Vedanta Opt. Maths Book 10 Final (2078)

vedanta Excel In Opt. Mathematics - Book 10

c.f., just greater than 12 is 13.

So, Q1 lies in (20 - 30)

L = 20, c.f. = 7, f = 6, i = 10

N - c.f 12 - 7 50
4 6 6
Q1 =L+ × i = 20 + × 10 = 20 +
f
= 20 + 8.33 = 28.33

To find the upper quartile (Q3), 3N = 3 × 48 = 36
4 4
c.f. just greater 36 is 40. So, Q3 lies in (60 - 70)

L = 60, c.f. = 32, f = 8, i = 10

3N - c.f 36 - 32 × 10
4 8 - 28.33) =
Q3 =L+ ×i = 60 +
Quartile f 65 Q1) =
= 60 + 4 × 10 = (Q3 -
8 1
Deviation (Q.D) = 2 1 (65 18.34
2
Q3 - Q1 65 - 28.33
coefficient of quartile diviation = Q3 + Q1 = 65 + 28.33 = 0.393

Example 5. Calculate the quartile deviation and its coefficient from the given data :

x 20-29 30-39 40-49 50-59 60-69 70-79 80 - 89
f 4 6 8 12 7 6 5

Solution: To calculate quartile deviation and its coefficient, given inclusive data is
changed into exclusive. The correction factor is given by using the formula.

Correction factor = Lower limit of the 2nd class – upper limit of first class

30 - 29 2
2
= = 0.5

So, the exclusive class of above data are given by

xf cf
4
19.5 - 29.5 4 10
18
29.5 - 39.5 6 30
37
39.5 - 49.5 8 43
48
49.50 - 59.5 12

59.5 - 69.5 7

69.5 - 79.5 6

79.5 - 89.5 5

N = 48

To find the lower quartile (Q1), N = 48 = 12
4 4

c.f., just greater 12 is 18.

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? Q1 lies in class (39.5 - 49.5)

L = 39.5, c.f. = 10, f = 8, i = 10

N - c.f 12 -10
4 8
Now, Q1 =L+ × i. = 39.5 + × 10
f
2
= 39.5 + 4 × 10 = 39.5 + 5 = 44.5

To find upper quartile (Q3) , 3N = 3 × 48 = 36
4 4

c.f., just greater 36 is 37.

Hence, Q3 lies in the class (59.5 - 69.5)

L = 59.5, c.f. = 30, f = 7, i = 10

3N - c.f 36 - 30
4 7
Now, Q3 =L+ × i. = 59.5 + × 10
f
= 59.5 + 8.57 =68.07

Q.D. = 1 (Q3 - Q1) = 1 (68.,07 - 44.5) = 11.785
2 2
Q3 - Q1
coefficient of Q.D. = Q3 + Q1

= 68.07 - 44.5 = 0.2094
68.07 + 44.5

Exercise 12.1

Short Questions

1. (a) Define measure of dispersion.

(b) List methods of measure of dispersion.

(c) Define quartile deviation.

(d) Write down formula to calculate Q.D.

(e) Write down formula to calculate coefficient of Q.D.

(f) Write a difference between Q.D. and its coefficient.

2. Calculate quartile deviation from following:

(a) Q1 = 25, Q3 = 75 (b) Q1 = 25, Q3 = 65

(c) Q1 = 40, Q3 = 80 (d) Q1 = 5, Q3 = 75

3. Calculate the coefficient of quartile deviation from the following:

(a) Q1 = 20, Q3 = 40 (b) Q1 = 40, Q3 = 80

(c) Q1 = 2, Q3 = 22 (d) Q1 = 2.5, Q3 = 22

4. (a) In a set of continuous data, if Q1 = 35, and quartile deviation is 20, find Q3.

(b) The coefficient of quartile diviation of a grouped data is 0.25 and the upper quartile
is 60. Find the value of lower quartile.

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(c) The coefficient of quartile deviation for continuous data is 0.39 and upper quartile
49.75. Find the first quartile.

