RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 9
Chapter NameMean, Median, Mode of Grouped Data
Exercise9 A
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A

Question 1:
Table is as given below:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 1.1
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 1.2

Question 2:
We have

ClassFrequency fiMid Value xi fixi
0-1010-2020-3030-4040-5050-607561282515253545553575150420360110
 ∑fi=40 ∑fixi=1150
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 2.1

Question 3:
We have

ClassFrequency fiClass Mark xifixi
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 701115203014101525354555651653757001350770650
 ∑fi=100 ∑fixi=4010

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 3.1
Question 4:
We have

ClassMid value fiFrequency xi fixi
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8015253545556575681373219020045531516513075
∑fi=40  ∑fixi=1430

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 4.1
Question 5:
We have

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

Mean,RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 5.1
Question 6:
We have

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

Mean,RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 6.1
Question 7:
We have

ClassFrequency fiMid Value xi fixi
0 – 1010 – 2020 – 3030 – 4040 – 50152035p105152535457530087535p450
∑fi=80+p ∑fixi=1700+35p

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 7.1
Question 8:
We have
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 8.1

ClassFrequency fiMid Value xi fixi
0 – 201710170
20 – 40 f13030f1
40 – 6032501600
60 – 8052 -f1703640 – 70f1
80 – 10019901710
∑fi=120∑fixi=7120−40f1

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 8.2
Question 9:
We have
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 9.1

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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 9.2
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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 10.1
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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 11.1
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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 12.1
Hence, Mean = 146.25
Question 13:
Let A = 50 be the assumed mean, we have

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

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 13.1
Question 14:

MarksFrequency fiMid value xiui=(xi−Ah)f× 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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 14.1
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) f× 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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 15.1
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 fiMid value xi  ui=(xi−Ah)  f× 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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 16.1
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)  f× 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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 17.1
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 fiMid value xi ui=(xi−Ah)  f× 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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 18.1
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) f× 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
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 19.1
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) f× 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
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 20.1
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) f× 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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 21.1
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) f× 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
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 22.1
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) f× 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 Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 23.1
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) f× 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
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A 24.1
Hence, Mean = 34.87 years.

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