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

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:
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 1
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
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 2
⇒ y = 29 × 2\(\sqrt{3}\) = 58 \(\sqrt{3}\) m.
Hence, the distance travelled by the balloon during the interval is 58 \(\sqrt{3}\) m.

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

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
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 3
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,
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 4
⇒ 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:
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 5
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°.

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

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
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 6
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\(\sqrt{3}\)

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

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
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 7
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)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Optional Exercise 8
(But in text book, it is asked as \(\frac{\sqrt{3}(1+\sqrt{5})}{2}\) which is wrong).

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

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° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 1
\(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) ………….. (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.

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

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
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 13
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
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 3

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

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
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 4
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
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 5
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.

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

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.
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 6
Let AX = a + b and AY = b
Angles of elevations are ∠BXA = 30° and ∠BYA = 60°
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 7
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 mC
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 8
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.

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

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° = \(\frac{\mathrm{AB}}{\mathrm{BE}}\)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 9
⇒ \(\frac{\sqrt{3}}{1}\) = \(\frac{\mathrm{AB}}{\mathrm{x}}\)
⇒ 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.

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

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

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

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.
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.2 11
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.

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

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

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

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

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

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}

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

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)
TS 10th Class Maths Solutions Chapter 13 Probability InText Questions 1
= \(\frac{1}{6}\)

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

ii) Number of outcomes favourable to D = 1
Probability of getting D
= P(D)
TS 10th Class Maths Solutions Chapter 13 Probability InText Questions 2
= \(\frac{1}{6}\)

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) = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

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.

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

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 = \(\frac{2}{4}\) = \(\frac{1}{2}\)

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions

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)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

TS 10th Class Maths Solutions Chapter 13 Probability Optional Exercise

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 :
TS 10th Class Maths Solutions Chapter 13 Probability Optional Exercise 1
TS 10th Class Maths Solutions Chapter 13 Probability Optional Exercise 2
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 – \(\frac{1}{5}\) = \(\frac{4}{5}\)

TS 10th Class Maths Solutions Chapter 13 Probability Optional Exercise

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.

TS 10th Class Maths Solutions Chapter 13 Probability Optional Exercise

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.

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 13 Probability Ex 13.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 13 Probability Exercise 13.2

Question 1.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball selected is (i) red ? (ii) not red ? (AS1)
Solution:
i) Total number of balls in the bag = 3 red + 5 black = 8 balls
Number of total outcomes when a ball is selected at random = 3 + 5 = 8
Now, number of favourable outcomes of red ball = 3
∴ Probability of getting a red ball
P(E) = \(\frac{\text { No. of favourable outcome }}{\text { No. of total outcomes }}\)
= \(\frac{3}{8}\)

ii) If P(\(\overline{\mathrm{E}}\)) is the probability of selecting no red balls, then
P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(\(\overline{\mathrm{E}}\)) = 1 – P(E) = 1 – \(\frac{3}{8}\) = \(\frac{5}{8}\)

Question 2.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white ? (iii) not green ? (AS1)
Solution:
There are 5 red marbles, 8 white marbles and 4 green marbles in a bag.
∴ Total number of marbles in the bag = 5 + 8 + 4= 17
∴ Number of all possible outcomes = 17
i) Let E be the event that the marble taken out will be red.
Total number of red marbles in the bag = 5
∴ Number of outcomes favourable to E = 5
P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{5}{17}\)

ii) Let E be the event that the marble taken out will be white.
Total number of white marbles in the bag = 8
∴ Number of outcomes favourable to E = 8
P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{8}{17}\)

iii) Let E be the event that the marble taken out will be green.
Total number of green marbles in the bag = 4
∴ Number of outcomes favourable to E = 4
P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{4}{17}\)
∴ Probability that the marble taken out will not be green.
= 1 – P (Probability that the marble taken out will be green)
= 1 – P(E) = 1 – \(\frac{4}{17}\) = \(\frac{13}{17}\)

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Question 3.
A Kiddy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin ? (ii) will not be a ₹ 5 coin ? (AS1)
Solution:
i) Number of 50p coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins = 20
Number of ₹ 5 coins = 10
∴ Total number of coins
= 100 + 50 + 20 + 10 = 180
Number of total outcomes for a coin to fall down = 180
Number of outcomes favourable to 50p coins to fall down = 100
∴ Probability of a 50p coin to fall down No. of favourable outcomes
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{100}{180}\) = \(\frac{5}{9}\)

ii) Let P(E) be the probability for a ₹ 5 coin to fall down.
= \(\frac{1}{2}\) (P(\(\overline{\mathrm{E}}\))) = \(\frac{1}{2}\) (P(\(\overline{\mathrm{E}}\)))
No. of outcomes favourable to ₹ 5 coin = 10
∴ Probability for a ₹ 5 coin to fall down
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{10}{180}\) = \(\frac{1}{18}\)
Then P(\(\overline{\mathrm{E}}\)) is the probability of a coin which fall down is not a ₹ 5 coin.
Again P(E) + P(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\)) = 1 – P(E)
= 1 – \(\frac{1}{18}\)
= \(\frac{17}{18}\)

Question 4.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (See figure). What is the probability that the fish taken out is a male fish ? (AS4)
TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2 1
Solution:
Number of male fish in the acquarium = 5
Number of female fish in the acquarium = 8
Total number of fish in the acquarium = 5 + 8 = 13
∴ Number of all possible outcomes = 13
Let E be the event that the fish taken out is a male fish.
Number of outcomes favourable to E = 5
∴ P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{5}{13}\)

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Question 5.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See figure) and these are equally likely outcomes. What is the probability that it will point at
TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2 2
i) 8?
ii) an odd number ?
iii) a number greater than 2 ?
iv) a number less than 9 ?
Solution:
i) The figure shows the numbers from 1 to 8.
Let E be the event that the arrow comes to rest pointing at 8.
Number of outcomes favourable to E = 1
Number of all possible outcomes = 8
∴ P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{1}{8}\)

ii) The odd numbers shown in the figure are 1, 3, 5 and 7 = 4.
Let E be the event that the arrow will point an odd number.
Number of outcomes favourable to E = 4
Number of all possible outcomes = 8
∴ P(E) = \(\frac{\text { Number of outcomes favourable to E}}{\text { Number of all possible outcomes }}\)
= \(\frac{4}{8}\) = \(\frac{1}{2}\)

iii) The numbers which are greater than 2 as per the figure given are 3, 4, 5, 6, 7 and 8 = 6
Let E be the event that the arrow will point a number greater than 2.
Number of outcomes favourable to E = 6
Number of all possible outcomes = 8
∴ P(E) = \(\frac{\text { Number of outcomes favourable to E}}{\text { Number of all possible outcomes }}\)
= \(\frac{6}{8}\) = \(\frac{3}{4}\)

iv) The numbers less than 9 are 1, 2, 3. 4, 5, 6, 7,8
Let E be the event that the arrow will point a number less than 9.
Number of outcomes favourable to E = 8
Number of all possible outcomes = 8
∴ P(E) = \(\frac{\text { Number of outcomes favourable to E}}{\text { Number of all possible outcomes }}\)
= \(\frac{8}{8}\) = 1

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Question 6.
One card is selected from a well-shuffled deck of 52 cards. Find the probability of getting
i) a king of red colour
ii) a face card
iii) a red face card
iv) the jack of hearts
v) a spade
vi) the queen of diamonds.
Solution:
Total number of cards = 52
∴ Number of all possible outcomes in selecting a card at random = 52
i) Number of out comes favourable to the kings of red colour = 2(♥k, ♥k)
∴ Probability of getting the king of red colour
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

ii) Number of face cards in a deck of cards = 4 × 3 = 12(K, Q, J)
No. of outcomes favourable to select face card = 12
∴ Probability of getting a face card
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)

iii) Number of red face cards = 2 × 3 = 6
∴ No. of outcomes favourable to select a red face card = 6
∴ Probability of getting a red face card
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{6}{52}\) = \(\frac{3}{26}\)

iv) No. of outcomes favourable to the jack of hearts = 1
∴ Probability of getting the jack of hearts.
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{52}\)

v) No. of spade cards = 13
∴ No. of outcomes favourable to ‘a spade card’ = 13
∴ Probability of getting a spade card
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{13}{52}\) = \(\frac{1}{4}\)

vi) No. of outcomes favourable to the queen of diamonds = 1
∴ Probability of getting the queen of diamonds
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{52}\)