Long Questions :

5. Calculate the quartile deviation and its coefficient from the following data :

(a) Marks 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No. of students 4 12 16 10 8 6 2

(b) Marks 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No. of students 3 5 6 8 4 4 3

(c) Volume of water (l) 5-10 10-15 15-20 20-25 25-30 30-40
No. of families 4 12 16 6 2 1

(d) Age (years) 0-5 5-10 10-15 15-20 20-25 25-30
No. of persons 5 20
15 20 10 2

(e) Age (years) 60-65 65-70 70-75 75-80 80-85 85-90
No. of students 7 5 8 4 3 3

(f) x 4-8 8-12 12-16 16-20 20-24 24-28 28-32 32-36 30-40
f 6 10 18 30 15 12 10 6 5

6. Calculate quartile deviation and its coefficient from the following data:

(a) Marks Less than 30 30-40 40-50 50-60 60-70 70 and above

No. of students 3 6954 2

(b) x Below 25 25-30 30-35 35-40 40-45 45 and above

f 5 12 22 25 20 9

(c) x Below 30 30-40 40-50 50-60 60-70 70 and above

f 2 4686 4

(d) Income (Rs.) Below 50 50-70 70-90 90-110 110- 130-150 150 above
130

Workers 5 10 20 25 18 12 10

7. Calculate the quartile deviation and its coefficient from the given data:

(a) Class interval 20-29 30-39 40-49 50-59 60-69 70-79
Frequency 100 80 75 95 70 40

(b) Marks 0-9 10-19 20-29 30-39 40-49 50-59

No. of students 1 4 8 10 5 3

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8. (a) Calculate the semi-interquartile range and its coefficient of the following data:
(prepare a frequency distribution table taking class interval 10)

70 55 51 42 57 45 60 47 63 53 33 65 39
82 55 64 58 61 65 42 50 52 54 45 45 25
36 59 63 39 65 45 49 54 64 75 42 41 52
35 30 35 15 48 26 20 40 55 46 18

(b) Prepare a frequency distribution table taking (0 - 4) as one of the class. The calculate
the quartile deviation. 7, 3, 10, 4, 1, 9, 11, 18, 8, 5, 6,4, 13, 17,6,8, 12,17,19, 5, 3,
17, 16, 3, 2, 14, 13, 4, 10

2. (a) 25 (b) 20 (c) 20 (d) 35
3. (a) 0.33 (b) 0.33 (c) 0.83 (d) 0.7959
4. (a) 75 (b) 36 (c) 21.83
5. (a) 11.56, 0.23 (b) 13.18, 0.25 (c) 3.51, 0.22 (d) 5.13, 0.38
(f) 5.5, 0.27
(e) 6.31, 0.088 (b) 5.008, 0.137 (c) 10.83, 0.20 (d) 23.33, 0.22
6. (a) 10.21, 0.215 (b) 8.035, 0.251
7. (a) 13.79, 0.30 (b) 3.97, 0.3683
8. (a) 9.62, 0.1924

12.2 Mean Deviation

Mean deviation is a measure of dispersion that is based on all the values of a set of data. It
is defined as the arithmetic mean of the absolute deviations taken from central value (mean,
median or mode). It shows the variation of the items from an average. Therefore, it is also
called average deviation. Mean deviation is denoted by MD.

While computing mean deviation, the deviation of items can be taken from mean, median
or mode. However, it gives the best results when deviations are taken from the median. In
calculation of mean deviation, negative sings are ignored and all the values are treated as
positive.

Calculation of Mean Deviation for Continuous Series :

Let m be the mid values of corresponding classes in continuous series. Then,

Mean Deviation from mean = ∑ f|m- x|
Mean Deviation from median =
∑ Nf|m-Md|
N
∑ f|m-Mo|
Mean Deviation from mode = N

Here, we study only mean deviation from mean or median

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Mean deviation is an absolute measure of dispersion. For computing two or more series
having different units, we calculate coefficient of mean deviation. The relative measure of
dispersion based on mean deviation is called coefficient of mean deviation.

Coefficient of M.D. from mean = M.D. from mean
mean
M.D. from median
Coefficient of M.D. from median = median

Worked out Examples

Example 1. In a grouped data if ∑f|m-x| =400, x = 20, N = 50, find M.D. and its
Solution: coefficient.

Example 2. Here, N = 50, x = 20, ∑f|m-x| =400

∑ f|m - x | 400
Now, Mean Deviation from mean = 50 = 50 = 8
M.D. from mean 8
coefficient of M.D. from mean = = 20 = 0.4
Mean

Calculate the mean deviation and it coefficient from the following data :

(i) deviations from mean

(ii) deviations from median.

