Oasis Math 9

(a) Less than Ogive /cumulative frequency curve

In this case we start from lower to higher values of variables in different classes
maintaining the same class magnitude. Frequency of previous class will be added to the
next class continuously to get the cumulative frequency of different classes.

In this curve, upper limit of each class interval is taken as x coordinates and corresponding
cumulative frequency as y coordinates.

i.e.,

Marks 0 – 10 10 – 20 20 – 30 30– 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100
Numbers of students 2 53 4
61534 2
Solution:
Here,

Marks Upper Limit Frequency Less than C.F.
0 – 10 10 2 2
10 – 20 20 5
20 – 30 30 3 2+5=7
30 – 40 40 4 7 + 3 = 10
40 – 50 50 6 10 + 4 = 14
50 – 60 60 1 14 + 6 = 20
60 – 70 70 5 20 + 1 = 21
70 – 80 80 3 21 + 5 = 26
80 – 90 90 4 26 + 3 = 29
90 – 100 100 2 29 + 4 = 33

33 + 2 = 35

Y Less than ogive Steps:

c.f. 35 • Plot the points tak-
30 ing upper limit of
each class interval on
25 X co–ordinates and
corresponding c.f. as
20 Y co–ordinates.
15
10 • Join the points by
5 free hand.

X' O X

10 20 30 40 50 60 70 80 90 100

Y' Marks

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(b) More than Ogive/cumulative frequency curve

When we start considering the frequency upto the lower limit of the given class, the
cumulative frequency curve so plotted against the variable classes is called more than
cumulative frequency curve. In the graph cumulative frequency will be plotted against
the lower limit of given class variables. The following example shows more than
cumulative frequency curve. In this curve, lower limit of each class interval is taken as
X- co–ordinates and corresponding c.f. as Y co–ordinates.

Marks 0–10 10–20 20–30 30–40 40–50
No. of students 2 5 7 3 8

Solution:
Here,

Marks Lower Frequency More than cumulative Start to add
0 – 10 2 frequency from the
limit frequency (C.F.) bottom.
0 23+2=25

10 — 20 10 5 18+5=23

20 — 30 20 7 11+7=18

30 — 40 30 3 8+3=11

40 — 50 40 8 8
Σf = 25

More than c.f. curve
Graph change

Y

25 Steps:
Plot all the points
20 taking lower limit as
More than ogive X co–ordinates and
corresponding c.f.
15 as Y co–ordinates.
Join all the points by
10 freehand.

5

X' O X

10 20 30 40 50 60 70

Y'

296 Oasis School Mathematics-9

Exercise 16.2

1. Study the given line graph and answer the questions given below:

Y



70

60

Number of students 50

40

30

20

10 X

O 10 20 30 40 50 60 70 80
Marks

(i) How many students get 10 marks?

(ii) How many get 40 marks?

(iii) How many get 60 marks?

2. (a) In a pie-chart, what magnitude of angle represents whole data?
(b)
(c) What does the width of a histogram represent?
(d)
(e) What does the height of a histogram represent?

What are the types of cumulative frequency? Y
(f)
What type of cumulative frequency

curve is this? 25

20

c. f. c. f. 15

10

5

X' O 10 20 30 40 50 60 70 X
X
Y' Class

Y

What type of cumulative frequency 25
curve is this? 20
15
10
5

X' 10 20 30 40 50 60 70
O

Y' Class

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3. Draw a line graph for the following data.

(a) Years 2015 2016 2017 2018 2019 2020
No of Students 150 160 180 200 250 240

(b) Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday

AverageTemperature 120° 18° 22° 25° 20° 21° 22°

4. From the following data construct a pie–chart.

(a) Subjects English Nepali Maths Science Social Studies
80 50
Marks 70 65 85

(b) Items Food Clothing Rent Education Others

Expenditure (in %) 30 20 15 20 15

5. The votes secured by four candidates A, B, C and D B A
are shown in the pie–chart. If A secured 1000 vote, 80º
90º 100º
(i) how many votes were there?
C D
(ii) how many votes did B, C and D get?

6. The monthly budget of a family is shown in the table. Represent it in a
pie-chart.

Heading Food Education House rent Transportation Miscellaneous Total
Expenses 1500 500 1500 500 1000 5000

7. Draw a histogram to represent the following data:

(a) Daily Exp. (In Rs.) 100 –125 125 – 150 150–175 175 – 200 200 – 225
10 5
Number of persons 5 8 7

(b) Marks 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50
Number of students 2 5 3 4 6

8. Draw a less than ogive curve form the following data.

(a) Height (In cm) 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

Number of persons 15 10 12 7 18 3

298 Oasis School Mathematics-9

(b) Weight ( in kg) 40 – 45 45 – 50 50 – 55 55–60 60 – 65
6 5
Number of students 7 3 4

9. Draw a more than ogive curve:

(a) Daily Earning (in Rs.) 1000–1200 1200–1400 1400–1600 1600–1800 1800–2000

Number of shops 35 20 15 25 30

(b) Age (in years) 0–5 5 – 10 10 – 15 15 – 20 20 – 25
Number of persons 5 3 4 9 4

10. (a) Study the given less than Ogive curve and fill the c.f.

Less than Ogive Y

Marks Less than c.f. 70

Less than 10 60
50
Less than 20 40
30
Less than 30 c.f. 20
c.f.10
Less than 40
O 10 20 30 40 50 60 70 80
Less than 50 Marks X

Less than 60


(b) Study the given less than Ogive curve and fill the c.f.

Less than Ogive Y

Wages Less than c.f. 60
50
Less than 50 40
30
Less than 100 20
10
Less than 150
O 50 100 150 200 250 300
Less than 200 wages

Less than 250 X

Less than 300

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11. (a) Study the given more than Ogive curve and fill the c.f.

More than Ogive

Class More than c.f. Y

More than 0 60
50
More than 10 40
30
More than 20 20
10
More than 30 c.f.
c.f. O 10 20 30 40 50 60 70
More than 40 Class

More than 50 X

More than 60


(b) Study the given less than Ogive curve and fill the c.f.

Less than Ogive Y

Marks More than c.f. 70

More than 0 60
50
More than 20 40
30
More than 40 20
10
More than 60
O 20 40 60 80 100 120 140
More than 80 Marks X

More than 100

12. (a) Study the given ogive curve and answer the following questions.