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Question 7.
Five cards-the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is selected at random.
i) What is the probability that the card is the queen ?
ii) If the queen is selected and put aside (without replacement), what is the probability that the second card selected is
(a) an ace ? (b) a queen ?
Solution:
Total number of cards = 5
Well – Shuffled with their face downwards.
i) Let E be the event that the card is the queen. Therefore, the number of outcomes favourable to E = 1
So, P(E) = \(\frac{\text { Number of outcomes favourable to E}}{\text { Number of all possible outcomes }}\)
= \(\frac{1}{5}\)

ii) If the queen is selected & put a side then the number of remaining cards is 4 (i.e.,) (5 – 1 = 4)
∴ No. of all possible outcomes = 4

a) Let E be the event that the second card picked up is an ace.
Then, the number of outcomes favourable toE = 1
So. P(E)
= \(\frac{\text { Number of outcomes favourable to E}}{\text { Number of all possible outcomes }}\)
= \(\frac{1}{4}\)

b) Let E be the event that the second card selected is a queen.
Then, the number of outcomes favourable to E is 0 (∵ there is no queen)
So, P(E)
= \(\frac{\text { Number of outcomes favourable to E}}{\text { Number of all possible outcomes }}\)
= \(\frac{0}{4}\) = 0

Question 8.
12 defective pens are accidentally mixed 10. with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens = 12
Number of good pens = 132
∴ Total number of pens = Number of defective pens + number of good pens
= 12 + 132 = 144
Let E be the event that the pen taken out is a good one,
Then, the number of outcomes favourable to
E = 132
So, P(E) = \(\frac{\text { Number of outcomes favourable to E}}{\text { Number of all possible outcomes }}\)
= \(\frac{132}{144}\) = \(\frac{11}{12}\)

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Question 9.
A lot of 20 bulbs contain 4 defective ones. One bulb is selected at random from the lot. What is the probability that this bulb is defective ? Suppose the bulb selected in previous case is not defective and is not replaced. Now one bulb is selected at random from the rest. What is the probability that this bulb is not defective ? (AS1, AS4)
Solution:
Total number of bulbs = 20
∴ No. of all possible outcomes = 20
i) Let E be the event that the bulbs drawn at random from the lot is defective.
Then, the number of outcomes favourable to E = 4
∴ P(E) = \(\frac{\text { Number of outcomes favourable to E}}{\text { Number of all possible outcomes }}\)
= \(\frac{4}{20}\) = \(\frac{1}{5}\)

ii) As one bulb is selected at random from the rest,
Total number of bulbs = 20 – 1 = 19
Number of defective bulbs = 4
Let E be the event that the bulb selected is not defective.
Then, the number of outcomes favourable to E is 15 since, now there are 19 – 4 = 15 bulbs which are not defective.
∴ P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{15}{19}\)

Question 10.
A box contains 90 discs which are numbered from 1 to 90. If one disc is selected at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5. (AS1)
Solution:
Number of discs in the box = 90
∴ Number of all possible outcomes = 90
i) Let E be the event that the disc bears a two-digit number.
One digit numbers are 1, 2, 3, 4, 5, 6, 7, 8 and 9. These are 9 in numbers.
Then, the number of outcomes favourable to E = 90 – 9 = 81
∴ P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{81}{90}\) = \(\frac{9}{10}\)

ii) Let E be the event that the disc bears a perfect square number.
The perfect square numbers from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81. These are 9.
Then, the number of outcomes favourable to E = 9
∴ P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{9}{90}\) = \(\frac{1}{10}\)

iii) Let E be the event that the disc bears a number divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90. These are 18.
Then, the number of outcomes favourable to E = 18
∴ P(E) = \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{18}{90}\) = \(\frac{1}{5}\)

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Question 11.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1m ? (AS4)
TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2 3
Solution:
Length of the rectangular region = 3 m
Breadth of the rectangular region = 2m
Area of the rectangular region =
Length × Breadth = 3 × 2 = 6m2
Diameter of the circle = 1 m
∴ Radius of the circle = \(\frac{1}{2}\) m
∴ Area of the circle = πr2
= \(\frac{22}{7}\) \(\frac{1}{2}\) \(\frac{1}{2}\) = \(\frac{11}{14}\) m2
∴ Probability that the dice will land inside the circle
= \(\frac{\frac{11}{14}}{6}\)
= \(\frac{11}{14}\) \(\frac{1}{6}\) = \(\frac{11}{84}\)

Question 12.
A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that
(i) She will buy it ?
(ii) She will not buy it ? (AS4)
Solution:
Total number of ball pens = 144
i) ∴ Number of all possible outcomes = 144
Number of defective ball pens = 20
∴ Number of good ball pens
= 144 – 20 = 124
∴ Probability that Sudha will buy it
= \(\frac{124}{144}\) = \(\frac{31}{36}\)

ii) Probability that Sudha will not buy it = 1 – (Probability that Sudha will buy it)
= 1 – \(\frac{31}{36}\) = \(\frac{5}{36}\)

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Question 13.
Two dice are rolled simultaneously and counts are added (i) complete the table given below :

Event: Sum on 2 dice’23456789101112
Probability\(\frac{1}{36}\)\(\frac{5}{36}\)\(\frac{12}{36}\)

i) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of
them has a probability \(\frac{1}{11}\). Do you agree with this argument ? Justify your answer. (AS3)
Solution:
When two dice are rolled simultaneously there are 36 possible out comes. So n(s) = 36.
i) Let E3 denotes that event that the sum on two dice is 3 the outcomes favourable to the event E3 = (1, 2), (2, 1)
No. of favourable out comes n(E3) = 2
Probability = \(\frac{\mathrm{n}\left(\mathrm{E}_3\right)}{\mathrm{n}(\mathrm{S})}=\frac{2}{36}=\frac{1}{18}\)

ii) Let E4 denotes the event that the sum on two dice is 4 the outcomes favourable to the event E4 = (1,3), (2, 2), (3, 1).
No. of favourable outcomes n(E4) = 3
probability = \(\frac{\mathrm{n}\left(\mathrm{E}_4\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

iii) Let E5 denotes the event that the sum on two dice is 5 the outcomes favourable to the event E5 = (1, 4), (2, 3), (3, 2), (4, 1)
No. of favourable outcomes n(E5) = 4
Probability = \(\frac{\mathrm{n}\left(\mathrm{E}_5\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

iv) Let E6 denotes the event that the sum on two dice is 6 the outcomes favourable to the event E6 = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
No. of favourable outcomes n(E6) = 5
Probability = \(\frac{\mathrm{n}\left(\mathrm{E}_6\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{5}{36}\)

v) Let E7 denotes the event that the sum on two dice is 7 the outcomes favourable to the event E7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
No. of favourable outcomes n(E7) = 6
Probability = \(\frac{\mathrm{n}\left(\mathrm{E}_7\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

vi) Let E8 denotes the event that the sum on two dice is 8
probability = \(\frac{\mathrm{n}\left(\mathrm{E}_8\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{5}{36}\)

vii) Let E9 denotes the event that the sum on two dice is 9 the outcomes favourable to the event E9 = (3, 6), (4, 5), (5, 4), (6, 3), (7,2), (8,1)
No. of favourable outcomes n (E9) = 8
probability = \(\frac{\mathrm{n}\left(\mathrm{E}_9\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

viii) Let E10 denotes the event that the sum on two dice is 10 the outcomes favourable to the event E10 = (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6. 4), (7, 3), (8, 2), (9, 1).
No. of favourable outcomes n(E10) = 3
Probability = \(\frac{\mathrm{n}\left(\mathrm{E}_10\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

ix) Let E11 denotes the event that the sum on two dice is 11 the outcomes favourable to the event E11 = (5, 6), (6, 5)

Question 14.
A game consists of tossing a one rupee coin 3 times and recording its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Let H : Head, T : Tail
When we toss a one rupee coin 3 times,
S = {H, T} × (H, T} × {H, T} where ‘S’ denotes the cartesian product.
So, n(S) = 2 × 2 × 2 = 8
Let E be the event that Hanif will win the game
i.e., either three heads or three tails will come.
So, E = {(H, H, H), (T, T, T)}
n(E) = 2
So, probability that Hanif will win the game,
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\)
So. probability that Hanif will lose the game,
P(\(\overline{\mathrm{E}}\)) = 1 – P(E) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\).

