
Excluded number = 280 â€“ 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 â€“ sum of wrong items + sum of right items
By substituting the values
Correct sum of items = 90000 â€“ (52 + 8) + (152 + 88)
On further calculation
Correct sum of items = 9000 â€“ 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 â€“ total sum of 31 results
By substituting the values
16^{th} result = 928 + 992 â€“ 1860
On further calculation
16^{th} result = 1920 â€“ 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 â€“ total sum of 11 numbers
By substituting the values
6^{th} number = 276 + 222 â€“ 462
So we get
6^{th} number = 498 â€“ 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 â€“ total weight of 25 students
By substituting the values
Weight of 13^{th} student = 624 + 715 â€“ 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 â€“ 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 Ã· x/6
It can be written as
Average sailing speed for the whole journey = (2x Ã— 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 â€“ 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 â€“ total weight of 10 girls
By substituting the values
Total weight of 40 boys = 2200 â€“ 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 â‚¹ 18720 during the first 3 months, â‚¹ 20340 during the next 4 months and â‚¹ 21708 during the last 5 months of a year. If the total savings during the year be â‚¹ 35340 find the average monthly income of the family.
Solution:
We know that
Total earnings of the year = â‚¹ (3 Ã— 18720 + 4 Ã— 20340 + 5 Ã— 21708 + 35340)
On further calculation
Total earnings of the year = â‚¹ (56160 + 81360 + 108540 + 35340)
So we get
Total earnings of the year = â‚¹ 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 = â‚¹ 23450
Therefore, the average monthly income of the family is â‚¹ 23450.
29. The average weekly payment to 75 workers in a factory is â‚¹ 5680. The mean weekly payment to 25 of them is â‚¹ 5400 and that of 30 others is â‚¹ 5700. Find the mean weekly payment of the remaining workers.
Solution:
It is given that
Average weekly payment of 75 workers = â‚¹ 5680
So the total payment = â‚¹ (75 Ã— 5680) = â‚¹ 426000
Mean weekly payment of 25 workers = â‚¹ 5400
So the total payment = â‚¹ (25 Ã— 5400) = â‚¹ 135000
Mean weekly payment of 30 workers = â‚¹ 5700
So the total payment = â‚¹ (30 Ã— 5700) = â‚¹ 171000
We know that number of remaining workers = 75 â€“ 25 â€“ 30 = 20
So the total weekly payment of remaining 20 workers = Total weekly payment of 75 workers â€“ total weekly payment of 25 workers â€“ total weekly payment of 30 workers
By substituting the values
Total weekly payment of remaining 2 workers = 426000 â€“ 135000 â€“ 171000 = â‚¹ 120000
So the mean weekly payment of remaining 20 workers = 120000/20 = â‚¹ 6000
Therefore, the mean weekly payment of the remaining workers is â‚¹ 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 â‚¹ 45900. If the managerâ€™s salary is added, the average salary becomes â‚¹ 49200 per month. Whatâ€™s managerâ€™s monthly salary?
Solution:
It is given that
Mean monthly salary of 20 workers = â‚¹ 45900
So the total monthly salary = â‚¹ (20 Ã— 49500) = â‚¹ 918000
Mean monthly salary of 20 workers and manager = â‚¹ 49200
So the total monthly salary = â‚¹ (21 Ã— 49200) = â‚¹ 1033200
We know that
Managerâ€™s monthly salary = Total monthly salary of 20 workers and manager â€“ Total monthly salary of 20 workers
By substituting the values
Managerâ€™s monthly salary = 1033200 â€“ 918000 = â‚¹ 115200
Therefore, the managerâ€™s monthly salary is â‚¹ 115200.
Access other exercise solutions of Class 9 Maths Chapter 18: Mean, Median and Mode of Ungrouped Data
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 Exercise 18A
RS Aggarwal Solutions Class 9 Maths Chapter 18 Mean, Median and Mode of Ungrouped Data Exercise 18A consists of various methods and formulas which can be used in finding mode of given data.
Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18A
- The solutions are provided by experienced faculty after vast research conducted on each topic.
- The exercise wise problems can be solved using the PDF as a guide to score better marks in the board exam.
- The answers are accurate in a stepwise manner based on the latest CBSE guidelines and weightage for each concept.
- The main aim of preparing solutions is to improve confidence among students to appear for board exams.