Class 10

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year :

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. (AS₃, AS₄)
Solution:
Maximum number of patients joined in the age group 35 – 45
∴ Modal class is 35 – 45.
Lower limit of the modal class ‘l’ = 35
class size, h = 10
Frequency of modal class. f₁ = 23
Frequency of the class preceding the modal class f₀ = 21
Frequency of the class succeding the modal class f₂ = 14

∴ Mode = l + \(\frac{\left(f_1-f_0\right)}{2 f_1-f_0-f_2}\) × h
= 35 + \(\left[\frac{23-21}{2 \times 23-21-14}\right]\) × 10
= 35 + \(\left[\frac{2}{46-35}\right]\) × 10
= 35 + \(\frac{2}{11}\) × 10 = 35 + 1.81818 ….
= 36.8 years

Mean \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = \(\frac{2830}{80}\) = 35.37 years.

Interpretation : Mode age is 36.8 years, Mean age = 35.37 years.
Maximum number of patients admitted in the hospital are the age 36.8 years, while on an average the age of patients admitted to the hospital is 35.37 years. Mode is less than the mean.

Question 2.
The following data gives the information on the observed life times (in hours) of 225 electrical components : (AS₄)

Lifetimes (in hours)	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100	100 – 120
Frequency	10	35	52	61	38	29

Determine the modal life times of the components.
Solution:

Class interval	Frequencies
0-20	10
20-40	35
40-60	52
60-80	61
80 – 100	38
100 – 120	29

Since the maximum frequency 61 is in the class 60 – 80, this is the required modal class.
Modal class frequency, f₁ = 61
Frequency of the class preceding the modal class f₀ = 52
Frequency of the class succeding the modal class f₂ = 38
lower boundary of the modal class l = 60
Height of the class, h = 20
∴ Mode (z) = l + \(\frac{\left(f_1-f_0\right)}{2 f_1-\left(f_0+f_2\right)}\) × h
= 60 + \(\left[\frac{61-52}{2 \times 61-(52+38)}\right]\) × 20
= 60 + \(\frac{9}{122-90}\) × 20
= 60 + \(\frac{9}{32}\) × 20
= 60 + 5.625 = 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of Gummadidala village. Find the modal monthly expenditure of the families. Also find the mean monthly expenditure. (AS₄)

Expenditure (in rupees)	1000-1500	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of families	24	40	33	28	30	22	16	7

Solution:

Assumed mean (a) = 3250
Σf_i = 200; Σf_iu_i = – 235
Mean monthly income = \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
= 3250 – \(\frac{235}{200}\) × 500
= 3250 – 587.5
= ₹ 2662.50
Since the maximum families 40 lies in the class 1500 – 2000, this is the required modal class.
Lower boundary of modal class (l) = 1500
Frequency of the modal class (f₁) = 40
Frequency of the class preceding the modal class f₀ = 24
Frequency of the class succeding the modal class f₂ = 33
Height of the class, h = 500
Mode (z) = l + \(\frac{\left(f_1-f_0\right)}{2 \times f_1-\left(f_0+f_2\right)}\) × h
⇒ 1500 + \(\frac{40-24}{2 \times 40-(24+33)}\) × 500
= 1500 + \(\frac{16 \times 500}{80-57}\) = 1500 + \(\frac{8000}{23}\)
⇒ 1500 + 347.826
= ₹ 1847.83
Hence, modal monthly income = ₹ 1847.83

Question 4.
The following distribution gives the state-wise, teachers, student ratio in higher second’ ary schools of India. Find the mode and mean of this data. Interpret the two measures. (AS₃, AS₄)

Number of students	15-20	20-25	25-30	30-35	35-40	40-45	45-50	50-55
Number of States	3	8	9	10	3	0	0	2

Solution:

Mean (\(\overline{\mathrm{x}}\)) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
a – assumed mean = 32.5; h – height of the class = 5
∴ \(\overline{\mathrm{x}}\) = 32.5 – \(\frac{22}{35}\) × 5 = 32.5 – 3.28 = 29.22
Since the maximum number of states ‘10′ lies in the class interval 30 – 35, this is the modal class.
Lower boundary of the modal class, l = 30
Frequency of the modal class, f₁ = 10
Frequency of the class preceding the modal class, f₀ = 9
Frequency of the class succeding the modal class, f₂ = 3
Height of the class h = 5
Mode (z) = l + \(\left[\frac{f_1-f_0}{\left(f_1-f_0\right)+\left(f_1-f_2\right)}\right]\) × h
⇒ 30 + \(\frac{10-9}{(10-9)+(10-3)}\) × 5
= 30 + \(\frac{1 \times 5}{1+7}\) = 30 + \(\frac{5}{8}\)
= 30 + 0625 ⇒ 30.625.
Mode of states have a teacher students ratio 30.625 and on an average of this ratio is 29.22.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one international cricket matches.

Runs	3000-4000	4000-5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000	10000-11000
Number of batsmen	4	18	9	7	6	3	1	1

Find the mode of the data. (AS₄)
Solution:

Class interval	Frequencies
3000 – 4000	4
4000 – 5000	18
5000 – 6000	9
6000 – 7000	7
7000 – 8000	6
8000 – 9000	3
9000 – 10000	1
10000- 11000	1

Maximum no.of batsmen are in the class 4000 – 5000
Modal class is 4000 – 5000
Lower boundary of the modal class ‘l’ = 4000
frequency of the modal class, f₁ = 18
frequency of the class preceding the modal class f₀ = 4
frequency of the class succeeding the modal class f₂ = 9
Size of the class, h = 1000
Mode(z) = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{\left(\mathrm{f}_1-\mathrm{f}_0\right)+\left(\mathrm{f}_1-\mathrm{f}_2\right)}\right]\) × h
Mode (z) = 4000 + \(\frac{18-4}{(18-4)+(18-9)}\) × 1000
= 4000 + \(\frac{14}{14+9}\) × 1000
= 4000 + \(\frac{14000}{23}\) = 4000 + 608.695
= 4608.69
≅ 4608.7 runs

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes and summarised this in the table given below.

Number of cars	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80
Frequency	7	14	13	12	20	11	15	8

Find the mode of the data. (AS₄)
Solution:

No.of cars	Frequency
0 – 10	7
10 – 20	14
20 – 30	13
30 – 40	12
40 – 50	20
50 – 60	11
60 – 70	15
70 – 80	8

Since, the maximum frequency is 20, the modal class is 40 – 50.
Lower boundary of the modal class, ‘l’ = 40
Frequency of the modal class, f₁ = 20
Frequency of the class preceding the modal class f₀ = 12
Frequency of the class succeeding the modal class f₂ = 11
Height of the class, h = 10;
Mode (z) = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{\left(\mathrm{f}_1-\mathrm{f}_0\right)+\left(\mathrm{f}_1-\mathrm{f}_2\right)}\right]\) × h
= 40 + \(\frac{(20-12)}{(20-12)+(20-11)}\) × 10
= 40 + \(\frac{8}{8+9}\) × 10
= 40 + \(\frac{80}{17}\)
= 40 + 4.70588
= 44.705
≅ 44.7 cars.

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rupees)	250-300	300-350	350-400	400-450	450-500
Number of workers	12	14	8	6	10

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive. (AS₅)
Solution:
For less than ogive, we take the upper class limits on X-axis and their corresponding cumulative frequencies on Y – axis, choosing a convenient scale.

Daily income (in Rupees)	250-300	300-350	350-400	400-450	450-500
Number of workers	12	14	8	6	10

Class Interval (Upper Limits)	frequency	cumulative frequency	Points
300	12	12	(300, 12)
350	14	26	(350, 26)
400	8	34	(400, 34)
450	6	40	(450, 40)
500	10	50	(500, 50)

The points to be plotted are (300, 12) (350, 26) (400, 34) (450, 40) and (500, 50)
Scale :
On X-axis : 1 cm = 50 units
On Y-axis : 1 cm = 5 units

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows : (AS₅)

Weight (in kg)	Number of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
The points to be plotted on a graph paper are : (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)
Scale :
On X-axis : 1 cm = 2 units
On Y-axis : 1 cm = 2 units

Number of observations = 35
Here \(\frac{\mathrm{n}}{2}\) = \(\frac{35}{2}\) = 17.5
Locate the point 17.5 on the Y – axis. From the point, draw a line parallel to the X – axis cutting the curve at a point. From the point, draw a perpendicular to the X – axis.

