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

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 9
Chapter NameMean, Median, Mode of Grouped Data
Exercise9E
CategoryRS Aggarwal Solutions

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

Question 1:
From the given table, we may prepare the cumulative frequency table as shown below:

Class intervalC.F
Less than 2207
Less than 24010
Less than26016
Less than 28024
Less than 30026
Less than 32030

On a graph paper, we plot the point (220, 7), (240, 10), (260, 16), (280, 24), (300, 26), (320, 30)
We join these points successively with a freehand to get the cumulative frequency curve or an ogive
Here N = 30, N2= 15
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E 1.1
Take point (0, 15) on the y-axis and draw AP || x-axis meeting the curve at P. Draw PM ⊥  x -axis intersecting the x-axis at M. Then OM = 258 and hence median = 258.

Question 2:
Cumulative frequency table is as follows:

MarksC.F
Marks less than 103
Marks less than 2011
Marks less than 3028
Marks less than 4048
Marks less than 5070

Scale :
Horizontal: 1 small div = 10 marks
Vertical : 1 small div = 10 students
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E 2.1
The points (10, 3), (20, 11), (30, 28), (40, 48), (50, 70) are plotted and joined as shown above. This is required cumulative curve.
N = 70, N2= 35
Ob vertical line OY, take OA = 35
Through A, a horizontal line AP is drawn meeting the graph at P
Through P, a vertical line PM is drawn.
Now, OM = 34, therefore Median = 34

Question 3:
The following table gives height in metre of 360 tress:

Height in mNumber of trees
Less than 7Less than 14Less than 21Less than 28Less than 35Less than 42Less than 49Less than 56254595140235275320360

The points (7, 25), (14, 45), 21, 95), (28, 140), (35, 235), (42, 275), (49, 320), (56, 360) are plotted and joined one by one as shown in the figure.
This is the required cumulative curve
N = 360, N2= 180
At y = 180, affix A.
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E 3.1
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 31.
Hence median = 31

Question 4:
Less than series

Less thanCapital(in lakh of Rs)No. of Companies(C.F.)
Less than 10Less than 20Less than 30Less than 40Less than 50Less than 60Less than 70Less than 8025122338454750

Scale: Horizontal : 1 small div = capital of 1 lakh of Rs
Vertical: 1 small div. = 1 company.
Plot the points (10, 2), (20, 5), (30, 12), (40, 23), (50, 38), (60, 45), (70, 47) and (80, 50).
Through A(0, 25), AP is drawn parallel to OX and PM  OX, OM = 41. Hence median = 41.
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E 4.1

Question 5:
More than series

ScoreNo of Candidates
More than 400230
More than 450210
More than 500175
More than 550135
More than 600103
More than 60079
More than 70052
More than 75034

We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E 5.1
Hence, N = 230, N2= 115
Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM x-axis intersecting x-axis at M
Then, OM = 590
Hence median = 590


Question 6:
(i) Less than series:

MarksNumber of Students
Less than 5Less than 10Less than 15Less than 20Less than 25Less than 30Less than 35Less than 40Less than 45Less than 502713213156769498100

Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E 6.1
Join these points free hand to get the curve representing “less than” cumulative curve.
(ii) From the given table we may prepare the ‘more than’ series as shown below

MarksNumber of Students
More than 45More than 40More than 35More than 30More than 25More than 20More than 15More than 10More than 5More than 02624446979879398100

Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)
Join these points free hand to get required curve
Here N = 100, N2= 50
Two curves intersect at point P(28, 50)
Hence, the median = 28


Question 7:
We may prepare less than series and more than series
(i) Less than series

Height in (cm)Frequency
Less than 140Less than 144Less than 148Less than 152Less than 156Less than 160Less than 164Less than 168Less than 172Less than 176Less than 18003123667109173248330416450

Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)
(ii) More than series

Height in (cm)C.F.
More than 140More than 144More than 148More than 152More than 156More than 160More than 164More than 168More than 172More than 176More than 180450447438414383341277202120340

Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E 7.1
The curves intersect at (167, 225).
Hence, 167 is the median.

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