Class 0 - 10 10 -20 20 -30 30 -40 40 -50
Frequency 5 8 15 10 6

Solution: (i) To calculate Mean Deviation from mean.

Class Mid-values Frequency (f) fm |m-x| f|m - x|
(m)
104.5
0-10 5 5 25 20.9 87.2
13.5
10-20 15 8 120 10.9 91.0
114.6
20-30 25 15 375 0.9 6f|m - x| =
410.9
30-40 35 10 350 9.1

40-50 45 6 270 19.1

N = 44 6fm = 1140

Mean ( x ) = ∑fm = 1140 = 25.9
N 44
∑f|m-x |
Mean deviation from mean = 410N.9
= =
44 9.34

Coefficient of M.D. from mean = M.D. from mean
mean
9.34
= 25.9 = 0.36

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(ii) Calculation of mean deviation from median:

Class Mid-values Frequency (f) cf |m-md| f|m -md|
(m)

0-10 5 5 5 21 105

10-20 15 8 13 11 88

20-30 25 15 28 1 15

30-40 35 10 38 9 90

40-50 45 6 44 19 114

N = 44 6f|m - md| =
412

To calculate median,

N = 44 = 22
2 2

c.f just greater than 22 is 28. Hence median class is (20 -30).

L = 20, c.f. = 13, f = 15, i = 10

N - c.f
2
Median (Md) =L+ ×i
f
22 - 13
= 20 + 15 × 10

= 20 + 6 = 26

Now, M.D. from median = ∑f |m-Md|
N
412
= 44 = 9.36

Exercise 12.2

Short Question

1. (a) Define Mean Deviation

(b) Write down the formula to compute mean deviation for continuos series of data:
(i) from mean

(ii) from median

(c) Write down the formula to compute the coefficient of mean deviation:

(i) from mean

(ii) from median
2. (a) In a grouped data, if ∑ f|m- x | = 400, N = 80, x = 40, find the M.D. and its

coefficient .

(b) In a continuous series of data, if ∑f |m-Md| = 250, Md = 22, N = 25, find the M.D.
and its coefficient.

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Long questions
3. Calculate the mean, deviation from (i) mean (ii) median for the following data :

(a) Class interval 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 2 3 4 9 2 4

(b) Class interval 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 10 12 25 35 40 50

(c) Marks 0-10 10-20 20-30 30-40 40-50

Number of students 9 6 4 12 9

(d) Class interval 0≤x≤10 10≤x≤20 20≤x≤30 30≤x≤40
Number of students 15 12 7 8

4. Calculate mean deviation from median and its coefficient for the following grouped
data :

(a) Marks 20-30 30-40 40-50 50-60 60-70

No. of students 5 7 8 6 4

(b) Marks 0-20 20-40 40-60 60-80 80-100
No. of students 5 6 7 10 12

(c) Marks 0-10 10-20 20-30 30-40 40-50

No. of students 5 8 15 16 6

(d) Profit (Rs) 0-10 10-20 20-30 30-40 40-50 50-60
No. of persons 10 12 25 35 40 50

5. Compute mean deviation and its coefficient from mean of the following date :

(a) Mid point 2 6 10 14 18 22 26
Frequency 7
13 10 5 8 4 3

(b) Mid-value 5 15 25 35 45 55
No. of students 2 46864

6. Construct a frequency distribution table taking a class interval of 10 and calculate the
M.D. from median.

28, 49 37, 5, 18, 14, 24, 7, 38, 46, 30, 21, 16, 31,

45, 27, 10, 4, 17, 29, 35, 36, 41, 47, 44, 33, 34, 17, 18, 20.

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7. Compute the mean deviation its coefficients, from (i) mean and (ii) median from the
given data :

Age (years) Number of people
Less then 10 10
Less than 20 25
Less than 30 40
Less than 40 45
Less than 50 70
Less than 60 85
Less than 70 100
Less than 80 110
Less than 90 120

2. (a) 5, 0.125 (b) 10, 0.45 (b) (i) 12.57, 0.33 (ii) 12.45, 0.28
3. (a) (i) 11.46, 0.35 (ii) 11.25, 0.34 (d) (i) 9.59, 0.56 (ii) 9.05, 0.60
(ii) 13.21, 0.43 (c) 9.56, 0.34 (d) 12.45, 0.30
(c) (i) 13.43. 0.51 (b) 23.60, 0.37
4. (a) 10.58, 0.24 (b) 11.73, 0.36 (ii) 19.75, 0.43
5. (a) 6.05, 0.36 7.(i) 16.32, 0.37
6. 12.67, 0.63

12.3 Standard Deviation

Standard deviation is the positive square root of the mean of the square of the deviations
about mean. It is also known as "Root mean square deviation" Standard deviation is denoted
by Greek letter V (read as sigma).