TS 10th Class Maths Solutions Chapter 13 Probability Ex 13.2

Question 15.
A dice is thrown twice. What is the probability that (i) 5 will not come up either time ? (ii) 5 will come up at least once ? [Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
Solution:
When a dice is thrown twice.
S = {1, 2, 3, 4, 5, 6} × (1, 2, 3, 4, 5, 6}
So, n (S) = 6 × 6 = 36
Let E denote the event that 5 will come up at least once.
Then, E = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)}.
∴ n(E) = 11
i) Probability that 5 will not come up either time = P(\(\overline{\mathrm{E}}\)) = 1 – P(E)
= 1 – \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\)
= 1 – \(\frac{11}{36}\)
= \(\frac{25}{36}\)

ii) Probability that 5 will come up at least once,
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{11}{36}\)

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Exercise 4.1

Question 1.
By comparing the ratios \(\frac{a_1}{a_2}\), \(\frac{b_1}{b_2}\), \(\frac{c_1}{c_2}\) find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or coincident.
a) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
b) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
c) 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
a) The given pair of linear equations are
5x – 4y + 8 = 0 ———- (1)
7x + 6y – 9 = 0 —— (2)
Comparing equations (1) and (2) with standard pair of linear equations (i.e.,)
a1x + b1y + c1 = 0 and
a2x + b2y + c2= 0, we get
a1 = 5; b1 = -4; c1 = 8
a2 = 7; b2 = 6; c2 = -9
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{5}{7}\); \(\frac{b_1}{b_2}\) = \(\frac{-4}{6}\); \(\frac{c_1}{c_2}\) = \(\frac{8}{-9}\)
Since, \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\), the pair of linear equations will represent intersecting lines.

b) The given pair of linear equations are
9x + 3y + 12 = 0 —- (1)
18x + 6y + 24 = 0 —- (2)
Here, a1 = 6; b1 = 3; c1 = 12
a2 = 18; b2 = 6; c2 = 24
\(\frac{a_1}{a_2}\) = \(\frac{9}{18}\) = \(\frac{1}{2}\) ;
\(\frac{b_1}{b_2}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\) ; \(\frac{c_1}{c_2}\) = \(\frac{12}{24}\) = \(\frac{1}{2}\)
Since \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\) = \(\frac{1}{2}\), The pair of linear equations will represent coincident lines.

c) The given pair of linear equations are
6x – 3y + 10 = 0
2x – y + 9 = 0
Here,
a1 = 6; b1 = -3; c1 = 10
a2 = 2; b2 = -1; c2 = 9
\(\frac{a_1}{a_2}\) = \(\frac{6}{2}\) = \(\frac{3}{1}\) ; \(\frac{b_1}{b_2}\) = \(\frac{-3}{-1}\) = \(\frac{3}{1}\);
\(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{10}{9}\)
Since \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\), the pair of linear equations will represent parallel lines.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

Question 2.
Check whether the following equations are consistent or inconsistent. Solve them graphically.
a) 3x + 2y = 5; 2x – 3y = 7
b) 2x – 3y = 8; 4x – 6y = 9
c) \(\frac{3}{2}\)x + – y = 7; 9x – 10y = 14
d) 5x – 3y = 11; -10x + 6y = -22
e) \(\frac{4}{3}\)x + 2y = 8; 2x + 3y = 12
f) x + y = 5; 2x + 2y = 10
g) x – y = 8; 3x – 3y = 16
h) 2x + y – 6 = 0; 4x – 2y – 4 = 0
i) 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
Solution:
a) 3x + 2y – 5 = 0
2x – 3y – 7 = 0
Here
a1 = 3; b1 = 2; c1 = -5
a2 = 2; b2 = -3; c2 = -7
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{3}{2}\) ; \(\frac{b_1}{b_2}\) = \(\frac{2}{-3}\) ; \(\frac{c_1}{c_2}\) = \(\frac{-5}{-7}\) = \(\frac{5}{7}\)
⇒ \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)
∴ The given equations are consistent.
They intersect at one point. There is an unique solution.
The given equations are 3x + 2y = 5 and 2x – 3y = 7
3x + 2y = 5 —– (1)
⇒ 2y = 5 – 3x
⇒ y = \(\frac{5-3 x}{2}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 1
2x – 3y = 7 —- (2)
⇒ 3y = 2x – 7
⇒ y = \(\frac{2 x-7}{3}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 2

Scale :
X – axis: 1 unit = 1 cm
Y – axis: 1 unit = 1 cm
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 3
The given lines intersect at one point. The solution is (2,-1).
x = \(\frac{29}{13}\) ; y = \(\frac{-11}{13}\)

b) 2x – 3y = 8
4x – 6y = 9
Solution:
2x – 3y – 8 = 0
4x – 6y – 9 = 0
Here,
\(\frac{a_1}{a_2}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) ; \(\frac{b_1}{b_2}\) = \(\frac{-3}{-6}\) = \(\frac{1}{2}\) ; \(\frac{c_1}{c_2}\) = \(\frac{-8}{-9}\) = \(\frac{8}{9}\)
\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
∴ The given equations are inconsistent.
They are parallel lines. There is no solution.
The given equations are 2x – 3y = 8 and
4x — 6y = 9
2x – 3y = 8 —– (1)
⇒ 3y = 2x – 8
⇒ y = \(\frac{2 x-8}{3}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 4
4x – 6y = 9
⇒ 6y = 4x – 9
⇒ y = \(\frac{4 x-9}{6}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 5
The given lines parallel to each other.
There is no solution.
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 6
Scale: X – axis: 1 unit = 1 cm
Y – axis: 1 unit = 0.5 cm

c) \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7
9x – 10y = 14
Solution:
\(\frac{3 x}{2}\) + \(\frac{5 y}{3}\) = 7
\(\frac{6(3 x)}{2}\) + \(\frac{6(5 y)}{3}\) = 7 × 6
(Multiplying each term by 6, we get)
⇒ 9x + 10y = 42
⇒ 9x + 10y – 42 = 0 —- (1)
⇒ 9x – 10y – 14 = 0 —- (2)
Here, a1 = 9; b1 = 10; c1 = -42
a2 = 9; b2 = -10; c2 = -14
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 7
∴ The given equations are consistent.
They intersect at one point. There is an unique solution.
The given equations are \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 and 9x – 10y = 14
\(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 —– (1)
⇒ 9x + 10y = 42 ⇒ 10y = 42 – 9x
⇒ y = \(\frac{42-9 x}{10}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 8
9x – 10y = 14
⇒ 10y = 9x – 14
y = \(\frac{9 x-14}{10}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 9
The given lines intersect at one point.
x = \(\frac{28}{9}\) and y = \(\frac{7}{5}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 10
Scale : X-axis: 1 unit = 1 cm
Y-axis: 1 unit = 0.5 cm

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

d) 5x – 3y = 11
-10x + 6y = -22
Solution:
5x – 3y – 11 = 0
-10x + 6y + 22 = 0
Here, a1 = 5; b1 = -3; c1 = -11
a2 = -10: b2 = 6; c2 = 22
\(\frac{a_1}{a_2}\) = \(\frac{5}{-10}\) = \(\frac{-1}{2}\) ; \(\frac{\mathrm{b}_1}{\mathrm{~b}_2}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\);
\(\frac{c_1}{c_2}\) = \(\frac{-11}{22}\) = \(\frac{-1}{2}\)
⇒ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
∴ The given equations are dependent and consistent. There are infinitely many solutions.
The given equations are 5x – 3y = 11 and -10x + 6y = -22
5x – 3y = 11 —– (1)
⇒ 3y = 5x – 11
⇒ y = \(\frac{5 x-11}{3}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 11
⇒ 10x + 6y = -22 —– (2)
⇒ 6y = 10x – 22
⇒ y = \(\frac{10 x-22}{6}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 12
The given lines are coincident.
There are infinitely many solutions.
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 13
Scale: X – axis: 1 cm = 1 unit
Y – axis: 1 cm = 1 unit

e) \(\frac{4}{3}\)x + 2y = 8
2x + 3y = 12
Solution:
4x + 6y = 24
Multiplying each term by (3), we get
⇒ 4x + 6y – 24 = 0 —— (1)
2x + 3y – 12 = 0 ——- (2)
a1 = 4; b1 = 6; c1 = -24
a2 = 2; b2 = 3; c2 = -12
Here,
\(\frac{a_1}{a_2}\) = \(\frac{4}{2}\) = 2; \(\frac{b_1}{b_2}\) = \(\frac{6}{3}\) = 2 ; \(\frac{c_1}{c_2}\) = \(\frac{-24}{-12}\) = 2
⇒ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
∴ The given equations are dependent and consistent.
There are infinitely many solutions.
The given equations are \(\frac{4}{3}\)x + 2y = 8 and 2x + 3y = 12
\(\frac{4}{3}\)x + 2y = 8 —– (1)
⇒ 4x + 6y = 24
⇒ 6y = 24 – 4x
⇒ y = \(\frac{24-4 x}{6}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 14
2x + 3y = 12 —– (2)
⇒ 3y = 12 – 2x
⇒ y = \(\frac{12-2 x}{3}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 15
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 16
The given lines are coincident.
There are infinitely many solutions.

