Vedanta Excel in Mathematics Book 9 Final (2078)

Statistics

The mark 18 is repeated only one time. So, 1 is the frequency of 18.
Thus, a frequency is the number of times a value occurs. Data and their frequencies
can be presented in a table called frequency table.

Marks Tally marks Frequency

10 || 2
12 |||| 4
15 ||| 3
18 | 1

Tallying is a system of recording and counting results using diagonal lines grouped in
fives. Each time five is reached, a horizontal line is drawn through the tally marks to
make a group of five. The next line starts a new group.
For example, the tally marks of frequency 5 is ||||, frequency 6 is |||| |, 7 is |||| ||,
and so on.

17.4 Grouped and continuous data

Let’s consider the following marks obtained by 20 students in a test of mathematics.

25, 38, 21, 18, 7, 28, 49, 17, 27, 36

26, 45, 32, 16, 24, 39, 20, 33, 40, 35

The above mentioned data are called individual or discrete data. Another way of
organising data is to present them in a grouped form. For grouping the given data, we
should first see the smallest value and the largest value. We have to divide the data
into an appropriate class–interval. The numbers of values falling within each class–
interval give the frequency. For example,

Marks Tally marks Frequency

0 – 10 | 1
10 – 20 ||| 3
20 – 30 |||| || 7
30 – 40 |||| | 6
40 – 50 ||| 3

Total 20

In the above series, 7 is the smallest value and 49 is the largest value. So, the data are
grouped into the interval of 0 – 10, 10 – 20, ... 40 – 50, so that the smallest and the
largest values should fall in the lowest and the highest class–interval respectively.

Let’s consider a class–interval 10 – 20.

Here, 10 is called the lower limit and 20 is the upper limit of the class–interval. The
difference between the two limit is called the length or height of each class–interval.
For example,

in 10 – 20 the length of the class–interval is 10.

Again, let’s take class–intervals, 0 – 10, 10 – 20, 20 – 30, …

Here, the upper limit of a pervious class–interval has repeated as the lower limit of the
consecutive next class–interval. Such an arrangement of data is known as grouped
and continuous data.

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17.5 Cumulative frequency table

The word ‘cumulative’ is related to the word ‘accumulated’, which means to ‘pile up’.
The table given below shows the marks obtained by 20 students in a mathematics test
of full marks 50 and the corresponding cumulative frequency of each class–interval.

Marks Frequency Cumulative 1 student obtained marks 0 to less than 10.
(f) frequency (c.f) 4 students obtained marks 0 to less than 20.
0 – 10 11 students obtained marks 0 to less than 30.
10 – 20 1 1 17 students obtained marks 0 to less than 40.
20 – 30 3 1+3=4 20 students obtained marks 0 to less than 50.
30 – 40 7 4 + 7 = 11
40 – 50 6 11 + 6 = 17
Total 3 17 + 3 = 20

20

Thus, cumulative frequency corresponding to a class–interval is the sum of all
frequencies up to and including that class–interval.

17.6 Graphical representation of data

We have already discussed to present data in frequency distribution tables.
Alternatively, we can also present data graphically. Different types of diagrams are
used for this purpose. Here, we shall discuss three types of diagrams: histogram, line
graph, and pie chart.

Histogram
Statistical data can be represented by various types of diagrams, such as, bar diagram,
pie-chart, line graph, etc. A histogram is a graphical representation of a continuous
frequency distribution. For this purpose, we draw rectangular bars with class intervals
bases and corresponding frequencies as heights.
Study the following steps to construct a histogram.

(i) If the given frequency distribution has inclusive classes, change it into exclusive.
(i.e. continuous) classes.

(ii) Choose a suitable scale and mark the class intervals on the x-axis and the
corresponding frequencies on the Y-axis.

(iii) Draw adjacent rectangles on x-axis with class interval as base and the
corresponding frequencies as height.

Worked-out examples

Example 1: The table given below shows the marks obtained by 36 students in
mathematics. Represent the data by drawing a histogram.

Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80

No. of students 2 3 5 8 10 5 3

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Solution: Y Histogram of marks obtained by 45 students

16

15

14
13

12

No. of Students 11

10

9

8
7

6

5

4

3

2

1
0
10 20 30 40 50 60 70 80 X

Marks

In the case of mid-points of class intervals are given, we should first find the class intervals
of each mid-point.

Example 2 Construct a histogram from the data given in the table below.

Mid-point 7.5 12.5 17.5 22.5 27.5 32.5 37.5

Solution: Frequency (f) 3 5 7 9 11 4 1

The class intervals of each mid-point are given in the table below.

Class interval 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40

Frequency (f) 4 7 10 8 13 6 2

Y Histogram of the given data
17
Frequency 16
15
10 15 20 25 30 35 40 X
14 Class interval
13

12
11

10
9
8
7

6
5
4

3
2
1
0

5

In the case of inclusive frequency distribution, we should first change it into exclusive
(or continuous) class intervals.

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Example 3: The table given below shows the heights of 75 plants of a garden. Construct
a histogram to represent the data.

Heights (in inch) 1 – 9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69

No. of plants 15 6 18 11 14 9 2
Solution:

The given frequency distribution is inclusive type. So, it should be changed into exclusive
(or continuous) class interval with the help of a correction factor.

Correction factor = Lower limit of a class interval – Upper limit of previous class interval
2
10 – 9 1
= 2 = 2 = 0.5

Then, the correction factor 0.5 is subtracted from the lower limit and added to the upper

limit of each class interval.

Height in inch (Inclusive class) Height in inch (Exclusive class) No. of plants
1–9 0.5 – 9.5 15
9.5 – 19.5 6
10 – 19 18
20 – 29 19.5 – 29.5 11
30 – 39 29.5 – 39.5 14
40 – 49 39.5 – 49.5 9
50 – 59 49.5 – 59.5 2
60 - 69 59.5 – 69.5

19 No. of plants Histogram of heights of 75 plants
18
17 0 . 5 9 . 5 19.5 29.5 39.5 49.5 59.5 69.5
16 Heights (in cm)
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0

Line graphs

A line graph, also known as a line chart, is a type of graph used to visualise the value
of something that changes over time.

We can also present the data by plotting the corresponding frequencies in the graph
paper. The line so obtained by joining the points is called the line graph.

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While constructing a line graph, the frequencies of the items are plotted along the
y–axis. The line graph given in the following example shows the daily sales of a shop
for six days of a week.

Example 4: The table given below shows the daily sales of a shop for six days of a week.

Day Sun Mon Tues Wed Thurs Fri
Sales (in Rs) 3000 4500 2000 4000 2500 6000

(i) Construct a line graph for the frequency table.

(ii) On which days were the sales above Rs 4,000.

Solution:
(a)

6000

5000
4000
3000

2000
1000

0
Sun Mon Tue Wed Thus Fri

b) The sales were above Rs 4,000 on Monday and Friday.

EXERCISE 17.1

General section

1. a) Write a few paragraphs about the application of statistics.
b) Define primary, secondary, raw, and array data.
c) What do you mean by frequency of an observation in a data?
d) Define cumulative frequency.

e) What is histogram?

2. a) From the marks given below obtained by 20 students in mathematics, construct a
frequency distribution table with tally marks.

15, 18, 12, 16, 18, 10, 15, 16, 15, 12
10, 12, 15, 12, 16, 18, 12, 15, 12, 16

b) The amount of money (in Rs) brought by 30 students of class IX for tiffin is given
below. Construct the frequency distribution table with tally bars.

30, 20, 50, 40, 25, 50, 40, 35, 20, 50
40, 50, 35, 25, 30, 40, 50, 25, 40, 30
35, 25, 40, 50, 40, 30, 25, 50, 35, 40

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3. a) The marks obtained by 40 students in mathematics in the final examination of
class 9 are given below. Group the data into the class intervals of length 10 and
construct a cumulative frequency distribution table.