Among all the methods of finding out dispersion, standard deviation is regarded as the best
because of the following reasons.

1. Its value of based on all the variate values.
2. The deviation of each variate is taken from the mean.
3. Algebraic sign of each deviation is considered.

Calculation of Standard Deviation for Continuous Series :
To calculate standard deviation for a continuous series, the following methods can be applied.

(a) Direct Method

In this method, the formulae used is

( )V =
∑fm2 - ∑fm 2
N N ..............................(I)

where, m is mid-value of each class.

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To use the above formula, follow the following steps.
1. Find the mid value of each class which is denoted by m.
2. Find ∑f = N
3. Find the products fm and ∑fm
4. Multiply fm by m to get fm2 and ∑fm2,
5. Substitute all above values of ∑fm, ∑fm2, N in formula (I), we get standard deviation of

given data.

(b) Actual Mean Method

In this method, deviations of each mid point of class intervals are taken from the

arithmetic mean. The formula used in this method is V = ∑f(m - x )2 ...(II)
N
Note :
Above formulae (I) and (II) represent same direct method

To use this formula, follow the steps below:

1. Find the mid point of each class which is denoted by m

2. Find arithmetic mean, x = ∑fm where N= ∑f
N
3. Take deviations of each mid point from the arithmetic mean. i.e. (m - x )

4. Square each of (m - x ), to get (m - x )2

5. Multiply each of (m - x )2 with their corresponding frequencies to get f(m - x )2 and
obtain ∑f(m - x )2

6. Substitute the values of each of ∑f (m- x )2 and N in formulae (II), we get standard
deviation.

(c) Short-cut Method

In this method, deviation of each of mid-point the class is taken from assumed mean.

The fromula used in this method is

( )V =
∑fd2 - ∑fd 2
N N ......... (III)

Where, d = m - a

a = assumed mean

m = mid point of each class .

The following step is used in this method:
1. Find the mid-point of each class which is denoted by m.

2. Take middle figure of mid points as an assumed mean and denote it by a. Take deviations
of each mid point from the assumed mean. Denote it by d i.e. d = m - a

3. Multiply each of d with the corresponding frequency f, to get fd and obtain ∑fd.

4. Multiply each of fd with the corresponding d to get fd2 and then find ∑fd2.

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5. Substitute all the values of ∑fd2 , ∑fd, N in formula (III), we get standard deviation.

(d) Step Deviation Method :

This method is used mostly in the cases, when each of variety value has a common
factor. In case of continuous series, the common factor, is h, the class-size. The formula
used in this method is

( )V = ∑fd'2 - ∑fd' 2
N N
× h ........... (IV)

Where, d' = m -a
h
a = assumed mean

m = mid point of each class

h = class size

The following steps are used in this method:
1. Find the mid point of each class which is denoted by m.

2. Take an assumed mean 'a' from the mid figure of mid-points.

3. Find deviation of each mid point from the assumed mean i.e. (m - a)

4. Divide each of (m - a) by h, denote it by d' = m -a
h
5. Multiply each of d' by the corresponding frequency f to get fd' and find ∑fd'

6. Multiply each of fd' by d' and denote thus by fd'2 and find ∑fd'2

7. Substitute the values ∑fd' , ∑fd', ∑fd'2, N in formula (IV) to get standard deviation.

Coefficient of Standard Deviation

The relative measure of standard deviation is known as the coefficient of standard deviation.

? Coefficient of S.D. = standard deviation = V
mean x

If the coefficient of standard deviation is multiplied by 100%, then it is called coefficient of

variation (C.V.).

? Coefficient of variation (c.v) = V × 100%
x

Varience

The square of standard deviation is called varience. If standard deviation is V, the varience
is V

Example : If standard deviation of a set of data is 4,

then the varience is V2 = 16.

i,e, S.D. (V)= 4,

varience (V2) = 42 = 16

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Worked out Examples

Example 1. Calculate the standard deviation from given data by direct method.