f) x + y = 5 (A.P. Jun.15)
2x + 2y = 10
Solution:
x + y – 5 = 0
2x + 2y – 10 = 0
Here, a1 = 1; b1 = 1; c1 = -5
a2 = 2; b2 = 2; c2 = -1O
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{1}{2}\) ; \(\frac{b_1}{b_2}\) = \(\frac{1}{2}\) ; \(\frac{c_1}{c_2}\) = \(\frac{-5}{-10}\) = \(\frac{1}{2}\)
⇒ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
∴ The given equations are dependent and consistent. There are infinitely many solutions.
The given equations are x + y = 5 and 2x + 2y = 10
x + y = 5 —— (1)
⇒ y = 5 – x
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 17
2x + 2y = 10 —– (2)
⇒ 2y = 10 – 2x
⇒ y = \(\frac{10-2 \mathrm{x}}{2}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 18
The given lines are coincident.
There are infinitely many solutions.
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 19

g) x – y = 8
3x – 3y = 16
Solution:
x – y – 8 = 0
3x – 3y – 16 = 0
Here, a1 = 1; b1 = -1; c1 = -8
a2 = 3; b2 = -3; c2 = -16
⇒ \(\frac{a_1}{a_2}\) = \(\frac{1}{3}\) ; \(\frac{\mathrm{b}_1}{\mathrm{~b}_2}\) = \(\frac{-1}{-3}\) = \(\frac{1}{3}\);
\(\frac{\mathrm{c}_1}{\mathrm{c}_2}\) = \(\frac{-8}{-16}\) = \(\frac{1}{2}\)
⇒ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
The given equations are inconsistent.
They are parallel lines. There is no
solution.
The given equations are x – y = 8 and 3x – 3y = 16
x – y = 8 —– (1)
⇒ y = x – 8
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 20
3x – 3y = 16 —- (2)
⇒ 3y = 3x – 16
⇒ y = \(\frac{3 x-16}{3}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 21
The given lines are parallel to each other. There is no solution.
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 22
Scale: X – axis : 1 unit = 1 cm
Y – axis: 1 unit = 1 cm

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

h) 2x + y – 6 = 0
4x – 2y – 4 = 0
Solution:
2x + y – 6 = 0
4x – 2y – 4 = 0
Here, a1 = 2; b1 = 1; c1 = -6
a2 = 4; b2 = -2; c2 = -4
\(\frac{a_1}{a_2}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) ; \(\frac{b_1}{b_2}\) = \(\frac{1}{-2}\) ; \(\frac{c_1}{c_2}\) = \(\frac{-6}{-4}\) = \(\frac{3}{2}\)
⇒ \(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) ≠ \(\frac{b_1}{b_2}\)
∴ The given equations are consistent.
They intersect at one point. There is a unique solution.
The given equations are 2x + y – 6 = 0 and 4x – 2y – 4 = 0
2x + y – 6 = 0 —– (1)
⇒ y = 6 – 2x
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 23
4x – 2y – 4 = 0 —– (2)
⇒ 2y = 4x – 4
⇒ y = \(\frac{4 x-4}{2}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 24
Scale: X-axis : 1 unit = 1 cm
Y-axis : 1 unit = 1 cm
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 25
The given lines intersect at one point (2, 2).
The solution is x = 2 and y = 2.

i) 2x – 2y – 2 = 0
4x – 4y – 5 = 0
Solution:
2x – 2y – 2 = 0
4x – 4y – 5 = 0
Here, a1 = 2; b1 = -2; c1 = -2
a2 = 4; b2 = -4; c2 = -5
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) ; \(\frac{\mathrm{b}_1}{\mathrm{~b}_2}\) = \(\frac{-2}{-4}\) = \(\frac{1}{2}\)
\(\frac{\mathrm{c}_1}{\mathrm{~c}_2}\) = \(\frac{-2}{-5}\) = \(\frac{2}{5}\)
⇒ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
∴ The given equations are consistent.
They are parallel lines. There is no solution.
The given equations are 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
2x – 2y – 2 = 0 —– (1)
⇒ 2y = 2x – 2
⇒ y = \(\frac{2 x-2}{2}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 26
4x – 4y – 5 = 0 —– (2)
⇒ 4y = \(\frac{4 x-5}{4}\)
⇒ y = \(\frac{4 x-5}{4}\)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 27
The given lines are parallel to each other.
There is no solution.
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 28
Scale: X – axis: 1 unit = 1 cm
Y – axis: 1 unit = 1 cm

Question 3.
Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased”. Help her friend to find how many pants and skirts Neha bought.
Solution:
Let the number of pants Neha bought be x.
Let the number of skirts Neha bought be y.
By problem,
y = 2x – 2
⇒ 2x – y = 2 —— (1)
y = 4x – 4
⇒ 4x – y = 4 —– (2)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 29
Substitute x = 1 in (1) or (2), we get
⇒ 2(1) – y = 2
⇒ 2 – y = 2
⇒ -y = 2 – 2 = 0
∴ y = o
Number of pants purchased = 1
∴ Number of skirts purchased = 0

Question 4.
10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.
Solution:
Number of boys who took part in the quiz = x
Number of girls who took part in the quiz = y
It is given that 10 students took part in the quiz.
∴ x + y = 10 ——— (1)
Since the number of girls is 4 more than the number of boys, we have
y = x + 4
⇒ -x + y = 4 —– (2)
Adding (1) and (2)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 30
Substitute y = 7 in equation (1), we get
x + 7 = 10
x = 10 – 7 = 3
Therefore, Number of boys who took part in the quiz = 3
Number of girls who took part in the quiz = 7

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

Question 5.
5 pencils and 7 pens together cost ₹ 50 whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
Let the cost of one pencil = ₹ x
Cost of one pen = ₹ y
Total cost of 5 pencils and 7 pens = 5x + 7y
Total cost of 7 pencils and 5 pens = 7x + 5y
By problem, 5x + 7y = 50 —- (1)
7x + 5y = 46 —– (2)
We equate the coefficients of ‘x’ in (1) and (2).
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 31
Substitute y = 5 in equation (1), we get
5x + (7 × 5) = 50
5x + 35 = 50
5x = 50 – 35 =15
∴ x = \(\frac{15}{5}\) = 3
Therefore, cost of one pencil = ₹ 3
Cost of one pen = ₹ 5

Question 6.
Half of the perimeter of a rectangular garden, whose length is 4m more than its width is 36m. Find the dimensions of the garden.
Solution:
Let the length of the garden be x metres.
Let the breadth of the garden be y metres.
Perimeter of the garden = 2(x + y) metres.
Half of the perimeter of the garden =
\(\frac{2(x+y)}{2}\) = x + y
By problem, x + y = 36 —— (1)
It is given that the length is 4 metres more than its width.
(i.e.,) x = y + 4
⇒ x – y = 4 —– (2)
Solving (1) and (2)
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 32
Substitute, x = 20 in equation (1),
20 + y = 36
⇒ y = 36 – 20 = 16
Therefore, length of the garden = 20 metres
Breadth of the garden =16 metres

Question 7.
We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair is formed intersecting lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
Solution:
The given linear equation is 2x + 3y – 8 = 0.

  1. The required linear equation in two variables such that it is formed an intersecting line with the given one, is 6x – 5y – 10 = 0.
  2. The linear equation in two variables (i.e.,) 4x + 6y – 10 = 0 forms a parallel line with the given one.
  3. The linear equation in two variables (i.e.,) 6x + 9y – 24 = 0 forms a coincident line with the given one.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1

Question 8.
The area of a rectangle gets reduced by 80 sq units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq units. Find the length and breadth of the rectangle.
Solution:
Let the length of the rectangle be x units and the breadth be y units.
Then the area of the rectangle = length × breadth = x × y = xy
If the length is reduced by 5 units and breadth is increased by 2 units, then its area is reduced by 80 sq. units.
∴ (x – 5) (y + 2) = xy – 80
⇒ xy + 2x – 5y – 10 = xy – 80
⇒ 2x – 5y = -70 —– (1)
When the length is increased by 10 units and breadth is decreased by 5 units, the area is increased by 50 sq. units.
∴ (x + 10) (y – 5) = xy + 50
⇒ xy – 5x + 10y – 50 = xy + 50
⇒ -5x + 10 y = 100 —– (2)
Equation (1) × 2
Solving (1) and (2), we get
4x – 10y = -140
-5x + 10y = 100 Adding
-x = -40
∴ x = 40
Substitute x = 40 in (2), we get
⇒ (-5 × 40) + 10y = 100
⇒ – 200 + 10y = 100
⇒ 10y = 100 + 200 = 300
∴ y = \(\frac{300}{10}\) = 30
Length of the rectangle = 40 units
Breadth of the rectangle = 30 units

Question 9.
In class X, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.
Solution:
Let the number of students in class X be x.
Let the number of benches in that class be y.
If three students sit on each bench, one student will be left.
∴ 3y = x – 1
⇒ x – 3y = 1 —- (1)
If four students sit on each bench, one bench will be left.
∴ 4(y – 1) = x
⇒ 4y – 4 = x
⇒ x – 4y = -4 —– (2)
Solving (1) and (2), we get
TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 33
Substitute y = 5 in (1), we get
x – (3 × 5) = 1
⇒ x – 15 = 1
∴ x = 1 + 15 = 16
∴ Number of students in class X = 16
Number of benches in that class = 5
Form a pair of linear equations for each of the following problems and find their solution.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 12 Applications of Trigonometry Ex 12.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Exercise 12.1

Question 1.
A tower stands vertically on the ground. From a point which is 15 meter away from the foot of the tower, the angle of elevation of the top of the tower is 45°. What is the height of the tower ? (AS4)
Solution:
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 1
In ∆ ABC, ∠B = 90°
AB represents the height of the tower and ∠ACB = 45°
tan 45° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
1 = \(\frac{\mathrm{AB}}{15}\)
⇒ AB = 15 meters.
Therefore, the height of the tower = 15 meters.

