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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.4

Exercise 1.4 Class 9 Maths Samacheer Question 1.
If P= {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}, then find
(i) (P ∪ Q) ∪ R
(ii) (P ∩ Q) ∩ S
(iii) (Q ∩ S) ∩ R
Solution:
(i) (P ∪ Q) ∪ R
(P ∪ Q) = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11} = {1, 2, 3, 5, 7, 9, 11}
(P ∪ Q) ∪ R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9} = {1, 2, 3, 4, 5, 7, 9, 11}

(ii)(P ∩ Q) ∩ S
(P ∩ Q) = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11} = {2, 5, 9}
(P ∩ Q) ∩ S = {2, 5, 9} ∩ {2, 3, 4, 5, 8} = {2, 5}

(iii) (Q ∩ S) ∩ R
(Q ∩ S) = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8} = {2, 3, 5}
(Q ∩ S) ∩ R = {2, 3, 5} ∩ {3, 4, 5, 7, 9} = {3, 5}

Exercise 1.4 Class 9 Maths Solution Samacheer Question 2.
Test for the commutative property of union and intersection of the sets
P = {x : x is a real number between 2 and 7} and
Q = {x : x is an irrational number between 2 and 7}
Solution:
Commutative Property of union of sets
(A ∪ B)’ = (B ∪ A)
Here P = {3, 4, 5, 6}, Q = {\(\sqrt{3}, \sqrt{5}, \sqrt{6}\)}
P ∪ Q = {3, 4, 5, 6} ∪ {\(\sqrt{3}, \sqrt{5}, \sqrt{6}\)} = {3, 4, 5, 6, \(\sqrt{3}, \sqrt{5}, \sqrt{6}\)} …………. (1)
Q ∪ P = {\(\sqrt{3}, \sqrt{5}, \sqrt{6}\)} ∪ {3, 4, 5, 6}= {\(\sqrt{3}, \sqrt{5}, \sqrt{6}\), 3, 4, 5, 6} ………… (2)
(1) = (2)
∴ P ∪ Q = Q ∪ P
∴ It is verified that union of sets is commutative.
Commutative Property of intersection of sets (P ∩ Q) = (Q ∩ P)
P ∩ Q = {3, 4, 5, 6} ∩ {\(\sqrt{3}, \sqrt{5}, \sqrt{6}\)} = { } ………. (1)
Q ∩ P = {\(\sqrt{3}, \sqrt{5}, \sqrt{6}\)} ∩ {3, 4, 5, 6} = { } …………. (2)
From (1) and (2)
P ∩ Q = Q ∩ P
∴ It is verified that intersection of sets is commutative.

Samacheer Kalvi 9th Maths Exercise 1.4 Question 3.
If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets.
Solution:
Associative Property of union of sets
A ∪ (B ∪ C) = (A ∪ B) ∪ C)
B ∪ C = {m, n, q, s, t} ∪ {m, n,p, q, s}= {m, n, p, q, s, t}
A ∪ (B ∪ C) = {p, q, r, s} ∪ {m, n, p, q, s, t} = {m, n, p, q, r, s, t} ………..… (1)
(A ∪ B) = {p, q, r, s} ∪ {m, n, q, s, t} = {p, q, r, s, m, n, t}
(A ∪ B) ∪ C = {p, q, r, s, m, n, t} ∪ {m, n, p, q, s} = {p, q, r, s, m, n, t} …………… (2)
From (1) & (2)
It is verified that A ∪ (B ∪ C) = (A ∪ B) ∪ C

9th Standard Maths Exercise 1.4 Question 4.
Verify the associative property of intersection of sets for A = {-11, \(\sqrt{2}, \sqrt{5}\) ,7}, B = {\(\sqrt{3}, \sqrt{5}\), 6, 13} and C = {\(\sqrt{2}, \sqrt{3}, \sqrt{5}\), 9}.
Solution:
Associative Property of intersection of sets A ∩ (B ∩ C) = (A ∩ B) ∩ C)

From (1) and (2), it is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C

9th Maths Exercise 1.4 Question 5.
If A = {x : x = 2ⁿ, n ∈ W and n < 4}, B = {x: x = 2 n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets.
Solution:
A = {x : x = 2ⁿ, n ∈ W, n < 4}
⇒ x = 2°= 1
x = 2¹ = 2
x = 2² = 4
x = 2³ = 8
∴ A = {1, 2, 4, 8}

B = {x : x = 2n, n ∈ N and n ≤ 4}
⇒ x = 2 × 1 = 2
x = 2 × 2 = 4
x = 2 × 3 = 6
x = 2 × 4 = 8
∴ B = {2, 4, 6, 8}
C = {0, 1, 2, 5, 6}
Associative property of intersection of sets
A ∩ (B ∩ C) = (A ∩ B) ∩ C
B ∩ C = {2, 6}
A ∩ (B ∩ C) = {1, 2, 4, 8} ∩ {2, 6} = {2} ………..… (1)
A ∩ B = {1, 2, 4, 8} ∩ {2, 4, 6, 8} = {2, 4, 8}
(A ∩ B) ∩ C = {2, 4, 8} ∩ {0, 1, 2, 5, 6} = {2} …………….. (2)
From (1) and (2), It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 6.2 மழைச்சோறு Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 6.2 மழைச்சோறு

கற்பவை கற்றபின்

Question 1.
உங்கள் பகுதியில் பாடப்படும் மழை தொடர்பான நாட்டுப்புறப் பாடல்களைத் தொகுத்து எழுதுக.
Answer:
பாடல் – 1

ஆத்தா மகமாயி வந்திடம்மா
ஆத்தா மகமாயி வந்திடம்மா
உனக்கு எத்தனையோ பூச செஞ்சோம்
உனக்கு எத்தனையோ பூச செஞ்சோம்
வாம்மா வாம்மா வந்து மழைய குடும்மா
குடும்மா கருத்தம்மா
பசி வயிறு புடுங்கு தம்மா
மழை பெய்யச் சொல்லம்மா
மழை பெய்யச் சொல்லம்மா

பாடல் – 2

மழையப்பா மழையப்பா
கொஞ்சம் வாப்பா
இத்தனை நாள் வயல்
காணாதது போதாதா?
என்ன அப்பா கோபம்
மகன்கள் பண்ண
தப்ப மன்னிக்க மாட்டியா?
மன்னிச்சு வாப்பா
மானங்காக்க வாப்பா
மனமிரங்கி வாப்பா
மழையப்பா மழையப்பா

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
கனத்த மழை என்னும் சொல்லின் பொருள் ………………………
அ) பெருமழை
ஆ) சிறு மழை
இ) எடை மிகுந்த மழை
ஈ) எடை குறைந்த மழை
Answer:
அ) பெருமழை

Question 2.
‘வாசலெல்லாம்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………
அ) வாசல் + எல்லாம்
ஆ) வாசல் + எலாம்
இ) வாசம் + எல்லாம்
ஈ) வாசு + எல்லாம்
Answer:
அ) வாசல்+எல்லாம்

Question 3.
‘பெற்றெடுத்தோம்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………
அ) பெறு + எடுத்தோம்
ஆ) பேறு + எடுத்தோம்
இ) பெற்ற + எடுத்தோம்
ஈ) பெற்று + எடுத்தோம்
Answer:
ஈ) பெற்று + எடுத்தோம்

Question 4.
கால் + இறங்கி என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் …………………..
அ) கால்லிறங்கி
ஆ) காலிறங்கி
இ) கால் இறங்கி
ஈ) கால்றங்கி
Answer:
ஆ) காலிறங்கி

குறுவினா

Question 1.
மழைச்சோறு பாடலில் உழவர் படும் வேதனை எவ்வாறு கூறப்படுகிறது?
Answer:
(i) கடலைச் செடி, முருங்கைச் செடி, கருவேலங்காடு, காட்டுமல்லி என அனைத்தும் மழையில்லாமல் வாடிப்போனது. பெற்றெடுத்த குழந்தைகளின் பசியைத் தீர்க்க முடியவில்லை .

(ii) கலப்பை பிடிப்பவரின் கை சோர்ந்து விட்டது, ஏற்றம் இறைப்பவரின் மனம் தவிக்கிறது என்றும் இதற்குக் காரணம் மழை இல்லாமையே இன்று உழவர் வேதனைப் படுகின்றனர்.

Question 2.
மக்கள் ஊரைவிட்டு வெளியேறக் காரணம் என்ன?
Answer:
மழை இல்லாததால் உழவுத் தொழில் செய்ய முடியவில்லை. எனவே மக்கள் ஊரை விட்டு வெளியேறுகின்றனர்.

சிறுவினா

Question 1.
கோலம் கரையாத நிலையை மழைச்சோறு பாடல் எவ்வாறு விளக்குகிறது?
Answer:

• வாளியில் கரைத்த மாவால் வாசலில் கோலம் போட்டனர்.
• இந்தக் கோலத்தைக் கரைக்க மழை வரவில்லை !
• பானையில் மாவைக் கரைத்து, பாதை எல்லாம் கோலம் போட்டனர்.
• அந்தக் கோலம் கரைக்கவும் மழை வரவில்லை .

Question 2.
மழையின்மையால் செடிகள் வாடிய நிலையை விளக்குக.
Answer:

• கல் இல்லாத காட்டில் கடலைச் செடி நட்டு வளர்த்தார்கள். அதற்கும் மழை பெய்யவில்லை.
• முள் இல்லாத காட்டில் முருங்கைச் செடி நட்டு வளர்த்தார்கள். அதற்கும் மழை வரவில்லை .
• கருவேலங்காடும் மழையில்லாமல் பூக்கவில்லை.
• மழை இல்லாததால் காட்டு மல்லியும் பூக்கவில்லை.

Question 3.
மழைச்சோறு எடுத்தபின் எவ்வாறு மழை பெய்தது?
Answer:

• மழைச் சோறு எடுத்தபின், பேய் மழையாக ஊசிபோல கால் இறங்கி உலகமெல்லாம் பெய்கிறது.
• சிட்டுப் போல மின்னி மின்னி ஊரெங்கும் பெய்கிறது.
• ஊரெங்கும் செல்ல மழை பெய்கிறது.

சிந்தனை வினா

Question 1.
மழைவளம் பெருக நாம் செய்ய வேண்டுவன யாவை?
Answer:
மழை வளம் பெருக அதிகப்படியான மரங்களை நட்டு வளர்க்க வேண்டும். மரங்களை நட்டால் மட்டும் போதாது. அதனை நன்கு பராமரிக்க வேண்டும். எங்காவது மரங்கள் வெட்டப்படும் போது, அதனைத் தடுக்க வேண்டும். ஒவ்வொரு வீட்டிலும் மழைநீர் சேமிப்புத் தொட்டி கட்டாயம் வைக்க வேண்டும். மழை பெய்யும் காலங்களுக்கு முன் குளங்கள் குட்டைகளை தூர்வார வேண்டும்.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
ஒரு நாட்டின் வளத்திற்கு அடிப்படையாக விளங்குவது …………………….
அ) மழை
ஆ) உணவு
இ) உடை
ஈ) பணம்
Answer:
அ) மழை

Question 2.
கல் இல்லாக் காட்டில் …………………….. போட்டனர்.
அ) முருங்கைச் செடி
ஆ) கடலைச் செடி
இ) கருவேல மரம்
ஈ) காட்டு மல்லி
Answer:
ஆ) கடலைச் செடி

Question 3.
முள்ளில்லா காட்டில் …………………. போட்டனர்.
அ) முருங்கைச் செடி
ஆ) கடலைச் செடி
இ) கருவேல மரம்
ஈ) காட்டு மல்லி
Answer:
அ) முருங்கைச் செடி

Question 4.
‘வனவாசம் சென்று விடுவோம்’ என்று கூறியவர் …………………..
அ) புலவர்
ஆ) குறவர்
இ) உழவர்
ஈ) மறவர்
Answer:
இ) உழவர்

குறுவினா

Question 1.
எங்கெல்லாம் கோலம் இடப்பட்டது?
Answer:
வாசல் மற்றும் பாதைகளில் கோலம் இடப்பட்டது.

Question 2.
கடலைச் செடி வாடக் காரணம் யாது?
Answer:
மழை இல்லாததால் கடலைச் செடி வாடியது.

Question 3.
எவற்றை உழவர்கள் தலையில் வைத்துச் செல்கின்றனர்?
Answer:
மழைச் சோறு வாங்கிய பானை, அகப்பை, பழைய முறம் ஆகியவற்றை உழவர்கள் தலையில் வைத்துச் செல்கின்றனர்.

Question 4.
சிட்டு போல மின்னியது எது?
Answer:
சிட்டு போல மின்னியது மழை.

சிறுவினா

Question 1.
மழைச் சோற்று நோன்பு பற்றிக் குறிப்பிடுக.
Answer:
(i) மழையில்லாமல் பஞ்சம் ஏற்படும் நேரங்களில் சிற்றூர் மக்கள் வீடு வீடாகச் சென்று உப்பில்லாத சோற்றை ஒரு பானையில் வாங்குவார்கள். ஊர்ப் பொது இடத்தில் வைத்து அனைவரும் பகிர்ந்து உண்பர்.

(ii) கொடிய பஞ்சத்தைக் காட்டும் அடையாளமாக இது நிகழும். இதனைக் கண்டு மனம் இரங்கி மழை பெய்யும் என்பது மக்களின் நம்பிக்கை. இதனை மழைச்சோற்று நோன்பு என்று கூறுவார்கள்.

சொல்லும் பொருளும்

பாதை – வழி
கனத்த – மிகுந்த
பெண்டுகளே – பெண்களே
சீமை – ஊர்

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Tamilnadu Samacheer Kalvi 10th Social Science Civics Solutions Chapter 3 State Government

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State Government Textual Exercise

I. Choose the correct answer.

State Government Lesson Questions And Answers Question 1.
The Governor of the State is appointed by the …………..
(a) Prime Minister
(b) Chief Minister
(c) President
(d) Chief Justice
Answer:
(c) President

Samacheer Kalvi Guru 10th Social Question 2.
The Speaker of a State is a:
(a) Head of State
(b) Head of Government
(c) President’s agent
(d) None of these
Answer:
(d) None of these

Civics Chapter 3 Question 3.
Which among the following is not one of the powers of the Governor?
(a) Legislative
(b) Executive
(c) Judicial
(d) Diplomatic
Answer:
(d) Diplomatic

Samacheer Kalvi.Guru 10th Social Question 4.
Who can nominate one representative of the Anglo-Indian Community to the State Legislative Assembly?
(a) The President
(b) The Governor
(c) The Chief Minister
(d) the Sneaker of State legislature
Answer:
(b) The Governor

Chapter 3 Civics Class 10 Question 5.
The Governor does not appoint …………….
(a) Chief Minister
(b) Chairman of the State Public Service Commission
(c) Advocate General of the State
(d) Judges of the High Court
Answer:
(d) Judges of the High Court

Question 6.
The Chief Minister of a State is appointed by:
(a) The State Legislature
(b) The Governor
(c) The President
(d) The Speaker of State Legislative Assembly
Answer:
(b) The Governor

Question 7.
The State Council of Ministers is headed by …………….
(a) The Chief Minister
(b) The Governor
(c) The Speaker
(d) The Prime Minister
Answer:
(a) The Chief Minister

Question 8.
The Legislative Council:
(a) Has a term of five years
(b) Has a term of six years
(c) Is a permanent house
(d) Has a term of four years
Answer:
(b) Has a term of six years

Question 9.
The minimum age for the membership of the Legislative Council is ……………
(a) 25 years
(b) 21 years
(c) 30 years
(d) 35 years
Answer:
(c) 30 years

Question 10.
The members of the Legislative Council are:
(a) Elected by the Legislative Assembly
(b) Mostly nominated
(c) Elected by local bodies, graduates, teachers, Legislative Assembly etc.
(d) Directly elected by the people
Answer:
(c) Elected by local bodies, graduates, teachers, Legislative Assembly etc.

Question 11.
Which one of the following States does not possess a bicameral legislature?
(a) Andhra Pradesh
(b) Telangana
(c) Tamil Nadu
(d) Uttar Pradesh
Answer:
(c) Tamil Nadu

Question 12.
The High Courts in India were first started at:
(a) Calcutta, Bombay, Madras
(b) Delhi and Calcutta
(c) Delhi, Calcutta, Madras
(d) Calcutta, Madras, Delhi
Answer:
(a) Calcutta, Bombay, Madras

Question 13.
Which of the following States have a common High Court?
(a) Tamil Nadu and Andhra Pradesh
(b) Kerala and Telangana
(c) Punjab and Haryana
(d) Maharashtra and Gujarat
Answer:
(c) Punjab and Haryana

II. Fill in the Blanks.

1. Governor of the state government surrenders his resignation to …………….
2. Members of the Legislative assembly (MLAs) are elected by the ………….
3. …………….. is the first women Governor of Tamil Nadu.
4. ………….. acts as the chancellor of universities in the state.
5. The Seventh Amendment Act of ………….. authorized the Parliament to establish a common high court for two or more states.
6. The Chairman and Members of the State Public Service Commission can be removed only by the …………….
Answers:
1. President
2. People
3. M. Fathima Bheevi
4. Governor
5. 1956
6. President

III. Match the Following.

Answer:
1. (b)
2. (a)
3. (d)
4. (e)
5. (c)

IV. Choose the Correct Statement.

Question 1.
(i) Only some States in India have Legislative Councils.
(ii) Some members of Legislative Councils are nominated.
(iii) Some members of Legislative Councils are directly elected by the people.
(a) ii and iv are correct
(b) iii and iv are correct
(c) i and ii are correct
(d) i, ii, and iii are correct
Answer:
(c) i and ii are correct

Question 2.
Assertion (A): There are limitations on the Legislative authority of the State Legislature.
Reason (R): Certain bills on the State List can be introduced in the State Legislature only with the President’s approval.
(a) (A) is false but (R) is true
(b) (A) is true but (R) is false
(c) Both (A) and (R) are true and (R) is the correct reason for (A)
(d) Both (A) and (R) are true and (R) is not the correct reason for (A)
Answer:
2. (c) Both (A) and (R) are true and (R) is the correct reason for (A)

V. Answer the following.

Question 1.
How the State of Jammu and Kashmir differ from the other states of India?
Answer:

1. The Constitution of India grants special status to Jammu and Kashmir among Indian states.
2. The constitution of Jammu and Kashmir was adopted on the 17th November 1957 and came into force on 26th January 1957.
3. It is the only state in India to have a separate constitution.
4. The Directive principles of state policy and Fundamental duties are not applicable to Jammu and Kashmir.
5. Right to property which is denied as a fundamental right to rest of the India is still guaranteed in Jammu and Kashmir.