Class 40-50 50-60 60-70 70-80 80-90 90-100
20 14
Frequency 6 10 20 30

Solution: To calculate standard deviation by direct method :

Example 2. Class Mid-value f fm fm2
Solution: 12150
40-50 45 6 270 30250
84500
50-60 55 10 550 168750
144500
60-70 65 20 1300 126350
6fm2 = 566500
70-80 75 30 2250

80-90 85 20 1700

90-100 95 14 1330

N = 100 6fm = 7400

( )∑fm2- ∑fm 2
N N
Standard deviation (V) =

( )=
566500 - 7400 2
100 100

= 5665 - 5476 = 13.74

Find the standard deviation from the following data by
(a) Actual mean (or direct Method) (b) Short -cut method
(c) Step deviation method
Also find the coefficient of S.D. and coefficient of variation.

x 0 -10 10-20 20-30 30-40 40-50 50-60
F 4 6 10 20 6 4

(a) Actual mean method (or direct method)
To calculate standard deviation by actual mean method (or direct method).

x Mid-value f fm |m-x| |m-x|2 f|m - x|2
4
0-10 5 6 20 -26 676 2704
10-20 15 10 90 -16 256 1536
20-30 25 20 250 -6 36 360
30-40 35 6 700 4 16 320
40-50 45 4 270 14 196 1176
50-60 55 220 24 576 2304
N = 50 6fm = 6fm = 6f|m - x|2
7400 1550 = 8400

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Mean (x ) = ∑fm = 1550 = 31
N 50
∑f(m - x )2
Standard deviation =
N

= 8400 168 = 12.96
50 =

coefficient of S.D. (V) = V × 100%
x

= 12.96 × 100% = 41.81%
31

(b) To calculate standard deviation by shortcut method.

Let = 35

x Mid-value f d=m–d fd fd2

0-10 5 4 -3 -120 3600
-120 2400
10-20 15 6 -20 -100 1000

20-30 25 10 -10

30-40 35 20 0 00
40-50 45 6 10 60 600

50-60 55 4 20 80 1600

N = 50 6fd = -200 6fm = 1550

( )∑fd2– ∑fd 2
N N
Standard deviation (V) =
= ( )9200– – 200 2
50 50

= 184 – 16 = 168 = 12.96

(c) To calculate standard deviation by step-deviation method

x Mid-value (m) f m – 35 fd' fd'2
10

0-10 5 4 -3 -12 36

10-20 15 6 -2 12 24

20-30 25 10 -1 -10 10

30-40 35 20 0 0 0

40-50 45 61 6 6

50-60 55 4 2 8 16

N = 50 6fd' = -20 6fd'2 = 92

We have from above table, N = 50, ∑fd' = –20, ∑fd'2

Standard deviation (V) = ( )∑fd'2–∑fd' 2
N N ×h

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( )= 92 – –20 2 × 10
50 50

= 46 – –4 × 10
25 25

= 42 × 10 = 1.296 × 10 = 12.96
25

Example 3. Calculate the standard deviation from the given data

Class 0 -10 10-20 20-30 30-40 40-50
Frequency 4 10 15 18 20

Solution: To calculate standard deviation, the given data can be written in the following table:

x Mid-value f d = m– a fd fd2
h
16
0-10 5 4 -2 -8 6
0
10-20 15 6 -1 -6 3
8
20-30 25 = a 5 0 0 6fd2 = 38

30-40 35 31 3

40-50 45 22 4

N = 50 6fd = -7

standard deviation (V) = ( )∑fd2 – ∑fd 2
NN ×h

where, h = 10, ∑fd'2 = 33, ∑fd' = –7

( )= 33 – –7 2
20 20 × 10

? V = 1.65 – 0.1225 × 10 = 12.36

Exercise 12.3

Very Short Question
1. (a) Define standard deviation

(b) Write the formulas to calculate standard devition for continuous series of data by
(i) direct method
(ii) actual mean method
(iii) Short-cut method
(iv) step-deviation method

(c) Define coefficient of variation with formulae
(d) Define varience

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Short questions

2. Calculate standard deviation and its coefficient for continuous series, if

(a) ∑fm2 = 566500, N = 100, ∑fm = 7400, x = 31
(b) ∑fd2 = 993, N = 60, x = 37, ∑fd = –130
(c) ∑fd' = –20, ∑fd'2 = 92, h = 10, x = 37, N = 100
(d) ∑fd' = –4, ∑fd'2 = 28, N = 29, x = 24.5, i = 10
Long Questions