42, 68, 80, 45, 92, 36, 8, 17, 49, 30 5, 26, 98, 74,

53, 65, 72, 28, 55, 46 86, 70, 62, 27, 16, 44, 85, 59,

51, 73 66, 78, 38, 81, 97, 77, 69, 45, 33, 67.
b) The weights (in kg) of 30 teachers of a school are given below. Group the data in the

class interval of length 5 and construct the frequency distribution table.

71, 53, 60, 50, 64, 74, 59, 67, 61, 58, 65, 70, 73, 51, 63,
66, 70, 50, 60, 70, 56, 62, 52, 67, 59, 77, 64, 78, 57, 57

4. a) The given histogram shows the distribution of No. of students 30
marks obtained by the students of class - IX in 25
mathematics mock test. Answer the following 20
questions. 15
10
(i) Find the number of students who obtained
marks more than 70. 5

(ii) Find the number of students who obtained 40 50 60 70 80 90
marks between 50 and 80 marks. Marks obtained

(iii) Find the number of students who obtained at
most 60 marks.

b) The adjoining histogram is drawn against the 60
number of students and their weights. Study 50
it and answer the following question.

(i) How many students are there who have the No. of students 40
weight less than 30 kg? 30

(ii) In which weight group are there maximum 20
number of students and what is this number? 10

(iii) In which weight group are there minimum 20 25 30 35 40 45
number of students and what is this number? Weight (in kg)

Creative section

5. a) Draw a histogram to represent the data given below in the table.

Age (in years) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of people 6 10 16 20 12 7 3

b) The marks obtained by 70 students in mathematics are given below. Construct a
histogram to represent the marks.

Marks 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90

No. of Students 4 7 12 18 10 8 9 2

c) The daily wages of 60 workers are shown in the table given below.

Wages (in Rs) 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99
No. of people 10 15 6 17 8 4

Construct a histogram to represent the wages of the workers.

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d) The values of diastolic blood pressure of 80 people of 20-50 years are shown in the
table given below.

Blood pressure (mmHg) 70 – 75 75 – 80 80 – 85 85 – 90 90 – 95 95 – 100

No. of people 10 17 23 16 9 5

Construct a histogram to show the above data.

e) Draw a histogram to represent the data given below in the table.

Mid-point 12 16 20 24 28 32 36 40
Frequency 5 11 17 13 15 20 12 8

6. a) The heights (in cm) of 30 pupils are given below:
135, 140, 120, 160, 110, 107, 150, 136 102, 113,
116, 109, 129, 124, 131, 147, 142, 123, 118, 152
128, 143, 127, 134, 115, 126, 136, 147, 129, 139

Construct a frequency distribution table of continuous class with length 10 and
draw a histogram to represent the data.
b) The daily wages (in Rs) of 40 employees of a factory are given below:
415, 440, 405, 412, 402, 418, 446, 406, 435, 422,
425, 430, 428, 416, 445, 421, 449, 437, 441, 419,
432, 426, 443, 436, 424, 438, 423, 448, 414, 427,
443, 412, 448, 447, 425, 415, 445, 434, 440, 428

Construct a frequency distribution table of continuous class with length 10 and
draw a histogram to represent the data.

7. a) The adjoining line graph
shows the average rain fall in
mm during 6 months of the
year 2020.

(i) On which month was
the minimum average
rainfall? On which
month was it maximum?

(ii) How much was the
average rainfall recorded
in August?

(iii) Write a paragraph about
the common trend of
rain fall during these
months.

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b) The table given below shows the velocity of a bus at different interval of time. Draw
a line graph to show the velocity–time graph.

Time (in second) 5 10 15 20 25 30
Velocity (in m/s) 5 15 20 30 10 25

c) The table given below shows the number of students who secured A+ grade in
mathematics of a school in SEE from 2072 B.S. to 2076 B.S.

Year (in B. S.) 2074 2075 2076 2077 2078

No. of students 10 3 7 12 5

(i) Construct a line graph for the above frequency table.

(ii) In which years the number of students who secured A+ grade were les than 10?

d) The table given below shows the daily expense of a family.

Day Sunday Monday Tuesday Wednesday Thursday Friday
500
Expense (in Rs) 400 600 1300 1000 1200

(i) Construct a line graph for the above data.

(ii) On which days were the expenses more than Rs 1,000?

Project work

8. a) Measure the weights of every students in your class. Show the weights in a
frequency distribution table of continuous class of length 10. Then, present the
data in a histogram.

b) Show the marks obtained by the students of your class in mathematics in recently
conducted exam in a frequency distribution table of continuous class of length 5.
Then, show it in a histogram.

c) Collect the data of the number students who passed SEE from the last five years to
the recent year in your school. Construct a line graph to show the data.

d) Collect the data from the reliable website about the number deaths in the SAARC
countries due to Corona Virus during 2020. Show the data in a line graph.

Pie-charts

A pie-chart is also a graphical presentation of data in which a circle is divided into
sectors and the sectors represent the frequencies.
We draw a pie-chart by following the steps given below:

Add all the frequencies and write each frequency as a fraction of the total
frequency.

Change each fraction into a number of degrees by multiplying 360q. (There are

360q in a circle) corresponding frequency
Total frequency
i.e., number of degrees of each fraction = × 360°

Tabulate the angles in ascending or descending order.

Draw a circle of the convenient size. Then, draw a radius as a starting point.

Use a protractor to construct the angles at the centre corresponding to each
sector.

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Worked-out examples

Example 1: The table given below shows the number of students of a school who
obtained different grades in mathematics in SEE 2078. Represent the data
in a pie chart.

Grades A+ A B+ B C+

No. of students 10 16 5 63

Solution:

Here, total number of students = 10 + 16 + 5 + 6 + 3 = 40

Number of degrees of the sector of A+ grade = 10 u 360q = 90q
40

Number of degrees of the sector of A grade = 16 u 360q = 144q
40

Number of degrees of the sector of B+ grade = 5 u 360q = 45q A-grade
40

Number of degrees of the sector of B grade = 6 u 360q = 54q A+-grade 144°
40
90° 27° C-grade
Number of degrees of the sector of C grade = 3 u 360q = 27q B+-grade
40
B-grade
Now, tabulating the angles in descending order 45°

Grades A A+ B B+ C 54°

Angle 144q 90q 54q 45q 27q

Example 2: The monthly budget of a family is shown in
the given pie chart. If the total expenditure is
Rs 14,400, calculate the expenditure on each
item and show in a table.

Solution:

Here, total expenditure = Rs 14,400

Expenditure on food = Rs 14400 u 120 = Rs 4,800 Whole circle represents the
360

Expenditure on education = Rs 14400 u 95 = Rs 3,800 total expenditure.
360
i.e., 360q represents Rs 14,400
14400 14400
Expenditure on rent = Rs 360 u 75 = Rs 3,000 1q represents Rs 360

Expenditure on fuel = Rs 14400 u 40 = Rs 1,600 120q represents Rs13464000u 120
360

Expenditure on miscellaneous = Rs 14400 u 30 = Rs 1,200
360

Now, the table given below shows the actual expenditure.

Items Food Education Rent Fuel Miscelleneous
Expenditure (Rs)
4800 3800 3000 1600 1200

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EXERCISE 17.2

General section

1. a) Define pie-chart.

b) The monthly expenditure of a man is Rs 20,000. If he spends Rs 6,000 on food, what
is the central angle of the sector of pie-chart representing expense on food?

c) The Government of Nepal allocated 10% of total budget in Education, Science and
Technology (EST) sector in the fiscal year 2076 B. S. What is the central angle of the
sector of pie-chart denoting the EST?

d) There are 400 students in a school. The central angle of sector representing the
students going by bus in a pie-chart is 135°. Find the number of students who go to
the school by bus.

Creative section

2. a) The table given below shows the monthly income of a family from different sources.
Represent the data in a pie chart.