Question 2.
What is the importance of the Governor of a state?
Answer:
The Governor is the Constitutional head of the State Executive. The administration of a State is carried on in the name of the Governor. He directly rules a State when there is the imposition of the President’s rule in the State. He is an integral part of the State legislative.

Question 3.
What are the qualifications for the appointment of a Governor?
Answer:

1. Article 157 and Article 158 of the constitution of India specify the eligibility requirements for the post of Governor. They are:
2. He should be a citizen of India
3. He must have completed 35 years of age.
4. He should not be a member of parliament or of any state legislature.
5. If he is a member of any of legislature he has to vacate his seat on assuming office.
6. He should not hold any other profitable occupation.

Question 4.
What is the original jurisdiction of the High Court?
Answer:
In their judicial capacity, the High Courts of the Presidency towns – Bombay, Calcutta and Madras have both original and appellate jurisdictions, while other High Courts have mostly appellate jurisdiction.
Only in matters of admiralty, probate, matrimonial and contempt of court, they have original jurisdiction.

Question 5.
What do you understand by the ‘Appellate Jurisdiction’ of the High Court?
Answer:

1. As courts of appeal, all High Courts can hear appeals in civil and criminal cases from their subordinate courts as well as on their own.
2. They cannot exercise Jurisdiction over Tribunals established under the laws relating to Armed Forces of the country.

VI. Answer in Detail.

Question 1.
What are the powers and functions of the Chief Minister?
Answer:
The powers and the functions of the Chief Minister are:

1. The Chief Minister is the head of the Council of Ministers. He recommends the persons who can be appointed as ministers by the Governor. He allocates the portfolio among the ministers.
2. He presides over the meetings of the council of Ministers and influences its decisions.
3. The Chief Minister is the principal channel of communication between the Governor and the Council of Ministers.
4. He announces the Government policies on the floor of the House. He can introduce the Bills in the Legislative Assembly.
5. For smooth functioning of the State and for good center-state relations, he has to develop a rapport with the Union Government.

Question 2.
Describe the various powers and functions of the Governor.
Answer:
Powers and functions of the Governor: The Governor is the head of the State Executive .He is guided by the aid and advice of the Chief Minister and the Council of Ministers.

Executive power:

1. Governor is the constitutional head of the state. All the administration of the state is carried on in his name.
2. Being the executive head he makes various appointment as follows.
3. He appoints the leader of the majority party in the State Legislature as the Chief Minister of the state.
4. On the advice of the Chief Minister he appoints the other members of the Council of Ministers.
5. He appoints the advocate General of the state and determine his remuneration.
6. He appoints the Chairman and Members of the State Public Service Commission and the State Election Commissioner and determine the service and tenure of office.
7. He acts as the Chancellor of Universities in the State. He appoints the Vice-Chancellors of Universities in the State.
8. When the State Emergency is proclaimed by the President the Governor becomes the real executive.

Legislative powers:

1. The Governor is an integral part of the State Legislature.
2. He has the right to summon, prorogue the State Legislature and dissolve the State Legislative Assembly if no confidence motion is passed against the ruling party.
3. Commencement of the first session of the Legislature after the general election and the first session of each year is addressed by the Governor.
4. Relating to a bill pending in the legislature he can send messages to the houses.
5. When the office of the Speaker and the Deputy Speaker falls vacant he can appoint any member of the Legislative Assembly to preside over.
6. He nominates 1 member from the Anglo-Indian community to the State Legislative Assembly and 1/6th of the members to the State Legislative Council from amongst the persons having special knowledge or practical experience in literature, science art, co-operative movement and social service.
7. In consultation with the Election Commission he decides on the question of disqualification of members of the State Legislature.
8. Every bill passed by the State legislature will become law only after his signature.
9. He has the option to give his assent to the bill or withhold his assent to the bill or return the bill for the reconsideration of the Legislature.
10. If any bill endangers the position of the State High Court he can reserve the bill passed by the Legislature for the consideration of the President.
11. Under Article 213 the Governor has the power to promulgate ordinances when the legislature is not in session.
12. These ordinances should be approved by the legislature within six months. He can also withdrew an ordinance at any time.
13. He has to lay before the State Legislature the Annual Reports of State Finance Commission, the State Public Service Commission and the Comptroller and Auditor General relating to the accounts.

Financial power:

1. Money bills can be introduced in the state Legislature only with his prior recommendation.
2. He causes the Annual Budget of the state to be presented in the Legislative Assembly.
3. If there is a need he presents through the Minister of Finance the state supplementary Budget.
4. No demand for any grant can be made except on his recommendation.
5. He can make advances out of the state contingency fund to meet any unforeseen expenditure.
6. He constitutes a Finance Commission after every five years to review the financial position of the panchayats and the municipalities.

Judicial powers:

1. He appoints the Attorney General of the State.
2. The Chief Justice of the High Court in the state is appointed by the President in consultation with him.
3. He makes appointment, postings and promotions of the District Judges in consultation with the State High Court.
4. He appoints judges to the subordinate courts in the state.
5. He can pardon, commute or reprieve punishment on receipt of appeals of mercy.

Discretionary power: (special power)

1. He seeks information from the Chief Minister relating to the administrative and legislative matters of the state.
2. The Governor can reserve a bill for the consideration of the President.
3. He can call the leader of any party to form ministry in the state when no party get majority in the general elections.
4. If the Council of Ministers has lost its majority he can dissolve the Assembly.

Emergency Powers: If the Governor feels that the Government of the state is not carried on in accordance with the provisions of the constitution under Article 356 recommend to the President to impose President rule. At that time the (imposition of President Rule) administration of the state is carried on by the Governor as the representative of the President.

Question 3.
Briefly discuss the Functions of the State Legislature.
Answer:
The powers and functions of the State Legislature are almost the same as that of the Parliament.
(i) The State Legislature can pass laws on all subjects mentioned in the State List as per the constitutions. It can also pass laws on concurrent subjects.

(ii) The Legislature controls the finances of the State. The Lower House enjoys greater power than the Upper House in money matters. Money Bills can be introduced only in the Lower House of the Assembly.

(iii) The Legislature controls the Executive. The council of Ministers is responsible to the Assembly. The ministers have to answer questions asked by the members of the Legislature.

(iv) The Council cannot vote for grants.

(v) No new tax can be levied without the sanction and permission of the Assembly.

Question 4.
Critically examine the functions and powers of the Council of Ministers.
Answer:
Article 163 of our constitution provides for Council of Ministers to aid and advice the Governor. Article 163 (1) states that there shall be a Council of Ministers with the Chief Minister at the head to aid and advice the Governor in the exercise of his function.

1. All the Ministers work as a team under the Chief Minister.
2. The Council of Ministers (cabinet) are responsible to formulate and decide the policies of the State and implements them effectively,
3. They are the instrumental in framing the policies of the State.
4. It decides the Legislative programmes of the Assembly and sponsors all important bills.
5. It decides all the bills or money bills to be introduced in the Legislative Assembly.
6. It chalks out programmes and schemes for the socio-economic changes so that the state makes head way in various inter related field.
7. It controls the financial policy and decides the structure.
8. It frames the proposal for incurring expenditure out of state reserves each Ministry is responsible in this matter.
9. Each Ministry of the Council of Ministers supervises, controls and co-ordinates the department concerned.
10. Annual Financial Statement called the Budget is finalised by the Council of Ministers.

The smooth functioning of the state lie in the hands of the Council of Ministers.

Question 5.
Describe the powers and functions of the High Court.
Answer:
The powers and functions of the High Court:

1. The High Court enjoys original jurisdiction in matters of admiralty, probate, matrimonial and contempt of court.
2. It hears appeals in both civil and criminal cases against the decisions of the subordinate courts and can review their judgments.
3. It can issue writs such as habeas corpus, mandamus, prohibition, quo warranto and certiorari not only for the enforcement of the Fundamental Rights but also for other purposes.
4. It has the power of supervision over all courts and tribunals functioning in its territorial jurisdiction (except military courts or tribunals).
5. It controls and supervises the working of the lower courts and maintains records of its proceedings and decisions.

State Government Additional Questions

I. Choose the correct answer.

Question 1.
The administration of a State is carried on in the name of the ……………
(a) Chief Minister
(b) Governor
(c) President
(d) Council of Ministers
Answer:
(b) Governor

Question 2.
In part ……………… of our constitution lays down a uniform structure for the State Government.
(a) VI
(b) IV
(c) IV (A)
(d) VII
Answer:
(a) VI

Question 3.
Who determines the number of judges of the High Court?
(a) The President
(b) The Chief Justice of the Supreme Court
(c) Governor
(d) The Attorney General of India
Answer:
(a) The President

Question 4.
The ……………… commission was set up to suggest and review the centre state relations.
(a) Mandal (commission)
(b) Public service (commission)
(c) Sarkaria (commission)
(d) Election (commission)
Answer:
(c) Sarkaria (commission)

Question 5.
The Legislative Assembly of Tamil Nadu consists of ……………..
(a) 235 members
(b) 240 members
(c) 245 members
(d) 250 members
Answer:
(a) 235 members

Question 6.
The ……………… is the rapport between the Governor and the Council of Ministers.
(a) Chief Minister
(b) Speaker
(c) Deputy Speaker
(d) Attorney General
Answer:
(a) Chief Minister

Question 7.
The ………….. is the real executive head of the State administration.
(a) Chief Minister
(b) Minister
(c) IAS Officer
Answer:
(a) Chief Minister

Question 8.
If Chief Minister resigns it implies that the ……………… also has to resign.
(a) Governor
(b) Commission
(c) Council of Ministers
(d) Advocate General
Answer:
(c) Council of Ministers

Question 9.
The ……………….. is the presiding officer of the Upper House.
(a) Speaker
(b) Chairman
(c) Chief Minister
Answer:
(b) Chairman

Question 10.
Every bill passes through ……………… readings before it becomes an Act.
(a) 2
(b) 3
(c) 1
(d) 5
Answer:
(b) 3

II. Fill in the Blanks.

1. The Tamil Nadu Legislative Council was abolished by Tamil Nadu Legislative Council (Abolition) Bill in …………..
2. Every Member of Legislative Council serves for ……………… term.
3. The …………….. Amendment Act of 1976 curtailed the judicial review power of the High Court.
4. The writ of …………… is issued against a person who claims or usurps a public offence.
5. The Governor can reserve a bill for the consideration of the …………………
6. The …………….. is the Constitutional head of the State.
7. The Legislative Assembly elects two of its members as the ……………. and ……………..
8. The …………… is the highest judicial authority in a State.
9. The President of India appoints the ……………… of a High Court in consultation with the Chief Justice of Supreme Court and the Governor of that particular State.
10. The High Court has administrative control over the …………..
Answers:
1. 1986
2. Six years
3. 42^nd
4. Quo Warranto
5. President
6. Governor
7. Speaker and Deputy Speaker
8. High Court
9. Chief Justice
10. Subordinate Courts

III. Match the Following.

Answer:
1. (e)
2. (c)
3. (a)
4. (b)
5. (d)
6. (j)
7. (h)
8. (f)
9. (g)
10. (i)

IV. Choose the Correct Statement.

Question 1.
(i) The institution of High Court originated in India in 1900.
(ii) The High Court of Guwahati is common for seven northeastern states
(iii) The High Courts are the highest courts at the State level.
(iv) Delhi, though not a State, has its own separate High Court.
(a) i, ii and in are correct
(b) ii, iii and iv are correct
(c) ii, iii and iv are correct
(d) i, ii and iv are correct
Answer:
(c) ii, iii and iv are correct

Question 2.
(i) The Speaker vacates his office, if he cannot continue to be a member of the Assembly.
(ii) The Speaker cannot be removed from his office in any circumstances.
(iii) The Speaker does not vacate his office, when the Assembly is dissolved.
(iv) In the absence of the Speaker, the Deputy Speaker performs his functions.
(a) i, ii and iii are correct
(b) i, ii and iv are correct
(c) ii, and iii are correct
(d) i, iii and iv are correct
Answer:
(d) i, iii and iv are correct

V. Answer in Brief

Question 1.
Write a brief note on the governance of the state as per the Indian constitution.
Answer:

1. The constitution of India contains provisions for the governance of both the Union and the States.
2. In part IV of the constitution from Article 152 to 237 is applicable to all . the states except Jammu and Kashmir.
3. The structure of the State Government consists of three branches to govern.
4. The Executive, the Legislature and the Judiciary.

Question 2.
What is Habeas Corpus?
Answer:
Habeas Corpus is a writ issued to a detaining authority, ordering the detainer to produce the detained person in the issuing court, along with the cause of his/her detention. If the detention is found to be illegal, the court issues an order to set the person free.

Question 3.
Name some states that have Bi-Cameral state legislature? What type of legislature Tamil Nadu has?
Answer:
Bihar, Karnataka, Maharashtra, Uttar Pradesh, Andhra Pradesh, Telengana and Jammu and Kashmir are some states having Bi-Cameral legislature. Legislative Assembly and Legislative Council. In Tamil Nadu the State legislature has only one house i.e. Legislative Assembly. It has Uni Cameral Legislature.

Question 4.
What do you know about the composition of the Legislative Assembly of Tamil Nadu?
Answer:
The Legislative Assembly of Tamil Nadu consists of 235 members out of which 234 members are directly elected by the people from the constituencies on the basis of adult franchise and one member is nominated by the Governor from the Anglo-Indian community. However, seats shall be reserved in the House for the SCs and STs.

Question 5.
When was the legislative council abolished in Tamil Nadu?
Answer:
The Tamil Nadu legislative council was abolished by Tamil Nadu legislative council (Abolition) Bill 1986. The Act came into force on 1st November 1986.

Question 6.
Mention one function of the judiciary.
Answer:
The judiciary settles disputes.

Question 7.
Name the states that have common High Court.
Answer:

1. The states of Punjab and Haryana and the Union Territory of Chandigarh have a common High Court situated at Chandigarh.
2. The High Court of Guwahati is common for seven North Eastern states of Assam, Nagaland, Manipur, Meghalaya, Mizoram, Tripura and Arunachal Pradesh.

Question 8.
The seven north-east states have a common High Court. Where is this located?
Answer:
It is located at Guwahati.

Question 9.
Write a note on Court of Record.
Answer:
All the decisions and decrees issued by the High Court are printed and are kept as a record for future references by the court as well as by the lawyers. Thus it also acts as a Court of Record.

Question 10.
What is a judicial system?
Answer:
A judicial system is a mechanism of Court that a citizen can approach when a law is violated.

Question 11.
Name the branches of legal systems?
Answer:
(i) Civil and (ii) Criminal

Question 12.
Differentiate between the Supreme Court and the High Court.
Answer:
The Supreme Court of India is the apex Court in India. It hears both civil and criminal cases as well as those associated with the interpretation of the Constitution. At the state level, the High Court is the highest Court of appeal. Generally each state has a High Court, but two or more States may have a common High Court.

VI. Answer in Detail.

Question 1.
Mention the powers and functions of the Chief Minister relating to the Council’of Ministers.
Answer:
As the head of the Council of Ministers, th,e Chief Minister performs the following functions:

• The Chief Minister recommends the persons who can be appointed as Ministers by the Governor.
• He allocates the portfolio among the Ministers.
• He shuffles and reshuffles his ministry.
• He can ask a minister to resign or to advise the Governor to dismiss him in case of difference of opinion.
• He presides over the meetings of the Council of Ministers and influences its decisions
• He can bring about the collapse of the Council of Ministers by resigning from office
• He guides, directs, controls and coordinates the activities of all the Ministers.

Question 2.
Write a note on the appointment of the Chief Minister?
Answer:

1. The Chief Minister is appointed by the Governor of the state.
2. The leader of the majority party or majority group in the State Legislative Assembly is appointed as the Chief Minister.
3. In case no party commands absolute majority in the Legislative Assembly or the majority fails to elect its leader the Governor can use his discretionary power and invite the leader of the other largest party to form the ministry.
4. He has to prove the confidence (majority support) in the Legislative Assembly within the period stipulated by the Governor.
5. The term of the Chief Minister in general is five years .But it is not fixed.
6. He may remain as the Chief Minister as long as he enjoys the support of the Legislative Assembly.
7. When he losses the confidence of the majority in the Assembly he has to resign.

Question 3.
Write about the Judicial Review.
Answer:
Judicial review is the power of a High Court to examine the constitutionality of legislative enactments and executive orders of both the Central and State Governments. Though the phrase judicial review has no where been used in the Constitution, the provisions of Articles 226 and 227’ explicitly confer the power of judicial review on a High Court.

The 42^nd Amendment Act of 1976 curtailed the judicial review power of High Court. It debarred the High Courts’ from considering the Constitutional validity of any central law. However, the 43rd Amendment Act of 1977 restored the original position.