3. Calculate the standard deviation and its coefficient from the following data by

(i) direct method (ii) short-cut method (iii) step-deviation

(a) Class Interval 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 10 8 32 40 22 18

(b) Marks 0-10 10-20 20-30 30-40 40-50

No. of students 7 12 24 10 7

(c) Wages (Rs.) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

No. of Workers 12 18 35 42 50 45 20 8

(d) Age (yrs.) 20-25 25-30 30-35 35-40 40-45 45-50

No. of persons 170 110 80 45 40 45

4. Calculate the standard deviation and coefficient of variation:

(a) Mark Secured 0-20 20-40 40-60 60-80 80-100

No. of Students 2 8 16 10 4

(b) Height (cm) 0-8 8-16 16-24 24-32 32-40
No. of plants 6
7 10 8 9

5. Calculate the standard deviation and coefficient of variation from the following:

(a) x 0≤x<10 10≤x<20 20≤x<30 30≤x<40 40≤x<50
f 7 10 14 12 6

(b) Mid-value 4 8 12 16 20 24

Frequency 10 15 11 16 14 5

(c) x less than 10 less than 20 less than 30 less than 40 less than 50
f 12 19 24 33 40

(d) x above 20 above 40 above 60 above 80 above 100 and less than 120

f 50 42 30 18 7

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(e) Marks 0-10 0-20 0-30 0-40 0-50
54
No. of students 7 18 30 42

6. From the data given below which series is more variable (inconsistent).

Variable 10-20 20-30 30-40 40-50 50-60 60-70
Section A 10 18 32 40 22 18

Frequency 18 22 40 32 20 10
Section B

Project work

List the marks obtained by the students in your class in optional mathematics and compulsory
mathematics. Find the standard deviation, coefficient of standard deviation and coefficient
of variation (C.V.). Compare the marks obtained in these two subjects in terms of C.V.

2. (a) 13.74, 0.4432 (b) 3.44, 0.093 (c) 9.34, 0.2535 (d) 9.7, 0.3959
3. (a) 13.72, 0.3156 (b) 11.39, 0.4617 (c) 17.25, 0.426 (d) 8.22, 0.269
4. (a) 20.27, 38.24 (b) 10.86, 50.75
5. (a) 12.29, 49.16% (b) 6.07, 45.47% (c) 15.03, 75.15% (d) 25.79, 37.41%

(e) 12.87, 50.27% CV(A') = 33.34%, CV(B) = 37.02%
6. V (A) = 14.05, V (B) = 14.10

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Curriculum Development Centre, Sanothimi, Bhaktapur

Syllabus of Optional Mathematics - Grade 10

S.N. Content Course Title Tentative
1. Algebra periods

2. Continuity Functions: Algebraic and trigonometrical functions with 35
3. Matrix
4. Coordinate their graphs of the types y = mx + c, y = ax2, a ≠ 0; y = ax3,

Geometry a ≠ 0; y = sinA, y = cosA, y = tanA(–2S ≤ A ≤ 2S).

Composite function of two functions and inverse
functions with their presentation in arrow diagrams.

Polynomials: Simple operation of polynomials,
Synthetic division, Remainder and Factor theorem and
their applications; Solutions of polynomial equations
upto degree three using Remainder and Factor theorems.

Sequence and Series: Arithmetic sequence - Introduction,
General term, Mean, Sum of the terms of an arithmetic
sequence, Sum of the first n natural numbers including
odd and even natural numbers.

Geometric Sequence: Introduction, General term, Mean,
Sum of the finite terms of a geometric sequence.

Linear Programming: Introduction, Linear inequalities,
Detriment of inequality by graph, Maximization and
minimization of linear programming problems.

Quadratic Equations and Graphs: Graphs of quadratic 10
and cubic functions, Solution of quadratic equations
using graph, Solution of simultaneous linear and 15
quadratic equations by graphical and substitution 30
method.

Simple Concept of Continuity: Investigation of
continuity in set of real numbers, Investigation of
continuity and discontinuity in different types of set of
numbers, Investigation of the continuity of the function
using graph, Symbolic representation of the continuity
of the function.

Determinant of matrix of order 2 × 2, Inverse of a
2 × 2 matrix, Solution of simultaneous linear equations
using matrix method, Cramer's rule, and its application
upto a determinant of order 2 × 2.