Source Salary House rent Business Agriculture Others
Income (Rs) 5500
4000 4100 2400 2000

b) The monthly expenditure of a family in different headings is given below.
Represent the data in a pie chart.

Item Education House rent Food Others
Amount (Rs)
8000 14000 10000 4000

c) The monthly budget of a family is given below.

Food – Rs 2100 Clothing – Rs 3300

Miscellaneous – Rs 2250 Saving – Rs 3150

Represent the above data in a pie chart.

d) The monthly salary sheet of staffs of an office is given below. Represent the data in
a pie-chart with radius at least 3 cm.

Designation Manager Officer Clerk Security guard Sweeper
Salary (Rs) 30000 25000 8000
15000 12000

e) A shoe store sold the shoes of the following brands in a month. Draw a pie-chart to
show the data.

Brand No. of pair of shoes
Addidas 260
Nike 240
Goldstar 180
Fila 80
Converse 40

f) The present data of world’s consumption of energy is given below.

Coal – 25 % Natural gas – 20 %

Hydro – 10 % Oil – 35 %

Nuclear – 5 % Others – 5%

Represent the above data in a pie chart.

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3. a) A person’s monthly expenditure is Rs 16,200. Rent
The diagram on the right is a pie chart showing his
expenditure on different headings. Work out how Food 75°
much was spent under each heading.
150° Miscellaneous
25°

Education Transportation
50°
60°

b) The diagram on the right is a pie chart showing the Salaries
expenses of a small manufacturing firm. The total
expenses were Rs 3,60,000 in a month. Calculate the 150°
expenditure on each heading.
Raw Miscellaneous

materials 40°

120° Fuel
50°

c) The pie chart alongside shows the percentage of types of Walking
transportation used by 400 students to come to school.
Bus 40%
(i) How many students come to school by bus?
30% Van
(ii) How many students come to school by bicycle? Bicycle 10%

(iii) How many students do not come to school by school
van?

d) The pie chart given alongside shows the votes secured by X Z
three candidates X, Y, and Z in an election. If X secured 120° 140°
5760 votes,
Y
i) how many votes did Z secure?

ii) who secured the least number of votes? How many
votes did he secure?

e) The given pie chart shows the composition of different Polyester
materials in a type of cloth in percentage. 144°

i) Calculate the percentage of each material found in the Cotton Nylon
cloth. 90° 54°

ii) Calculate the weight of each material contained by a Others
bundle of 50 kg of cloth. 72°

Project work

4. a) Make a group of your friend. Conduct a survey to find the number of students from
class 6 to 10 in your school. Present the data you obtained in a pie-chart.

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b) Collect the number of students of your school who secured A+, A, B+, B, etc. grades
in Mathematics in SEE last year and show the data in pie-chart.

c) Find the number of students who come to school by bus, van, bike, bicycle,
walking, etc. and show the data in a pie chart.

d) Visit the available website and search the budget allocated by government of Nepal
in three bigger sectors this year and show the data in pie-chart.

17.7 Ogive (Cumulative frequency curve)

The sum of the frequencies of all the values up to a given value is known as
cumulative frequency. It is denoted by c.f. Ogive is a graphic presentation of the
cumulative frequency distribution of continuous series.

In the case of a continuous series, if the upper limit (or the lower limit) of each
class interval is taken as x-coordinate and its corresponding c.f. as y-coordinate and
the points are plotted in the graph, we obtain a curve by joining the points with
freehand. Such curve is known as ogive or cumulative frequency curve.

Since ‘less than’ c.f. and ‘more than’ c.f. are the two types of cumulative frequency
distributions, there are two types of ogive. They are less than ogive and more than
ogive.

(i) Less than ogive (or less than cumulative frequency curve)
When the upper limit of each class interval is taken as x-coordinate and its
corresponding frequency as y-coordinate, the ogive so obtained is known as
less than ogive (or less than cumulative frequency curve). Obviously, less than
ogive is an increasing curve, sloping upwards from left to right and has the
shape of an elongated S.

(ii) More than ogive (or more than cumulative frequency curve)

When the lower limit of each class interval is taken as x-coordinate and its
corresponding frequency as y-coordinate, the ogive so obtained is known as
more than ogive (or more than cumulative frequency curve). More than ogive
is a decreasing curve sloping downwards from left to right and has the shape of
an elongated S, upside down.

17.8 Construction of less than ogive and more than ogive

Study the following steps to construct a less than ogive.
(i) Make a less than cumulative frequency table.
(ii) Choosethesuitablescaleandmarktheupperclasslimitsofeachclassintervalalong

x-axis and cumulative frequencies along y-axis.
(iii) Plot the coordinates of (the upper limit, less than c.f.) on the graph.
(iv) Join the point by freehand and obtain a less than ogive.

In the case of more than ogive, we should prepare the more than cumulative
frequency table. The lower class limits of each class interval are marked on x-axis.
Then, the process is similar to the construction of less than ogive.

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Worked-out examples

Example 1: Look at the given ‘less than’ ogive and Cumulative 50
answer the following questions. No. of students 40
(i) How many students are there in all? 30
(ii) How many students obtained less 20
than 20 marks? 10
(iii) In which class interval of obtained
marks, there are the maximum 10 20 30 40 50
number of students? Marks obtained

Solution:
(i) Total number of students is 40

(ii) Number of students who obtained less than 20 marks is 10.

(iii) The class interval of marks obtained by maximum number of students is (20 - 30).

Example 2: The table given below shows the marks obtained by 50 students in
Mathematics. Construct (i) less than ogive and (ii) more than ogive.

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of students 3 8 10 17 6 2 4
Solution:

Less than cumulative frequency table.

Marks No. of students (f) Upper limit less than c.f. Coordinates (x, y)

0 – 10 3 10 3 (less than 10) (10, 3)

10 – 20 8 20 11 (less than 20) (20, 11)

20 – 30 10 30 21 (less than 30) (30, 21)

30 – 40 17 40 38 (less than 40) (40, 38)

40 – 50 6 50 44 (less than 50) (50, 44)

50 – 60 2 60 46 (less than 60) (60, 46)

60 - 70 4 70 50 (less than 70) (70, 50)

The less than ogive

Less than cumulative frequencies 60

50 (70, 50)
(60, 46)

40 (50, 44)
(40, 38)

30

20 (30, 21)

10 (20, 11)

0 (10, 3)
10 20 30 40 50 60 70

Upper limits

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More than cumulative frequency table

Marks No. of students (f) Lower limit more than c.f. Coordinates
(0, 50)
0 – 10 3 0 50 (more than 0) (10, 47)
(20, 39)
10 – 20 8 10 50 – 3 = 47 (more than 10) (30, 29)
(40, 12)
20 – 30 10 20 47 – 8 = 39 (more than 20) (50, 6)
(60, 4)
30 – 40 17 30 39 – 10 = 29 (more than 30)

40 – 50 6 40 29 – 17 = 12 (more than 40)

50 – 60 2 50 12 – 6 = 6 (more than 50)

60 - 70 4 60 6 – 2 = 4 (more than 60)

More than cumulative frequencies The more than ogive

60
(0, 50)

50 (10, 47)
40 (20, 39)

30 (30, 29)

20
(40, 12)

10 (50, 6) (60, 4)

0 10 20 30 40 50 60 70
Lower limits

If we draw both the less than ogive and more than ogive of a distribution on the same graph
paper, they intersect at a point. The foot of the perpendicular drawn from the point of
intersection of two ogives to the x-axis gives the value of median of the distribution.