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Students can Download Maths Chapter 1 Number System Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Intext Questions

(Try These Textbook Page No. 1)

Samacheer Kalvi 7th Maths Book Answers Question 1.
Write the following integers in ascending order: -5,0,2,4, -6,10, -10
Solution:
Plotting the points on the number line, we get

The numbers are placed in an increasing order from left to right.
∴ Ascending order: -10 < -6 < -5 < 0 < 2 < 4 < 10

7th Maths Number System Question 2.
If the integers -15, 12, -17, 5, -1, -5, 6 are marked on the number line then the integer on the extreme left is _____ .
Solution:
The least number will be on the extreme left.
∴ -17 will be on the extreme left.

Number System 7th Standard Question 3.
Complete the following pattern:
50, ___ 30, 20, _, 0, -10, _, _, -40, _, ___.
Solution:
The difference between the consecutive number is 10.
50, 40, 30,20, 10, 0, -10, -20, -30, -40, -50, -60

Samacheer Kalvi 7th Maths Books Answers Question 4.
Compare the given numbers and write “<”, “>” or in the boxes.

Solution:
(a)  A positive number is greater than a negative number.
(b) 1000,0 is less than all positive integers.
(c)

7th Maths Exercise 1.1 Samacheer Kalvi Question 5.
Write the given integers in descending order, -27, 19, 0, 12, -4, -22, 47, 3, -9, -35.
Solution:
Separating positive and the negative integers, we get -27, -4, -22, -9, -35
Arranging the numbers in descending order -4 > -9 > -22 > -27 > -35
The positive numbers are 19,12,47, 3
Arranging in descending order, we get 47 > 19 > 12 > 3
0 stands in the middle.
∴ Descending order: 47 > 19 > 12 > 3 > 0 > -4 > -9 > -22 > -27 > -35

(Try This Text Book Page No. 3)

7th Maths Guide Try These Question 1.
Find the value of the following using the number line activity.
(i) (-4) + (+3)
(ii) (-4) + (-3)
(iii) (+4) + (-3)
Solution:
(i) (-4) + (+3)
To find the sum of (-4) and (+3), we start at zero facing positive direction continuing in the same direction and move 4 units backward to represent (-4).
Since the operation is addition we maintain the same direction and move three units forward to represent (+3)
We land at -1
So (-4) + (+3) = -1

(ii) (-4) + (-3)
From zero move 4 steps backward to represent (-4)
From the same direction again move 3 units backward to represent (-3)
We land at -7 So (-4) + (-3) = -7

(iii) (+4) + (-3)
We start at zero facing positive direction and move 4 steps forward to represent (+4) Since the operation is addition we maintain the same direction and move three units backward to represent (-3).
We land at +1.
So (+4) + (-3) = +1

(Properties of Addition Textbook Page No. 6)

Samacheer Kalvi 7th Maths Solutions Question 1.
Complete the given table and check whether the sum of two integers is an integer or not?
(i) 7 + (-5) = (+2)
(ii) (-6)+ (-13) = (-19)
(iii) 25 + 9 = 34
(iv) (-12) + 4 = -8
(v) 41 + 32 = 73
(vi) (-19) + (-15) = (-34)
(vii) 52 + (-15) = (+37)
(viii)(-7) + 0 = (-7)
(ix) 0 + 12 = 12
(x) 14 + 0 = 14
(xi) (-6) +(-6) = (-12)
(xii) (-27) + 0 = -27
Solution:
The sum of two integers is an integer.

(Try These Textbook Page No. 7)

Samacheer Kalvi Guru 7th Maths Question 1.
Fill in the blanks:
(i) 20 + (-11) = -(11)+ 20 [∵ Addition is commutative]
(ii) (-5) + (-8) = (-8) + (-5) [∵ Addition is commutative]
(iii) (-3) +12 =12 + (-3) [∵ Addition is commutative]

7th Standard Number System Question 2.
Say True or False.
(i) (-11) + (-8) = (-8) + (-11)
(ii) -7 + 2 = 2 + (-7)
(iii) (-33) + 8 = 8 + (-33)
Solution:
(i) True, because addition is commutative for intergers
(ii) True, by commutative property on intergers
(iii) True, by commutative property on intergers

Samacheer Kalvi 7th Maths Question 3.
Verify the following.
(i) [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
(ii) [7 + (-8)] + (-5) = 7 + [(-8) + (-5)]
(iii) [(-11) + 5]+ (-14) = (-11) + [5 + (-14)]
(iv) (-5) + [(-32) +(-2)] = [(-5) + (-32) + (-2)]
Solution:
(i) [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
[(-2) + (-9)] + 6 = (-11) + 6 = -5
Also (-2) + [(-9) + 6] = (-2) + (-3) = -5
Both the cases the sum is -5.
∴ – [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]

(ii) [7 + (-8)] +(-5) = 7 + [(-8) + (-5)]
Here [7 + (-8)] + (-5) = (-1) + (-5) = -6
Also 7 + [(-8) + (-5)] = 7 + (-13) = 7 – 13 = -6
In both the cases the sum is -6.
∴ [7 + (-8)] + (-5) = 7 + [(-8) + (-5)]

(iii) [(-11) + 5] + (-14) = (-11) + [5 + (-14)]
Here [(-11) + 5] + (-14) = (-6) + (-14) = (-20)
(-11) + [5 + (-14)] = (-11) +(-9) = (-20)
In both the cases the sum is -20.
∴ [(-11) + 5] + (-14) = (-11) + [5 + (-14)]

(iv) (-5) + [(-32) + (-2)] = [(-5) + (-32)] + (-2)
(-5) + [(-32) + (-2)] = (-5) + (-34) = -39
Also [(-5) + (-32)] + (-2) = (-37) + (-2) = -39
In both the cases the sum is -39.
∴ (-5)+ [(-32) +(-2)] = [(-5)+ (-32)] +(-2)

Samacheer Kalvi 7th Maths Book Answers Pdf Question 4.
Find the missing integers:
(i) 0 + (-95) = -95
(ii) -611 + 0 = -611
(iii) ____ + 0 = _____ Any integer; the same integer
(iv) 0 + (-140) = -140

Samacheer Kalvi 7th Maths Book Solutions Question 5.
Complete the following:
(i) -603 + 603 = 0
(ii) 9847+ (-9847) = 0
(iii) 1652 + (-1652) = 0
(iv) -777 + 777 = 0
(v) –5281 +5281 = 0

Exercise 1.2

Subtraction of Integers

(Try These Text book Page No. 11)

Samacheer Kalvi 7th Maths Answers Question 1.
Do the following by using number line.
(i) (-4) – (+3)
Solution:
We start at zero facing positive direction move 4 units backward to represent (-4). Then turn towards negative side and move 3 units forward.

We reach -7.
∴ (-4) – (+3) = -7.

(ii) (-4) – (-3)
Solution:
We start at zero facing positive direction. Move 4 units backward to represent -4. Then turn towards the negative side and move 3 units backwards.

We reach at-1.
∴ (-4) – (-3) = -1.

Samacheer Kalvi Guru 7th Standard Maths Question 2.
Find the values and compare the answers.
(i) (-6) – (-2) and (-6) + 2
Solution:
(-6) – (-2) = -6 + (Additive inverse of-2)
= -6 + (+2) = -4
Also (-6)+ 2 = -4
∴ (-6) – (-2) = (-6) + 2

(ii) 35 – (-7) and 35 + 7.
Solution:
35 – (-7) = 35 + (Additive inverse of -7) = 35 + (+7) = 42
Also 35 + 7 = 42 ; 35 – (-7) =35 + 7

(iii) 26 – (+10) and 26 + (-10)
Solution:
26 – (+10) = 26 + (Additive inverse of +10) = 26 + (-10) = 16
Also 26 + (-10) = 16; 26 – (+10) = 26 + (-10)

Number System 7th Class Question 3.
Put the suitable symbol <, > or = in the boxes.

Solution:
(i) -10 – 8 = -18 & -10 + 8 = – 2
(ii) (-20) + 10 = -10 & (-20) – (-10) = -10
(iii) -70 – 50 = (-70) + (-50) = -20
(iv) 100 – (+100) = 0 & 100 – (-100) = 100 + (+100) = 200
(v) -50 – 30 = -50 + (-30) = -80 Also -100 + 20 = – 80

(Try These Text book Page No. 14)

Samacheer Kalvi Guru 7th Maths Solutions Question 1.
Fill in the blanks.
(i) (-7) – (-15) = +8
-7 – (-15) = -7 + (Additive inverse of-15)
= -7 + 15 = +8
(ii) 12 – (-7) = 19          12 – (-7) = 19
(iii) -4 – (-5) = 1

Question 2.
Find the values and compare the answers.
(i) 15 – 12 and 12 – 15
(ii) -21 – 32 and -32 – (-21)
Solution:
(i) 15 – 12 = 3 & 12-15 = 12 +(-15) = -3

(ii) -21 – 32 = (-21) + (-32) = -53
Also -32 – (-21) = (-32) + (+21) = -11 ; -53 < -11

Question 3.
Is associative property true for subtraction of integers. Take any three examples and check.
Solution:
Consider the numbers 1,2 and 3. Now (1 – 2) – 3 = -1 – 3 = -4
Also 1 – (2 – 3) = 1 – (-1) = 1 + 1 = 2
∴ (1 – 2) – ≠ 1 – (2 – 3)
∴ Associative property is not true for subtraction of integers.

Exercise 1.3

Multiplication of Integers

(Try These Textbook Page No. 16)

Question 1.
Find the product of the following
(i) (-20) × (-45) = +900 [As we know the product of two negative integers is positive, the answer is +900.]
(ii) (-9) × (-8) = 72 [ ∵ Product of two negative integers is positive]
(iii) (-30) × 40 × (-1) = (+1200) [Product of two integers with opposite sings is negative integer.
(-30) × 40 × (-1) = (-1200) × (-1) = +1200)]
(iv) (-50) × 2 × (-10) = -1000 [Product of two integers with opposite signs is negative.
(+50) × 2 × (-10) = 100 × (-10) = -1000)]

Question 2.
Complete the following table by multiplying the integers in the corresponding row and column headers.

Solution:
We know that
(i) product of two positive integers is positive
(ii) product of two negative integers is
(ii) product of two negative integers is positive
(iii) product of integers with opposite sign is negative.
∴ The table will be as follows:

Question 3.
Which of the following is incorrect?
(i) (-55) × (-22) × (-33) < 0
(ii) (-1521) × 2511 < 0
(iii) 2512 – 1525 < 0
(iv) (1981) × (+2000) < 0
Solution:
(iii) and (iv) are incorrect because 2512 – 1252 is a positive integer.
Also (+1981) × (+2000) is a positive integer.

(Try These Textbook Page No. 18)

Question 1.
Find the product and check for equality
(i) 18 × (-5) and (-5) × 18
Solution:
Here 18 × (-5) = -90 Also (-5) × 18 = -90
∴ 18 × (-5)= (-5) × 18

(ii) 31 × (-6) and (-6) × 31
Solution:
Here 31 × (-6) = -186 Also (-6) × 31 =-186
∴ 31 × (-6)= (-6) × 31

(iii) 4 × 51 and 51 × 4
Solution:
Here 4 × 51 = 204 Also 51 × 4 = 204
∴ 4 × 51 = 51 × 4

Question 2.
Prove the following.
(i) (-20) × (13 × 4) = [(-20) × 13] × 4
Solution:
LHS = (-20) × (13 × 4) = (-20) × 52 = -1040
RHS = [(-20) × 13] × 4 = (-260) × 4 = -1040
LHS = RHS
∴ (-20) × (13 × 4) = [(-20) × 13] × 4

(ii) [(-50) × (-2)] × (-3) = (-50) × [(-2) × (-3)]
Solution:
LHS = [(-50) × (-2)] × (-3) = 100 × (-3) = -300
RHS = (-50) × [(-2) × (-3)] = (-50) × 6 =-300
LHS = RHS
∴ [(-50) × (-2)] × (-3) = (-50) × [-2) × (-3)]

(iii) [(-4) × (-3)] × (-5) = (-4) × [(-3) × (-5)]
Solution:
LHS = [(-4) × (-3)] × (-5) = 12 × (-5) = -60
RHS = (-4) × [(-3) × (-5)] = (-4) × 15 = -60
LHS = RHS
∴ [(-4) × (-3)] × (-5) = (-4) × [(-3) × (-5)]

(Try These Textbook Page No. 19)

Question 1.
Find the values of the following and check for equality:
(i) (-6) × (4 + (-5)) and ((-6) × 4) + ((-6) × (-5))
Solution:
(-6) × (4 + (-5)) = (-6) × (-1) = 6 .
((-6) × 4) + ((-6) × (-5)) = (-24)+ 30 = 6
Hence (-6) × (4 + (-5)) = ((-6) × 4) + ((-6) × (-5))

(ii) (-3) × [2 + (-8)] and [(-3) × 2] + [(-3) × 8]
Solution:
(-3) × [2 + (-8)] = (-3) × (-6) = 18
Also [(-3) × 2] + [(-3) × 8] = (-6)+ (-24) = -30
(-3) × [2 + (-8)] ≠ [(-3) × 2] + [(-3) × 8]

Question 2.
Prove the following.
(i) [(-5) × (-76)] + [(-5) × 8]
Solution:
LHS = (-5) × [(-76) + 8] = (-5) × (-68)
= +340
RHS = [(-5) × (-76)] + [(-5) × 8]
= +380 + (-40) = +380 – 40
= +340
LHS = RHS
∴ (-5) × [(-76) + 8] = [(-5) × (-76)] + [(-5) × 8]

(ii) (42 × 7) + [42 × (-3)]
Solution:
LHS = 42 × [7 + (-3)]
= 168
RHS = (42 × 7) + [42 × (-3)] = 294 – 126
= 168
LHS = RHS
∴ 42 × [7 + (-3)] = (42 × 7) + [42 × (-3)]

(iii) [(-3) × (-4)] + [(-3) × (-5)]
Solution:
LHS = (-3) × [(-4) + (-5)] = (-3) × (-9)
= +27
RHS = [(-3) × (-4)] + [(-3) × (-5)] = 12 + 15 = 27
LHS = RHS
∴ (-3) × [(-4) + (-5)] = [(-3) × (-4)] + [(-3) × (-5)]

(iv) 103 × 25 = (100 + 3) × 25 = (100 × 25) + (3 × 25)
Solution:
First consider 103 × 25 = 2575
Now (100 + 3) × 25 = 103 × 25 = 2575
Also (100 × 25) + (3 × 25) = 2500 + 75
= 2575
∴ All the three are same. 103 × 25 = (100 + 3) × 25 = (100 × 25) +(3 × 25)

Exercise 1.4

Division of Integers

(Try These Text book Page No. 22)

Question 1.
(i) (-32) ÷ 4 = _____
(ii) (-50) ÷ 50 = ____
(iii) 30 ÷ 15 = ______
(iv) -200 ÷ 10 = _____
(v) -48 ÷ 6 = ______
Solution:
(i) -8
(ii) -1
(iii) 2
(iv) -20
(v) -8

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.5

12th Maths Exercise 5.5 Samacheer Kalvi Question 1.
A bridge has a parabolic arch that is 10m high in the centre and 30m wide at the bottom. Find the height of the arch 6m from the centre, on either sides.
Solution:
From the diagram, equation of the parabolic arch

∴ The required height =10 – y₁ = 10 – 1.6 = 8.4m.

12th Maths Exercise 5.5 Question 2.
A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16m, and the height at the edge of the road must be sufficient for a truck 4m high to clear if the highest point of the opening is to be 5m approximately. How wide must the opening be?
Solution:
From the diagram,
AA’ = 16 m, OA = 8m, OB = 5m
∴ Equation of the ellipse is

∴ The required wide for the opening is 2y₁ = 2(4.8) = 9.6 m

12th Maths Exercise 5.2 Samacheer Kalvi Question 3.
At a water fountain, water attains a maximum height of 4m at horizontal distance of 0.5m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75m from the point of origin.
Solution:
From the diagram
Equation of the path of water

The refined height = 4 – y₁ = 4 – 1 = 3 m

10th Maths Exercise 5.5 Samacheer Kalvi Question 4.
An engineer designs a satellite dish with a parabolic cross section. The dish is 5m wide at the opening, and the focus is placed 1.2m from the vertex
(a) Position a coordinate system with the origin at the vertex and the x-axis on the parabola’s axis of symmetry and find an equation of the parabola.
(b) Find the depth of the satellite dish at the vertex.
Solution:
From the diagram,

(a) Consider the satellite dish is open rightward parabola
y² = 4 ax ……….. (1)
Clearly a = 1.2m
(1) ⇒ y² = 4(1.2)
y² = 4.8x
(b) Use the point (x₁, 2.5) in (1)
(2.5)² = 4(1.2)x₁
\(\frac{(2.5)^{2}}{4(1.2)}\) = y₁
x₁ = 1.3 m
∴ The depth of the satellite dish at vertex is 1.3 m

Two Dimensional Analytical Geometry Pdf Question 5.
Parabolic cable of a 60m portion of the roadbed of a suspension bridge are positioned as shown below. Vertical Cables are to be spaced every 6m along this portion of the roadbed. Calculate the lengths of first two of these vertical cables from the vertex.
Solution:
From the diagram,

Equation of the suspension bridge
(x – h)² = 4a(y – k)
But V (0, 3)
x² = 4a (y – 3)
Use the point (30, 16) in (1)
30² = 4a (16 – 3) ⇒ 900 = 13 × 4a
\(\frac{900}{13 \times 4}\) = a

(i) The length of the first vertical cable from the vertex is
Use (6, y₁) in (2)
(2) ⇒ (6)² = \(\frac{900}{13}\) (y₁ – 3)
\(\frac{36 \times 13}{900}\) = y₁ – 3
0.52 = y₁ – 3
y₁ = 3.52 m
(ii) The length of the second vertical cable from the vertex is
Use the point (12, y₂) in (2)
(2) ⇒ (12)² = \(\frac{900}{13}\) (y₂ – 3)
\(\frac{144 \times 13}{900}\) = y₂ – 3
0.52 = y₂ – 3
y₂ = 3.52 m

12th Samacheer Maths Solutions Question 6.
Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation \(\frac{x^{2}}{30^{2}}-\frac{y^{2}}{44^{2}}\) = 1. The tower is 150m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower.
Solution:
From the diagram,equation of hyperbola is

Samacheer Kalvi 12th Maths Guide Question 7.
A rod of length 1.2m moves with its ends always touching the coordinate axes. The locus of a point P on the rod, which is 0.3m from the end in contact with x-axis is an ellipse. Find the eccentricity.
Solution:
From the diagram,
(i) ∆^le OAB be a right angle triangle.
(ii) ∠APD and ∠PBC are corresponding angles, so corresponding angles are equal.

