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AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.3

10th Class Maths 11th Lesson Trigonometry Ex 11.3 Textbook Questions and Answers

Question 1.
Evaluate:
i) \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)
Answer:
Given that \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)
= \(\frac{\tan 36^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}\) [∵ cot (90 – θ) = tan θ]
= \(\frac{\tan 36^{\circ}}{\tan 36^{\circ}}\)
= 1

ii) cos 12° – sin 78°
Answer:
Given that cos 12° – sin 78°
= cos 12° – sin(90° – 12°) [∵ sin (90 – θ) = cos θ]
= cos 12° – cos 12° = 0

iii) cosec 31° – sec 59°
Answer:
Given that cosec 31° – sec 59°
= cosec 31° – sec (90° – 31°) [∵ sec (90 – θ) = cosec θ]
= cosec 31° – cosec 31° = 0

iv) sin 15° sec 75°
Answer:
Given that sin 15° sec 75°
= sin 15° . sec (90° – 15°)
= sin 15° . cosec 15° [∵ sec (90 – θ) = cosec θ]
= sin 15° . \(\frac{1}{\sin 15^{\circ}}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

v) tan 26° tan 64°
Answer:
Given that tan 26° tan 64°
= tan 26° . tan (90° – 26°)
= tan 26° . cot 26° [∵ tan (90 – θ) = cot θ]
= tan 26° . \(\frac{1}{\tan 26^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]
= 1

Question 2.
Show that
i) tan 48° tan 16° tan 42° tan 74° = 1
Answer:
L.H.S. = tan 48° tan 16° tan 42° tan 74°
= tan 48°. tan 16° . tan(90° – 48°) . tan(90° – 16°)
= tan 48° . tan 16° . cot 48° . cot 16° [∵ tan (90 – θ) = cot θ]
= tan 48° . tan 16° . \(\frac{1}{\tan 48^{\circ}}\) . \(\frac{1}{\tan 16^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]
= 1 = R.H.S.
∴ L.H.S. = R.H.S.

ii) cos 36° cos 54° – sin 36° sin 54° = 0
Answer:
L.H.S. = cos 36° cos 54° – sin 36° sin54°
= cos (90° – 54°) . cos (90° – 36°) – sin 36° . sin 54° [∵ cos (90 – θ) = sin θ]
= sin 54° . sin 36° – sin 36° . sin 54°
= 0 = R.H.S.
∴ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle. Find the value of A.
Answer:
Given that tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90 – θ)]
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
⇒ A = \(\frac{108^{\circ}}{3}\) = 36°
Hence the value of A is 36°.

Question 4.
If tan A = cot B where A and B. are acute angles, prove that A + B = 90°.
Answer:
Given that tan A = cot B
⇒ cot (90° – A) = cot B [∵ tan θ = cot (90 – θ)]
⇒ 90° – A = B
⇒ A + B = 90°

Question 5.
If A, B and C are interior angles of a triangle ABC, then show that \(\tan \left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)=\cot \frac{\mathbf{C}}{2}\)
Answer:
Given A, B and C are interior angles of right angle triangle ABC then A + B + C = 180°.
On dividing the above equation by ‘2’ on both sides, we get 180°

On taking tan ratio on both sides

Hence proved.

Question 6.
Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0° and 45°.
Answer:
We have sin 75° + cos 65°
= sin (90° – 15°) + cos (90° – 25°)
= cos 15° + sin 25° [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.2

10th Class Maths 11th Lesson Trigonometry Ex 11.2 Textbook Questions and Answers

Question 1.
Evaluate the following.
i) sin 45° + cos 45°
Answer:
sin 45° + cos 45°
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= \(\frac{1+1}{\sqrt{2}}\)
= \(\frac{2}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}\)
= √2

ii)

Answer:

iii)

Answer:

iv) 2 tan² 45° + cos² 30° – sin² 60°
Answer:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) – \(\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= \(\frac{2}{1}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\)
= \(\frac{8+3-3}{4}\)
= \(\frac{8}{4}\)
= 2

v)

Answer:

Question 2.
Choose the right option and justify your choice.
i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) sin 60°
b) cos 60°
c) tan 30°
d) sin 30°
Answer:

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) tan 90°
b) 1
c) sin 45°
d) 0
Answer:
\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-(1)^{2}}{1+(1)^{2}}\)
= \(\frac{0}{1+1}\) = \(\frac{0}{2}\) = 0

iii) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\)
a) cos 60°
b) sin 60°
c) tan 60°
d) sin 30°
Answer:

Question 3.
Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°). What can you conclude?
Answer:
Take sin 60°.cos 30° + sin 30°.cos 60°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{(\sqrt{3})^{2}}{4}\) + \(\frac{1}{4}\)
= \(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{3+1}{4}\)
= \(\frac{4}{4}\) = 1 …… (1)
Now take sin (60° + 30°)
= sin 90° = 1 …….. (2)
From equations (1) and (2), I conclude that
sin (60°+30°) = sin 60° . cos 30° + sin 30° . cos 60°.
i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.
Is it right to say cos (60° + 30°) = cos 60° cos 30° – sin 60° sin 30° ?
Answer:
L.H.S. = cos (60° + 30°)
cos 90° = 0
R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°.
= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0
∴ L.H.S = R.H.S
Yes, it is right to say
cos (60°+30°) = cos 60° . cos 30° – sin 60° . sin 30°.
i.e., cos (A + B) = cos A . cos B – sin A . sin B

Question 5.
In right angle triangle △PQR, right angle is at Q and PQ = 6 cms, ∠RPQ = 60°. Determine the lengths of QR and PR.
Answer:
Given that △PQR is a right angled triangle, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°.

tan 60° = \(\frac{\text { Opposite side to } \angle P}{\text { Adjacent side to } \angle P}\)
√3 = \(\frac{RQ}{6}\)
which gives RQ = 6√3 cm ……. (1)
To find the length of the side RQ, we consider

∴ The length of QR is 6√3 and RP is 12 cm.

Question 6.
In △XYZ, right angle is at Y, YZ = x, and XY = 2x then determine ∠YXZ and ∠YZX.
Answer:
Note: In the problem take
YX = x, and XZ = 2x.

Given that △XYZ is a right angled triangle and right angle at Y, and YX = x and XZ = 2x.
By Pythagoras theorem
XZ² = XY² + YZ²
(2x)² = (x)² + YZ²
4x² = x² + YZ²
YZ² = 4x² – x² = 3x²
YZ = \(\sqrt{3 x^{2}}\) = √3x
Now, from the △XYZ
tan X = \(\frac{XZ}{XY}\) = \(\frac{\sqrt{3} x}{x}\)
tan X = √3 = tan 60°
∴ Angle YXZ is 60°.
tan Z = \(\frac{XY}{YZ}\) = \(\frac{x}{\sqrt{3} x}\)
tan Z = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle YZX is 30°.
Hence ∠YXZ and ∠YZX are 60° and 30°.

Question 7.
Is it right to say that
sin (A + B) = sin A + sin B? Justify your answer.
Answer:
Let A = 30° and B = 60°
L.H.S = sin (A + B)
= sin (30° + 60°) = sin 90° = 1
R.H.S = sin 30° + sin 60°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
Hence L.H.S ≠ R.H.S
So, it is not right to say that sin (A + B) = sin A + sin B

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.1

10th Class Maths 11th Lesson Trigonometry Ex 11.1 Textbook Questions and Answers

Question 1.
In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.
Answer:
Given that
△ABC is a right angle triangle and Lengths of AB, BC and CA are 8 cm, 15 cm and 17 cm respectively.
Among the given lengths CA is longest.
Hence CA is the hypotenuse in △ABC and its opposite vertex having right angle.
i.e., ∠B = 90°.
With reference to ∠A, we have opposite side = BC = 15 cm
adjacent side = AB = 8 cm
and hypotenuse = AC = 17

sin A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{BC}{AC}\) = \(\frac{15}{17}\)
cos A = \(\frac{\text { Adjacent side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{8}{17}\)
tan A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Adjacent side of } \angle \mathrm{A}}\) = \(\frac{BC}{AB}\) = \(\frac{15}{8}\)
∴ sin A = \(\frac{15}{17}\);
cos A = \(\frac{8}{17}\)
tan A = \(\frac{15}{8}\)

Question 2.
The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠P = 90° respectively. Then find, tan Q – tan R.
Answer:

Given that △PQR is a right angled triangle and PQ = 7 cm, QR = 25 cm.
By Pythagoras theorem QR² = PQ² + PR²
(25)² = (7)² + PR²
PR² = (25)² – (7)² = 625 – 49 = 576
PR = √576 = 24 cm
tan Q = \(\frac{PR}{PQ}\) = \(\frac{24}{7}\);
tan R = \(\frac{PQ}{PR}\) = \(\frac{7}{24}\)
∴ tan Q – tan R = \(\frac{24}{7}\) – \(\frac{7}{24}\)
= \(\frac{(24)^{2}-(7)^{2}}{168}\)
= \(\frac{576-49}{168}\)
= \(\frac{527}{168}\)

Question 3.
In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.
Answer:
Given that ABC is a right angle triangle with right angle at B, and BC = a = 24 units, CA = b = 25 units and ∠BAC = θ.

By Pythagoras theorem
AC² = AB² + BC²
(25)² = AB² + (24)²
AB² = 25² – 24² = 625 – 576
AB² = 49
AB = √49 = 1
With reference to ∠BAC = θ, we have
Opposite side to θ = BC = 24 units.
Adjacent side to θ = AB = 7 units.
Hypotenuse = AC = 25 units.
Now
cos θ = \(\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{7}{25}\)
tan θ = \(\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}\) = \(\frac{BC}{AB}\) = \(\frac{24}{7}\)
Hence cos θ = \(\frac{7}{25}\) and tan θ = \(\frac{24}{7}\)

Question 4.
If cos A = \(\frac{12}{13}\), then find sin A and tan A.
Answer:
From the identity
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A
= 1 – \(\left(\frac{12}{13}\right)^{2}\)
= 1 – \(\frac{144}{169}\)
= \(\frac{169-144}{169}\)
= \(\frac{25}{169}\)
∴ sin A = \(\sqrt{\frac{25}{169}}\) = \(\frac{5}{13}\)

∴ sin A = \(\frac{5}{13}\); tan A = \(\frac{5}{12}\)

Question 5.
If 3 tan A = 4, then find sin A and cos A.
Answer:
Given 3 tan A = 4
⇒ tan A = \(\frac{4}{3}\)
From the identify sec² A – tan² A = 1
⇒ 1 + tan² A = sec² A

If cos A = \(\frac{3}{5}\) then from
sin² A + cos² A = 1
We can write sin²A = 1 – cos²A
= 1 – \(\left(\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}\)
⇒ sin² A = \(\frac{16}{25}\)
⇒ sin A = \(\frac{4}{5}\)
∴ sin A = \(\frac{4}{5}\); cos A = \(\frac{3}{5}\)

Question 6.
In △ABC and △XYZ, if ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.
Answer:

In the given triangle, cos A = cos X
⇒ \(\frac{AC}{AX}\) = \(\frac{XC}{AX}\)
⇒ AC = XC
⇒ ∠A = ∠X (∵ Angles opposite to equal sides are also equal)