For example,

Marks No. of students (f) Upper limit Less than c.f. Lower limit More than c.f.
0 – 10 3 10 30 50
10 – 20 8 20 11 10 47
20 – 30 10 30 21 20 39
30 – 40 17 40 38 30 29
40 – 50 6 50 44 40 12
50 – 60 2 60 46 50 6
60 – 70 4 70 50 60 4

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60

Less than cumulative frequencies 50 (0, 50) (70, 50)
(10, 47) (60, 46)
(50, 44)
40 (20, 39) (40, 38)

30 (30, 29)

20 (30, 21)

10 (20, 11) (40, 12)
(50, 6) (60, 4)
(10, 3)
0 10 20 30 40 50 60 70

Upper limits

From the graph, the perpendicular drawn from the point of intersection of two ogives meets
x-axis at 32.65 (approx.) units from the origin. So, the required median of the given
distribution is 32.65.

EXERCISE 17.3
General Section

1. a) Looking at the given less than cumulative Cumulative 60
frequency curve, answer the following numbers of people 50
questions. 40
30
(i) How many people are participated in the 20
survey? 10

(ii) How many people are there, who are less 10 20 30 40 50 60
than 30 years? Ages (in years)

(iii) How many people are there in the age group
(30 - 40) years?

b) Looking at the given less than ogive curve, Numbers of students 35 Y
answer the following questions. 30
25
(i) How many students are there who obtained 20
less than 50 marks? 15
10
(ii) In which class interval do the marks obtained
by the maximum member of students fall? 5

(iii) Find the number of students who obtained 10 20 30 40 50 60 70
marks in the interval (10 - 20)? Marks obtained less than

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2. a) Looking at the more than ogive curve given Numbers of students 24
alongside, answer the following questions. 20
16
(i) How many students are there? 12
8
(ii) How many students are there who obtained
more than 20 marks? 4

10 20 30 40 50 60
Marks obtained (more than)

b) Look at the given more than ogive and answer Numbers of workers 90
the following questions. 75
60
(i) How many workers are there altogether? 45
30
(ii) How many workers have monthly salary 15
more than Rs 15,000?

Creative Section 5 10 15 20 25 30 35
Monthly salary (in Rs 1,000)

4. a) Draw a ‘less than’ ogive from the data given below.

Data 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency 5
8 15 10 6 3

b) Construct a ‘more than’ ogive from the data given below.

Marks 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100

No. of students 6 10 16 15 10 8 4 3

c) The table given below shows the marks obtained by 60 students in mathematics.
Construct (i) less than ogive and (ii) more than ogive.

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students
4 10 20 15 6 5

5. a) Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find
the median of the distribution.

Data 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
Frequency 6 12 18 20 15 9

b) The table given below shows the daily wages 40 workers of a factory.

Wages (in Rs) 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 75 – 80
No. of workers 4 6 10 12 5 3

Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find
the median of the distribution.

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6. a) The following are the marks obtained by the students in mathematics in an
examination.

49, 23, 64, 35, 44, 60, 79, 20, 41, 50

32, 34, 25, 57, 40, 54, 23, 60, 63, 68

(i) Make a frequency distribution table of class interval 10.

(ii) Construct a less than ogive from the above data.

b) The following are the ages (in years) of patients admitted in a month in a hospital.

15, 12, 23, 35, 46, 57, 18, 12, 39, 51

32, 42, 59, 18, 38, 48, 40, 32, 54, 49

(i) Make a frequency distribution table of class size 10.

(ii) Construct a more than ogive from the above data.

17.9 Measures of central tendency

The measure of central tendency gives a single central value that represents the
characteristics of entire data. A single central value is the best representative of the
given data towards which the values of all other data are approaching.

Average of the given data is the measure of central tendency. There are three types
of averages which are commonly used as the measure of central tendency. They are:
mean, median and mode.

17.10 Arithmetic mean

Arithmetic mean is the most common type of average. It is the number obtained by
dividing the sum of all the items by the number of items.

sum of all the items
i.e., mean = the number of items

(i) Mean of non–repeated data

If x represents all the items and n be the number of items, then mean ( x ) = 6x
n

Worked-out examples

Example 1: Calculate the average of the following marks obtained by 7 students of a
class in mathematics.

87, 63, 45, 72, 95, 35, 79

Solution:

Here, 6x = 87 + 63 + 45 + 72 + 95 + 35 + 79 = 476 and n = 7

Now, mean ( x ) = 6x = 476 = 68
n 7

Example 2: If the average of the following wages received by 5 workers is Rs 400, find
the value of p.

300, 380, p, 420, 460.

Solution:

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Here, 6x = 300 + 380 + p + 420 + 460 = 1560 + p and n = 5

Now, average = 6x
n

or, 400 = 1560 + p or, 1560 + p = 2000 or, p = 440
5
Hence, the required value of p is Rs 440.

Example 3: In a discrete data, if 6fx = 400 + 20a and 6f = 20 + a, find the mean ( x ) .
Solution:
Here, 6fx = 400 + 20a and 6f = 20 + a, x = ?

Now, x = 6fx = 400 + 20a = 20(20 + a) = 20
6f 20 + a 20 + a

Example 4: In a discrete series, the mean is 50, 6fx = 100p + 150 and N = 4p – 15, find

the values of p and N.

Solution:
Here, mean ( x ) = 50, 6fx = 100 p + 150 and N = 4p – 15
6fx 100p + 150
Now, x = N or, 50 = 4p – 15

or, 50 (4p – 15) = 100p + 150
or, 200p – 750 = 100p + 150
or, 100p = 900
? p =9
Also, N = 4p – 15 = 4 × 9 – 15 = 21
Hence, the required values of p is 9 and N is 21.

(ii) Mean of individual repeated data (Mean of a frequency distribution)

In the case of repeated data, follow the steps given below to calculate the mean.

– Draw a table with 3 columns.

– Write down the items (x) in ascending or descending order in the first column
and the corresponding frequencies in the second column.

– Find the product of each item and its frequency (fx) and write in the third
column.

– Find the total of f column and fx column.

– Divide the sum of fx by the sum of f (total number of items), the quotient is the
required mean.

Example 5: From the table given below, calculate the mean mark.

Marks 14 17 28 35 40
5 10 15 8 12
No. of students
Solution: No. of students (f) fx
Calculation of mean marks:
5 70
Marks (x) 10 170
15 420
14 8 280
17 12 480
28
35 N = 50 6fx =1420
40

Total

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Now, mean mark ( x ) = 6fx = 1420 = 28.4
N 50
Hence, the required mean mark is 28.4.

Example 6: If the mean of the data given below be 32, find the value of p.

Solution: x 10 20 30 40 50 60
f 5 8 10 9 p 1

x f fx

10 5 50
20 8 160
30 10 300
40 9 360
50 p 50p
60 1 60

Total N = 33 + p 6fx = 930 + 50p

Now, mean ( x ) = 6fx
N

930 + 50p
or, 32 = 33 + p

or, 1056 + 32p = 930 + 50p

or, 18p = 126

or, p = 7

Hence, the required value of p is 7.

EXERCISE 17.4
General section

1. a) The marks obtained by 6 students out of 50 full marks are given below. Find the
average marks.

48, 26, 36, 14, 42, 38
b) The daily wages (in Rs) of 5 workers are given below. Calculate the average wage.

360, 500, 450, 540, 400
c) The ages of Anupa, Bimlesh, Chandra, Dipika and Elina are 12, 18, 13, 16 and 6

years respectively. Find their average age.
d) Find the mean from the data given below.

72, 46, 54, 80, 99, 62, 56
2. a) The average age of 5 students is 9 years. Out of them the ages of 4 students are 5, 7,

8 and 15 years. What is the age of the remaining student?

b) If 7 is the mean of 3, 6, a, 9 and 10, find the value of a.

c) Find x if the mean of 2, 3, 4, 6, x, and 8 is 5.
3. a) In a discrete series, 6fx = 15p + 800, 6f = p + 10 and mean ( x ) = 20, find the value

of p.
b) In a discrete data, if 6fx = 12 + 34n, N = 12 + 3n and x = 10, find the n and N.
c) In a discrete series, if 6fx = 45 + 60k and N = 9 + 12k, find the mean ( x ).
d) In a discrete series, if the sum of frequencies (6f) = a2 + 4a and 6fx = 25a2 + 100a,

find the mean ( x ).

