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


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RS Aggarwal Solutions Class 10 Chapter 9





Exercise 9A





Question 1:
Table is as given below:
rs-aggarwal-class-10-solutions-mean-median-mode-of-grouped-data-q1-1
rs-aggarwal-class-10-solutions-mean-median-mode-of-grouped-data-q1-2





Question 2:
We have





ClassFrequency fiMid Value xi fixi
0-1010-2020-3030-4040-5050-607561282515253545553575150420360110
 [latex]\sum { { f }_{ i } } =40  [/latex] [latex]\sum { { f }_{ i } } { x }_{ i }=1150  [/latex]




rs-aggarwal-class-10-solutions-mean-median-mode-of-grouped-data-q2




Question 3:
We have





ClassFrequency fiClass Mark xifixi
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 701115203014101525354555651653757001350770650
 [latex]\sum { { f }_{ i } } =100  [/latex] [latex]\sum { { f }_{ i } } { x }_{ i }=4010  [/latex]




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





ClassMid value fiFrequency xi fixi
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8015253545556575681373219020045531516513075
[latex]\sum { { f }_{ i } } =40  [/latex]  [latex]\sum { { f }_{ i } } { x }_{ i }=1430  [/latex]




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





ClassFrequency fiMid value xi fixi
25 – 3535 – 4545 – 5555 – 6565 – 7561081243040506070180400400720280
 [latex]\sum { { f }_{ i } } =40  [/latex] [latex]\sum { { f }_{ i } } { x }_{ i }=1980  [/latex]




Mean,Solution of RS Aggarwal Class 10 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
Question 6:
We have





ClassFrequency fiMid Value xi fixi
0 – 100100 – 200200 – 300300 – 400400 – 5006915128501502503504503001350375042003600
[latex]\sum { { f }_{ i } } =50  [/latex] [latex]\sum { { f }_{ i } } { x }_{ i }=13200  [/latex]




Mean,RS Aggarwal Solutions Class 10 2017 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
Question 7:
We have





ClassFrequency fiMid Value xi fixi
0 – 1010 – 2020 – 3030 – 4040 – 50152035p105152535457530087535p450
[latex]\sum { { f }_{ i } } =80+p  [/latex] [latex]\sum { { f }_{ i } } { x }_{ i }=1700+35p  [/latex]




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





ClassFrequency fiMid Value xi fixi
0 – 201710170
20 – 40 f13030f1
40 – 6032501600
60 – 8052 -f1703640 – 70f1
80 – 10019901710
[latex]\sum { { f }_{ i } } =120  [/latex][latex]\sum { { f }_{ i } } { x }_{ i }=7120-40{ f }_{ 1 }  [/latex]




RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
Question 9:
We have
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 9 Mean, Median, Mode of Grouped Data ex 9A





ClassFrequency fiMid Value xi fixi
0 – 2071070
20 – 40f13030f1
40 – 601250600
60 – 80f2=18 -f1701260 – 70f1
80 – 100890720
100 – 1205110550
[latex]\sum { { f }_{ i } } =50  [/latex] [latex]\sum { { f }_{ i } } { x }_{ i }=3200-40{ f }_{ 1 }  [/latex]




RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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
[latex]\sum { { f }_{ i } } =100  [/latex][latex]\sum { { (f }_{ i } } \times { d }_{ i })=300  [/latex]




RS Aggarwal Class 10 Book pdf download Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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
[latex]\sum { { f }_{ i } } =120  [/latex] [latex]\sum { { (f }_{ i } } \times { d }_{ i })=1250  [/latex]




RS Aggarwal Class 10 Book pdf download Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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
[latex]\sum { { f }_{ i } } =80  [/latex][latex]\sum { { (f }_{ i } } \times { d }_{ i })=300  [/latex]




Class 10 Maths RS Aggarwal Solutions Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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
[latex]\sum { { f }_{ i } } =200  [/latex] [latex]\sum { { (f }_{ i } } \times { d }_{ i })=2760 [/latex]




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





MarksFrequency fiMid value xi[latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h }  \right)   [/latex]f× ui
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 601218272017651525 = A354555-2-10123-24-180203418
[latex]\sum { { f }_{ i } } =100  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=30 [/latex]




