RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data

Class 10: Maths Chapter 18 solutions. Complete Class 10 Maths Chapter 18 Notes.

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data

RS Aggarwal 10th Maths Chapter 18, Class 10 Maths Chapter 18 solutions

Exercise 18A Solutions

Question 1:
Table is as given below:

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 1
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 1


Question 2:
We have

ClassFrequency fiMid Value xi fixi
0-1010-2020-3030-4040-5050-607561282515253545553575150420360110
 ∑fi=40 ∑fixi=1150
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 2

Question 3:
We have

ClassFrequency fiClass Mark xifixi
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 701115203014101525354555651653757001350770650
 ∑fi=100 ∑fixi=4010
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 3


Question 4:
We have

ClassMid value fiFrequency xi fixi
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8015253545556575681373219020045531516513075
∑fi=40  ∑fixi=1430
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 4


Question 5:
We have

ClassFrequency fiMid value xi fixi
25 – 3535 – 4545 – 5555 – 6565 – 7561081243040506070180400400720280
 ∑fi=40 ∑fixi=1980

Mean,

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 5


Question 6:
We have

ClassFrequency fiMid Value xi fixi
0 – 100100 – 200200 – 300300 – 400400 – 5006915128501502503504503001350375042003600
∑fi=50 ∑fixi=13200

Mean,

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 6


Question 7:
We have

ClassFrequency fiMid Value xi fixi
0 – 1010 – 2020 – 3030 – 4040 – 50152035p105152535457530087535p450
∑fi=80+p ∑fixi=1700+35p
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 7


Question 8:
We have

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 8
ClassFrequency fiMid Value xi fixi
0 – 201710170
20 – 40 f13030f1
40 – 6032501600
60 – 8052 -f1703640 – 70f1
80 – 10019901710
∑fi=120∑fixi=7120−40f1


Question 9:
We have

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 9
ClassFrequency fiMid Value xi fixi
0 – 2071070
20 – 40f13030f1
40 – 601250600
60 – 80f2=18 -f1701260 – 70f1
80 – 100890720
100 – 1205110550
∑fi=50 ∑fixi=3200−40f1
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 9


Question 10:
We have, Let A = 25 be the assumed mean

MarksFrequency fiMid value xiDeviation di=(xi-25)(fi × di)
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 601218272017651525 = A354555-20-100102030-240-1800200340180
∑fi=100∑(fi×di)=300
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 10


Hence mean = 28.
Question 11:
A = 100 be the assumed mean, we have

MarksFrequency fiMid value xiDeviation di=(xi-100)(fi × di)
0 – 4040 – 8080 – 120120 – 160160 – 20012203530232060100 = A140180-80-4004080-960-800012001840
∑fi=120 ∑(fi×di)=1250
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 11


Hence, mean = 110.67
Question 12:
Let the assumed mean be 150, h = 20

MarksFrequency fiMid value xiDeviation di = – 150(fi × di)
100 – 120120 – 140140 – 160160 – 180180 – 200102030155110130150=A170190-40-2002040-400-4000300200
∑fi=80∑(fi×di)=300
RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 12


Hence, Mean = 146.25
Question 13:
Let A = 50 be the assumed mean, we have

MarksFrequency fiMid value xi Deviation di=(xi-50) fi × di
0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120203552443831103050 = A7090110-40-200204060-800-700088015201860
∑fi=200 ∑(fi×di)=2760


Question 14:

MarksFrequency fiMid value xiui=(xi−Ah)fi × ui
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 601218272017651525 = A354555-2-10123-24-180203418
∑fi=100 ∑(fi×ui)=30

We have h = 10 and let assumed mean = 25.
A = 25, h = 10, ∑ fi= 100 and ∑(fi ×ui)= 30

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 14


Hence the mean of given frequency distribution is 28.
Question 15:
We have h = 4 and let assumed mean be A = 26. We have the table given below:

MarksFrequency fiMid value xi ui=(xi−Ah) fi × ui
4 – 88 – 1212 – 1616 – 2020 – 2424 – 2828 – 3232 – 362121525181213361014182226 = A3034-5-4-3-2-1012-10-48-45-50-180136
∑fi=100 ∑(fi×ui)=−152

A = 26, h = 4, ∑ fi= 100 and ∑(fi ×ui)= -152

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 15


Hence the mean of given frequency distribution is 19.92.
Question 16:
We have h= 30 and let A = 75 be the assumed mean. we have the table given below:

MarksFrequency fi Mid value xi  ui=(xi−Ah)  fi × ui
0 – 3030 – 6060 – 9090 – 120120 – 150150 – 180122134522011144575 = A105135165-2-10123-24-210524033
∑fi=150 ∑(fi×ui)=80