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Creative section
4. a) Find the mean of the data given in the table.

Marks obtained 40 50 60 70 80 90
No. of students 274863

b) The ages of the students of a school are given below. Find the average age.

Age (in years) 5 8 10 12 14 16

No. of students 20 16 24 18 25 15

c) The weight of the teachers of a school are given below. Find the average weight.

Weight (in kg) 50 55 60 65 70 75 80
No. of teachers 2 5 7 10 4 3 1

d) Compute the arithmetic mean from the following frequency distribution table.

Height (in cm) 58 60 62 64 66 68
No. of teachers 12 14 20 13 8 5

5. a) Find the missing frequency p for the following distribution whose mean is 50.

x 10 30 50 70 90
f 17 p 32 24 19

b) The mean of the data given below is 17. Determine the value of m.

x 5 10 15 20 25 30
f 2 5 10 m 4 2

c) The heights of plants planted by the students of class IX in the afforestation program
are given below. The average height of the plant is 36 cm.

Height (in cm) 5 15 25 35 45 55
Number of plants 8 5 12 p 24 16

(i) Find the value of p. (ii) Find the total number of plants.

d) The speeds of vehicles recorded in a highway during 15 minutes on a day is given
in the following table. The average speed of the vehicles was 54 km per hour.

Speed (km/hr) 10 30 50 70 90
No. of vehicles 7 k 10 9 13

(i) Find the value of k.
(ii) Find the total number of vehicles counted during the time.

6. a) Find the value of p, for the following distribution whose mean is 10.

x 3 5 8 p 16 21
f 8 4 9 10 3 6

b) Find the missing value of k if the mean of the following distribution is 21.

x 9 14 17 k 28 34
f 12 15 16 25 13 19

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c) If the mean of the following data is 14, find the value of m.

x m 10 15 20 25
f7 6 8 4m
d) If the mean of the following data is 56, determine the value of ‘a’.

x 10 30 50 70 89
f 7 8 a 15 a

17. 11 Median

Let’s take a series: 25, 13, 7, 10, 16, 22, 19

Now, arranging the series in ascending order.

7 10 13 16 19 22 25

Thus, when we arrange the given series in ascending order, the fourth item 16 falls
exactly at the middle of the series. Therefore, 16 is called the median of the series.
In this way, median is the value that falls exactly at the middle of an array data.

i.e., Median = the middle value of a set of arry data

(i) Median of ungrouped data

To find the median of an ungrouped data, arrange them in ascending or descending

order. Let the total number of observation be n. th

If n is odd, the median is the value of the n + 1 observation.
2
th th
If n is even, the median is the average of n n
and 2 + 1 the observations.
2

Worked-out examples

Example 1: The heights of 7 students (in cm) are given below. Find the median
height.
115, 119, 125, 110, 140, 135, 128

Solution:
Arranging the heights in ascending order, we have,

110, 115, 119, 125, 128, 135, 140

Here, n= 7 th th

Now, the position of median = n+1 term = 7+1 term = 4th term
2 2
Thus, median lies at the 4th position.

? Median = 125.

Example 2: The daily wages (in Rs) of 12 workers are given below. Calculate the median

wage.

400, 350, 450, 520, 250, 375, 480, 550, 380, 555, 600, 580
Solution:

Arranging the wages in ascending order, we have,

250, 350, 375, 380, 400, 450, 480, 520, 550, 555, 580, 600

Here, n = 12

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n+1 th 12 + 1 th
2 2
Now, the position of median = term = term = 6.5th term

6.5th term is the average of 6th and 7th terms.

? Median = 450 + 480 930
= 2 = 465
2

Hence, median wage is Rs 465.

Example 3: If the given data is in ascending order and median is 34, find the value of x.
10, 15, 26, 3x + 1, 39, 44, 53

Solution:
Here,

The given data in ascending order is 10, 15, 26, (3x + 1), 39, 44, 53

Number of observations (n) = 7 th th

Now, the position of median = n+1 term = 7 + 1 term = 4th term
2 2
? Median =3x + 1

But, by question median = 34

or, 3x + 1 = 34

or, 3x = 33

? x = 11

Hence, the required value of x is 11.

(ii) Median of Discrete series

To compute the median of a discrete series of frequency distribution, we should
display the data in ascending or descending order in a cumulative frequency table.
Then, the median is obtained by using the formula,

N + 1 th
2
Median = value of observation.

Example 4: Compute the median from the table given below,

Weight (in kg) 25 30 35 40 45 50
No. of students 4 7 10 4 3 2
Solution:

Cumulative frequency table:

Weight in kg (x) No. of students (f) c.f.

25 4 4
30 7 11
35 10 21
40 4 25
45 3 28
50 2 30

Total N = 30

N + 1 th 30 + 1 th
2 2
Now, position of median = term = term = 15.5th term

In c.f. column, the c.f. just greater than 15.5 is 21 and its corresponding values is 35.

? Median = 35 kg.

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17.12 Quartiles

Quartiles are the values that divide the data arranged in ascending or descending
order into four equal parts. Hence, there are three quartiles that divide a distribution
into four equal part.

x1 x2 x3 x4 x5 x6 x7

1st quartile (Q1) 2nd quartile (Q2) 3rd quartile (Q3)

The first or lower quartile (oQf1t)hies the point below which 25 % of the observations
lie and above which 75 % observations lie.

The second quartile (%Q2o)fisthteheobpsoeirnvtabtieolnoswliwe.hOicfhco5u0r%se,otfhtehseeocbonsedrvqautaiortnisleliies
and above which 50

the median.

lTieheanthdiradboorveupwpheircqhu2a5rt%ileo(Qf t3h) eisotbhseeprvoaintitobnesloliwe.which 75 % of the observations

If N be the number of observations in ascending (or in descending) order of a

distribution, then in the case of discrete data th

The position of the first quartile (Q1) = N + 1 term
4

2 (N + 1)th term = N + 1 th
4 2
The position of the second quartile (Q2) = term

The position of the third quartile (Q3) = 3 (N + 1) th term.
4

After finding the positions of a quartile, we apply the similar process of computing the

required quartile as like the process of computing median.

Example 5: Find the first quartile (Q1) and the third quartile (Q3) from the data given
below.

18, 24, 20, 19, 23, 34, 28

Solution:

Arranging the data in ascending order,

18, 19, 20, 23, 24, 28, 34

Here, N = 7 N+ 1 th term 7+1
4 4
The position of the first quartile (Q1) = = th term = 2nd item
T?hTuhseQf1irlsiet sqautatrhtiele2(nQd p1)o=sit1io9n. . (Q3) 1)
3(N +
Again, the position of the third quartile = 4 th term = 6th term

Thus, Q3 lies at the 6th position.
? The third quartile (Q3) = 28

Example 6: The observations 2x + 1, 3x – 1, 3x + 5, 5x – 1, 38, 7x – 4 and 50 are in
ascending order. If the first quartile is 20, what is the value of x? Also, find
the third quartile of the observations.

Solution: Here,

The given data in ascending order is 2x + 1, 3x – 1, 3x + 5, 5x – 1, 38, 7x – 4, 50
No. of items (n) = 7
N+1 th 7+1 th
Now, the position of the first quartile (Q1) = 4 4
term = = 2nd term

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The value of 2nd term is 3x – 1. So, the first quartile (Q1) = 3x – 1.
According to question, Q1 = 20

or, 3x – 1 = 20

or, 3x = 21

? x =7

Again, the position of the third quartile (Q3) = 3 n+1 th 7+1 th
4 4
term = 3 term = 6th term

The value of 6th term is 7x – 4.

? The third quartile (Q3) = 7x – 4 = 7 × 7 – 4 = 45.

Example 7: Compute the first and the third quartiles from the table given below.

Marks 40 50 60 70 80 90

Solution: No. of students 4 6 10 12 5 2

Cumulative frequency distribution table,

Marks (x) No. of students (f) c.f.