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 30° angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling. (AS4) (A.P. Mar. ’16)
Solution:
Let AB represents the height of the tree
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 2
AC represents the broken part.
AC = CD (∵ the broken part touches the ground)
Let BC = x meters, the height of the tree after it is broken
In ∆CBD, ∠B = 90° and ∠BDC = 30°
tan 30° = \(\frac{\mathrm{BC}}{\mathrm{BD}}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{x}{6}\)
⇒ \({\sqrt{3}}\) x = 6
⇒ x = \(\frac{6}{\sqrt{3}}\) (∵ \(\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{6 \sqrt{3}}{3}=2 \sqrt{3}\))
⇒ x = \(2 \sqrt{3}\)
sin 30° = \(\frac{\mathrm{BC}}{\mathrm{BD}}\)
\(\frac{1}{2}\) = \(\frac{2 \sqrt{3}}{C D}\)
⇒ CD = 2 × 2\({\sqrt{3}}\) = 4\({\sqrt{3}}\)
∴ The height of the tree before falling down
= AB
= AC + CB
= 4\({\sqrt{3}}\) + 2\({\sqrt{3}}\) (∵ AC = CD = 4\({\sqrt{3}}\))
= 6\({\sqrt{3}}\) m

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1

Question 3.
A contractor wants to set up a slide for the children to play in the park. He wants to set it up at the height of 2m and by making an angle of 30° with the ground. What should be the length of the slide ? (AS4)
Solution:
In the triangle ABC, ∠B = 90°
Let AC represents the length of the side
AB = 2 m
sin 30° = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
\(\frac{1}{2}\) = \(\frac{2}{\mathrm{AC}}\)
⇒ Ac = 4
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 3
Hence, length of the side = 4m

Question 4.
Length of the shadow of a 15 meter high pole is 5\({\sqrt{3}}\) m at 7 o’ clock in the morning. Then, what is the angle of elevation of the sun rays with the ground at the time ? (A.P. Mar.’15) (AS4)
Solution:
In ∆ABC, ∠B = 90°
AB represents the height of the pole.
AB = 15m
Let BC represents its shadow at 7 o’ clock in the morning.
BC = 5\({\sqrt{3}}\) m
Let the angle of elevation of the sun rays with the ground be ‘θ’.
Now, tan θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
= \(\frac{15}{5 \sqrt{3}}\)
= \(\frac{3}{\sqrt{3}}\)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 4
We know that tan 60° = \({\sqrt{3}}\)
∴ θ = 60°
Hence, the angle of elevation of the sun rays with the ground at the time = 60°.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1

Question 5.
You want to erect a pole of height 10 m with the support of three ropes. Each rope has to make an angle 30° with the pole. What should be the length of the rope ? (AS4)
Solution:
In the figure,
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 5
Let AB be the height of the pole = 10 m
Let AC be the length of the rope
Angle of elevation is 30°
From right angled ∆ ABC
cos 30° = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{10}{\mathrm{AC}}\)
⇒ AC = \(\frac{2 \times 10}{\sqrt{3}}\)
⇒ AC = \(\frac{20}{1.732}\) (∵ \({\sqrt{3}}\) = 1.732)
⇒ AC = 11.547 cm
Hence, the length of the rope is 11.547 m

Question 6.
Suppose you are shooting an arrow from the top of a building at a height of 6 m to a target on the ground at an angle of depression of 60°. What is the distance be-tween you and the object ? (AS4) (A.P. Mar.15)
Solution:
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 6
In figure,
Let BC be the height of a building = 6 m
‘C’ be the point of the observation and A’ be the target on the ground.
Angle of depression is ∠CAB = 60°
Let AC be the distance between me and the object
From the right angled ∆ABC,
sin 60° = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{6}{\mathrm{AB}}\)
⇒ AB = \(\frac{6 \times 2}{\sqrt{3}}=\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{12 \sqrt{3}}{3}\) = 4\({\sqrt{3}}\) m
Hence, the distance between me and the object is 4\({\sqrt{3}}\) m.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1

Question 7.
An electrician wants to repair an electric connection on a pole of height 9 m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use, when he climbs it at an angle of 60° with the ground ? What will be the distance between foot of the ladder and foot of the pole ? (AS4)
Solution:
In the figure,
Let AB be the height of a pole is 9 m
AC be the actual required height of the pole is 7.2 m
Angle of elevation is ∠CDA = 60°.
CD be the length of the ladder.
AD be the distance between foot of the ladder and foot of the pole.
From the right angled ∆ADC,
tan 60° = \(\frac{\mathrm{AC}}{\mathrm{AD}}\)
⇒ \({\sqrt{3}}\) = \(\frac{7.2}{\mathrm{AD}}\)
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 7
⇒ AD = \(\frac{7.2}{\sqrt{3}}=\frac{7.2}{1.732}\)
⇒ AD = 4.15692 m
Again, from the ∆ADC,
sin 60° = \(\frac{\mathrm{AC}}{\mathrm{CD}}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{7.2}{\mathrm{CD}}\)
⇒ CD = \(\frac{7.2 \times 2}{\sqrt{3}}\)
⇒ CD = \(\frac{7.2 \times 2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
CD = \(\frac{7.2 \times 2 \times \sqrt{3}}{3}\)
CD = 2.4 × 2 × \({\sqrt{3}}\)
CD = 4.8 × \({\sqrt{3}}\)
CD = 8.3138 m
Hence, the distance between foot of the ladder and foot of the pole is 4.15692 m and the length of the ladder is 8.3138 m.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1

Question 8.
A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to the stream of the river and travels a distance of 600 m to reach the another side of the river. What is the width of the river ? (AS4)
Solution:
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 8
In the figure,
Let A’ be the required point to reach the bank of the river.
Let ‘C’ be the present position of the boat (or) observation point.
AC be the distance travelled by the boat is 600 m
Angle of elevation is ∠ACB = 60°
AB be the actual width of the river
From the right angled ∆ABC,
sin 60° = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{\mathrm{AB}}{600}\)
⇒ AB = 600 × \(\frac{\sqrt{3}}{2}\)
⇒ AB = 300\({\sqrt{3}}\) m
⇒ Hence, the width of the river is 300\({\sqrt{3}}\) m

Question 9.
An observer of height 1.8m is 13.2m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45°. What is the height of palm tree ? (AS4)
Solution:
AB represents the height of the observer.
CD denotes the height of the palm tree.
AB = 1.8 m; BC = 13.2 m AE = 13.2 m (∵ ABCE is a rectangle so, opposite sides of BC and AE are equal)
(AB and EC are also opposite sides of the rectangle ABCE)
∴ AB = EC
In ∆AED, ∠E = 90°
tan 45° = \(\frac{\mathrm{DE}}{\mathrm{AE}}\)
1 = \(\frac{\mathrm{DE}}{13.2}\)
⇒ DE = 13.2 m
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 9
Therefore, the height of the palm tree CD
= DE + EC
= 13.2 + 1.8
= 15 meters.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1

Question 10.
In the adjacent figure
In ∆ABC, AC = 6 cm, AB = 5 cm and ∠BAC = 30°. Find the area of the triangle. (AS4)
Solution:
In ∆ ABC,
AB = 5 cm, AC = 6 cm
∠BAC = 30°
Draw BD ⊥ AC
TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry Ex 12.1 10
In ∆ ABD, ∠ADB = 90°
sin 30° = \(\frac{\mathrm{BD}}{\mathrm{AB}}\)
\(\frac{1}{2}\) = \(\frac{\mathrm{BD}}{5}\)
⇒ 2 × BD = 5
BD = \(\frac{5}{2}\) = 2.5 cm
Hence, area of the triangle ABC
= \(\frac{1}{2}\) × Base × Altitude
= \(\frac{1}{2}\) × AC × BD
= \(\frac{1}{2}\) × 6 × 2.5
= 3 × 2.5 = 7.5 cm2

TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.4

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets Exercise 2.4

Question 1.
State which of the following sets are empty and which are not ?
a) The set of straight lines passing through a point.
b) Set of odd natural numbers divisible by 2.
c) {x : x is a natural number, x < 5 and x > 7}
d) {x : x is a common point to any two parallel lines}
e) Set of even prime numbers.
Answer:
a) The number of straight lines passing through a point is infinite. So, the given set is non-empty.
b) We know the odd natural numbers are 1, 3, 5, 7,…. are not divisible by 2. Hence the given set is empty.
c) There is no natural number satisfying the given condition. Hence the given set is empty.
d) There is no common point to any two parallel lines because they do not meet when produced on either side. Hence the given set is empty.
e) 2 is the only even prime number. Therefore the set contains one element. Hence the given set is non-empty.