Samacheerkalvi.Guru 12th Maths Question 8.
Assume that water issuing from the end of a horizontal pipe, 7.5m above the ground, describes a parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position 2.5m below the line of the pipe, the flow of water has curved outward 3m beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground?
Solution:
From the diagram,

Equation of the water path is
x² = – 4 ay
Use the point (3, – 2.5) in (1)
(3)² = – 4a(- 2.5)
9 = 10a
a = \(\frac{9}{10}\) substituting in (1)
(1) ⇒ x² = -4\(\frac{9}{10}\)y …………. (2)
Use the point (x₁, -7.5) in (2)
(2) ⇒ x₁² = -4 \(\frac{9}{10}\)(-7.5) ⇒ x₁² = 30(\(\frac{9}{10}\))
x₁ = \(\sqrt{3 \times 9}\)
x₁ = \(3 \sqrt{3}\) m
∴ The water strikes the ground \(3 \sqrt{3}\) m beyond the vertical line.

12th Samacheer Maths Solution Question 9.
On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4m when it is 6m away from the point of projection. Finally it reaches the ground 12m away from the starting point. Find the angle of projection?
Solution:
From the diagram,

Equation of the parabolic path is
x² = -4ay
Use the point (6, -4) in (1)
(1) ⇒ (6)² = 16a
\(\frac{36}{16}\) = 10a
substitute a = \(\frac{9}{14}\) in (1)
(1) ⇒ x² = -4\(\left(\frac{9}{4}\right)\)y
x² = -9y ……….. (2)
Differentiate with respect to ‘x’

12th Maths 5th Chapter In Tamil Question 10.
Points A and B are 10km apart and it is determined from the sound of an explosion heard at those points at different times that the location of the explosion is 6km closer to A than B. Show that the location of the explosion is restricted to a particular curve and find an equation of it.
Solution:
From the diagram,

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.5 Additional Problems

Two Dimensional Analytical Geometry 2 Question 1.
If a parabolic reflector is 20cm in diameter and 5cm deep, find the distance of the focus from the centre of the reflector.
Solution:

By the property of the parabolic reflector, the position of the bulb should be placed at the focus.
By taking the vertex at the origin the equation of the reflector is y² = 4ax.
Let PQ be the diameter of the reflector P = (5, 10)
Since P (5, 10) lies on the parabola,
10² = 4a × 5
ie., 100 = 20a ⇒ a = 5
So the focus is at a distance of 5cm from the vertex and focus is (5, 0).

Exercise 5.5 Class 12 Question 2.
The focus of a parabolic mirror is at a distance of 8cm from its centre (vertex). If the mirror 25cm deep, find the diameter of the mirror.
Solution:

Let the vertex be at the origin.
VF = a = 8cm
The equation of the parabola is
Y² = 4ax = 4(8)x = 32x
Depth of the mirror = x₁ = 25cm.
So, radius is 0.
⇒ y² = 32(25) = 800
y = \(\sqrt{800}=10 \sqrt{8}=10 \times 2 \sqrt{2}=20 \sqrt{2}\) = Radius of the mirror
∴ Diameter of the mirror = 2 × 20\(\sqrt{2}\) = 40 \(\sqrt{2}\) cm of the mirror.

Two Dimensional Analytical Geometry Question 3.
A cable of a suspension bridge is in the form of a parabola whose span is 40 mts. The roadway is 5 mts below the lowest point of the cable. If an extra support is provided across the cable 30 mts above the ground level, find the length of the support if the height of the pillars are 55 mts.
Solution:

The lowest point on the cable is taken as the vertex and it is taken as the origin.
Let AB, CD be the pillars.
Span of parabola = 40 mts = distance between AB and CD
C’V = VA’ = 20 mts
Height of each pillar = 55 mts ⇒ AB = 55 mts
So, A’B = 55 – 5 = 50 mts
Thus, the point B is (20, 50).
Equation of the parabola is x² = 4ay
Here, B is a point on the parabola, x² = 4ay
(20)² = 4a (50) ⇒ 4a = \(\frac{20 \times 20}{50}\) = 8
∴ The equation is x² = 8y
Let PQ be the length of the extra support RQ.
RQ = 30, RR’ = 5 ⇒ R’Q = 25
Let VR’ be x₁ ∴ Q is (x₁, 25).
Q is a point on parabola
x₁² = 8 x 25 = 200
x₁ = \(\sqrt{200}=10 \sqrt{2}\)
The entire length, PQ = 2 x₁ = 20\(\sqrt{2}\).mts.

Samacheer Kalvi Guru 12th Maths Question 4.
A kho-kho player in a practice session while running realises that the sum of the distances from the kho-kho poles from him is always 8m. Find the equation of the path traced by him if the distance between the poles is 6m.
Solution:

Give FP + F’P = 4
ie., 2a = 8 and
FF’ = 2ae = 6
ie., a = 4 and ae = 3
e = \(\frac{a e}{a}=\frac{3}{4}\)
b² = a²(1 – e²) = 16(1 – \(\frac{9}{16}\)) = 7
So the equation of the path is an ellipse whose equation is \(\frac{x^{2}}{16}+\frac{y^{2}}{7}\) = 1.

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.1

12th Maths Exercise 5.1 Samacheer Kalvi Question 1.
Obtain the equation of the circles with radius 5 cm and touching x-axis at the origin in general form.

Solution:
Given radius = 5 cm and the circle is touching x axis
So centre will be (0, ± 5) and radius = 5
The equation of the circle with centre (0, ± 5) and radius 5 units is
(x – 0)² + (y ± 5)² = 5²
(i.e) x² + y² ± 10 y + 25 – 25 = 0
(i.e) x² + y² ± 10y = 0

12th Maths Exercise 5.1 Question 2.
Find the equation of the circle with centre (2, -1) and passing through the point (3, 6) in standard form.
Solution:
Centre = C = (2, -1); Passing through = A = (3, 6)
So radius = CA = \(\sqrt{(2-3)^{2}+(-1-6)^{2}}=\sqrt{1+49}=5 \sqrt{50}\)
Now centre = (2, -1) and radius = \(\sqrt{50}\)
So equation of the circle is
(i.e) (x – 2)² + (y + 1)² = \(\sqrt{50}^{2}\) ⇒ (x – 2)² + (y + 1)² = 50

12th Maths Exercise 5.2 Samacheer Kalvi Question 3.
Find the equation of circles that touch both the axes and pass through (-4, – 2) in general form.
Solution:
Since the circle touches both the axes, its centre will be (r, r) and radius wil be r.
Here centre = C = (r, r) and point on the circle is A = (-4, -2)
CA = r ⇒ CA² = r²
(i.e) (r + 4)² + (r + 2)² = r²
⇒ r² + 8r +16 + r² + 4r + 4 – r² = 0
(i.e) r² + 12r + 20 = 0
(r + 2) (r + 10) = 0
⇒ r = -2 or -10
When r = -2, the equation of the circle will be (x + 2)² + (y + 2)² = 2²
(i.e) x² + y² + 4x + 4y + 4 = 0
When r = -10, the equation of the circle will be (x + 10)² + (y + 10)² = 10²
(i.e) x² + y² + 20x + 20y + 100 = 0

12th Maths Chapter 5 Exercise 5.1 Question 4.
Find the equation of the circle with centre (2, 3) and passing through the intersection of the lines 3x – 2y -1 = 0 and 4x + y – 27 = 0 .
Solution:
To find the point of intersection of the two lines we have to solve the two equations.
Now solving them: 3x — 2y = 1 ……….. (1)
4x + y = 27 ……….(2)
(2) × 2 ⇒ 8x + 2y = 54 ……….. (3)
(1) ⇒ 3x – 2y = 1 ………. (1)
(3) + (1) ⇒ 11x = 55 ⇒ x = \(\frac{55}{11}\) = 5
Substituting x = 5 in (2) we get
20 + y = 27 ⇒ y = 27 – 20 = 7
∴ The point = A = (5, 7)
Given centre = C = (2, 3)
∴ radius = \(\sqrt{(5-2)^{2}+(7-3)^{2}}=\sqrt{9+16}=5\) = 5
Now centre = (2, 3) and radius = 5
So equation of the circle is
(x – 2)² + (y – 3)² = 5²
(i.e) x² + y² – 4x – 6y – 12 =0

Class 12 Maths Chapter 5 Exercise 5.1 Question 5.
Obtain the equation of the circle for which (3, 4) and (2, -7) are the ends of a diameter.
Solution:
The equation of a circle with (x₁ , y₁) and (x₂ , y₂ ) as end points of a diameter is
(x – x₁ )(x – x₂) + (y – y₁ )(y – y₂) = 0
Here the end points of a diameter are (3, 4) and (2, -7)
So equation of the circle is (x – 3 )(x – 2 ) + (y – 4) (y + 7.) = 0
x² + y² – 5x + 37 – 22 = 0

Two Dimensional Analytical Geometry 2 Question 6.
Find the equation of the circle through the points (1, 0), (-1, 0), and (0, 1).
Solution:
Let the required circle be
x² + y² + 2gx + 2fy + c = 0 …………. (A)
The circle passes through (1, 0), (-1, 0) and (0, 1)
(1, 0) ⇒ 1 + 0 + 2g(1) + 2f(0) + c = 0
2g + c = -1 ……………. (1)
(-1, 0) ⇒ 1 + 0 + 2g (-1) + 2f(0) + c = 0
-2g + c = -1 ……….. (2)
(0, 1) ⇒ 0 + 1 + 2g (0) + 2f(1) + c = 0
2f+ c = -1 ……….. (3)
Now solving (1), (2) and (3) .
2g + c = -1 ………… (1)
-2g + c = -1 ………….. (2)
(1) + (2) ⇒ 2c = -2 ⇒ c = -1
Substituting c = -1 in (1) we get
2g – 1 = -1
2g = -1 + 1 = 0 ⇒ g = 0
Substituting c = -1 in (3) we get
2f – 1 =-1 ⇒ 2f = -1 + 1 = 0 ⇒ f = 0
So we get g = 0, f= 0 and c = -1
So the required circle will be
x² + y² + 2(0) x + 2(0)y – 1 = 0
(i.e) x² + y² – 1 = 0 ⇒ x² + y² = 1

Exercise 5.1 Class 12 Question 7.
A circle of area 9n square units has two of its diameters along the lines x + y = 5 and x – y = 1. Find the equation of the circle.
Solution:
Area of the circle = 9π
(i.e) πr² = 9π
⇒ r² = 9 ⇒ r = 3
(i.e) radius of the circle = r = 3
The two diameters are x + y = 5 and x – y = 1
The point of intersection of the diameter is the centre of the circle = C
To find C: Solving x + y = 5 ……… (1)
x – y = 1 ………. (2)
(1) + (2) ⇒ 2x = 6 ⇒ x = 3
Substituting x = 3 in (1) we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Centre = (3, 2) and radius = 3
So equation of the circle is (x – 3)² + (y – 2)² = 3²
(i.e) x² + y² – 6x – 4y + 4 =0

Ex 5.1 Class 12 Question 8.
If y = \(2 \sqrt{2} x\) + c is a tangent to the circle x² + y² =16 , find the value of c .
Solution:
The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is
c² = a²(1 + m²)
Here x² + y² = 16 ⇒ a² = 16
y = \(2 \sqrt{2} x\) + c ⇒ m = \(2 \sqrt{2}\) and c = c
The condition is c² = a² (1 + m²)
(i.e) c² = 16(1 + 8) = 144
⇒ c = ± 12

Chapter 5 Maths Class 12 Question 9.
Find the equation of the tangent and normal to the circle x² + y² – 6x + 6y – 8 = 0 at (2, 2).
Solution:
The equation of the tangent to the circle x² + y² + 2 gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 9
So the equation of the tangent to the circle
x² + y² – 6x + 6y – 8 = 0 at (x₁, y₁) is
xx₁ + yy₁ – \(\frac{6\left(x+x_{1}\right)}{2}+\frac{6\left(y+y_{1}\right)}{2}\) – 8 = 0
(i.e) xx₁ + yy₁ – 3(x + x₁) + 3(y + y₁) – 8 = 0
Here (x₁, y₁) = (2, 2)
So equation of the tangent is
x(2) + y(2) – 3(x + 2) + 3(y + 2) – 8 = 0
(.i.e) 2x + 2y – 3x – 6 + 3y + 6 – 8 = 0
(i.e) -x + 5y – 8 = 0 or x – 5y + 8=0
Normal is a line ⊥r to the tangent
So equation of normal circle be of the form 5x + y + k = 0
The normal is drawn at (2, 2)
⇒ 10 + 2 + k = 0 ⇒ k = -12
So equation of normal is 5x + y – 12 = 0

Chapter 5 Class 12 Maths Question 10.
Determine whether the points (-2, 1), (0, 0) and (-4, -3) lie outside, on or inside the circle x² + y² – 5x + 2y – 5 = 0 .
Solution:
To find the position of a point with regard to a given circle, substitute the point in the equation of the circle if we get a positive value, the point lies outside the circle.
If we get a -ve value the point lies inside the circle and if we get O then the point lies on the circumference of the circle.
The given circle is x² + y² – 5x + 2y – 5 = 0 ……….. (1)
Substituting the point (-2, 1) in (1) we get
4 + 1 – 5(-2) + 2(1) – 5 = 5 + 10 + 2 – 5 = 12
⇒ (- 2, 1) lies outside the circle
Substituting the point (0, 0) in (1) we get
-5 < 0 ⇒ (0, 0) lies inside the circle Substituting the point (-4, -3) in (1) we get 16 + 9 + 20 – 6 – 5 = 34 >0
⇒ (- 4, -3) lies outside the circle

12th Maths 5th Chapter Samacheer Kalvi Question 11.
Find centre and radius of the following circles.
(i) x² + (y + 2)² = 0
(ii) x² + y² + 6x – 4y + 4 = 0
(iii) x² + y² – x + 2y – 3 = 0
(iv) 2x² + 2y² – 6x + 4y + 2 = 0
Solution:
(i) x² + (y + 2)² = 0
(i.e) x² + y² + 4y + 4 = 0
Comparing this equation with the general form x² + y² + 2gx + 2fy + c = 0
we get 2g = 0 ⇒ g = 0
2f = 4 ⇒ f= 2 and c = 4
Now centre = (-g, -f) = (0, -2)
Radius = r = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{0+4-4}\)
∴ Centre = (0, -2) and radius = 0

(ii) x² + y² + 6x – 4y + 4 = 0
Comparing with the general form we get
2g = 6, 2f = -4
⇒ g = 3, /= -2 and c = 4
Centre = (-g, -f) = (-3, 2)
Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{9+4-4}\)= 3
∴ Centre = (-3, 2) and radius = 3

(iii) x² + y² – x + 2y – 3 = 0
Comparing with the general form of the circle we get

(iv) 2x² + 2y² – 6x + 4y + 2 = 0
(÷ by 2) ⇒ x² + y² – 3x + 2y + 1 =0
Comparing this equation with the general form of the circle we get
2g = -3, 2f= 2
g = \(-\frac{3}{2}\), g= 1 and c = 1
So centre = (-g, -f) = (\(\frac{3}{2}\), -1)
and radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{\frac{9}{4}+1-1}=\frac{3}{2}\)
∴ Centre = (\(\frac{3}{2}\), -1) and radius = \(\frac{3}{2}\)

Two Dimensional Analytical Geometry Pdf Question 12.
If the equation 3x² + (3 -p) xy + qy² – 2px = 8 pq represents a circle, find p and q. Also determine the centre and radius of the circle.
Solution:
For a circle co-eff of x² = co-efif of y²
⇒ 3 = q
co-eff of xy = 0
⇒ 3 – p = 0 ⇒ p = 3
So p = q = 3
So the equation of the circle becomes 3x² + 3y² – 6x – 72 = 0
(÷ by 3) ⇒ x² + y² – 2x – 24 = 0
Comparing this equation with the general form of the circle we get
2g = -2, 2f = 0
g = -1, f = 0 and c = -24
So centre = (-g, -f) = (1, 0) and radius = \(\sqrt{g^{2}+f^{2}-c}\) = 5
∴ Centre =(1,0) and radius = 5

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.1 Additional Problems

Ex5 1 Class 12 Question 1.
Find the equation of the circle whose centre is (2, -3) and passing through the intersection of the line 3x – 2y = 1 and 4x + y = 27.
Solution:
Solving 3x – 2y = 1 and 4x+y = 27
Simultaneously, we get x = 5 and y = 7
∴ The point of intersection of the lines is (5, 7)

Now we have to find the equation of a circle whose centre is (2, -3) and which passes through (5,7)

∴ Required equation of the circle is
(x – 2)² + (y + 3)² = \((\sqrt{109})^{2}\)
⇒ x² + y² – 4x + 6y – 96 = 0

Class 12 Maths Chapter 5 Question 2.
Find the equation of a circle of radius 5 whose centre lies on x-axis and which passes through the point (2, 3).
Solution:
Let the centre of the circle be (h, k). Since the centre lies on x-axis, we have k = 0
Therefore, centre = (h, 0)
Also the circle has radius 5 units and passes through the point (2, 3).

