R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18  Mean, Median and Mode of Ungrouped Data
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.
R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18  Mean, Median and Mode of Ungrouped Data Page/Excercise MCQ
Correct option: (b)
If each observation of the data is decreased by 8 then their mean is also decreased by 8.
R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18  Mean, Median and Mode of Ungrouped Data Page/Excercise 18A
Prime numbers between 50 and 80 are as follows:
53, 59, 61, 67, 71, 73, 79
Total prime numbers between 50 and 80 = 7
Factors of 20 are: 1,2,4,5,10,20
_{ }
First seven multiples of 5 are: 5,10,15, 20, 25, 30, 35
Therefore, Mean =20
First ten odd numbers are:
1,3,5,7,9,11,13,15, 17, and 19
_{}
first eight natural numbers are:
1,2,3,4,5,6,7and 8
_{}
_{}
Sol.3 _{}
_{}
_{}
Total numbers of observations = 5
Thus, last three observations are (9 + 4), (9 + 6) and (9 + 8),
i.e. 13, 15 and 17
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)
=(288235)=53kg.
_{}
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)]
=(195023+43) =1970
_{}correct mean =
Let the given numbers be x_{1},X_{2......}X_{24}
_{}
Let the given numbers be x_{1}, x_{2}.....x_{20}
Then , the mean of these numbers =
_{}
Let the given numbers be x_{1}, x_{2}.......x_{15}
Then the mean of these numbers=27
_{}
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)
=(138100) =38
_{} The excluded number=38.
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
_{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 =(16451581)kg= 64kg}
_{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 =(410.2)kg = 40.8 kg }
_{Total weight of 35 students = 40.8x35 kg = 1428 kg.}
_{the weight of the student who left =(14761428)kg =48 kg.}
_{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 =(400.2)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}
_{= 15921560 = 32 kg}
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.}
_{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}
_{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 =}
_{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.}
_{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}
_{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+7151300 kg}
_{= 39 kg.}
Therefore, the weight of 13th student is 39 kg.
_{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 =}
_{ 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}
_{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.}
_{ }
_{Total number of students = 50}
_{Total number of girls = 5040 = 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 = 2200400 kg =1800 kg}
_{the average weight of the boys = }
Total earnings of the year
= Rs. (3 × 18720 + 4 × 20340 + 5 ×21708 + 35340)
= Rs. (56160 + 81360 + 108540 + 35340)
= Rs. 281400
Number of months = 12
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
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.
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
For calculating the mean , we prepare the following frequency table :
Weight (in kg) (X_{i}) 
No of workers (f_{i}) 
f_{i}X_{i} 
60 63 66 69 72

4 3 2 2 1 
240 189 132 138 72 

_{} 
771 
For calculating the mean , we prepare the following frequency table :
Age (in years) (X_{i}) 
Frequency (f_{i}) 
f_{i}X_{i} 
15 16 17 18 19 20 
3 8 9 11 6 3 
45 128 153 198 114 60 

_{} 
698 
For calculating the mean , we prepare the following frequency table :
Variable (X_{i}) 
Frequency (f_{i}) 
f_{i}X_{i} 
10 30 50 70 89 
7 8 10 15 10 
70 240 500 1050 890 


_{ } 
We prepare the following frequency table :
(X_{i}) 
(f_{i}) 
f_{i}X_{i} 
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
We prepare the following frequency distribution table:
(X_{i}) 
(f_{i}) 
f_{i}X_{i} 
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
We prepare the following frequency distribution table:
(X_{i}) 
(f_{i}) 
f_{i}X_{i} 
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 = 16601228
_{} 24p = 432
_{}
_{} the value of p =18
Let f_{1 } and f_{2 }be the missing frequencies.
We prepare the following frequency distribution table.
(X_{i}) 
(f_{i}) 
f_{i}x_{i} 
10 30 50 70 90

17 f_{1} 32 f_{2} 19

170 30f_{1} 1600 70f_{2} 1710 
Total 
120 
3480 + 30f_{1 }+ 70f_{2} 
Here,
Thus, .......(1)
Also,
Substituting the value of f_{1} in equation 1, we have,
f_{2}=52  28 = 24
Thus, the missing frequencies are f_{1} =28 and f_{2}=24 respectively.
R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18  Mean, Median and Mode of Ungrouped Data Page/Excercise 18C
Arranging the data in accending order, we have
2,2,3, 5, 7, 9, 9, 10, 11
Here n = 9, which is odd
_{}
Arranging the data in ascending order , we have
6, 8, 9, 15, 16, 18, 21, 22, 25
Here n = 9, which is odd
Arranging data in ascending order:
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here n = 11 odd
_{}
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
_{}
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}
_{}Arranging the data in ascending order , we have
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here n = 10, which is even
_{}
Arranging the data in ascending order , we have
9, 10, 17, 19, 21, 22, 32, 35
Here n = 8, which is even
_{ }
_{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.
_{}
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
_{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 }
_{}
_{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}
_{ }
_{ The ten observations in ascending order:}
_{10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41}
_{Here, n =10, which is even}
_{}
Total number of observations = n = 10 (even)
Median = 65
Total number of observations = n = 8 (even)
Median = 25
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
_{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}
_{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}
_{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}
_{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.
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.
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.
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)
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.
R. S. Aggarwal and V. Aggarwal  Mathematics  IX Class 9 Chapter Solutions
 Chapter 1  Number Systems
 Chapter 2  Polynomials
 Chapter 3  Factorisation of Polynomials
 Chapter 4  Linear Equations in Two Variables
 Chapter 5  Coordinate Geometry
 Chapter 6  Introduction to Euclid's Geometry
 Chapter 7  Lines And Angles
 Chapter 8  Triangles
 Chapter 9  Congruence of Triangles and Inequalities in a Triangle
 Chapter 10  Quadrilaterals
 Chapter 11  Areas of Parallelograms and Triangles
 Chapter 12  Circles
 Chapter 13  Geometrical Constructions
 Chapter 14  Areas of Triangles and Quadrilaterals
 Chapter 15  Volume and Surface Area of Solids
 Chapter 16  Presentation of Data in Tabular Form
 Chapter 17  Bar Graph, Histogram and Frequency Polygon
 Chapter 18  Mean, Median and Mode of Ungrouped Data
 Chapter 19  Probability
CBSE Class 9 Textbook Solutions
 NCERT Biology Class 9 Textbook Solution
 NCERT Chemistry Class 9 Textbook Solution
 NCERT Mathematics  IX Class 9 Textbook Solution
 NCERT Physics Class 9 Textbook Solution
 NCERT Beehive Class 9 Textbook Solution
 NCERT क्षितिज भाग  १ Class 9 Textbook Solution
 NCERT India and the Contemp. World  I Class 9 Textbook Solution
 NCERT Democratic Politics  I Class 9 Textbook Solution
 NCERT Contemporary India  I Class 9 Textbook Solution
 NCERT Economics Class 9 Textbook Solution
Browse Study Material
Browse questions & answers
TopperLearning provides stepbystep 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.