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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data

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Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 18 - Mean, Median and Mode of Ungrouped Data.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Correct option: (b)

If each observation of the data is decreased by 8 then their mean is also decreased by 8. 

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise 18A

Solution 1(v)

Prime numbers between 50 and 80 are as follows:

53, 59, 61, 67, 71, 73, 79

Total prime numbers between 50 and 80 = 7

Solution 1(iv)

Factors of 20 are: 1,2,4,5,10,20

        

Solution 1(iii)

First seven multiples of 5 are: 5,10,15, 20, 25, 30, 35

Therefore, Mean =20

Solution 1(ii)

First ten odd numbers are:

1,3,5,7,9,11,13,15, 17, and 19

Solution 1(i)

first eight natural numbers are:

1,2,3,4,5,6,7and 8

Solution 2

Solution 3

Sol.3

Solution 4

Solution 5

Total numbers of observations = 5

Thus, last three observations are (9 + 4), (9 + 6) and (9 + 8),

i.e. 13, 15 and 17

Solution 6

Mean weight of the boys =48 kg


Sum of the weight of6 boys =(48x6)kg =288kg

Sum of the weights of 5 boys=(51+45+49+46+44)kg=235kg

Weight of the sixth boy=(sum of the weights of 6 boys ) - (sum of the weights of 5 boys)

=(288-235)=53kg.

Solution 7

Calculated mean marks of 50 students =39

calculated sum of these marks=(39x 50)=1950

Corrected sum of these marks

=[1950-(wrong number)+(correct number)]

=(1950-23+43) =1970

correct mean =

Solution 8

Let the given numbers be x1,X2......X24

Solution 9

Let the given numbers be x1, x2.....x20

Then , the mean of these numbers =

Solution 10

Let the given numbers be x1, x2.......x15

Then the mean of these numbers=27

Solution 11

Solution 12

 

Solution 13

Mean of 6 numbers = 23

Sum of 6 numbers =(23x6 )=138

Again , mean of 5 numbers =20

Sum of 5 numbers=(20x 5 ) =100

The excluded number= (sum of 6 numbers )-(sum of 5 numbers)

=(138-100) =38

The excluded number=38.

Solution 14

Mean height of 30 boys = 150 cm

Total height of 30 boys = 150 × 30 = 4500 cm

Correct sum = 4500 - incorrect value + correct value

= 4500 - 135 + 165

= 4530

Solution 15

Mean weight of 34 students = 46.5 kg

Total weight of 34 students =(34x46.5)kg =1581 kg

Mean weight of 34 students and the teacher =(46.5+0.5)kg=47kg 

Total weight of 34 students and the teacher

=(47x35)kg =1645kg

Weight of the teacher =(1645-1581)kg= 64kg

Solution 16

Mean weight of 36 students = 41 kg

Total weight of 36 students = 41x 36 kg = 1476kg

One student leaves the class mean is decreased by 200 g.

New mean =(41-0.2)kg = 40.8 kg      

Total weight of 35 students = 40.8x35 kg = 1428 kg.

the weight of the student who left =(1476-1428)kg =48 kg.

Solution 17

Mean weight of 39 students =40 kg

Total weight of 39 students = 40x 39) = 1560 kg

One student joins the class mean is decreased by 200 g.

New mean =(40-0.2)kg = 39.8 kg              

Total weight of 40 students =(39.8x40)kg=1592 kg.

the weight of new student

= Total weight of 40 students - Total weight of 39 students

= 1592-1560 = 32 kg

Solution 18

The increase in the average of 10 oarsmen = 1.5 kg

Total weight increased =(1.5x10) kg=15 kg

Since the man weighing 58 kg has been replaced,

 Weight of the new man =(58+15)kg =73kg.

Solution 19

Mean of 8 numbers=35

Total sum of 8 numbers = 35x8 = 280

 Since One number is excluded, New mean = 35 - 3 = 32

Total sum of 7 numbers = 32x7 = 224

the excluded number = Sum of 8 numbers - Sum of 7 numbers

= 280 - 224 = 56

Solution 20

Mean of 150 items = 60

Total Sum of 150 items = 150x60 = 9000

Correct sum of items =[(sum of 150 items)-(sum of wrong items)+(sum of right items)]

= [9000 - (52 + 8) + (152 + 88)]

= [9000-(52+8)+(152+88)]

= 9180

Correct mean =

Solution 21

Mean of 31 results=60

Total sum of 31 results = 31x60 = 1860

Mean of the first 16 results =16x58=928

Total sum of the first 16 results=16x58=928

Mean of the last 16 results=62

Total sum of the last 16 results=16x62=992

The 16th result = 928 + 992 - 1860

                          = 1920 - 1860 = 60

The 16th result  = 60.