The point of intersection of this perpendicular with the X – axis determines the median of the given data as 46.8 kg.

Weight	Number of Students (c.f)	Frequency (f)
Below 38	0	0
38-40	3	3
40-42	5	2
42-44	9	4
44-46	14	5
46-48	28	14
18-50	32	4
50-52	35	3

Number of observations = n = 35
\(\frac{\mathrm{n}}{2}\) = \(\frac{35}{2}\) = 17.5
17.5 belongs to the class 46 – 48
∴ Median class = 46 – 48
l – lower boundary of class = 46
f – frequency of the median class = 14
c.f. = 14
Class size = 2
Median = l + \(\frac{\left(\frac{n}{2}-c . f .\right)}{f}\) × h
= 46 + \(\frac{17.5-14}{14}\) × 2
= 46 + \(\frac{3.5}{14}\) × 2
= 46 + \(\frac{7}{14}\) = 46 + \(\frac{1}{2}\) = 46.5
Hence, median is 46.5 either ways.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village. (AS₅)

Production yield (Qui/Hec)	50-55	55-60	60-65	65-70	70-75	75-80
Number of farmers	2	8	12	24	38	16

Change the distribution to a more than type distribution and draw its ogive.
Solution:
More than type distribution

Points to be plotted on the graph paper are:
(50, 100), (55, 98), (60, 90), (65. 78), (70, 54) and (75, 16)
Scale:
On X-axis: 1 cm = 5 units
On Y-axis: 1 cm = 10 units

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. (AS₄)

Monthly consumption	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Number of consumers	4	5	13	20	14	8	4

Median :
Solution:

Here n = 68; \(\frac{\mathrm{n}}{2}\) = \(\frac{68}{2}\) = 34
The median lies in the class 125 – 145.
Lower limit (l) of the median class = 125
Frequency of the median class (f) = 20 of (cumulative frequency of the class 105 – 125) = 22
class size (h) = 20
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 125 + \(\frac{\left[\frac{n}{2}-c f\right]}{f}\) × 20
= 125 + 12
= 137 units.

Mean :

Mean (\(\overline{\mathrm{X}}\)) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
Here a = 135, h = 20, Σf_i = 68, Σf_iu_i = 7
= 135 + \(\left[\frac{7}{68}\right]\) × 20
= 135 + \(\frac{35}{17}\)
= 135 + 2.05
= 137.05

Mode :
Since the maximum number of consumers have their monthly consumption (in units) in the interval 125 – 145
∴ lower limit of the modal class (l) = 125
class size (h) = 20
Frequency of the modal class (f₁) = 20
Frequency of the class preceding the modal class (f₀) = 13
Frequency of the class succeeding the modal class (f₂) = 14
Modal = l + \(\frac{\left(f_1-f_0\right)}{\left(2 f_1-f_0-f_2\right)}\) × h
= 125 + \(\frac{(20-13)}{(2 \times 20-13-14)}\) × 20
= 125 + \(\frac{7}{(40-13-14)}\) × 20
= 125 + \(\frac{7 \times 20}{13}\) = 125 + \(\frac{140}{13}\)
= 125 + 10.76 = 135.76 units

Comparison :
In this case, the three measures i.e., mean, median and mode are approximately equal.

Question 2.
If the median of 60 observations, give below is 28.5, find the value of x and y. (AS₁)

Class interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	5	x	20	15	y	5

Solution:

Hence, n = 60 (given)
\(\frac{\mathrm{n}}{2}\) = \(\frac{60}{2}\) = 30
⇒ 45 + x + y = 60
⇒ x + y = 60 – 45 = 15 ……………… (1)
The median is given as 28.5.
If lies in the class 20 – 30.
So, l = 20;
Frequency of the median class (f) = 20
cf (cumulative frequency of the class preceding the median class 10 – 20) = 5 + x
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
⇒ 28.5 = 20 + \(\left[\frac{30-(5+x)}{20} \times 10\right]\)
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
⇒ \(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
⇒ 25 – x = 2 × 8.5
⇒ 25 – x = 17
⇒ x = 25 – 17 = 8 ……………… (2)
from (1) and (2) : we get
8 in x + y = 15, 8 + y = 15
∴ y = 15 – 8 = 7
Hence, x = 8 and y = 7

Question 3.
A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. (policies are given only to persons having age 18 years onwards but less than 60 years.) (AS₄)

Age (in years)	Below 20	Below 25	Below 30	Below 35	Below 40	Below 45	Below 50	Below 55	Below 60
Number of policy holders	2	6	24	45	78	89	92	98	100

We find the class intervals and their corresponding frequencies to calculate the median age.
Solution:

Hence, n = 100: \(\frac{\mathrm{n}}{2}\) = \(\frac{100}{2}\) = 50
The median lies in the class 35 – 40. So l = 35
frequency of the median class (f) = 33
cf (cumulative frequency of the class preceeding the median dass 30 – 35) = 45
class size(h) = 5
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 35 + \(\frac{(50-45)}{33}\) × 5
= 35 + \(\frac{5 \times 5}{33}\) = 35 + \(\frac{25}{33}\)
= 35 + 0.76 = 35.76
Hence, the median age = 35.76 years.

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre and the data obtained Is represented in the following table: (AS₄)

Length (in mm)	118-126	127-135	136-144	145-153	154-162	163-171	172-180
Number of leaves	3	5	9	12	5	4	2

Find the median length of the leaves (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5 ………. 171.5 – 180.5)
Solution:

Here, n = 40; \(\frac{\mathrm{n}}{2}\) = \(\frac{40}{2}\) = 20
The median class is 144.5 – 153.5
Lower limit (l) of the median class = 144.5
cf (cumulative frequency of class preceding the median class 144.5 – 153.5) = 17
f (frequency of the median class) = 12
h (class size) = 9
using the formula, median
= l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 144.5 + \(\frac{(20-17) \times 9}{12}\) × 9
= 144.5 + \(\frac{3 \times 9}{12}\) = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25
= 146.75
∴ Median length = 146.75 mm.

Question 5.
The following table gives the distribution of the life-time of 400 neon lamps

Life time (in hours)	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of lamps	14	56	60	86	74	62	48

Find the median life time of a lamp. (AS₄)
Solution:

Life time (in hours) class intervals (C.I)	No.of lamps (f)	cumulative frequency (cf)
1500 – 2000	14	14
2000 – 2500	56	70
2500 – 3000	60	130
3000 – 3500	86	216
3500 – 4000	74	290
4000 – 4500	62	352
4500 – 5000	48	400

Here, n = 400; \(\frac{\mathrm{n}}{2}\) = \(\frac{400}{2}\) = 200
The median class is 3000 – 3500
lower limit (1) of the median class = 3000
cf (cumulative frequency of the class preceding the median class 2500 – 3000) = 130
f (frequency of the median class) = 86
h(class size) = 500
using the formula, Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 3000 + \(\frac{(200-130) \times 500}{86}\)
= 3000 + \(\frac{70 \times 500}{86}\)
= 3000 + 406.98
= 3406.98 hours
∴ Median life = 3406.98 hours.