Question 2.
Which of the following sets are finite or infinite ?
a) The set of months in a year.
b) {1, 2, 3,…….., 99, 100}
c) The set of prime numbers less than 99.
Answer:
a) There are 12 months in a year. The set of months in a year contains 12 elements. Hence the set is finite.
b) Obviously, the given set contains 100 elements. Hence the set is finite.
c) We can count the prime numbers less than 99. Hence the set is finite.

TS 10th Class Maths Solutions Chapter 1 Sets Ex 2.4

Question 3.
State whether each of the following sets is finite or infinite.
a) The set of letters in the english alphabet.
b) The set of lines which are parallel to the x – axis.
c) The set of numbers which are multiples of 5.
d) The set of circles passing through the origin (0, 0).
Answer:
a) The set of letters in the english alphabet contains 26 elements. Hence, the set is finite.
b) We cannot count the number of parallel lines drawn to the x – axis. Hence the set is infinite.
c) The set of numbers which are multiples of 5 is {5, 10, 15, 20, 25, …}. Hence the set contains infinite number of elements. Hence, the set is infinite.
d) The number of circles that can be drawn through the origin (0, 0) is countless. Hence the set is infinite.

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.1

Students can practice 10th Class Maths Solutions Telangana Chapter 3 Polynomials Ex 3.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.1

Question 1.
If p(x) = 5x7 – 6x5 + 7x – 6, find
(i) Coefficient of x5
(ii) degree of p(x)
(iii) constant term.
a) If P(x) = 5x7 – 6x5 + 7x – 6
Solution:
i) coefficient of x5 is -6
ii) degree of p(x) = highest degree of x = 7
iii) constant term is -6

Question 2.
State which of the following statements are true and which are false ? Give reasons for your choice.

i) The degree of the polynomial
\(\sqrt{2}\)x2 – 3x + 1 is \(\sqrt{2}\).
Solution:
The given statement is false because \(\sqrt{2}\) is the coefficient of x2 but not its degree. The degree of the polynomial is 2.

ii) The coefficient of x2 in the polynomial p(x) = 3x3 – 4x2 + 5x + 7 is 2.
Solution:
The given statement is false because the coeffi-cient of x2 in the polynomial is -4 but not 2.

iii) The degree of a constant term is zero.
Solution:
The given statement is true. Because 3x0 = 3 the degree is 0

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.1

iv) \(\frac{1}{x^2-5 x+6}\) is a quadratic polynomial.
Solution:
The given statement is false because the variable ‘x’ appears in the denominator.

v) The degree of a polynomial is one more than the number of terms ¡n it.
Solution:
The given statement is false. There is a no relationship between the degree of the polynomial and the number of terms in it.

Question 3.
If p(t) = t3 – 1, find the values of p(1), p(-1), p(0), p(2). p(-2) (A.P.Mar. ’15)
Solution:
Given that p(t) = t3 – 1
∴ p(1) = (1)3 – 1 = 1 – 1 = 0
p(-1) = (-1) – 1 = -1 – 1 = -2
p(0) = (0) – 1 = 0 – 1 = -1
p(2) = (2) – 1 = 8 – 1 = 7
= (-2) – 1 = -8 – 1 = -9

Question 4.
Check whether -2 and 2 are the zeroes of the polynomial x4 – 16.
Solution:
p(x) = x4 – 16
p(-2) = (-2)4 – 16 = 16 – 16 = 0
p(2) = (24) – 16 = 16 – 16 = 0
Yes, -2 and 2 are zeroes of the polynomial x4 – 16.

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.1

Question 5.
Check whether 3 and -2 are the zeroes of the polynomial p(x) when p(x) = x2 – x – 6
Solution:
p(x) = x2 – x – 6
= (3)2 – 3 – 6 = 9 – 3 – 6 = 9 – 9 = 0
p(-2) = (-2)2 – (-2) – 6 = 4 + 2 – 6 = 6 – 6 = 0
Yes, 3 and -2 are zeroes of the polynomial
p(x) = x2 – x – 6

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Do This

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions 1

Question 1.
List the teeth under each of the following type.
i) Incisors
Solution:
Central incisors = 4
Lateral incisors = 4
TS 10th Class Maths Solutions Chapter 2 Sets InText Questions 2
Total incisors = 8

ii) Canines
Solution:
Total Canines = 4

iii) Pre-molars
Solution:
First premolars = 4
Second premolars = 4
TS 10th Class Maths Solutions Chapter 2 Sets InText Questions 3
Total premolars = 8

iv) Molars
Solution:
First molars = 4
Second molars = 4
Third molars = 4
TS 10th Class Maths Solutions Chapter 2 Sets InText Questions 4
Total molars = 12

Question 2.
Identify and write the “common property” of the following collections. (Page No. 26)

1) 2, 4, 6, 8 ….
2) 3, 5, 7, 11, ….
3) 1, 4, 9, 16, ………
4) January, February, March, April,………..
5) Thumb, index finger, middle finger, ring finger, pinky.
Solution:
1) For given integers 2, 4, 6, 8 …….. 2n, n = 1, 2, 3, 4,…. is any positive integer.
2) 2, 3, 5, 7, 11, …. prime numbers.
3) For given integers 1, 4, 9, 16, …. n2, where ‘n’ is any positive integer and square of the numbers.
4) January, February, March, April,…. months of the every year.
5) Thumb, index finger, middle finger, ring finger, pinky fingers of the human hand.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 3.
Write the following sets. (Page No. 27)
1) Set of the first five positive integers.
2) Set of multiples of 5 which are more than 100 and less than 125.
3) Set of first five cubic numbers.
4) Set of digits in the Ramanujan number.
Solution:
1) { + 1, +2, +3, +4, +5}
2) {105, 110, 115, 120}
3) {1, 8, 27, 64,125} = {13, 23, 33, 43, 53}
4) Ramanujan Number = 1729. So the set = {1, 2, 7, 9}

Question 4.
Some numbers are given below. Decide the numbers to which number sets they belong to and does not belong to and express with correct symbols. (Page No. 27)
i) 1
ii) 0
iii) -4
iv) \(\frac{5}{6}\)
v) \(\text { 1. } \overline{3}\)
vi) \(\sqrt{2}\)
vii) log 2
viii) 0.03
ix) π
x) \(\sqrt{-4}\)
Solution:
Set of natural numbers = N
set of integers = Z
Set of rational numbers = Q
Set of real numbers = R
i) 1 ∈ (N, Z, Q, R}
ii) 0 ∉ N, 0 ∈ {Z, Q, R}
iii) -4 ∉ N, -4 ∈ (Z, Q, R}
iv) \(\frac{5}{6}\) ∉ (N, Z} But \(\frac{5}{6}\) ∈ {Q, R}
v) \(1 . \overline{3}\) ∉ {N, Z} But \(1 . \overline{3}\) ∈ {Q, R}
vi) \(\sqrt{2}\) ∉ {N, Z} But \(\sqrt{2}\) ∈ {Q, R}
vii) log2 ∉ N,Z But log 2 ∈ {Q, R}
viii) 0.03 ∉ {N, Z}But 0.03 ∈ {Q, R}
ix) π ∉ {N, Z} But π ∈ (Q, R}
x) \(\sqrt{-4}\) ∉ (N, Z, Q} But \(\sqrt{-4}\) ∈ R.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 5.
List the elements of the following sets.
i) G = {all the factors of 20}
ii) F = {the multiples of 4 between 17 and 61 which are divisible by 7}
iii) S = {x : x is a letter in the word MADAM’}
iv) P = {x : x is a whole number between 3.5 and 6.7} (Page No. 29)
Answer:
i) G = {1, 2, 4, 5, 10, 20}
ii) Multiples of 4 between 17 and 61.
x = {20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60}
F = {28, 56}
iii) S = {M, D, A}
iv) P = {4, 5, 6}

Question 6.
Write the following sets in the roaster form. (Page No. 29)
i) B is the set of all months in a year having 30 days.
ii) P is the set of all prime numbers less than 10.
iii) X is the set of colours of the rainbow.
Answer:
i) B = {April, June, September, November}
ii) P = {2, 3, 5, 7}
iii) X = {Violet, Indigo, Blue, Green, Yellow, Orange, Red}

Question 7.
A is the set of factors of 12. Which one of the following is not a member of A ? (Page No. 29)
A) 1
B) 4
C) 5
D) 12
Answer:
[C]

Think – Discuss

Question 1.
Observe the following collections and prepare as many as generalised statements you can describing their more properties. (Page No. 26)
i) 2, 4, 6, 8, ….
ii) 1, 4, 9, 16
Answer:
i) 3, 6, 9, 12, ….
Property : multiples of 3.

ii) 1, 8, 27, 64,….
Property : cube of the numbers, i.e., 13, 23, 33, 43, ….