Hence the centre is (6, 0) or (-2, 0)
∴ The equation of the circle is
(x – 6)² + (y – 0)² = 5² ⇒ x² + y² – 12x+ 11 = 0 (Or) (x + 2)² + (y – 0)² = 5² ⇒ x² + y² + 4x – 21 =0

Two Dimensional Analytical Geometry Question 3.
Find the centre and radius of the following circles:
x² + y² – 2x + 4y – 4 = 0 and 2x² + 2y² + 16x – 28y + 32 = 0. Also find the ratio of their diameters.
Solution:
Comparing the equation x² + y² – 2x + 4y – 4 = 0 with x² + y² + 2gx + 2 fy + c = 0, we get
2g = -2 ⇒ g = -1, 2f = 4 ⇒ f = 2, c = -4

Comparing with x² + y² + 2gx + 2fy + c = 0
2g = 8 ⇒ g = 4, 2f = -14 ⇒ f = -7, c = 16

Diameter = 2r₂ = 2 × 7 = 14 units
Ratio of their diameters = 6 : 14 = 3 : 7.

Class 12 Maths Ex5.1 Question 4.
Find the equation of the circle whose radius is 4 and which is concentric with the circle x² + y² + 2x – 6y = 0
Solution:
x² + y² + 2x – 6y = 0 …(1)
Here 2g = 2 ⇒ g = 1, 2f = -6 ⇒ f = -3
Centre of the circle = (-g, -f) = (-1, 3)
Since the required circle is concentric with (1), its centre is also (-1, 3).
∴ The equation of the circle whose centre is (-1, 3) and radius 4 is
(x + 1)² + (y – 3)² = 4²
⇒ x² + y² + 2x – 6y – 6 = 0

Exercise 5.1 Class 12 Solutions Question 5.
Show that the four points (1, 0), (2, -7), (8,1) and (9, 6) are concyclic.
Solution:
Let the equation of the circle be Since it passes through (1,0) ⇒ x² + y² + 2gx + 2fy + c = 0 …(1)

Exercise 5.1 Class 12 Maths Question 6.
Find the equation of a circle which passes through the points (1, -2) and (4, -3) and whose centre lies on the line 3x + 4y = 0
Solution:
Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0 …(1)
Since it passes through (1, -2) and (4, -3)
∴ 5 + 2g – 4f + c = 0 …(2)
and 25 + 8g – 6f + c = 0 …(3)
Also (-g, -f) centre of circle (1) lies on 3x + 4y = 7
-3g – 4f = 7 …(4)
Subtracting (2) from (3), we get
20 + 6g – 2f = 0 …(5)
Solving (4) and (5), we get

Substituting these values of g and f in (2)

∴ From (1), we get,

Which is the required equation on the circle.

Math Solution Class 12 Chapter 5 Question 7.
Find the equation (s) of the circle passing through the points (1, 1) and (2, 2) and whose radius is 1.
Solution:
Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0 …. (1)
Since it passes through (1,1)
∴ 1 + 1 + 2g + 2f + c = 0
⇒ 2g + 2f + c + 2 = 0 ….(2)
Again it passes through (2, 2)
∴ 4 + 4+ 4g + 4/+ c = 0
⇒ 4g + 4f + c + 8 = 0 …(3)

Adding (2) and (4)
g² + 2g + f² + 2f + 2 = 1
g² + 2g + (-g – 3)² + 2(-g – 3) + 1 = 0
⇒ 2g² + 6g + 4 = 0 ⇒ g² + 3g + 2 = 0
⇒ (g + 1)(g + 2) = 0 ⇒ g = -1, -2
when g = -1, from (5), f = – (-1) – 3 = 1 – 3 = -2
when g = -2, from (5), f = – (-2) – 3 = 2 – 3 = -1
Substituting g = -1, f = -2 in (2), we get
2(-1) + 2(-2) + c + 2 =0
⇒ – 6 + c + 2 = 0 ⇒ c = 4
Now putting g = -1, f = -2 and c = 4 in (1), we get
x² + y² – 2x – 4y + 4 = 0
Again putting g = -2, f = -1 in (2), we get
2(-2) + 2(-1) + c + 2 =0
⇒ -4 – 2 + c + 2 =0 ⇒ c = 4
Putting g = -2, f = -1 and c = 4 (in) (1), we get
x² + y² – 4x – 2y + 4 = 0
Hence the required equation (s) of the circle are
x² + y² – 2x – 4y + 4 =0 and x² + y² – 4x – 2y + 4 = 0.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.4

10th Maths Exercise 6.4 Samacheer Kalvi Question 1.
From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree. (\(\sqrt{3}\) = 1.732)
Solution:

Ex 6.4 Class 10 Samacheer Question 2.
A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill. (\(\sqrt{3}\) = 1.732)
Solution:


∴ The height of the hill = 120 + 40 = 160 m
The distance of the hill from the ship is AC = x = \(40 \sqrt{3}\) m = 69.28 m

10th Maths Exercise 6.4samacheer Kalvi Question 3.
If the angle of elevation of a cloud from a point ‘h’ metres above a lake is θ₁ and the angle of depression of its reflection in the lake is θ₂. Prove that the height that the cloud is located from the ground is \(\frac{h\left(\tan \theta_{1}+\tan \theta_{2}\right)}{\tan \theta_{2}-\tan \theta_{1}}\)
Solution:

Let AB be the surface of the lake and let p be the point of observation such that AP = h meters.
Let C be the position of the cloud and C’ be its reflection in the lake. Then CB = C’B.
Let PM be ⊥^r from P on CB
Then ∠CPM = θ₁, and ∠MPC = θ₂
Let CM = x.
Then CB = CM + MB = CM + PA
= x + h



Hence proved

10th Maths Exercise 6.4 Question 4.
The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30°. If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms.
Solution:


Since 150m > 120m, yes the height of the above mentioned tower meet the radiation norms.

10th Maths Trigonometry Exercise 6.4 Question 5.
The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 600 and 30° respectively. Find
(i) The height of the lamp post.
(ii) The difference between height of the lamp post and the apartment.
(iii) The distance between the lamp post and the apartment. (\(\sqrt{3}\) = 1.732)
Solution:


= 114.312 m

10th Maths Exercise 6.3 Samacheer Kalvi Question 6.
Three villagers A, B and C can see each T other across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30° . Calculate :

(i) the vertical height between A and B.
(ii) the vertical height between B and C. (tan20° = 0.3640, \(\sqrt{3}\) = 1.732)
Solution:

Students can Download Computer Applications Chapter 6 Word Processor Basics (OpenOffice Writer) Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Applications Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Applications Solutions Chapter 6 Word Processor Basics (OpenOffice Writer)

Samacheer Kalvi 11th Computer Applications Word Processor Basics (OpenOffice Writer) Text Book Back Questions and Answers

I. Choose The Correct Answer

11th Computer Application Chapter 6 Book Back Answers Question 1.
Which is the opening screen of OpenOffice?
(a) Star desktop
(b) Star Center
(c) Star Screen
(d) Star window
Answer:
(b) Star Center

Samacheer Kalvi Guru 11th Computer Application Question 2.
Which option allows you to assign text, tables, graphics and other items to a key or key combination?
(a) Automatic
(b) Autoformat
(c) Auto Text
(d) Autographies
Answer:
(c) Auto Text

Samacheer Kalvi 11th Computer Application Question 3.
Which menu contains the Numbering option?
(a) File
(b) Edit
(c) Tools
(d) Format
Answer:
(c) Tools

Samacheer Kalvi Computer Application Question 4.
There are types of hyperlinks displayed on the left pane of the dialog box?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

11th Computer Application Samacheer Kalvi Question 5.
………………….. menu contains the Numbering option.
(a) File
(b) Edit
(c) Tools
(d) Format
Answer:

11th Computer Applications Samacheer Kalvi Question 6.
The option inserts an image that overlaps the text which as a result will be hidden.
(a) Font style
(b) Wrap Through
(c) Font group
(d) Shadow
Answer:
(b) Wrap Through

Computer Application Samacheer Kalvi Question 7.
……………………….. is the default setting, with which the table occupies the entire width of the text area.
(a) Default
(b) Right
(c) Left
(d) Automatic
Answer:
(a) Default

Question 8.
……………………… icon on the drawing toolbar gets you a text box.
(a) Text icon
(b) Text box icon
(c) Draw icon
(d) Draw box icon
Answer:
(a) Text icon

Question 9.
……………………. allows you to assign text, tables, graphics and other items to a key or key combination.
(a) Auto
(b) Automatic
(c) AutoText
(d) Default
Answer:
(c) AutoText

Question 10.
……………………… is the shortcut key for finding and replacing text in a document.
(a) Ctrl + F
(b) Ctrl + F4
(c) Ctrl + F5
(d) Ctrl + F7
Answer:
(a) Ctrl + F

II. Short Answers

Question 1.
How do you insert pictures in to your document?
Answer:
Open office Writer has the ability to insert and edit images in a more simple way.

1. Place the insertion pointer where you want the image to appear.
2. Select Insert → Picture From file.
3. The insert picture dialog box appears where the picture gallery opens from which the desired picture can be selected.
4. Click on the Open button.
5. The selected picture is inserted into the document.

Question 2.
What are Hyperlinks?
Answer:
There are four types of hyperlinks displayed on the left pane of the dialog box:

1. Internet: A web address, normally starting with http://
2. Mail and News: For example an email address.
3. Document: A hyperlink that points to another document or to another place in the presentation.
4. New document: The hyperlink creates a new document.

Question 3.
What is auto text in writer?
Answer:
AutoText allows you to assign text, tables, graphics and other items to a key or key combination. For example, rather than typing “Tamil Nadu” every time you use that phrase, you might just type “tn” and press F3. You can also save a formatted Tip as AutoText and then insert a copy by typing “tip” and pressing F3.

Question 4.
How do you merge cells in a table?
Answer:
To merge a group of cells:

1. Select the cells to merge.
2. Right click and choose Cell → Merge or choose Table → Merge Cells from the menu bar.

Question 4.
What is the use of Word Art in writer?
Answer:
Word Art is a text modifying feature in the open office writer. It includes effects such as shadows, outlines, colors, gradients, and 3D effects that can be added to a word or phrase. Word Art helps to apply Special effects and change the appearance of the text to make it more presentable and attractive.

III. Explain in Brief

Question 1.
Write about the drawing toolbar?
Answer:
Open office writer uses the drawing tools, to create various shapes by using the Drawing toolbar.
To use th e drawing tools repeatedly, you can move this toolbar to a convenient place on the window. The drawing toolbar can be obtained by clicking View → Toolbars → Drawing.

Question 2.
What are comments, footnotes and endnotes in writer?
Answer:
Footnotes appear at the bottom of the page on which they are referenced. Endnotes are collected at the end of a document. To work effectively with footnotes and endnotes.
→ Insert footnotes.
→ Define formate of footnotes.
→ Define the location of footnotes on the page.
To insert a footnote on an endnote, then select Insert → Footnote / Endnote from the Menu bar or click directly by icon Insert footnote or Insert endnote.

Question 3.
How do you insert cells, rows and columns?
Answer:
To insert rows or columns inside a table:

• Place the insertion pointer in the row or in the column where you would like to add new rows or columns and right-click.
• Choose Row → Insert – to insert a row or Column → Insert – to insert a column. A dialog box will appear, from which you can select the number of rows or columns to insert. You can also set the position of the new rows or columns to Before or After.
• Click OK to close the dialog box.

Question 4.
Write the steps to insert line numbers in writer?
Answer:
Line numbering puts automatic line numbers in the margin. The use of line numbers make it easier to identify specific locations in the document. The most common method of assigning numbers to lines is to assign every line a unique number, starting at 1 for the first line, and incrementing by 1 for each successive line. To apply line numbering:
Click Tools → Line Numbering and select the Show numbering option in the top left comer. Then click OK.

Question 5.
What is a Watermark?
Answer:
Watermarks are images or text displayed in transparency across the text. To create a watermark, it is best to use a Font works object wrapped in the background. The Wrap Through option inserts an image that overlaps the text which as a result will be hidden. To make the text appear, change the transparency of the picture, although the words under the image become visible, they may be difficult to read and will appear lighter than the rest of the text.

IV. Explain in detail

Question 1.
What are the different methods to change margin in writer?
Answer:
Page margins are the white space around the top, bottom, left, and right of your document. Margins let Writer know where to start placing the text at the top of a document, when to move on to the next page at the bottom, where to start typing text on the left side, and where to stop • and move to the next line on the right.

Changing or setting page margins in open office writer can be done in two ways:

1. Using the Page rulers-quick and easy, but does not have precise values.
2. Using the Page Style dialog box- can specify precise values for the margins.

Changing page margins – using Ruler:

1. The shaded sections of the rulers are the margins.
2. Hold the mouse pointer over the line between the gray and white sections.
3. The mouse pointer turns into a double-headed arrow.
4. Hold down the left mouse button and drag the mouse to move the margin and release it at the required point.
5. The new margin is set.

Using the Page Style dialog box:
To change margins using the Page Style dialog box.

1. Right-click anywhere on the page and select Page from the popup menu or select page tab of page style dialog box.
2. In the Margins boxes, specify the values for left, right, top and bottom margins.
3. Click on ok button.

Question 2.
What are Header and Footer? How do you insert page numbers?
Answer:
Header:
It is a section of the document that appears in the top margin, Which displays the title or chapter name, author name of a document.

Footer:
It is a section of the document that appears in the bottom margin of the page which displays the page number, date, time etc., which gets displayed on all the pages automatically.

Inserting Header and Footer:

1. Select from the main menu Insert → Header → Default
2. The header text area is separated from the normal text area.
3. Enter the text that is to be repeated in all pages or Select Insert → Fields → Title.

Similarly to insert a Footer, the steps are as given below:

1. Select from the main menu Insert → Footer → Default
2. Place the insertion pointer in the footer part of the page.
3. Select Insert → Fields → Date to insert date in all the pages.
4. Once the headers and footers are given in the first page, the same text appears in all the pages.

Inserting and Formatting page numbers:
Once the Header / footer area is enabled, the page numbers can be inserted by performing the following steps:

1. position the cursor where you want to insert the number
2. choose Insert → Fields → Page Number
3. The page number appears with a gray background

Question 3.
Write about the drawing toolbar?
Answer:
Open office writer uses the drawing tools, to create various shapes by using the Drawing toolbar. To use the drawing tools repeatedly, you can move this toolbar to a convenient place on the window. The drawing toolbar can be obtained by clicking View → Toolbars → Drawing.

To use a drawing tool, the steps are as given below:

1. Click in the document where you want the drawing to be anchored.
2. Select the tool from the Drawing toolbar. The mouse pointer changes to a drawing- functions pointer.
3. Move the cross-hair pointer to the place in the document where you want the graphic to appear and then click-and-drag to create the drawing object.
4. Release the mouse button. The selected drawing function remains active, so you can draw another object of the same type.
5. To cancel the selected drawing function, press the Esc key or click on the Select icon on the Drawing toolbar.
6. You can now change the properties (fill color, line type and weight, anchoring, and others) of the drawing object using either the Drawing Object Properties toolbar or the choices in the dialog box.

Question 4.
Explain Page formatting in writer?
Answer:

The most important thing in a word processor is how to format the page with elements such as margins, numbering, page layout, headers and footers. Formatting your pages makes them look more attractive and makes them easier to read.

Changing Page Size:
The default page size in writer is 8.5 × 11”, the same as that of a standard A4 printing paper. However, for different types of documents, you may need to change the page size. To change the paper size:

1. Select the page whose page size is to be changed
2. Select Format → Page, the page style dialog box.
3. Select Page Tab.
4. In the paper format group, select the format like A4, legal ….
5. Or the width and height option can be used to set the page size.