Question 6.
loo surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters ¡n the English alphabet in the surnames was obtained as follows:

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Determine the median number of letters In the surnames. Find the mean number of letters in the surnames P Also, find the modal size of the surnames. (AS₄)
Solution:
Median:

No.of letters class intervals (C.I)	No.of surnames (f)	cumulative frequency (cf)
1-4	6	6
4-7	30	36
7-10	40	76
10-13	16	92
13-16	4	96
16-19	4	100

Here, n = 100; \(\frac{\mathrm{n}}{2}\) = \(\frac{100}{2}\) = 50
So, the median lies in the class 7 – 10
lower limit (1) of the median class = 7
cf (cumulative frequency of the class preceding median class 7 – 10) = 36
f (frequency of the median class) = 40
h (class size) = 3
using the formula. Median
= l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h × h
= 7 + \(\frac{(50-36) \times 3}{40}\)
= 7 + \(\frac{14 \times 3}{40}\) = 7 + \(\frac{21}{20}\)
= 7 + 1.05 = 8.05
Hence, the median of letters in the surnames = 8.05

Mean :

Mean (\(\overline{\mathrm{x}}\)) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\right]\)
Here a = 8.5, h = 3, Σf_i = 100, Σf_iu_i = -6
= 8.5 – \(\frac{6 \times 3}{100}\)
= 8.5 – \(\frac{18}{100}\)
= 8.5 – 0.18
= 8.32
Hence, the mean of the surnames = 8.32

Mode:

Class intervals (C.I)	Frequency (f)
1-4	6
4-7	30
7-10	40
10-13	16
13-16	4
16 -19	4

Since, the maximum number of surnames have number of letters in the interval 7 – 10, the modal class is 7 – 10
Lower limit (1) of the modal class = 7
frequency of the modal class (f₁) = 40
Frequency of the class preceding the modal class (f₀) = 30
frequency of the class succeeding the modal class (f₂) = 16
class size(h) = 3
∴ Mode = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right]\) × h
= 7 + \(\frac{(40-30)}{(2 \times 40-30-16)}\) × 3
= 7 + \(\left[\frac{10 \times 3}{80-30-16}\right]\)
= 7 + \(\frac{30}{34}\)
= 7 + \(\frac{15}{17}\)
= 7 + 0.88 = 7.88
Hence, the modal size of the surnames = 7.88

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students. (AS₄) (Mar ’16 (A.P))

Weight (in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75
Number of students	2	3	8	6	6	3	2

Solution:

Class Intervals weight in kgs (C.I)	No.of students (f)	cumulative frequency (cf)
40-45	2	2
45-50	3	5
50 -55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

Hence, n = 30; \(\frac{\mathrm{n}}{2}\) = \(\frac{30}{2}\) = 15
So, the median lies in the class 55 – 60
∴ l = 55
frequency of the median class (f) = 6
cf (cumulative frequency, of the class 50 – 55) = 13
class size (h) = 5
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 55 + \(\frac{(15-13) \times 5}{6}\)
= 55 + \(\frac{2 \times 5}{6}\)
= 55 + \(\frac{5}{3}\) = 55 + 1.67
= 56.67
Hence, the median weight of the students = 56.67 kgs.

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants	0 – 2	2 – 4	4-6	6 – 8	8 – 10	10 – 12	12-14
Number of houses	1	2	1	5	6	2	3

Solution:

∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = \(\frac{162}{20}\) = 8.1
!! Since f_i and x_i are of small values we use direct method.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory. (AS₄)

Daily wages in Rupees	200 – 250	250 – 300	300 – 350	350 – 400	400- 450
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

Here, the x_i are of large numerical values in 250 – 300. So, a = 275.
So we use Assumed Mean method then,
\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
Here, the Assumed mean is taken as 275.
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = 275 + \(\frac{1900}{50}\) = 275 + 38 = 313

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f. (AS₄)

Daily pocket allowance(in Rupees)	11 – 13	13 – 15	15 – 17	17 – 19	19 – 21	21 – 23	23 – 25
Number of children	7	6	9	13	f	5	4

Solution:

\(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
\(\overline{\mathrm{x}}\) = 18 (given)
⇒ 18 = \(\frac{752+20 f}{(44+f)}\)
⇒ 18(44 + f) = 752 + 20f ⇒ 20f – 18f = 792 – 752
⇒ 2f = 40
∴ f = \(\frac{40}{2}\) = 20

Question 4.
Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method. (AS₄)

Number of heart beats/minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Solution:

\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 75.5 + \(\frac{12}{30}\) = 75.5 + 0.4 = 75.9

Question 5.
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges. (AS₄)

Number of oranges	10-14	15-19	20-24	25-29	30-34
Number of baskets	15	110	135	115	25

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose ?
Solution:

Here, we use step division method where a = 22, h = 5
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h
⇒ 22 + \(\frac{25}{400}\) × 5
⇒ 22 + 0.31 = 22.31

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rupees)	100-150	150-200	200-250	250-300	300-350
Number of house holds	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method. (AS₄)
Solution:

Here, a = 225, h = 50
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h
= 225 + \(\frac{(-7)}{25}\) × 50 = 225 – 14 = 211
The average daily expenditure on food = ₹ 211

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collect for 30 localities in a certain city and is presented below.

Concentration of SO2 in ppm	0.00-0.04	0.04-0.08	0.08-0.12	0.12-0.16	0.16-0.20	0.20-0.24
Frequency	4	9	9	2	4	2

Find the mean concentration of SO₂ in the air. (AS₄)
Solution:

∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2.96}{30}\) = 0.0986666 …….
≅ 0.099 ppm

Question 8.
A class teacher has the following attendence record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term. (AS₄)

Number of days	35-38	38-41	41-44	44-47	47-50	50-53	53-56
Number of students	1	3	4	4	7	10	11

Solution:

Here, a = 51.5
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 51.5 – \(\frac{99}{40}\)
⇒ 51.5 – 2.475 = 49.025 ≅ 49 days

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. (AS₄)

Literacy rate in %	45-55	55-65	65-75	75-85	85-95
Number of cities	3	10	11	8	3

Solution:

Here a = 70, h = 10
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
⇒ \(\overline{\mathrm{x}}\) = 70 – \(\frac{2}{35}\) × 10
= 70 – \(\frac{20}{35}\) = 70 – 0.5714
= 69.4285 ≅ 69.43%

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.3

Question 1.
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients.

i) x² – 2x – 8
ii) 4s² – 4s + 1
iii) 6x² – 3 – 7x
iv) 4u² + 8u
v) t² – 15
vi) 3x² – x – 4
Solution:
i) Let p(x) = x² – 2x – 8
= x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
The zeroes of p(x) are given by P(x) = 0
⇒ (x + 2) (x – 4) = 0
⇒ x + 2 = 0 (or) x – 4 = 0
⇒ x = -2 (or) x = 4
Hence the zeroes of x² – 2x – 8 are -2 and 4.
Sum of the zeroes = -2 + 4 = 2

Product of the zeroes = -2 × 4 = -8

ii) Let p(s) = 4s² – 4s + 1
= (2s)² – 2(2s) (1) + (1)²
= (2s – 1)²
= (2s – 1) (2s – 1)
The zeroes of p(s) are given by p(s) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1 =0 (or) 2s – 1 = 0
⇒ s’ = \(\frac{1}{2}\) (or) s = \(\frac{1}{2}\)
Hence the zeroes of 4s² – 4s + 1 are \(\frac{1}{2}\), \(\frac{1}{2}\)
Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1;

(iii) Let p(x) = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1 (2x – 3)
= (3x + 1) (2x – 3)
The zeroes of p(x) are given by p(x) = 0
⇒ (3x + 1) (2x – 3) = 0
⇒ 3x + 1 = 0 (or) 2x – 3 = 0
⇒ 3x = -1 (or) 2x = 3
⇒ x = \(\frac{-1}{3}\) (or) x = \(\frac{3}{2}\)
Hence the zeroes of 6x – 7x – 3 are \(\frac{-1}{3}\) and \(\frac{3}{2}\)
Sum of the zeroes = \(\frac{-1}{3}\) + \(\frac{3}{2}\)

iv) Let p(u) = 4u² + 8u
= 4u(u + 2)
The zeroes of p(u) are given by
⇒ p(u) = 0
⇒ 4u(u + 2) = 0
⇒ 4u = 0 (or) u + 2 = 0
⇒ u = 0 (or) u = -2
Hence the zeroes of 4u + 8u are 0 and -2.
Sum of the zeroes = 0 + (-2) = -2

v) Let p(t) = t² – 15
The zero of p(t) is given by p(t) = 0
⇒ t² – 15 = 0 ⇒ t² = 15

vi) Let P(x) = 3x²
The zero of p(x) is given by P(x) = 0
⇒ 3x² – x – 4 = 0
⇒ 3x² + 3x – 4x – 4 = 0
⇒ 3x(x + 1) – 4(x + 1) = 0
⇒ (x + 1) (3x – 4) = 0
⇒ x + 1 = 0 (or) 3x – 4 = 0
⇒ x = -1 (or) x = 4/3
Hence the zeroes of 3x² – x – 4 are -1 (or) 4/3
Sum of the zeroes of 3x² – x – 4 are -1 (or) 4/3.
Sum of the zeroes =

Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

i) \(\frac{1}{4}\), -1
ii) \(\sqrt{2}\), \(\frac{1}{3}\)
iii) 0, \(\sqrt{5}\)
iv) 1, 1
v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
vi) 4, 1 (A.P. Mar. ’15)
Solution:
i) \(\frac{1}{4}\), -1
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = \(\frac{1}{4}\)
Product of the zeroes = αβ = -1
The required quadratic polynomial will be
k[x² – x (α + β) + αβ] where k is a constant
⇒ k[x² – x(\(\frac{1}{4}\)) + (-1)]
⇒ k(x² – \(\frac{x}{4}\) – 1)
If k = 4, then the polynomial will be 4(x² – \(\frac{x}{4}\) – 1) = 4x² – x – 4

ii) \(\sqrt{2}\), \(\frac{1}{3}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = \(\sqrt{2}\)
Product of the zeroes = αβ = \(\frac{1}{3}\)
∴ The required quadratic polynomial will be
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(\(\sqrt{2}\)) + \(\frac{1}{3}\)]
⇒ k[x² – \(\sqrt{2}\)x + \(\frac{1}{3}\)]
when k = 3, then the polynomial will be
3[x² – \(\sqrt{2}\)x + \(\frac{1}{3}\)] (or) 3x² – 3\(\sqrt{2}\)x + 1

iii) 0, \(\sqrt{5}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = \(\sqrt{5}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ where k is a constant
⇒ k[x² -x(0) + \(\sqrt{5}\)]
⇒ k[x² + \(\sqrt{5}\)].
Where k = 1, the polynomial will be x² + \(\sqrt{5}\).

iv) 1, 1
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = 1
Product of the zeroes = αβ = 1
∴ The required quadratic polynomial will be
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(α – β) + 1]
⇒ k[x² – x + 1]
Where k = 1, the polynomial will be x² – x + 1.

v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = –\(\frac{1}{4}\)
Product of the zeroes = αβ = \(\frac{1}{4}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(-\(\frac{1}{4}\)) + \(\frac{1}{4}\)]
⇒ k[x² + \(\frac{x}{4}\) + \(\frac{1}{4}\)]
Where k = 4, the polynomial will be
4[x² + \(\frac{x}{4}\) + \(\frac{1}{4}\)] = 4x² + x + 1

Question 3.
Find the quadratic polynomial for the zeroes α, β given in each case. (A.P.Mar.’16)

i) 2, -1
ii) \(\sqrt{3}\), –\(\sqrt{3}\)
iii) \(\frac{1}{4}\), -1
iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Solution:
i) 2,-1
Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = 2, β = -1
Sum of the zeroes = α + β = 2 + (-1) = 1
Product of the zeros = αβ = 2 × (-1)
= -2
∴ The required quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
⇒ k [x² – x(1) + (-2)]
⇒ k[x² – x – 2]
Where k = 1, the quadratic polynomial will be x² – x – 2.

ii) \(\sqrt{3}\), –\(\sqrt{3}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\sqrt{3}\) , β = –\(\sqrt{3}\)
Sum of the zeroes = α + β
= \(\sqrt{3}\) + (-\(\sqrt{3}\)) = o
Product of the zeros
= αβ = \(\sqrt{3}\) × –\(\sqrt{3}\) = -3
Therefore the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
⇒ k [x² – x(0) + (-3)]
⇒ k[x² – 3]
Where k = 1, the quadratic polynomial will be [x² – 3].

iii) \(\frac{1}{4}\), -1
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\frac{1}{4}\), β = -1
Sum of the zeroes = α + β
= \(\frac{1}{4}\) + (-1) = \(\frac{-3}{4}\)
Product of the zeroes = αβ = \(\frac{1}{4}\) × (-1)
= –\(\frac{1}{4}\)
Therefore, the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
Where k = 4, the quadratic polynomial will be [4x² + 3x – 1]

iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\frac{1}{2}\), β = \(\frac{3}{2}\)
Sum of the zeroes = α + β
= \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{4}{2}\) = 2
Product of the zeroes = αβ
= \(\frac{1}{2}\) × \(\frac{3}{2}\) = \(\frac{3}{4}\)
Therefore the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant

Where k = 4, the quadratic polynomial will be [4x² – 8x + 3].

Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficients. (A.P. Mar. ’15)
Solution:
The given polynomial is x³ + 3x² – x – 3
Comparing the given polynomial with ax³ + bx² + cx + d
We get a = 1, b = 3, c = -1, d = -3
Let p(x) = x³ + 3x² – x – 3
Then p(1) = (1)³ + 3(1)² – 1 – 3
= 1 + 3 – 1 – 3
= 4 – 4 = 0
p(-1) = (-1)³ + 3(-1)² – (-1) – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)³ + 3(-3)² – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore 1, -1 and -3 are the zeroes of
x³ + 3x² – x – 3.
So α = 1, β = -1, γ = -3
α + β + γ = 1 – 1 – 3 = -3 = -3/1 = \(\frac{-b}{a}\)
αβ + βγ + αγ = 1(-1) + (-1)(-3) + (-3)(1)
= -1 + 3 – 3 = -1 = -1/1 = c/a
αβγ = (1)(-1)(-3) = – \(\left(\frac{-3}{1}\right)\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.2

Question 1.
The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes
of p(x).
Solution:

Solution:
Six figures are given and they are numbered.
In the first figure, there are no zeroes.
In the second figure, there is one zero.
In the third figure, there are three zeroes.
In the fourth figure, there are two zeroes.
In the fifth figure, there are four zeroes.
In the sixth figure, there are three zeroes.

Question 2.
Find the zeroes of the given polynomials.
(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x⁴ – 16
Solution:
i) p(x) = 3x
p(0) = 3 × 0 = 0
∴ The zero of p(x) = 3x is 0

ii) p(x) = x² + 5x + 6
= x² + 3x + 2x + 6
= x(x + 3) + 2(x + 3) = (x + 2) (x + 3)
To find zeroes, let p(x) = 0
⇒ (x + 2) (x + 3) = 0
So, x + 2 = 0 (or) x + 3 = 0
⇒ x = -2 (or) x = -3
Therefore, the zeroes of x + 5x + 6 are – 2 and -3.

iii) p(x) = (x + 2) (x + 3)
⇒ x² + 2x + 3x + 6
⇒ x² + 5x + 6
To find zeroes, let p(x) = 0
⇒ (x + 2) (x + 3) = 0
So, x + 2 = 0 (or) x + 3 = 0
⇒ x = -2 (or) x = -3
∴ The zeroes of (x + 2) (x + 3) are -2 and -3.

iv) p(x) = x⁴ – 16
= (x²)² – (4)² = (x² + 4) (x² – 4)
= (x² + 4)(x + 2)(x – 2)
To find zeroes, let p(x) = O
⇒ (x² + 4) (x + 2)(x – 2) = 0
(i.e.,) x² + 4 = 0 (or) x + 2 = 0 (or) x – 2 = 0
If x² + 4 = 0, then x² = -4 ⇒
x = ± \(\sqrt{-4}\)
If x + 2 = 0, then x = -2
If x – 2 = 0, then x = 2
∴ The zeroes of the polynomial x⁴ – 16 are -2, 2 and ± \(\sqrt{-4}\).