Straight Line: Angle between two straight lines,
Condition for two lines to be parallel and perpendicular.

Pair of Straight Lines: Equations of the lines given
by the homogeneous equation of degree two and the
angle between them. Conditions for the two lines to be
coincident and perpendicular.

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vedanta Excel In Opt. Mathematics - Book 10

Conic Section: Introduction, Different types of conic

section formed by the intersection of a cone and a plane.

Circle: Definition, Equation of the circle of the types:

x2 + y2 = r2, (x – h)2 + (y – k)2 = r2, (x – x1) (x – x2) +
(y – y1) (y – y2) = 0 and x2 + y2 + 2gx + 2fy + c = 0 and
the related problems.

5. Trigonometry Trigonometric Identities: Trigonometric ratios of 35
18
multiple and submultiple angles (sine, cosine and
15
tangent only), Transformation of identities into the sum, 12

difference and the product form (in sine and cosine only)

Conditional Identities: Solution of the trigonometric

identities under the condition A + B + C = Sc

Trigonometric Equations: Solution of trigonometric

equation (upto degree two) (0 ≤ T < 2Sc)

Height and Distance: Word problems of height and

distance consisting of two angle - angle of elevation and

angle of depression.

6. Vectors Scalar product of two vectors (dot product), Condition

for two vectors to be perpendicular.

Vector Geometry: Mid-point theorem, Section formula.

Theorems:

(i) The straight line joining the middle points of two

sides of a triangle is parallel to and half of the third

side.

(ii) The straight line joining the vertex and the middle

point of the base is perpendicular to the base.

(iii) The straight lines joining the middle points of

the sides of a quadrilateral taken in order is a

parallelogram.

(iv) The diagonals of a parallelogram bisect each other.

(v) The diagonals of a rectangle are equal.

(vi) The diagonals of a rhombus bisect each other at

right angles.

(vii) The angle in the semi-circle is a right angle.

(viii)The middle point of the hypotenuse of a right

angled triangle is equidistant from the vertices.

7. Transformation Coposite transformation of any of two transformations.

Reflection, Rotation, Translation and Enlargement,

Inversion transformation, Inverse circle, Use of matrix

in transformation.

8. Statistics Dispersion: Quartile deviation and its coefficient

(continuous series only), Mean deviation from mean

and median and its coefficient (continuous series only),

Standard deviation and its coefficient and analysis

(continuous series only), coefficient of variation (C.V.).

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vedanta Excel In Opt. Mathematics - Book 10

SEE Specification Grid 2076

Issued by CDC

K U A HA

SN Contents Topics Each Each Each Each TQ TM
of 1 of 2 of 4 of 5

Mark Marks Marks Marks

1. Algebra Function

Polynominals 2 3 2 1 8 21
Sequence and Series

Quadratic Equation and Graph

2. Limit and Numbers and Continuity

Continuity Discontinuity in Graph 1 - 1 - 25
Notational Representation of

Continuity

3. Matrix Determinant and Inverse of

Matrix

Solving Equations by Matrix 12 1 - 49

Method

Cramer's Rule

4. Coordinate Angle between two lines

Geometry Pair of straight lines 2 2 1 1 6 15
Conic Sections

Circle

5. Trigonometry Multiple and sub-multiple angles

Transformation of
Trigonometric Identies

Conditional Trigonometric 23 3 - 8 20

Identities

Trigonometric Equations

Height and Distance

6. Vectors Scalar Product 1 2 - 1 4 10
Vector Geometry

7. Transformation Combined Transformation

Inversion Transformation and 1 - 1 1 3 10
Inversion Circle

Matrix Transformation

8. Statistics Quartile Deviation

Mean Deviation -12 - 3 10
Standard Deviation and

Coefficient of Variation

Total 10 13 11 4 38 100

INDEX U = Understanding A = Application
K = Knowledge

HA = Higher Ability TQ = Total no. of Questions TM = Total Marks

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vedanta Excel In Opt. Mathematics - Book 10

SEE New Model Questions 2076

Issued by CDC-2076 (2019)

Group A (10 × 1 = 10)

1. (a) Define trigonometric function.

(b) What is the arithmetic mean between two numbers 'a' and 'b'?

2. (a) Write a set of numbers which is continuous in number line.