Samacheer Kalvi 11th Computer Applications Solutions Word Processor Basics (OpenOffice Writer) Additional Questions and Answers

I. Choose The Correct Answer

Question 1.
Which is the leading open-source office software to works on all common computers and available in many languages.
(a) OpenOffice
(b) StarOffice
(c) MSOffice
(d) None of these
Answer:
(a) OpenOffice

Question 2.
Which company has developed by OpenOffice writer?
(a) Microsoft
(b) Kingsoft
(c) Apache
(d) Sun micro systems
Answer:
(c) Apache

Question 3.
The open screen of the OpenOffice is known as:
(a) starcenter
(b) desktop
(c) window
(d) screen
Answer:
(a) starcenter

Question 4.
The keyboard shortcut key can be used to open a new text document in OpenOffice:
(a) Ctrl + O
(b) Ctrl + A
(c) Ctrl + N
(d) Ctrl + S
Answer:
(c) Ctrl + N

Question 5.
The bar, that displayed at the top most part of the window:
(a) Toolbar
(b) Title bar
(c) Menu bar
(d) Scroll bar
Answer:
(b) Title bar

Question 6.
Which is called a flashing vertical bar appears at the beginning of the screen?
(a) Insertion pointer
(b) Cursor
(c) Horizontal ruler
(d) Vertical ruler
Answer:
(a) Insertion pointer

Question 7.
The keyboard shortcut key can be used to close a document in OpenOffice:
(a) Ctrl + N
(b) Ctrl + A
(c) Ctrl + W
(d) Ctrl + Q
Answer:
(c) Ctrl + W

Question 8.
The keyboard shortcut key can be used to open an existing document in OpenOffice:
(a) Ctrl + 0
(b) Ctrl + A
(c) Ctrl + N
(d) Ctrl + S
Answer:
(a) Ctrl + 0

Question 9.
Which key is used to delete the character left of the insertion pointer?
(a) Backspace key
(b) Delete key
(c) Shift key
(d) Home key
Answer:
(a) Backspace key

Question 10.
Which key is used to delete the character right of the insertion pointer?
(a) Backspace key
(b) Delete key
(c) Shift key
(d) Home key
Answer:
(b) Delete key

Question 11.
The key is used toggle between Insertion mode and overwritten mode:
(a) Insert key
(b) Home key
(c) Shift key
(d) Ctrl key
Answer:
(a) Insert key

Question 12.
The OpenOffice writer provides help window, which function key is used to getting the help window?
(a) F1
(b) F2
(c) F3
(d) F4
Answer:
(a) F1

Question 13.
Which option is used to removed by selecting highlighted text?
(a) Cancel
(b) No fill
(c) Change
(d) None of these
Answer:
(b) No fill

Question 14.
For selecting whole document which shortcut key is used?
(a) Ctrl + A
(b) Ctrl + W
(c) Ctrl + M
(d) Ctrl + S
Answer:
(a) Ctrl + A

Question 15.
Which shortcut key is used to clear the formatting document?
(a) Ctrl + A
(b) Ctrl + W
(c) Ctrl + M
(d) Ctrl + A
Answer:
(c) Ctrl + M

Question 16.
A first-line indentation indents value is:
(a) positive
(b) negative
(c) zero
(d) none of these
Answer:
(a) positive

Question 17.
The hanging indent value can be entered as:
(a) positive
(b) negative
(c) zero
(d) none of these
Answer:
(b) negative

Question 18.
In page orientation, the height of the document is more than the width is called:
(a) portrait
(b) landscape
(c) legal
(d) A4 size
Answer:
(a) portrait

Question 19.
In page orientation-the width of the document is more than height is called:
(a) portrait
(b) landscape
(c) legal
(d) A4 size
Answer:
(b) landscape

Question 20.
Which shortcut key is used to find the replace the text?
(a) Ctrl + F
(b) Ctrl + B
(c) Ctrl + P
(d) Ctrl + S
Answer:
(a) Ctrl + F

Question 21.
Which menu option is choosed to open AutoCorrect dialog box?
(a) Tools → Change
(b) Tools → AutoCorrect
(c) Tools → Add
(d) Tools → Ignore
Answer:
(b) Tools → AutoCorrect

Question 22.
Which command is used to deselect the size of the picture?
(a) Crop
(b) Format
(c) Cut
(d) Delete
Answer:
(a) Crop

Question 23.
Which option is used to inserts an image that overlaps the text which result will be hidden?
(a) Drawing toolbar
(b) Wrap through
(c) Format
(d) Transparency
Answer:
(b) Wrap through

Question 24.
Which key is assign the AutoText shortcut?
(a) Ctrl + FI
(b) Ctrl + F2
(c) Ctrl + F3
(d) Ctrl + F4
Answer:
(c) Ctrl + F3

Question 25.
A specified number of rows and columns are called:
(a) Table
(b) Cell
(c) Column
(d) Row
Answer:
(a) Table

Question 26.
Which combination key is used to insert table dialog box?
(a) Ctrl + F10
(b) Ctrl + F11
(c) Ctrl + F11
(d) Ctrl + F12
Answer:
(d) Ctrl + F12

Question 27.
Multiple copies of a document to send to a list of different recipients is called:
(a) Mail merge
(b) Copy
(c) address book
(d) all of the above
Answer:
(a) Mail merge

Question 28.
Which opition can choose to print all colour text and graphics as gray scale in OpenOffice?
(a) Tools > options > OpenOffice.org > print
(b) Tools > options > OpenOffice.org > ok
(c) Tools > options > OpenOffice.org > Next
(d) None of the above
Answer:
(a) Tools > options > OpenOffice.org > print

Question 29.
When using Tab key in OpenOffice writer by default moving insertion point is:
(a) 1 inch
(b) ^1/4 inch
(c) ^1/2 inch
(d) 2 inch
Answer:
(c) ^1/2 inch

Question 30.
Which rules you can add to decorate any or all the four sides of a paragraph?
(a) Highlighted
(b) Borders
(c) Shading
(d) Bold
Answer:
(b) Borders

Question 31.
Which can you use as background for enhancing the appearance of a text or paragraph?
(a) Highlighted
(b) Borders
(c) Shading
(d) Bold
Answer:
(c) Shading

Question 32.
The section’of the document that appears in the top margin:
(a) header
(b) footer
(c) margin
(d) window
Answer:
(a) header

Question 33.
The section of the document that appears in the bottom margin:
(a) header
(b) footer
(c) margin
(d) window
Answer:
(b) footer

Question 34.
Which is the shortcut key to insert a comment within the text?
(a) Ctrl + Alt + C
(b) Ctrl + Alt + I
(c) Ctrl + Alt + V
(d) Ctrl + Alt + C
Answer:
(a) Ctrl + Alt + C

Question 35.
Which shortcut key is used to undo last action?
(a) Ctrl + C
(b) Ctrl + F
(c) Ctrl + Z
(d) Ctrl + E
Answer:
(c) Ctrl + Z

Question 36.
Which shortcut key is used to center aligned text?
(a) Ctrl + C
(b) Gtrl + F
(c) Ctrl + Z
(d) Ctrl + E
Answer:
(d) Ctrl + E

Question 37.
Which combination of shortcut is used subscript a number?
(a) Ctrl + Shift + B
(b) Ctrl + Shift + V
(c) Ctrl + Shift + P
(d) Ctrl + Shift + S
Answer:
(a) Ctrl + Shift + B

Question 38.
Which combination of shortcut key is used superscript a number?
(a) Ctrl + Shift + B
(b) Ctrl + Shift + V
(c) Ctrl + Shift + P
(d) Ctrl + Shift + S
Answer:
(c) Ctrl + Shift + P

Question 39.
Which shortcut key is used to inserting a new paragraph without numbering?
(a) Alt + Shift
(b) Alt + Enter
(c) Alt + Tab
(d) Ctrl + Enter
Answer:
(b) Alt + Enter

Question 40.
Which shortcut key is used to apply default paragraph style?
(a) Ctrl + 0
(b) Ctrl + 1
(c) Ctrl + 2
(d) Ctrl + 3
Answer:
(a) Ctrl + 0

Question 41.
……………………. is the pictorial representation of data.
(a) Object
(b) Charts
(c) Shapes
(d) Colour
Answer:
(b) Charts

II. Short Answers

Question 1.
What is meant by word processing?
Answer:
Word processor is a computer software to create, edit, manipulate, transmit, store and retrieve a text document. The above said activities are called as “Word Processing”. In other words, Word processing is an activity carried out by a computer with suitable software to create, edit, manipulate, transmit, store and retrieve text documents.

Question 2.
What are the familiar open source word processing software?
Answer:
OpenOffice Writer, LibreOffice Writer, Abiword etc.

Question 3.
What are the familiar Tamil word processor?
Answer:
Tamil OpenOffice writer, Tamil Libre, Kamban 3.0, Mentamizh 2017 there are familiar word processors exclusively for Tamil Language.

Question 4.
How to open the OpenOffice Writer?
Answer:
A new OpenOffice Writer document can be created by various methods. From windows, select
Start → All Programs → OpenOffice → OpenOffice Writer.
(or)
From Star Center (Welcome Screen): Double-click on “OpenOffice” icon the desktop.

Question 5.
What is meant by star center?
Answer:
In OpenOffice, open screen is called as “Star Center”. Writer is one of the components of OpenOffice. So, it may be invoked from the “Star Center” by simply clicking on the “Text Document” icon.

Question 6.
How can you close the OpenOffice application?
Answer:
This button is called as “Close” button,when you click this button, the application is closed and OpenOffice returns back to the desktop. So, the red colored close button may be called as “Exit” or “Quit”.

There is another (‘x’) mark on the right most comer of the menu bar. This is actually used to close your document. When you click this (‘x’) mark, your document will be closed, OpenOffice will be still opened.

Question 7.
What are the types of toolbar available in the menu bar?
Answer:
Under the menu bar, there are two toolbars available by default.
They are:

1. Standard Tool bar
2. Formatting Toolbar.

Question 8.
What are the types of ruler scale?
Answer:
The ruler is a scale below the formatting toolbar which shows the margins. There are two set of rulers

1. Horizontal ruler and
2. Vertical ruler. Horizontal rules is used to set left and right margins of a page and vertical ruler for top and button setting.

Question 9.
What is called insertion pointer?
Answer:
The work space is the blank area which is used to type the content of the file. A flashing vertical bar appears at the beginning of the screen which is called as “Insertion pointer”.

Question 10.
Which is known as word wrap?
Answer:
When the text reaches the end of the line, the word is automatically wrapped to the next line. This feature in any word processor is known as “Word Wrap”.

Question 11.
How to save the document?
Answer:
One can select the drive and folder where the file will be stored. To save a document for a first time,the following steps are used:
Click File → Save → (or) File → Save As (or) Ctrl + S.

Question 12.
How to set a password to protect your document while saving a file?
Answer:
In OpenOffice writer, a document can be protected with a password. You can set a password to protect your document while saving a file. To save a file with password, click on “Save with Password”check box and then click “Save” button. Immediately it shows “Set Password” dialog box. Enter a password in “Enter Password to open” text box and retype the same password in “Confirm Password” box for confirmation. Finally click ‘OK’ button.

Question 13.
What is the use of Insert key?
Answer:
To insert a text in between if something is left out, the insertion can be made by taking the insertion pointer to the current location and Press the Insert Key the newly typed text is inserted, so that the existing text moves to the right. Press the Insert Key again, the text is over written on the existing text. This is called ‘Type over mode’. You can toggle between the insert mode and type over mode by pressing the‘Insert key’.

Question 14.
What is the purpose of using paste special option?
Answer:
When you move or copy information, the paste option is used to send the information as a whole. But, to move or copy only some aspects of the data, like only its formatting or only value, the Paste Special option is used. To use the paste special, select the text and apply move or copy, then at the destination location,
Click Edit → Paste Special (or) press Ctrl + Shift + V, (or) Alt + E + S the Paste Special dialog box opens.

Question 15.
What is difference between paste and paste special?
Answer:
Paste:
Paste is a feature that lets a user cut or copy items from Cells and transfer them to another completely.

Paste Special:
Paste Special allows the items being transferred to be formatted in several different ways. Paste Special is a feature found in software like Microsoft Word, Microsoft Excel and Open Office.

Question 16.
What is meant by Text formatting?
Answer:
A font is a set of characters in a particular style. Changing the default appearance of the text like changing the font type, size, color, style etc., are called as Text formatting.

Question 17.
What are the short cut commands using to make the text Formatting?
Answer:

Question 18.
What is the use of change case option of format menu?
Answer:
Normally any text can be typed in upper or lower case. The text can be changed to different cases like:

1. Uppercase (Capital letters)
2. Toggle case (reverse case),
3. Sentence case (first letter of each sentence Capital),
4. Capitalize every word (first letter of each word capital),
5. Lower case. (Small letters)
   1. This can be done by:
   2. Select the text to change case
   3. Select Format → Change case.

Question 19.
What is meant by highlighting?
Answer:
Highlighting is used to draw attention to important information in a text. Highlighting is beneficial because it first asks the reader to pick out the important parts, and then gives an effective way to review that information later.
Highlighting can be applied by selecting the text and click Highlighting icon.

Question 20.
What is meant by hanging indent?
Answer:
This is a special kind of indent where the first line of the paragraph alone hangs outside leaving the rest of the text. To apply Hanging indent, a negative value is given in the “first line” option of the paragraph dialog box.

Question 21.
What are the types different page orientation?
Answer:
Page orientation refers to hgw the document will be displayed on screen and printed. There are two different orientations:
Landscape:
The width of the document is more than the height. This is best suited for displaying professional photos, invitations, albums, tables etc.,

Portrait:
This is the most common orientation. Here, the height of the document is more than the width. Normally books, newspapers will be displayed in this format.

Question 22.
What is the function of AutoCorrect?
Answer:
Auto Correct function has the facility to correct the common misspellings and typing errors, automatically. For example, “hte” will be changed to “the” which can be done through the menu option, Tools → AutoCorrect.

Question 23.
What is meant by spell check program?
Answer:
Spell check is a software program that corrects spl74568f.dfml d,589f fjidflfefelling errors in ewfword processing, email and documents, Spell check identifies and corrects misspelled words and the grammatical error is displayed with a green wavy line under the wrong word.

Question 24.
What is meant by table?
Answer:
A table is a grid with a specified number of rows and columns. Tables can often be used as an alternative to spreadsheet to organize materials. A well-designed table can help readers understand better what you are trying to convey.

Question 25.
What is meant by formatting table?
Answer:
Formatting a table involves formatting of the table layout, formatting of the table text, adjusting the size of the table, its position on the page, adding or removing rows or columns, merging and splitting cells, changing borders and the background.

III. Explain in Brief

Question 1.
What are the common software packages in OpenOffice?
Answer:
OpenOffice is a productive office suite with a collection of different software packages such as
OpenOffice Writer – Word Processor to create text documents
OpenOffice Calc – Spreadsheet to create worksheets
OpenOffice Base – Database
OpenOffice Impress – Presentation software
OpenOffice Draw – Drawing Software
OpenOffice Formula – Create formula and equations

Question 2.
What are the important features which provides OpenOffice writer?
Answer:

1. Templates and styles.
2. Page layout methods, including frames, columns, and tables.
3. Embedding or linking of graphics, spreadsheets, and other objects.
4. Built-in drawing tools.
5. Master documents to group a collection of documents into a single document.
6. Change tracking during revisions.
7. Database integration, including a bibliography database.
8. Export to PDF, including bookmarks.

Question 3.
Draw a OpenOffice writer window? With mark the important components?
Answer:
OpenOffice Writer window such as Title bar, Menu bar, Standard Toolbar, Formatting Toolbar, Ruler, Work space and Status bar. The components of a OpenOffice writer window are explained.

Question 4.
How to select a text continuous and discontinuous?
Answer:
In any word processor, the text has to be selected for performing any operation like copying, moving, formatting etc. This text selection can be done by two methods;

1. Selecting the continuous text.
2. Selecting the non continuous text.

Continuous text:
To select the text continuously take the insertion pointer to the starting of the text.

Selecting the continuous text:
To select the text continously take the insertion pointer to the starting of the text.

1. Hold the SHIFT key and drag the mouse across until the required text is selected and then release the SHIFT key.
2. The select text can be used for any operation.

Selecting the non continuous text:
To select the text not continuously, take the insertion pointer to the starting of the text,

1. Hold the CTRL key and drag across it till the required text is selected and release the CTRL key.
2. The required text is selected for any operation.

Question 5.
How to use the help system in open writer?
Answer:
Open Office Writer provides several forms of help. By pressing FI or select Help from the menu bar the help window appears.

1. To activate tool tips, extended tips, and the help Agent, click Tools → Options → OpenOffice.
2. For a more detailed explanation, select Help → What’s This? and hold the mouse pointer over the icon where you want more help with.

Question 6.
How can you change the line spacing adjustment?
Answer:
Line spacing determines the amount of vertical space between lines of text in a paragraph. By default, the lines are single-spaced, that is the spacing accommodates the largest font in that line, plus a small amount of extra space. In Open Office, setting line spacing is quite easy through the context menu, select the line or word or phrase, right-click → line spacing, select the type single, 1.5 or double. There are seven different types of line spacing.

1. Select the entire document by Edit → Select All.
2. Format → paragraph.
3. The paragraph dialog box appears, click Indents and spacing tab.
4. In the line spacing option, select the type and click ok button.

Question 7.
What is a indent? What are the types of Indent?
Answer:
Indent is the distance from the left and right margin of a paragraph. It is used to improve the efficiency and readability of the paragraph and makes the paragraph look more attractive.
There are four types of indents:

1. Left Indent
2. Right Indent
3. First Line Indent and
4. Hanging Indent.

Question 8.
How to change the page colour and borders?
Answer:
Changing the page color is not quite common. To do so, in the page style dialog box, in the Background option, click on color and select the “color” from the color palette or select “graphic” to apply an image as a page background.

Borders can be applied to an entire document, an entire page, paragraph, or just to certain sections of the document. From the page style dialog box, select the Border tab, the user defined area helps to define the area of borders, the line style of borders, color of borders can be selected.

Question 9.
Howto insert special characters and equations?
Answer:
Many symbols which are used in a mathematical equation like alpha (α), beta (β), pi (π) etc.,, are not available on the standard keyboard. However, representing these characters are very much essentia in mathematical equations. To insert such characters, the procedure given below
is folio-wed:

1. Place the cursor in your document where you want the character to appear.
2. Click on the Insert → Special character.
3. The Special character dialog box appears from which the desired symbol can be selected by clicking on the character.
4. As you select each character, it is shown on the lower right, along with the numerical code for that character.
5. If you do not find a particular special character you want, try changing the font selection.
6. Click the Ok button and the character is inserted at the current location.

Question 10.
How to inserted the chart in the worksheet?
Answer:
Charts are pictorial representations of data. The data can be either typed in cells in writer or it can be inserted.
Click inside the Writer table.