Question 3.
Draw the graphs of the given polynomials and find the zeroes. Justify the answers. (A.P.June’ 15)
(i) p(x) = x² – x – 12
(ii) p(x) = x² – 6x + 9
(iii) p(x) = x² – 4x + 5
(iv) p(x) = x² + 3x – 4
(v) p(x) = x² – 1
Solution:
i) p(x) = x² – x – 12

∴ The zeros of x² – x – 12 are 4 and -3.

Scale:
On X-axis: 1 cm = 1 unit
OnY-axis: 1 cm = 2 units

The graph intersects the X-axis at (4, 0) and (-3, 0)

ii) p(x) = x² – 6x + 9


The graph intersects one the X-axis at only point (3, 0)

iii) p(x) = x² – 4x + 5

Scale:
On X-axis: 1 cm = 1 unit
OnY-axis: 1 cm = 2 units

The graph does not intersect at the X-axis.
There are no zeroes of the polynomial x² – 4x + 5

iv) p(x) = x² + 3x – 4


∴ The graph intersects the X-axis at (1, 0) and (-4, 0)
∴ The zeroes of the polynomial x² + 3x – 4 are 1 and -4.

v) p(x) = x² – 1


The graph intersects the X-axis at (-1, 0) and (1, 0)
∴ The zeroes of the polynomial x² – 1 are -1 and 1.

Question 4.
Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x² + 3x – 1?
Solution:
p(x) = 4x² + 3x – 1
∴ p(\(\frac{1}{4}\)) = 4(\(\frac{1}{4}\))² + 3(\(\frac{1}{4}\)) – 1
= \(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{1}{1}\)
= \(\frac{1+3-4}{4}\) = \(\frac{0}{4}\) = 0
∴ p(-1) = 4(-1)² + 3(-1) – 1
= 4(1) – 3 – 1
= 4 – 3 – 1
= 4 – 4 = 0
Since p(\(\frac{1}{4}\)) and p(-1) are equal to zero.
\(\frac{1}{4}\) and -1 are the zeroes of the polynomial.

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 12 Applications of Trigonometry InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions

Do This

Question 1.
Draw diagram for the following situations :
i) A person is flying a kite at an angle of elevation a and the length of thread from his hand to kite is ‘l’.
ii) A person observes two banks of a river at angles of depression θ₁ and θ₂ (θ₁ < θ₂) from the top of a tree of height ‘h’ which is at a side of the river. The width of the river is ‘d’. (AS₅) (Page No. 297)
Solution:
i)

In the figure
A is the position of the person.
B is the position of the kite.
AB is the length of thread.

ii)

In the figure,
‘D’ is the position of person
CD is the height of the tree
AB is the width of the river
Angles of depression are θ₁ and θ₂.

Think – Discuss

Question 1.
You are observing top of your school building at an angle of elevation a from a point which is at ‘d’ meter distance from foot of the building. Which trigonometric ratio would you like to consider to find the height of the building.
(AS₃) (Page No. 297)
Solution:
Rough sketch

I would like to use the “Tangent” to find the height of the building by the “Rough sketch”.

Question 2.
A ladder of length ‘x’ meter is leaning against a wall making angle θ with the ground. Which trigonometric ratio would you like to consider to find the height of the point on the wall at which the ladder is touching ? (AS₃) (Page No. 297)
Solution:

I would like to use the “sin θ” to find the height of the point on the wall at which the ladder is touching.

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 12 Applications of Trigonometry Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise

Question 1.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the ballon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.
Solution:

From the figure,
Let AD be the height of a tall girl standing on the horizontal line AB is 1.2 m.
Let FH = EB = 88.2 m be the height of balloon from the line AB.
At the eyes of the girl D, the angle of elevations are ∠FDC = 60° and ∠EDC = 30°
Now, FG = EC = 88.2 – 1.2 = 87 m.
Let the distance travelled by the balloon,
HB = y m and AH = x m.
∴ DG = x m and GC = ym
In right angled ∆FGD
tan 60° = \(\frac{\mathrm{FG}}{\mathrm{DG}}\)
⇒ \(\sqrt{3}\) = \(\frac{87}{\mathrm{x}}\)
⇒ x = \(\frac{87}{\sqrt{3}}\) ………. (1)
Again, in right angled ∆ECD,
tan 30° = \(\frac{\mathrm{EC}}{\mathrm{DC}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{87}{\mathrm{DG}+\mathrm{GC}}\)
⇒ x + y = 87\(\sqrt{3}\) …………. (2)
∴ From equation (1), substituting
x = \(\frac{87}{\sqrt{3}}\) in equation (2), we get

⇒ y = 29 × 2\(\sqrt{3}\) = 58 \(\sqrt{3}\) m.
Hence, the distance travelled by the balloon during the interval is 58 \(\sqrt{3}\) m.

Question 2.
The angles of elevation of the top of a lighthouse from 3 boats A, B and C in a straight line of same side of the lighthouse are a, 2a, 3a respectively. If the distance between the boats A and B is x meters, find the height of lighthouse.
Solution:
From the figure,
Let PQ be the height of the lighthouse = h m

A = First point of observation
B = Second point of observation
C = Third point of observation
Given, AB = x and BC = y (Not given in the text)
Exterior angle = Sum of the opposite interior angles
∠PBQ = ∠BQA + ∠BAQ and
∠PCQ = ∠CBQ + ∠CQB
∴ AB = x = QB
By applying the sine rule,

⇒ h² = \(\frac{x^2}{4 y^2}\) (3y – x)(x + y)
∴ h = \(\frac{x}{2 y} \sqrt{(3 y-x)(x+y)}\)
Height of lighthouse
= \(\frac{x}{2 y} \sqrt{(3 y-x)(x+y)}\) meters.

Question 3.
Inner part of a cupboard is in the cuboidical shape with its length, breadth and height in the ratio 1 : \(\sqrt{2}\) : 1. What is the angle made by the longest stick which can be inserted cupboard with its base inside ?
Solution:

Inner part of a cupboard is in the cuboidical shape.
Ratio of length, breadth and height are
1 : \(\sqrt{2}\) : 1
In the figure,
Let AB be the length and BC be the height of the cupboard.
AC be the length of the stick.
‘θ’ be the angle of elevation of the stick making with base of the cupboard.
In ∆ABC,
tan θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ tan θ = \(\frac{1}{1}\)
⇒ tan θ = tan 45°
θ = 45°
∴ The angle of elevation is 45°.

Question 4.
An iron spherical ball of volume 232848 cm³ has been melted and converted into a cone with the vertical angle of 120°. What are its height and base ?
Solution:
Given :
AC = Slant height = l
AB = Vertical height = h
BC = radius = r
Volume of iron spherical ball
V = 232848 cm³

As per the problem,
Volume of spherical ball = Volume of cone
∴ \(\frac{1}{3}\) πr²h = 232848
In ∆ABC,
tan 60° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
\(\sqrt{3}\) = \(\frac{\mathrm{r}}{\mathrm{h}}\) ⇒ r = \(\sqrt{3}\) h
\(\frac{1}{3}\) π(\(\sqrt{3}\) h)² × h = 232848
\(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × h² × h = 232848
h³ = \(\frac{232848 \times 7}{22}\)
h³ = 10584 × 7 = 74088
h³ = 42³
h = 42
But r = h\(\sqrt{3}\)
∴ r = 42\(\sqrt{3}\)

Question 5.
A right circular cylindrical tower, height ‘h’ and radius ‘r’, stands on the ground. Let ‘P’ be a point in the horizontal plane ground and ABC be the semi-circular edge of the top of the tower such that B is the point in it nearest to P. The angles of elevation of the points A and B are 45° and 60° respectively. Show that \(\frac{\mathrm{h}}{\mathrm{r}}=\frac{\sqrt{3}(1+\sqrt{5})}{2}\).
Solution:
As shown in the figure

OA = BD = h (height of cylinder)
OD = r (radius of cylinder)
‘O’ is the centre.
ABC is the edge of semi-circle (in the top of cylinder).
B is nearer to the point R So B should be at the outer edge of diameter. That means just above ‘D’.
∠DPB = 60°, ∠OPA = 45° (given)
In ∆BDP, tan P = \(\frac{\mathrm{BD}}{\mathrm{DP}}\) (P = 60°, BD = h)
tan 60° = BD/DP
⇒ \(\sqrt{3}\) = \(\frac{\mathrm{h}}{\mathrm{DP}}\)
⇒ h = \(\sqrt{3}\) DP …………… (1)
In ∆OAP, tan P = \(\frac{\mathrm{OA}}{\mathrm{OP}}\) (here P = 45°, OA = h)
⇒ tan 45° = \(\frac{\mathrm{OA}}{\mathrm{OP}}\) = 1 ⇒ OA = OP
So OA = h = OP = OD + DP
So h = r + DP …………. (2)
From the equation (1) & (2)
h = \(\sqrt{3}\) DP, h = r + DP
∴ \(\sqrt{3}\) DP = r + DP
So r = \(\sqrt{3}\) DP – DP
r = DP(\(\sqrt{3}\) – 1) …………. (3)

(But in text book, it is asked as \(\frac{\sqrt{3}(1+\sqrt{5})}{2}\) which is wrong).