Solution 22

Mean of 11 numbers = 42

Total sum of 11 numbers = 42x11 = 462

Mean of the first 6 numbers = 37

Total sum of first 6 numbers = 37x6 = 222

Mean of the last 6 numbers = 46  

Total sum of last 6 numbers = 6x46 = 276

The 6th number= 276 + 222 - 462

                          = 498 - 462 = 36

The 6th number = 36

Solution 23

Mean weight of 25 students = 52kg

Total weight of 25 students = 52x25 kg=1300 kg

Mean of the first 13 students = 48 kg

Total weight of the first 13 students = 48x13 kg = 624kg

Mean of the last 13 students = 55 kg

Total weight of the last 13 students = 55x13 kg = 715 kg

The weight of 13th student

= Total weight of the first 13 students +  Total weight of the last 13 students - Total weight of 25 students

= 624+715-1300 kg

= 39 kg.

Therefore, the weight of 13th student is 39 kg.

Solution 24

Mean score of 25 observations = 80

Total score of 25 observations = 80x25 = 2000

Mean score of 55 observations = 60

Total score of 55 observations = 60x55 =3300

Total no. of observations = 25+55 =80 observations

Total score = 2000+3300 = 5300

Mean score =

Solution 25

Average marks of 4 subjects = 50

Total marks of 4 subjects = 50x4 = 200

36 + 44 + 75 + x = 200

155  + x = 200

 x = 200 - 155 = 45

The value of x = 45

Solution 26

Let the distance of mark from the staring point be x km.

Then , time taken by the ship reaching the marks=

Time taken by the ship reaching the starting point from the marks =

Total time taken =

Total distance covered =x+x=2x km.


 

Solution 27

Total number of students = 50

Total number of girls = 50-40 = 10

Average weight of the class = 44 kg

Total weight of 50 students= 44x 50 kg = 2200kg

Average weight of 10 girls = 40 kg

Total weight of 10 girls = 40x10 kg = 400 kg

Total weight of 40 boys = 2200-400 kg =1800 kg

the average weight of the boys = 

Solution 28

Total earnings of the year

= Rs. (3 × 18720 + 4 × 20340 + 5 ×21708 + 35340)

= Rs. (56160 + 81360 + 108540 + 35340)

= Rs. 281400

Number of months = 12

Solution 29

Average weekly payment of 75 workers = Rs. 5680

Total weekly payment of 75 workers = Rs. (75 × 5680) = Rs. 426000

 

Mean weekly payment of 25 workers = Rs. 5400

Total weekly payment of 25 workers = Rs. (25 × 5400) = Rs. 135000

 

Mean weekly payment of 30 workers = Rs. 5700

Total weekly payment of 30 workers = Rs. (30 × 5700) = Rs. 171000

 

Number of remaining workers = 75 - 25 - 30 = 20

Therefore, Total weekly payment of remaining 20 workers

= Rs. (426000 - 135000 - 171000)

= Rs. 120000

Solution 30

Let the ratio of number of boys to the number of girls be x : 1.

Then,

Sum of marks of boys = 70x

Sum of marks of girls = 73 × 1 = 73

And, sum of marks of boys and girls = 71 × (x + 1)

70x + 73 = 71(x + 1)

70x + 73 = 71x + 71

x = 2

Hence, the ratio of number of boys to the number of girls is 2 : 1.

Solution 31

Mean monthly salary of 20 workers = Rs. 45900

Total monthly salary of 20 workers = Rs. (20 × 45900) = Rs. 918000

 

Mean monthly salary of 20 workers + manager = Rs. 49200

Total monthly salary of 20 workers + manager = Rs. (21 × 49200) = Rs. 1033200

 

Therefore, manager's monthly salary = Rs. (1033200 - 918000) = Rs. 115200

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise 18B

Solution 1

Solution 2

For calculating the mean , we prepare the following frequency table :

Weight (in kg)

(Xi)

No of workers

(fi)

fiXi

60

63

66

69

72

4

3

2

2

1

240

189

132

138

72

 

771

 

Solution 3

Solution 4

For calculating the mean , we prepare the following frequency table :

Age (in years)

(Xi)

Frequency

(fi)

fiXi

15

16

17

18

19

20

3

8

9

11

6

3

45

128

153

198

114

60

 

698

Solution 5

For calculating  the mean , we prepare  the following frequency table :

Variable

(Xi)

Frequency

(fi)

fiXi

10

30

50

70

89

7

8

10

15

10

70

240

500

1050

890

 

  

Solution 6

Solution 7

Solution 8

We prepare  the following frequency table :

 

(Xi)

(fi)

fiXi

3

5

7

9

11

13

6

8

15

P

8

4

18

40

105

9P

88

52

 

 

303 + 9p = 8(41+p)

303 + 9p= 328 + 8p

9p - 8p = 328 -303

P=25

the value of P=25

Solution 9

We prepare the following frequency distribution table:

(Xi)

(fi)

fiXi

15

20

25

30

35

40

8

7

P

14

15

6

120

140

25p

420

525

240

 

1445 + 25p = (28.25)(50+p)

1445 + 25p = 1412.50 + 28.25p

-28.25p + 25p = -1445 + 1412.50

-3.25p = -32.5

 

the value of p=10

Solution 10

We prepare the following frequency distribution table:

(Xi)

(fi)

fiXi

8

12

15

P

20

25

30

12

16

20

24

16

8

4

96

192

300

24p

320

200

120

 

 

1228 + 24p = 1660

24p = 1660-1228

24p = 432

 

the value of p =18

Solution 11

Solution 12

Let f1  and f2 be the missing frequencies. 