1. Choose Insert → Object → Chart, You see a chart preview and the Chart Wizard.
2. The chart toolbar at the top of the page which contains the formatting tools.
3. The insert menu shows the various attributes of the chart like Title, Legends, Axis, grid, label etc.,

Question 11.
What are the steps needed to create watermark?
Answer:

1. Insert the image or text of your choice
2. Anchor the image to the page and
3. Select the wrap through option form the Format Wrap menu or right-clicking on the image and selecting Wrap
4. Wrap Through from the pop-up menu
5. Move the image into the desired position
6. The Picture toolbar should be displayed when the image is selected
7. Change the transperency to a suitable value so that the text can be read.

Question 12.
What are the steps follows to assign Auto Text?
Answer:
To assign AutoText shortcut to some text, the steps arc as follows:

1. Type the text into your document.
2. Select the text so that it is highlighted.
3. Select Edit → AutText (or press Ctrl+F3).
4. Enter a name for your shortcut. Writer will suggest a one-letter shortcut, which you can change.
5. Click the AùtoText button on the right and select New (text only) from the menu.
6. Click Close to return to your document.

Question 13.
How can you merging and splitting cells?
Answer:
To merge a group of cells:

1. Select the cells to merge.
2. Right click and choose Cell → Merge or choose Table → Merge Cells from the menu bar.
   1. To split a cell:
   2. Place the insertion pointer inside the cell.
   3. Right click and choose Cell → Split, or choose Table → Split Cells from the menu bar.
   4. Select the direction of the split, horizontally (for rows), or vertically (for columns), as well as the total number of cells to create.

Question 14.
What are the basic types of graphics in OpenOffice writer?
Answer:

1. Image files, including photos, drawings, scanned images, and others.
2. Diagrams created using OOo’s drawing tools.
3. Charts created using OOo’s Chart facility.

IV. Explain in detail

Question 1.
Explain the menu bar in OpenOffice writer?
Answer:
Menu Bar:
The menu bar is just below the title bar which comprises of various menus consisting of various options.

File:
The File menu contains the commands of all file management tasks like, Create a new file, Open an existing file, Close the current file, Save a file, Save As another file, print file, Export file etc.,

Edit:
The Edit menu contains the editing commands like, cut, copy, paste, Undo, Redo etc.,

View:
The View menu contains the commands which are used to modify the environment of write like display of toolbars, web layout, print layout, navigator etc.,

Insert:
The Insert menu contains commands for inserting various elements such as pictures, tables, charts, comments, headers, footers, special characters, cross reference etc.,

Format:
The Format menu contains the commands of various text and page formatting features like page size, layout, font characteristics, bullets and numbering etc.,

Tables:
The Table menu contains various tools to manage and manipulate tables such as create tajfie, insert rows, insert columns, s’plit cells, merge cells etc.,

Tools:
The Tools menu contains various tools and functions such as spell check, macros, mail merge, end notes/footnotes etc.,

Window:
The window menu shows display options such as New Window, Close Windows, Split and Freeze. .

Help:
The Help menu lists out the inbuilt help features available with OpenOffice.

Question 2.
What are the different shortcut keys of moving within a document?
Answer:
There are different ways of moving within a document. There are many shortcut keys given in Table which are used to move easily within a document.

Question 3.
Explain the various editing operations using a document?
Answer:
The various editing operations like cut, copy and paste.
Moving: To move a text from one location to another.

1. Select the text to he moved.
2. Click Ctrl + X or Cut Icon or Edit → Cut.
3. The text is removed from the source location and placed in the clipboard.
4. Take the insertion pointer to the new location to be moved and
5. Click Ctrl + V or Paste Icon or Edit → Paste.

The required text is moved to the required location.
Copying: To copy a text from one location to another.

1. Select the text to be copied .
2. Click Ctrl + C or Copy Icon or Edit → Copy
3. A duplicate copy of the text is made and send to the clipboard
4. Take the insertion pointer to the new location to be copied and
5. Click Ctrl + V or Paste Icon or Edit → Paste

The required text is copied to the required location.
The Editing shortcut keys are:

Question 4.
What are the various alignment in formatting paragraph?
Answer:
A paragraph is any text that ends with a hard return. A hard return is accomplished anytime you press the Enter key. Paragraph Alignment or justification refers to the way in which the lines of a paragraph are aligned. Paragraph alignment lets you control the appearance of individual paragraphs.

There are four types of alignment available in Open office Writer – left-alignment, Right-alignment, Center-alignment, and Justify-alignment. The paragraph formatting can also be done by icons using the formatting toolbar.

Question 5.
How can you use Bullets and Numbering in a document?
Answer:
Bullets:
This is a paragraph level attribute that applies a bullet character to the start of the paragraph. In bulleted lists, each paragraph begins with a bullet character. This is suitable when the text has to be presented as a list of items preceded by a bullet symbol and no sequence has to be followed. Bullets are quickly created by clicking on the bullet icon.

Numbering:
This attribute applies a numeral to the start of the paragraph. Numbering is more suitable when the text has to be presented as a sequence. In numbered list, each paragraph begins with an expression that includes a number or letter and a separator such as a period or parenthesis. The numbers in a numbered list are updated automatically when you add or remove paragraphs in the list. Numbering is quickly created by clicking on the numbering icon.

Style of Bullets and Numbering:
The default type of bullet is (•) and the default type of numbering is (1,2, 3 ……………….). The style of bullets and numbering can be changed by applying the following steps:

1. Select the text to be bulleted.
2. Format → Bullets and Numbering.
3. Select Bullets Tab.
4. The bullets and Numbering dialog box where different styles of bullets are displayed.
5. Click on the required style.
6. Click OK button.
7. The selected text is bulleted

To apply Numbering:

1. Select the text to be numbered.
2. Format → Bullets and Numbering.
3. Select Numbering Type Tab.
4. The Bullets and Numbering dialog box where different styles of numbering are displayed.
5. Click on the particular style.
6. Click Ok button.
7. The selected text is numbered.

Question 6.
How can you find a word in Find and Replace text?
Answer:
OpenOffice Writer has a Find and Replace feature that helps to locate for a text inside a document and replace it with another word. In addition to finding and replacing words and phrases, you can also use wildcards and regular expressions to perform advanced search. To search a word

1. Click Edit → Find & Replace (or) Ctrl + F
2. The Find & Replace dialog box appears.

Steps to find & replace a text

• Type the text you want to find in the Search for box.
  Eg: To search a word “Bombay” in a document and replace with “Mumbai”, enter the word “Bombay” in the Search for box.
• To replace the text with different text, type the new text in the Replace with box.
  Enter the word “ Mumbai” in the Replace with box and Click Find button, to start the search, the found word is highlighted and the first occurance of “Bombay” is highlighted.
  
• To replace text, click Replace button. The highlighted word is replaced with the word given in the Replace with box.
• Click Find All, Writer selects all instances of the search text in the document. All occurrences of Bombay are highlighted.
• Click Replace All button, Writer replaces all matches. This will replace all occurances of “Bombay” with “Mumbai”.
• Enable Match case to perform the search case sensitively so that uppercase and lower cases are distinguished separately.
• Enable Whole Words only to make the search more specific to words used separately alone.

Question 7.
Discuss the Auto spell check?
Answer:
Auto spell check option checks each word as it is typed and displays a wavy red line under any misspelled words. Once the word is corrected, the red wavy line disappears. This can be done through clicking the icon.

To perform a separate spelling check on the document (or a text selection) click the Spelling and Grammar button. This checks the document or selection and opens the Spelling dialog box if any misspelled words are found, This can be achieved by clicking the icon

Here are some more features of the spelling tool:

• Right-click on a word with a wavy underline, to open a powerful context menu. Correct words can be selected from the suggested words on the menu. The selection will replace the misspelled word with correct word. Other menu options are discussed below.
• The dictionary language can be changed (For example, Spanish, French, or German) from the Spelling dialog box.
• The new words can be added to a dictionary. Click Add in the Spelling dialog box and pick the dictionary to add it to.

Question 8.
How can you using different techniques to insert tables?
Answer:
Different techniques to insert tables:
To insert a new table, position the insertion pointer where you want the table to appear, then use any of the following methods to open the Insert Table dialog box.

There are two methods to create a table:

1. Table Icon:
To insert a table quickly from the standard tool bar:

1. Place the insertion pointer where you want the table to appear.
2. Click the arrow to the right side of the Table icon
3. In the drop down grid, select the number of rows and columns for the table. (iv) The table will appear at the location of your insertion pointer.

2. Insert table dialog box:
To insert a table with more control over the settings and properties, use the Insert Table dialog box.
To open the dialog box: Select Table → Insert → Table (or) Ctrl +F12 or left-click the Table icon. From this dialog box, you can:

1. Select the number of rows and columns of the table.
2. Give a Name to the table to later distinguish it in the Navigator.
3. Select the Heading option to define the first row in the table as the heading.
4. Select the Repeat heading option to repeat the heading row if the table more than one page.
5. Select the Don’t split table option to prevent the table from spanning more than one page.
6. Select the Border option to surround each cell of the table with a border.

Students who are preparing for the Science exam can download this Tamilnadu State Board Solutions for Class 10th Science Chapter 19 from here for free of cost. These Tamilnadu State Board Textbook Solutions PDF cover all 10th Science Origin and Evolution of Life Book Back Questions and Answers.

All these concepts of Chapter 19 Origin and Evolution of Life are explained very conceptually by the subject teachers in Tamilnadu State Board Solutions PDF as per the prescribed Syllabus & guidelines. You can download Samacheer Kalvi 10th Science Book Solutions Chapter 19 Origin and Evolution of Life State Board Pdf for free from the available links. Go ahead and get Tamilnadu State Board Class 10th Science Solutions of Chapter 19 Origin and Evolution of Life.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 19 Origin and Evolution of Life

Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 19th Science Chapter 19 Origin and Evolution of Life Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 19 easily after studying the Tamilnadu State Board Class 19th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 19 Origin and Evolution of Life solutions of Class 19th by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 10th Science Origin and Evolution of Life Textual Evaluation Solved

I. Choose the Correct Answer.

Samacheerkalvi.Guru Science Question 1.
Biogenetic law states that _____.
(a) Ontogeny and phylogeny go together.
(b) Ontogeny recapitulates phylogeny.
(c) Phylogeny recapitulates ontogeny.
(d) There is no relationship between phylogeny and ontogeny.
Answer:
(b) Ontogeny recapitulates phylogeny.

Samacheer Kalvi Guru Science Question 2.
The ‘use and disuse theory’ was proposed by _____.
(a) Charles Darwin
(b) Ernst Haeckel
(c) Jean Baptiste Lamarck
(d) Gregor Mendel.
Answer:
(c) Jean Baptiste Lamarck

10th Science Solution Samacheer Kalvi Question 3.
Paleontologists deal with _____.
(a) Embryological evidences
(b) Fossil evidences
(c) Vestigial organ evidences
(d) All the above.
Answer:
(b) Fossil evidences

10th Science Solutions Samacheer Kalvi Question 4.
The best way of direct dating fossils of recent origin is by _____.
(a) Radio – carbon method
(b) Uranium lead method
(c) Potassium – argon method
(d) Both (a) and (c).
Answer:
(a) Radio – carbon method

Samacheerkalvi.Guru 10th Science Question 5.
The term Ethnobotany was coined by _____.
(a) Khorana
(b) J.W. Harshberger
(c) Ronald Ross
(d) Hugo de Vries.
Answer:
(b) J.W. Harshberger

II. Fill in the blanks.

Question 1.
The characters developed by the animals during their life time, in response to the environmental changes are called _____.
Answer:
Acquired characters.

Question 2.
The degenerated and non-functional organs found in an organism are called _____.
Answer:
Vestigial organ.

Question 3.
The forelimbs of bat and human are examples of ______ organs.
Answer:
Homologous.

Question 4.
The theory of natural selection for evolution was proposed by _____.
Answer:
Charles Darwin.

III. State whether True or False. If false, write the correct statement.

Question 1.
‘The use and disuse theory of organs’ was postulated by Charles Darwin?
Answer:
False.
Correct statement: ‘The use and disuse theory of organs’ was postulated by Jean Baptiste Lamarck.

Question 2.
The homologous organs look similar and perform similar functions but they have different origin and developmental pattern?
Answer:
False.
Correct statement: The homologous organs look dissimilar and perform dissimilar functions, but they have the same origin and developmental pattern.

Question 3.
Birds have evolved from reptiles.
Answer:
True.

IV. Match the following:

Question 1.

Answer:

1. (c) rudimentary tail and thick hair on the body
2. (a) caudal vertebrae and vermiform appendix
3. (d) a wing of a bat and a wing of an insect
4. (b) a forelimb of a cat and a bat’s wing
5. (f) Thiruvakkarai
6. (e) radiocarbon dating.

V. Answer in a word or Sentence

Question 1.
A human hand, a front leg of a cat, a front flipper of a whale and a bat’s wing look dissimilar and adapted for different functions. What is the name given to these organs?
Answer:
Homologous organs.

Question 2.
Which organism is considered to be the fossil bird?
Answer:
Fossil bird Archaeopteryx.

Question 3.
What is the study of fossils called?
Answer:
Palaeontology.

VI. Short Answers Questions

Question 1.
The degenerated wing of a kiwi is an acquired character. Why is it an acquired character?
Answer:
Kiwi does not have the need to fly that is why they do not have wings. The characters developed by the animals during their life in response to environmental changes. The vestigial wings are so small (invisible) under the bristly, hair-like two-branched feathers. So the degenerated wing of a kiwi is an acquired character.

Question 2.
What is the study of fossils called?
Answer:
Palaeontology is called as the study of fossils.

Question 3.
Define Ethnobotany and write its importance.
Answer:
Ethnobotany is the study of a region’s plants and their practical uses through the traditional knowledge of the local culture of people.
Importance of Ethnobotany:

• It provides traditional uses of the plant.
• It gives information about certain unknown and known useful plants.
• The ethnomedicinal data will serve as a useful source of information for the chemists, pharmacologists and practitioners of herbal medicine.
• Tribal communities utilize ethnomedicinal plant parts like bark, stem, roots, leaves, flowers, flower bud, fruits, seeds, oils, resins, dyes and gum for the treatment of diseases like diarrhoea, fever, headache, diabetes, jaundice, snakebites and leprosy, etc.

Question 4.
How can you determine the age of the fossils?
Answer:
The age of fossils is determined by radioactive elements present in it. They may be carbon, uranium, lead or potassium.

VII. Long Answer Questions

Question 1.
Natural selection is a driving force for evolution-How?
Answer:
Darwin published his observations under the name “origin of species”. It elaborates on the theory of natural selection for evolutionary transformation.
The principles of Darwinism tells that natural selection is a driving force for evolution.

1. Overproduction: Living beings have the ability to reproduce and have the capacity to multiply in a geometrical manner.

2. Struggle for existence: Due to overproduction, a geometric ratio of increase in population occurs. The space to live and food available for the organisms remain the same. This creates a competition among the organisms, for food and space, leading to struggle.

• The competition may be among the individuals of the same species (Intraspecific struggle).
• Competition between the organisms of different species living together (Interspecific struggle).
• Natural conditions like extreme heat or cold drought and floods can affect the existence of organisms (Environmental struggle).

3. Variations: Small variations are important for evolution. According to Darwin, favourable variations are useful to the organisms and unfavourable variations are harmful or useless to the organisms.

4. Survival of the fittest or Natural selection: During the struggle for existence, the organisms which can overcome the challenging situation, survive and adapt to the surrounding environment. Organisms, which are unable to face the challenges, are unfit to survive and disappear. The process of selection of organisms with favourable variation is called Natural selection.

5. Origin of species: According to Darwin, new species originates by the gradual accumulation of favourable variations for a number of generations.

Question 2.
How do you differentiate homologous organs from analogous organs?
Answer:

Question 3.
How does fossilization occur in plants?
Answer:
A plant fossil is any preserved part of a plant that has died long back. Fossils may be a prehistoric impression that may be hundred to millions of years old. Majority of the plant fossils are disarticulated parts of plants, it is rare to find plants to be preserved as whole.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
Arun was playing in the garden. Suddenly he saw a dragonfly sitting on a plant. He observed the wings of it. He thought it looked similar to a wing of a crow. Is he correct? Give a reason for your answer.
Answer:
No. He is not correct. Both crow and dragonfly have the same function of flying with wings. But it’s the origin (basic structure) is different. Dragonfly wing is the membranous extension. But the wing of the crow is the modification of forelimb.

Question 2.
Imprints of fossils tell us about evolution- How?
Answer:
Fossil records show that evolution has taken a gradual process from simple to complex organisms. The study of fossils helps us to understand the link of evolution. The origin of modem birds is supported by the evidence of palaeontology.

Question 3.
Octopus, cockroach and frog all have eyes. Can we group these animals together to establish a common evolutionary origin? Justify your answer.
Answer:
Convergent evolution is the process, by which the independently evolved features, may similar to each other, but can arise through different developmental pathways. So the octopus, cockroach and frog all have eyes. The independently evolved eye may similar in each other, but can arise through different developmental pathways.

Samacheer Kalvi 10th Science Origin and Evolution of Life Additional Questions Solved

I. Fill in the blanks.

Question 1.
The history of life has two aspects namely ______ and ______.
Answer:
Origin of life; The evolution of life.

Question 2.
The theory which postulates that life originates from pre – existing life is ______.
Answer:
Biogenesis.

Question 3.
_____ is the gradual change occurring in living organisms over a period of time.
Answer:
Evolution.

Question 4.
Sexual reproduction, which involves meiosis helps in the recombination of ______ during gametic fusion.
Answer:
Genes.