Y

70

60

50

40 (i) What kind of ogive curve is this?
30 (ii) How many students are there?
20 (iii) How many students got less than 20 marks?
10 (iv) How many students got less than 40 marks?

X' O 10 20 30 40 50 60 70 X

Y'

300 Oasis School Mathematics-9

(b) Study the given ogive curve and Number of workers 60Y X
answer the questions given below:
50
(i) Find the number of workers whose wages 40
are less than Rs. 40. 30
20
(ii) Find the number of workers whose wages 10
are more than Rs. 30.
O 10 20 30 40 50 60
Answer Wages (in Rs.)
Consult your teacher.

16.4 Measures of Central Tendency

The single value which represents the given set of variables is called central tendency.

In our daily uses we often ask what is the standard of the school? What is the market
price of vegetables ? Hence, we often talk about the average to know about the strength
of class, price of vegetables for the daily budget, etc.

There are mainly three types of measurement of central tendency. They are:

(i) Mean

(ii) Median

(iii) Mode

(i) Mean (Arithmetic mean)

The most popular and widely used measure of central tendency is the mean which is
also called arithmetic mean. It is denoted by the symbol x .

Calculation of arithmetic mean (x) for individual series

Direct method: If we have the set of variables, we first add them all and divide the sum
by the number of variables to get A.M. ∴ Mean ( x ) = Σnx

where,

Σx = sum of the variables

n = number of variables

Calculation for arithmetic mean (x) for discrete distribution

If we have the set of variables with their respective frequencies, first of all we

find the product of frequency and corresponding variables. Then we find their

sum and divide by the number of items. We can find the value of mean by using

the formula,

mean ( x ) = Σfx . Remember !
N
N = Σf


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Combined Mean

If n1 = number of first variables

x1 = mean of the first variables
n2 = number of second variables

x2 = mean of the second variables

Then the combined mean of both variables is denoted by x12 and is defined by,

x12 = n1x1+n2x2
n2+n2



Worked Out Examples

Example: 1

Find the mean of the following: 2, 5, 6, 10, 15.

Solution:

Here, Sum of variables = (Σx)

= 2 + 5 + 6 + 10 + 15 = 38

Number of variables (n) = 5

∴ Mean ( x ) = Σnx = 358 = 7.6

Example: 2

If the mean of the data 4, 6, 9, 10 and m is 8, find the value of m.

Solution:

Here, Mean ( x ) = 8
= 4 + 6 + 9 + 10 + m
Some of the variables (Σx)

= 29 + m.

Number of terms (n) = 5

We have, x = Σnx

8 = 29 + m
40 5

= 29 + m

or, m = 40 – 29
= 11

302 Oasis School Mathematics-9

Example: 3

Find the mean of the given data.

Height in cm 100 110 120 150 170
Number of persons 2 4 31 5
Solution:
fx
Height (x) No of person (f) 200
100 2 440
110 4 360
120 3 150
150 1 850
170 5 Σfx = 2000

Σf = 15

Now, Mean (X) = Σfx .
Σf

= 210050 = 133.33 cm

Exercise 16.3

1. (a) Write the formula to calculate mean in an individual series.

(b) Write the formula to calculate the arithmetic mean in discrete series.

(c) The mean of n1 items is x1 and the mean of another n2 items is x2, find the
mean of n1 + n2 items.

(d) ∑x = 250 and n = 10, what is the value of x?
2. Find the arithmetic mean of the following data.
(a) 30, 40, 45, 50, 90, 100
(b) 2 kg, 5 kg, 4 kg, 7 kg, 8 kg, 10 kg
(c) 60°, 65°, 70°, 80°, 90°, 110°
3. (a) Find the arithmetic mean of all odd numbers between 90 and 100.
(b) Find the arithmetic mean of all the multiples of 3, less than 20.
4. (a) If the mean of 10,19, 17, 27, a, 29, 33 is 23, find the value of a.
(b) If the mean of 10, 20, 30, 40, 50, 100, k, 500, 550 and 700 is 220, find the value of k.
(c) The arithmetic mean of 24, 25, 26, 30, 34, m and 35 is 30, find the value

of m.

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5. (a) In a discrete data, x = 6, Σfx = 5p+18, N = p+2, find the value of 7p.

(b) In a series, x = 22, Σfx = 400 + 20a, and Σf = 18 + a. Find the value of a.

(c) In a series, x = 10, Σfx = 700 + 5 m and Σf = 40 + 3 m. Calculate the value of m.

6. (a) If the mean of first 3 numbers is 3 and the mean of next 3 numbers is 30,
find the mean of all 6 numbers.

(b) The mean of 5 different numbers is 6 and the mean of 4 different numbers
is 15. Find the mean of 9 numbers.

(c) The mean of a, b, c, d is 9 and e and f is 6, find the mean of a, b, c, d, e and f.

(d) The average marks of 25 boys of class IX is 40 and that of 15 girls is 45. Find
the average marks of the class.

(e) The average weight of 20 girls is 42 kg and that of 16 boys is 48 kg. Find the
average weight of boys and girls.

7. (a) The mean of a, b, c, d, e and f is 12 and mean of e and f is 6. Find the mean
of a, b, c and d.

(b) Mean of 9 numbers is 10. If the mean of 5 different numbers is 6. Find the
mean of remaining numbers.

8. Find the arithmetic mean of the following data.

(a) x 10 12 14 16 18 20

f 5 3 7 4 11 2

(b) x 10 20 30 40 50 60
f 3 8 12 2 6 1

(c) x 100 200 300 400 500
f 2 5 4 10 8

9. (a) If the arithmetic mean of the following distribution is 31.43, find the value

of p.

x 30 31 p 33 34 20

f 8 10 5 89 2

(b) If the value of mean of given data is 28.5, find the value of 'a'.

x 15 21 27 a 39 45

f 2 4 5 62 1

(c) If the mean of the following data is 17, find the value of m.

x5 10 15 20 25 30
f2 5 10
m4 2

304 Oasis School Mathematics-9

Answer

1. Consult your teacher 2. (a) 59.17 (b) 6 kg (c) 79.17° 3. (a) 95° (b) 10.5

4. (a) 26 (b) 200 (c) 36 5. (a) 6 (b) 2 (c) 12 6. (a) 16.5 (b) 10 (c) 8 (d) 41. 87 (e) 44.67
7. (a) 15 (b) 15 8. (a) 15.19 (b) 30.94 (c) 358.62 9. (a) 32 (b) 33 (c) 7

Median

The data of the variables are generally collected randomly. One must arrange them
systematically in order. Suppose the data are arranged in ascending order of magni-
tude. The data which lies exactly in the middle is called median. It is denoted by Md.
Hence, positional median is the partition value which divides whole data into two equal
parts.

