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

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Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 9
Chapter Name	Mean, Median, Mode of Grouped Data
Exercise	9 A
Category	RS Aggarwal Solutions

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

Question 1:
Table is as given below:

Question 2:
We have

Class	Frequency f_i	Mid Value x_i	f_ix_i
0-1010-2020-3030-4040-5050-60	7561282	51525354555	3575150420360110
	∑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

Class	Frequency f_i	Class Mark x_i	f_ix_i
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 70	111520301410	152535455565	1653757001350770650
	∑fi=100		∑fixi=4010

Question 4:
We have

Class	Mid value f_i	Frequency x_i	f_ix_i
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80	15253545556575	68137321	9020045531516513075
	∑fi=40		∑fixi=1430

Question 5:
We have

Class	Frequency f_i	Mid value x_i	f_ix_i
25 – 3535 – 4545 – 5555 – 6565 – 75	6108124	3040506070	180400400720280
	∑fi=40		∑fixi=1980

Mean,
Question 6:
We have

Class	Frequency f_i	Mid Value x_i	f_ix_i
0 – 100100 – 200200 – 300300 – 400400 – 500	6915128	50150250350450	3001350375042003600
	∑fi=50		∑fixi=13200

Mean,
Question 7:
We have

Class	Frequency f_i	Mid Value x_i	f_ix_i
0 – 1010 – 2020 – 3030 – 4040 – 50	152035p10	515253545	7530087535p450
	∑fi=80+p		∑fixi=1700+35p

Question 8:
We have

Class	Frequency f_i	Mid Value x_i	f_ix_i
0 – 20	17	10	170
20 – 40	f₁	30	30f₁
40 – 60	32	50	1600
60 – 80	52 -f₁	70	3640 – 70f₁
80 – 100	19	90	1710
	∑fi=120		∑fixi=7120−40f1

Question 9:
We have

Class	Frequency f_i	Mid Value x_i	f_ix_i
0 – 20	7	10	70
20 – 40	f₁	30	30f₁
40 – 60	12	50	600
60 – 80	f₂=18 -f₁	70	1260 – 70f₁
80 – 100	8	90	720
100 – 120	5	110	550
	∑fi=50		∑fixi=3200−40f1

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

Marks	Frequency f_i	Mid value x_i	Deviation d_i=(x_i-25)	(f_i× d_i)
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60	12182720176	51525 = A354555	-20-100102030	-240-1800200340180
	∑fi=100			∑(fi×di)=300

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

Marks	Frequency f_i	Mid value x_i	Deviation d_i=(x_i-100)	(f_i× d_i)
0 – 4040 – 8080 – 120120 – 160160 – 200	1220353023	2060100 = A140180	-80-4004080	-960-800012001840
	∑fi=120			∑(fi×di)=1250

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

Marks	Frequency f_i	Mid value x_i	Deviation d_i = – 150	(f_i× d_i)
100 – 120120 – 140140 – 160160 – 180180 – 200	102030155	110130150=A170190	-40-2002040	-400-4000300200
	∑fi=80			∑(fi×di)=300

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

Marks	Frequency f_i	Mid value x_i	Deviation d_i=(x_i-50)	f_i× d_i
0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120	203552443831	103050 = A7090110	-40-200204060	-800-700088015201860
	∑fi=200			∑(fi×di)=2760

Question 14:

Marks	Frequency f_i	Mid value x_i	ui=(xi−Ah)	f_i× u_i
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60	12182720176	51525 = A354555	-2-10123	-24-180203418
	∑fi=100			∑(fi×ui)=30

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

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:

Marks	Frequency f_i	Mid value x_i	ui=(xi−Ah)	f_i× u_i
4 – 88 – 1212 – 1616 – 2020 – 2424 – 2828 – 3232 – 36	21215251812133	61014182226 = A3034	-5-4-3-2-1012	-10-48-45-50-180136
	∑fi=100			∑(fi×ui)=−152