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 2.
Can you write the set of rational numbers listing elements in it ? (Page No. 28)
Solution:
No

Try This

Question 1.
Write some sets of your choice, involving algebraic and geometrical ideas. (Page No. 29)
Answer:
1) The natural number greater than three and less than twelve.
2) A two digit number such that the sum of its digits is 8.
3) The set of quadrilaterals.
4) The set of all triangles in a plane.

Question 2.
Match roaster forms with the set builder form.
i) {P, R, I, N, C, A, L}
ii) {0}
iii) {1, 2, 3, 6, 9, 18}
iv) {3, -3} (Page No. 29)
a) {x : x is a positive integer and is a divisor of 18}
b) {x : x is an integer and x2 – 9 = 0}
c) {x : x is an integer and x + 1 = 1}
d) {x: x is a letter of the word PRINCIPAL}
Answer:
i) d
ii) c
iii) a
iv) b

Do This

Question 1.
A = {1, 2, 3, 4}, B = {2, 4}, C = {1, 2, 3, 4, 7}, F = { } (Page No. 33)
Fill in the blanks with ⊂ or ⊄.
i) A………B
ii) C……..A
iii) B……..A
iv) A……..C
v) B……..C
vi) F……B
Answer:
i) AB
ii) CA
iii) BA
iv) AC
v) BC
vi) FB

Question 2.
State which of the following statements are true. (Page No. 33)
i) { } = ϕ
ii) ϕ = 0
iii) 0 = {ϕ}
Answer:
i) True (T)
ii) False (F)
iii) False (F)

Question 3.
Let A = {1, 3, 7, 8} and B = {2, 4, 7, 9}. Find A∩B.(Page No. 37)
Solution:
Given sets
A = {1, 3, 7, 8} and B = {2, 4, 7, 9}
A ∩ B = {1, 3, 7, 8} ∩ {2, 4, 7, 9}
= {7}

Question 4.
If A = {6, 9, 11}; ϕ = { }, find A ∪ ϕ. (Page No. 37)
Solution:
Given sets A = {6, 9, 11} and B = {2, 4, 7, 9}
A ∪ ϕ = {6, 9, 11} ∪ {ϕ}
= {6, 9, 11} = A
∴ A ∪ ϕ = A

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 5.
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
B = {2, 3, 5, 7}. Find A ∩ B. (Page No. 37) (June ’15(AP))
Solution:
Given sets A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}
A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {2, 3, 5, 7}
= {2, 3, 5, 7} = B
∴ A ∩ B = B

Question 6.
If A = {4, 5, 6}; B = {7, 8}, then show that A ∪ B = B ∪ A. (Page No. 37)
Solution:
Given sets are
A = (4, 5, 6} and B = (7, 8}
A ∪ B = {4, 5, 6} ∪ {7, 8}
= {4, 5, 6, 7, 8}
B ∪ A = {7, 8} ∪ {4, 5, 6}
= {4, 5, 6, 7, 8}
∴ A ∪ B = B ∪ A

Question 7.
If A = {1, 2, 3, 4, 5}; B = {4, 5, 6, 7} then find A-B and B-A. Are they equal ? (Page No. 38)
Solution:
Given sets are A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}
A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7} = {1, 2, 3}
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5} = {6, 7}
No, A – B ≠ B – A

Question 8.
If V = {a, e, i, o, u} and B = {a, i, k, u}, find V – B and B – V. (Page No. 38)
Solution:
Given sets are
V = {a, e, i, o, u} and B = {a, i, k, u}
V – B = {a, e, i, o, u} – {a, i, k, u}
= {e, o}
B – V = {a, i, k, u} – {a, e, i, o, u}
= {k}

Try This

Question 1.
A = {set of quadrilaterals}, B = {square, rectangle, trapezium, rhombus}
State whether
A ⊂ B or B ⊂ A. Justify your answer. (Page No. 33)
Answer:
A ⊄ B
B ⊂ A every element of B is also an element of A.

Question 2.
If A = {a, b, c, d}. How many subsets does the set A have? (Remember null set and equal sets). (Page No. 33)
A) 5
B) 6
C) 16
D) 65
Solution:
A = {a, b, c, d}
Subsets of A = {a}, {b}, {c}, {d};
n(A) = 4
Number of subsets for a set, which is having ‘n’ elements is 2n.
So n(A) = 4
Number of subsets for A is 24 =16.
Answer:
(C)

Question 3.
P is the set of factors 5, Q is the set of factors of 25 and R is the set of factors of 125.
Which one of the following is false?
A) P ⊂ Q
B) Q ⊂ R
C) R ⊂ P
D) P ⊂ R (Page No. 33)
Solution:
P = {1, 5}
Q = {1, 5, 25}
R = {1, 5, 25, 125}
Answer:
(C)

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 4.
A is the set of prime numbers less than 10, B is the set of odd numbers < 10 and C is the set of even numbers < 10. How many of the following statements are true ? (Page No. 33)
i) A ⊂ B
ii) B ⊂ A
iii) A ⊂ C
iv) C ⊂ A
v) B ⊂ C
vi) X ⊂ A
Solution:
A = {2, 3, 5, 7}
B = {1, 3, 5, 7, 9}
C = {2, 4, 6, 8}
The given all statements are false.

Question 5.
List out some sets A and B and choose their elements such that A and B are disjoint. (Page No. 37)
Solution:
A and B are disjoint sets.
i) A = {2, 3, 5} B = {4, 6, 8}
ii) A = {1, 2, 3} B = {4, 5, 6}
iii) A = {1, 3, 5, 7} B = {2, 4, 6, 8}

Question 6
If A = {2, 3, 5}, find A∪ϕ and ϕ∪A and compare. (Page No. 37)
Solution:
A = {2, 3, 5}; ϕ = { }
A∪ϕ = {2, 3, 5} ∪ { } = {2, 3, 5}
ϕ∪A = { } ∪ {2, 3, 5} = {2, 3, 5}
∴ A∪ϕ = ϕ∪A

Question 7.
If A = {1, 2, 3, 4}; B = {1, 2, 3, 4, 5, 6, 7, 8} then find A ∪ B, A ∩ B. What do you notice about the result ? (Pg. No. 37)
Solution:
A = {1, 2, 3, 4};
B = {1, 2, 3, 4, 5, 6, 7, 8}
A∪B = {1, 2, 3, 4} ∪ {1, 2, 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}
The common elements in both A and B sets are 1, 2, 3, 4.
A∩B = {1, 2, 3, 4}
Result :
i) A∪B = B
ii) A∩B = A
iii) A ∩ B ⊂ A ∪ B

Question 8.
A = {1, 2, 3, 4, 5, 6}; B = {2, 4, 6, 8, 10}. Find the intersection of A and B. (Page No. 37)
Solution:
A = {1, 2, 3, 4, 5, 6}
B = {2, 4, 6, 8, 10}
The common elements in both A and B are 2, 4, 6.
∴ A ∩ B = {2, 4, 6}

Think : Discuss

Question 1.
Is empty set subset to every set? (Page No. 34)
Solution:
Yes.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 2.
Is any set subset to itself? (Page No. 34)
Solution:
Yes.

Question 3.
You are given two sets such that a set is not a subset of the other. If you have to prove this, how do you prove ? Justify your answers. (Page No. 34)
Solution:
The above statement is true only in the two sets does not have common elements.
Ex : A = {1, 2, 3}; B = {4, 5, 6}
So, AB.