Calculation of median in individual series

Let there be n observations of variables. Then positional value of median is given by

n+1

( (2
( (value.
thwhere n is number of observations. From this, we can easily find the median
i.e. Median
(Md) = n+1 th term.
2

Calculation of median in discrete distribution

If the variables are given in terms of their respective frequencies, we will find
the cumulative frequency of each variable. The median position will be obtained
by using the formula,

( (Median = N+1 th term
Where, 2
N = Σf

= total number of observations

Worked Out Examples

Example: 1

Find the median size of shoes from the given data: 2, 5, 7, 1, 4, 3, 6, 8, 10.
Solution:

Arranging in ascending order of magnitude,

Here,

1, 2, 3, 4, 5, 6, 7, 8, 10

Number of observations (n) = 9

We have, ( (= n+1 th term
Median 2

( (= 9+1th term = 5th term
∴ Median (Md) 2
= 5

Oasis School Mathematics-9 305

Example: 2

Find the median from the following data:

10, 12, 15, 11, 16, 20

Solution:

Arranging in ascending order of magnitude,

10, 11, 12, 15, 16, 20

Here,
Number of observations (n) = 6

We have, ( (Median = n+1 th term
2
∴
( ( = 6+1 th term = 3.5th term
2
Alternatively
Median (Md) = 12+15
Example: 3 2 Md = 12 + 0.5 (15 – 12)
= 12 + 0.5 × 3
= 227

= 13.5 = 12 + 1.5

= 13.5

If x – 5, x, x + 5, x + 20 and x + 40 are in ascending order and their median is 15. Find the

value of x.

Solution:

x – 5, x, x + 5, x + 20 , x + 40

Here,
Number of observations (n) = 5

W e have, ( ( Median = n+1 th term
2

( ( = 5+1 th term
2
∵
By the question, = 3rd term

Median = (x + 5)
∴
Median = 15
x + 5 = 15

x = 10

Example: 4

Find the median from the data given:

Height (in cm) 50 60 70 80 90 100
10
Number of persons 5 6 7 8 9

Solution:

306 Oasis School Mathematics-9

Height (in cm) Number of persons (f) Cumulative frequency
50 5 5
60 6
70 7 5 + 6 = 11
80 8 11 + 7 = 18
90 9 18 + 8 = 26
100 10 26 + 9 = 35
35 + 10 = 45
N = 45

( (We have, Median Position = N+1th term
2

( (= 452+1 th term

= 23rd term

Here, c.f. just greater than 23 is 26, which is corresponding to the variable 80.
Hence, median height = 80 cm

Exercise 16.4

1. (a) Write the formula to calculate the median (md) in an individual series.
(b) Write the formula to calculate the median (md) in a discrete series.
(c) If the median of the series 23, 12, 16, 28, 31, is the third term, find the median.
(d) If the median of the given distribution is 15.5th term, find the median.

x f c.f.
766
9 8 14
12 10 24
15 6 30

N = 30

2. Find the median from the following data.
(a) 2, 5, 7, 10, 15, 18, 25, 30 and 35 .
(b) 5 kg, 9 kg, 12 kg, 16 kg, 20 kg, and 22 kg.
(c) 1.5 m, 1.6 m, 1.7 m, 1.8 m, 1.9 m, 2 m. and 2.2 m.
(d) 100, 50, 110, 70, 90, 150

3. (a) If the median of 2, 10, 15, x – 5, 24, 27 and 30, taken into order is 22, find the value
of x.

(b) If the median of 10, 19, (x + 10), (x + 20), 32 and 38 taken in order is 25, find the
value of x.

Oasis School Mathematics-9 307

(c) (x+1), (2x+1), (x+7) and (3x+4) are in ascending order. If the median is 13, find the
value of x.

(d) If 15, 18, 2x – 1, 2x + 1, 30 and 35 are in ascending order and the median is 20, find
the value of x.

4. Calculate the median from the following data:

(a)

Monthly income (in Rs.) 26000 31000 36000 41000 46000
No. of families 20 25 28 15 10

(b)

x 15 25 35 45 55 65
f 5 8 10 9 7 1
(c)

Marks 20 30 40 50 60 70 80 90
5
No. of students 3 4 15 10 8 7 3

Answer 2. (a) 27 (b) 10
1. (a) 15 (b) 14 kg (c) 1.8 m (d) 95 (c) 50
(c) 6 (d) 10 3. (a) 36,000 (b) 35
4. (a) 36,000 (b) 35 (c) 50

Mode

Marks obtained by 10 students of class IX in a test of 30 full marks is given below.

25, 20, 25, 22, 25, 21, 18, 25, 20, 18

Which marks is repeated maximum number of times?

25 4 times

Here, 25 is repeated maximum number of times i.e. 4 times.

∴ Mode of this data is 25.

Since the frequency of an item represents the number of repetitions, the variate value
having highest frequency is the mode in the discrete series.

Worked Out Examples

Example: 1

Find the mode from the following distribution: 10, 12, 18, 15, 12, 14, 12, 15,
Solution:

308 Oasis School Mathematics-9

Given data 10, 12, 18, 15, 12, 14, 12, 15.
Here, the value 12 is repeated maximum number of items.
Hence, the mode is 12.
Here the value which repeats maximum number of times is 4. Hence modal value = 4.

Example: 2

Find the mode from the following distribution.

x 10 20 30 40 50 60 70 80
f 43537846

Solution:
Here, the value 60 has the largest frequency 8 i.e. 60 is repeated maximum number of
times i.e. 8 times.
Hence, Mode (Mo) = 60.