Question 7.
Given cot θ = \(\frac{7}{8}\), then evaluate

Answer:

cot² θ = (cot θ)²
= \(\left(\frac{7}{8}\right)^{2}\) = \(\frac{49}{64}\) …… (1)

= sec θ + tan θ
So cot θ = \(\frac{7}{8}\)
⇒ tan θ = \(\frac{8}{7}\)
⇒ tan² θ = \(\left(\frac{8}{7}\right)^{2}\) = \(\frac{64}{49}\)
From sec² θ – tan² θ = 1
⇒ 1 + tan² θ = sec² θ

Question 8.
In a right angle triangle ABC, right angle is at B, if tan A = √3, then find the value of
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C
Answer:
Given, tan A = \(\frac{\sqrt{3}}{1}\)
Hence \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{\sqrt{3}}{1}\)
Let opposite side = √3k and adjacent side = 1k

In right angled △ABC,
AC² = AB² + BC²
(By Pythagoras theorem)
⇒ AC² = (1k)² + (√3k)²
⇒ AC² = 1k² + 3k²
⇒ AC² = 4k²
∴ AC = \(\sqrt{4 k^{2}}\) = 2k

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.2

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.2 Textbook Questions and Answers

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.
Answer:

Let the height of the tower = h mts say
Width of the road be = x m.
Distance between two points of observation = 10 cm.
Angles of elevation from the two points = 60° and 30°.
From the figure
tan 60° = \(\frac{h}{x}\)
√3 = \(\frac{h}{x}\)
⇒ h = √3x …….(1)
Also tan 30° = \(\frac{h}{10+x}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{10+x}\)
⇒ h = \(\frac{10+x}{\sqrt{3}}\) ………(2)
From equations (1) and (2) h
h = √3x = \(\frac{10+x}{\sqrt{3}}\)
∴ √3x = \(\frac{10+x}{\sqrt{3}}\)
√3 × √3x = 10 + x
⇒ 3x – x = 10
⇒ 2x = 10
⇒ x = \(\frac{10}{2}\) = 5m
∴ Width of the road = 5 m
Now Height of the tower = √3x = 5√3 m.

Question 2.
A 1.5 m tall boy is looking at the top of a temple which is 30 metre in height from a point at 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.
Answer:

Height of the temple = 30 m
Height of the man = 1.5 m
Initial distance between the man and temple = d m. say
Let the distance walked = x m.
From the figure
tan 30° = \(\frac{30-1.5}{d}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{d}\)
∴ d = 28.5 × √3m ………(1)
Also tan 60° = \(\frac{28.5}{d-x}\)
⇒ √3 = \(\frac{28.5}{d-x}\)
⇒ √3(d-x) = 28.5
⇒ √3(28.5 × √3-x) = 28.5
⇒ 28.5 × 3 – √3x = 28.5
⇒ √3x = 3 × 28.5-28.5
⇒ √3x = 2 × 28.5 = 57
∴ x = \(\frac{57}{\sqrt{3}}=\frac{19 \times 3}{\sqrt{3}}\) = 19√3
= 19 × 1.732
= 32.908 m.
∴ Distance walked = 32.908 m.

Question 3.
A statue stands on the top of a 2 m 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.
Answer:

Height of the pedestal = 2 m.
Let the height of the statue = h m. Angle of elevation of top of the statue = 60°.
Angle of elevation of top of the pedestal = 45°.
Let the distance between the point of observation and foot of the pedestal = x m.
From the figure
tan 45° = \(\frac{2}{x}\)
1 = \(\frac{2}{x}\)
∴ x = 2 m.
Also tan 60° = \(\frac{2+h}{x}\)
⇒ √3 = \(\frac{2+h}{x}\)
⇒ 2√3 = 2 + h
⇒ h = 2√3 – 2
= 2(√3-1)
= 2(1.732 – 1)
= 2 × 0.732
= 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 7 m, then find the height of the tower.
Answer:

Angle of elevation of the top of the tower = 60°.
Angle of depression to the foot of the tower = 45°.
Distance between tower and building = 7 m.
Let the height of the building = x m and tower = y m.
From the figure
tan 45° = \(\frac{x}{7}\)
1 = \(\frac{x}{7}\)
∴ x = 7 m.
Also tan 60° = \(\frac{y-x}{7}\)
⇒ √3 = \(\frac{y-x}{7}\)
⇒ 7√3 = y – 7
∴ y = 7 + 7√3
= 7 (√3 + 1)
= 7(1.732 + 1)
= 2.732 × 7
= 19.124 m.

Question 5.
A wire of length 18 m had been tied with electric pole at an angle of eleva¬tion 30° with the ground. As it is 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 ?
Answer:

Length of the wire = 18 m
Let the length of the wire removed = x
Height of the pole be = h
From the figure
sin 30° = \(\frac{h}{18}\)
⇒ \(\frac{1}{2}\) = \(\frac{h}{18}\)
⇒ h = \(\frac{18}{2}\) = 9 m
Also sin 60° = \(\frac{h}{18-x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{9}{18-x}\)
√3(18-x) = 9 × 2
18√3 – √3x = 18
√3x = 18√3 – 18
√3x = 18(√3-1)
x = \(\frac{18(\sqrt{3}-1)}{\sqrt{3}}\)
= \(\frac{6 \times 3(\sqrt{3}-1)}{\sqrt{3}}\)
= 6√3(√3-1)
= 6(3-√3)
= 18 – 6√3
= 18 – 10.392
= 7.608 m.

Question 6.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.
Answer:

Height of the tower = 30 m
Angle of elevation of the top of the tower = 60°.
Angle of elevation of the top of the building = 30°.
Let the distance between the foot of the tower and foot of the building be d m and height of the building be x m.
From the figure
tan 60° = \(\frac{30}{d}\)
√3 = \(\frac{30}{d}\)
⇒ d = \(\frac{30}{\sqrt{3}}=\frac{10 \times 3}{\sqrt{3}}\) = 10√3m
Also tan 30° = \(\frac{x}{d}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{x}{10 \sqrt{3}}\)
⇒ x = \(\frac{10 \sqrt{3}}{\sqrt{3}}\) = 10 m
∴ Height of the building = 10 m

Question 7.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Answer:

Width of the road = 120 f.
Angle of elevation of the top of the 1st tower = 60°.
Angle of elevation of the top of the 2 tower = 30°.
Let the distance of the point from the 1st pole = x.
Then the distance of the point from
the 2nd pole = 120 – x.
and height of each pole = h say.
From the figure
tan 60° = \(\frac{h}{x}\)
⇒ √3 = \(\frac{h}{x}\)
⇒ h = √3x ……..(1)
Also tan 30° = \(\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ h = \(\frac{120-x}{\sqrt{3}}\)
From (1) and (2)
√3x = \(\frac{120-x}{\sqrt{3}}\)
⇒ √3.√3x = 120-x
⇒ 3x = 120 – x
⇒ 3x + x = 120
⇒ 4x = 120
⇒ x = \(\frac{120}{4}\) = 30 ft
Now h = √3x = √3 × 30 = 1.732 x 30 = 51.960 feet
∴ Distances of the poles = 30 ft. and 120 – 30 fts = 90 ft.
Height of each pole = 51.96 ft.

Question 8.
The angles of elevation of the top of a tower from two points at a distance of 4 m 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.
Answer:

Let the height of the tower = h m.
Angles of elevation of the top of the tower from two points = x° and (90° – x)
From the figure
tan x = \(\frac{h}{4}\) ……. (1)
Also tan (90° – x) = \(\frac{h}{9}\)
⇒ cot x = \(\frac{h}{9}\)
⇒ \(\frac{1}{\tan x}\) = \(\frac{h}{9}\)
∴ tan x = \(\frac{9}{h}\) …….. (2)
From (1) and (2)
tan x = \(\frac{h}{4}\) = \(\frac{9}{h}\)
∴ \(\frac{h}{4}\) = \(\frac{9}{h}\)
h × h = 9 × 4
⇒ h² = 36
⇒ h = 6 m

Question 9.
The angle of elevation of a jet plane from a point A on the ground is 60°. After 4 flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane. (√3 = 1.732)
Answer:

Height of the plane from the ground PM = RN = 1500√3 m.
Angle of elevation are 30° and 60°.
From the figure
tan 60° = \(\frac{PM}{QM}\)
√3 = \(\frac{1500 \sqrt{3}}{\mathrm{QM}}\)
QM = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\) = 1500 m
Also tan 30° = \(\frac{RN}{QN}\)
\(\frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{\mathrm{QM}+\mathrm{MN}}\)
QM + MN = 1500√3 × √3
1500 + MN = 1500 × 3
MN = 4500 – 1500
MN = 3000 mts.
∴ Distance travelled in 15 seconds = 3000 mts.
∴ Speed of the jet plane = \(\frac{\text { distance }}{\text { time }}=\frac{3000}{15}\) = 200 m/s
= 200 × \(\frac{18}{5}\) kmph
= 720 kmph
Speed = 200 m/sec. or 720 kmph.

AP SSC 10th Class Maths Textbook Solutions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.1

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.1 Textbook Questions and Answers

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?
Answer:
Let the height of the tower = h m
Distance of the point of observation from the foot of the tower =15 cm.
Angle of elevation of the top of the tower = 45°
From the figure tan θ = \(\frac{\text { opp. side }}{\text { adj. side }}\)
tan 45° = \(\frac{h}{15}\)
⇒ 1 = \(\frac{h}{15}\)
∴ h = 1 × 15 = 15 m

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 down.
Answer:
Distance between the foot of tree and the point of contact of the top of the tree on the ground = 6 cm.
Let the length of the remaining part be = h m.
Let the length of the broken part be = x m.
Angle made by the broken part with the ground = 30°.
From the figure
tan 30° = \(\frac{h}{6}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{6}\)
∴ h = \(\frac{6}{\sqrt{3}}=\frac{3 \times 2}{\sqrt{3}}\) = 2√3 m
Also cos 30° = \(\frac{6}{x}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6 \times 2}{\sqrt{3}}\) = \(\frac{3 \times 2 \times 2}{\sqrt{3}}\) = 4√3
∴ Height of the tree = broken part + remaining part
= x + h
= 2√3 + 4√3 = 6√3 m
= 6 × 1.732
≃ 10.392 m.

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 2 m and by making an angle of 30° with the ground. What should be the length of the slide?
Answer:
Height of slide = 2 m
Let the length of the slide = x m.
Angle made by the slide with the ground = 30°
From the figure
sin 30° = \(\frac{2}{x}\)
⇒ \(\frac{1}{2}\) = \(\frac{2}{x}\)
⇒ x = 2 × 2 = 4 m
Length of the slide = 4 m.

Question 4.
Length of the shadow of a 15 meter high pole is 5√3 meters at 7 o’clock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time?
Answer:
Height of the pole = 15 m
Length of the shadow = 5√3 m
Let the angle of elevation be ‘θ’.
Then from the figure
tan θ = \(\frac{15}{5 \sqrt{3}}=\frac{5 \times \sqrt{3} \times \sqrt{3}}{5 \times \sqrt{3}}\) = √3
tan θ = √3 = tan 60°
∴ θ = 60°
∴ Angle of elevation of Sun rays with the ground = 60°.