40 4 4
50 6 10
60 10 20
70 12 32
80 5 37
90 2 39

Total N = 39

Now, the position of the first quartile (Q1) = N + 1 th term = 39 + 1 th term = 10th term
44
In c.f. column, the corresponding value of the c.f. 10 is 50.

? The first quartile (Q1) = 50 3 (N + 1) th term = 30th term
and i4ts corresponding value
Again, the position of the third quartile (Q3) = is 70.
In c.f. column, the c.f. just greater than 30 is 32

? The third quartile (Q3) = 70

17.13 Mode

Let’s take a set of ordered observations. 2, 5, 5, 7, 7, 7, 7, 10, 10, 10, 15
Here, the observation ‘7’ appears maximum number of times. Therefore, 7 is the
mode of the given observations. Thus, the mode of a set of data is the value with the
highest frequency. A distribution that has two modes is called bimodal. The mode
of a set of data is denoted by Mo.

(i) Mode of discrete data

In the case of discrete data, mode can be found just by inspection, i.e., just by
taking an item with highest frequency.

Example 8: Find the mode for the following distribution, 10

12, 10, 18, 12, 10, 12, 15, 12,
Solution:

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Arranging the data in ascending order
10, 10, 10, 12, 12, 12, 12, 15, 18

Here, 12 has the highest frequency.
? Mode = 12

Example 9: Find the mode of the given distribution:

Marks 27 30 36 40 45 50

No. of students 4 10 16 9 6 5
Solution:

Here, the marks 36 has the highest frequency, i.e., 16.
? Mode = 36.

(ii) Mode of grouped and continuous data

In the case of grouped and continuous data, the class with highest frequency

is observed and it is taken as the model class. Then, by using the following

formula, mode can be computed.

Mode (M0) = L + f1 – f0 f2 uc
2f1 –f0 –

Where, L = the lower limit of the model class

f0 = the frequency of the class preceding the model class
f1 = the frequency of the model class
f2 = the frequency of the class succeeding the model class
c = the width of the class interval

Alternatively, mode can also be computed by the following empirical relation.

Mode = 3 Median – 2 Mean

Example 10: Find the model class from the data given below:

x 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50

f 6 8 12 7 3
Solution:

Here, 12 is the highest frequency and its corresponding class is 20 – 30.

? Model class is 20 – 30.
Example 9: Compute the mode from the data given below:

Age in years 3 – 5 5 – 7 7 – 9 9 – 11 11 – 13 13 – 15

Solution: No. of pupils 10 35 70 35 12 6

Here, 70 is the highest frequency and its corresponding class is 7 – 9.

? Model class is 7 – 9.

Now, L = 7, f1 = 70, f0 = 35, f2 = 35 and c = 2

? Mode (M0) = L + f1 – f0 f2 uc = 7 + 70 – 35 35 × 2 = 7 + 35 × 2 = 8
2f1 –f0 – 140 – 35 – 70

So, the required mode is 8 years.

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EXERCISE 17.5

General section
1. a) Define median of a set of observations.

b) Define the first quartile (Q1) and the third quartile (Q3) of set of observations.
c) Write the formula to find the position of median of a descrete series.
d) Write the formulae to find Q1 and Q3 of a descrete series.
e) Define mode. Write the formula to find mode of a grouped and continuous data.

2. a) Find the medians of the following sets of data.
(i) 18, 16, 27, 20, 25
(ii) 21, 28, 14, 42, 35
(iii) 15, 30, 35, 25, 20, 45, 40
(iv) 22, 16, 14, 26, 32, 30
(v) 16, 13, 10, 14, 11, 12, 15

b) The weights of five high school students are given below. Find their median weight.
50 kg, 54 kg, 45 kg, 63 kg, 48 kg

c) Find the median age of a group of 7 people whose ages in years are as follows.
47, 61, 13, 34, 56, 22, 8

d) What is the position of the median marks in the table given below?

Marks 18 27 32 40 46
No. of students 2 3 10 9 5

3. a) Find the first quartiles (Q1) of the following sets of data.
(i) 14, 12, 17, 23, 20, 16, 10

(ii) 16, 25, 10, 30, 35, 8, 12

(iii) 40, 20, 30, 10, 16, 12, 8

b) Find the third quartiles (Q3) of the following sets of data.
(i) 15, 9, 21, 33, 27, 39, 45

(ii) 18, 26, 14, 22, 30, 38, 34

(iii) 30, 20, 50, 80, 40, 60, 70

4. a) If the given data is in ascending order and the median is 40, find the value of x.

20, 30, 3x + 5, 50, 60
2
b) If the given data is in ascending order and the median is 200, find the value of p.

100, 150, 5p + 10, 250, 300
4
c) a + 1, 2a – 1, a + 7 and 3a + 4 are in ascending order. If the median is 17, find the

value of a.

d) 5, k + 3, 2k + 1, 3k – 2, 26 and 31 are in ascending order. If its median is 17, find
the value of k.

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5. a) 20, x + 6, 35, 40, 47, 51, 58 are in ascending order. If the first quartile of the data is

24, find the value of x.

b) 110, 2x + 1, 120, 125, 130, 135, and 140 are in ascending order. If the first quartile
3

of the data is 115, find the value of x.

c) 9, 11, 12, 14, 18, 8m + 1, 28 are in ascending order. If the third quartile is 25, find
the value of m.

d) x + 1, x + 3, x + 5, 2x, 3x – 5, 4x – 10 and 3x – 1 are in ascending order. If the upper
quartile of the data is 18, find the value of x.

6. a) Find the modes of the following distributions.
(i) 7, 9, 5, 7, 10, 9, 7, 12
(ii) 15 kg, 21 kg, 17 kg, 21 kg, 28 kg, 21 kg, 15 kg, 21 kg

b) In a class, there are 15 students of 16 years, 14 students of 17 years, and 16 students
of 18 years. What is the modal age of the class?

c) In a factory, the daily wages of 30 labourers is Rs 25, 35 labourers is Rs 36, and
20 labourers is Rs 45. What is the modal wage of the factory?

d) The age of 5000 students of a school are shown in the table. What is its modal class?

Age in years No. of students

4 to 8 1050
8 to 12 2856
12 to 16 1094

e) In a factory, number of labourers and their remuneration are as follows. Find the
modal class.

Remuneration (in Rs) 1500 – 2000 2000 – 2500 2500 – 3000

No. of labourers 220 215 120

f) Find the mode of the following distribution.

(i) a, c, b, b, c, a, c, a, b, c, a, c, (ii) d, f, e, f, e, d, f, d, e, f, d, f
g) Find the modal size of the shoes from the data given below.

Size of shoe 5 6 7 8 9 10
No. of men 10 15 30 25 18 12

h) Find the mode from the following data:

Daily wages (in Rs) 70 90 110 130 150 170
No. of workers 4 12 15 18 20 12

Creative section

7. Find the mode from the following distribution.

a) Age in years 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
50 40
No. of people 150 200 140 90

b) Daily wages (in Rs) 400 – 500 500 – 600 600 – 700 700 – 800 800 – 900
No. of workers 35 50 30 25 18

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8. Find the median marks from the data given below:

a) Marks 24 36 50 65 78

No. of students 2 4 12 11 6

b) Marks 10 15 20 25 30 35

No. of students 3 7 15 12 7 3

c) Wage per hour (in Rs) 45 55 65 75 85 95

No. of workers 20 25 24 18 15 7

d) Weight (in kg) 36 40 25 44 30 49
No. of students 8 10 3 65 4

9. a) Calculate the first quartile (Q1) from the data given below:

Marks obtained 32 36 40 44 48 52
No. of students 259632

b) Compute the third quartile (Q3) from the following table:

Wages per hour (in Rs) 50 60 70 80 90 100

No. of workers 6 10 15 13 8 3

c) Find the first quartile (Q1) and the third quartile (Q3) from the following distribution:

Ages (in years) 22 27 32 37 42
No. of people 35 42 40 30 24

d) Find the quartiles from the following distribution:

Height (in cm) 90 100 110 120 130 140
No. of students 20 28 24 40 35 18

Project work

10. a) Measure the height of your any 10 friends and compute the mean, median, and
mode height.

b) Write the number of students in class 4 to class 10 of your school. Collect the data
of monthly fees of each students of these classes. Show these information in table.
Then compute the mean, median and quartiles of the data.