Exercise 16.5


1. Find the mode from the given data.
(a) 35, 40, 42, 40, 55, 60, 35, 45, 38, 40.
(b) 5 kg, 7 kg, 9 kg, 10 kg, 3 kg, 7 kg, 8 kg, 1 kg,
(c) 2 m, 12 m, 8 m, 4 m, 8 m, 6 m, 2 m, 7 m, 2 m
(d) 7, 18, 15, 20, 15, 19, 20, 15

2. Find the modal–value from the following data.
(a)

Marks 5 10 15 20 25 30 35 40
No of students 3 5 23 6 91 4
(b)

Income (in Rs.) 5,000 6,000 7,000 8,000 9,000 10,000
9 12 21
No of workers 15 20 7

(c)

Height (in m) 5 6 7 8 9 10
3 12
No. of trees 2 54

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3. Find the modal class from the following distribution:
(a)

x 0 – 20 20 – 40 40–60 60 – 80 80 – 100

f 534 8 2
(b)

x 0 – 20 20 – 40 40 – 60 60 – 80 80 –100 100 – 120

f 194 7 8 3

Answer (b) 7 kg (c) 2 m (d) 15 m 2. (a) 30 (b) 10,000
1. (a) 40 3. (a) 60–80 (b) 20 – 40
(c) 6

16.5 Partition Values

Let's take 7 letters
A B C D E F G

First quartile Second quartile Third quartile
(Q1) (Q2) (Q3)

Let's select three partition values which divide the given data into 4 equal parts.

By inspection we can say that position of B, D and F divide the data into 4 equal parts.

Hence, B, D and F are quartiles in this data.

The data when arranged in ascending or descending order of magnitude can be divided
into many equal parts called partition values. If the arrangement is divided into four
equal parts, the values are called quartiles. The value which divides the lower half into
two equal parts is called lower quartile and it is denoted by Q1 and the value which
divides the upper half into two equal parts is called upper quartile and it is denoted by
Q3. The median (Md) is called second quartile and it is denoted by (Q2).

Calculation of Q1, and Q3 for individual series Third quartile (Q3)
Second quartile (Q2)
Lower quartile (Q1) is obtained by using the formula First quartile (Q1)

( (Q1 n+1 th term
= 4

Upper quartile (Q3) is obtained by using the formula
( )Q3
= 3 n+1 th term
2

Where, n is the total number of terms.

310 Oasis School Mathematics-9

Calculation of Q1 and Q3 in discrete series

In discrete series, Q1 and Q3 can be obtained by using the formula,

( (Q1 = N+1 th term
4

( (Q3 = 3 N+1 th term
4

Where, N = Σf = total number of terms

Note:
First quartile is also known as lower quartile and third quartile also as
upper quartile.

Worked Out Examples

Example: 1

Find the lower and upper quartiles from the following data: 12, 14, 15, 11, 16, 19 and 17.
Solution:

Arranging into ascending order, 11, 12, 14, 15, 16, 17, 19,

Here,

Number of observations (n) = 7

We have, ( (Q1 = n+1 th term Again, We have, 3(n+1) th
4 4

( ( = 7+1 th term Q3 = term
4
= 2nd term
= 3×(7+1)th term
∴ Q1 = 12 4

Example: 2 = 6th term

Find Q1 and Q3 from the given data: 20, 25, 28, 40∴ Q1 = 17
30, 34,

Solution:

Given data 20, 25, 28, 30, 34, 40

Here, = 6
Number of observations (n)

We have, Q1 ( (= n+1 th term
4
th
( (= 6+1 th term = 7 term
4 4

= 1.75th term

∴ Q1 = 1st term + 0.75 (2nd term – 1st term)

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= 20 + 0.75 (25 – 20)

= 20 + 0.75 × 5
= 20 + 3.75 = 23.75

Again, Q3 = 3(n+1) th term
4
= (3 7 )th
= 3(64+1) th term × 4 term

= (241)th term = 5.25th term
∴
Q3 = 5th term + 0.25 (6th term – 5th term)

= 34 + 0.25 (40 – 34)

Example: 3 = 34 + 0.25 × 6

= 34 + 1.5 = 35.5

Find the lower and upper quartile values.

x 30 32 34 36 38 40
25 64 5 3
f
f c.f. Q1 = 6.5 th item.
Solution: 2 2 • See c.f. which is just
5 2+5=7
x 6 7 + 6 = 13 greater than 6.5.
30 4 13 + 4 = 17 • It is 7.
32 5 17 + 5 = 22 • See its corresponding
34 3 22 + 3 = 25
36 Σf = 25 value.
38 • It is 32.
40

Here,

Number of observations (N) = 25
( ( ( )Q1
We have, = N+1 th term = 25+1 th term
4 4
( ) th term
= 26
4

= 6.5th term

C.f. just greater than 6.25 is 7, which corresponds to the value of variable 32.

∴ Q1 = 32

Again, Q3 = 3(n+1) th term
4
3(25+1) th term = 19.5th term
= 4

c.f. just greater than 19.5 is 22 which corresponds to the variable 38.

∴ Q3 = 38

312 Oasis School Mathematics-9

Exercise 16.6



1. (a) Write the formula to calculate lower quartile (Q1) in an individual series.
(b) Write the formula to calculate upper quartile (Q3) in an individual series.
(c) Write the formula to calculate lower quartile (Q1) in a discrete series.
(d) Write the formula to calculate upper quartile (Q3) in a discrete series.
(e) If the Q1 of the data 16, 12, 19, 20, 25, 10 is 1.75th term, find its Q1.
(f) If Q3 of the data 15, 25, 30, 10, 35, 40 is 5.25th term, find Q3.
(g) If Q1 and Q3 of given distribution are 9th and 28th term, find Q1 and Q3.

x f c.f.

12 7 7

18 3 10

24 9 19

30 12 31

36 4 35

N = 35

2. Find the lower quartile:
(a) 50, 100, 150, 200, 300, 400 and 500.
(b) 27, 23, 26, 17, 15, 24, 12, 10, 29, 30, 32.
(c) 102, 106, 94, 96, 98, 104, 100.
(d) 2, 5, 9, 20, 30 and 50 .
(e) 24, 25, 32, 37, 20.

3. Find the upper quartile :
(a) 40, 50, 60, 70, 90, 110 and 160 .
(b) 46, 34, 25, 44, 46, 30, 40, 35, 41, 36, 45.
(c) 400, 300, 500, 200, 700 and 800.
(d) 24, 28, 35, 37, 30

4. (a) If 22, 23, 25, 28, 32, x and 42 are in ascending order and Q3 = 36, find the
value of x.

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(b) If 8, 10, 12, 16, 20, 5a + 5, and 40 are in ascending order and Q3 = 30, find the value of a.

(c) If 7, 2x +1, 13, 19, 21, 23 and 28 are in ascending order and Q1 = 11, find the value of x.