A = 26, h = 4, ∑ f_i= 100 and ∑(f_i×u_i)= -152

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:

Marks	Frequency f_i	Mid value x_i	ui=(xi−Ah)	f_i× u_i
0 – 3030 – 6060 – 9090 – 120120 – 150150 – 180	122134522011	144575 = A105135165	-2-10123	-24-210524033
	∑fi=150			∑(fi×ui)=80

Thus, A = 75, h = 30, ∑ f_i= 150 and ∑(f_i×u_i)= 80

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:

Marks	Frequency f_i	Mid value x_i	ui=(xi−Ah)	f_i× u_i
0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120120 – 140	1218152526159	10305070 = A90110130	-3-2-10123	-36-36-150263027
	∑fi=150			∑(fi×ui)=−4

Thus, A = 70, h = 20, ∑ f_i= 120 and ∑(f_i×u_i)= -4

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:

Marks	Frequency f_i	Mid value x_i	ui=(xi−Ah)	f_i× u_i
0 – 1414 – 2828 – 4242 – 5656 – 70	721351116	72135 = A4963	-2-1012	-14-2101132
	∑fi=90			∑(fi×ui)=8

Thus, A = 35, ∑ f_i= 90, h = 14 and ∑(f_i×u_i)= 8

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:

Marks	Frequency f_i	Mid value x_i	ui=(xi−Ah)	f_i× u_i
10 – 1515 – 2020 – 2525 – 3030 – 3535 – 40	5681263	12.517.522.5 = A27.532.537.5	-2-10123	-10-6012129
	∑fi=40			∑(fi×ui)=17

Thus, A = 22.5 and h = 5
∑ f_i= 40 and ∑(f_i×u_i)= 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.

Age	Frequency f_i	Mid value x_i	ui=(xi−Ah)	f_i× u_i
18 – 2424 – 3030 – 3636 – 4242 – 4848 – 54	6812842	212733 = A394551	-2-10123	-12-80886
	∑fi=40			∑(fi×ui)=2

Thus, A = 33, h = 6, ∑ f_i= 40 and ∑(f_i×u_i)=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	f_i	x_i	ui=(xi−Ah)	f_i× u_i
84 – 9090 – 9696 – 102102 – 108108 – 114114 – 120	152220182025	879399=A105111117	-2-10123	-30-220184075
	∑fi=120			∑(fi×ui)=81

Thus, A = 99, h = 6 and ∑ f_i= 120, ∑(f_i×u_i)=2

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

Age	Frequency f_i	Mid value x_i	ui=(xi−55020)	f_i× u_i
500 – 520520 – 540540 – 560560 – 580580 – 600600 – 620	1495435	510530550 = A570590610	-2-10123	-27-904615
	∑fi=40			∑(fi×ui)=−12

Thus, A = 550, h = 20, and ∑ f_i= 40, ∑(f_i×u_i)=-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

Class	Frequency f_i	Mid value x_i	ui=(xi−425)	f_i× u_i
24.5 – 29.529.5 – 34.534.5 – 39.539.5 – 44.544.5 – 49.549.5 – 54.554.5 – 59.5	4142216653	27323742 = A475257	-3-2-10123	-12-28-2206109
	∑fi=70			∑(fi×ui)=−37

Thus, A = 42, h = 5, ∑ f_i= 70 and ∑(f_i×u_i)=-37

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

class	Frequency f_i	Mid value x_i	ui=(xi−29.510)	f_i× u_i
4.5 – 14.514.5 – 24.524.5 – 34.534.5 – 44.544.5 – 54.554.5 – 64.5	6112123145	9.519.529.5=A39.549.559.5	-2-10123	-12-110232815
	∑fi=80			∑(fi×ui)=43

Thus, A = 29.5, h = 10, ∑ f_i= 80 and ∑(f_i×u_i)=43

Hence, Mean = 34.87 years.

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

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