Objective Questions
1. In a frequency distribution, if the lower limit of a class is 20 and width of the class is 4,

what is its upper limit?

(A) 12 (B) 18 (C) 22 (D) 24

2. The mid-value of a class interval of width 20 is 50, what is the lower limit of the class is:

(A) 20 (B) 40 (C) 60 (D) 80

3. The correction factor of a data with classes 10 – 19, 20 – 29, 30 – 39, … is

(A) 0.5 (B) 1 (C) 1.5 (D) 2

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4. A teacher wishes to get a single value representing the heights of the students of a class.
Whichmeasureofcentraltendencywouldbemostappropriate ifthedataisprovidedtohim?
(A) Mean (B) Median (C) Mode (D) all of the above

5. Of all mathematics textbooks from different publications in a stationery, a student
has to purchase the textbook which is most liked by students and teachers. What
measure of central tendency would be most appropriate, if the data is provided to her?
(A) Mean (B) Mode (C) Median (D) Any of the three

6. The mean of three different natural numbers is 20. If lowest number
is 7, what could be highest possible number of remaining two numbers?

(A) 45 (B) 40 (C) 48 (D) 52

7. Which of the following has the same mean, median and mode?
(A) 1, 10, 11, 16, 10 (B) 15, 8, 10, 10, 7 (C) 10, 9, 3, 18, 10 (D) 4, 10, 20, 7, 9

8. Ram said to Sita - “When I was married 10 years ago my wife was the 6th member of
the family. Today, my father is died and a baby girl is born to me. The average age of my
family during my marriage was same as today.” What is the age of Ram’s father when he
died?

(A) 85 years (B) 69 years (C) 94 years (D) 96 years

9. The average age of 5 children is 6 years and that of 4 teenagers is 15 years, what is the
average age of all 9?

(A) 9 years (B) 10 years (C) 11 years (D) 12 years

10. In a discrete data, the mean of 100 observations was 60. If the frequency of the observation
60 is mistakenly written as 25 instead of 15 then what will be exact mean?

(A) 60 (B) 56 (C) 54 (D) 49

11. The mean of a data with 11 observations is 42. If the mean of first 6 observations is 39
and that of last 6 observations is 44, what is the sixth observation?

(A) 35 (B) 36 (C) 38 (D) 43

12. If the median of the numbers 10, 13, x +7, 2x – 3, 21, and30 in ascending order is 20,
what is the value of x?

(A) 12 (B) 15 (C) 19 (D) 21

13. What is the name of the quartile which divides the data below 25%?

(A) Q1 (B) Q2 (C) Q3 (D) both (A) and (C)

14. In what ratio does the upper quartile divide the data arranged in ascending order from
bottom?

(A) 1:2 (B) 1:3 (C) 3:1 (D) 2:1

15. What percent of observations are there below Q3 of a data?

(A) 25% (B) 50% (C) 60% (D) 75%

Vedanta Excel in Mathematics - Book 9 294 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Ministry of Education, Science and Technology ,
Curriculum Development Centre
Curriculum of Class 9 (2071)

Area Subject Matters No. of
Periods

1. Set * Review of set 8
* Union, intersection, difference of two sets, complement

of set
* Problem solving of simple verbal problems in two sets

by using Venn-diagram

* Problems related to profit and loss

* Problems related to Commission, discount, tax, 20
dividend 24
2. Arithmetic

* Household arithmetic (Electricity bill, water bill,
telephone bill, taximeter)

3. Mensuration * Simple problems related to area (carpeting the floor,
paving stones, construction paths, colouring and
* plastering, etc.) and expense on it.
* Surface area, area of cross-section and volume of prism,
cube, cuboid
Project work and problem collections related to
mensuration in real life situations

4. Algebra * Factorization of the expressions of the form
a4 + a2b2 + b4
30
* Problems related to indices (Simplification of negative
and fractional indices)

* Exponential equations (excluding the quadratic form)
* Simple problems related to ratio and proportion
* Solution of simultaneous linear equations (substitution

method, elimination method and graphical method)
* Solution of quadratic equations (factorization method,

completing square method and by using formula)

5. Geometry Verification of properties of triangles (use of
experimental and theoretical verifications)
* The sum of three angles of triangle is two right angles 55
(only theoretical proof)
* The exterior angle formed by producing a side of a
triangle is equal to the sum of two opposite interior
angles (only theoretical proof)

Vedanta Excel in Mathematics - Book 9 354 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Area Subject Matters No. of
Periods
* The base angles of an isosceles triangle are equal (only
theoretical proof) 44

* The sides opposite to two equal angles of a triangle are
also equal (Theoretical proof)

* The angular bisector of vertical angle of an isosceles
triangle is perpendicular to the base and bisects the
base and its converse theorem (Theoretical proof)

* The sum of any two sides of a triangle is greater than
the third side (experimental verification)

* In any triangle, the side opposite to the greater angle is
longer than the side opposite to the smaller angle and
its converse theorem (Experimental verification)

* Of all straight line segments drawn to a given line from
an external point, the perpendicular is the shortest one
(Experimental verification)

* Pythagoras Theorem (Experimental verification)
* Discussion on the interrelationship of the properties of

sides and angles of triangle
* A straight line segment joining the mid-points of any

two sides of a triangle is parallel to the third side and
it is equal to half of the length of the third side and its
conversion theorem (only theoretical proof)
Verification of properties of parallelograms
* The straight line segments that join the ends of two
equal and parallel line segments towards the same
sides are also equal and parallel (Theoretical proof)
* The opposite sides of a parallelogram are equal
(Theoretical proof)
* The opposite angles of a parallelogram are equal
(Theoretical proof)
* The diagonals of a parallelogram bisect each other
(Theoretical proof)
* The theoretical proofs of the converse of theorems
mentioned above
* Discussion of the interrelationship of quadrilaterals
(parallelogram, rectangle, square, rhombus)
Construction
* Construction of quadrilaterals (square, rectangle,
rhombus, parallelogram and trapezium)
Similarity
* Review of similarity
* Simple problems on similar polygons

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 355 Vedanta Excel in Mathematics - Book 9

Area Subject Matters No. of
Periods
Circle
* The perpendicular drawn from the centre of a circle to

a chord, bisects the chord (Theoretical proof)
* The line joining the mid-point of a chord and the centre

of a circle is perpendicular to the chord (Theoretical
proof)
* Equal chords of a circle are equidistant from the centre
and its converse theorem (Theoretical proof)

6. Trigonometry * Introduction of trigonometric ratios on the basis of 12
* right-angled triangle 13
* Measurement and problem solving related to the ratios
sine, cosine and tangent
Problems related to sine, cosine and tangent of the
angles 0°, 30°, 45°, 60° and 90° in right-angled triangles.