(b) If matrix A = p q , what is the value of |A|?
r s
3. (a) If the slopes of two straight lines are m1 and m2 respectively and T be the angle
between them, write the formula for tanT.

(b) Which geometric figure is formed if a plane intersects a cone parallel to its base?

4. (a) Express sin2A in terms of tanA.

(b) Define angle of elevation.

5. (a) What is the scalar product of two vectors oa and ob if the angle between them is T?

(b) In an inversion transformation if P' is image of P and r is radius of inversion circle
with centre O, write the relation of OP.OP' and r.

Group B (13 × 2 = 26)

6. (a) Find f-1(x) if f(x) = 4x + 5.

(b) If g(x) = 2x – 1 and f(x) = 4x, find the value of gof(x).

(c) What are the points of intersection of the curve f(x) = x2 – 1 and f(x) = 3?

7. (a) If A = 2 –1 , find |A| and write if A–1 is defined.
3 1
(b) According to Cramer's rule, find the values of D1 and D2 for ax + by = c and
px + qy = r.

8. (a) Find the slopes of two straight lines 3x + 4y + 5 = 0 and 6x + 8y + 7 = 0 and
write the relationship between them.

(b) Find the single equation for the pair of lines represented by 3x + 2y = 0 and
2x – 3y = 0.

9. (a) Convert sin6A . cos4A into sum or difference of sine or cosine.

(b) Express 1 sinA in terms of sub-multiple angle of tangent.
+ cosA

(c) If 2sin2T = 3, find the value of T. (0° ≤ T ≤ 180°)

10. (a) Find the angle between two vectors oa and ob if |oa | = 2, |ob | = 12 and oa .ob = 12.
A
(b) From the adjoining figure, find AoP and express op in terms
of oa and ob . oa op P

(c) If the standard deviation of a set of data is 0.25, find its O ob B
variance.

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vedanta Excel In Opt. Mathematics - Book 10

Group C (11 × 4 = 44)

11. Solve : x3 – 3x2 – 4x + 12 = 0

12. Optimize P = 5x + 4y under the given constraints : x – 2y ≤ 1, x + y ≤ 4, x ≥ 0, y ≥ 0

13. For a real valued function f(x) = 2x + 3

(a) Find the values of f(2.95), f(2.99), f(3.01), f(3.05), and f(3).

(b) Is this function continuous at x = 3?

14. By using matrix method, solve the following system of equations: 3x + 5y = 11, 2x – 3y = 1

15. Find the single equation of pair of straight lines passing through the origin and
perpendicular to the lines represented by 2x2 – 5xy + 2y2 = 0.

16. Find the value of : sin20° . sin30° . sin40° . sin80°

17. If A + B + C = Sc, prove that sin2A – sin2B + sin2C = 2sinA.cosB.sinC

18. From a point at the ground level infront of a tower, the angle of elevations of the top
and bottom of flagstaff 6 m high situated at the top of a tower are observed 60° and
45° respectively. Find the height of the tower and the distance between the base of the
tower and point of observation.

19. Find the 2 × 2 matrix which transforms a unit square to a parallelogram 0 34 1 .
0 01 1
20. Find the mean deviation from mean and its coefficient from given data:

Marks obtained 0-10 10-20 20-30 30-40 40-50
No. of students 2 3 6 5 4

21. Find the standard deviation and coefficient of variation from given data:

Age 0-4 4-8 8-12 12-16 16-20 20-24
No. of students 7
7 10 15 7 6

Group D (4 × 5 = 20)

22. A contractor on construction job specifies a penalty for delay of completion beyond
a certain date as: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the
third day and so on. The penalty for each succeeding day is Rs. 50 more than that of
the preceding day. How much money does the contractor have to pay as penalty, if he
delays the work by 30 days?

23. On a wheel, there are three points (5, 7), (–1, 7) and (5, –1) located such that the distance
from a fixed point to these points is always equal. Find the coordinate of the fixed point
and then derive the equation, representing the locus that contains all three points.

24. By using vector method, prove that the quadrilateral formed by joining the mid-points
of adjacent sides of a quadrilateral is a parallelogram.

25. The coordinates of vertices of a quadrilateral ABCD are A(1, 1), B(2, 3), C(4, 2),
and D(3, –2). Rotate this quadrilateral about origin through 180°. Reflect this image
of quadrilateral about y = –x. Write the name of transformation which denotes the
combined transformation of above two transformations.

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