RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9E Class 10 Maths
Chapter Name  RS Aggarwal Chapter 9 Mean, Mode and Median 
Book Name  RS Aggarwal Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 9E Solutions
1. Find the median of the following data by making a ‘less than ogive’.
Marks 
0  10 
10  20 
20  30 
40  50 
50  60 
60  70 
70  80 
80  90 
90 100 
Number of students 
5 
3 
4 
3 
3 
4 
7 
9 
8 
Solution
The frequency distribution table of less than type is given as follows:
Marks (upper class limits) 
Cumulative Frequency (cf) 
Less than 10 
5 
Less than 20 
5 + 3 = 8 
Less than 30 
8 + 4 = 12 
Less than 40 
12 + 3 = 15 
Less than 50 
15 + 3 = 18 
Less than 60 
18 + 4 = 22 
Less than 70 
22 + 7 = 29 
Less than 80 
29 + 9 = 38 
Less than 90 
38 + 7 = 45 
Less than 100 
45 + 8 = 53 
Taking upper class limits of class intervals on xaxis and their respective frequencies on yaxis, its ogive can be drawn as follows:
Here, N = 53 ⇒ N/2 = 26.5
Mark the point A whose ordinate is 26.5 and its xcoordinate is 66.4.
Thus, median of the data is 66.4.
2. The given distribution shows the number of wickets taken by the bowlers in oneday international cricket matches:
Number of wickets 
Less than 15 
Less than 30 
Less than 45 
Less than 60 
Less than 75 
Less than 90 
Less than 105 
Less than 120 
Number of bowlers 
2 
5 
9 
17 
39 
54 
70 
80 

Solution
Taking upper class limits of class intervals on xaxis and their respective frequencies on yaxis, its ogive can be drawn as follows:
Here, N = 80 ⇒ N/2 = 40.
Mark the point a whose ordinate is 40 and its xcoordinate is 76.
Thus, median of the data is 76.
3. Draw a ‘more than’ ogive for the data given below which gives the marks of 100 students.
Marks 
0  10 
10  20 
20  30 
30  40 
40  50 
50  60 
60  70 
70  80 
No. of students 
4 
6 
10 
10 
25 
22 
18 
5 
Solution
The frequency distribution table of more than type is as follows:
Marks (upper class limits) 
Cumulative frequency (cf) 
More than 0 
96 + 4 = 100 
More than 10 
90 + 6 = 96 
More than 20 
80 + 10 = 90 
More than 30 
70 + 10 = 80 
More than 40 
45 + 25 = 70 
More than 50 
23 + 22 = 45 
More than 60 
18 + 5 = 23 
More than 70 
5 
Taking lower class limits of on xaxis and their respective cumulative frequencies on yaxis, its ogive can be drawn as follows:
Height (in cm) 
135  140 
140  145 
145  150 
150  155 
155  160 
160  165 
No. of students 
5 
8 
9 
12 
14 
2 
Draw a ‘more than type’ ogive for the above data.
Solution
The frequency distribution table of more than type is as follows:
Height (in cm) (lower class limit) 
Cumulative frequency (cf) 
More than 135 
5 + 45 = 50 
More than 140 
8 + 37 = 45 
More than 145 
9 + 28 = 37 
More than 150 
12 + 16 = 28 
More than 155 
14 + 2 = 16 
More than 160 
2 
Monthly Consumption (in units) 
140  160 
160  180 
180  200 
200  220 
220  240 
240  260 
260  280 
Number of Families 
3 
8 
15 
40 
50 
30 
10 
Prepare a ‘more than type’ ogive for the given frequency distribution.
Solution
The frequency distribution table of more than type is as follows:
Height (in cm) (lower class limit) 
Cumulative frequency (cf) 
More than 140 
3 + 153 = 156 
More than 160 
8 + 145 = 153 
More than 180 
15 + 130 = 145 
More than 200 
40 + 90 = 130 
More than 220 
50 + 40 = 90 
More than 240 
30 + 10 = 40 
More than 260 
10 
Taking the lower class limits of an xaxis and their respective cumulative frequencies on yaxis, its ogive can be drawn as follows:
Production Yield (kg/ha) 
50  55 
55  60 
60  65 
65  70 
70  75 
75  80 
Number of farms 
2 
8 
12 
24 
238 
16 
Change the distribution to a ‘more than type’ distribution and draw its ogive. Using ogive, find the median of the given data.
Solution
The frequency distribution table of more than type is as follows:
Production yield (kg/ha) (lower class limits) 
Cumulative frequency (cf) 
More than 50 
2 + 98 = 100 
More than 55 
8 + 90 = 98 
More than 60 
12 + 78 = 90 
More than 65 
24 + 54 = 78 
More than 70 
38 + 16 = 54 
More than 75 
16 
Taking the lower class limits on xaxis and their respective cumulative on yaxis, its ogive can be drawn as follows:
Here, N = 100
⇒ N/2 = 50.
Mark the point A whose ordinate is 50 and its xcoordinate is 70.5.
Thus, median of the data is 70.5.
7. The table given below shows the weekly expenditures on food of some households in a locality.
Weekly expenditure (in ₹) 
No. of house holds 
100  200 
5 
200  300 
6 
300  400 
11 
400  500 
13 
500  600 
5 
600  700 
4 
700  800 
3 
800  900 
1 
Draw a ‘less than type ogive’ and a ‘more than type ogive’ for this distribution.