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?
Answer:
Height of the pole = 10 m
Let the length of each rope = x
Angle made by the rope with the pole = 30°

From the figure
cos 30° = \(\frac{10}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{10}{x}\)
⇒ x = \(\frac{10 \times 2}{\sqrt{3}}=\frac{20}{\sqrt{3}}\)
∴ Length of each rope = \(\frac{20}{\sqrt{3}}\)m
= 11.546 m.

∴ Total length of the rope = 3 × \(\frac{20}{\sqrt{3}}\)
= 20√3
= 20 × 1.732
≃ 34.64 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 between you and the object?
Answer:
Height of the building = 6 m
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = \(\frac{6}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6 \times 2}{\sqrt{3}}=\frac{2 \times \sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\) = 4√3
∴ Distance = 4√3 m or
4 × 1.732 = 6.928 m.

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?
Answer:
Height of the pole = 9m
Height of the point from the ground where he reaches the pole = 9 – 1.8 = 7.2 m
Angle of elevation = 60°
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = \(\frac{7.2}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{7.2}{x}\)
⇒ x = \(\frac{7.2 \times 2}{\sqrt{3}}=\frac{3 \times 2.4 \times 2}{\sqrt{3}}=\frac{\sqrt{3} \times \sqrt{3} \times 4.8}{\sqrt{3}}\)
⇒ x = 1.732 × 4.8
≃ 8.31 m
Also tan 60° = \(\frac{7.2}{d}\)
√3 = \(\frac{7.2}{d}\)
⇒ d = \(\frac{7.2}{\sqrt{3}}=\frac{2.4 \times 3}{\sqrt{3}}\) = 2.4 × √3 = 2.4 × 1.732
∴ d ≃ 4.1568 m

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?
Answer:
Let the width of the river = AB = x m
Angle made by the boat = 60°
Distance travelled = AC = 600 m
From the figure
cos 60° = \(\frac{x}{600}\)
\(\frac{1}{2}\) = \(\frac{x}{600}\)
⇒ x = \(\frac{600}{2}\) = 300 m.
In the figure
A = Boat’s place
C = Reach place of another side (or) Point of observation.
AC = Travelling distance of the boat ∠AC = 60°
AB = width of the river AB
In △ABC, sin 60° = \(\frac{AB}{AC}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{AB}{600}\)
⇒ AB = 600 × \(\frac{\sqrt{3}}{2}\) = 300√3

Question 9.
An observer of height 1.8 m is 13.2 m 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 the palm tree?
Answer:
Height of the observer = 1.8 m
Distance of the observer from the palm tree = 13.2 m
From the figure
tan 45° = \(\frac{x}{13.2}\)
⇒ 1 = \(\frac{x}{13.2}\)
⇒ x = 13.2 m
∴ Height of the palm tree = 13.2 + 1.8 = 15 m.

Question 10.
In the given figure, AC = 6 cm, AB = 5 cm and ∠BAC = 30°. Find the area of the triangle.
Answer:
Draw a perpendicular BD to AC
∴ BD ⊥ AC
Now let AD = 6 – x and DC = x
Given AB = 5 cm and ∠BAD = 30° then in △ABD
sin 30° = \(\frac{BD}{AB}\) = \(\frac{BD}{5}\) = \(\frac{1}{2}\)
⇒ BD = \(\frac{5}{2}\) = 2.5 cm
and cos 30° = \(\frac{AD}{AB}\) = \(\frac{6-x}{5}\) = \(\frac{\sqrt{3}}{2}\)
⇒ 6 – x = \(\frac{5 \sqrt{3}}{2}\)
⇒ x = 6 – \(\frac{5 \sqrt{3}}{2}\) = 6 – \(\frac{5(1.732)}{2}\)
∴ x = 1.67
∴ Area of △ABC = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) × AC × BD
= \(\frac{1}{2}\) × 6 × 2.5
= 7.5 cm²

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.2

10th Class Maths 13th Lesson Probability Ex 13.2 Textbook Questions and Answers

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 drawn is (i) red ? (ii) not red?
Answer:
i) Total number of balls in the bag = 3 red + 5 black = 8 balls.
Number of total outcomes when a ball is drawn at random = 3 + 5 = 8
Now, number of favourable outcomes of red ball = 3.
∴ Probability of getting a red ball = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{3}{8}\)
ii) If P( E) is the probability of drawing no red ball, 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?
Answer:
Total number of marbles in the box = 5 red + 8 white + 4 green = 5 + 8 + 4= 17
Number of total outcomes in drawing a marble at random from the box =17.
i) Number of red marbles = 5
Number of favourable outcomes in drawing a red ball = 5
∴ Probability of getting a red ball P(R) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
P(R) = \(\frac{5}{17}\)

ii) Number of white marbles = 8
Number of favourable outcomes in drawing a white marble = 8
∴ Probability of getting a white marble
P(W) = \(\frac{8}{17}\)

iii) Number of ‘non-green’ marbles = 5 red + 8 white = 5 + 8 = 13
Number of outcomes favourable to drawing a non-green marble =13.
∴ Probability of getting a non- green marble
P(non – green) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
P(non – green) = \(\frac{13}{17}\)
Probability of getting a green ball = \(\frac{\text { No. of green balls }}{\text { Total no. of balls }}\) = \(\frac{4}{17}\)
Now P(G) = \(\frac{4}{17}\)
and P(G) + P(\(\overline{\mathrm{ G}}\)) = 1
∴ P(\(\overline{\mathrm{G}}\)) = 1 – P(G)
= 1 – \(\frac{4}{17}\)
= \(\frac{13}{17}\)

Question 3.
A Kiddy bank contains hundred 50p coins, fifty Rs. 1 coins, twenty Rs. 2 coins and ten Rs. 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 50 p coin? (ii) will not be a Rs. 5 coin?
Answer:
i) Number of 50 p coins = 100
Number of Rs. 1 coins = 50
Number of Rs. 2 coins = 20
Number of Rs. 5 coins = 10
Total number of coins = 180
Number of total outcomes for a coin to fall down = 180.
Number of outcomes favourable to 50 p coins to fall down = 100.
∴ Probability of a 50 p coin to fall down = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{100}{180}\) = \(\frac{5}{9}\)

ii) Let P(E) be the probability for a Rs. 5 coin to fall down.
Number of outcomes favourable to Rs. 5 coin = 10.
∴ Probability for a Rs. 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 Rs. 5 coin.
Again P(E) + P(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\))= l-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?
Answer:
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 m + 8 f
= 13 fishes.
∴ Number of total outcomes in taking a fish at random from the aquarium =13.
Number of male fish = 5
∴ Number of outcomes favourable to male fish = 5.
∴ The probability of taking a male fish = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{5}{13}\)
= 0.38

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
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Answer:
Number of total outcomes are (1,2,……….., 8) = 8

i) Number of outcomes favourable to 8 = 1.
∴ P(8) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{8}\)

ii) Number of ‘odd numbers’ on the spinning wheel = (1, 3, 5, 7) = 4
∴ Number of outcomes favourable to an odd number.
∴ Probability of getting an odd number = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{4}{8}\) = \(\frac{1}{2}\)

iii) Number greater than 2 are (3, 4, 5, 6, 7, 8)
Number of outcomes favourable to ‘greater than 2’ are = 6.
Probability of pointing a number greater than 2
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{6}{8}\) = \(\frac{3}{4}\)

iv) Number less than 9 are: (1,2, 3, 4, 5, 6, 7, 8 …… 8)
∴ Number of outcomes favourable to pointing a number less than 9 = 8.
∴ Probability of a number less than 9
P(E) = \(\frac{\text { No. of outcomes favourable to less than } 9}{\text { No.of total outcomes }}\)
= \(\frac{8}{8}\) = 1
Note : This is a sure event and hence probability is 1.

Question 6.
One card is drawn 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.
Answer:
Total number of cards = 52.
∴ Number of all possible outcomes in drawing a card at random = 52.
i) Number of outcomes favourable to the king of red colour = 2(♥ K, ♦ K)
∴ Probability of getting the king of red colour
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

ii) Number of face cards in a deck of cards = 4 × 3 = 12 (K, Q, J)
Number of outcomes favourable to select a 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.
∴ Number of outcomes favourable to select a red face card = 6.
∴ Probability of getting a red face
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{6}{52}\) = \(\frac{3}{26}\)

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

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

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

Question 7.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
i) What is the probability that the card is the queen?
ii) If the queen is drawn and put aside, what is the probability that the second card picked is (a) an ace? (b) a queen?
Answer:
Total number of cards = 5.
∴ Number of total outcomes in picking up a card at random = 5.
i) Number of outcomes favourable to queen = 1.
∴ Probability of getting the queen
= \(\frac{\text { No.of outcomes favourable to the ‘Q’ }}{\text { No.of total outcomes }}\)
= \(\frac{1}{5}\)

ii) When queen is drawn and put aside, remaining cards are four.
∴ Number of total outcomes in drawing a card at random = 4.
a) Number of favourable outcomes to ace 1
Probability of getting an ace
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{4}\)

b) Number of favourable outcomes to ‘Q’ = 0 (as it was already drawn and put aside)
∴ Probability that the card is Q = \(\frac{0}{4}\) = 0
After putting queen aside, selecting the queen from the rest is an impossible event and hence the probability is zero.

Question 8.
12 defective pens are accidentally mixed 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.
Answer:
Number of good pens = 132
Number of defective pens = 12
Total number of pens = 132 + 12 = 144
∴ Total number of outcomes in taking a pen at random = 144.
No. of favourable outcomes in taking a good pen = 132.
∴ Probability of taking a good pen
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{132}{144}\) = \(\frac{11}{12}\)

Question 9.
A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in previous case is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer:
Given :
4 out of 20 bulbs are defective
(i.e.) Number of defective bulbs = 4
Number of non-defective bulbs = 20 – 4 = 16
If a bulb is drawn at random, the total outcomes are = 20
Number of outcomes favourable to ‘defective bulb’ = 4
∴ Probability of getting a defective bulb
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{4}{20}\) = \(\frac{1}{5}\)
Suppose a non-defective bulb is drawn and not replaced, then the bulbs remaining are = 19
∴ Total outcomes in drawing a bulb from the rest = 19
Number of favourable outcomes in drawing non-defective bulb from the rest = 16 – 1 = 15
∴ Probability of getting a non-defective bulb in the second draw
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{15}{19}\)

Question 10.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn 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.
Answer:
Total number of discs in the box = 90
∴ Number of total outcomes in drawing a disc at random from the box = 90.

i) Number of 2-digit numbers in the box (10, 11,….., 90) = 81
i.e., Number of favourable outcomes in drawing a 2 – digit numbers = 81
∴ Probability of selecting a disc bearing a 2 – digit number
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{81}{90}\) = \(\frac{9}{10}\) = 0.9

ii) Number of perfect squares in the box (1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64 and 9² = 81) = 9
i.e., Number of favourable out-comes in drawning a disc bearing a perfect square = 9
∴ Probability of drawning a disc with a perfect square
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{9}{90}\) = \(\frac{1}{10}\)

iii) Number of multiples of 5 from 1 to 90 are (5, 10, 15, ……….., 90) = 18
i.e., Number of favourable outcomes in drawing a disc with a multiple of 5 = 18
∴ Probability of drawing a disc bearing a number divisible by 5
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{18}{90}\) = \(\frac{1}{5}\)

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 1 m?