We prepare the following frequency distribution table.

 

(Xi)

(fi)

fixi

10

30

50

70

90

17

f1

32

f2

19

170

30f1

1600

70f2

1710

Total

120

3480 + 30f+ 70f2

Here, 

Thus,  .......(1)

Also,

 

Substituting the value of f1 in equation 1, we have,

f2=52 - 28 = 24

Thus, the missing frequencies are f1 =28 and f2=24 respectively.

Solution 13

Solution 14

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise 18C

Solution 1(i)

Arranging the data in accending order, we have

2,2,3, 5, 7, 9, 9, 10, 11

Here n = 9, which is odd

 

Solution 1(ii)

Arranging the data in ascending order , we have

6, 8, 9, 15, 16, 18, 21, 22, 25

Here n = 9, which is odd

Solution 1(iii)

Arranging data in ascending order:

6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25

Here n = 11 odd

Solution 1(iv)

Arranging the data in ascending order , we have

0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10

Here n = 13, which is odd

Solution 2(iii)

Arranging the data in ascending order , we have

3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81 Here n = 12, which is even


Solution 2(ii)

Arranging the data in ascending order , we have

29, 35, 51, 55, 60, 63, 72, 82, 85, 91

Here n = 10, which is even

Solution 2(i)

Arranging the data in ascending order , we have

9, 10, 17, 19, 21, 22, 32, 35

Here n = 8, which is even

 

Solution 3

Arranging the data in ascending order , we have

17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40

Here n = 15, which is odd

Thus, the median score is 23.


Solution 4

Total number of students = n = 9 (odd)

Arranging heights (in cm) in ascending order, we have

144, 145, 147, 148, 149, 150, 152, 155, 160

Solution 5

Arranging the weights of 8 children in ascending order, we have

9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

Here , n= 8 , which is even

Solution 6

Arranging the ages of teachers in ascending order , we have

32, 34, 36, 37, 40, 44, 47, 50, 53, 54

Here, n =10, which is even

Solution 7

  The  ten observations in ascending order:

10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41

Here, n =10, which is even

Solution 8

Total number of observations = n = 10 (even)

Median = 65

  

Solution 9

Total number of observations = n = 8 (even)

Median = 25

  

Solution 10

Total number of observations = n = 11 (odd)

Arranging data in ascending order, we have

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Now, 41 and 55 are replaced by 61 and 75 respectively. 

Arranging new data in ascending order, we have

33, 35, 46, 58, 61, 64, 75, 77, 87, 90, 92

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise 18D

Solution 1

Arrange the given data in ascending order we have

0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6

Let us prepare the following table:

Observations(x)

0

1

2

3

4

5

6

Frequency

2

1

1

1

1

2

4

As 6 ocurs the maximum number of times i.e. 4, mode = 6

Solution 2

Arranging the given data in ascending order , we have:

15, 20, 22, 23, 25, 25, 25, 27, 40

The frequency table of the data is :

Observations(x)

15

20

22

23

25

27

40

Frequency

1

1

1

1

3

1

1

As 25 ocurs the maximum number of times i.e. 3, mode = 25

Solution 3

Arranging the given data in ascending order , we have:

1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9,

The frequency table of the data is :

Observations(x)

1

2

3

4

5

6

7

8

9

Frequency

2

1

2

1

2

2

1

1

5

As 9, occurs the maximum number of times i.e. 5, mode = 9

Solution 4

Arranging the given data in ascending order , we have:

9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60

The frequency table of the data is :

Observations(x)

9

19

27

28

30

32

35

50

60

Frequency

1

1

1

1

1

1

1

4

1

As 50, ocurs the maximum number of times i.e. 4, mode = 50

Thus, the modal score of the cricket player is 50.

Solution 5

Total number of observations = n = 7

Mean = 18

Thus, data is as follows:

3, 21, 25, 17, 24, 19, 17

The most occurring value is 17.

Hence, the mode of the data is 17.

Solution 6

Number of values = n = 9 (odd)

Numbers in ascending order:

52, 53, 54, 54, (2x + 1), 55, 55, 56, 57 

  

Thus, we have

52, 53, 54, 54, 55, 55, 55, 56, 57

The most occurring number is 55.

Hence, the mode of the data is 55. 

Solution 7

Mode of the data = 25

So, we should have the value 25 occurring maximum number of times in the given data.

That means, x + 3 = 25

x = 22

Thus, we have 24, 15, 40, 23, 27, 26, 22, 25, 20, 25.

Arranging data in ascending order, we have

15, 20, 22, 23, 24, 25, 25, 26, 27, 40

Number of observations = 10 (even)

Solution 8

Total number of observations = n = 9 (odd)

Median = 45

  

Thus, we have

42, 43, 44, 44, 45, 45, 45, 46, 47

The most occurring value is 45.

Hence, the mode of the data is 45.

TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 18 - Mean, Median and Mode of Ungrouped Data now.

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