We have h = 10 and let assumed mean = 25.
A = 25, h = 10, ∑ fi= 100 and ∑(fi×ui)= 30
Class 10 Maths RS Aggarwal Solutions Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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 [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h }  \right)   [/latex] f× ui
4 – 88 – 1212 – 1616 – 2020 – 2424 – 2828 – 3232 – 362121525181213361014182226 = A3034-5-4-3-2-1012-10-48-45-50-180136
[latex]\sum { { f }_{ i } } =100  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=-152 [/latex]




A = 26, h = 4, ∑ fi= 100 and ∑(fi×ui)= -152
Maths RS Aggarwal Class 10 Solutions Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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  [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h }  \right)   [/latex]  f× ui
0 – 3030 – 6060 – 9090 – 120120 – 150150 – 180122134522011144575 = A105135165-2-10123-24-210524033
[latex]\sum { { f }_{ i } } =150  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=80 [/latex]




Thus, A = 75, h = 30, ∑ fi= 150 and ∑(fi×ui)= 80
RS Aggarwal Solutions Class 10 2018 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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 [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h }  \right)   [/latex]  f× ui
0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120120 – 140121815252615910305070 = A90110130-3-2-10123-36-36-150263027
[latex]\sum { { f }_{ i } } =150  [/latex][latex]\sum { { (f }_{ i } } \times { u }_{ i })=-4 [/latex]




Thus, A = 70, h = 20, ∑ fi= 120 and ∑(fi×ui)= -4
RS Aggarwal Solutions Class 10 2018 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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 [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h }  \right)   [/latex]  f× ui
0 – 1414 – 2828 – 4242 – 5656 – 7072135111672135 = A4963-2-1012-14-2101132
[latex]\sum { { f }_{ i } } =90  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=8 [/latex]




Thus, A = 35, ∑ fi= 90, h = 14 and ∑(fi×ui)= 8
RS Aggarwal Solutions Class 10 2018 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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 [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h }  \right)   [/latex] f× ui
10 – 1515 – 2020 – 2525 – 3030 – 3535 – 40568126312.517.522.5 = A27.532.537.5-2-10123-10-6012129
[latex]\sum { { f }_{ i } } =40  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=17 [/latex]




Thus, A = 22.5 and h = 5
∑ fi= 40 and   ∑(fi×ui)= 17
RS Aggarwal Solutions Class 10 2018 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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 [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h }  \right)   [/latex] f× ui
18 – 2424 – 3030 – 3636 – 4242 – 4848 – 546812842212733 = A394551-2-10123-12-80886
[latex]\sum { { f }_{ i } } =40  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=2 [/latex]




Thus, A = 33, h = 6, ∑ fi= 40 and ∑(fi×ui)=2
RS Aggarwal Solutions Class 10 2018 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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 [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h }  \right)   [/latex] f× ui
84 – 9090 – 9696 – 102102 – 108108 – 114114 – 120152220182025879399=A105111117-2-10123-30-220184075
[latex]\sum { { f }_{ i } } =120  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=81 [/latex]




Thus, A = 99, h = 6 and ∑ fi= 120, ∑(fi×ui)=2
RS Aggarwal Solutions Class 10 2017 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
Hence, Mean = 103.05.
Question 22:
Let h = 20 and assume mean = 550, we prepare the table given below:





AgeFrequency fiMid value xi [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-550 }{ 20 }  \right)   [/latex] f× ui
500 – 520520 – 540540 – 560560 – 580580 – 600600 – 6201495435510530550 = A570590610-2-10123-27-904615
[latex]\sum { { f }_{ i } } =40  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=-12 [/latex]




Thus, A = 550, h = 20, and ∑ fi= 40, ∑(fi×ui)=-12
RS Aggarwal Solutions Class 10 2017 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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 [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-42 }{ 5 }  \right)   [/latex] 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
[latex]\sum { { f }_{ i } } =70  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=-37 [/latex]




Thus, A = 42, h = 5, ∑ fi= 70 and ∑(fi×ui)=-37
RS Aggarwal Solutions Class 10 2017 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
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 [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-29.5 }{ 10 }  \right)   [/latex] 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
[latex]\sum { { f }_{ i } } =80  [/latex] [latex]\sum { { (f }_{ i } } \times { u }_{ i })=43 [/latex]




Thus, A = 29.5, h = 10, ∑ fi= 80 and ∑(fi×ui)=43
RS Aggarwal Solutions Class 10 2017 Chapter 9 Mean, Median, Mode of Grouped Data ex 9A
Hence, Mean = 34.87 years.





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