Answer:
3 m.
Length of the given rectangle = 3 m.
and its breadth = 2 m.
Area of the rectangle
= length × breadth = 3 × 2 = 6 m²
∴ Total area of the region for landing = 6 m².
Diameter of the given circle = 1 m.
Area of the circle = \(\frac{\pi \mathrm{d}^{2}}{4}\)
= \(\frac{22}{7} \times \frac{1 \times 1}{4}\left[\text { or } \pi r^{2}=\frac{22}{7} \times \frac{1}{2} \times \frac{1}{2}\right]\)
= \(\frac{22}{28}\)
∴ Probability of the coin to land on the circle
= \(\frac{\frac{22}{28}}{6}\)
= \(\frac{22}{28×6}\)
= \(\frac{11}{28×3}\)
= \(\frac{11}{84}\)

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?
Answer:
Given : 20 out of 144 are defective i.e., no. of defective ball pens = 20
no. of good ball pens = 144 – 20 = 124
∴ Total outcomes in drawing a ball pen at random = 144.

i) Sudha buys it if it is not defective / a good one.
No. of outcomes favourable to a good pen = 124.
∴ Probability of buying it
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{124}{144}\) = \(\frac{31}{36}\)

ii) Sudha will not buy it-if it is a defective pen
No. of outcomes favourable to a defective pen = 20
∴ Probability of not buying it
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{20}{144}\) = \(\frac{5}{36}\)

!! (not buying) = 1 – P (buying)
= 1 – \(\frac{31}{36}\) = \(\frac{5}{36}\)

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

(ii) 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.
Answer:
When two dice are rolled, total number of outcomes = 36 (see the given table).
(i)
(ii) The above (given) argument is wrong [from the above table].
The sum 2, 3, 4, ………… and 12 have different no. of favourable outcomes, moreover total number of outcomes are 36.

Question 14.
A game consists of tossing a one rupee coin 3 times and noting 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.
Answer:
When a coin is tossed for n – times, the total number of outcomes = 2ⁿ.
∴ If a coin is tossed for 3 – times, then the total number of outcomes = 2³ = 8
Note the following :
TTT
TTH
THT
HTT
HHT
HTH
THH
HHH
Of the above, no. of outcomes with different results = 6.
Probability of losing the game
= \(\frac{\text { No. of favourable outcomes to lose }}{\text { No. of total outcomes }}\)
= \(\frac{6}{8}\) = \(\frac{3}{4}\)

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 atleast once? [Hint : Throwing a dice twice and throwing two dice simultaneously are treated as the same experiment].
Answer:
If a dice is thrown n-times or n-dice are thrown simultaneously then the total
number of outcomes = 6×6×6….×6
(n – times) = 6ⁿ.
No. of total outcomes in throwing a dice for two times = 6² = 36.
i) Let E be the event that 5 will not come up either time, then the favourable outcomes are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3/2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6) = 25.
∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{25}{36}\)

ii) Let E be the event that 5 will come up atleast once.
Then the favourable outcomes are (1,5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) = 11 No. of favourable outcomes = 11
∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{11}{36}\)

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.1

10th Class Maths 13th Lesson Probability Ex 13.1 Textbook Questions and Answers

Question 1.
Complete the following statements:
i) Probability of an event E + Probability of the event ‘not E’ = 1 .
ii) The probability of an event that cannot happen is zero.
Such an event is called an impossible event.
iii) The probability of an event that is certain to happen is  1  such an event is called sure or certain event.
iv) The sum of the probabilities of all the elementary events of an experiment is 1 .
v) The probability of an event is greater than or equal to zero and less than or equal to 1 .

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
i) A driver attempts to start a car. The car starts or does not start.
Answer:
Equally likely. Since both have the same probability \(\frac{1}{2}\).

ii) A player attempts to shoot a basket-ball. She/he shoots or misses the shot.
Answer:
Equally likely. Since both have the same probability \(\frac{1}{2}\).

iii) A trial is made to answer a true-false question. The answer is right or wrong.
Equally likely. Since both have the same probability \(\frac{1}{2}\).

iv) A baby is born. It is a boy or a girl.
Equally likely. Since both the events have the same probability \(\frac{1}{2}\).

Question 3.
If P(E) = 0.05, what is the probability of not E?
Answer:
Given: P(E) = 0.05
Hence, P(E) + P(\(\overline{\mathrm{E}}\)) = 1, where P(\(\overline{\mathrm{E}}\)) is the probability of ‘not E’
0.05 + P(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\)) = 1 -0.05 = 0.95.

Question 4.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
i) an orange flavoured candy?
ii) a lemon flavoured candy?
Answer:
Bag contains only lemon flavoured candies.
i) Taking an orange flavoured candy is an impossible event and hence the probability is zero.
ii) Also taking a lemon flavoured candy is a sure event and hence its probability is 1.

Question 5.
Rahim removes all the hearts from the cards. What is the probability of
i. Picking out an ace from the remaining pack.
ii. Picking out a diamond.
iii. Picking out a card that is not a heart.
iv. Picking out the Ace of hearts.
Answer:
Total number of cards in the deck = 52.
Total number of hearts in the deck of cards =13.
When Hearts are removed, remaining cards = 52 – 13 = 39.
i)Picking out an Ace:
Number of outcomes favourable to Ace = 3 [∵ ♦ A, ♥ A, ♠ A, ♣ A]
Total number of possible outcomes from the remaining cards = 39
– after removing Hearts.
Probability = P(A)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{39}\) = \(\frac{1}{13}\)

ii) Picking out a diamond:
Number of favourable outcomes to diamonds (♦) = 13
Total number of possible outcomes = 39
∴ p(♦) = \(\frac{13}{39}\) = \(\frac{1}{3}\)

iii) Picking out a card that is ‘not a heart’:
As all hearts are removed, the remain-ing cards are all non-heart cards. So the picked card will be definitely a non-heart card. So this is a sure event.
Hence its probability is one
P(E) = \(\frac{39}{39}\) = 1

iv) Picking out the Ace of Hearts:
a) As all the heart cards are removed the left over cards will have three suits (i) spades, (ii), clubs, (iii) dia¬monds of each 13.
Hence total outcomes = 3 × 13 = 39 But among them there is no Ace of heart. So number of favourable outcomes for picking Ace of heart = zero.
∴ Probability P(E) = \(\frac{0}{39}\) = 0
So it is an impossible event.

b) If picking from the rest of the cards, it is an impossible event and hence probability is zero.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992.

Question 6.
What is the probability that the 2 students have the same birthday?
Answer:
Let P(E) = The probability that two students not having the same birthday = 0.992
Then P(\(\overline{\mathrm{E}}\)) = The complementary event of E, i.e., two students having the same birthday Also, P(E) + p(\(\overline{\mathrm{E}}\)) = 1
∴ The probability that two students have the same birthday P(\(\overline{\mathrm{E}}\)) = 1 – P(E)
= 1 – 0.992 = 0.008

Question 7.
A die is thrown once. Find the probability of getting
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Answer:
i) When a die is thrown for one time, total number of outcomes = 6
No. of outcomes favourable to a prime number (2, 3, 5) = 3
∴ Probability of getting a prime = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

ii) No. of outcomes favourable to a number lying between 2 and 6 (3, 4, 5) = 3
∴ Probability of getting a number between 2 and 6
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

iii) Number of outcomes favourable to an odd number (1, 3, 5) = 3
∴ Probability of getting an odd number P(odd)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 8.
What is the probability of drawing out a red king from a deck of cards?
Answer: Number of favourable outcomes to red king (♥ K, ♦ K) = 2.
Number of total outcomes = 52
(∵ Number of cards in a deck of cards = 52)
∴ Probability of getting a red king P (Red king)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 9.
Make 5 more problems getting probability using dice, cards or birthdays and discuss with friends and teacher about their solutions.
Answer:
Class-room activity.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.4

10th Class Maths 14th Lesson Statistics Ex 14.4 Textbook Questions and Answers

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

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Answer:
Since the curve is a less than type graph the data changes to

X – axis – upper limits 1 cm = 50 units.
Y – axis – less than c.f. 1 cm = 5 units.

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

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Answer:
Given: Upper limits of the classes and less than cumulative frequencies. Therefore required points are (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35)
X – axis – upper limits 1 cm = 2 units.
Y – axis – less than c.f. 1 cm = 4 units.

Number of observations = 35
∴ \(\frac{N}{2}\) = \(\frac{35}{2}\) = 17.5
Locate the point on the ogive whose ordinate is 17.5.
The x – coordinate of this point is the required median.
From the graph, median = 46.5.

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

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

Change the distribution to a more than type distribution, and draw its ogive.
Answer:
The given data is to be changed to more than frequency distribution type.

A graph is plotted by taking the lower limits on the X – axis and respective of Y – axis.
Scale:
X – axis: 1 cm = 5 units
Y – axis: 1 cm = 5 units

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.3

10th Class Maths 14th Lesson Statistics Ex 14.3 Textbook Questions and Answers

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

Answer:

Sum of the frequencies = 68
∴ \(\frac{n}{2}\) = \(\frac{68}{2}\) = 34
Hence median class = 125 – 145
Lower boundary of the median class, l = 125
cf – cumulative frequency of the class preceding the median class = 22
f – frequency of the median class = 20
h = class size = 20
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 125 + \(\frac{[34-22]}{20}\) × 20
∴ Median = 125 + 12 = 137
Maximum number of consumers lie in the class 125 – 145
Modal class is 125 -145
l – lower limit of the modal class =125
f₁ – frequency of the modal class = 20
f₀ – frequency of the class preceding the modal class =13
f₂ – frequency of the class succeeding the modal class =14
h – size of the class = 20
Mode (Z) = \(l+\frac{f_{1}-f_{0}}{\left(f_{1}-f_{0}\right)+\left(f_{1}-f_{2}\right)} \times h\)
Mode (Z) = 125 + \(\frac{20-13}{(20-13)+(20-14)} \times 20\)
= 125 + \(\frac{7}{7+6}\) × 20
= 125 + \(\frac{140}{13}\)
= 125 + 10.76923
∴ Mode = 135.769
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
a = assumed mean = 135
∴ \(\overline{\mathbf{x}}\) = 135 + \(\frac{7}{68}\)
= 135 + 0.102941
≃ 135.1
Mean, Median and Mode are approximately same in this case.

Question 2.
If the median of 60 observations, given below is 28.5, find the values of x and y.