Question 4.
The intersection of any two disjoint sets is a null set. Justify your answer. (Page No. 37)
Answer:
Yes, this statement is true because A ∩ B = ϕ when A and B are disjoint sets.

Question 5.
The sets A – B, B – A and A ∩ B are mutually disjoint sets. Use examples to observe if this is true. (Page No. 38)
Solution:
Let the sets are
A = {1, 2, 3, 4} and B = {5, 6, 7, 8}
A – B = {1, 2, 3, 4} – {5, 6, 7, 8} = {1, 2, 3, 4}
B – A = {5, 6, 7, 8} – {1, 2, 3, 4} = {5, 6, 7, 8}
A ∩ B = {1, 2, 3, 4} ∩ {5, 6, 7, 8} = { } = ϕ
∴ A – B, B – A and A ∩ B are disjoint sets.

Do This

Question 1.
Which of the following are empty sets? Justify your answer. (Page No. 44)
i) Set of integers which lie between 2 and 3.
ii) Set of natural numbers that are less than 1.
iii)Set of odd numbers that have remainder zero, when divided by 2.
Answer:
i) This is null set. We know that there is no integer that lie between 2 and 3.
ii) This is also a null set. We know that there is natural number less than ‘1’.
iii) This is a null set. We know that odd numbers do not leave remainder zero when divided by 2.

Question 2.
State which of the following sets are finite and which are infinite. Give reasons for your answers.
i) A = {x: x ∈ N and x < 100}
ii) B = {x : x ∈ N and x ≤ 5}
iii)C = {12, 22, 32, ………… )
iv)D = {1, 2, 3, 4)
v) {x : x is a day of the week) (Page No. 44)
Answer:
i) A = {1, 2, 3, ………., 98, 99}
This set is finite, because there are 99 numbers possible to count.

ii) B = {1, 2, 3. 4, 5}
This set is finite, because there are 5 numbers possible to count.

iii) C = {12, 22, 32,………)
This set is infinite, because there are infinite numbers.

iv) D = {1, 2, 3, 4}
This set is finite because there are 4 numbers that are possible to count.

v) E = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday)
This set is finite, because there are 7 days in a week possible to count.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 3.
Tick the set which is infinite
A) The set of whole numbers < 10
B) The set of prime numbers < 10
C) The set of integers < 10
D) The set of factors of 10 (Page No. 44)
Answer:
[C]
The set of integers < 10
{……., -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Try This

Question 1.
Which of the following sets are empty sets? Justify your answer.
i) A = {x : x2 = 4 and 3x = 9}
ii) The set of all triangles in a plane having the sum of their three angles less than 180. (Page No. 44)
Answer:
i) Empty set.
To satisfy these equations same x value is not possible.
ii) Empty set.
The sum of the three angles of a triangle is equal to 180°.

Question 2.
B = {x : x + 5 = 5} is not an emptyset. Why? (Page No. 44)
Solution:
x + 5 = 5
x = 5 – 5
x = 0
For x = 0 it is true
Only one element is there.
So it is not an empty set.

Think — Discuss

Question 1.
An empty set is a finite set. Is this statement true or false? Why? (Page No. 44)
Answer:
Yes, it is a finite set because there is finite number i.e., ‘0’ elements it consists.

Question 2.
What is the relation between n(A), n(B), n(A ∩ B) and n (A ∪ B)? (Page No. 45)
Solution:
n(A) = elements in set A, n(B) = elements in set B
n(A ∩ B) = set of all elements which are common to both A and B.
n(A ∪ B) = elements in set A and set B (or) union of sets A and B.
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 3.
If A and B are disjoint sets, then how can you find n(A ∪ B)? (Page No. 45)
Solution:
n(A) = elements in set A.
n(B) = elements in set B.
n(A ∩ B) = elements in set A and set B
Here it is ‘0’ (∴ A and B are disjoint sets)
n(A ∪ B) = elements in set A or set B.
∴ n(A ∪ B) = n(A) ÷ n(B) – n(A ∩ B)
= n(A) + n(B) – 0
= n(A) + n(B).

TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.3

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets Exercise 2.3

Question 1.
Which of the following sets are equal ?
a) A = {x : x is a letter in the word FOLLOW’}
b) B = {x : x is a letter in the word ‘FLOW’}
c) C = {x : x is a letter in the word ‘WOLF’}
Answer:
a) Writing the given set in the roaster form, we have A = {F, O, L, W}
b) Writing the given set in the roaster form, we have B = {F, L, O, W}
c) Writing the given set in the roaster form, we have C = {W, O, L, F}
Therefore, A, B, C are equal sets.
[∴ The sets A, B, C have exactly the same elements]

Question 2.
Consider the following sets and fill up the blank in the statement given below with = or ≠ so as to make the statement true.
A = {1, 2, 3};
B = {The first three natural numbers}
C = {a, b, c, d};
D = {d, c, a, b}
E = {a, e, i, o, u};
F = {set of vowels in English Alphabet}
i) A ……… B
ii) A …….. E
iii) C ……. D
iv) D …… F
v) F ……. A
vi) D …… E
vii) F ……. B
Answer:
i) A = B
ii) A ≠ E
iii) C=D
iv) D ≠ F
v) F ≠ A
vii) D ≠ E
viii) F ≠ B

TS 10th Class Maths Solutions Chapter 1 Sets Ex 2.3

Question 3.
In each of the following, state whether
A = B or not.
i) A = {a, b, c, d}; B = {d, c, a, b}
ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}
iii) A = (2, 4, 6, 8, 10)
B = {x: x is a positive even integer and x ≤ 10}
iv) A = {x: x ¡s a multiple of 10};
B = {10, 15, 20, 25, 30,……. }
Answer:
i) A = B because A and B have exactly the same elements i.e., a, b, c, d.
ii) A ≠ B because A and B have not exactly the same elements.
iii) A = B because A and B have exactly the same elements.
Writing B in roaster form, we have
B = {2, 4, 6, 8, 10}
iv) A = {10, 20, 30, 40,……..}
B = {10, 15, 20, 25,……..}
A ≠ B because A and B have not exactly the same elements.

Question 4.
State the reasons for the following:
i) {1,2, 3,…, 10} ≠ {x : x ∈ N and 1 < x < 10}
ii) {2, 4, 6, 8, 10} ≠ {x : x = 2n + 1 and x ∈ N}
iii) {5, 15, 30, 45} ≠ {x : x is a multiple of 15
iv) {2, 3, 5, 7, 9} ≠ {x : x is a prime number}
Solution:
The first set is {1, 2, 3, ……, 10}
Writing the second set in roaster form, we have {2, 3, 4, ……, 9}
The first set and the second set have not exactly the same elements.
∴ {1, 2, 3,……10} ≠ {x : x ∈ N and 1 < x < 10}

ii) The first set is {2, 4, 6, 8, 10}
Writing the second set in roaster form, we have {3, 5, 7, 9, ….}
∴ {2, 4, 6, 8, 10} ≠ {3, 5, 7, 9, ….}
x = 2n + 1 means x is odd.

iii) The first set is {5, 15, 30, 45}
Writing the second set in roaster form, we have {15, 30, 45, 60, …}
∴ {5, 15, 30, 45} ≠ {15, 30, 45, 60,…}
5 does not exist, since x is multiple of 15.

iv) The first set is {2, 3, 5, 7, 9}
Writing the second set in roaster form, we have {2, 3, 5, 7, 11, 13,…}
∴ {2, 3, 5, 7, 9} ≠ {2, 3, 5, 7, 11, 13 }
9 is not a prime number.

TS 10th Class Maths Solutions Chapter 1 Sets Ex 2.3

Question 5.
List all the subsets of the following sets.
i) B = {p, q}
ii) C = {x, y, z}
iii) D = {a, b, c, d}
iv) E = {1, 4, 9, 16}
v) F = {10, 100, 1000}
Solution:
i) {p}, {q}, {p, q}, { ϕ }
ii) {x}, {y}, {z}, {x, y}, {y, z}, {x, z}, {x, y, z}, {ϕ}
iii) {a}, {b}, {c}, {d}, {a, b}, {b, c}, {c, d}, {a, c}, [a, d}, {b, d}, {a, b, c}, {b, c, d}, {a, b, d}, {a, c, d}, {a, b, c, d}, {ϕ}
iv) {1}, {4}, {9}, {16}, {1, 4}, {4, 9}, {9, 16}, {1, 9}, {1, 16}, {4, 16}, {1, 4, 9}, {4, 9, 16}, {1, 4, 16}, {1, 9, 16}, {1, 4, 9, 16}, {ϕ}
v) {10}, {100}, {1000}, {10, 100}, {100, 1000}, {10, 1000}, {10, 100, 1000}, {ϕ}