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

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E 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 9E Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E Maths book pdf download. Now you will get step by step solution to each question.

### 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:

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

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:

Scale :
Horizontal: 1 small div = 10 marks
Vertical : 1 small div = 10 students

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:

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.

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

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.

Question 5:
More than series

We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)

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:

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 the curve representing “less than” cumulative curve.
(ii) From the given table we may prepare the ‘more than’ series as shown below

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

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

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)

The curves intersect at (167, 225).
Hence, 167 is the median.

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