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 12 Applications of Trigonometry Ex 12.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Exercise 12.2

Question 1.
A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the road. (AS₄)
Solution:
In the adjacent figure,
AB denotes the height of the tower.
BC denotes the width of the road.
CD = 10 cm
∠ACB = 60° and ∠ADC = 30°
In ∆ACB, ∠B = 90°
tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)

\(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) ………….. (1)
In ∆ ABD, ∠ABD = 90°
tan 30° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}+\mathrm{CD}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}+10}\) ………….. (2)
dividing eq. (1) by eq. (2) we have
\(\frac{\mathrm{eq} \mathrm{(1)}}{\mathrm{eq} \mathrm{(2)}}=\frac{\sqrt{3} \times \sqrt{3}}{1}=\frac{\mathrm{AB}}{\mathrm{BC}} \times \frac{\mathrm{BC}+10}{\mathrm{AB}}\)
\(\frac{3}{1}\) = \(\frac{\mathrm{BC}+10}{\mathrm{BC}}\)
3BC = BC + 10
3BC – BC = 10
2BC = 10
BC = \(\frac{10}{2}\) = 5 …………… (3)
from (1),
\(\sqrt{3}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
\(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{5}\)
⇒ AB = 5\(\sqrt{3}\)
Hence, the height of the tower = 5 \(\sqrt{3}\) m and the width of the road = 5 m.

Question 2.
A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height from a point of certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple. (AS₄)
Solution:
CE denotes the height of the boy CE = 1.5 m
AF denotes the height of the temple AF = 30 m
CB || EF
CEFB is a rectangle
∴ CE = 8F = 1.5 m
∴ AB = AF – BF
= 30 – 1.5
= 28.5 m

In ∆ ABC, ∠B = 90°
∴ tan 30° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{\mathrm{BC}}\)
⇒ BC = 28.5 × \(\sqrt{3}\) …………… (1)
In ∆ ABD, ∠B = 90°
tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)
⇒ \(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)
⇒ \(\frac{\sqrt{3}}{1}\) = \(\frac{28.5}{\mathrm{BD}}\)
⇒ BD × \(\sqrt{3}\) = 28.5
⇒ BD = \(\frac{28.5}{\sqrt{3}}\)
Therefore, the distance, the boy walked towards the temple is CD
CD = BC – BD

Question 3.
A statue stands on the top of a 2m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue. (AS₄)
Solution:
In ∆ABD, ∠B = 90° and ∠DAB = 45°
∴ tan 60° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
\(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
AB × \(\sqrt{3}\) = BC

AB = \(\frac{\mathrm{BC}}{\sqrt{3}}\) …………….. (1)
In ∆ABD, ∠B = 90° and ∠DAB = 45°
∴ tan 45° = \(\frac{\mathrm{DB}}{\mathrm{AB}}\)
\(\frac{1}{1}\) = \(\frac{2}{\mathrm{AB}}\)
⇒ AB = 2 m ………………. (2)
from (i), \(\frac{\mathrm{BC}}{\sqrt{3}}\) = \(\frac{2}{1}\)
⇒ BC = 2\(\sqrt{3}\)
= 2 × 1.732 = 3.464 m
Therefore, the height of the statue
CD = BC – BD
= 3.464 – 2 = 1.464 m

Question 4.
From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7m, then find the height of the tower. (AS₄)
Solution:
From the figure,
Let AB be the height of the tower.
CD be the height of the building.
Distance between the building from the tower is 7m

Angles of elevation and depression are
∠BDE = 60° and ∠EDA = 45° and DE = AC = 7m and DC = AE
From the right angled ∆BDE,
We have
tan 60° = \(\frac{\mathrm{BE}}{\mathrm{DE}}\)
⇒ \(\sqrt{3}\) = \(\frac{\mathrm{BE}}{7}\)
⇒ BE = 7 \(\sqrt{3}\) ……………….. (1)
From the right angled ∆ADC, we have
tan 45° = \(\frac{\mathrm{CD}}{\mathrm{AC}}\)
1 = latex]\frac{\mathrm{CD}}{7}[/latex]
⇒ CD = 7 ………………. (2)
From the equations (1) and (2)
Height of the tower = AB = AE + BE
= 7 + 7 \(\sqrt{3}\) = 7(1 + \(\sqrt{3}\))
= 7(1 + 1.732)
= 7(2.732) = 19.124 m
Hence, the height of the tower is 19.124 m.

Question 5.
A wire of length 18 m had been tied with electric pole at an angle of elevation 30° with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ? (AS₄)
Solution:
In the figure,
Let AB be the height of the electric pole = h m.
BC be the actual length of the wire = 18 m.
X and Y are the first and second points of observations.

Let AX = a + b and AY = b
Angles of elevations are ∠BXA = 30° and ∠BYA = 60°

Again from the ∆ ABY
Cos 60° = \(\frac{\mathrm{AY}}{\mathrm{BY}}\)
\(\frac{1}{2}\) = \(\frac{\mathrm{b}}{\mathrm{BY}}\)
BY = 2b
BY = 2(3\(\sqrt{3}\))
= 6\(\sqrt{3}\)
= 6 (1.732)
∴ BY = 10. 3920
∴ BY = 10.39230
The length of the cut wire = BX – BY
= 18 – 10.39230
= 7.6076 m
= 7.608 m

Question 6.
The angle of elevation of the top of a build¬ing from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of building is 60°. If the tower is 30 m high, find the height of the building. (AS₄)
Solution:
From the figure,
Let ‘BC’ be the height of the tower is 30 m.
Let ‘AD’ be the height of the building is h m_C

Angle of elevation, from the bottom of building and tower as well as
∠BAC = 60° and ∠ABD = 30°
Also, let AB = x be the distance between foot of the tower and building.
In right angled ∆ABD, we have
tan 30° = \(\frac{\mathrm{AD}}{\mathrm{AB}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{h}}{\mathrm{x}}\)
⇒ h = \(\frac{\mathrm{x}}{\sqrt{3}}\) ………….. (1)
Again, in right angled ∆BAC, we have
tan 60° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ \(\sqrt{3}\) = \(\frac{30}{\mathrm{x}}\)
x = \(\frac{30}{\sqrt{3}}\) m
Substituting x = \(\frac{30}{\sqrt{3}}\) in equation (1) we get
h = \(\frac{30}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\) = \(\frac{30}{3}\) = 10 m
Hence, the height of the building is 10 m.

Question 7.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles. (AS₄)
Solution:
AB and CD are two poles of equal height.
Let BE = x, then ED = (120 – x)
In ∆ABE, ∠B = 90°, ∠AEB = 60°
∴ tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BE}}\)

⇒ \(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{\mathrm{x}}\)
⇒ AB = \(\sqrt{3}\) x ……………… (1)
In ∆CDE, ∠D = 90° and ∠CED = 30°
∴ tan 30° = \(\frac{\mathrm{CD}}{\mathrm{ED}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{CD}}{120-\mathrm{x}}\)
⇒ CD × \(\sqrt{3}\) = 120 – x
⇒ CD = \(\frac{120-x}{\sqrt{3}}\) ………………… (2)
from (1) & (2), we have
\(\frac{120-x}{\sqrt{3}}\) = \(\frac{\sqrt{3} x}{1}\) (∵ AB = CD)
\(\sqrt{3}\) × \(\sqrt{3}\)x = 120 – x
3x = 120 – x
3x + x = 120
4x = 120
x = \(\frac{120}{4}\) = 30
AB = \(\sqrt{3}\) x
= \(\sqrt{3}\) × 30
= 30\(\sqrt{3}\) = 30 × 1.732
= 51.96 ft
The height of the pole = 51.96 ft.
The distance of one pole AB from the point E = 30 ft.
The distance of another pole CD from the point E = 120 – 30 = 90 ft.

