The solutions of Class 10 for Chapter 9 lets you practice different questions on a regular basis and is the key to scoring good marks in the exam. The RS Aggarwal Solutions Class 10 Chapter 9 Mean Median Mode Ex 9D isÂ the best study material that you can refer to while preparing for your Class 10 exam. All the solutions mentioned below are solved in a step-by-step way which helps you to understand the logic behind such questions easily.

These solutions can be referred by students whenever they have any doubt or get stuck while solving any particular question. To excel in the exams, you should practice the RS Aggarwal Class 10 SolutionsÂ for Maths regularly. Firstly, practice the difficult topics to clear your concepts and develop your problem-solving skills. It will help you solve questions in quick-time and score good marks.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 9â€“Â Mean Median Mode Ex 9D (9.4)

Q1: Find the median of the following data by making a â€˜less than giveâ€™

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |

No. of students | 5 | 3 | 4 | 3 | 3 | 4 | 7 | 9 | 7 | 8 |

Sol:

The frequency distribution table of less than type is given as follows:

Maks (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,

\(\Rightarrow \frac{N}{2}= 26.5\)

Mark the point A whose ordinate is 26.5 and its x-coordinate is 66.4.

The median of the data is 66.4.

Q2: The given distribution shows the number of wickets taken by the bowlers in one-day international cricket matches:

No. 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 |

No. of bowlers | 2 | 5 | 9 | 17 | 39 | 54 | 70 | 80 |

Sol:

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,

\(\Rightarrow \frac{N}{2}= 40 \)

Mark the point A whose ordinate is 40 and its x-coordinate is 76.

The median of the data is 76.

Q3: Draw a â€˜more thanâ€™ 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 |

Sol:

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

Marks (lower 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 the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows:

Q4: The height 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 girls | 5 | 8 | 9 | 12 | 14 | 2 |

Draw a â€˜more than typeâ€™ ogive for the above data.

Sol:

The frequency distribution table of more than type is as follows:

Height (in cm) (lower class limits) | Cumulative frequency (cf) |

More than 135 | 45+5=50 |

More than 140 | 37+8=45 |

More than 145 | 28+9=37 |

More than 150 | 16+12=28 |

More than 155 | 14+2=16 |

More than 160 | 2 |

Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows:

Q5: 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 |

No. of families | 3 | 8 | 15 | 40 | 50 | 30 | 10 |

Sol:

The frequency distribution table of more than type is as follows:

Height (in cm) (lower class limits) | 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 on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows:

Q6: 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 |

No. of farms | 2 | 8 | 12 | 24 | 38 | 16 |

Change the distribution to â€˜a more than typeâ€™ distribution and draw its ogive. Using ogive, find the median of the given data.

Sol:

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 |

Here, N=100,

\(\Rightarrow \frac{N}{2}= 50 \)

Mark the point A whose ordinate is 50 and its x-coordinate is 70.5.

The median of the data is 70.5

Q7: The table below shows the weekly expenditures on food of some household in a locality:

Weekly introduction (in Rs.) | No. of household |

100-200 | 5 |

200-300 | 6 |

300-400 | 11 |

400-500 | 13 |

500-600 | 5 |

600-700 | 4 |

700-800 | 3 |

800-900 | 2 |

Draw a â€˜less than type ogiveâ€™ and a â€˜more than type ogiveâ€™ for this distribution.

Sol:

The frequency distribution table of less than type is as follows:

Weekly expenditure ( in Rs)
(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 |

Now,

The frequency distribution table of more than type is as follows:

Weekly expenditure (in Rs) ( lower class limits) | Cumulative frequencies (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 |

Q8: From the following frequency distribution , prepare the â€˜more than ogiveâ€™.

Score | No. 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.

Sol:

From the given table, we may prepare the â€˜more thanâ€™ 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(700,50), 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 â€˜more than typesâ€™ series.

Here N= 230

\(\Rightarrow \frac{N}{2}= 115\)

From P(0,115), draw PQ meeting the curve at Q.

Draw QM meeting at M.

Clearly, QM = 590 units

Hence, median = 590 units

Q9: The marks obtained by 100 students of a class in an examination are given below:

Marks | No. 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) â€˜Less thanâ€™ series and

(ii) â€˜More thanâ€™ series

Hence find the median.

Sol:

(i) From the given table, we may prepare the â€˜less thanâ€™ frequency table as shown below:

Marks | No. 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), 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 â€˜less than typesâ€™ series.

(ii) More than series:

Marks | No. 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 â€˜more than typeâ€™ series.

The two curves intersect at point L. Draw \(LM \perp OX\) cutting the x-axis at M. Clearly, M= 29.5.

Hence, Median = 29.5

Q10: From the following data, draw two types of cumulative frequency curves and determine the median

Height ( in cm) | 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 |

Sol:

(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 | 330 |

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), 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 â€˜less than typesâ€™ series.

(ii) More than series:

Marks | No. 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 Â \(A_{1}(140,4510) ,B_{1}(144,447), C_{1}(148,447), D_{1}(152,414), E_{1}(156,383), F_{1}(160,341),G_{1}(164,277), H_{1}(168,202),I_{1}(172,120) and J_{1}(176,34)\)

Join, \(A_{1}B_{1}, B_{1}C_{1}, C_{1}D_{1},D_{1}E_{1},E_{1}F_{1},F_{1}G_{1},G_{1}H_{1},H_{1}I_{1},I_{1}J_{1}\) with a free hand to get â€˜more than typeâ€™ series.

The two curves intersect at point L. Draw \(LM \perp OX\) cutting x-axis at M. Clearly, M= 166 cm.

Hence, Median = 166 cm.

### Key Features of RS Aggarwal Class 10 Solutions Chapter 9 â€“Â Mean Median Mode Ex 9D (9.4)

- It will improve your speed and accuracy along with overall academic performance.
- RS Aggarwal Maths SolutionsÂ helps in practicing different types of questions.
- It gives an idea to solve difficult and tricky questions in a simple way.
- These solutions are based on the recent syllabus of the CBSE Class 10.