(d) 18, 3x +12, 24, 27, 30, 33 and 36 are in ascending order and if Q1 = 21, find the value of x.
5. From the following distribution compute Q1 and Q3.

(a) Marks 1 2 3 4 5 30

Number of Students 20 12 10 15 25 1

(b) Income (In Rs.) 10,000 11,000 12,000 13,000 14,000 15,000 16,000
Number of Persons 2 4 8 1 6 7 3

(c) Ages (in yr) 25 30 35 40 45 30
Number of People 8 796 5 1

(d) x 20 30 15 10 25

f 4 632 4

75 100 125 150 175 200
(e) Wages

Number of Workers 4 63 2 4 1

Answer (e) Q1 = 100, Q3 = 175

1. Consult your teacher

2. (a) 100 (b) 15 (c) 96 (d) 4.25 (e) 24.5
3. (a) 110 (b) 45 (c) 725 (d) 36
4. (a) 36 (b) 5 (c) 5 (d) 3
5. (a) Q1 = 1 and Q3 = 5 (b) Q1 = 12,000 and
Q3 = 15,000 (c) Q1 = 30, Q3 = 40 (d) Q1 = 20, Q3 = 30

314 Oasis School Mathematics-9

Assessment Test Paper

Attempt all the questions Full Marks : 22

Group 'A' [2 × 1 = 2]

1. (a) Find the median from the data 13, 16, 18, 25, 30.

(b) Write the formula to calculate mean (x) in the discrete data.

Group 'B' [4×2= 8]

2. (a) Marks obtained by 30 students of class IX is given below. Make the class
internal of 10 and prepare a frequency distribution table using tally
bars.

27, 22, 47, 18, 38, 34, 59, 60, 71, 70,
8, 21, 15, 17, 24, 23, 28, 35, 37, 46,
54, 62, 7, 16, 28, 34, 38, 37, 45, 52

(b) Draw the line graph from the given data

x 10 15 20 25 30 35 40 45
f 5 10 20 25 15 10 5 10

3. (a) If the mean of the given data is 23, find the value of 'a'
10, 17, 19, a, 27, 29, 33
(b) Find the third quartile (Q3) from the given data:
50, 40, 55, 60, 61, 70, 46

Group 'B' [4×3 = 12]
4. Draw the less than ogive curve from the given information:

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency 5 10 20 25 15 10

5. If the mean of the given data is 17, find the value of x:

Marks 5 10 15 20 25 30

Number of students 2 5 10 x 4 2

6. Given pie chart represents the votes obtained A B
by 4 candidates A, B, C, and D in an election. 126º 36º

Find the percentage of votes received by each candidate. 144º 5C4º
D

Project work

Get into eight different groups of 3-4 students. Ask each group to select different subjects and collect
the raw data of marks obtained by the class in a terminal examinations. Construct the frequency
distribution table by taking suitable class interval. Based on the table construct

(i) less than ogive curve (ii) more than ogive curve and also find the class on which Q1, Md and Q3 fall.

Oasis School Mathematics-9 315

Probability

7Estimated Teaching Hours

Contents

• Term related to probability
• Probability scale
• Theoretical probability and empirical probability

Expected Learning Outcomes

At the end of this unit, students will be able to develop the
following competencies:
• To define and understand the terms related to probability
• To understand the probability scale
• To find the theoretical and empirical probability

Teaching Materials

• Flash cards, A4 paper, Playing cards, Number cards, Dice,
Coin, etc.

316 Oasis School Mathematics-9

Unit

17 Probability

17.1 Introduction

In everyday life, we come across statements such as:
(i) Most probably, it will rain today.
(ii) Chances are high that the price of petrol will go up.
(iii) I doubt that he will win the race.
The words 'most probably', 'chance', 'doubt', etc., show uncertainty or probability of

occurrence of an event.
Though, probability started with gambling, it is now used extensively in science,

commerce, biological sciences, weather forecasting, etc.

17.2 Some Terms Related to Probability

Experiment
An operation which can produce some well defined outcomes, is called an experiment.

Each outcome is called an event. Example, heating of potassium chlorate is an experiment
and occurring of oxygen and potassium chloride is an event.

Random experiment
An experiment in which all possible outcomes are known and the exact outcome cannot

be predicted in advance is called a random experiment.
For example, throwing of a dice is a random experiment because there are six faces of a

cubical dice, all faces are equally likely so either first or second or ………. or sixth face
will turn up but we cannot predict in advance which one will turn up.

Trial
By a trial, we mean performing a random experiment. Example, tossing a coin is trial

and getting either head or tail on a coin is an event.
Sample space
The set of all possible outcomes of a random experiment is called sample space. When

we toss a coin once, the sample space S = {H, T}, where 'H' denotes turning up head and
'T' denotes the turning up of tail.
Similarly, if we toss a coin two times we will get S = { TT, HT, TH, HH} and the cardinal
number of the set S is n(S) = 4. If we want to know the chance of getting both head, the
event is E = {HH}. Then, n(E) = 1.
The event of getting at least one head is E = {HT, TH, HH}

Oasis School Mathematics-9 317

Then n(E) = 3.
The event of getting both tail is E = {T T} then n(E) = 1

Dice

Possible outcomes on rolling a dice is {1, 2, 3, 4, 5, 6} .

Playing cards

♣◊♥♠
Club (13 cards) Diamond (13 cards) Heart (13 cards) Spade (13 cards)

Red Cards ⇒ All the cards of Diamond and Heart (26 cards)

Black Cards ⇒ All the cards of Club and Spade (26 cards)

Jack Queen King Ace

Face cards ⇒ Jack, Queen and King

318 Oasis School Mathematics-9

There are 12 face cards in playing cards.
Equally likely cases

Two or more events are said to be equally likely if all of these have equal chance of
occurring . For example, if we toss an unbiased coin, the outcomes, 'Head' or 'Tail' are
equally likely.

Probability of an event

The probability P(E) of an event is defined as the ratio of the number of favorable
outcomes to the number of possible outcomes.

In symbol, we can write
number of favourable outcomes
P(E) = number of possible outcomes

P(E) = n(E)
n(S)

Probability of any event E is a number lying between 0 and 1.

So, we must have 0 ≤ P(E) ≤ 1. This is a classical probability.

Note:
If E be the probability of happening of an event E and E be the probability of
not–happening an event E, the total probability is unity i.e.