Question 8.
The angles of elevation of the top of a tower from two points are at a distance of 4m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary. (AS₄)
Solution:
From the figure,
AB be the height of a tower = h m
Let the two points on the ground are ‘C’ and
‘D’, such that they make a distance 4 m and
AC = 4 m and AD = 9 m

Angles of elevation are ∠ACB = θ and ∠ADB = 90 – θ
In the right angled ∆ABC, we have
tan θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ tan θ = \(\frac{\mathrm{h}}{4}\)
Again from the right angled ∆ABD, we have
tan (90 – θ) = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)
⇒ cot θ = \(\frac{\mathrm{h}}{9}\)
⇒ \(\frac{1}{\tan \theta}\) = \(\frac{\mathrm{h}}{9}\)
⇒ tan θ = \(\frac{9}{\mathrm{h}}\) …………….. (2)
From the equations (1) and (2) :
\(\frac{\mathrm{h}}{4}\) = \(\frac{9}{\mathrm{h}}\)
⇒ h² = 36
h = \(\sqrt{36}\) = 6 m
The height of the tower is 6 m.

Question 9.
The angle of elevation of jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500 \(\sqrt{3}\) meters, find the speed of the jet plane. (\(\sqrt{3}\) = 1.732) (AS₄)
Solution:
From the figure,
Let P and Q be the two positions of the plane.

Let A’ be the point of observation.
Let ABC be the horizontal line through A.
It is given that angle of elevation of the plane Q
from a point A’ are 60° and 30° respectively.
∴ ∠PAB = 60°, ∠QAB = 30°
Constant height of jet plane = 1500 \(\sqrt{3}\) m
In the right angled ∆ABP we have
tan 60° = \(\frac{\mathrm{BP}}{\mathrm{AB}}\)
⇒ \(\sqrt{3}\) = \(\frac{1500 \sqrt{3}}{\mathrm{AB}}\)
⇒ AC = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\) = 1500 m
In the right angled ∆ACQ, we have
tan 30° = \(\frac{\mathrm{CQ}}{\mathrm{AC}}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{1500 \sqrt{3}}{\mathrm{AC}}\)
⇒ AC = 1500 × \(\sqrt{3}\) × \(\sqrt{3}\)
= 1,500 × 3 = 4,500 m
From the figure
PQ = BC = AC – AB
= 4500 – 1500 = 3000 m
Thus, the plane travels 3000 m in 15 seconds.
Hence, speed of a plane = \(\frac{3000}{15}\)
= 200m/sec.

Question 10.
The angle of elevation of the top of a tower from the foot of the building is 30° and the angle of elevation of the top of the building from the foot of the tower is 60°. What is the ratio of heights of tower and building ? (AS₄)
Solution:
Let the height of the tower = x m
Let the height of the building = y m

Distance between the tower and building = d m.
Angle of elevation of the top of the tower = 30°
From the figure,
tan 30° = \(\frac{\mathrm{x}}{\mathrm{d}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{x}}{\mathrm{d}}\)
d = \(\sqrt{3}\)x …………….. (1)
Also tan 60° = \(\frac{\mathrm{y}}{\mathrm{d}}\)
\(\sqrt{3}\) = \(\frac{\mathrm{y}}{\mathrm{d}}\)
d = \(\frac{y}{\sqrt{3}}\) …………….. (2)
From (1) and (2)
\(\sqrt{3}\)x = \(\frac{y}{\sqrt{3}}\)
\(\frac{x}{y}=\frac{1}{\sqrt{3} \cdot \sqrt{3}}=\frac{1}{3}\)
∴ x : y = 1 : 3
∴ The ratio of heights of tower and building = 1 : 3.

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 13 Probability InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

Do This

(a) Outcomes of which of the following experiments cure equally likely. (AS₃)(Page No. 307)

Question 1.
Getting a digit 1, 2, 3, 4, 5 or 6 when a die is rolled.
Solution:
Equally likely

Question 2.
Selecting a different colour ball from a bag of 5 red balls, 4 blue balls and 1 black ball.
Note : Picking two different colour balls, i.e., Picking a red, a blue (or) black ball from a ………
Solution:
Not equally likely

Question 3.
Winning in a game of carrom.
Solution:
Not equally likely

Question 4.
Units place of a two digit number selected may be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.
Solution:
equally likely

Question 5.
Selecting a different colour ball from a bag of 10 red balls, 10 blue balls and 10 black balls.
Solution:
equally likely

Question 6.
Raining on a particular day of July.
Solution:
equally likely

(b) Are the outcomes of every experiment equally likely ?
Solution:
Outcomes of all experiments need not necessarily be equally likely.

(c) Give examples of 5 experiments that have equally likely outcomes and five more examples that do not have equally likely outcomes.
Solution:
Equally likely events :
a) Getting an even or odd number when a die is rolled.
b) Getting tail or head when a coin is tossed.
c) Getting an even (or) odd number when a card is drawn at random from a pack of cards numbered from 1 to 10.
d) Picking a green or black ball from a bag containing 8 green balls and 8 black balls.
e) Selecting a boy or girl from a class of 20 boys and 20 girls.
f) Selecting a red or black card from a deck of cards.

Events which are not equally likely :
a) Getting a prime (or) composite number when a die is thrown.
b) Getting an even or odd number when a card is drawn at random from a pack of cards numbered from 1 to 5.
c) Getting a number which is a multiple of 3 (or) not a multiple of 3 from numbers 1, 2, …….., 10.
d) Getting a number less than 5 or greater than 5.
e) Drawing a white ball or green ball from a bag containing 5 green balls and 8 white balls.

Think of 5 situations with equally likely events and find the sample space.
a) Tossing a coin : Getting a tail or head when a coin is tossed.
Sample space = {T, H}
b) Getting an even (or) odd number when a die is rolled.
Sample space = {1, 2, 3, 4, 5, 6}
c) Winning a game of shuttle Sample space = {win, loss}
d) Picking a black (or) blue ball from a bag containing 3 blue balls and 3 black balls = {blue, black}
e) Drawing a red coloured card or black coloured card from a deck of cards = {black, red}

d) Is getting a head complementary to getting a tail ? Give reasons.
Solution:
Number outcomes favourable to head = 1
Probability of getting a head = \(\frac{1}{2}\) [P(E)]
Number of outcomes not favourable to head = 1
Probability of not getting a head = \(\frac{1}{2}\) (P (\(\overline{\mathrm{E}}\)))
Now P(E) + P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1
∴ Getting a head is complementary to getting a tail.

e) In case of a die is getting a 1 complementary to events getting 2, 3, 4, 5, 6 ? Give reasons for your answer.
Solution:
Yes, complementary events
∴ Probability of getting a 1 = \(\frac{1}{6}\) [P(E)]
Probability of getting 2, 3, 4, 5, 6
= P (\(\overline{\mathrm{E}}\)) = \(\frac{5}{6}\)
p(E) + P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{6}\) + \(\frac{5}{6}\) = \(\frac{6}{6}\) = 1

f) Write of five new pair of events that are complementary.
Solution:
a) When a die is thrown, getting an even number is complementary to getting an odd number.
b) Drawing a red card from a deck of cards is complementary to getting a black card.
c) Getting an even number is complementary to getting an odd number from numbers 1, 2, ……….., 8.
d) Getting a Sunday is complementary to getting any day other than Sunday in a week.
e) Winning a running race is complementary to loosing it.