Solution
The frequency distribution table of less than type is as follows:
Weekly expenditure (in ₹) (upper class limits) 
Cumulative frequency (cf) 
Less than 200 
5 
Less than 300 
5 + 6 = 11 
Less than 400 
11 + 11 = 22 
Less than 500 
22 + 13 = 35 
Less than 600 
35 + 5 = 40 
Less than 700 
40 + 4 = 44 
Less than 800 
44 + 3 = 47 
Less than 900 
47 + 2 = 49 
Taking the lower class limits on xaxis and their respective cumulative frequencies on yaxis, its ogive can be obtained as follows:
Now, the frequency distribution table of more than type is as follows:
Weekly expenditure (in ₹) (lower class limits) 
Cumulative frequency (cf) 
More than 100 
44 + 5 = 49 
More than 200 
38 + 6 = 44 
More than 300 
27 + 11 = 38 
More than 400 
14 + 13 = 27 
More than 500 
9 + 5 = 14 
More than 600 
5 + 4 = 9 
More than 700 
2 + 3 = 5 
More than 800 
2 
Taking the lower class limits on xaxis and their respective cumulative frequencies on yaxis, its ogive can be obtained as follows:
8. From the following frequency, prepare the ‘more than’ ogive.
Score 
Number of candidates 
400  450 
20 
450  500 
35 
500  550 
40 
550  600 
32 
600  650 
24 
650 – 700 
27 
700 – 750 
18 
750 – 800 
34 
Total 
230 
Also, find the median.
Solution
From the given table, we may prepare than ‘more than’ frequency table as shown below:
Score 
Number of candidates 
More than 750 
34 
More than 700 
52 
More than 650 
79 
More than 600 
103 
More than 550 
135 
More than 500 
175 
More than 450 
210 
More than 400 
230 
We plot the points A(750, 34), B(700, 52), C(650, 79), D(600, 103), E(550, 135), F(500, 175), G(450, 210) and H(400, 230).
Join AB, BC, CD, DE, EF, FG, GH and HA with a free hand to get the curve representing the ‘more than type’ series.
Here, N = 230
⇒ N/2 = 115
From P(0, 115), draw PQ meeting the curve at Q. Draw QM meeting at M.
Clearly, OM = 590 units
Hence, median = 590 units.
9. The marks obtained by 100 students of a class in an examination are given below:
Marks 
Number of students 
0  5 
2 
5  10 
5 
10 – 15 
6 
15 – 20 
8 
20  25 
10 
25 – 30 
25 
30  35 
20 
35  40 
18 
40 – 45 
4 
45  50 
2 
Draw cumulative frequency curves by using (i) ‘less than’ series and (ii) ‘more than’ series. Hence, find the median.
Solution
(i) From the given table, we may prepare the ‘less than’ frequency table as shown below:
Marks 
Number of students 
Less than 5 
2 
Less than 10 
7 
Less than 15 
13 
Less than 20 
21 
Less than 25 
31 
Less than 30 
56 
Less than 35 
76 
Less than 40 
94 
Less than 45 
98 
Less than 50 
100 
We plot the points A(5, 2), B(10, 7), C(15, 13), D(20, 21), E(25, 31), F(30, 56), G(35, 76) and H(40, 94), I(45, 98) and j(50, 100).
Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing the ‘less than type’ series.
(ii) More than series:
Marks 
Number of students 
More than 0 
100 
More than 5 
98 
More than 10 
93 
More than 15 
87 
More than 20 
79 
More than 25 
69 
More than 30 
44 
More than 35 
24 
More than 40 
6 
More than 45 
2 
Now, on the same graph paper, we plot the points (0, 100), (5, 98), (10, 94), (15, 76),, (20, 56), (25, 31), (30, 21), (35, 13), (40, 6) and (45, 2).
Join with a free hand to get the ‘more than type’ series.
The two curves intersect at point. Draw LM ⊥ OX cutting the xaxis at M.
Clearly, M = 29.5
Hence, Median = 29.5
10. From the following data, draw the two types of cumulative frequency curves and determine the median:
Marks 
Frequency 
140  144 
3 
144  148 
9 
148  152 
24 
152  156 
31 
156  160 
42 
160  164 
64 
164 – 168 
75 
168 – 172 
82 
172 – 176 
86 
176  180 
34 
Solution
(i) Less than series:
Marks 
No. of students 
Less than 144 
3 
Less than 148 
12 
Less than 152 
36 
Less than 156 
67 
Less than 160 
109 
Less than 164 
173 
Less than 168 
248 
Less than 172 
230 
Less than 176 
416 
Less than 180 
450 
We plot the points A(144, 3), B(148, 12), C(152, 36), D(156, 67), E(160, 109), F(164, 173), G(168, 248) and H(172, 330), I(176, 416) and J(180, 450).
Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing the ‘less than type’ series.
(ii) More than series:
Marks 
Number of students 
More than 140 
450 
More than 144 
447 
More Than 148 
438 
More than 152 
414 
More than 156 
383 
More than 160 
341 
More than 164 
277 
More than 168 
202 
More than 172 
120 
More than 176 
34 
Now, on the same graph paper, we plot the points A_{1} (140, 150), B_{1}(144, 447), C_{1}(148, 438), D_{1}(152, 414), E_{1}(156, 383), F_{1}(160, 277), H_{1}(168, 202), I_{1}(172, 120) and J_{1} (176, 34),
Join A_{1}B_{1}, B_{1}C_{1}, C_{1}D_{1}, D_{1}E_{1}, E_{1}F_{1}, F_{1}G_{1}, G_{1}H_{1}, H_{1}I_{1} and I_{1}J_{1 }with a free hand to get the ‘more than type’ series.
The two curves intersect at point L. Draw LM ⊥ OX cutting the xaxis at M. Clearly M = 166 cm
Hence, median = 166 cm