# 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 Exercise 9AExercise 9BExercise 9CExercise 9DExercise 9F 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 x-axis and their respective frequencies on y-axis, 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 x-coordinate 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 one-day 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
Draw a ‘less than type’ ogive from the above data. Find the median.

Solution

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows:

Here, N = 80 ⇒ N/2 = 40.

Mark the point a whose ordinate is 40 and its x-coordinate 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 x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows:

4. The heights of 50 girls of Class X of a school are recorded 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
Taking lower class limits of an x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows:

5. The monthly consumption of electricity (in units) of some families of a locality is given in the following frequency distribution:
 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 x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows:

6. The following table gives the production yield per hectare of wheat of 100 farms of a village.
 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 x-axis and their respective cumulative on y-axis, its ogive can be drawn as follows:

Here, N = 100

⇒ N/2 = 50.

Mark the point A whose ordinate is 50 and its x-coordinate 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 x-axis and their respective cumulative frequencies on y-axis, 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 x-axis and their respective cumulative frequencies on y-axis, 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 x-axis 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 A1 (140, 150), B1(144, 447), C1(148, 438), D1(152, 414), E1(156, 383), F1(160, 277), H1(168, 202), I1(172, 120) and J1 (176, 34),

Join A1B1, B1C1, C1D1, D1E1, E1F1, F1G1, G1H1, H1I1 and I1J1 with a free hand to get the ‘more than type’ series.

The two curves intersect at point L. Draw LM ⊥ OX cutting the x-axis at M. Clearly M = 166 cm

Hence, median = 166 cm