P(E) + P(E) = 1 or, P(E) = 1 – P(E) i. e. P(not E) = 1 – P(E)

17.3 Theoretical Probability and Empirical Probability

Probability given by theory is called theoretical probability. There are six faces in a

cubical dice, all faces have equal chances to turn up, so the probability of getting 1, 2, 3,

4, 5 or 6 in the dice is 1 each. This is the probability given by theory. It means that when
6

a dice is rolled 6 times, each face comes once. But in real practice it may not be so.

A coin is tossed six times. The result is shown in the table.

Face 1 2 3 4 5 6
1 0 0
Turning up 0 2 3

From the table,

probability of getting 1st face = 0
6
2
probability of getting 2nd face = 6

probability of getting 3rd face = 3
6
1
probability of getting 4th face = 6

probability of getting 5th face = 0
6
0
probability of getting 6th face = 6

Oasis School Mathematics-9 319

From the above example, it is clear that the probability given by theory and practical is
different. So the probability given by actual experiment is called Empirical Probability.

Range of Probability

If a dice is rolled, probability of getting all faces of a dice is,

P(1) = 1 , P(2)= 1 , P(3) = 1 , P(4) = 1 , P(5) = 1 and P(6) = 1 .
6 6 6 6 6 6

And, total probability P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 + 1 + 1 + 1 + 1 + 1 = 1.
6 6 6 6 6 6

Again, events of getting a number divisible by 3 or 6.

Now, probability of getting 3 or 6 is = 2
6
4 2
Probability of getting the number other than 3 and 6 is 6 = 3 .

Hence, p = probability of happening an event = P (E)

q = probability of not happening an event = P ( E )

Then, p + q = 1.

i.e. P(E) + P ( E ) = 1

Again, if a dice is rolled, what is the probability of getting an event 0?

In this case, '0' cannot occur.

So, the probability of getting '0', P(E) = 0.

A bag contains 2 red balls. A ball is drawn at random.

What is the probability of getting a red ball?

In this case, P(E) = 2 = 1.
2

∴ Probability of sure event = 1.



Worked Out Examples


Example: 1

A card is drawn from a deck of well-shuffled 52 cards, what is the probability that
the card will be a club ?

Solution:

Let 'C' be the event of getting club.

Number of favorable outcomes n(C) = 13

Number of possible outcomes n(S) = 52
n(C)
We have, P(C) = n(S) = 13 = 1
52 4

320 Oasis School Mathematics-9

Example: 2
A bag contains 8 red, 5 white, and 7 green balls. Write down the sample space and
find the probability that the ball drawn is red.

Solution:

Let, R, W and G be the events of getting red, white and green balls respectively.

Total number of balls = 8 + 5 + 7 = 20

∴ n(S) = 8 + 5 + 7 = 20

and, number of favourable cases n(R) = 8

So, probability of drawing red balls = n(R) = 280 = 2
n(S) 3

Example: 3

A card is down from the pack of number cards 1 to 20. Find the probability of get-
ting a card which is a prime number.

Solution:

Let 'S' be the sample space, then

S = {1,2,3,4, …………., 19, 20}

n(S) = 20

Let 'E' be the event of selecting a prime number, then E = { 2, 3, 5, 7, 11, 13, 17, 19}

n(E) = 8, P(E) = ?
We have,
P(E) = n(E) = 8 = 2
Example: 4 n(S) 20 5

A card is drawn from a pack of 52 cards. Find the probability that,
(i) the card drawn is black.
(ii) the card drawn is an ace.

(iii) the card drawn is black ace.

(iv) the card drawn is a club.
(v) the card drawn is either a jack, a king or a queen.
Solution:

Let's be the sample space. Then, n(S) = 52

(i) Let B be the event of drawing black card. As number of black cards is 26.

we have, n(B) = 26

∴ P(probability of getting a black card) = P(B) = n(B) = 26 = 1
n(S) 52 2

(ii) Let 'A' be the event of drawing an ace. As number of ace is 4, i.e. n(A) = 4

∴ P(probability of getting an ace) = P(A) = n(A) = 4 = 113
n(S) 52

Oasis School Mathematics-9 321

(iii) Let 'E' be the event of getting black ace. Then, n(E) = 2.

∴P (a black ace) = n(E) = 2 = 216
n(S) 52

(iv) Let C be the event of drawing a club. As number of club = 13, then n(C) = 13

∴P(C) = n(C) = 13 = 1
n(S) 52 4

(v) Let 'E' be the event of drawing either a jack, a queen or a king. There are 4 cards of each,

so, we have,

n(E) = 4 + 4 + 4 = 12

∴ P(E) = n(E) = 12 = 3
n(S) 52 13

Example: 5

Two unbiased coins are tossed simultaneously. Write down the sample space, and
find the probability of getting,

(i) exactly two tails.

(ii) one head or one tail.
(iii) exactly two heads or two tails.

Solution:

Let S be the sample space, when two coins are tossed, then H and T are the events of
getting head and tail respectively.

Now,

S = {HH, HT, TH, TT} i.e. n(S) = 4

(i) Let E be the event of getting exactly two heads.

E = {HH} and n(E) = 1

We know that, P(E) = n(E)
n(S)

∴ P(E) = 1
4

(ii) Let, E1 be the event of getting one head and one tail.

Now,

E1 = {TH, HT} and n(E1) = 2

∴ P(E1) = n(E1) = 2 = 1
n(S) 4 2

(iii) Let E2 be the event of getting exactly two heads or exactly two tails.

Now,

E2 = {HH, TT} and n{E2) = 2

We have,

∴ P(E2) = nn((ESr)) = 2 = 1
4 2

322 Oasis School Mathematics-9

Example: 6

A coin is tossed 3 times. Find the probability of the event.

(i) one head. (ii) at least one tail. (iii) no heads.

Solution:
A coin is tossed 3 times. The sample space of the experiment is,

S = {HHH, HHT,HTH, HTT, THH, THH, THT, TTH, TTT}

∴ n (S) = 8

(i) If A be the event of getting exactly one head,

Then, the favourable cases of A are {TTH, HTT, THT} i.e. n(A) = 3
n(A)
∴ Probability of the event getting one head P(A) = n(S) = 3
8

(ii) If B be the event of getting at least one tail, then, the favourable cases of B are

{HHT, HTH, THH, HTT, THT, TTH, TTT} i.e. n(B) = 7
∴ n(B)
(iii) The probability of the event getting at least one tail = n(S) = 7
8

If C be the event of getting no heads,

Then, the favourable cases of C is {TTT}.i.e. n(C) = 1. n(C)
n(S)
∴ The probability of the event getting no heads P(C) = = 1 .
8

Example: 7

A dice is thrown 150 times and its numbers of faces were recorded as follows:

Outcomes 1 2 3 45 6
Numbers of repetitions 22 25 24 29 30 20

What is the probability of getting

(i) each number? (ii) prime number? (iii) odd number?

Solution: Oasis School Mathematics-9 323

(i) Here, total number of observations = 150.

Probability of getting 1 = P(1) = 22
150
25
Probability of getting 2 = P(2) = 150

Probability of getting 3 = P(3) = 24
150

Probability of getting 4 = P(4) = 29
150

Probability of getting 5 = P(5) = 30
150

Probability of getting 6 = P(6) = 20
150

(ii) 2, 3 and 5 are prime numbers,

Probability of getting prime numbers = P (2 or 3 or 5)

P ( 2 or 3 or 5) = 25 + 24 + 30
150 150 150

= 79
150

(iii) 1, 3 and 5 are odd numbers,

Probability of getting (1 or 3 or 5)

= 22 + 24 + 30
150 150 150
38
= 75

= 38 .
75

Exercise 17.1

1. (a) A coin is tossed. Find the probability of getting:

(i) head (ii) tail

(b) A dice is rolled, what is the probability of getting:

(i) 2 (ii) 4

(iii) an even number (iv) an odd number

(c) An ordinary fair dice is rolled at random. Find the probability of getting

(i) 5 (ii) not 5 (iii) greater than 3

(iv) 1 or 3 or 6 (v) not prime number (vi) less than 1

2. A card is drawn from a deck of well–shuffled 52 cards. Find is the probability of
getting?

(i) an ace (ii) a club (iii) a black card (iv) a diamond

3. (a) A bag contains 4 white, 8 black 6 red and 2 green balls. Find the probability
of getting:

(i) white ball (ii) red ball (iii) not black

(b) A basket contains 3 red, 4 yellow and 5 white flowers. A flower is chosen at random.
Find the probability of getting

(i) red flower (ii) yellow flower (iii) White flower

(c) A bag contains 4 white and 6 black balls. A ball is drawn at random from the
bag, find the probability of getting a ball.

(i) black (ii) not black.

324 Oasis School Mathematics-9

4. (a) From the pack of numbered cared 1 to 20, a card is drawn. Find the
probability of getting :

(i) even numbered card (ii) prime numbered card

(iii) card which is the multiple of 3.

(b) From the pack of numbered card 10 to 40, a card is drawn. Find the
probability that the card so drawn is

(i) multiple of 5 (ii) multiple of 7 (iii) no a multiple of 3.

5. (a) A card is drawn from a well shuffled pack of 52 cards. What is the probability
that the card will be king or queen?

(b) A card is drawn from the pack of 52 cards, find the probability of getting

(i) a diamond or a club (ii) a jack or an ace (iii) a spade or a heart.

6. (a) From the set of cards numbered 1 to 30, one card is drawn randomly. Find
the probability of getting a card divisible by 5 or 7.

(b) From the set of cards numbered 11 to 50, one card is drawn randomly. Find
the probability that the drawn card is divisible by 9 or 11.

7. (a) Two coins are tossed once. Find all possible outcomes and then represent
them by a sample space. Find the probability of getting:

(i) both heads (ii) both tails

(iii) one head and one tail (iv) at least one head.

(b) Two unbiased dices are thrown at random. Write down the sample space.
Also, find the probability of getting,

(i) a number 6 of the first dice.

(ii) a multiple of 3 as the sum.

(iii) sum of total number on the dice is 8.

8. (a) A bag contains 25 cards numbered from 1 to 5 as follows:

Card Number. 1 2 3 4 5

No. of repetitions 4 8 3 6 4

Find the probability of:
(i) even numbered card
(ii) prime numbered card

(b) In the adjoining figure a circle is divided into eight 2 1
equal sectors. If a grain of rice is dropped out the 3 8
circle, (Note: If the grain falls outside, the circle
drops the grain again) find the probability that the 4 7
rice drops on, 5 6

Oasis School Mathematics-9 325

(i) number 5 sector. (ii) not number 5 sector.

(iii) sectors of even number only. (iv) sectors of odd numbers only.

9. (a) A coin is tossed 50 times. The result of tossing is shown in the diagram.

outcomes H T
frequency 15 35

Find the empirical probability of getting (i) head (ii) tail

(b) A dice is rolled 50 times. The result is shown in the table

outcomes 1 2 3 4 56
12 10
frequency 12 6 3 7

Find the empirical probability of getting

(i) 1 (ii) less than 3

(iii) more than 5 (iv) even number.





Answer

1 1 1 1 1 1
1. (a) (i) 2 (ii) 2 (b) (i) 6 (ii) 6 (iii) 2 (iv) 2

(c) (i) 1 (ii) 5 (iii) 1 (iv) 1 (v) 1 (vi) 0
6 6 2 2 2

2. (i) 1 (ii) 1 (iii) 1 (iv) 1 3. (a) (i) 1 (ii) 3 (iii) 3 (b) (i) 1 (ii) 1 (iii) 5
13 4 2 4 5 10 5 4 3 12

(c) (i) 3 ( ii) 2 4. (a) (i) 1 (ii) 2 (iii) 3 (b) (i) 7 (ii) 4 (iii) 21
5 5 2 5 10 31 31 31

5. (a) 2 (b) (i) 1 (ii) 2 (iii) 1 6. (a) 1 (b) 1
13 2 13 2 3 5

7. (a) 1 , 1 , 1 , 3 (b) 1 , 1 , 356 8. (a) (i) 14 (ii) 3
4 4 2 4 6 5 25 5

(b) (i) 1 (ii) 7 (iii) 1 (iv) 1 9. (a) (i) 3 (ii) 7 (b) (i) 6 (ii) 9 (iii) 1 (iv) 23
8 8 2 2 10 10 25 25 5 50

326 Oasis School Mathematics-9

Specification Grid

Subject: Compulsory Mathematics Full marks: 100 Time : 3 hours

S.N. Area/Topic Unit Knowl- Concept Applica- Higher Total Remarks
edge (K) (C) tion (A) ability Marks
Sets (HA)
1. Sets 1 Profit and Loss, - - 1×4 4
Discount 1×4 14
Commission 1×1
2. Arithmetic 2 Taxation and At least
Dividend question of
3. Household two marks
Arithmetic should be
2×2 1×4 1×5 asked from
each unit
4.

3. Mensura- 5. Area of Plane 3×2 1×4 1×5 At least
tion Figure question of
16 three marks
4. Algebra 6. Area of 4 Walls 1 × 1 should be
and Area of asked from
Pathways each unit

7. Prism and Cost At least
Estimation question of
three marks
8. Factorization 1×1 5×2 2×4 1×5 should be
9. Indices 24 asked from
10. Ratio and each unit
At least
Proportion question of
11. Equation 25 four marks
should be
5. Geometry 12. Triangle 2×1 3×2 3×4 1×5 asked from
13. Parallelogram each unit
14. Construction 6

15. Circle 7
4
6. Trigonom- 16. Trigonometry - 1×2 1×4 - 37
100
etry 1×2 1×4 -
2×2 - -
7. Statistics 17. Statistics 1×1 17 10 4
34 40 20
8. Probability 18. Probability -

Total number of questions 6

Total marks 6

Oasis School Mathematics-9 327

Model Test Paper

Class: IX Time: 3 hours Full marks : 100
Attempt all the questions Group A [3×(1+1) = 6]

1. (a) Write the formula of profit percent if cost price (C.P) and selling price (S.P) are

given.

(b) Find the lateral surface area of a prism whose perimeter the base is 30cm and the

height is 10cm.

2. (a) If am × an = ax, find the value of x interns of m, n and p.
ap
A
(b) Find the arithmetic mean of: 8, 10, 12, 15, and 17 B

3. (a) In the given figure,ABCD is a parallelogram. IfAO = 6cm and

OB = 5cm, find the length of OC and OD. O
C
D

(b) In the given figure, O is the centre of the circle, OM⊥AB, O
AB = 8cm, find the length of AM. A MB
Group B [4×(2+2)+3×(2+2+2)=34]

4. (a) An article is sold at Rs. 600 at the profit of 20%, find its cost price.
(b) The minimum cost of electricity upto 20 units is Rs 80 and Rs. 7.50 per unit above

the consumption of 20 units. Find the bill to be paid if the total consumption is
30 units.
5. (a) Find the area of 4 walls of a room having length bre4adth and height 12 ft, 10 ft
and 8 ft. respectively.
(b) Find the volume of given prism.

10cm 20cm
10cm

10cm 50cm

(c) How many bricks of size 10cm × 5cm × 4cm are required to construct a wall of
dimension 10m × 2m × 10cm?

6. (a) Factorize: a4 + 64.

(b) Solve: 4x - 4x-1 = 192

7. (a) If 2x - 3y = 1 , find x: y.
2x + 3y 2

(b) Simplify: 9x + 2 + 10 × 9x .
9x + 1 × 11 - 8 × 9x

(c) Solve: 3x² + 5x - 8 = 0.

328 Oasis School Mathematics-9

AC

8 (a) Find the value of x from the given figure.

800° E 250° F
B
x
E D
yA

(b) In the given figure, AB = AC = CD,

∠ADC = 32°, find the value of x and y. x
CD
B A

(c) In the given figure, PQ⊥BC, BC = 4cm, QC = 2.5cm, P
PQ = 1.5cm, find the length of AB.
1.5 cm

B Q 2.5cm C
4cm

A

9. (a) In a right angled triangle ABC,∠ACB = 30°, BC = 20cm,
find the length of AB and AC.

30° 20cm B
C

(b) The following pie chart shows the monthly expenditure of a Food
household in different title. If the total expenditure is Rs. 7200, 1200
find the expenditure in each item. 1050° 700
Cloth 650° Rent

Miscellaneous

10. (a) A card is drawn from the pack of 52 cards, find the probability of getting (i) an

ace (ii) a black face card.

(b) A dice is rolled 40 times. The result is shown in the table.

Face 1234 5 6
4 5
Number of repetition 8 12 6 5

Group C [ 10 × 4 = 40]

11. In a survey among 200 people, the number of people who like milk only is twice the
number of people who like curd only. If 80 people like both milk and curd, 30 people
like neither milk nor curd, find the number of people in the survey.

12. A man sold an article on its marked price at a gain of 25%. But allowing Rs. 600 discount
there would have been a gain of Rs. 200. Find the cost price and the marked price.

13. A square room has 150m³ of air. It costs Rs. 1800 to plaster its 4 walls at the rate of Rs.

15 per square meter, find the height of the room.

a+b-c b+c-a xc c + a - b
xa
. .
( ) ( ) ( )14. Simplify:xaxb
xb xc

Oasis School Mathematics-9 329

( )15. If y - z 3
x = y = z , prove that x³ - y³ = b - c
a b c a³ - b³ .

16. Prove that the line joining the same sides of two equal and parallel lines are themselves

equal and parallel.

17. Construct a trapezium ABCD where BC||AD, BC = 7cm, AD = 5cm, AB = 6.5cm and

∠ABC = 60°.

18. Verify experimentally that equal chords are equidistant from the centre.

A

19. In the given figure, AD⊥BC, AD = 8cm, DC = 6cm, BD = 10cm, α

find sinα, tanα, sinθ and cosθ. 8cm C

θ

B 10cm D 6cm

20. Find Q1 and Q3 of :

x 6 12 24 18 30 36 42

f 2746597

Group C [4 × 5 = 20]

21. Monthly income of Kiran Kumar Shrestha is Rs. 75,000. He deposits 10% of his
income in P.F and 20% on C.I.F. Using the tax rate of Government of Nepal, calculate
the tax to be paid by him, if he is a married person.

22. A room is 3 times as long as it is broad and its height is 4.6m. If the cost of carpeting
its floor at the rate of Rs. 240 per square meter is Rs. 18,000, find the cost of painting
on its walls at Rs. 24 per square meter.

23. Solve: x-6 - x - 12 = 5 . A MB
x - 12 x-6 6 PQ
NC
24. In the given figure, ABCD is a parallelogram M and N

are the mid points at AB and DC respectively, prove that

AP = PQ = QC D

330 Oasis School Mathematics-9