Thus, A = 75, h = 30, ∑ fi= 150 and ∑(fi ×ui)= 80

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 16


Hence, the mean of the given frequency distribution is 91.
Question 17:
We ahve h = 20 and let A = 70 be the assumed mean. We have the table given below:

MarksFrequency fiMid value xi ui=(xi−Ah)  fi × ui
0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120120 – 140121815252615910305070 = A90110130-3-2-10123-36-36-150263027
∑fi=150∑(fi×ui)=−4

Thus, A = 70, h = 20, ∑ fi= 120 and ∑(fi ×ui)= -4

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 17


Hence the mean of given frequency distribution is 69.33.
Question 18:
We have h = 14 and let A = 35 be the assumed mean.
For calculating the mean, we prepare the table given below:

MarksFrequency fi Mid value xi ui=(xi−Ah)  fi × ui
0 – 1414 – 2828 – 4242 – 5656 – 7072135111672135 = A4963-2-1012-14-2101132
∑fi=90 ∑(fi×ui)=8

Thus, A = 35, ∑ fi= 90, h = 14 and ∑(fi ×ui)= 8

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 18


Hence, Mean = 36.24
Question 19:
Let h = 5 and let A = 22.5 be the assumed mean.
For calculating the mean, we prepare the table given below:

MarksFrequency fiMid value  xi ui=(xi−Ah) fi × ui
10 – 1515 – 2020 – 2525 – 3030 – 3535 – 40568126312.517.522.5 = A27.532.537.5-2-10123-10-6012129
∑fi=40 ∑(fi×ui)=17

Thus, A = 22.5 and h = 5
∑ fi= 40 and   ∑(fi ×ui)= 17


Hence the mean of given frequency distribution is 24.625.
Question 20:
We have h = 6 and let assume mean A = 33. For calculating the mean we prepare the table.

AgeFrequency fiMid value xi  ui=(xi−Ah) fi × ui
18 – 2424 – 3030 – 3636 – 4242 – 4848 – 546812842212733 = A394551-2-10123-12-80886
∑fi=40 ∑(fi×ui)=2

Thus, A = 33, h = 6, ∑ fi= 40 and ∑(fi ×ui)=2


Hence, Mean = 33.3 years
Question 21:
We have h = 6 and let assumed mean A = 99. For calculating the mean we prepare the table:

Class fi xi ui=(xi−Ah) fi × ui
84 – 9090 – 9696 – 102102 – 108108 – 114114 – 120152220182025879399=A105111117-2-10123-30-220184075
∑fi=120 ∑(fi×ui)=81

Thus, A = 99, h = 6 and ∑ fi= 120, ∑(fi ×ui)=2

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 21


Hence, Mean = 103.05.
Question 22:
Let h = 20 and assume mean = 550, we prepare the table given below:

AgeFrequency fiMid value xi ui=(xi−55020) fi × ui
500 – 520520 – 540540 – 560560 – 580580 – 600600 – 6201495435510530550 = A570590610-2-10123-27-904615
∑fi=40 ∑(fi×ui)=−12

Thus, A = 550, h = 20, and ∑ fi= 40, ∑(fi ×ui)=-12


Hence the mean of the frequency distribution is 544.
Question 23:
The given series is an inclusive series, making it an exclusive series, we have

ClassFrequency fiMid value xi ui=(xi−425) fi × ui
24.5 – 29.529.5 – 34.534.5 – 39.539.5 – 44.544.5 – 49.549.5 – 54.554.5 – 59.5414221665327323742 = A475257-3-2-10123-12-28-2206109
∑fi=70 ∑(fi×ui)=−37

Thus, A = 42, h = 5, ∑ fi= 70 and ∑(fi ×ui)=-37

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data Exercise 18A Question 23


Hence, Mean = 39.36 years.
Question 24:
The given series is an inclusive series making it an exclusive series, we get

classFrequency fiMid value xi ui=(xi−29.510) fi × ui
4.5 – 14.514.5 – 24.524.5 – 34.534.5 – 44.544.5 – 54.554.5 – 64.561121231459.519.529.5=A39.549.559.5-2-10123-12-110232815
∑fi=80 ∑(fi×ui)=43

Thus, A = 29.5, h = 10, ∑ fi= 80 and ∑(fi ×ui)=43


Hence, Mean = 34.87 years.

RS Aggarwal Solutions for Class 10 Maths Chapter 18: Download PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data

Download PDF: RS Aggarwal Solutions for Class 10 Maths Chapter 18–Mean, Median, Mode of Grouped Data PDF

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About RS Aggarwal Class 10 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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