Answer:

Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
It is given that ∑f = n = 60
So, 45 + x + y = 60
x + y = 60 – 45 = 15
x + y = 15 ….. (1)
The median is 28.5 which lies be-tween 20 and 30.
Median class = 20 – 30
Lower boundary of the median class ‘l’ = 20
\(\frac{N}{2}\) = \(\frac{60}{2}\) = 30
cf – cumulative frequency = 5 + x
h = 10
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
⇒ 28.5 = 20 + \(\frac{30-5-x}{20}\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
\(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
25 – x = 2 × 8.5
x = 25- 17 = 8
also from (1); x + y = 15
8 + y = 15
y = 7
∴ x = 8; y = 7.

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

Answer:

The given distribution being of the less than type, 25, 30, 35, give the upper limits of corresponding class intervals. So the classes should be 20 – 25, 25 – 30, 30 – 35, ………. 55 – 60.
Observe that from the given distribution 2 persons with age less than 20.
i.e., frequency of the class below 20 is 2.
Now there are 6 persons with age less than 25 and 2 persons with age less than 20.
∴ The number of persons with age in the interval 20 – 25 is 6 – 2 = 4.
Similarly, the frequencies can be calculated as shown in table.
Number of observations = 100
n = 100
\(\frac{n}{2}\) = \(\frac{100}{2}\) = 50, which lies in the class 35-40
∴ 35 – 40 is the median class and lower boundary l = 35
cf = 45;
h = 5;
f = 33
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 35 + \(\frac{50-45}{33}\) × 5
= 35 + \(\frac{5}{33}\) × 5
= 35 + 0.7575
= 35.7575
∴ Median ≃ 35.76

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

Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5,…, 171.5 – 180.5.)
Answer:
Since the formula, Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\) assumes continuous classes assumes continuous class, the data needs to be converted to continuous classes.
The classes then changes to 117.5 – 126.5; 126.5 – 133.5, …… 171.5 – 180.5.

∑f_i = n = 40
\(\frac{n}{2}\) = \(\frac{40}{2}\) = 20
\(\frac{n}{2}\)th observation lie in the class 144.5- 153.5
∴ Median class = 144.5 – 153.5
Lower boundary, l = 144.5
Frequency of the median class, f = 12
c.f. = 17
h = 9
∴ Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 144.5 + \(\frac{20-17}{12}\) × 9
= 144.5 + \(\frac{3}{12}\) × 9
= 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25
∴ Median length = 146.75 mm.

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

Find the median life-time of a lamp.
Answer:

Total observations are n = 400
\(\frac{n}{2}\)th observation i.e \(\frac{400}{2}\) = 200
200 lies in the class 3000 – 3500
∴ Median class = 3000 – 3500
Lower boundary l = 3000
frequency of the median class f = 86
c.f = 130
Class size, h = 500
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 3000 + \(\frac{200-130}{86}\) × 500
= 3000 + \(\frac{70}{86}\) × 500
= 3000 + 406.977
= 3406.98
∴ Median life ≃ 3406.98 hours

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

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames ? Also, find the modal size of the surnames.
Answer:
Number of letters in the surnames.
Also find the modal size of the surnames.

Total observations are n = 100
\(\frac{n}{2}\) = \(\frac{100}{2}\) = 50
50 lies in the class 7 – 10
∴ Median class = 7 – 10
l – lower boundary = 7
f – frequency of the median class = 40
cf = 36
Class size h = 3
Median:
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 7 + \(\frac{50-36}{40}\) × 3
= 7 + \(\frac{14}{40}\) × 3
= 7 + \(\frac{42}{40}\)
= 7 + 1.05
= 8.05
∴ Median = 8.05.

Mean:
Assumed mean, a = 8.5
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 8.5 + \(\frac{(-18)}{100}\)
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32.

Mode:
Maximum number of surnames = 40
∴ Modal class = 7-10
l – lower boundary of the modal class = 7
Mode (Z) = \(l+\frac{f_{1}-f_{0}}{\left(f_{1}-f_{0}\right)+\left(f_{1}-f_{2}\right)} \times h\)
l = 7; f₁ = 40, f₀ = 30, f₂ = 16, h = 3
Mode (Z) = 7 + \(\frac{40-30}{(40-30)+(40-16)}\) × 3
= 7 + \(\frac{10}{10+24}\) × 3
= 7 + \(\frac{30}{34}\)
= 7 + 0.882
= 7.882

Median = 8.0.5; Mean = 8.32; Modal size = 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

Number of observations (n) = ∑f_i
\(\frac{n}{2}\) = \(\frac{30}{2}\) = 15
15 lies in the class 50 – 55
∴ Median class = 50-55
l – lower boundary of the median class = 55
f – frequency of the median class = 8
c.f = 5
Class size h = 6
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 50 + \(\frac{15-5}{8}\) × 6
= 50 + 7.5
= 57.5
= 50 + 7.5 = 57.5
∴ Median weight = 57.5 kg.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.2

10th Class Maths 14th Lesson Statistics Ex 14.2 Textbook Questions and Answers

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

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

∴ Mode = \(l+\frac{\left(f_{1}-f_{0}\right)}{2 f_{1}-f_{0}-f_{2}} \times h\)
\(\begin{array}{l}
=35+\left(\frac{23-21}{2 \times 23-21-14}\right) \times 10 \\
=35+\left(\frac{2}{46-35}\right) \times 10
\end{array}\)
= 35 + \(\frac{2}{11}\) × 10
= 35 + 1.81818……
= 36.8 years.
Mean x = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2830}{80}\)
= 35.37 years.
Interpretation: Mode age is 36.8 years, Mean age = 35.37 years.
Maximum number of patients admitted in the hospital are of the age 36.8 years, while on an average the age of patients admitted to the hospital is 35.37 years. Mode is less than the Mean.

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

Determine the modal lifetimes of the components.
Answer:

Since the maximum frequency 61 is in the class 60 – 80, this is the required modal class.
Modal class frequency, f₁ = 61.
Frequency of the class preceding the modal class f₀ = 52
Frequency of the class succeeding the modal class f₂ = 38
Lower boundary of the modal class, l = 60
Height of the class, h = 20
∴ Mode (Z) = \(l+\frac{\left(f_{1}-f_{0}\right)}{2 f_{1}-f_{0}-f_{2}} \times h\)
\(=60+\left[\frac{61-52}{2 \times 61-(52+38)}\right] \times 20\)
= 60 + \(\frac{9}{122-90}\) × 20
= 60 + \(\frac{9}{32}\) × 20
= 60 + 5.625
= 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Answer:

Since the maximum families 40 lies in the class 1500 – 2000, this is the required modal class.
Lower boundary of the modal class (l) = 1500
Frequency of the modal class (f₁) = 40
Frequency of the class preceding the modal class f₀ = 24
Frequency of the class succeeding the modal class f₂ = 33
Height of the class, h = 500
Hence, modal monthly income = Rs. 1847.83.
Assumed mean (a) = 3250
∑f_i = 200, ∑u_if_i = -235
Mean monthly income = \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
= 3250 – \(\frac{235}{200}\) × 500
= 3250 – 587.5
= Rs. 2662.50

Question 4.
The following distribution gives the state-wise, teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Answer:

Since the maximum number of states ‘10’ lies in the class interval 30-35, this is the modal class.
Lower boundary of the modal class, l = 30
Frequency of the modal class, f₁ = 10
Frequency of the class preceding the modal class = f₀ = 9
Frequency of the class succeeding the modal class = f₂ = 3
Height of the class, h = 5
∴ Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
\(=30+\frac{10-9}{(10-9)+(10-3)} \times 5\)
= 30 + \(\frac{1×5}{1+7}\)
= 30 + \(\frac{5}{8}\)
= 30 + 0.625
= 30.625
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
a = assumed mean = 32.5; h = height of the class = 5
∴ x = 32.5 – \(\frac{23}{35}\) × 5
= 32.5 – 3.28
= 29.22
Mean = 30.625
Mode = 29.22
Mode states have a students – teacher ratio 29.22 and on an average this ratio is 30.625.

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

Find the mode of the data.
Answer:

Maximum number of batsmen are in the class 4000 – 5000.
∴ Modal class is 4000 – 5000.
Lower boundary of the modal class ‘l’ = 4000
Frequency of the modal class, f₁ = 18
Frequency of the class preceding the modal class, f₀ = 4
Frequency of the class succeeding the modal class, f₂ = 9
Height of the class, h = 1000
Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
Mode (Z) = \(4000+\frac{18-4}{(18-4)+(18-9)} \times 1000\)
= 4000 + \(\frac{14}{14+9}\) × 1000
= 4000 + \(\frac{14000}{23}\)
= 4000 + 608.695
= 4608.69
≃ 4608.7 runs

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

Find the mode of the data.
Answer:

Since the maximum frequency is 20, the modal class is 40 – 50.
Lower boundary of the modal class ‘l’ = 40
Frequency of the modal class, f₁ = 20
Frequency of the class preceding the modal class, f₀ = 12
Frequency of the class succeeding the modal class, f₂ = 11
Height of the class, h = 10;
Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
Mode (Z) = \(40+\frac{(20-12)}{(20-12)+(20-11)} \times 10\)
= 40 + \(\frac{8}{8+9}\) × 10
= 40 + \(\frac{80}{17}\)
= 40 + 4.70588
= 44.705
≃ 44.7 cars

 

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.1

10th Class Maths 14th Lesson Statistics Ex 14.1 Textbook Questions and Answers

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

Answer:

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

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

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

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

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

Answer:

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

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

Answer:

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

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

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

Here we use step deviation method where a = 135, h = 5,a multiple of all d_i
\(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)
= 22 + \(\frac{25}{400}\) × 5
= 22 + 0.31
= 22.31

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.
Answer:

Here a = 125, h = 50, ∑f_iu_i = 43
Now
\(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)
= 125 + \(\frac{43}{25}\) × 50
= 125 + (43 × 2)
= 125 + 86
= 211.
NOTE: If we consider first value as “a” then we dont get negative values in u_i, f_iu_i columns. Then it becomes easy for calculation.

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

Find the mean concentration of SO₂ in the air.
Answer:

∴ \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2.96}{30}\)
= 0.00986666…….
≃ 0.099

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

Answer:

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

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

Answer:

a = 70; h = 10
∴ \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{u}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}} \times \mathbf{h}\)
\(\Rightarrow \bar{x}=70-\frac{2}{35} \times 10\)
= 70 – \(\frac{2}{35}\) × 10
= 70 – \(\frac{20}{35}\)
= 70 – 0.57142
= 69.4285
≃ 69.43%

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 10 Natural Resources.

AP State Syllabus SSC 10th Class Biology Important Questions 10th Lesson Our Environment

10th Class Biology 10th Lesson Natural Resources 1 Mark Important Questions and Answers

Question 1.
Recently a new programme was launched in our state known as “Vanam – Manain”. Prepare any two slogans to promote the programme.
Answer:
a) Save forest, forest will save you.
b) Grow the plants and get the fresh air.

Question 2.
Suggest any two practices suitable to farmers with less water resources.
Answer:

1. Construction of percolation tanks (or) Soak pits
2. Irrigation techniques like drip irrigation and usage of sprinklers.

Question 3.
Why should we conserve forests? Give two reasons.
Answer:
a) Forests serve as lungs for the world. They purify the air and protect the earth from greenhouse effect and global warming,
b) Forests are rich habitats for plants

Question 4.
Ravi observed symbol on the plastic water bottle purchased by him. What does this symbol indicate? and animals.
Answer:
The symbol on the plastic bottle indicates that the bottle is made from recycled plastic and after its use it can be recycled.

Question 5.
Write any two suggestions for the conservation of biodiversity at your village.
Answer:

1. Protecting and preserving the natural habitats of birds and animals.
2. Replace the wood products with alternative sources.
3. Using Recycled products and following the 4’R Principle in day to day life.

Question 6.
Suggest any two activities to check soil erosion in your school.
Answer:

1. Observe the school ground after the rain.
2. Conduct a field project on soil erosion.

Question 7.
To create awareness on “Water conservation” in your locality, what slogan you will suggest?
Prepare two slogans on ‘Save Water’ propaganda.
Answer:
“Don’t Waste Water”.
“Save every drop”.
“Water is life”.

Question 8.
The symbol is there on the item you bought. What it instructs? (OR)
What does the given logo indicate? What does it mean?

Answer:
It is the Recycle logo. It indicates that the item we bought is prepared from recycled materials or the item can be recycled after use.

Question 9.
What happens if the forest area decreases rapidly?
Answer:
If the forest area decreases

1. It destroys wildlife habitat.
2. It increases soil erosion.
3. It releases green house gases into the atmosphere contributing to global warming.
4. It also harms people who relay on forest for their survival, hunting and gathering, harvesting forest products or using timber and firewood.

Question 10.
Write two activities which you are performing to save electricity.
(OR)
Write any two measures vou take in your home to reduce the consumption of electricity.
Answer:

1. We can reduce the consumption of electricity by putting off the fans and lights when there is no need.
2. We can use LED (Lighting Emitting Device) bulbs to save electricity.
3. To shut down laptops and computers when they are not in use.

Question 11.
Prepare two slogans on protecting non-renewable resources.
Answer:

1. Use Biofuel – Reduce Fossil Fuel.
2. Use alternative resources – Save the environment.

Question 12.
Write two examples for non-renewable resources.
Answer:
Examples for non-renewable resources are coal, petroleum and natural gas.

Question 13.
What is sustainable development? Is it needful for us?
Answer:
When we use the environment In ways that ensure we have resources for the future, It Is called sustainable development. It Is needed because development and conservation can coexist In harmony.

Question 14.
What are examples for natural resources?
Answer:
Examples for natural resources are water, soil, forests, flora, fauna, etc.

Question 15.
What are percolation tanks?
Answer:
Percolation tanks are normally earthen dams with masonry structures where water may overflow.

Question 16.
What are Kharif crops?
Answer:
Crops grown In the rainy season are termed as Kharif crops, e.g: Paddy, maize, millet and cotton crops.

Question 17.
What are Rabi crops?
Answer:
The crops that are grown only in winter season are generally called Rabi crops, e.g.: Wheat, Gram and Mustard.

Question 18.
What is the average fall of ground water level in Andhra Pradesh state during the period of 1998 – 2002?
Answer:
The average fall of ground water level In Andhra Pradesh state during the period of 1998 – 2002 Is 3 meters.

Question 19.
Which agency in villages of Warangal district helped in recharging wells that were being dried up?
Answer:
Centre for water solidarity (Secundrabad, T.S.) helped In recharging wells that were drying up In the villages.

Question 20.
Give examples for micro irrigation techniques.
Answer:
Drip irrigation, sprinklers are the examples for micro irrigation techniques.

Question 21.
Mow did the boundaries between the villages were fixed in ancient times?
Answer:
In ancient times village boundaries were decided upon a water shed (Land between water sources usually of two rivers or streams) basis fixed at the common point of the drainage system In between two villages by the expert farmers In the village.

Question 22.
Expand ICRISAT.
Answer:
International Crop Research Institute for Semi-Arid Tropics.

Question 23.
What is the other name for Sri Rama Sagar Project?
Answer:
Sri Rama Sagar Project also known as the Pochampadu project on the Godavari river,

Question 24.
What is the use qf planting Gliricidia on field bunds?
Answer:
Planting Gliricidia on field bunds help In strengthen them and make the soli nitrogen-rich.

Question 25.
What is the micro irrigation system that can reduce water consumption by 70%?
Answer:
Drip irrigation can reduce water consumption by 70%.

Question 26.
Who predicted that by 2025, 1.8 billion people will be living in countries or regions with absolute water scarcity ?
Answer:
The Food and Agriculture Organisation (FAO) of the united nations has predicted that by 2025, 1.8 billion people will be living in countries or regions with absolute water scarcity.

Question 27.
What happens if we use resources wisely?
Answer:
If resources are used wisely and efficiently they will last much longer. Through conservation people can reduce waste and manage natural resources wisely.

Question 28.
Give an example of country where restrictions on water usage were imposed.
Answer:
In Australia restrictions were imposed on activities like, watering lawns by using sprinkler systems, washing vehicles, using house pipes to clean paved areas, and refilling swimming pools.

Question 29.
Why are the natural resources used up quickly?
Answer:
The population of human beings has grown enormously in the past two centuries. Billions of people use up resources quickly as they eat food, build houses, produce goods and burn fuel for transportation and electricity.

Question 30.
What happens if we damage a forest resource?
Answer:
Harm to animals that may be forced to find new habitats. If we damage a forest resource indiscriminately the depletion of resources occur and we may have to face problem for water and timber in future.

Question 31.
What are the results of deforestation?
Answer:
Deforestation destroys wild life habitats and increases soil erosion and also releases green house gases into atmosphere, contributing to global warming.

Question 32.
How the people in China and Mexico recycle paper? (OR)
Give an example of recycling paper by the people. What is the use of recycling paper?
Answer:
People in China and Mexico reuse much of their waste paper, including writing paper, wrapping paper and card board.

Question 33.
How the soil is important for us ? How the soil is importane for us?
Answer:
Soil is vital to food production and also important to plants that grow in the wild.

Question 34.
What are the reasons for depletion of nutrients in soil?
Answer:
Poor farming methods, such as repeatedly planting the same type of crop in the same place cause depletion of nutrients in the soil.

Question 35.
What is biodiversity?
Answer:
Biodiversity is the variety of living things that populate the Earth.

Question 36.
How are people speeding up the loss of biodiversity?
Answer:
Through hunting, pollution, habitat destruction, people are speeding up the loss of biodiversity.

Question 37.
How many plant species are being used by us for medicines world wide?
Answer:
We use between 50,000 to 70,000 plant species for medicines world wide.

Question 38.
What is selective harvesting?
Answer:
The practice or removing individual plants or small groups of plants leaving other plants standing to anchor the soil is called selective harvesting.

Question 39.
How are fossil fuels produced?
Answer:
The fuels that are produced from the remains of ancient plants and animals are called fossil fuels. They include coal, petroleum and natural gas.

Question 40.
What are the alternate sources of energy?
Answer:
The alternate sources of energy are sun, wind and water.

Question 41.
What are the other products made from petroleum?
Answer:
Plastic, synthetic, rubber, fabrics like nylon, medicines, cosmetics, waxes, cleaning products, medical devices, etc., are the other products made from petroleum.

Question 42.
Which plant’s seeds are used for the production of bio-fuel?
Answer:
Seeds from the Jatropa Curcas plant are used for the production of bio-fuel.

Question 43.
How does the mining method, Mountain Top Removal mining (MTR) devastate the environment?
Answer:
The mining method Mountain Top Removal mining devastate the environment. They destroy soil, plants and animal habitats.

Question 44.
In which country car manufacturers recycle many raw materials used in making automobiles?
Answer:
In Japan car manufacturers recycle many raw materials used in making automobiles.

Question 45.
In which country nearly one third of the iron produced comes from recycled automobiles?
A. In the United States, nearly one-third of the iron produced comes from recycled automobiles.

Question 46.
What does the Indian tradition teach us?
Answer:
The Indian tradition teaches us that all forms of life – human, animal and plant are so closely inter linked that disturbance of one gives rise to imbalance in the other.

Question 47.
Expand IUCN.
Answer:
IUCN stands for International Union for the Conservation of Nature.

Question 48.
How is IUCN planning to protect wild life and habitats?
Answer:
IUCN monitors the status of endangered wild life, threatened national parks and preserves.

Question 49.
What are the four R’s to save the environment?
Answer:
Reduce, Reuse, Recycle, Recover are the four R’s to save the environment.

Question 50.
How did Amritha Devi and her daughters protest against the king’s order?
Answer:

1. Amritha Devi and her daughters, followed by villagers, who clung to trees in the forest surrounding their village and laid down their lives to save them.
2. They protested against the king’s order to collect wood for the construction of his palace.

Question 51.
Write a method of soil conservation.
Answer:
One soil conservating method is called contour strip cropping. Several crops such as corn, wheat and clover are planted to alternating strips across a slope or across the path of the prevailing wind.

Question 52.
What is the rate of extinction by the estimation of the scientists?
Answer:

1. Scientists estimate that the current rate of extinction is 1,000 times the natural rate through hunting, pollution, habitat destruction.
2. Based on various estimates of the number of species on Earth, we could be losing anywhere from 200 to 1,00,000 species each year.

Question 53.
What is the need to protect biodiversity?
Answer:
We need to protect biodiversity to ensure plentiful and varied food sources. Biodiversity is important for more than just food because many plant species are being used for medicines.

Question 54.
Mention two ways in which water harvesting can be undertaken?
Answer:
The two ways by which water harvesting can be undertaken are

1. Capturing run off water from, rooftops.
2. Capturing run off water from local catchments.

Question 55.
On the basis of the issues raised in the chapter management of natural resources, what changes you in corporate in your lifestyle in a move towards a sustainable use of our resources?
Answer:
I would incorporate the maximum of four R’s i.e., reduce, recycle, reuse and recover in my lifestyle in a move towards a sustainable use of our resources.

10th Class Biology 10th Lesson Natural Resources 2 Marks Important Questions and Answers

Question 1.
Rahul remarked that different human activities are responsible for global warming.
What might be the reasons for his statement?
Answer:

1. Deforestation.
2. Industrialisation and urbanization.
3. Conversion of agriculture lands into residential areas.
4. Home appliances like A/C, refrigerators, vehicle pollution.
5. Population explosion.

Question 2.
What steps do you take to improve natural resources?
Answer:

1. Motivate the people to conserve water.
2. Try to avoid wastage of water whenever possible.
3. Plantation in the vacant lands.
4. Educating the farmer regarding proper utilization of water for irrigation.
5. Encourage the people to recycle the water wherever possible.

Question 3.
Proper utilisation of natural resources is the way to show gratitude to our nation.
Can you support this statement? Give your argument.
Answer:

1. Natural resources of a nation influence its economical and social development.
2. Natural resources are freely available in nature and help in many activities and development of people.
3. The generation of natural resources take a lot of time.
4. They disappear by indiscriminate usage.
5. So proper utilization of natural resources is the way to show gratitude to our nation.

Question 4.
The humans are utilising natural resources indiscriminately. These resources are decreasing more rapidly. Guess what will be the consequences in future?
Answer:
Indiscriminate usage of natural resources causes the following consequences.

1. Reduction in rainfall
2. Drought will occur.
3. Atmospheric temperature becomes increase.
4. The rare species become extinct.

Question 5.
Write any four slogans on the conservation of natural resources.
Answer:
Slogans:

1. Waste water today – live in desert tomorrow
2. Practice eco-friendly methods.
3. Use natural resources judiciously.
4. Save nature – Save future.

Question 6.
There is an increase in the atmospheric temperature year by year. If it continues, guess and write what would be the consequences?
Answer:
If the temperature on earth increases, the consequences would be as follows.
a) All the glaciers and the frozen ice in the polar region start melting leading to rise in the sea water levels.
b) It results in the submergence of low lying coastal areas throughout the world. Millions of people of those areas would lost their homes.
c) Changes in rainfall patterns take place and it result in the occurance of droughts and decrease in crop production.
d) Global warming results in climate change which cause the breakout of climatic sensitive diseases like Malaria, Dengue, Diarrhoea, etc.

Question 7.
There is water scarcity in Ravi’s village during summer. He wants to conduct a rally to create awareness regarding conservation of water. Write any four slogans required to conduct this rally.
Answer:

1. Water is life.
2. Save water – Save a life.
3. Today’s rain water is tomorrow’s life saver.
4. No matter your occupation, water conservation is your obligation.

Question 8.
What steps you take to conserve the biofuels in your daily life?
Answer:

1. Development and usage of alternative energy resources in place of bio-fuels.
2. Minimise the usage of bio-fuels whenever possible.
3. Use public transport, ride by bicycle and walking regularly.
4. Use and purchase energy efficient appliances to save bio-fuels.

Question 9.
Why do we use fossil fuels judiciously?
Answer:

1. Fossil fuels were produced from the remains of ancient plants and animals.
2. They include coal, petroleum (oil) and natural gas.
3. We need to use fossil fuels judiciously because they are non – renewable resources.
4. We need to conserve fossil fuels so we don’t run out of them.
5. The pollution caused by them when burnt, to limit our fossil fuel use.
6. Future generations may not get these resources.
7. Balance in the nature will be disturbed.
8. Electricity production will be stopped.
9. Vehicles running with fossil fuels become useless.

Question 10.
Write two suggestions to create awareness on groundwater conservation.
Answer:

1. We need to adapt different methods to Improve the quality and increase the quantity of groundwater.
2. We should dig water harvesting pits for every house.
3. We should clean the silt, mud fill in tanks and ponds.
4. We should prohibit the establishment of borewells for extraction of groundwater for agricultural and Industrial use.
5. These measures will improve quality and quantity of groundwater.

Question 11.
What is the importance of 4R’s in achieving the goal of “Swachh Bharat”?
Answer:

1. Reduce the production of garbage.
2. Reuse the garbage for the production of manure and electricity.
3. Recycle the garbage by separating It as dry and wet garbage.
4. Recover the plants.

Question 12.
Suggest four measures to conserve fossil fuels.
Answer:
Measures to conserve fossil fuels:

1. Usage of alternatives to fossil fuel.
2. Minimise the usage of fossil fuel.
3. Walk, ride by bicycle and use public transportation whenever possible.
4. Purchase energy efficient appliances.
5. Turn off light and other electronics when you are not using them.

Question 13.
The indiscriminate digging of Borewells may result in what type of consequences in future?
Answer:

1. Due to over drilling of borewells and pulling out water by electric motors, the ground water level Is decreasing day by day.
2. It Is goes on without recharging, ground water becomes scarce.
3. It shows impact on agriculture and the productivity will decrease.
4. Fluorine levels In ground water will increase.
5. Sometimes, saline water may intrude Into the interior places of land and water becomes unfit for consumption.
6. Farmers have to drill the bore wells to more depths which Increase the losses for them.

Question 14.
Ramaiah made broad bed furrow around his field under employment guarantee scheme. Guess the reasons for if. If all the farmers of your village work together, will their water scarcity meet?
Answer:
The reason for Ramaiah making broad bed furrow around his field was, it is useful to conserve soil and water, fertilizer application weeding operations. It also conserves rain water.

The farmers are over coming the water scarcity by sharing water available in the village. They formed groups of farmer including large and small ones who would use the same water resource. Farmers were also motivated to use irrigation techniques like drip irrigation.

Question 15.
What are renewable sources and non-renewable resources?
Answer:
Renewable resources: Resources that can be replaced after they are used are called renewable resources.
Ex: Air, water and soil.
Non-renewable resources: Some other resources, cannot be replaced at all: Once they are used up they are gone forever and are called non renewable resources.
Ex: Coal, Petroleum, Natural gas (fossil fuels).

Question 16.
How do people waste natural resources?
Answer:

1. People often waste natural resources.
2. Animals are over hunted, forests are cleared, exposing land to wind and water damage.
3. Fertile soil is exhausted and lost to erosion because of poor farming practices.
4. Fuel supplies are depleted.
5. Water and air are polluted.
6. Water resources is indiscriminately used for crop growth.

Question 17.
How do people use the forest resources differently?
Answer:

1. The need to conserve resources often conflicts with other needs.
2. For some people, a forest area may be a good place to put a farm.
3. A timber company may want to harvest the area’s trees for construction materials.
4. A business company may want to build a factory or a shopping mall on the land.

Question 18.
What are die effects of deforestation?
Answer:

1. Deforestation destroys wild life habitats and increases soil erosion.
2. It also releases green house gases into the atmosphere, contributing to global warming,
3. Deforestation accounts for 15 per cent of the world’s green house gas emissions.
4. Deforestation also harms the people who rely on forests for their survival, hunting and gathering, harvesting forest products, or using the timber for firewood.

Question 19.
In your opinion What are the causes for soil erosion?
Answer:

1. Soil erosion is caused by poor farming methods such as repeatedly planting the same type of crop in the same place.
2. These methods deplete nutrients in the soil.
3. Soil erosion Is also caused by water and wind currents.
4. When farmers plough up and down hills, soil erosion occurs.
5. Overgrazing by cattle also causes soil erosion.
6. Natural floods causes the extensive damage to the top layer of the soil.

Question 20.
What is Biodiversity? Explain.
(OR)
What is the importance of biodiversity?
Answer:

1. Biodiversity is the variety of living things that populate the earth.
2. The products and benefits we get from nature rely on biodiversity.
3. We need to protect biodiversity to ensure plentiful and varied food sources.
4. Biodiversity is important for more than just food. For instance we use between 50,000 to 70,000 plant species for medicines world wide.

Question 21.
How can we use the fossil fuels carefully?
Answer:
We can use the fossil fuels carefully by taking the following measures.

1. Turn off lights and other electronics when we are not using them.
2. Purchase energy-efficient appliances.
3. Walk, ride a bicycle, if the distance is less.
4. Use public transportation whenever possible.
5. It is better to prefer public transport system like bus or train, instead of travel in personal vehicles.

Question 22.
Why the prices of aluminium and iron are expensive?
Answer:
Earth’s supply of raw material resources is in danger. Many mineral deposits that have been located and mapped have been depleted. As the ores for minerals like aluminium and iron become harder to find and extract, their prices go up.
This makes tools and machinery more expensive to purchase and operate.

Question 23.
What are the effects of mining?
Answer:

1. Many mining methods such as Mountain Top Removal mining (MTR) devastate the environment.
2. They destroy soil, plants and animal habitats.
3. Many mining methods also pollute water and air, as toxic chemicals leak into the surrounding ecosystem.

Question 24.
What did Smt. Indira Gandhi said, while launching the world conservation strategy in India on 6th March 1980?
Answer:
“The interest in conservation is not a sentimental one but the discovery of a truth well known to our ancient stages. The Indian tradition teaches us that all forms of life- human, animal and plant – are so closely inter-linked that disturbance in one gives rise to imbalance in the other” said by Smt. Indira Gandhi.

Question 25.
What are the steps taken by the government to conserve resources?
Answer:

1. Government enacts laws defining how land should be used and which areas should be set aside as parks and wild life preserves.
2. The government enforces laws designed to protect the environment from pollution, such as requiring factories to install pollution control devices and also provide incentives for conserving resources.

Question 26.
What is the necessity of sustainable management of natural resources? Out of the two methods reuse and recycle which one would you suggest to practice and why?
Answer:

1. Sustainable management of natural resources is necessary to Increase the over all life of natural resources specially non renewable resources and also to control the environmental pollution.
2. Both reuse and recycle are the good choice.
3. Reuse: If we reuse something then the cost of recycle will be saved.
4. Recycle: It is not necessary that each and everything can be reused, so after getting recycled the life of the resource will be enhanced.

Question 27.
“Burning fossil fuels is a cause of global warming”. Justify this statement?
Answer:

1. Fossil fuels are composed of carbon, hydrogen, nitrogen and sulphur.
2. When these are burnt they produce CO₂, H₂O, Oxides of Nitrogen and Sulphur.
3. Incomplete combustion of fossil fuels produces green house gases such as CO₂,
4. If huge amount of fossil fuels are burnt, It would produce high amount of CO₂ resulting intense global warming.

Question 28.
Can you suggest some changes in your school which would make it environment friendly?
Answer:
The changes that would make my school environment friendly are

1. Save energy by turning off lights that we are not using.
2. I will suggest to buy recycled paper for decoration and other purposes.
3. Use writing paper on both the sides.
4. Growing trees and plants all around the play ground.

Question 29.
What is the necessity of replenishment of forest? State four reasons.
Answer:
The replenishment of forest is necessary because of the following reasons.

1. It is used to conserve soil.
2. It provides shelter to wild animals.
3. It reduces atmospheric pollution.
4. It controls flood and increases frequency of rainfall.

10th Class Biology 10th Lesson Natural Resources 4 Marks Important Questions and Answers

Question 1.
Forest is renewable resource. But, each year, the Earth loses about 36 million acres of forest. In this type of situation, what suggestions do you give to save forests from turning into non-renewable resources ?
Answer:

1. Forests are the lungs of the world. So I will suggest the following measures to save forests from turning into non-renewable resources.
2. Sustainable forestry practices for ensuing resources into the future.
3. Low impact logging practices, harvesting with natural regeneration in mind. Prevention of removing all the high value trees or all the largest trees from the forests. Recycling methods should be adopted.
4. Replace wood products with alternative sources.
5. Preventing forest fires.
6. Implementing methods like agro forestry, social forestry crop rotation, green plantation, etc. are essential.

Question 2.
What are four R’s? Explain how they help to conserve the environment?
(OR)
Write about the 4 ‘R’s needed for the protection and conservation of environment.
Answer:
By pursuing the maximum of four R’s i.e., Reduce, Reuse, Recycle and Recover, we can save the environment in an effective way.

1. Reduce : It means to use less, I would save electricity by switching off unnecessary lights and fans, prefer walking or cycling than using a vechicle, turn off the engine of car at red light, repair leaky taps and would not waste food.
2. Recycle: It means to collect used things like plastic, paper, galss and metal items and recycle these materials to make required things instead of synthesising or extracting fresh plastic, paper, glass or metal.
3. Reuse: It refers to use things again and again. For example instead of throwing away used envelops, they can be used by pasting new labels.
4. Recover: We should implement ‘recover’ to prevent environmental threat. For example when we cut trees to construct industries or roads for transportation, it is important to grow trees in another areas.

Question 3.
What steps you would like to follow on your part to conserve bio-diversity?
Answer:

1. Biodiversity is the variety of living things that populate the earth.
2. To conserve biodiversity we should avoid hunting.
3. Sustainable forest conservation methods should be followed.
4. I will actively participate Vana Mahosthavam programmes.
5. I will educate and encourage people and make them participate in conservation programmes.
6. Create awareness programmes in and around school.
7. Writing slogans and also make some posters about conservation of biodiversity
8. Judicious use of electricity wherever possible.
9. Finding out of various alternative sources of energy.
10. Plant the saplings in the habitat.
11. Encouraging of social forestry.

Question 4.
Observe the pie diagram showing water resources available in our state for agriculture and answer the given questions.
A) Which water resource is using more for agriculture?
B) What are the consequences of excess utilization of underground water?
C) Which water resource should be utilized for agriculture?
D) What are the alternative ways to increase the underground water resources?

Answer:
A) Ground water.
B) Underground water table will be depleted and scarcity of drinking water will arise.
C)

1. Tanks should be constructed to harvest with rain water.
2. Projects should be constructed across the rivers to store water that can be utilized for agriculture.

D)

1. Construction of rain water storage structures on large scale.
2. Constructing soaking or percolation pits.
3. Contour field bunding.
4. Recharge of wells by building dykes or barriers in the nalla.
5. Plantation in waste lands.
6. Adapting micro irrigation techniques.
   (Any two points you can write)

Question 5.
Forests are renewable resource. Write four sentences supporting this.
(OR)
“Forest is a renewable resource”. Do you agree? Justify.
Answer:

1. Forests are rich habitat for plants and animals. Forests serve as lungs for the world and a bed of nutrients for new fife to prosper.
2. Forest’s pure air protects the earth from green house effect by removing carbon dioxide from the atmosphere and converts it into oxygen.
3. Many fruits, medicines, dyes, sandle wood and bamboo is obtained from forest by local people.
4. Forest provide employment to large number of people and also help in generating revenue.

Question 6.
Observe the above table and answer the following questions.

Village	Type of Farmer	Income per acre on Crops
		Paddy	Cotton	Mirchi	Maize
A	Small	7,500	9,300	5,200	5,000
	Large	26,700	38,000	16,700	12,900
B	Small	7,200	8,750	4,900	5,100
	Large	32,900	42,000	18,400	13,700′

1. Which crop is most suitable to cultivate for small farmer in both the villages?
Answer:
Cotton, paddy

2. If you are a large farmer, which crop do you select to cultivate?
Answer:
Cotton, paddy, mirchi

3. What similarities you have identified in village A and village B?
Answer:
Small and large fanners cultivated same type of crop in both villages. Large farmer gets more income per acre on crops than small farmer in both the villages.

4. Which is the lowest income crop ?
Answer:
Mirchi.

5. Is there any relationship between production of crops and income ? How ?
Answer:
Commercial crops are good for income. Income may or may not related to production of crop. It depends upon demand of the market.

Question 7.
Read the given information and answer the following questions.

A survey was conducted in two villages, Vanaparthy and Vaddicherla of Warangal district in Telangana State. The first with no scarcity and the second with scarce groundwater. Well census was carried out in the villages in order to get a complete picture of well irrigation and its status as well as availability of water. There are no alternative sources of supply as against wells in Vaddicherla, where there is an existing tank that has been converted into a percolation tank, so that the water situation is much better in Vanaparthy.

i) Why did they conduct survey?
Answer:
A compartive study on available water resources irigation method in the Vaddicharla and Wanaparthi of Warangal Dist of Telangana State.

ii) What are irrigation resources in Telangana State?
Answer:
Lakes, wells, canals and ground water etc…,

iii) In which village, do you suggest drip irrigation?
Answer:
Vaddicherla.

iv) Why the water situation is much better in Vanaparthy village compared to Vaddicherla?
Answer:
Existing tank has been converted into a percolation tank.

Question 8.
Observe the Pie diagram and answer the following questions.
i) Identify the fossil fuels from the above diagram.
Answer:
Coal, natural gas, oils are fossil fuels.

ii) Why wastes should be considered as primary energy source in future?
Answer:
The fossil fuels may be exhausted in future. So we may be considered that wastes are primary alternative energy resources.

iii) Why can’t we depend on fossil fuels forever?
Answer:
We can’t depend on fossil fuels forever because fossil fuels are non-renewable resources.

iv) What are the alternatives for fossil fuels?
Answer:
Solar energy, wind energy, tide energy, nuclear energy, energy from waste materials.

Question 9.
Explain the importance and implementation of community based interventions and farmer based interventions for water management.
Answer:
Community based interventions:

1. For water harvesting, there is an urgent need to construct earthen and masonry dams. They help us to store rain water during rainy seasons. They are help in increasing the ground water table.
2. Construction of percolation pits and field bunding are very helpful in the harvesting every rain drop.
3. Open dry wells near nalla canal were recharged by building dykes or barriers in the nalla and maintaining the run – off rain water. The ground water is recharged by these community based interventions.
   Farmer based interventions:
4. Broad Bed Furrow (BBF) land form and contour planting methods are very useful to conserve soil, water and fertilizer application and weeding operations.
5. Planting Gliricidia, a leguminous plant adapted to grow in dry areas on field bunds to strengthen them and make the soil nitrogen rich.
6. Farmers were encouraged to use water resource jointly and irrigate land using micro irrigation methods like sprinklers and drip irrigation.

Question 10.
Explain the farmer based and community based interventions to conserve soil and water resources.
Answer:

Water Management	Collected information
Farmer based water management	1. Farmer based water management implemented individual fields were Broad Bed Furrow (BBF) land form and Contour Planting to conserve in situ soil and water. 2. Use of tropiculator for planting, fertilizer appli­cation and weeding operations. Planting Gliricidia on field bunds to strengthen bunds conserve rain water and supply nitrogen rich organic matter for in situ application to crops. 3. Farmers will obtain 250 kg more pigeon pea and 50 kg more maize per hectare using broad bed furrows and micro irrigation techniques.
Community based water management	1. Fourteen water storage structures (one earthen and 13 masonry dams) with water storage capac­ity of 300 to 2000 m3 were to be constructed in Kothapally village of Rangareddy district. 2. More than 250 rain harvesting structures such as checkdams mini percolation pits, sunken pits and gully plugs were erected in watershed throughout the topo – sequence. 3. Farmers were encouraged for water sharing methods. They formed groups of farmers including large and small ones who would use the same water resource. 4. Farmers have to motivated to use irrigation techni- quies like drip irrigation, sprinklers, etc. 5. Construction of soak pits will help to tap rain water optimally should carry out as community effort.

Question 11.
“The humans who were developed by using the natural resources, today has become the reason for destroying them”. Explain analytically.
Answer:
“The humans who were developed by using the natural resources, today has become the reason for destroying them” – This statement is absolutely true.

1. Primitive man lived in forests and hills. He used the natural resources for his livelyhood. He worshipped nature and used them wisely for his development.
2. After his development, he becomes greedy and using the natural resources indiscriminately and held responsible for their destruction.
3. To meet the needs of growing population, industrialization, urbanization, and huge constructive activities, man utilised natural resources Indiscriminately. At the same time, he did not planned for their revival.
4. But now he realised the importance of natural resources and taken up steps for their conservation. The concept of “Sustainable development” is being implemented in natural resource management.
5. He focussed on development of alternatives for fossil fuels, conservation of water and soil at community level and farmer based interventions.
6. Now he is so keen on conserving forests, wild life and biodiversity.
7. He is so cautious in minimising the utilization of natural resources by following 4’R principle in the day to day life [R – Reduce, R – Reuse, R – Recycle, R – Recover]
8. Now, he is adopting micro-irrigation methods like sprinklers and drip Irrigation to minimise the water usage in low water available areas.
9. He is very interested in following eco-friendly techniques, natural farming methods, using biofertilizers, vermicompost and natural pest control methods in place of toxic chemical pesticides.

Question 12.
The wells and tanks in your village become dry. Ground water levels decreased. Assume the causes for this. Will there be no water scarcity if all the farmers of your village work collectively?
Answer:
Causes for decreasing ground water levels:

1. Varying monsoon behaviour in recent years, there is a pressure on ground water utilization.
2. Indiscriminate tapping of ground water in our village by too much drilling and construction of deep tube wells and bore wells have resulted in over exploitation and depletion of ground water resources.
3. There will be no water scarcity if all the farmers of our village work collectively. Farmers in our village were encouraged to use water resource jointly and irrigate land using micro irrigation techniques. By using micro irrigation techniques farmers in our village obtained more crop yield. Farmers in our village follow the micro irrigation method i.e. drip irrigation and can reduce water consumption by 70% in our village.

Question 13.
Whom do you meet to collect the information of the methods of farmer based, community based water management? Prepare information table to note down your observation.
Answer:
I will meet officials of International Crop Research Institute for Semi – Arid Tropics (ICRISAT) located at Hyderabad to collect information of the methods of farmer based and community based water management.
I also collect information from Central Research Institute for Dry Land Agriculture (CRIDA), National Remote Sensing Agency (NRSA), District Water Management Agency (DWMA) and M Venkatarangaiah Foundation (MVF) and NGO.
The information I gathered from these institutions is summarised below.
Information table:
For Information table See Q.No. 10 in 4 Marks.

Question 14.
Think that there is much scarcity of water for drinking and cultivation in your village. What advice do you give to prevent this?
(OR)
How do you overcome the problem of water scarcity in your village?
Answer:

1. Motivate the people to conserve water.
2. I will educate the people to avoid wastage of water whenever possible.
3. Construction of recharge pits in the house, school and in the open areas to increase the underground water level.
4. Planting trees wherever possible in the village particularly in the vacant lands.
5. Educate the farmers about the micro irrigation system like drip irrigation, sprin¬klers, etc.
6. Encourage the farmers to form groups to share available water among themselves.
7. Construction of percolation tanks in the low lying areas of the village.

Question 15.
What type of fossil fuels are used in your house? What measures do you take to conserve them?
Answer:
Fossil fuels are sources of energy for cooking, heating and burning in our households. Petrol and diesel are being used in our house for transport and running generators and water pumps.
Measures to be taken to conserve fossil fuels in my house :

1. I will put the food material to be cooked on the stove only after arranging all the things which are necessary for cooking.
2. By using pressure cookers 20% gas on rice and 41.5% on meat would be saved when compared to Other cooking means.
3. We must reduce the flame as soon as the boiling process starts in a pressure cooker. This process saves nearly 35% of fuel.
4. I will soak the food material before cooking. It saves 22% of fuel.
5. I will cook food in broad and low depth vessel.
6. I will keep lid on the cooking vessel. If not, it takes more time to cook.
7. For short distances to travel I will go by walk to save fuel for longer distance. I use public transport.
8. Encourage people to use solar water heater and solar cooker.