Question 5.
The major concept in astrobiology is the ______.
Answer:
Habitable zone.

Question 6.
The organisms, which live in extreme environmental conditions on Earth are called ______.
Answer:
Extremophiles.

Question 7.
Fossil records show that evolution has taken a gradual process from simple to _______ organisms.
Answer:
Complex.

II. Match the following:

Question 1.

Answer:

1. (e) origin of the universe
2. (d) preserve hard and soft parts
3. (f) fossil bird
4. (a) study of regions and practical use of plants
5. (b) organisms buried and cause depression
6. (c) other names for astrobiology.

Question 2.

Answer:

1. (d) Life from a chemical reaction
2. (i) Biogenetic law
3. (e) Father of palaeontology
4. (a) use and disuse theory
5. (h) Origin of species
6. (b) Mutation theory
7. (c) Father of Indian palaeobotany
8. (f ) Radioactive carbon
9. (g) Ethnobotany.

III. Choose the correct answer.

Question 1.
The theory of idea embodies that life on Earth is a divine creation ______.
(a) chemical reaction
(b) special creation
(c) spontaneous generation
(d) Biogenesis.
Answer:
(b) special creation

Question 2.
Palaeontology deals with the study of ______.
(a) analogous organs
(b) fossils
(c) gradual change
(d) homologous organ.
Answer:
(c) gradual change

Question 3.
The other name for continuous variation is ______.
(a) germinal variation
(b) somatic variation
(c) discontinuous variation
(d) fluctuating variation.
Answer:
(d) fluctuating variation.

Question 4.
It is a branch of palaeontology that deals with the recovery and identification of plant remain of geological past ______.
(a) palaeobotany
(b) Embryology
(c) mutation
(d) palaeontology.
Answer:
(a) palaeobotany

Question 5.
Book ‘Philosophic Zoologique’ published in the year 1809 was written by:
(a) Darwin
(b) Lamarck
(c) Wallace
(d) Mendel
Answer:
(b) Lamarck

IV. Answer the following shortly.

Question 1.
What is Abiogenesis?
Answer:
Abiogenesis or spontaneous generation theory states that life originated spontaneously from lifeless matter. It was believed that fishes originated from mud, frogs from moist soil and insects from decaying matter.

Question 2.
What is the cosmic origin?
Answer:
Cosmic origin of extraterrestrial theories states that life came from outer space. The unit of life called spores (panspermia) were transferred to different planets including Earth.

Question 3.
What is the chemical evolution of life?
Answer:
Chemical evolution of life theory states that life arose by a series of sequential chemical reactions. The first form of life could have come from pre – existing non – living inorganic molecules, which gave rise to the formation of diverse organic molecules, which are transformed into a colloid system to produce life.

Question 4.
Name the evidence of evolution What do they support?
Answer:

• Evidence from morphology and Anatomy.
• Evidence from embryology.
• Evidence from palaeontology.

Evolution is best understood by observing the relationship between the existing organisms and the similarities of the extinct organisms. This evidence supports the concept that all organisms have evolved from common ancestors.

Question 5.
Explain the evidence of evolution from Embryology.
Answer:
The embryos from fish to mammals are similar in their early stages of development. The study of comparative embryology of different animals supports the concept of evolution. The differentiation of their special characters appears in the later stages of development. Ontogeny recapitulates phylogeny. The stages of development of the individual animal repeat the evolutionary history of the entire race of the animal.

Question 6.
Represent a flow chart showing the postulates of Lamarckism?
Answer:
The postulates of Lamarckism:

Question 7.
Explain the evidence of evolution from palaeontology.
Answer:
Palaeontology deals with the study of fossils. The study of fossils helps us to understand the line of evolution of many invertebrates and vertebrates. Fossil records show that evolution has taken a gradual process from simple to complex organisms.

The origin of modem birds is supported by the evidence from palaeontology. Archaeopteryx is the oldest known fossil bird. It is considered to be a connecting link between reptiles and birds. It had wings with feathers, like a bird. It had a long tail, clawed digits and conical teeth like a reptile.

Question 8.
What is evolution?
Answer:
Formation of new species due to changes in specific characters over several generations as a response to natural selection is called evolution. Evolution is the gradual change occurring in living organisms over a period of time.

Question 9.
Represent a flow chart showing the postulates of Darwinism?
Answer:
The postulates of Darwinism:

Question 10.
Explain the types of variation?
Answer:
Variations are the differences found among individuals of the same species and the offspring of the same parent. New species originate by the gradual accumulation of variations. Evolution would not be possible without variation.
Somatic variation and germinal variation are the two types of variations:

1. Somatic variation: The variation which affects the body (somatic) cells of the organisms are called somatic variation, which is not heritable. They occur due to environmental factors.
2. Geminal variation: The variations, which are produced in germ cells and inherited are called germinal variation. They are classified into two types.
   • Continuous variation: Small variations which occur among individuals of a population, and occur by gradual accumulation is called continuous variation. They are also called fluctuating variation, e.g. skin colour, height and weight and colour of the eye, etc.
   • Discontinuous variation: These changes are sudden, which occur in an organism, due to mutations. These large variations are not useful for evolution, e.g. short – legged Ancon sheep and six or more digits (fingers) in human, etc.

Question 11.
Explain the importance of fossils?
Answer:
Importance of fossils:

1. They throw light on phylogeny and evolution of plants.
2. Fossil plants give a historical approach to the plant kingdom.
3. Fossils are useful in the classification of plants.
4. Fossil plants can be used in the field of descriptive and comparative anatomy.

Question 12.
What are Extremophiles?
Answer:
The organisms which live in extreme environmental conditions on Earth are called extremophiles within our own solar system, there are many areas that are different from the Earth. We may find the presence of life similar to extremophile bacteria.

Question 13.
What is Astrobiology or Exobiology? What does it deal?
Answer:
Astrobiology or exobiology is the science which looks for the presence of extraterrestrial in the universe.
Astrobiology deals with the origin, evolution and distribution of life in the universe and to investigate the possibility of living in another world.

The major concept in astrobiology is the habitable zone. Astrobiology explains that any planets can support the existence of life if it fulfils two criteria:

• It must have the right mass to retain an atmosphere.
• It must have an orbit at the right distance from Sun, that it allows liquid water to exist. The distance needs to be neither too hot nor too cold and is often called the Goldilocks zone for life.

V. Answer the following in detail.

Question 1.
Explain in detail with diagrams, the evidence of evolution, through morphology and anatomy.
Answer:
The comparative study of morphology and anatomy of organisms have evolved from a common ancestor.

(i) Homologous organs: The homologous organs are those which have inherited from common ancestors with similar developmental pattern in embryos. The forelimbs of mammals are homologous structures. A human hand, a front leg of a cat, flipper of a whale and a bat’s wing look dissimilar and adapted for different functions. Their mode of development and the basic structure of bone are similar.

(ii) Analogous organs: The analogous organs look similar and perform similar functions but they have different origin and developmental pattern. The function of the wings of a bat, the wings of a bird and wings of an insect are similar, but their basic structures are different.

(iii) Vestigial organs: The degenerated and non¬functional organs of animals are called vestigial organs. The same organs are found to be well- developed and functional, in some of the related forms. Some of the vestigial organs in man are a vermiform appendix, nictitating membrane, caudal vertebra and coccyx, etc.

(iv) Atavism: The reappearance of ancestral characters in some individuals is called atavism, e.g. Presence of rudimentary tail in newborn babies, the presence of thick hair on the human body.

Question 2.
(a) Name the popular names of Lamarckism.
(b) Explain the principles of Lamarckism.
Answer:
(a) Lamarckism is the hypothesis that an organism can pass on characteristics, that it has acquired through use or disuse, during it’s lifetime to the offspring. Lamarck’s theory of evolution was published in ‘Philosophic Zoologique’ in the year 1809. Lamarckism is popularly known as ‘Theory of inheritance of Acquired Characters” or “Use and Disuse theory”

(b) (i) internal vital force: Due to the inherent ability of the living organisms or their component parts tend to increase in ‘ size continuously.

(ii) Environment and new needs: A change in the environment brings about changes in the need of the organisms. In response to the changing environment, the organisms develop certain adaptive characters. The adaptations may be in the form of development of new parts of the body.

(iii) Use and disuse theory: Lamarck’s use and disuse theory state that, if an organ is used constantly, the organ develops well and gets strengthened. When an organ is not used for a long time, it gradually degenerates. The ancestors of Giraffe were provided with a short neck and short forelimbs. Due to a shortage of grass, they are forced to feed on leaves from trees. The continuous stretching of their neck and forelimbs resulted in the development of long neck and long forelimbs, which is an example fcc constant use of an organ. The degenerated wing of Kiwi is an example for organ of disuse.

(iv) Theory of Inheritance of acquired characters: Animals respond to the changes when there is a change in the environment. The develop adaptive structures. The characters, developed during their lifetime, in response to the environmental changes are called acquired characters. The acquired characters are transmitted to the offspring by the process of inheritance.

VI. Higher-Order Thinking skills [HOTS] Questions

Question 1.

1. What is the reason for the long neck of a giraffe.
2. Name the theory based on the shown figure.

Answer:

1. Due to a shortage of grass, they were forced to feed on leaves from trees. The continuous stretching of their neck and forelimb as resulted in the development of a long neck and long forelimbs.
2. Theory of inheritance of acquired characters.

Question 2.
What are living fossils?
Answer:
Living fossils are living organisms that are similar in appearance to their fossilized distant ancestors and usually have no extinct close features, eg. Ginko Biloba.

Question 3.
What is Mars 2020 Astrobiology?
Answer:
NASA is developing the Mars 2020, Astrobiology to investigate an astrobiological relevant ancient environment on Mars, its surface geological processes and the possibility of past life on Mars and preservation of biosignatures within accessible geological materials.

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You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.3

11th Maths Exercise 2.3 Question 1.
Represent the following inequalities in the interval notation:
Solution:

⇒ x ∈ [-1, 4)

[] closed interval, end points are included
() ➝ open interval
end points are excluded

(ii) x ≤ 5 and x ≥ -3[i] x ≤ 5 and x ≥ -3
Solution:

x ∈ [-3, 5)

(iii) x < -1 or x < 3
Solution:

x ∈ (-∞, -1) or x ∈ (-∞, 3)

(iv) – 2x > 0 or 3x – 4 < 11
Solution:
-2x > 0 ⇒ 2x < 0 ⇒ x < 0
x ∈ (-∞, 0)
3x – 4 < 11
⇒ 3x – 4 + 4 < 11 + 4

11th Maths Exercise 2.3 Answers Question 2.
Solve 23x < 100 when
(i) x is a natural number,
(ii) x is an integer.
Solution:
23x <100

(i.e.,) x > 4.3
(i) x = 1, 2, 3, 4 (x ∈ N)
(ii) x = …. -3, -2, -1, 0, 1, 2, 3, 4 (x ∈ Z)

11th Maths Chapter 2 Solution Samacheer Kalvi Question 3.
Solve -2x ≥ 9 when
(i) x is a real number,
(ii) x is an integer,
(iii) x is a natural number.
Solution:
-2x > 9 ⇒ 2x ≤ -9

(ii) x = …. -3, -2, -1, 0, 1, 2, 3, 4
(iii) x = 1, 2, 3, 4

Samacheer Kalvi 11th Maths Solution Question 4.

Solution:

(ii)

Solution:

Samacheer Kalvi 11th Maths Example Sums Question 5.
To secure A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If one scored 84, 87, 95, 91 in first four subjects, what is the minimum mark one scored in the fifth subject to get A grade in the course?
Solution:
Required marks = 5 × 90 = 450
Total marks obtained in 4 subjects = 84 + 87+95 + 91 = 357
So required marks in the fifth subject = 450 – 357 = 93

Samacheer Kalvi 11th Maths Solutions Question 6.
A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?
Solution:
12% solution of acid in 600 l ⇒ 600 × \(\frac{12}{100}\) = 72 l of acid
15% of 600 l ⇒ 600 × \(\frac{15}{100}\) = 90 l
18% of 600 l ⇒ 600 × \(\frac{18}{100}\) = 108 l
Let x litres of 18% acid solution be added

(600 + x)15 ≥ 7200 + 30x
9000+ 15x ≥ 7200 + 30x
1800 ≥ 15x
x ≤ 120
Let x litres of 18% acid solution be added

10800 + 18 ≤ 7200 + 30x
3600 ≤ 12x
x > 300
The solution is 120 ≤ x > 300

Samacheer Kalvi Class 11 Maths Solutions Question 7.
Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40.
Solution:
Let the two numbers be x and x + 2
x + x + 2 < 40
⇒ 2x < 38

⇒ x< 19 and x > 10
so x = 11 ⇒ x + 2 = 13
x = 13 ⇒ x + 2 = 15
x = 15 ⇒ x + 2 = 17
When x = 17 ⇒ x + 2 = 19
So the possible pairs are (11, 13), (13, 15), (15, 17), (17, 19)

Samacheer Kalvi 11 Maths Solutions Question 8.
A model rocket is launched from the ground. The height h of the rocket after t seconds from lift off is given by h(t) = -5t² + 100t; 0 ≤ r ≤ 20. At what time the rocket is 495 feet above the ground?
Solution:
h(t) = -5t² + 1oot
at t = 0, h(0) = 0
at t = 1, h(1) = -5 + 100 = 95
at t = 2, h(2) = -20 + 200 = 180
at t =3, h(3) = -45 + 300 = 255
at t = 4, h(4) = -80 + 400 = 320
at t = 5, h(5) = -125 + 500 = 375
at t = 6, h(6) = – 180 + 600 = 420
at t = 7, h(7) = -245 + 700 = 455
at t = 8, h(8) = – 320 + 800 = 480
at t = 9, h(9) = -405 + 900 = 495
So, at 9 secs, the rocket is 495 feet above the ground.

Samacheer Kalvi 11th Maths Answers Question 9.
A Plumber can be paid according to the following schemes: In the first scheme he will be paid Rs. 500 plus Rs.70 per hour, and in the second scheme he will be paid Rs. 120 per hour. If he works x hours, then for what value of x does the first scheme give better wages?
Solution:
I scheme with x hr
500 + (x- 1) 70 = 500 + 70x – 70
= 430 + 70x
II scheme with x hours
120x
Here I > II
⇒ 430 + 70x > 120x
⇒ 120x – 70x < 430
50x < 430
\(\frac{50 x}{50}<\frac{430}{50}\)
x < 8.6 (i.e.) when x is less than 9 hrs the first scheme gives better wages.

Samacheer Kalvi Guru 11th Maths Question 10.
A and B are working on similar jobs but their annual salaries differ by more than Rs 6000. If B earns Rs. 27000 per month, then what are the possibilities of A’s salary per month?
Solution:
A’s monthly salary = ₹ x
B’s monthly salary = ₹ 27000
Their annual salaries differ by ₹ 6000
A’s salary – 27000 > 6000
A’s salary > ₹ 33000
B’s salary – A’s salary > 6000
27000 – A’s salary > 6000
A’s salary < ₹ 21000
A’s monthly salary will be lesser than ₹ 21,000 or greater than ₹ 33,000

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.3 Additional Questions

11th Maths Solutions Samacheer Kalvi Question 1.

Solution:

Multiplying both sides by 30, we get 15x ≥ 2(4x – 1) ⇒ 15x ≥ 8x – 2⇒ 15x – 8x ≥ -2

11th Maths Samacheer Kalvi Solutions Question 2.
Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution:
Let x be the marks obtained by Ravi in the third test.

⇒ 145 + x ≥ 180 ⇒ x >180 – 145
⇒ x ≥ 35
Thus, Ravi must obtain a minimum of 35 marks to get an average of at least 60 marks.
Note. A minimum of 35 marks.
⇒ Marks greater than or equal to 35.

Samacheerkalvi.Guru 11th Maths Question 3.
To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get Grade ‘A’ in the course.
Solution:
Let x be the marks obtained by Sunita in the fifth examination. Then,

⇒ 368 + x ≥ 450 ⇒ x ≥ 450 – 368
⇒ x ≥ 82
Thus, Sunita must obtain marks greater than or equal to 82,
i. e., a minimum of 82 marks.

Samacheer Kalvi Guru 11th Maths Solution Question 4.
Find the pairs of ceonsecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Solution:
Let x be the smaller of the two consecutive odd positive integers, then the other is x + 2 .
According to the given conditions.
x < 10, x + 2 < 10 and x + (x + 2) > 11
⇒ x < 10, x < 8

From (1) and (2), we get 9
\(\frac{9}{2}\) < x < 8
Also, x is an odd positive integer. x can take values 5 and 7.
So, the required possible pairs will be (x, x + 2) = (5, 7), (7, 9)

Samacheer Kalvi 11th Maths Question 5.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Solution:
Let x be the smaller of the two consecutive even positive integers, then the other is x + 2. According to the given conditions.

Also, x is an even positive integer.
x can take the values 6, 8 and 10.
So, the required possible pairs will be (x, x + 2) = (6, 8), (8, 10), (10, 12)

Maths Solution Class 11 Samacheer Kalvi Question 6.
Forensic Scientists use h = 61.4 + 2.3F to predict the height h in centimetres for a female whose thigh bone (femur) measures F cm. If the height of the female lies between 160 to 170 cm find the range of values for the length of the thigh bone?
Solution:
Given h = 61.4 + 2.3 F
Given h = 160 ⇒ 160 = 61.4 + 2.3 F
⇒ 2.3 F = 160 – 61.4 = 98.6

Given h = 170 ⇒ 170 = 61.4 + 2.3 F
⇒ 170 – 61.4 = 2.3 F
2.3F = 108.6

So the ranges of values are 42.87 < x < 47.23