Try This

Question 1.
A child has a die whose six faces show the letters A, B, C, D, E and F. The die is thrown once. What is the probability of getting (i) A ? (ii) D ? (AS₄)(Page No. 312)
Solution:
Total number of outcomes [A, B, C, D, E and F] = 6
i) Number of favourable outcomes to A = 1
Probability of getting A = P(A)

= \(\frac{1}{6}\)

ii) Number of outcomes favourable to D = 1
Probability of getting D
= P(D)

= \(\frac{1}{6}\)

Question 2.
Which of the following cannot be the probability of an event ? (AS₃)(Page No. 312)
a) 2.3
b) – 1.5
c) 15%
d) 0.7
Solution:
a) 2.3 – Not possible
b) -1.5 – Not possible
c) 15% – May be the probability
d) 0.7 – May be the probability

Question 3.
You have a single deck of well shuffled cards. Then,
i) What is the probability that the card drawn will be a queen ? (AS₄)(Page No. 313)
Solution:
Number of all possible outcomes
= 4 × 13 = 1 × 52 = 52
Number of outcomes favourable to Queen
= 4[♥Q ♥Q ♥Q ♥Q]
Probability P(E)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no.of outcomes }}\)
= \(\frac{4}{52}\)
= \(\frac{1}{13}\)

ii) What is the probability that it is a face card ? (Page No. 314)
Solution:
Face cards are J, Q, K.
∴ Number of outcomes favourable to face cards = 4 × 3 = 12
No. of all possible outcomes = 52
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)

iii) What is the probability that it is a spade ? (Page No. 314)
Solution:
Number of spade cards = 13
Total number of cards = 52
Probability = \(\frac{\text { Number of outcomes favourable to spade }}{\text { Number of all outcomes }}\)
= \(\frac{13}{52}\) = \(\frac{1}{4}\)

iv) What is the probability that is the face cards of spades ? (Page No. 314)
Solution:
Number of outcomes favourable to face cards of spades = (K, Q, J) = 3
Number of all outcomes = 52 3
∴ P(E) = \(\frac{3}{52}\)

v) What is the probability it is not a face card ? (Page No. 314)
Solution:
Probability of a face card = \(\frac{12}{52}\)
∴ Probability that the card is not a face card
= 1 – \(\frac{12}{52}\) [P (\(\overline{\mathrm{E}}\)) = 1 – P(E)]
= \(\frac{52-12}{52}\)
= \(\frac{40}{52}\) = \(\frac{10}{13}\)

(Or)

Number of favourable outcomes = 4 × 10 = 40
Number of all outcomes = 52
∴ Probability = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{40}{52}\) = \(\frac{10}{13}\)

Think – Discuss

Question 1.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of any game ? (Page No. 312)
Solution:
Probability of getting a head is \(\frac{1}{2}\) and a tail is
\(\frac{1}{2}\) = 1
Hence, tossing a coin is a fair way.

Question 2.
Can \(\frac{7}{2}\) be the probability of an event ? Explain. (AS₃) (Page No. 312)
Solution:
\(\frac{7}{2}\) can’t be the probability of any event. Since probability of any event should lie between 0 and 1.

Question 3.
Which of the following arguments are correct and which are not correct ? Given reasons. (Page No. 312)

i) If two coins are tossed simultaneously, there are three possible outcomes – two heads, two tails (or) one of each. Therefore, for each. If these outcomes, the probability is \(\frac{1}{3}\)
Solution:
False
Reason :
All possible outcomes are 4. They are HH, HT, TH, TT
Thus, probability of two heads = \(\frac{1}{4}\)
Probability of two tails = \(\frac{1}{4}\)
Probability of one each = \(\frac{2}{4}\) = \(\frac{1}{2}\)

ii) If a die is thrown, there are two possible outcomes – an odd number (or) an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Solution:
True
Reason :
All possible outcomes = (1, 2, 3, 4, 5, 6) = 6
Outcomes favourable to an odd number = (1, 3, 5) = 3
Outcomes favourable to an even number = (2, 4, 6) = 3
∴ Probability (odd number)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 13 Probability Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 13 Probability Optional Exercise

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day ?
(ii) consecutive days ?
(iii) different days ?
Solution:
i) Shyam and Ekta can visit the shop in the following combination :


Number of Total outcomes
= 5 × 5 = 5² = 25 (also from the above table)
Number of favourable outcomes to that of visiting on the same day
(Tu, Tu), (W, W), (Th, Th), (I; F), (S, S) = 5
∴ Probability that visiting the shop on the same day
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{5}{25}\) = \(\frac{1}{5}\)

ii) Number of outcomes favourable to consecutive days
(Tu, W), (W, Th), (Th, F), (F, S) (W, Tu), (Th, W), (F, Th), (S, F) = 8
∴ Probability of visiting the shop on consecutive days = \(\frac{8}{25}\)

iii) If P(E) is the probability of visiting the shop on the same day, then P(\(\overline{\mathrm{E}}\)) is the probability of visiting the shop not on the same day. i.e., P(\(\overline{\mathrm{E}}\)) is the probability of visiting the shop on different days such that P(E) + P(\(\overline{\mathrm{E}}\)) =1
P(\(\overline{\mathrm{E}}\)) = 1 – P(E) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)

Question 2.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Number of red balls in the bag = 5
As the probability of blue balls is double the probability of red balls, we have that number of blue balls is double the number of red balls.
∴ Blue balls = 5 × 2 = 10.

!! Let the number of blue balls = x
Number of red balls = 5
Total no. of balls = x + 5
Total outcomes in drawing a ball at random = x + 5
Number of outcomes favourable to red ball = 5
∴ P(R) = \(\frac{5}{x+5}\)
from the problem.
P(B) = 2 × \(\frac{5}{x+5}\) = \(\frac{10}{x+5}\)
Also, \(\frac{5}{x+5}\) + \(\frac{10}{x+5}\) = 1
[∵ P(\(\overline{\mathrm{E}}\)) + P(E) = 1]
⇒ \(\frac{5+10}{x+5}\) = 1
⇒ \(\frac{15}{x+5}\) = 1 ⇒ x + 5 = 15
⇒ x = 15 – 5 = 10

Question 3.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball ? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
Number of black balls = x
Total number of balls in the box = 12
Probability of drawing a black ball
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{x}{12}\) ……………. (1)
When 6 more black balls are placed in the box, number of favourable outcomes to black ball becomes = x + 6.
Total number of balls in the box becomes = 12 + 6 = 18.
Now the probability of drawing a black ball becomes = \(\frac{x+6}{18}\) …………… (2)
By problem,
\(\frac{x+6}{18}\) = 2 . \(\frac{x}{12}\)
⇒ \(\frac{x+6}{18}\) = \(\frac{x}{6}\)
⇒ 6(x + 6) = 18(x)
⇒ 6x + 36 = 18x ⇒ 18x – 6x = 36
⇒ 12x = 36 ⇒ x = \(\frac{36}{12}\) = 3
Check:
Equation (1) ⇒ \(\frac{x}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
Equation (2) ⇒ \(\frac{x+6}{18}\) = \(\frac{3+6}{18}\)
= \(\frac{9}{18}\) = \(\frac{1}{2}\)
Equation (1) × 2 = \(\frac{1}{2}\) 2 = \(\frac{1}{2}\) and hence proved.

Question 4.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is 2 green is \(\frac{2}{3}\). Find the number of blue marbles 3 in the jar.
Solution:
Total number of marbles in the jar = 24
Let the number of green marbles = x
Then number of blue marbles = 24 – x.
Probability of drawing a green marbles
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{x}{24}\)
By Problem, \(\frac{x}{24}\) = \(\frac{2}{3}\)
⇒ 3 × x = 24 × 2
x = \(\frac{24 \times 2}{3}\) = 16
∴ Number of green marbles = 6
Number of blue marbles = 24 – 16 = 8

!! P(G) = \(\frac{2}{3}\)
P(G) + P(B) = 1
∴ P(B) = 1 – P(G) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Number of blue marbles in the jar
= \(\frac{1}{3}\) × 24 = 8.