7. Statistics * Introduction and construction of line graph, pie-chart,
histogram and ogive by using the collected data

* The other information of the data from histogram and
ogive

* Calculation of mean, median, mode and quartiles of
ungrouped data

8. Probability * Introduction to probability
*
* Definition of probability and basic concepts

Use of simple probability and introduction of
probability scale (0 - 1)

* Introduction and use of formula to estimate probability

a) Empirical probability 8

P = Number of observed outcomes
Total number of experiments

b) Probability of an event
Favourable number of cases
P = Exhaustive number of cases

Vedanta Excel in Mathematics - Book 9 356 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 357 Subject: Compulsory Mathematics Specification grid Time: 3 hours
Secondary Level (Class 9) HA Total Remarks

Full Marks: 100 -4

Cognitive Units KC A
domain 1×4
1×4
Subject area
1×4
1 Set 1 Set --
2×4
2 Profit and loss
3×4
2 Arithmetic 3 Commission, discount, tax, 1×1 2×2 1×4 1×5 14
dividend 1×4

4 Household arithmetic -
10
5 Area of plane surface 40

3 Mensuration 6 Surface area and area of 1×1 3×2 1×5 16
cross-section and volume of
prisms, cube and cuboid

7 Factorisation

8 Indices

4 Algebra 9 Exponential equation 1×1 5×2 1×5 24

Vedanta Excel in Mathematics - Book 9 10 Ratio and proportion

11 Equations

12 Triangle and parallelogram

5 Geometry 13 Construction 2×1 3×2 1×5 25
14 Similarity

15 Circle

6 Trigonometry 16 Trigonometry - 1×2 -6

7 Statistics 17 Statistics 1×1 1×2 -7
8 Probability 18 Probability - 2×2 -4
6 4 37
Total no. of question 6 17 20 100
Total marks 34

MODEL QUESTION SET Maximum Marks: 100
Time: 3 hrs.
Class: IX
Subject: Compulsory Mathematics

Attempt all the questions

Group-A (6 × 1 = 6 )
1. (a) Write down the formula to find the rate of discount when the marked price and
selling price after discount of an item are given.
(b) From the figure alongside, what is the area of path? d
b

2. (a) If (am × an) ÷ ap = ax then express x in terms of m, n and p. l

(b) The adjoining line graph depicts the average Rainfall in (mm)100
rainfall in mm during 5 months of 2020. In April80
which month was the maximum average rainfall? May60
How much was the average rainfall recorded in June40
August? July20
August
Months
A

3. (a) In the ∆ABC alongside, M and N are the middle MN
points of sides AB and AC respectively, write BC
the relation between MN and BC.

(b) In the given figure; O is the centre of circle. If OM P MQ
A PQ, ON A XY and OM = ON. Write the relation O
between the chord PQ and XY. N

Group-B (17 × 2 = 34) X

4. (a) A noodle factory made a profit of Rs. 2750000 in the last year. If the management
decided to distribute 16% bonus from the profit to its 80 employees equally, find
the bonus amount received by each employee.

(b) The meter reading of Rajani’s house on Falgun1st was 7410 and that on Chaitra 1st
was 7439. If the electricity charge up to 20 units is Rs. 3 per unit and Rs 7 from
21 to 30 units, find the amount of the bill to be paid of the month of Falgun with
electricity service charge of Rs 30.

5. (a) The frame of the window given alongside is made of a rectangle 140cm
with a semi-circular top. Find its perimeter.
210cm
(b) How many bricks each of 15cm × 10cm × 5cm are required to 2cm
construct a wall of 20m long, 3m high and 20 cm wide? Find it. 3cm

(c) Find the volume of the prism given alongside. 2cm

4cm

Vedanta Excel in Mathematics - Book 9 358 5cm
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

6. (a) Resolve into factors: x4 + 2x2 + 9

(b) Simplify: 3 (x + y)–8 ÷ (x + 2

y) 3

7. (a) If a = 9m, b = 9n and a n b m = 81, prove that mn = 1.

(b) What must be subtracted from each term of the ratio 3:4 so that the result may be
equal to 4:3?

(c) Solve the given simultaneous equations: 3x + y = 5, x –2y = 4 A

8. (a) In the figure alongside, ‘ABC = 6x, ‘BAC = 5x and ‘ACD = 110° 5x
1100, find the actual measure of ‘ABC. 6x
DC B

M

(b) In the adjoining figure; ∆MAN a ∆MEN, ‘MAN = ‘NME. Find
the length of MN.

A 5cm E 4cm N

(c) In the given circle, O is the centre and OW A XY. If XY= 6cm and O

OX = 5cm then find the length of radius OW of the circle. X WY

P

9. (a) In the adjoining figure alongside; ‘PQR = 900, ‘PRQ = T, 30cm
PQ = 30 cm and QR = 40 cm, find the value of cosT.
T R
(b) Find the first quartile from the data given below: Q 40cm

5kg, 12kg, 33kg, 56kg, 24kg, 15kg, 74kg

10. (a) A bag contains a dozen of pencils of same shape and size. Out of them 4 are red

and the rest are yellow. If a pencil is taken from the bag at random, what is the

probability that the pencil so chosen is yellow?
(b) A card is drawn from a well shuffled deck of 52 plying cards. What is the probability
that the card drawn will be a queen?

Group-C (10 × 4 = 40)
11. In a place, 400 people were asked whether they like television shows ‘Comedy Champion’
or ‘Nepal Idol’. 200 liked ‘Comedy Champion’, 250 liked ‘Nepal Idol’ and 10 disliked both
the shows.

(i) How many people liked both the shows?

(ii) How many people liked ‘Nepal Idol’ only?

(iii) Draw a Venn-diagram to show the above information.

12. A man bought two school-bags for Rs. 2020. He sold one of them at 20% profit and the
other at 20% loss. Find his gain or loss percent in this transaction if the selling prices of
both bags are same.

13. A rectangular park is 60 m long and 40 m broad. Two crossing paths at right angle each
of 3m wide are running across the middle of the park. Calculate the cost of gravelling
the path at Rs. 45 per square meter.

14. If a = b = c , prove that: a3 + b3 + abc = a
bcd b3 + c3 + bcd d
x–4 + x+4 331
15. Solve the given equations: x+4 x–4 =

16. Prove that the opposite sides of a parallelogram are equal.
17. Explore experimentally the relation between the sum of the length of any two sides of a
triangle and that of the third side. (Two figures of different measurements are necessary)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 359 Vedanta Excel in Mathematics - Book 9

18. Construct a quadrilateral ABCD in which AB = BC = 5.5 cm, CD = AD = 4.8 cm and
‘ABC = 600.
A

19. In the given figure, AED is a right angled triangle and BCDE 62cm
is a rectangle. Find the length of CD.

20. The heights of plants planted by the students of class- IX in an E 45°
afforestation program are given below. The average height of D
plants is 36 cm.
B 60m C

Height (in cm) 5 15 25 35 45 55
Number of plants 8 5 12 p 24 16

(i) Find the value of p. (ii) Find the total number of plants.

Group-D (4 × 5 = 20)

21. The present rate of income tax fixed by Inland Revenue Department (IRD) is given below.

Status: Unmarried Status: Married

Income slab Tax rate Income slab Tax rate

Up to Rs 400000 1% Up to Rs 450000 1%

Rs 400000 – Rs 500000 10% Rs 450000 – Rs 10%

550000

Rs 500000 – Rs 700000 20% Rs 550000 – Rs 20%

750000

The monthly salary of an unmarried manager of an office and a married cashier of a
bank is Rs 55000 each. Then according to the above table, who pays the more income
tax and by how much? Find it.

22. Express the area of four walls of a room in terms of perimeter of its floor (P) and the
height (h). The cost of carpeting a squared hall at the rate of Rs 200 per sq. metres is
Rs 64800. If the cost of plastering its walls at Rs 50 per sq. metre is Rs 16200, find the
length and height of the room. By how much percentage is the height of the room less
than its length? Find it.
111

23. The expression 1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a was given to
Chhiring, Amrita and Rohan to evaluate in a mathematics speed test. Chhiring’s answer

was 0, Amrita’s answer was 1 and Rohan’s answer was x a+ b + c. Simplify it and find
whose answer was correct.

A

24. In the figure alongside, X and Y are the centres of two intersecting P Y
circles. If AB intersects the circle with centre X at P and Q, and
AB A XY at E, prove that: XE
(i) AP = BQ (ii) AQ = BP Q
B

Vedanta Excel in Mathematics - Book 9 360 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur