Laxmi

Students can Download Social Science Civics Term 3 Chapter 2 The Judiciary Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Social Science 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 8th Social Science Civics Solutions Term 3 Chapter 2 The Judiciary

Samacheer Kalvi 8th Social Science Civics The Judiciary Textual Evaluation

I. Choose the best answer:

Question 1.
The highest and final judicial tribunal of India is-
(a) President
(b) Parliament
(c) Supreme Court
(d) Prime Minister
Answer:
(c) Supreme Court

Question 2.
Judicial system provides a mechanism for resolving disputes between-
(a) Citizen
(b) Citizen and the government
(c) Two State governments
(d) all the above
Answer:
(d) all the above

Question 3.
Dispute between States of India comes to the Supreme Court under-
(a) Original jurisdiction
(b) Appellate jurisdiction
(c) Advisory jurisdiction
(d) none of these
Answer:
(a) Original jurisdiction

Question 4.
Which of the following state/ Union territories have a common High Court?
(a) Punjab and Jammu Kashmir
(b) Assam and Bengal
(c) Panjab, Haryana and Chandigarh
(d) Uttar Pradesh and Bihar
Answer:
(c) Panjab, Haryana and Chandigarh

Question 5.
The System of Public Interest Litigation has been introduced in India by-
(a) Supreme Court
(b) Parliament
(c) Political parties
(d) Constitutional amendments
Answer:
(a) Supreme Court

Question 6.
How many courts are there in apex level in India?
(a) One
(b) Two
(c) Three
(d) Four
Ans:
(a) One

Question 7.
Supreme court is located at-
(a) Chandigarh
(b) Bombay
(c) Calcutta
(d) New Delhi
Answer:
(d) New Delhi

Question 8.
FIR means-
(a) First Information Report
(b) First information Result
(c) First Incident Report
(d) None of these
Answer:
(a) First Information Report

Question 9.
The court that hear criminal cases are called-
(a) District court
(b) Sessions court
(c) Family court
(d) Revenue court
Answer:
(b) Sessions court

II. Fill in the Blanks:

1. The ________High Court is the oldest High Court in India.
2. The framers of the Constitution established ________and _______ judiciary in India.
3. _______ a famous French philosopher propounded the idea of an independent judiciary.
4. ________ deals with disputes over money, property and social matters.
5. During ancient times, most of the Kings’ courts dispensed justice according to _______

Answer:

1. Calcutta
2. independent, impartial
3. Montesquieu
4. Civil Law
5. Dharmal

III. Match the following:

1. Supreme Court – social duties
2. High Court – speedy justice
3. Lok Adalat – highest court of appeal
4. Sir Elijah Impey – highest court in the States
5. Smiritis – chief justice

Answer:

1. Supreme Court – highest court of appeal
2. High Court – highest court in the States
3. Lok Adalat – speedy justice
4. Sir Elijah Impey – chief justice
5. Smiritis – social duties

IV. State True or False:

1. The Supreme Court of India was inaugurated on 28th January 1951.
2. During the Tughlaq period, the code of procedure was written in Arabic.
3. The Regulating Act of 1773 made provision for the formation of Supreme Court.
4. Sadar Diwani Adalat was a Criminal court of appeal.
5. The Allahabad High Court is the largest court in India.
6. The Constitution of India secures justice to all its citizen.

Answer:

1. False
2. True
3. True
4. False
5. True
6. True

V. Choose the correct statement:

Question 1.
Consider the following statements.

(i) A law commission was setup by Macaulay.
(ii) It codified the Indian Laws.

Which of the statements given above is / are correct?
(a) (i) only
(b) (ii) only
(c) Both (i) and (ii)
(d) Neither (i) nor (ii)
Answer:
(c) Both (i) and (ii)

Question 2.
Consider the following statements.

(i) An Indian Penal Code was prepared in 1860.
(ii) The Calcutta High Court was established in 1862.
(iii) The Government of India Act, 1935 created Federal Court.

Which of the statements given above is/are correct?
(a) (i) only
(b) (ii), (iii) only
(c) (i),(iii) only
(d) All the above
Answer:
(d) All the above

Question 3.
Which of the following statement is not true about India’s Supreme Court?
(i) The Supreme Court of India is the highest court of the land.
(ii) It was established by part V under Chapter IV of Indian Constitution.
(iii) Supreme court cannot transfer cases from one High court to another
(iv) Its decisions are binding on all courts
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer:
(c) iii

Question 4.
Assertion (A): The Supreme Court is a Court of Record.
Reason (R): It maintains records of the court proceedings and its decisions are finding upon the lower courts.
(a) A is correct and R is wrong.
(b) Both A and R are Wrong
(c) A is correct and R explains A
(d) A is correct and R does not explain A
Answer:
(c) A is correct and R explains A

Question 5.
Agree or disagree
(a) Every citizen of India can approach the Supreme court.
(b) Rich and powerful people control the judiciary system.
(c) Every citizen has a right to get justice through the courts.
(d) Politicians can not control over judges
Answer:
We agree with a, c, and d. Disagree with b

VI. Answer the following in one or two sentences:

Question 1.
Why do we need judicial system?
Answer:

1. Judiciary plays an important role as an organ of the government.
2. It administers justice, settles disputes, interprets laws, protects fundamental rights and acts as guardian of the Constitution.
3. So we need a judiciary to be efficient in order to ensure proper justice.

Question 2.
What are the different levels of courts in India?
Answer:
The different levels of courts in India are

1. Subordinate Courts.
2. District Courts.
3. High Court and
4. Supreme Court

Question 3.
Differentiate Law and Judiciary.
Answer:
Law:
It is a system of rules imposed through a government or institution to govern people.

Judiciary:
In law, the Judiciary or Judicial system is the system of courts which administers Justice in the name of the State.

Question 4.
Write a note on Lok Adalat.
Answer:

1. Lok Adalat was set up to provide speedy justice.
2. It hears and settles the disputes in the language of the people in the public presence.
3. A Lok Adalat is presided over by a retired judge along with a lawyer and a social worker.
4. Cases are put forward without advocates.
5. The first Lok Adalat was held on 1982 at Junagadh of Gujarat.

Question 5.
What are the advantages of mobile courts?
Answer:

1. Mobile courts will be of great relief to the rural people.
2. It would create greater awareness about the judicial system among the rural masses, cut costs for them and render justice at their doorstep.

VII. Answer the following in detail:

Question 1.
Write about the role of judiciary.
Answer:
Role of Judiciaiy can be classified as,
1. Dispute Resolution:
Role of Judiciary can be classified as, The Judicial system provides a mechanism for resolving disputes between Citizen, Citizens and the government, Two State governments,the Centre and the State governments

2. Judicial Review:
The Judicial has the power to declare a law unconstitutional if it believes that the law passed by the Parliament is violating the basic structure of the Constitution.

3. Upholding the Law and Enforcing Fundamental Rights:
Every citizen of India can approach the Supreme Court or High Court if they believe that their fundamental rights have been violated

Question 2.
Differentiate Civil Law and Criminal Law.
Answer:
Civil Law:

1. It deals with disputes over money property and social matters.
   Eg. Disputes relating to land, rent, marriage dispute etc.,
2. A petition has to be filed before the relevant court by the affected parties.
3. Sentences awarded in the form of money remuneration and as per the petitioner’s claim.

Criminal Law:

1. It deals with conducts or acts that the law defines as offences.
   Eg. Theft, murder, women harassment etc.,
2. It usually begins with lodging of an FIR (First Information Report) with police investigation after which a case is filed in the court.
3. Punishment is awarded if proven guilty, the accused is sent to jail.

Question 3.
Descrbe jurisdiction of Suppreme Court.
Answer:
1. Jurisdiction of Supreme Court:

• The Supreme Court can hear dispute for the first time in that court only.
• It is mainly concerned disputes between the Union and one or more states and between two or more states.

2. Appellate Jurisdiction:
It can hear appeals against the decisions made by the High Court on the granting of certificate by the High Court.

3. Advisory Jurisdiction: The Court can report to the President its opinion about a question of the public importance referred to it by the President.

4. Writs Jurisdiction:
The Writs are issued by the Supreme Court under Article 32 and by the High Courts under Article 226 of the Constitution of India.

5. Court of Record:
it maintains records of the court proceedings and its decisions are finding upon the lower courts.

6. Special Powers:
it supervises the functioning of the lower courts.

VIII. Project and Activity:

Question 1.
Discuss: Is it necessary to have an independent judiciary? List two reasons.
Answer:
Activity to be done by the students themselves

Question 2.
Organize your classroom for a lockup courtroom session. (Take a case with the help of the teacher)
Answer:
Activity to be done by the students themselves

Samacheer Kalvi 8th Social Science Civics The Judiciary Additional Questions

I. Choose the correct answer:

Question 1.
Supreme Court administers ………….
(a) justice
(b) settles disputes
(c) protects fundamental rights
(d) all of the abovel
Answer:
(d) all of the above

Question 2.
Fatawa – i – Alamgiri written in ………….
(a) 1670
(b) 1680
(c) 1660
(d) 1690
Answer:
(a) 1670

Question 3.
The East India Company established a Mayor’s Court in ………….
(a) Madras
(b) Bombay
(c) Calcutta
(d) All of the above
Answer:
(d) All of the above

Question 4.
Circuit Court was set up at ………….
(a) Calcutta
(b) Patna
(c) Dacca
(d) All of the above
Answer:
(d) All of the above

Question 5.
The cases dealing with land records are dealt with by ………….
(a) Sessions Court
(b) Panchayat Courts
(c) Revenue Courts
(d) LokAdalat
Answer:
(c) Revenue Courts

Question 6.
Fast Track courts were established in the year ………….
(a) 2000
(b) 1950
(c) 2010
(d) 1999
Answer:
(a) 2000Tele Law service is for the people in

Question 7.
Tele Law service is for the people in ………. areas.
(a) industrial
(b) rural
(c) Urban
(d) Coastal
Answer:
(b) rural

Question 8.
In 2005 these courts were established ………….
(a) Mobile courts
(b) Family Courts
(c) E – courts
(d) None of the above
Answer:
(c) E – courts

II. Fill In the blanks:

1. In ancient period Dharmic laws governed not only the individual but also the ………….
2. Amongst the Vajjis, there was a board of …………. Kulikas for the investigation of criminalcases.
3. A under the British a supreme court was established in 1824 in ………….
4. During the period of …………. four circuit courts were abolished.
5. The founding fathers of India constitution accorded highest place to ………….
6. Montesquieu was a …………. philosopher.
7. District courts hear …………. at the district level.
8. A Lok Adalat is presided over by a …………. along with a lawyer and a social worker.
9. The first Lok Adalat was held on 1982 at …………. of Gujarat.
10. The legal services authorities was passed in ………….
11. The special power of the Supreme Court is to supervises the functioning of the ………….
12. The Chief architect of the Indian constitution is …………..

Answer:

1. Society
2. eight
3. Bombay
4. William Bentinck
5. Justice
6. French
7. civil cases
8. retired judge
9. Junagadh
10. 1987
11. lower courts
12. Dr. B.R. Ambedkar

III. Match the following:

1. Ancient India – (a) Murshidabad
2. Fort William – (b) Law Commission
3. Circuit Court – (c) Weaker Sections
4. Macaulay – (d) Smiritis
5. NALSA – (e) Calcutta

Answer:

1. d
2. e
3. a
4. b
5. c

IV. State True or False:

1. Appeal proceeded from Gana Courts to Kula Courts
2. A supreme Court was established in Madras in 1801
3. Lord Cornwallis reorganised the civil and criminal judicial system.
4. Supreme Court can not transfer judges of High Courts.
5. The highest court of India is located in New Delhi.
6. An effective judiciary should be independent as well as accountable.

Answer:

1. False
2. True
3. True
4. False
5. True
6. True

V. Choose the correct Statement:

Question 1.
Consider the following Statements:

(i) A Supreme court was established for the first time at Fort Williams in Calcutta.
(ii) Sir Elijah Imphey was appointed its Chief Justice.

Which of the Statement/s given above is / are correct
(a) (i) only
(b) (ii) only
(c) Both (i) and (ii)
(d) Neither (i) nor (ii)
Answer:
(c) Both (i) and (ii)

Question 2.
Consider the following Statements:

(i) Family Courts specifically handle family law matters.
(ii) Mobile courts will be of great relief to the rural people.
(iii) The E-Court project was established in 2000 Which of the Statements given above is / are correct

(a) (i) only
(b) (i) and (ii) only
(c) (i) and (iii) only
(d) (iii) only
Answer:
(b) (i) and (ii) only

Question 3.
Which of the following statements is not true about different levels of courts in our country.

(i) Subordinate courts consider cases of civil and criminal nature.
(ii) District Courts can issue writs for restoring fundamental rights.
(iii) Supreme Court can transfer judges of High Court.
(iv) High Court can deal with cases within the jurisdiction of the state.

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer:
(b) (ii)

Question 4.
Assertion (A): Judiciary plays an important role as an organ of the government.
Reason (R): It administer justice, settles disputes, interprets laws protects fundamental rights.
(a) A is correct and R is wrong
(b) Both A and R are Wrong
(c) A is correct and R explains A
(d) A is correct and R does not explain A
Answer:
(c) A is correct and R explains A

VI. Answer the following in one or two sentences:

Question 1.
Who was Montesquieu?
Answer:

1. Montesquieu was the first political philosopher, who propounded the idea of an independent judiciary
2. He was famous French philosopher.
3. He believed in the theory of separation of powers of the three branches of the Government – Legislature, Executive and Judiciary.

Question 2.
What are Fast Track Court?
Answer:
Fast Track courts:
These courts were established in the year 2000 with an aim to clear the long pending Sessions and other lower judicial cases.

Question 3.
Explain the term Tele Law Initiative.
Answer:

1. To provide legal aid and service to the people in rural areas, the Tele Law Initiative was launched by the Ministry of Law and Justice in collaboration with the Ministry of Electronics and Information Technology.
2. People can seek legal advice from the lawyers through video conferencing available at the common service centre on the ‘Tele Law’ Portal.

Question 4.
Mention the role of Family Court.
Answer:

1. Family courts specifically handle family law matters.
2. They are civil courts and are utilised for various family related claims such as Child custody, Divorce, Adoption, Juvenile issues etc.,

Quetion 5.
State the importance of E- Courts?
Answer:

1. The E-Courts project was established in 2005.
2. According to the project all the courts will get computerised.
3. Judicial service centre is the part of e-court.
4. The public as well as the advocates can ask directly the case status and next hearing dates free of cost.

Question 6.
What is NALSA?
Answer:

1. The National Legal Service Authority has been constituted under the Legal Services Authorities Act, 1987 to provide free Legal Services to the weaker sections of the society.
2. It also organise Lok Adalats for amicable settlement of disputes.

Question 7.
Explain the term PIL.
Answer:

1. Public Interest Litigation (PIL):
   PIL is a litigation filed in a court of law for the protection of public interest.
2. The Supreme court introduced this šystem which allows a person to approach the court with his case.
3. PIL can be filed for the following reasons such as violation of basic human rights, religious rights, pollution, and road safety.
4. This could be done by a written letter stating the case. This concept is unique to the Indian Judiciary.

VII. Answer the following in detail:

Question 1.
Narrate the evolution of Indian Judiciary.
Answer:
1. Judiciary during ancient period:
During ancient times, the concept of justice was linked with religion. The king was regarded as the fountain head of Justice. Most of the Kings’ courts dispensed justice according to ‘dharma’, (righteousness and duty) a set of conventional laws, Dharmic laws governed not only the individual but also the society.

2. Judiciary during Medieval period:
The Tughlaq period saw the compilation of the code of civil procedure. It was called Fiqha-e-Feroze Shahi. The code prescribed details of the procedure and the law in several matters. It was written in Arabic and was translated into Persian.

3. Judiciary during Modern period:
(i) The Indian judicial system and laws we have today was formed by the British during the colonial rule in the country. The beginning of Indian codified common law is traced back to 1727 (Under the Charter of 1 726) when the Mayor ‘s court in Madras, Bombay and Calcutta were established by the East India Company.

(ii) Lord Cornwallis reorganised the civil and criminal judicial system. During tile period of William Bentinck four Circuit courts were abolished.

(iii) The Government of’ India Act 1935 created Federal Court.

Question 2.
Explain the Structure of Courts in India.
Answer:
There are three different levels of courts in our country.
1. Supreme Court of India:

• Its decisions are binding on all courts.
• Can transfer judges 6f High Courts.
• Can move cases from any courts to itself.
• Can transfer cases from one High Court to another.

2. High Court:

• Can hear appeals from lower courts.
• Can issue writs for restoring Fundamental Rights.
• Can deal with cases within the jurisdiction of the State.
• Exercises superintendence and control over courts below it.

3. District Court:

• Deals with cases arising in the District.
• Considers appeals on decisions given by lower courts.
• Decides cases involving serious criminal offences.

4. Subordinate Courts:

• Consider cases of civil and criminal nature.

Question 3.
Write a note on

1. District Court
2. Sessions Court
3. The Panchayat Courts
4. The Revenue Courts.

Answer:

1. District Courts – The Courts that hear civil cases at the district level are called District Courts.
2. Sessions Court – The Courts that hear criminal cases are called the Sessions Court.
3. The Panchayat Courts – The Panchayat Courts deal with Civil and Criminal cases at- the village level.
4. The Revenue courts – The Revenue courts deal with the cases of land records. It also assesses and collects land revenue from the land holders.

Students can Download English Lesson 6 On the Rule of the Road Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English 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 12th English Solutions Prose Chapter 6 On the Rule of the Road

Warm Up

(a) From the pictures given below, identity the actions that may cause inconvenience and discomfort to others. Discuss.

Answer:
Using mobile phones, listening to loud music and driving recklessly will cause inconvenience and discomfort to others.

(b) Classify these pictures to show what they depict-Personal freedom/Public liberty.

Answer:

Samacheer Kalvi 12th English On the Rule of the Road Textual Questions

1. Answer the following questions in one or two sentences each.

Question (a)
Why did the lady think she was entitled to walk down the middle of the road?
Answer:
The lady thought that they got liberty. She could walk anywhere she liked even if it is the middle of the road.

Question (b)
What would be the consequence of the old lady’s action?
Answer:
If the pedestrian takes to the road, the cars have to take the pavement to drive on. The lady’s action would result in a universal chaos.

Question (c)
What does the ‘rule of the road’ mean?
Answer:
The road of the rule means that in order that the liberties of all may be preserved, the liberty’ of everyone must be curtailed.

Question (d)
Why should individual liberty’ be curtailed?
Answer:
The individual liberty must be curtailed in order that everyone may enjoy a social order which makes the individual liberty a reality.

Question (e)
How would a reasonable person react when his actions affect other person’s liberty?
Answer:
A reasonable person would understand that if the traffic police doesn’t interfere with the liberty of individuals at signal points, there will be a huge traffic jam and none will be able to drive on the road.

Question (f)
Define Liberty as perceived by the author.
Answer:
Liberty is an accommodation of interests. It is a social contract rather than a personal affair,

Question (g)
According to the author, what are we more conscious of?
Answer:
We are more conscious of the imperfections of others than of our own.

Question (h)
What is the foundation of social conduct?
Answer:
A reasonable consideration for the rights or feelings of others is the foundation of social conduct.

Question (i)
How can we sweeten our life’s journey?
Answer:
We can sweeten our life’s journey with the-little habits of common place interactions accommodating the interests of fellow human beings.

Question (J)
What does the traffic policeman symbolize?
Answer:
The traffic police man is a symbol of liberty and not tyranny.

Additional Questions

Question (a)
How does individual liberty become social anarchy?
Answer:
When an individual does not follow rules of the road, he interferes with others. Everyone will get into everyone else’s way and thus individual liberty will become social anarchy.

Question (b)
What will happen if the police man does not interfere with traffic?
Answer:
If the traffic police does not interfere with the driving in any way, the result would be chaos. Piccadilly Circus would be a maelstrom which no one would be able to cross.

Question (c)
When can one enjoy a social order?
Answer:
When one submits to a curtailment of private liberty, one can enjoy a social order. Such a social order makes one’s liberty a reality.

Question (d)
Why does the author say liberty is a social contract?
Answer:
Liberty is not a personal affair but a social contract. It is an accommodation of interests.

Question (e)
What are we liable to forget?
Answer:
We are liable to forget that there are a lot of people in the world and we have to accommodate our liberties to their liberties.

Question (f)
What are we often conscious of?
Answer:
We are often conscious of the imperfections of others than of our own when it comes to accommodation of interests.

Question (g)
How do we declare ourselves civilised or uncivilised?
Answer:
It is in the small matters of conduct, in the observance of the rules of the road, that we pass judgement upon ourselves and declare if we are civilised or uncivilised.

2. Answer the following questions in three or four sentences each.

Question (a)
What is ‘liberty’ according to the old lady?
Answer:
According to the Russian old lady, liberty is the right to walk in the middle of the road. She is liberty-drunk. She wasn’t aware that if the pedestrian chooses to walk down the middle of the road, then the end of such liberty would be a universal chaos.

Question (b)
How would ‘liberty’ cause universal chaos?
Answer:
When liberty is used without accommodating the interests of others, it results in universal chaos. Everybody would be getting in everybody else’s -way. Nobody would get anywhere. Individual liberty would become a social anarchy.

Question (c)
Why is there a danger of the world getting ‘liberty drunk’?
Answer:
Similar to the Russian old lady many people are liberty-drunk. Many think that a traffic policeman is a symbol of tyranny. Being in a hurry, the people who are crazy driving their cars . at break-neck speed think that the traffic policeman interferes with their right to free use of highway.

Question (d)
‘Curtailment of private liberty is done to establish social order’ – Do you agree?
Answer:
Yes, curtailment of private liberty is done to establish social order. Liberty is not a personal affair only, it is a social contract. There are a lot of people in the world who need to accommodate their liberty to the liberties of others. A reasonable consideration of the rights or feelings of others in the foundation of social conduct. When one has submitted to the curtailment of private liberty, one may enjoy a social order which makes his liberty, a reality.

3. Answer each of the following in a paragraph of 100-150 words.

Question (a)
What do you infer from Gardiner’s essay ‘On the rule of the Road’?
Answer:
A.G. Gardiner’s essay ‘On the rule of the Road’ is a treatise on “liberty”. He starts the essay with an anecdote. A liberty-drunk Russian lady starts walking down the middle of Highway frustrating car drivers, bus drivers and the traffic police. When questioned about her behaviour, she just replied that she now has the liberty to walk anywhere she liked. The author observes that if a- pedestrian gives up the pavement in preference to the road, cars will be forced to move on to the pavement. This would result in universal chaos. Everybody would be getting in everybody’s way. Nobody would get anywhere. Individual liberty would have become social anarchy.

Under such circumstances the world is in the danger of getting liberty-drunk. The rule of the road reminds the readers that in order that liberties of all may be preserv ed, the liberty of everybody must be curtailed. Thus the traffic police at Piccadilly Circus is not a symbol of tyranny but of liberty. He doesn’t hinder but help the smooth flow of traffic. One has to allow curtailment of one’s liberty to enjoy the fruits of a social order.

Question (b)
Explain in your own words, “What freedom means?”
Answer:
Freedom is not the absolute right of an individual. It is not a personal affair only but a social contract. It is an accommodation of interests. One may exercise freedom in matters pertaining to choice of hair-cut, clothes, sandals and sleeping habits. There are thousand little things for which one can use freedom and be wise or otherwise. But beyond that kingdom of freedom, one’s freedom of action is qualified by other’s freedom. One might use trombone from midnight till three in the morning provided it is done in the Mount Everest and not where one’s own family members are asleep. One’s right to play on trombone must not interfere with the liberty of neighbours to sleep in quiet. A reasonable consideration for the rights or feelings of fellow humans is the foundation of social conduct.

Question (c)
“My right to swing my fist ends, where your nose begins.” Elucidate with reference to, ‘On the Rule of the Road’.
Answer:
Rights are not completely individual affairs. In order to enjoy one’s rights one has to respect the rights of others too. The lady in Petrograd had the right to walk on the pavement. The right to move stops when the other person’s right to drive starts. A person may have a walking stick and roll it too. But his right just stops where the other person’s nose begins. No one has the right to violate the rights of others. The right, one exercises, must not affect or erode the rights of others. One should not think of one’s own rights but also the rights of others.

A.G.Gardiner beautifully illustrates this idea by emphasising the metaphor of traffic rules. Rules of road are in fact rules of politeness and unselfishness. One may have absolute freedom in the choice , of food, religion, fashionable dress, up keep of hair, funny hairsty le, etc. But one must be conscious of the rights of others. So, the statement “my right to swing my fist ends, where your nose begins” fits well with the central theme of the essay ‘ On the Rule of the Road’.

Question (d)
Civilization can only exist when the public collectively accepts constraints on its freedom of action – Explain.
Answer:
Human beings, by nature are quick to find fault with the imperfections of others. They don’t usually realize the truth that they have to accommodate their liberty to the liberty’ of others. A reasonable consideration for the rights and feelings of others is the foundation of social conduct. It is in small matters of conduct, in the observance of the rules of road, that we pass judgement on ourselves. We assess ourselves as civilized or uncivilized.

We are civilized if we enjoy our rights without violating the rights of others or eroding into the privacy of others. It is the little habits of common place interaction that make up the great sum of life and sweeten or make bitter the journey.

In a civilized society, public collectively accepts the fact that freedom is an accommodation of interests of others. It means curtailing a part of one’s own liberty to the liberty of others. So, it is true that civilization can only exist and prosper when the public collectively accepts constraints to its freedom of action.

Vocabulary

Pronunciation – Functional Stress

(a) Now the teacher will read the following words. Listen carefully to the stress in each word. Write against each word whether it is a noun or a verb and mark the stress.

Answer:

American And British English

(c) Some British English words are given in column ‘A’. Write their corresponding American English word is Column ‘B.’

Answer:

(d) Similarly there is a difference in the spelling of certain words between American and British English. In Column ‘A’ words are spelled in American. Write down the corresponding British English spelling for those words in column ‘B\ (The first one is done for you)

Answer:

Listening Activity

Listen carefully to the story being read out and answer the questions.

A boy was flying a kite with his father. He was excited to watch the kite soar high and asked his father, “What kept the kite up?” Dad replied “The String”. The boy said “No dad, the string is holding the kite down.” The father suddenly broke the string, the kite came crashing down. Now the boy understood, what was holding the kite up. This is true in life as well. We are , advised to follow a set of rules or a prescribed code of conduct to enable us to reach our full potential. However, very often we hear the phrase “I want to be free.” If you take the train off the track, it is free. But where would it go? If each one of us follow our own traffic rules and drive on any side of the road what would you call it? Freedom or chaos? By observing rules we are actually gaining freedom. This is what.discipline is all about.

Question 1.
According to the father what kept the kite up?
(a) the kite itself
(b) air
(c) the string
(d) father’s skills
Answer:
(c) the string

Question 2.
The string helped the kite to ______
(a) withstand the pressure
(b) break free
(c) crash
(d) soar high
Answer:
(d) soar high

Question 3.
To realize our full potential, we should ______
(a) obey our elders
(b) follow a set of rules
(c) work hard
(d) plan well
Answer:
(b) follow a set of rules

Question 4.
A train should go on the ______ to reach its destination.
(a) track
(b) line
(c) road
(d) path
Answer:
(a) track

Question 5.
Taking freedom in our own hands will lead to ______
(a) self-discipline
(b) chaos
(c) a code of conduct
(d) freedom
Answer:
(b) chaos

Question 6.
The main idea of the passage is ______
(a) the art of flying kites
(b) father’s advice to his son
(c) the importance of discipline
(d) the right to freedome
Answer:
(c) the importance of discipline

Speaking Activity

(a) Everybody enjoys holidays but, when it comes to safety ‘There is no holiday for Safety’. Discuss in groups what safety measures one should take while driving a car/ two wheeler.

Group A:
While riding a bike, the rider should wear a helmet. If there is a pillion rider, he/she should also wear a helmet. When the signal is yellow, one must stop the vehicle as, red signal will follow soon. Don’t ride or drive with a breakneck speed. If somebody tries to cross the road suddenly, there could be a fatal accident. Before leaving the home, check the tyre pressure, adequacy of petrol or diesel or electricity if it is an electric bike/scooter. Thus, you can ensure safety by adopting the above strategies.

Group B:
If you plan to travel by a car, leave at least one hour early to avoid peak hour traffic. Wear seat belt. Before overtaking ensure if the vehicle ahead of you wishes to take a ‘U’ turn or not. If you wish to take a right turn or a left turn use the indicator to warn the following vehicle of your intended change in the direction of travel. If you tend to travel near a school, don’t speed up, look on either side to verify if children are trying to cross the road. Don’t honk near hospitals and schools. If there is an unmanned level crossing, park your car 10 metres away from the level crossing, check for the movement of train on either side. Even if the train is seen far away, wait for the train to pass by.

(b) Your school has arranged for a road safety campaign, Share a few ideas with your classmates which can be used during the campaign. For example ‘Speed thrills but kills’.

The following are some interesting slogans to be used in the road safety campaign.

• Drive as if every child on the road is your own.
• A spill, a slip, a hospital trip
• Start early, drive slowly, reach safely
• An accident would cripple your income
• Safety protects people
• Safety starts within you
• Risk prevention is better than cure
• Drowsy driving is impaired driving
• Fast drive’ could be the last drive
• Don’t lose your head to gain a minute, you need your head your brains are in it
• Hug your kids at home but belt them in the car
• Safety first because accident last
• Chance takers are accident makers
• Accidents hurts but safety doesn’t
• Speed kills, slow down
• Be cautious, nearly 40% victims in road accidents are pedestrians
• Sober up, drunk driving kills

Reading

Cyber Safety
Given below is a text on “Cyber safety” developed from the inputs received from Crime- Branch Crime Investigation Department (CBCID), Tamil Nadu dated: 05.12.2018. Read the text and answer the questions that follow.

Question 1
How should teenagers guard against cyber crimes?
Answer:

• Desist from interacting with strangers on social media and never meet strangers in person pursuant to social media chat/interaction.
• Avoid having your picture as profile picture. Even if you choose to have one, do so while restricting those persons who can see your profile picture.
• Do not download software’s/Apps from unknown sources. Do not download unnecessary apps in the devices. Be a member of social media group or whatsapp group only if you know majority of members in it and also if it is relevant to be part of the group. Do not believe and blindly share message in social media without verifying the facts.
• Keep front camera of mobile phones, laptops etc., closed when not used. For Example, Stickers may be used to close the camera and to avoid remote access of front camera.

Question 2
What should we do to ensure Safe Surfing?
Answer:

• Use a secure browser.
• Do not surf unsafe websites. A website with URL starting with https:// is a secure website. If the website URL starts only as http://, it is unsecure.
• Avoid clicking on links from unknown mails/pop ups.
• Do not enter passwords when connected to a public network (WiFi in Railway station or Airports)
• Always use a computer in which updated Anti Virus is installed.

Question 3
What are the details not to be revealed in public domain ?
Answer:

• Any Passwords
• Bank account /credentials
• Credit card/Debit card details
• Personal mobile number
• Date of Birth
• Any details which would help to track your routine activities

Question 4
What should parents do to ensure the safety of children in cyber space ?
Answer:

• Set a fixed time during which children are allowed to surf the internet.
• Always place the computer in that part of the house, which is visited most often by everyone. Don’t keep the computer in a secluded part. Talk with the children and educate them on the websites that they are allowed / not allowed to visit.
• Be “Friends” with the children’s social media account so that their activities are monitored.
• Install an Anti Virus with parental control in the computer.
• Make it a habit to check browsing history/hidden files, apps installed etc., on the computer/ device to monitor activity of the children.

Question 5
Should children be discouraged from playing online games?
Answer:
Yes, the following steps can be followed to discourage children from playing online games.

• Encourage children to play outdoor games.
• Monitor closely if they are found playing a single game for long hours.
• Watch out for early signs of addiction to games such as
  1. Decreasing Academic activity 4 Less time spent with family
  2. Loss of interest in things previously enjoyed.
  3. Lack of sleep / Sore Eyes
  4. Headaches
  5. Create awareness about the ill-effects of dangerous online games.

Questions:

Question 1.
Which of the following should one avoid while using social media? Tick against the correct options.
Answer:

• Interacting with strangers on social media
• Avoid posting your picture as profile picture
• Forwarding Whatsapp messages without verifying facts
• Use a secure browser
• Entering passwords while using public network

Question 2.
How can we identify insecure websites?
Answer:
If the website URL starts only as http://, it is insecure.

Question 3.
Mention any three details that should not be revealed in public domain.
Answer:

1. Password
2. Bank account details
3. Personal mobile number

Question 4.
What should parents do to ensure cyber safety for their wards ? (Any three points)
Answer:
Set a fixed time for viewing internet
Keep the computer in a place from where the content being viewed can be monitored by anyone Educate children on safe and unsafe sites.

Question 5.
Why is it not advisable to play online games?
Answer:

1. They could swindle the family fortune
2. Online games distract children from studies
3. Children tend to spend less quality time with family members.
4. Children tend to develop eye problems

Question 6.
Pick out words which mean the same as
Answer:
(a) stop doing something (para 1) – desist/avoid
(b) place or fix (para 2) – having
(c) not protected (para 2) – insecure
(d) keep a check on (para 3) – monitor

Grammar

Agreement of the subject with the verb
Task 1
I. Choose the correct option and complete the sentences.

1. Manoj ________ (was, were) present along with his parents.
2. Each of these boys ________ (has, have) passed.
3. Neither Lekha nor Leela ________ (has, have) been selected.
4. Every man, woman and child ________ (was, were) happy.
5. One of the machines ________ (is, are) defective.
6. A number of books ________ (is, are) missing.
7. Seker or his brothers ________ (has, have) done it.
8. To make a promise and then not to keep it ________ (is, are) dishonesty.
9. One or the other of those men ________ (has, have) lodged a complaint.
10. TO. Each leaf and each flower ________ (was, were) stripped off the tree.

Answers

1. was
2. has
3. 3. has
4. were
5. is
6. are
7. have
8. is
9. has
10. was

Task 2
II. Identify the errors in each of the following sentences and rewrite them Correctly.

Question 1.
Either Shyam or Ram have to pay the fine.
Answer:
Either Shyam or Ram has to pay the fine.

Question 2.
Abdul as well as Karim deserve praise.
Answer:
Abdul as well as Karim deserves praise.

Question 3.
Ten thousand rupees a month are an insufficient income.
Answer:
Ten thousand rupees a month is an insufficient income.

Question 4.
Many a student were awarded at the function.
Answer:
Many a student was awarded at the function.

Question 5.
Neither Veena nor her sisters has been informed of the accident.
Answer:
Neither Veena nor her sisters have been informed of the accident.

Question 6.
Mithra as well as her daughters enjoy singing.
Answer:
Mithra as well as her daughters eniovs singing.

Question 7.
You, who is my friend, should help me.
Answer:
You, who are my friend, should help me.

Question 8.
My scissors is missing.
Answer:
My scissors are missing.

Question 9.
A variety of pleasing objects charm the eye.
Answer:
A variety of pleasing objects charms the eye.

Question 10.
Sixty miles are a long distance.
Answer:
Sixty miles is a long distance.

Writing

Describing a process
Task 1:

Preparation of apple juice
You plan to delight your parents and sister, serving them chilled apple juice. Here is the process:

(Complete the sentences with the right form of the verbs)
Four or five apples (1) ______ (take) and (2) ______ (wash) well. They (3) ______ (wipe) dry and cut into pieces of medium size. The seeds (4) ______ (remove). Then the apples (5) ______ (put) into the mixer. Some milk (6) ______ (add) .The apples(7) ______ (crush) and a fine liquid (8) ______ (obtain). This liquid (9) ______ (filter) and the juice (10) ______ (store) in the refrigerator. It is (11) ______ (take ) out whenever needed, and after adding sugar, it (12) ______ (sen e) in cups.
Answer:

1. should be taken
2. washed
3. should be wiped
4. should be removed
5. should be put
6. should be added
7. should be crushed
8. will be obtained
9. should be filtered
10. should be stored
11. taken
12. is served

Task 2:
Installing a computer

The description of installing a computer in your study room is given in the form of jumbled sentences. Rearrange the sentences in the right order and form a coherent paragraph.

1. Once you connect the CPU, connect the keyboard and mouse.
2. Before turning on the power, check that all parts are connected to the CPU.
3. First open the box and take out the computer parts.
4. Plug both the computer and the monitor with a power cord.
5. Set the computer on a table or flat surface.
6. Finally turn on the power.

Task 3:
Attempt a description of the following processes, in about 100 words each, either using the imperative or the passive.

Question 1.
Preparing your favourite dish
Answer:

1. Finely chop one medium size onion, 1 to 2 green chillies, half inch ginger and some coriander leaves.
2. One must keep aside 8 to 10 curry leaves, 10 to 12 cashew nuts, 1 teaspoon chana dal and 1 teaspoon of urud dal.
3. Heat a pan or kadai. Add one cup rava and roast at a low flame, until the aroma of roasted rava arises.
4. Shift the roasted rava to another plate.
5. In a pan, heat 2 tablespoons of ghee or oil and add mustard seeds. Wait until it crackles.
6. Then add cumin seeds and chana dal and fry them until brown.
7. Add cashews and roast till they become golden colour. Now add finely chopped onions and.saute them till they become translucent.
8. Now add green chilly, ginger, curry’ leaves. Saute for a minute. Add tw’o and a half cups of,water and add salt to taste.
9. Wait until the water boils and add the roasted rava in four or five batches. Keep stirring and mix well under low flame.
10. The rava will absorb the water and will be cooked.
11. Keep on stirring until all the water is absorbed by the rava and then turn off the flame.
12. Serve the delicious upma with coconut chutney or sugar.

Question 2.
Organising a birthday party in your house
Answer:
Organising a birthday party especially for young children could be great fun.

1. Shortlist the names of invitees.
2. Write or print as many invitation cards as necessary,
3. Place them inside the cover and write the names on each cover.
4. Decide on the items to be served to the invitees such as juice, cakes, ice-cream, candies etc.
5. Decorate the home with colour flags, balloons, etc.
6. Decide on the games to be played.
7. Along with food items procure gift items to be given to children who come to honour the child in the birthday party.
8. Either have a photographer or videographer to record the event.
9. Now go ahead with the plan, you w ill have a wonderful birthday party.

Question 3.
Sending a letter by courier service
Answer:

1. Write the letter, fold it and place it inside an envelope.
2. Paste the right end of the envelope and w’rite your address at the left hand bottom.
3. Write the addressee’s name and address in the middle of the envelope.
4. Take it to the courier office.
5. Write on top of the envelope “By Courier”.
6. Get it weighed.
7. The front desk personnel will ask you to pay an amount according to the weight of the envelope and distance it has to travel
8. Pay the money and get the invoice.
9. Now the courier is on its way to the intended destination.

Question 4.
Obtaining a demand draft from a bank
Answer:

1. It is a very simple process.
2. Go to any nationalised bank and ask for a challan to fill in details such as beneficiary’s name,date, amount, commission to be paid.
3. One can turn the challan leaf to find the Tamil version of the challan to be filled,
4. Total.the amount and the commission and fill the challan.
5. Draw a self-drawm cheque and attach it along with the DD challan.
6. The bank will deduct the money from your account and give a demand draft to you. If you don’t have an account, the commission amount may be a little extra.

On the Rule of the Road About The Author

Alfred George Gardiner was a British journalist and author. He was a prolific essayist and his style and subject matter easily qualified him to be categorized as what the English would call a very civilized gentleman. His essays include ‘On Habits’, ‘On Being Tidy’ and ‘On Talk and Talkers’. ‘On the Rule of the Road’, was included in one of Gardiner’s compilations titled “Leaves in the Wind’ and was published under his pseudonym “Alpha of the Plough”.

On the Rule of the Road Summary in English

Introduction
On the rule of the road, A. G. Gardiner emphasises the necessity of certain constraints on individual liberty if society is to function in a genuinely civilised manner.

Social awareness
A plump old lady was walking with her basket in the middle of the road in Petrograd. It created great confusion to the traffic. Even her own life was in danger. Somebody pointed out to her that the pavement was meant for the pedestrians. Her reply was that she enjoyed liberty and could walk anywhere she liked. She did not realise that her liberty or such unrestrained freedom could lead to an all-round disorder. If everybody interferes with the liberties of others, nobody would be able to enjoy liberty. This could lead to social anarchy.

Liberty drunk public
In the modem world, people are liberty dmnk like that old lady. Therefore, it is important to remind them what the mle of the road means. It means that to preserve the liberties of all, everybody’s liberties must be reduced. He makes his point clear with an example of a traffic policeman. The traffic policeman is not a symbol of tyranny but of liberty. An unreasonable man will be annoyed by the interference of the traffic policeman. If the policeman stops a man in a hurry, he will feel that his liberty has been injured by violence. A reasonable man will surely think him to be the right man to do so. If he does not interfere with him, he would interfere with none. It would lead to chaos. Then nobody would be able to cross the road. Therefore, one’s liberty has to be curtailed to create social order and make liberty real.

Personal liberty vs Public liberty

Liberty is not only a personal affair but a social contract too. It requires an adjustment of our interests. There are matters of a private kingdom which does not touch anybody else’s liberty’. In this kingdom, one can follow one’s own fancy and ask no one’s permission. For example, one may follow any religion or one may marry a dark or fair lady. One may dye one’s hair or do whatever one may like with his hairstyle. These actions do not interfere with the liberties of others. But w’hen one comes out of the kingdom of one’s personal liberty of action, becomes qualified by other people’s liberty. For example, one may like to play upon a guitar. But one should do the practice mildly to accommodate one’s liberty to others liberties. But most often people forget this. People look to the faults of others but forget their own. A reasonable consideration for the rights or feelings of others is the foundation of social conduct

Equal rights for small and big people

Rights and liberties of small people are as important as those of small nationalities. Gardiner here condemns the attack of Germany on Belgium. In World War-II, Germany, a super military pow’er overran the small and poorly equipped Belgium. He also criticises the aggressive and bullying nature of some people and nations of the w orld. He called them an ugly specimen of a civilised world.

Small actions with big impact
Small actions of conduct count more. By these actions, like observing or not the rules of the road, we can be civilised or uncivilised. It is the little habits of common place intercourse that make up the great sum of life sweeter or make the journey bitter.

Conclusion
A.G. Gardiner illustrates with interesting examples of how the social anarchy could be prevented by adhering to rules of the road. In a way, adhering to rules of the road will not only retain a social order but also prevent fatal accidents on the road and embarrassing traffic jams.

On the Rule of the Road Summary in Tamil

முன்னுரை
_ ‘On the Rule of the Road’ (“சாலை விதியைப் பற்றி”) என்ற பாடத்தின் வாயிலாக கார்டினர், நம் தனி சுதந்திரத்தில் சில கட்டுப்பாடுகளை வகுத்துக் கொண்டால் நம் சமுதாயம் உண்மையாக நாகரிகம் நிறைந்ததாக செயல்படும் என்கிறார்.

சமூக விழிப்புணர்வு:
பெட்ரோகாட் நகரத்தில் பருமனான வயது | முதிர்ந்த பெண்மணி ஒருவர் தன் கையில் ஒரு | கூடையுடன் நடுரோட்டில் நடந்து கொண்டிருந்தார். இது சாலை போக்குவரத்தை பெரிய குழப்பத்திற்கு | உண்டாக்கியது. அவர் உயிருக்கே ஆபத்தாக இருந்தது. | நடைபாதையில் நடக்குமாறு சிலர் கூறிய போது, தான் | சுதந்திரத்தை விரும்புதால் தான் எங்கும் நடப்பதற்கு | உரிமை உண்டு எனப் பதிலளித்தார். அத்தகைய சுதந்திரம் மொத்தத்தில் குழப்பத்தையே ஏற்படுத்தும் என்பதை அவர் உணரவில்லை. ஒவ்வொருவரும் அடுத்தவரின் தனி சுதந்திரத்தில் தலையிட்டால் எவராலும் சுதந்திரத்தை அனுபவிக்க முடியாது. இது சமூக அராஜகத்தை உண்டாக்கும்.

சுதந்திர போதையில் தள்ளாடும் பொதுமக்கள்:
தனிச் சுதந்திரம் என்ற அபாய நோக்கமானது இக்காலத்தினரிடயே அந்த வயதான மூதாட்டி போல் தொற்றிக் கொண்டுள்ளது. ஆதலால், சாலை ஒழுக்கத்தை அவர்களுக்கு நினைவுபடுத்துவது முக்கியமான கடமையாகிறது. அதாவது அனைவரின் சுதந்திரத்தை பாதுகாக்கும் பொருட்டு அவர்களின் சுதந்திரத்தில் எல்லை வரையறுக்கப்படுகிறது. ஆசிரியர் இந்த கருத்தை தெளிவாக்க சாலை பாதுகாப்பு காவல்காரரை உதாரணமாகக் கொள்கிறார். சாலை பாதுகாப்பு காவலர் எதேச்சதிகாரத்தின் அடையாளம் அல்ல. அவர் உரிமையின் சின்னமாவார். இதைப் புரிந்து கொள்ளாதவர் காவலர் குறுக்கிடுதலை இடையூறாக கருதுவார். அவசரமாக காவலர் தடுத்து நிறுத்தினால் அவர் நம்மை வன்முறைக்கு உள்ளாக்குகிறார் என நினைக்கக்கூடும். புரிந்துகொண்டவர்களுக்கு அவர் நம் நன்மைக்கே செய்கிறார் என்பது விளங்கும். அவர் நம்மிடம் தலையிடாவிட்டால் அவரால் யாரிடமும் தலையிட முடியாது. அது குழப்பத்தையே உருவாக்கும். ஆகவே, வெறும் சாலையை கடக்க முடியாது. உங்களின் தனிச் சுதந்திரத்தை கட்டுப்படுத்துவது உங்களை பொது விதிகளை அனுசரிக்கச் செய்து உண்மை சுதந்திரத்தை உணரச் செய்வதற்கே ஆகும்.

தனிப்பட்ட சுதந்திரம் மற்றும் பொதுச் சுதந்திரம்

சுதந்திரம் என்பது தனி நபரைச் சார்ந்தது மட்டும் அல்ல. அது ஒரு பொதுக் கோட்பாடு ஆகும். இதில் நம் உரிமைகளை பொருந்தச் செய்யவேண்டும். தனிப்பட்ட சுதந்திரக் கோட்பாடுகள் வேறு எவரின் சுதந்திரத்தையும் பாதிப்பதில்லை. இந்த நாட்டில் ஒருவர் தன் மனம் செல்லும் வழியில் செல்ல யாரிடமும் உத்தரவு கேட்பதில்லை. உதாரணத்திற்கு ஒருவர் எந்த மதத்தையும் பின்பற்றலாம் அல்லது கறுப்பு நிறத்தவரையோ, வெள்ளை நிறத்தவரையோ திருமணம் செய்து கொள்ளலாம். அவர் தலைக்கு சாயம் அடித்துக் கொள்ளலாம். எதுவாயினும் செய்யலாம். இந்தச் செயல்கள் அடுத்தவரின் சுதந்திரத்தைப் பாதிப்பதில்லை. ஆனால் இந்த தனிப்பட்ட சுதந்திர வட்டத்தை விட்டு வெளியேறும் போது மற்றவரின் சுதந்திரத்தை மதித்து அனுமதிக்க வேண்டியுள்ளது. அதாவது ஒருவர் கிடார் (guitar) வாசிக்க விரும்பலாம். ஆனால், அடுத்தவரின் சுதந்திரத்தை அனுசரிக்கும் வகையில் தன் சுதந்திரம் இருக்க அவர் மெதுவாக வாசிக்க வேண்டும். பெரும்பான்மையான மக்கள் இதை மறந்து விடுகிறார்கள். அடுத்தவர் குறையைக் காணும் மக்கள் தன் குறையை மறந்து விடுகிறார்கள். அடுத்தவரின் உரிமை அல்லது உணர்வுகளுக்கு இடமளிப்பதே சமூக நடத்தையின் அடித்தளம் ஆகும்.

எளியவருக்கும், வலியவருக்கும் ஒரே விதி:

சிறிய நாடுகளில் வாழும் மக்களைப் போன்ற சிறுபான்மைப்பட்ட மக்களின் விதிகளும், உரிமைகளும் முக்கியமானதே. கார்டினர் (Gardiner) ஜெர்மனி நாடானது பெல்ஜியத்தின் மேல் போர் தொடுத்ததை வன்மையாகக் கண்டிக்கிறார். இரண்டாம் உலகப் போரின் போது சிறுபான்மையான மற்றும் போதிய இராணுவ வசதிகள் இல்லாத பெல்ஜிய நாட்டை பெரும் இராணுவப் படையுடைய ஜெர்மனி தன் வசப்படுத்தியது. சில மக்கள் மற்றும் உலக நாட்டவரின் ஆக்ரமிக்கும் குணம் மற்றும் அகங்காரம் குறித்தும் ஆசிரியர் விமர்சனம் செய்கிறார். அவர்களை நாகரிக உலகத்தின் வெறுக்கத்தக்க இனம் எனக் குறிப்பிடுகிறார்.

சிறிய நடவடிக்கை கொடுத்த பெரிய விளைவு:
சிறிய சிறிய நம் செயல்கள் பெரிய தாக்கத்தை உருவாக்குகின்றன. இத்தகைய சிறிய செயல்களான விதிமுறைகளைக் கடைப்பிடிப்பது அல்லது | தவிர்ப்பது வழியாகவே நாம் நாகரிகமானவரா அல்லது (நாகரிகமற்றவரா என வரையறுத்துக் கொள்ளலாம். | இத்தகைய சிறிய பொது இடக் கோட்பாடுகளே நம் பெரும்பான்மையான வாழ்வை இனிமையாக்குவது அல்லது கசப்பாக்குவது என்பதை முடிவு செய்கின்றன. இத்தகைய சிறிய பொது இடக் கோட்பாடுகளே நம் பெரும்பான்மையான வாழ்வை இனிமையாக்குவது அல்லது கசப்பாக்குவது என்பதை முடிவு செய்கின்றன.

முடிவுரை:
சாலை விதிகளைக் கடைபிடிப்பதன் மூலம் சமூக அராஜகத்தை எப்படி தடை செய்யலாம் என்பதை ஏ.ஜி. கார்டினர் சுவாரஸ்யமான உதாரணங்களுடன் விளக்குகிறார். ஒரு வழியில், சாலையின் விதிகள் கடைபிடிக்கப்படுவது ஒரு சமூக ஒழுங்கைத் தக்கவைத்துக் கொள்வது மட்டுமல்லாமல், சாலையில் விபத்துகள் ஏற்படுவதை தடுக்கவும், சங்கடமான போக்குவரத்து நெரிசலைத் தடுக்கவும் உதவுகிறது.

On the Rule of the Road Glossary

Textual:

Additional:

On the Rule of the Road Synonyms

Find out the synonym of the underlined word in each of the following sentences.

Question 1.
A stout old lady walking down the middle of street caused a great confusion.
(a) vision
(b) clarity
(c) mess
(d) mesh
Answer:
(c) mess

Question 2.
She caused great peril to herself.
(a) safety
(b) frill
(c) shrill
(d) danger
Answer:
(d) danger

Question 3.
One who walks on the pavement is called,
(a) stalker
(b) walker
(c) pedestrian
(d) martian
Answer:
(c) pedestrian

Question 4.
We have got liberty now.
(a) tyranny
(b) freedom
(c) avarice
(d) slavery
Answer:
(b) freedom

Question 5.
The end of such liberty would be universal chaos.
(a) brotherhood
(b) love
(c) peace
(d) confusion
Answer:
(d) confusion

Question 6.
Individual liberty would become social anarchy.
(a) cooperation
(b) vanity
(c) civility
(d) lawlessness
Answer:
(d) lawlessness

Question 7.
Liberties of all may be preserved.
(a) excluded
(b) severed
(c) mainteained
(d) pointed
Answer:
(c) mainteained

Question 8.
Liberties of everybody must be curtailed.
(a) enhanced
(b) increased
(c) levelled
(d) reduced
Answer:
(d) reduced

Question 9.
He is a symbol of tyranny.
(a) plutocracy
(b) democracy
(c) aristocracy
(d) autocracy
Answer:
(d) autocracy

Question 10.
We may choose to be wise or ridiculous.
(a) serious
(b) funny
(c) earnest
(d) critical
Answer:
(b) funny

Question 11.
You can be conventional or odd.
(a) common
(b) usual
(c) strength
(d) popular
Answer:
(c) strength

Question 12.
If I play trombone in my bedroom, my family will object.
(a) accept
(b) approve
(c) oppose
(d) oppese
Answer:
(c) oppose

Question 13.
Liberty is a social contract.
(a) contact
(b) encounter
(c) confront
(d) agreement
Answer:
(d) agreement

Question 14.
I have to accommodate the interests of others.
(a) reject
(b) decline
(c) accomplish
(d) fit in with
Answer:
(d) fit in with

Question 15.
A reasonable consideration for the rights or feelings for the others is the foundation of social conduct.
(a) thoughtfulness
(b) rejection
(c) confrontation
(d) argument
Answer:
(a) thoughtfulness

On the Rule of the Road Antonyms

Find out the antonym of the underlined word in each of the following sentences.

Question 1.
A stout old lady was walking.
(a) fat
(b) obese
(c) brittle
(d) lean
Answer:
(d) lean

Question 2.
We’ve got liberty now.
(a) freedom
(b) slavery
(c) right
(d) democracy
Answer:
(b) slavery

Question 3.
Such liberty would be universal chaos.
(a) disorder
(b) confusion
(c) order
(d) mess
Answer:
(c) order

Question 4.
Individual liberty would have become social anarchy.
(a) order
(b) disorder
(c) confusion
(d) mess
Answer:
(a) order

Question 5.
Liberties of all may be preserved.
(a) conserved
(b) damaged
(c) retained
(d) maintained
Answer:
(b) damaged

Question 6.
Liberties of everybody must be curtailed.
(a) reduced
(b) constrained
(c) restricted
(d) increased
Answer:
(d) increased

Question 7.
The policeman is not a symbol of tyranny.
(a) democracy
(b) autocracy
(c) plutocracy
(d) aristocracy
Answer:
(a) democracy

Question 8.
If you are a reasonable person, you would understand liberty.
(a) sound
(b) fair
(c) logical
(d) unreasonable
Answer:
(d) unreasonable

Question 9.
I have liberty to be indifferent to you.
(a) caring
(b) disinterested
(c) heedless
(d) unconcerned
Answer:
(a) caring

Question 10.
We may choose to be harsh or easy.
(a) rude
(b) soft
(c) rough
(d) severe
Answer:
(b) soft

Question 11.
I have to accommodate my liberty to other’s liberties.
(a) adjust
(b) hinder
(c) fit in with
(d) give up
Answer:
(b) hinder

Question 12.
We are conscious of much more imperfections of others than of our own.
(a) frailties
(b) follies
(c) perfections
(d) shortcomings
Answer:
(c) perfections

Question 13.
A reasonable consideration for the rights or feelings for the others is the foundation of social conduct.
(a) thoughtfulness
(b) heedfulness
(c) heedlessness
(d) sensitivity
Answer:
(c) heedlessness

Question 14.
I must not interfere with their liberty to sleep in guiet.
(a) solemnity
(b) quietude
(c) tranquillity
(d) noise
Answer:
(d) noise

Question 15.
The lady walked down the middle of the road with great peril to herself.
(a) danger
(b) safety
(c) horror
(d) criticality
Answer:
(b) safety

Students can Download Maths Chapter 3 Algebra Ex 3.3 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 3 Chapter 3 Algebra Ex 3.3

Miscellaneous Practice problems

Question 1.
Using identity, find the value of
(i) (4.9)²
(ii) (100.1)²
(iii) (1.9) × (2.1)
Solution:
(i) (4.9)²
(4.9)² = (5 – 0.1)²
Substituting a = 5 and b = 0.1 in
(a – b)² = a² – 2ab + b², we have
(5 – 0.1)² = 5² – 2(5) (0.1) + (0.1)²
(4.9)² = 25 – 1 + 0.01 = 24 + 0.01
(4.9)² = 24.01

(ii) (100.1)²
(100.1)² = (100 + 0.1)²
Substituting a = 100 and b = 0.1 in
(a + b)² = a² + 2ab + b², we have
(100 + 0.1)² = (100)² + 2(100) (0.1) + (0.1)²
(100.1)² = 10000 + 20 + 0.01
(100.1)² = 10020.01

(iii) (1.9) × (2.1)
(1.9) × (2.1) = (2 – 0.1) × (2 + 0.1)
Substituting a = 100 and b = 0.1 in
(a – b) (a + b) = a² – b² we have
(2 – 0.1) (2 + 0.1) = 2² – (0.1)²
(1.9) × (2.1) = 4 – 0.01
(9.9) (2.1) = 3.99

Question 2.
Factorise: 4x² – 9y²
Solution:
4x² – 9y² = 22 x² – 3² y² = (2x)² – (3y)²
Substituting a = 2x and b = 3y in
(a² – b²) = (a + b) (a – b), we have
(2x)² – (3y)² = (2x + 3y) (2x – 3y)
∴ Factors of 4x² – 9y² are (2x + 3y) and (2x – 3y)

Question 3.
Simplify using identities
(i) (3p + q) (3p + r)
(ii) (3p + q) (3p – q)
Solution:
(i) (3p + q) (3p + r)
Substitute x = 3p,a = q and b = r in
(x + a) (x + b) = x² + x(a + b) + ab
(3p + q)(3p + r) = (3p)² + 3p (q + r) + (q × r)
= 3² p² + 3p (q + r) + qr
(3p + q)(3p + r) = 9p² + 3p(q + r) + qr

(ii) (3p + q) (3p – q)
Substitute a = 3p and b = q in
(a + b) (a – b) = a² – b², we have
(3p + q) (3p – q) = (3p)² – q² = 32 p² – q²
(3P + q) (3p – q) = 9p² – q²

Question 4.
Show that (x + 2y)² – (x – 2y)² = 8xy.
Solution:
LHS = (x + 2y)² – (x – 2y)²
= x² + (2 × x × 2y) + (2y)² – [x² – (2 × x × 2y) + (2y)²]
= x² + 4xy + 4y² – [x² – 4xy + 2²y²]
= x² + 4xy + 4y² – x² + 4xy – 4y²
= x² – x² + 4xy + 4xy + 4y² – 4y²
= x² (1 – 1) + xy (4 + 4) + y² (4 – 4)
= 0x² + 8xy + 0y² = 8xy = RHS
∴ (x + 2y)² – (x – 2y)² = 8xy
[∵ (a + b)² = a² + 2ab + b² (a – b)² = a² – 2ab + b²]

Question 5
The pathway of a square paddy field has 5 m width and length of its side is 40 m. Find the total area of its pathway. (Note: Use suitable identity)
Solution:
Given side of the square = 40 m
Also width of the pathway = 5 m
∴ Side of the larger square = 40m + 2(5)m = 40m + 10m = 50m
Area of the path way = area of large square – area of smaller square
= 50² – 40²

Substituting a = 50 and b = 40 in
a² – b² = (a + b) (a – b) we have
50² – 40² = (50 + 40) (50 – 40)
Area of pathway = 90 × 10
Area of the pathway = 900 m²

Challenge Problems

Question 1.
If X = a² – 1 and Y = 1 – b², then find X + Y and factorize the same.
Solution:
Given X = a² – 1
Y = I – b²
X + Y = (a² – 1) + (1 – b²)
= a² – 1 + 1 – b²
We know the identity that a² – b² = (a + b) (a – b)
∴ X + Y = (a + b) (a – b)

Question 2.
Find the value of (x – y) (x + y) (x² + y²).
Solution:
We know that (a – b) (a + b) = a² – b²
Put a = x and b = y in the identity (1) then
(x – y) (x + y) = x² – y²
Now (x – y) (x + y)(x² + y²) = (x² – y²) (x² + y²)
Again put a = x² and b = y² in (1)
We have (x² – y²) (x² + y²) = (x²)² – (y²)² = x⁴ – y⁴
So (x – y) (x + y) (x² + y²) = x⁴ – y⁴

Question 3.
Simplify (5x – 3y)² – (5x + 3y)².
Solution:
We have the identities
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²
So (5x – 3y)² – (5x + 3y)² = (5x)² – (2 × 5x × 3y) + (3y)²
= 5²x² – 30xy + 3² y² – [5²x² – 30xy + 3² y²]
= 25x² – 30xy + 9y² – [25x² + 30xy + 9y²]
= 25x² – 30xy + 9y² – 25x² – 30xy – 9y²
= x² (25 – 25) – xy (30 + 30) + y² (9 – 9)
= 0x² – 60xy + 0y² = – 60 xy
∴ (5x – 3y)² – (5x + 3y)² = -60xy

Question 4.
Simplify : (i) (a + b)² – (a – b)²
(ii) (a + b)² + (a – b)²
Solution:
Applying the identities
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²

(i) (a + b)² – (a – b)² = a² + 2ab + b² – [a² – 2ab + b²]
= a² + 2ab + b² – a² + 2ab – b²
= a² (1 – 1) + ab (2 + 2) + b² (1 – 1)
= 0a² + 4 ab + 0b² = 4ab
(a + b)² – (a – b)² = 4ab

(ii) (a + b)² + (a – b)² = a² + 2ab + b² + (a² – 2ab + b²)
= a² + 2ab + b² + a² – 2ab + b²
= a² (1 + 1) + ab (2 – 2) + b² (1 + 1)
= 2a² + 0 ab + 2b² = 2a² + 2b² = 2 (a² + b²)
∴ (a + b)² – (a – b)² = 2 (a² + b²)

Question 5.
A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m², find the area of lawn.
Solution:
Let the side of the lawn = a m
then side Of big square = (a + 2(2)) m
= (a + 4)m

Area of the path – Area Of large square – Area of smaller square
136 = (a + 4)2 – a2
136 = a² + (2 × a × 4) + 4² – a²
136 = a² + 8a + 16 – a²
136 = 8a + 16
136 = 8 (a + 2)
Dividing by 8
17 = a + 2
Subtracting 2 on both sides
17 – 3 = a + 2 – 2
15 = a
∴ side of small square = 15 m
Area of square = (side × side) Sq. units
∴ Area of the lawn = (15 × 15)m² = 225 m²
∴ Area of the lawn = 225 m²

Question 6.
Solve the following inequalities.
(i) 4n + 7 > 3n + 10, n is an integer
(ii) 6(x + 6) > 5 (x – 3), x is a whole number.
(iii) -13 < 5x + 2 < 32, x is an integer.
Solution:
(i) 4n + 7 > 3n + 10, n is an integer.
4n + 7 – 3n > 3n + 10 – 3n
n(4 – 3) + 7 > 3n + 10 – 3n
n (4 – 3) + 7 > n (3 – 3) + 10
n + 7 > 10
Subtracting 7 on both sides
n + 7 – 7 > 10 – 7
n > 3
Since the solution is an integer and is greater than or equal to 3, the solution will be 3,
4, 5, 6, 7, …..
n = 3, 4, 5, 6,7, ….

(ii) 6 (x + 6) > 5 (x – 3), x is a whole number.
6x + 36 > 5x – 15
Subtracting 5x on both sides
6x + 36 – 5x > 5x – 15 – 5x
x (6 – 5) + 36 > x(5 – 5) – 15
x + 36 > -15
Subtracting 36 on both sides
x + 36 – 36 > -15 -36
x > -51
The solution is a whole number and which is greater than or equal to -51
∴ The solution is 0, 1, 2, 3, 4,…
x = 0,1,2, 3,4,…

(iii) -13 < 5x + 2 < 32, x is an integer.
Subtracting throughout by 2
-13 – 2 < 5x + 2 – 2 < 32 – 2
-15 < 5x < 30
Dividing throughout by 5
\(\frac { -15 }{ 5 } \) < \(\frac { 5x }{ 5 } \) < \(\frac { 30 }{ 5 } \)
– 3 < x < 6
∴ Since the solution is an integer between -3 and 6 both inclusive, we have the solution
as -3, -2, -1,0, 1,2, 3, 4, 5, 6.
i.e. x = -3, -2, 0, 1, 2, 3,4, 5 and 6.

Students can Download Maths Chapter 2 Percentage and Simple Interest Additional 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 3 Chapter 2 Percentage and Simple Interest Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
72% of 25 students are good at science. How many are not good at science?
Solution:
Number of students who are good at science
= 72% of 25 = \(\frac { 72 }{ 100 } \) × 25 = 18 students
∴ Number of students who are not good at science
= 25 – 18 = 7 students

Question 2.
A flower garden has 1000 plants. 5% of the plants are roses and 1% are daisy plants. What is the total number of other plants.
Solution:
Total plants = 1000
Number of rose plants = 5 % of 1000 = \(\frac { 5 }{ 100 } \) × 1000 = 50
Number of Daisy plants = 1 % of 1000 = \(\frac { 1 }{ 100 } \) × 1000 = 10
Total of rose and daisy = 50 + 10 = 60
Number of other plants = 1000 – 60 = 940

Question 3.
Find 135 % of 80 ₹.
Solution:
135 % of 80 = \(\frac { 135 }{ 100 } \) × 80 = ₹ 108

Exercise 2.2

Question 1.
Neka bought 72.3m of cloth from a role of 100m. Express the cloth bought in terms of percentage.
Solution:
Total length of the cloth = 100 m
Length of cloth bought = 72.3 m
Percentage of cloth bought = \(\frac { 72.3 }{ 100 } \) = 72.3 %

Question 2.
Convert (i) 88 % (ii) 1.86 % into decimals.
Solution:
(i) 88 % = \(\frac { 88 }{ 100 } \) = 0.88
(ii) 1.86 % = \(\frac { 1.86 }{ 100 } \) = 0.0186

Question 3.
Convert (i) 3.35 (ii) 0.5 into percentage.
Solution:
(i) 3.35 = \(\frac { 335 }{ 100 } \) × 100 % = 335 %
(ii) 0.5 = \(\frac { 5 }{ 10 } \) × 100 % = 50 %

Exercise 2.3

Question 1.
If Gayathri had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Suppose Gayathri had ₹ X in the beginning.
Then money Spend = 75 % of X = \(\frac { 75 }{ 100 } \) X = \(\frac { 3X }{ 4 } \)
Money left with her = X – \(\frac { 3X }{ 4 } \) = \(\frac { 4X-3X }{ 4 } \) = \(\frac { X }{ 4 } \)
But it is given that money left = ₹ 600
i.e. \(\frac { X }{ 4 } \) = 600
X = 600 × 4 = 2400
∴ Gayathri had ₹ 2,400

Question 2.
Mohan gets 98 marks in her exams. This amounts to 56% of the total marks, What are the maximum marks?
Solution:
Let the maximum marks be X. 56 % of X = 98
\(\frac { 56 }{ 100 } \) × (X) = 98
⇒ X = 98 × \(\frac { 100 }{ 56 } \)
X = 175
∴ Maximum marks = 175

Exercise 2.4

Question 1.
On what sum of money lent out at 9% per annum for 6 years does the simple interest amount to ₹ 810?
Solution:
Given Simple Interest I = ₹ 810
Let the sum of money (Principal) be P
Rate of interest r = 9 % Per annum.
Time n = 6 years
I = \(\frac { pnr }{ 100 } \)
810 = \(\frac{P \times 6 \times 9}{100}\)
P = \(\frac{810 \times 100}{6 \times 9}\)
P = ₹ 1500
Sum of money required = ₹ 1500

Question 2.
Find the simple interest on ₹ 1120 for 2 \(\frac { 2 }{ 5 } \) years at the rate of 5% per annum.
Solution:
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P = ₹ 1120
Time n = 2 \(\frac { 2 }{ 5 } \) years
= \(\frac { 12 }{ 5 } \) years
Rate of Interest r = 5 %

∴ I = 1120 × \(\frac { 12 }{ 5 } \) × \(\frac { 5 }{ 100 } \)
= \(\frac { 672 }{ 5 } \)
= ₹ 134.4
Simple interest = = ₹ 134.4

Students can Download Maths Chapter 2 Percentage and Simple Interest 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 3 Chapter 2 Percentage and Simple Interest Intext Questions

Exercise 2.1
Try These (Text book Page No. 28)

Question 1.
Find the percentage of children whose scores fall in different categories given in table below.

Solution:

Try These (Text book Page No. 29)

Question 1.
There are 50 students in class VII of a school. The number of students involved in these activities are :
Scout: 7
Red Ribbon Club : 6
Junior Red Cross : 9
Green Force : 3
Sports : 14
Cultural activity : 11
Find the percentage of students who involved in various activities.
Solution:

Try These (Text book Page No. 30)

Question 1.
Convert the fractions as percentage.
(i) \(\frac { 1 }{ 20 } \)
(ii) \(\frac { 13 }{ 25 } \)
(iii) (i) \(\frac { 45 }{ 50 } \)
(iv) \(\frac { 18 }{ 5 } \)
(v) \(\frac { 27 }{ 10 } \)
(vi) \(\frac { 72 }{ 90 } \)
Solution:
(i) \(\frac { 1 }{ 20 } \)
= \(\frac { 1 }{ 20 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 1 }{ 20 } \) × 100 %
= 5 %

(ii) \(\frac { 13 }{ 25 } \)
= \(\frac { 13 }{ 25 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 13 }{ 25 } \) × 100 %
= 52 %

(iii) \(\frac { 45 }{ 50 } \)
= \(\frac { 45 }{ 50 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 45 }{ 50 } \) × 100 %
= 90 %

(iv) \(\frac { 18 }{ 5 } \)
= \(\frac { 18 }{ 5 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 18 }{ 50 } \) × 100 %
= 360 %

(iv) \(\frac { 27 }{ 10 } \)
= \(\frac { 27 }{ 10 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 27 }{ 10 } \) × 100 %
= 270 %

(iv) \(\frac { 27 }{ 10 } \)
= \(\frac { 27 }{ 10 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 27 }{ 10 } \) × 100 %
= 270 %

(vi) \(\frac { 72 }{ 90 } \)
= \(\frac { 72 }{ 90 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 72 }{ 90 } \) × 100 %
= 80 %

Question 2.
Convert the following percentage as fractions.
(i) 50%
(ii) 75%
(iii) 250%
(iv) 30 \(\frac { 1 }{ 5 } \) %
(v) \(\frac { 7 }{ 20 } \) %
(vi) 90 %
Solution:
(i) 50 %
= \(\frac { 50 }{ 100 } \)
= \(\frac { 5 }{ 10 } \)
= \(\frac { 1 }{ 2 } \)

(ii) 75 %
= \(\frac { 75 }{ 100 } \)
= \(\frac { 3 }{ 4 } \)

(iii) 250 %
= \(\frac { 250 }{ 100 } \)
= \(\frac { 25 }{ 10 } \)
= \(\frac { 5 }{ 2 } \)

(iv) 30 \(\frac { 1 }{ 5 } \) %
= \(\frac{30 \frac{1}{5}}{100}=\frac{\left(\frac{151}{5}\right)}{100}\)
= \(\frac { 151 }{ 500 } \)

(v) \(\frac { 7 }{ 20 } \) %
= \(\frac{\frac{7}{20}}{100}=\frac{7}{20 \times 100}\)
= \(\frac { 7 }{ 2000 } \)

(vi) 90 % = \(\frac { 90 }{ 100 } \) = \(\frac { 9 }{ 10 } \)

Think (Text book Page No. 32)

Question 1.
What is the difference between 0.01 and 1%.
Solution:
0.01 = \(\frac { 1 }{ 100 } \) = 1%
0.01 and 1% are the same.

Question 2.
In a readymade shop there will be a board showing upto 50% off. Most of the people will realize that everything is half of its original price, Is that true?
Solution:
No. Only some of them are half of its original price.

Exercise 2.2
Try These (Text book Page No. 33)

Question 1.
Convert these decimals to percentage.
(i) 0.25
(ii) 0.07
(iii) 0.8
(iv) 0.375
(v) 3.75
Solution:
(i) 0.25
= \(\frac { 25 }{ 100 } \) = 25 %

(ii) 0.07
= \(\frac { 7 }{ 100 } \) = 7 %

(iii) 0.8
= \(\frac { 80 }{ 100 } \) = 80 %

(iv) 0.375
= \(\frac { 375 }{ 1000 } \)
= \(\frac { 375 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= 37.5 %

(v) 3.75
= \(\frac { 375 }{ 100 } \) = 375 %

Try These (Text book Page No. 34)

Question 1.
Write these percentage as decimals
(i) 3 %
(ii) 25 %
(iii) 80 %
(iv) 67 %
(v) 17.5 %
(vi) 135 %
(vii) 0.5 %
Solution:
(i) 3 %
= \(\frac { 3 }{ 100 } \) = 0.03

(ii) 25 %
= \(\frac { 25 }{ 100 } \) = 0.25

(iii) 80 %
= \(\frac { 80 }{ 100 } \) = 0.8

(iv) 67 %
= \(\frac { 67 }{ 100 } \) = 0.67

(v) 17.5 %
= \(\frac { 17.5 }{ 100 } \) = 0.175

(vi) 135 %
= \(\frac { 135 }{ 100 } \) = 1.35

(vii) 0.5 %
= \(\frac { 0.5 }{ 100 } \) = 0.005

Exercise 2.3
Try These (Text book Page No. 38)

Question 1.
Level of water in a tank is increased from 35 litres to 50 litres in 2 minutes, what is the Percentage of increase?
Solution:
Level of water in the tank originally = 35 litres.
Increase in the water level = amount of change = 50 – 35 = 15 litres

Exercise 2.4
Try These (Text book Page No. 41)

Question 1.
Arjun borrowed a sum of ₹ 5,000 from a bank at 5% per annum. Find the interest and amount to be paid at the end of three year.
Solution:
Here Principal (P) = ₹ 5,000
Rate of interest (r) = 5 % Per annum
Time (n) = 3 years
Simple Interest I = \(\frac { pnr }{ 100 } \)
= \(\frac{5000 \times 3 \times 5}{100}\)
= ₹ 750
Amount to be paid A = P + I = ₹ 5,000 + ₹ 750 = ₹ 5,750
I = ₹ 750 ; A = ₹ 5,750

Question 2.
Shanti borrowed ₹ 6,000/- from a Bank for 7 years at 12 % per annum. What amount will clear off her debt?
Solution:
Here principal (P) = ₹ 6,000
Rate of Interest (r) = 12 % Per annum
Time (n) = 4 Years
Simple Interest (I) = \(\frac { pnr }{ 100 } \) =
= \(\frac{6000 \times 7 \times 12}{100}\)
I = ₹ 5,040
Amount to be paid A = P + I = 6,000 + 5,040 = ₹ 11,040

Think (Text book Page No. 43)

Question 1.
In simple interest, a sum of money doubles itself in 10 years. In how many years it will get triple itself.
Solution:
Let the Principal be P and Rate of interest be r % per annum.
Here the number of years n = 10 years
Given in 10 years P becomes 2 P.
A = P + I
After 2 years A = 2P
i.e. 2P = P + I
2P – P = I

Now if the amount becomes triple then A = P + I = 3P
3P = P + I
3P – P = I
2P = I

∴ After 20 years the amount get tripled.

Students can Download Maths Chapter 3 Algebra Ex 3.2 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 3 Chapter 3 Algebra Ex 3.2

Question 1.
Given that x > y. Fill in the blanks with suitable inequality signs.
(i) y [ ] x
(ii) x+ 6 [ ] y + 6
(iii) x² [ ] xy
(iv) -xy [ ] – y²
(v) x – y [ ] 0
Answer:
(i) y [<] x
(ii) x+ 6 [>] y + 6
(iii) x² [>] xy
(iv) -xy [<] – y²
(v) x – y [>] 0

Question 2.
Say True or False.

(i) Linear inequation has almost one solution.
Answer:
False

(ii) When x is an integer, the solution set for x < 0 are -1, -2,..
Answer:
False

(iii) An inequation, -3 < x < -1, where x is an integer, cannot be represented in the number line.
Answer:
True

(iv) x < -y can be rewritten as – y < x
Ans :
False

Question 3.
Solve the following inequations.
(i) x < 7, where x is a natural number.
(ii) x – 6 < 1, where x is a natural number.
(iii) 2a + 3 < 13, where a is a whole number.
(iv) 6x – 7 > 35, where x is an integer.
(v) 4x – 9 > -33, where x is a negative integer.
Solution:
(i) x < 7, where x is a natural number.
Since the solution belongs to the set of natural numbers, that are less than or equal to 7, we take the values of x as 1,2, 3, 4, 5, 6 and 7.

(ii) x – 6 < 1, where x is a natural number.
x – 6 < 1 Adding 6 on the both the sides x – 6 + 6 < 1 + 6
x < 7
Since the solutions belongs to the set of natural numbers that are less than 7, we take the values of x as 1,2, 3, 4, 5 and 6

(iii) 2a + 3 < 13, where a is a whole number.
2a + 3 < 13
Subtracting 3 from both the sides 2a + 3 – 3 < 13 – 3
2a < 10
Dividing both the side by 2. \(\frac { 2a }{ 2 } \) < \(\frac { 10 }{ 2 } \)
a < 5
Since the solutions belongs to the set of whole numbers that are less than or equal to 5 we take the values of a as 0, 1, 2, 3, 4 and 5

(iv) 6x – 7 > 35, where x is an integer.
6x – 7 > 35 Adding 7 on both the sides
6x – 7 + 7 > 35 + 7
6x > 42
Dividing both the sides by 6 we get \(\frac { 6x }{ 6 } \) > \(\frac { 42 }{ 6 } \)
x > 7
Since the solution belongs to the set of integers that are greater than or equal to 7, we take the values of x as 7, 8, 9, 10…

(v) 4x – 9 > -33, where x is a negative integer.
4x – 9 > – 33 + 9 Adding 9 both the sides
4x – 9 + 9 > -33 + 9
4x > – 24
Dividing both the sides by 4
\(\frac { 4x }{ 4 } \) > \(\frac { -24 }{ 4 } \)
x > -6
Since the solution belongs to a negative integer that are greater than -6, we take values of u as -5, -4, -3, -2 and -1

Question 4.
Solve the following inequations and represent the solution on the number line:
(i) k > -5, k is an integer.
(ii) -7 < y, y is a negative integer.
(iii) -4 < x < 8, x is a natural number.
(iv) 3m – 5 < 2m + 1, m is an integer.
Solution:
(i) k > -5, k is an integer.
Since the solution belongs to the set of integers, the solution is -4, -3, -2, -1, 0,… It’s graph on number line is shown below.

(ii) -7 < y, y is a negative integer.
-7 < y
Since the solution set belongs to the set of negative integers, the solution is
-7, -6, -5, -4, -3, -2, -1.
Its graph on the number line is shown below

(iii) -4 < x < 8, x is a natural number.
-4 < x < 8
Since the solution belongs to the set of natural numbers, the solution is
1, 2, 3, 4, 5, 6, 7 and 8.
Its graph on number line is shown below

(iv) 3m – 5 < 2m + 1, m is an integer.
3m – 5 < 2m + 1
Subtracting 1 on both the sides
3m – 5 – 1 < 2m + 1 + 1
3m – 6 < 2m
Subtracting 2m on both the sides 3m- 6 – 2m < 2m -2m
m – 6 < 0
Adding 6 on both the sides m – 6 + 6 < 0 + 6
m < 6
Since the solution belongs to the set of integers, the solution is
6, 5, 4, 3, 2, 1, 0,-1,…
Its graph on number line is shown below

Question 5.
An artist can spend any amount between ₹ 80 to ₹ 200 on brushes. If cost of each brush is ₹ 5 and there are 6 brushes in each packet, then how many packets of brush can the artist buy?
Solution:
Given the artist can spend any amount between ₹ 80 to ₹ 200
Let the number of packets of brush he can buy be x
Given cost of 1 brush = ₹ 5
Cost of 1 packet brush (6 brushes) = ₹ 5 × 6 = ₹ 30
∴ Cost of x packets of brushes = 30 x
∴ The inequation becomes 80 < 30x < 200
Dividing throughout by 30 we get \(\frac { 80 }{ 30 } \) < \(\frac { 30x }{ 30 } \) < \(\frac { 200 }{ 30 } \)
\(\frac { 8 }{ 3 } \) < x < \(\frac { 20 }{ 3 } \) ;
2 \(\frac { 2 }{ 3 } \) < x < 6 \(\frac { 2 }{ 3 } \)
brush packets cannot get in fractions.
∴ The artist can buy 3 < x < 6 packets of brushes,
or x = 3, 4, 5 and 6 packets of brushes.

Objective Type Questions

Question 1.
The solutions set of the inequation 3 < p < 6 are (where p is a natural number)
(i) 4,5 and 6
(ii) 3,4 and 5
(iii) 4 and 5
(iv) 3,4,5 and 6
Answer:
(iv) 3,4,5 and 6

Question 2.
The solution of the inequation 5x + 5 < 15 are (where x is a natural number)
(i) 1 and 2
(ii) 0,1 and 2
(iii) 2, 1,0, -1,-2
(iv) 1, 2, 3..
Answer:
(i) 1 and 2
Hint: 5x + 5 < 15
5x < 15 – 5 = 10
x < \(\frac { 10 }{ 5 } \) = 2

Question 3.
The cost of one pen is ₹ 8 and it is available in a sealed pack of 10 pens. If Swetha has only ₹ 500, how many packs of pens can she buy at the maximum?
(i) 10
(ii) 5
(iii) 6
(iv) 8
Answer:
(iii) 6
Hint:
Price of 1 pen = ₹ 8
Price of 1 pack = 10 × 8 = 80
Number of packs Swetha can buy = x
80x < 500
8x < 50
x < \(\frac { 50 }{ 8 } \) = 6.25
x is a natural number x = 1, 2, 3, 4, 5, 6

Question 4.
The inequation that is represented on the number line as shown below is _______.

(i) -4 < x < 0
(ii) -4 < x < 0
(iii) -4 < x < 0
(iv) -4 < x < 0
(v) -4 < x < 2
Answer:
(v) -4 < x < 2

Students can Download Maths Chapter 4 Information Processing Ex 4.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2

Question 1.
Choose the correct answer

Question (i)
Common prime factors of 30 and 250 are
(a) 2 x 5
(b) 3 x 5
(c) 2 x 3 x 5
(d) 5 x 5
Answer:
(a) 2 x 5
Hint:
Prime factors of 30 are 2 x 3 x 5
Prime factors of 250 are 5 x 5 x 5 x 2
∴ Common prime factors are 2 x 5

Question (ii)
Common prime factors of 36, 60 and 72 are
(a) 2 x 2
(b) 2 x 3
(c) 3 x 3
(d) 3 x 2 x 2
Answer:
(d) 3 x 2 x 2
Hint:
Prime factors of 36 are 2 x 2 x 3 x 3
Prime factors of 60 are 2 x 2 x 3 x 5
Prime factors of 72 are 2 x 2 x 2 x 3 x 3
∴ Common prime factors are 2 x 2 x 3

Question (iii)
Two numbers are said to be co-prime numbers if their HCF is –
(a) 2
(b) 3
(c) 0
(d) 1
Answer:
(d) 1

Question 2.
Using repeated division method find HCF of the following:

1. 455 and 26
2. 392 and 256
3. 6765 and 610
4. 184, 230 and 276

1. 455 and 26 divisor
Solution:

• Step 1: The larger number should be dividend = 455 & smaller number should be divisor = 26
• Step 2: After 1st division, the remainder becomes new divisor & the previous divisor
• becomes next dividend.
• Step 3: This is done till remainder is zero.
• Step 4: The last divisor is the HCF

Answer:
The HCF is 13.

2. 392 and 256
256 is smaller, so it is the 1st divisor

∴ HCF = 8

3. 6765 and 610

∴ HCF = 5

4. 184, 230 and 276
First let us take 184 & 230

∴ 46 is the Hi of 184, and 230.
Now the HCF of the first two numbers is the dividend for the third number.

Answer:
HCF of 184, 230 & 276 is 46

Question 3.
Using repeated subtracting method find HCF of the following:

1. 42 and 70
2. 36 and 80
3. 280 and 420
4. 1014 and 654

1. 42 and 70
Solution:
Let number be m & n
m > n
We do m – n & the result of subtraction becomes new ‘m’. if m becomes less than n, we do n – m and then assign the result as n. We should do this till m = n. When m = n then ‘m’ is the HCF.
42 and 70 now m = 70, n = 42
70 – 42 = 28, now m = 42, n = 28
42 – 28 = 14, now m = 28, n = 14
28 – 14= 14, now m = 14, n = 14
we stop here as m = n
∴ HCF of 42 & 70 is 14

2. 36 and 80
Solution:
36 and 80 m = 80, n = 36
80 – 36 = 44, now n = 44, m = 36
Since n > m, we should do n m
44 – 36 = 8, now n = 8, m = 36
36 – 8 = 28
Similarly, processing, proceeding, we do repeated subtraction
till m = n
28 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 – 4 = 4 now m = n = 4
∴ HCF is 4

3. 280 and 420
Solution:
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140

4. 1014 and 654
Solution:
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294, n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
∴ m – n = 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6
now m = n
∴ HCF of 1014 and 654 is 6

Question 4.
Do the given problems in repeated subtraction method
1. 56 and 12

2. 320,120 and 95

1. 56 and 12
Solution:
56 & 12
Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 = 8 = 4
m – n = 8 – 44. now m = n
HCF of 56 & 12 is 4

2. 320, 120 and 95
Let us take 320 & 120 first m = 320, n = 120
m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = w = 40 → HCF of 320, 120
Now let us find HCF of 40 & 95
m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5
HCF of 40 & 95 is 5 10 – 5 = 5
HCF of 320, 120 & 95 is 5

Question 5.
On a school trip, 56 girls and 98 boys went to Kanyakumari. They were divided into as many groups as possible so that there were equal numbers of girls and boys in each group. Find the largest group possible? (To find the HCF using repeated division method)
Solution:
56 girls & 98 boys. Now, we need to find HCF of 56 & 98
Using repeated division method. So, we first divide. 98 by 56

Answer:
14.

Question 6.
Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168 mm and by 196 mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Solution:
Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
m = 196, n = 168
m – n = 196 -168 = 28 now n = 28, m = 168
m – n = 168 – 28 = 140 now m = 140, n = = 28
m – n = 140 – 28 = 112 now m = 112, n = 28
m – n = 112 -28 = 84 now m = 84, n = 28
m – n = 84 – 28 = = 56 now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28

Students can Download English Lesson 5 All Summer in a Day Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English 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 12th English Solutions Supplementary Chapter 5 All Summer in a Day

Warm Up

(a) What makes the earth the one and only living planet?
(b) Name a few things that make the earth a unique planet.

Answer:

(c) Do you enjoy watching movies? What type of movies do you like to watch? Here are the pictures of a few blockbuster movies.

Classify the movies based on the following categories: Historical/ Comedy/ Horror/ Cartoon/ Science Fiction

Discuss : Gravity is a science fiction movie.

Student A: Gravity, a 2013 block buster science fiction movie received an overwhelming response for its realistic depiction of the various conditions and challenges faced by astronauts to save themselves after their spare shuttle gets’ clobbered by space debris from a destroyed Russian satellite.

Student B: As much wre loved watching Gravity, its important that we get acquainted with various scientific aspects depicted in the movie.

Student C: Let me tel! you about the opening scene of the movie. Three astronauts including Dr. Ryan Stone (played by Sandra Bullock) and Lieutenant Matt Kowalski (played by George Clooney) are seen space walking and having a nice informal chat.

Student D: Allow me to explain what the movie tries to convey to us. The crew members of space shuttle explorer are servicing the Hubble space telescope. Suddenly they are informed by mission control in Houston that a cloud of debris (caused by the destruction of a Russian defunct spaceship) is headed towards their location. They are advised to abort the mission to avoid a nasty collision. But the collision happens. Dr. Stone and Kowalski head for ISS after their shuttle is hit. ISS is destroyed. Kowalski and Stone get separated as their tethers get twisted. She couldn’t dock the Soyuz with Chinese station. She ejects herself from Soyuz via

explosive decompression. She uses pressure from fire extinguisher to push herself towards Tiangong which is also abandoned in space. She enters Shenzhou capsule but is unable to separate the capsule from the space station. The space station begins to break up and the capsule gets separated. As the capsule falls to the earth, it corrects its position and descends through the atmosphere. Parachute opens up. Stone lands in a lake and comes out with unsteady legs.

Question (a)
Which aspect of the movie is real science?
Answer:
It is true that space debris can indeed cause uncontrolled collisions between objects present in lower earth orbit (LEO); the orbit that encloses every orbit below 2000 km, Kessler syndrome, excess of “trash” in the space orbit could cause problems. The world communication can be blacked out.

It is also true that space station can fall out of orbit. ISS has to be boosted to make sure that it maintains the orbit. The process doesn’t take place suddenly. It takes a few years, it is true that the earth’s atmosphere is relative to Earth’s size. In the film, the fantastic shots of earth enveloped a thin translucent layer of atmosphere which was truly mesmerising. The best thing is that they have correctly depicted the atmosphere’s thinness relative to the earth’s size without exaggeration.

Question (b)
Which aspect of the movie is fictional?
Answer:
Informal chats in a space walk is fictional. Every action during a space walk is premeditated in order to avoid unnecessary ambling and minimize the use of oxygen.

The possibility of the Hubble telescope and a destroyed Russian satellite colliding in the space is fictional because of the following reasons:

The Hubble telescope orbits at an altitude of 500 kilometres above the earth. The film claims that the Russian satellite is also in orbit at the same height. It is wrong because this type of communication satellites which are called TDRSS (Tracking and Data Relay Satellite System) stay in the geosynchronous orbit which is much higher (i.e.) almost 35000 km above the earth. The film showing the Hubble Telescope, ISS and Tiangong-1 located close to each other is fictional. In fact, these three man-made structures are not only at different heights but also lie in different orbits around the earth.

The scene in which Dr. Stone’s tears trickle down heT cheeks and then floats is fictional as it doesn’t happen that way in space.

Samacheer Kalvi 12th English All Summer in a Day Textual Questions

1. Based on your understanding of the story, answer the following questions in a sentence or two.

Question (a)
What do children get ready for at the beginning of the story?
Answer:
Children get ready to have a glimpse of Sun in the planet Venus after a passage of seven years.

Question (b)
How is life in the planet Venus described?
Answer:
It was raining continuously for seven years without Sunlight. Trees grew and perished in the planet.

Question (c)
Who is Margot? How is she different from the rest of the children?
Answer:
Margot had joined the children recently. She had spent her first five years in Ohio. She was sent to Venus by her parents. She remembered the appearance of Sun vividly. Other children had seen it only when they were 2 years old. Her knowledge of Sun made her different.

Question (d)
What does Margot like the most – the sun or the rain?
Answer:
Margot liked Sun the most.

Question (e)
What was Margot waiting for? Why did William say that it was a joke?
Answer:
Margot was waiting for the reappearance of the Sun. She believed in the prediction of reappearance of Sun on the 7th year. William hadn’t seen the Sun after (i.e.) for seven whole years. So, he called the possible reappearance of Sun as a joke.

Question (f)
Why does Margot wish to return to the earth?
Answer:
Margot is accustomed to the systematic change of day and night on the earth. She hated the wretched rain which lasted for seven years. She even refused to take a shower because she never felt like living a normal life in Venus. So, she wanted to go back to the earth.

Question (g)
Why did the children lock Margot in a closet?
Answer:
The children, like those in the novel, “Lord of the flies”, consider the logic and reasoning of Margot as stupid. Her faith in the definite appearance of the Sun is ridiculed by her classmates. She asserts that the prediction of the Sun will be true. She even describes Sun like a lemon but hot in nature. Fed up with her optimism, the children lock her up in a closet.

Question (h)
Margot could recall what the sun looked like while the other children could not. Why?
Answer:
Margot had come only 4 years ago to Venus. She had seen the sunlight almost everyday in Ohio. So, she could recall the appearance of the Sun vividly.

Question (i)
How long did the Sun shine on Venus?
Answer:
The Sun shined for only one hour in Venus.

Question (J)
Why did one of the girls wail?
Answer:
One of the girls wailed because rain resumed after one hour of sunlight. She caught a droplet of rain in her palm and reacted to the abrupt end of sunlight and resumption of rain.

2. Based on your understanding of the story, answer the following in three or four sentences.

Question (a)
What is the significance of the particular day described in the story “All summer in a day”?
That particular day all the children were clamouring for a glimpse of sunlight which was predicted for an hour. Only Margot had a clear memory of sunlight that she had seen five years ago. Others had not seen the Sun for about seven long years. Her explanation was an orb like a penny and its colour was that of a lemon, the children were annoyed and could not believe her words of experience.

Question (b)
What happens to Margot while the teacher is out of the classroom?
Answer:
Margot was gazing out of the glass window. William asked her what she was looking at. She didn’t speak. They shoved her and edged away from her. The children in the school hated Margot, the snow faced girl. They hated her for her knowledge of the Sun. She was firm that they would see sunlight that day as per the prediction of the scientists. The boys got annoyed. They seized her roughly and took her to a closet and locked her ignoring her pitiful cries of protest.

Question (c)
How did Margot describe the sun to others?
Margot explained that Sun looked like a penny. When other children objected she said, “It’s like a fire in the stove”. She even wrote a poem, “I think the Sun is a flower, that blooms for just one hour.” One of the boys even protested that Margot did not write the poem.

Question (d)
How did the children react when the sun came out after seven years?
Answer:
It was the colour of flaming bronze and it was very large. And the sky around it was blazing in blue tile colour. The jungle burned with sunlight and the children released from the spell, ran out, yelling into the spring time.

Question (e)
Why did William and the other children bully Margot?
Answer:
William and other children did not have deep living knowledge about the Sun. They had seen it for an hour seven years ago when they were only two years old. But Margot had the fortune to enjoy the warmth of Sunlight till she was 5 years old in Ohio. Others got used to live in the dark but Margot wasn’t comfortable living in the dark planet in the underground tunnel. She longed to see the Sun. She wrote poem on the Sun. She distanced herself from others . and spoke less. The feeling that she was different from them was a cause that made the other students in the school bully her.

Question (f)
What were their feelings towards Margot at the end of the story?
Answer:
At the end of the story, they felt sorry for Margot whom they had locked up. They realized that they had done a cruel thing to her. Their faces had become solemn and pale as they were feeling guilty and moved slowly towards the closet and opened the door to let her out.

Question (g)
What does the title of the story convey?
Answer:
The title of the story suggests, the summer in Venus does not last a few months. It lasts only for an hour in a day. The entire season is capsuled in a single day which is unusual.

3. Answer the following questions in a paragraph of about 100-150 words each.

Question (a)
What is the conflict between Margot and the other children in the story, “All Summer in a day”?
Answer:
Margot had a vivid memory of having seen the Sun till she was five years old on the planet earth. Though she had come to Venus planet school, her heart longed for the ‘Sunny day’. Preparations were going on to send her back to earth because one day she threw tantrums refusing to take a shower. Her intended visit could cost thousands of dollars to her parents. She drew paintings of the Sun and clarified the shape of the Sun and the warmth it generated. All the other students hated her superior understanding of the Sun and her possible return to the earth. She was not their “kind”. She kept her convictions. She refused to mix with them. This was the conflict with Margot and other children in the story.

Question (b)
How do the children react to the long awaited event in the story?
Answer:
Children get really excited. They want to see the touch of sunlight on all form of life in the planet. They want to tan themselves in the new found warmth of the Sun. Children persist that they be allowed to go out to the Sun. The teacher allows them with a warning that they must be back in two hours. They start running and turning their faces up to the sky and feeling the Sun on their cheeks like a warm iron. They take off their jackets to allow the Sun to bum their arms. They gladly shouted “Oh, its better than Sun lamps. Children stood in the great jungle in the Venus. The jungle was in the colour of stones as they had not seen the Sun for years.

The children lay out laughing on the jungle mattresses and heard it sigh and squeak under them resilient and alive. They ran among the trees, they slipped and fell and pushed one another. They played hide and seek. But most of them squinted at the sun until tears ran down their faces. They put their hands up to the yellowness and the amazing blueness, they breathed off the fresh air and listened to the silence in a blessed sea of sound lessness. They looked at everything and savoured everything. Like animals escaped from their caves, they ran shouting in circles. Their mirth continued for one hour.

Question (c)
The sun brought about a positive change in the attitude of the children. Illustrate the statement.
Answer:
After really enjoying the brief summer that lasted for an hour, they quietly returned to the tunnel. Suddenly they remembered that they had locked Margot in a closet and a sense of guilt pervaded among all the students. Margot, who was most anxiously waiting to see the Sun had been locked up. They had the opportunity to play in the brief summer but she had to stay in the dark, feeling lonely. She did not deserve that punishment. Realizing their folly, they went back and opened the door very slowly. There was silence. They let Margot out slowly. The arrogance in their behaviour was gone. They were sorry for Margot who they had punished just because she was sharing her real experiences of ‘sunlight’. She had lost the chance to see the Sun.

Question (d)
Did the children regret having locked Margot in a closet? Answer citing relevantly from the story.
Answer:
Yes, the children did regret. Young children can’t keep feelings like anger and hatred for long. The collective sense of guilt tormented them. The joy they experienced under the sunlight for an hour vanished. A kind of uncomfortable gloom and silence fell among them. They looked at one another with remorsefulness. They couldn’t meet each other’s glances. They felt as if someone had driven them like so many stakes into the floor.

One of the girls said, “Margot”. Another girl whispered, “go on.” They could perceive only silence behind the door. They unlocked the door even more slowly and let Margot out. After all, all her descriptions of the Sun turned out to be correct. They had unjustly punished her. Their arrogance and aversion had abandoned them. They were guilt-driven. They did not know how to appease her. They were penitent but had no words to seek Margot’s forgiveness.

Additional Questions

Question (a)
What kind of vegetation was seen in the forest of Venus? How was it different.
Answer:
The jungle continued to grow tumultuously. The children saw a nest of Octopi, clustering up great arms of flesh like weed, wavering and flowering in that brief spring. It was the colour of rubber and ash, this jungle, from the many years without sunlight was the colour of stones and white cheeses and ink. It was the colour of moon. The grass was not green. A thousand forests had been crushed under the rains. They have grown again up a thousand times to be crushed again. This was the way the jungle existed in Venus.

4. Based on your understanding of the story, complete the story map.

Answer:

5. Find out and encircle the following words in the word grid. (The words have been placed horizontally, vertically, diagonally and even back to front)

Answer:

Now read the sentences below. Complete them appropriately with the words you identified from the grid.

1. The scientists ______ that the Sun would come out on Venus that day after seven years.
2. The children are getting ready for the _____ event.
3. The children Margot as she _____ the Sun.
4. William and other children have bullied her and _____ her in a closet.
5. When the Sun comes out, the children _____ the sun.

Answers:

1. predicted
2. anticipated
3. discriminate, recalls
4. locked
5. experience

Speaking Activity

Pair Work
Ray Bradbury’s “All Summer in a Day” is a piece of science fiction. Discuss plots of similar stories with your partner and share your ideas with the class.

We read a story called “The thief’. Yarmuk is one of the top thieves of the solar system. He overhears a secret about planet ‘X’. There is no crime in planet ‘X’. There are no policemen. Most of the shops had no assistants. They were all honesty stores. Yarmuk wanted details about the location of the planet so that he could steal from a jewel store there and settle for life. He mixes a tranquillizer in his friend’s Makin’s second drink. When Makin loses his consciousness, he obtains the address of a man who knows the coordinates of the planet ‘X’. He steals into the house of that man. He uses a hallucinogen to get the information he wants from the man. He prowls the space parking lot. On visiting a space ship dealer, he collects information on different spacecraft and narrows down on MIG-31 spacecraft for his trip to planet ‘X’. He finds a newly married couple parking in a hotel. He takes a room just next to the couple.

Using a false key, he enters their room. He gasses them both and takes space-port parking lot pass; the electronic card which opens their spaceships air lock and controls its engines and the hyper-wave radio key. He vacates the room explaining that he got a video call from his wife that she was sick. Yarmuk enters the spaceport and explaining his wife’s sudden illness get permission to blast off. He reaches planet ‘X’. The parking charges in the space port are ridiculously cheap. To try to find out, if alarm goes off, he stealthily picks up socks in an unmanned textile showroom. Nothing happens. Emboldened by the attempt he goes to one of the posh jewellery shops.

He makes a mental note of which ever jewels he wants to steal the following day. He takes two big suitcases and stuffs all valuable jewels and rushes back to the hotel. Nobody checks him. He sleeps gladly. The next morning when he leaves the hotel, he is stopped by the security guards with stenguns. He does not understand how his theft got found out. Later he learns that planet ‘X’ is full of telepathists, mind readers and clairvoyants. Stray customers who had seen him at the jewellery and the hotel staff have read his thoughts and just communicated in a flash of a second through their minds. He is sentenced to 20 years in the alien’s jail.

All Summer in a Day About The Author

Ray Douglas Bradbury (August 22, 1920 – June 5, 2012) was an American author and screenwriter. He worked in a variety of genres, including fantasy, science fiction, horror, and fiction. Widely regarded as the most important figure in the development of science fiction as a literary genre, Ray Bradbury’s works evoke the themes of racism, censorship, technology, nuclear war, humanistic values and the importance of imagination. Ray Bradbury is well-known for his incredibly descriptive style.

He employs figurative language (mostly similes, metaphors, and personification) throughout the novel and enriches his story with symbolism. On April 16, 2007, Bradbury received a special citation from the Pulitzer Prize jury “for his distinguished, prolific, and deeply influential career as an unmatched author of science fiction and fantasy.” Bradbury also wrote and consulted on screenplays and television scripts, including Moby Dick and It Came from Outer Space. Many of his works were adapted to comic book, television, and film formats.

All Summer in a Day Summary in English

Introduction
Bradbury’s story “All Summer in a day” revolves around a day’s happening in planet Venus after it is colonized by humans. This story falls under the category of Science fiction. So far, Earth is the only planet supporting human beings and the survival of millions of plants and thousands of animals. But in this story we come across children who are settled in Venus in an underground tunnel.

Excitement of children
Scientist predicted that the Sun would be visible in planet Venus for an hour. Children, in the dark planet, were overjoyed about the event. Each tried to visualise how the Sun would appear. Except Margot, all the children had seen the Sun seven years ago. M&rgot had come only four years ago. Till 5, she was in Ohio and had seen and enjoyed many “sunny days” and summer lasting three months. She says it is an orb like a penny. It is a kind of a ball with yellow flame. Its burning like a stove. The more details she gives, the more she is hated by her peers. Realising the unspoken but obvious contempt in the eyes of the peers, Margot keeps quiet. William teases Margot. Her silence of deep understanding of Sun annoys peers.

Be different and be punished
Margot is different from the students in the school. One day she even threw tantrums and wanted to be sent back to her parents to the earth. Other children realised that it would cost a fortune for her parents. But she was helpless. She stopped talking to the peers who did not hide their hatred. They teased her. She ignored and started anxiously waiting for the brief encounter with the Sun. She deeply believed that as per the prediction of the scientists, they would see the sun for an hour in the Venus planet. Her knowing silence and unwillingness disturbed the peers so much that they forcefully locked her up in a room.

Vivid Memory
When other children had nightmarish dreams recalling how the Sun was 7 years ago, Margot remembered the yellow lemon-like ball of fire she had seen almost everyday in Ohio for 5 long years. She secretly disdained the inadequate knowledge of her peers.

Pride and Prejudice
Margot was in love with the Sun. Margot wrote a poem, “I think Sun is a flower, that blooms for just one hour.” She read out the poem when it was raining outside the classroom. She knew for sure that her friends were dreaming and remembering gold or yellow crayon or coin large enough to buy the world with. They were often woken up by the tatting drum of rainfall. Her unexpressed pride over her knowledge of the Sun had really caused a prejudice among children. They locked her up in the anticipated day of sunlight

All Summer in a Day Summary in Tamil

முன்னுரை:
‘All summer in a day’ (“ஒரு நாளில் கோடை |காலம்” என்ற பிராட்பரியின் இந்தக் கதை வீனஸ் கோளை மனிதன் ஆக்ரமித்தப் பிறகு நடக்கும் நிகழ்வுகளைத் தழுவி சுழல்கிறது. இது அறிவியல் புனைக் கதையாகும். இதுவரை நம் பூமி தான் பல கோடி மரங்களும் மற்றும் பல்லாயிரம் விலங்குகளும், மனிதனும் வாழ்வதற்கு துணையாக நின்றது. ஆனால், இந்தக் கதையில் நாம் காணுவது வீனஸ் கோளில் | குழந்தைகள் பாதாளத்தில் வாழும் அதிசயம்.

குழந்தைகளின் ஆரவாரம்:
விஞ்ஞானிகள் வீனஸ் கோளில் சூரியன் ஒரு மணி நேரம் தென்படும் எனக் கணித்து அறிவித்தனர். இருண்ட கோளில் இருந்த குழந்தைகளுக்கு ஒரே குதூகலம். சூரியன் எங்ஙனம் இருக்கும் என அவரவர் கற்பனை செய்தனர். மார்கோட்டைத் தவிர மற்ற அனைவரும் | சூரியனை ஏழு வருடங்களுக்கு முன்பு பார்த்திருந்தனர். | ஆனால் மார்கோட் 4 வருடம் முன்பு தான் இங்கு வந்ததால், ஐந்து வயது வரை ஓஹையோவில் இருந்த போது நிறைய வெயில் காலங்கள்

அதுவும் மூன்று மாதங்களுக்கு நீடிக்கக்கூடியதாக அமைய அதை அனுபவித்துள்ளாள். சூரியனை மார்கோட் ஒரு பைசா அளவு வட்டப் பந்து என சித்தரிக்கிறாள். அது மஞ்சள் | நிற தீப்பந்து எனவும் சித்தரிக்கிறாள். அடுப்பைப் | போன்று எறியக் கூடியதாக இருந்தது என்கிறாள். சூரியனைப் பற்றி அவள் அதிக தகவலைக் கொடுக்க கொடுக்க அவளின் குழு அவளை வெறுத்தது. குழுவினர் எதிர் பேச்சு பேசாவிடினும் அவர்கள் கண்களில் தெரிந்த வெறுப்பைக் கண்டு மௌனமானாள். வில்லியம், மார்கோட்டை கிண்டல் செய்தான். சூரியன் பற்றி அவள் அறிந்திருந்த ஆழமான தகவல்கள் அவர்களை வெறுப்புறச் செய்தது.

வேறுபட்டு இருந்ததால் தண்டனைக்கு ஆளாக்கப்பட்டாள்.
பள்ளிக்கூடத்தில் மார்கோட் மற்ற மாணவ மாணவியிடம் இருந்து வேறுபட்டு இருந்தாள். ஒரு நாள் தன்னை தன் பெற்றோரைக் காண பூமிக்கு அனுப்பி விடுமாறு ரகளையும் செய்தாள். மற்ற குழந்தைகள் அது மார்கோட்டின் பெற்றோருக்கு பெருத்த செலவினை உண்டாக்கும் என அறிந்தனர். ஆனால், மார்கோட் விடாப்பிடியாக இருந்தாள். தன் மீது வெறுப்பை வெளிப்படுத்திய தன் குழுவினரோடு பேச மறுத்தாள். அவர்கள் மார்கோட்டை கிண்டல் செய்தனர். அதை அவள் பொருட்படுத்தாமல் சூரியனை சற்று நேரம் தரிசிப்பதை எதிர் நோக்கி காத்திருந்தாள். விஞ்ஞானிகள் கூறியது போல் ஒரு மணி நேரம் சூரியனை வீனஸ் கோளில் தரிசிக்கப் போவது உறுதி என நம்பினாள். மார்கோட்டின் மௌனமும், மற்றவரிடத்தே காட்டிய அலட்சியப் போக்கும் வெகுவாக குழுவினரைப் பாதித்ததால் மார்கோட்டை அவர்கள் ஒரு அறையில் போட்டு பூட்டினர்.

தெளிவான நினைவு:
சூரியன் ஏழு வருடங்கள் முன் எங்ஙனம் இருந்தது என்பதைப் பற்றிய பயங்கரக் கனவுகள் குழந்தைகளுக்கு ஏற்பட, மார்கோட்டிற்கோ சூரியனை சிறிய எலுமிச்சை பந்தாய் ஐந்து ஆண்டுகள் ஓஹையோவில் தான் ஒவ்வொரு நாளும் கண்டது நினைவுக்கு வந்தது. குழவினர் அளித்த போதாத தகவல்களை அவள் இரகசியமாக புறக்கணித்தாள்.

மமதையும் பாரபட்சமும்:
மார்கோட் சூரியனிடத்தில் பிரியம் கொண்டிருந்தாள், மார்கோட் ஒரு கவிதை எழுதினாள். சூரியன் ஒரு பூப் போன்றது. ‘அது ஒரு மணி நேரம் தான் பூக்கும்’ என்று, இந்தக் கவிதையை மழை பெய்து கொண்டிருக்கும் போது வகுப்பறையில் வாசித்தாள். | கட்டாயமாக குழுவினர் தங்கம் அல்லது மஞ்சள் | நிறத்திலோ அல்லது உலகை வாங்கவல்ல பெரிய அளவிலான பைசாவையோ சூரியனுக்கு ஒப்பாக கற்பனை செய்து கொண்டிருக்கிறார்கள் என்பதை அறிந்தாள். அவர்கள் மழையோசைக்கு அடிக்கடி விழித்த வண்ணம் இருந்தனர். சூரியனின் பால் தான் அறிந்திருந்த தகவலின் பேரில் உண்டான கர்வம், குழுவினரிடையே பாரபட்ச உணர்வை உண்டாக்கியது. ஆதலால், சூரியன் உதிக்கும் என எதிர்நோக்கிய நாளன்று மார்கோட்டை ஒரு அறையில் தள்ளி பூட்டினர்

7 வருடம் பெய்த மழை:
ஏழு வருடங்களாக தொடர்ந்து மழை பெய்தது. தொடர்ந்து பெய்த மழையால் ஏற்பட்ட பெருத்த வெள்ளப் பெருக்கின் காரணமாக தீவைச் சுற்றி அலைகள் மோதிக் கொண்டிருந்தன. ஏழாயிரம் காடுகள் மழையால் அழிந்தன.

வீனஸ் கோளில் சூரிய உதயம்:
திடீரென்று மழை நின்றது. சூரியன் தோன்றியது. வெண்கல நிறத்தில் மிகப் பெரியதாகக் காணப்பட்டது. சூரியனைச் சுற்றி வானம் மிகுந்த நீல நிறமாய் காணப்பட்டது. காடு சூரிய ஒளியில் பிரகாசிக்க, மாய வசப்பட்டிருந்த குழந்தைகள் திடீரென்று விடுதலை பெற்றது போல் கூக்குரலுடன் ஓடினர். ஆசிரியர் அவர்களை நீண்ட தூரம் செல்லா வண்ணம் எச்சரித்தார். அவர்களுக்கு 2 மணி நேரம் வெளியே சென்று வர அனுமதி கிடைத்தது.

திடீரென்று மழை நின்றது.
சூரியன் தோன்றியது. வெண்கல நிறத்தில் மிகப் பெரியதாகக் காணப்பட்டது.) சூரியனைச் சுற்றி வானம் மிகுந்த நீல நிறமாய் காணப்பட்டது. காடு சூரிய ஒளியில் பிரகாசிக்க, மாய | வசப்பட்டிருந்த குழந்தைகள் திடீரென்று விடுதலை பெற்றது போல் கூக்குரலுடன் ஓடினர். ஆசிரியர் அவர்களை நீண்ட தூரம் செல்லா வண்ணம் எச்சரித்தார். அவர்களுக்கு 2 மணி நேரம் வெளியே சென்று வர அனுமதி கிடைத்தது.

All Summer in a Day Glossary

Textual:

Additional:

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.5 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 3 Chapter 2 Percentage and Simple Interest Ex 2.5

Miscellaneous Practice Problems

Question 1.
When Mathi was buying her flat she had to put down a deposit of \(\frac { 1 }{ 10 } \) of the value of the flat. What percentage was this?
Solution:
Percentage of \(\frac { 1 }{ 10 } \) = \(\frac { 1 }{ 10 } \) × 100 % = 10 %
Mathi has to put down a deposit of 10 % of the value of the flat.

Question 2.
Yazhini scored 15 out of 25 in a test. Express the marks scored by her in percentage.
Solution:
Yazhini’s score = 15 out of 25 = \(\frac { 15 }{ 25 } \)
Score in percentage = \(\frac { 15 }{ 25 } \) × 100% = 60%

Question 3.
Out of total 120 teachers of a school 70 were male. Express the number of male teachers as percentage.
Solution:
Total teachers of the school = 120
Number of male teachers = 70
∴ Percentage of male teacher = \(\frac { 70 }{ 120 } \) × 100 % = \(\frac { 700 }{ 12 } \) %
Score in percentage = 58.33%
Percentage of male teachers = 58.33%

Question 4.
A cricket team won 70 matches during a year and lost 28 matches and no results for two matches. Find the percentage of matches they won.
Solution:
Number of Matches won = 70
Number of Matches lost = 28
“No result” Matches = 2
Total Matches = 70 + 28 + 2 = 100
Percentage of Matches won = \(\frac { 70 }{ 100 } \) × 100 % = 70 %
The won 70% of the matches

Question 5.
There are 500 students in a rural school. If 370 of them can swim, what percentage of them can swim and what percentage cannot?
Solution:
Total number of students = 500
Number of students who can swim = 370
Percentage of students who can swim = \(\frac { 370 }{ 500 } \) × 100 % = 74 %
Number of students who cannot swim = 500 – 370 = 130
Percentage of students who cannot swim = \(\frac { 130 }{ 500 } \) × 100 % = 26 %
i.e. 74% can swim and 26% cannot swim

Question 6.
The ratio of Saral’s income to her savings is 4 : 1. What is the percentage of money saved by her?
Solution:
Total parts of money = 4 + 1 = 5
Part of money saved = 1
∴ Percentage of money saved = \(\frac { 1 }{ 5 } \) × 100% = 20%
∴ 20% of money is saved by Saral

Question 7.
A salesman is on a commission rate of 5%. How much commission does he make on sales worth ₹ 1,500?
Solution:
Total amount on sale = ₹ 1,500
Commission rate = 5 %
Commission received = 5 % of ₹ 1,500 = \(\frac { 5 }{ 100 } \) × 1500 = ₹ 75
∴ Commission received = ₹ 75

Question 8.
In the year 2015 ticket to the world cup cricket match was ₹ 1,500. This year the price has been increased by 18%. What is the price of a ticket this year?
Solution.
Price of a ticket in 2015 = ₹ 1500
Increased price this year = 18% of price in 2015
= 18 % of ₹ 1500 = \(\frac { 18 }{ 100 } \) × 1500
= ₹ 270
Price of ticket this year = last year price + increased price
= ₹ 1500 + ₹ 270 = ₹ 1770
Price of ticket this year = ₹ 1770

Question 9.
2 is what percentage of 50?
Solution:
Let the required percentage be x
x% of 50 = 2
\(\frac { x }{ 100 } \) × 50 = 2
x = \(\frac{2 \times 100}{50}\) = 4 %
∴ 4 % of 50 is 2

Question 10.
What percentage of 8 is 64?
Solution:
Let the required percentage be x
So x % of 8 = 64
\(\frac { x }{ 100 } \) × 8 = 64
x = \(\frac{64 \times 100}{8}\) = 800
∴ 800 % of 8 is 64

Question 11.
Stephen invested ₹ 10,000 in a savings bank account that earned 2% simple interest. Find the interest earned if the amount was kept in the bank for 4 years.
Solution:
Principal (P) = ₹ 10,000
Rate of interest (r) = 2%
Time (n) = 4 years
∴ Simple Interest I = \(\frac { pnr }{ 100 } \)
= \(\frac{10000 \times 4 \times 2}{100}\)
= ₹ 800
Stephen will earn ₹ 800

Question 12.
Riya bought ₹ 15,000 from a bank to buy a car at 10% simple interest. If she paid ₹ 9,000 as interest while clearing the loan, find the time for which the loan was given.
Solution:
Here Principal (P) = ₹ 15,000
Rate of interest (r) = 10 %
Simple Interest (I) = ₹ 9000
I = \(\frac { pnr }{ 100 } \)
9000 = \(\frac{15000 \times n \times 10}{100}\)
n = \(\frac{9000 \times 100}{15000 \times 10}\)
n = 6 years
∴ The loan was given for 6 years

Question 13.
In how much time will the simple interest on ₹ 3,000 at the rate of 8% per annum be the same as simple interest on ?4,000 at 12% per annum for 4 years?
Solution:
Let the required number of years be x
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P₁ = ₹ 3000
Rate of interest (r) = 8 %
Time (n₁) = n₁ years
Simple Interest I₁ = \(\frac{3000 \times 8 \times n_{1}}{100}\) = 240 n₁
Principal (P₂) = ₹ 4000
Rate of interest (r) = 12 %
Time n₂ = 4 years
Simple Interest I₂ = \(\frac{4000 \times 12 \times 4}{100}\)
I₂ = 1920
If I₁ = I₂
240 n₁ = 1920
n₁ = \(\frac { 1920 }{ 240 } \) = 8
∴ The required time = 8 years

Challenge Problems

Question 14.
A man travelled 80 km by car and 320 km by train to reach his destination. Find what percent of total journey did he travel by car and what per cent by train?
Solution:
Distance travelled by car = 80 km.
Distance travelled by train = 320 km
Total distance = 80 + 320 km = 400 km
Percentage of distance travelled by car = \(\frac { 80 }{ 400 } \) × 100 % = 20 %
Percentage of distance travelled by train = \(\frac { 320 }{ 800 } \) × 100 % = 40 %

Question 15.
Lalitha took a math test and got 35 correct and 10 incorrect answers. What was the percentage of correct answers?
Solution:
Number of correct answers = 35
Number of incorrect answers = 10
Total number of answers = 35 + 10 = 45
Percentage of correct answers = \(\frac { 35 }{ 45 } \) × 100 %
= 77.777 % = 77.78 %

Question 17.
The population of a village is 8000. Out of these, 80% are literate and of these literate people, 40% are women. Find the percentage of literate women to the total population?
Solution:
Population of the village = 8000 people
literate people = 80 % of population
= 80 % of 8000 = \(\frac { 80 }{ 100 } \) × 8000
literate people = 6400
Percentage of women = 40 %
Number of women = 40 % of literate people
= \(\frac { 40 }{ 100 } \) × 6400 = 2560
∴ literate women : Total population
= 8000 : 2560
= 25 : 8

Question 18.
A student earned a grade of 80% on a math test that had 20 problems. How many problems on this test did the student answer correctly?
Solution:
Total number of problems in the test = 20
Students score = 80 %
Number of problem answered = \(\frac { 80 }{ 100 } \) × 20 = 16

Question 19.
A metal bar weighs 8.5 kg. 85% of the bar is silver. How many kilograms of silver are in the bar?
Solution:
Total weight of the metal = 8.5 kg
Percentage of silver in the metal = 85%
Weight of silver in the metal = 85% of total weight
= \(\frac { 85 }{ 100 } \) × 8.5 kg
= 7.225 kg
7.225 kg of silver are in the bar.

Question 20.
Concession card holders pay ₹ 120 for a train ticket. Full fare is ₹ 230. What is the percentage of discount for concession card holders?
Solution:
Train ticket fare = ₹ 230
Ticket fare on concession = ₹ 120
Discount = Ticket fare – concession fare = 230 – 120 = ₹ 110

Percentage of discount = 47.83%

Question 21.
A tank can hold 200 litres of water. At present, it is only 40% full. How many litres of water to fill in the tank, so that it is 75 % full?
Solution:
Capacity of the water tank = 200 litres
Percentage of water in the tank = 40%
Percentage of water to fill = Upto 75%
Difference in percentage = 75 % – 40 % = 35 %
∴ Volume of water to be filled = Percentage of difference × total capacity
= \(\frac { 35 }{ 100 } \) × 200 = 70 l
70 l of water to be filled

Question 22.
Which is greater 16 \(\frac { 2 }{ 3 } \) or \(\frac { 2 }{ 5 } \) or 0.17 ?
Solution:
16 \(\frac { 2 }{ 3 } \) = \(\frac { 50 }{ 30 } \)
= \(\frac { 50 }{ 30 } \) × 100 % = 1666.67 %
⇒ \(\frac { 2 }{ 5 } \)
= \(\frac { 2 }{ 5 } \) × 100 = 40 %
0.17 = \(\frac { 17 }{ 100 } \) = 17 %
∴ 1666.67 is greater
∴ 16 \(\frac { 2 }{ 3 } \) is greater

Question 23.
The value of a machine depreciates at 10% per year. If the present value is ₹ 1,62,000, what is the worth of the machine after two years.
Solution:
Present value of the machine = ₹ 1,67,000
Rate of depreciation = 10 % Per annum
Time (n) = 2 years
For 1 year depreciation amount = \(\frac{1,62,000 \times 1 \times 10}{100}\) = ₹ 16,200
Worth of the machine after one year = Worth of Machine – Depreciation
= 1,67,000 – 16,200 = 1,45,800
Depreciation of the machine for 2nd year = 145800 × 1 × \(\frac { 10 }{ 100 } \) = 14580
Worth of the machine after 2 years = 1,45,800 – 14,580 = 1,31,220
∴ Worth of the machine after 2 years = ₹ 1,31,220

Question 24.
In simple interest, a sum of money amounts to ₹ 6,200 in 2 years and ₹ 6,800 in 3 years. Find the principal and rate of interest.
Solution:
Let the principal P = ₹ 100
If A = 6200
⇒ Principal + Interest for 2 years = 6200
A = ₹ 7400
⇒ Principal + Interest for 3 years = 7400
∴ Difference gives the Interest for 1 year
∴ Interest for 1 year = 7400 – 6200
I = 1200
\(\frac { pnr }{ 100 } \) = 1200 ⇒ \(\frac{P \times 1 \times r}{100}\) = 1200
If the Principal = 10,000 then
\(\frac{10,000 \times 1 \times r}{100}\) = 1200 ⇒ r = 12 %
Rate of interest = 12 % Per month

Question 25.
A sum of ₹ 46,900 was lent out at simple interest and at the end of 2 years, the total amount was ₹ 53,466.Find the rate of interest per year.
Solution:
Here principal P = ₹ 46900
Time n = 2 years
Amount A = ₹ 53466
Let r n be the rate of interest per year p
Intrest I = \(\frac { pnr }{ 100 } \)
A = P + I
53466 = 46900 + \(\frac{46900 \times 2 \times r}{100}\)
53466 – 46900 = \(\frac{46900 \times 2 \times r}{100}\)
6566 = 469 × 2 × r
r = \(\frac{6566}{2 \times 469}\) % = 7 %
Rate of interest = 7 % Per Year

Question 26.
Arun lent ₹ 5,000 to Balaji for 2 years and ₹ 3,000 to Charles for 4 years on simple interest at the same rate of interest and received ₹ 2,200 in all from both of them as interest. Find the rate of interest per year.
Solution:
Principal lent to Balaji P₁ = ₹ 5000
Time n₁ = 2 years
Let r be the rate of interest per year
Simple interest got from Balaji = \(\frac { pnr }{ 100 } \) ⇒ I₁ = \(\frac{5000 \times 25 \times r}{100}\)
Again principal let to Charles P₂ = ₹ 3000
Time (n₂) = 4 years
Simple interest got from Charles (I₂) = \(\frac{3000 \times 4 \times r}{100}\)
Altogether Arun got ₹ 2200 as interest.
∴ I₁ + I₂ = 2200
\(\frac{5000 \times 2 \times r}{100}+\frac{3000 \times 4 \times r}{100}\) = 2200
100r + 120r = 2200
220r = 2200 = \(\frac { 2200 }{ 220 } \)
r = 10 %
Rate of interest per year = 10 %

Question 27.
If a principal is getting doubled after 4 years, then calculate the rate of interest. (Hint: Let P = ₹ 100)
Solution:
Let the principal P = ₹ 100
Given it is doubled after 4 years
i.e. Time n = 4 years
After 4 years A = ₹ 200
∴ A = P + I
A – P = I
200 – 100 = I
After 4 years interest I = 100
I = \(\frac { pnr }{ 100 } \) ⇒ 100 = \(\frac{100 \times 4 \times r}{100}\)
4r = 100 ⇒ r = 25 %
Rate of interest r = 25 %

Students can Download Maths Chapter 3 Algebra Ex 3.1 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 3 Chapter 3 Algebra Ex 3.1

Question 1.
Fill in the blanks.
1. (p – q)² = _______
2. The product of (x + 5) and (x – 5) is _______
3. The factors of x² – 4x + 4 are _______
4. Express 24ab²c² as product of its factors is _______
Answers:
1. p² – 2pq + q²
2. x² – 25
3. (x – 2) and (x – 2)
4. 2 × 2 × 2 × 3 × a × b × b × c × c

Question 2.
Say whether the following statements are True or False.
(i) (7x + 3) (7x – 4) = 49 x² – 7x – 12
(ii) (a – 1)² = a² – 1.
(iii) (x² + y²)(y² + x²) = (x² + y²)²
(iv) 2p is the factor of 8pq.
Answers:
(i) True
(ii) False
(iii) True
(iv) True

Question 3.
Express the following as the product of its factors.
(i) 24ab²c²
(ii) 36 x³y²z
(iii) 56 mn²p²
Solution:
(i) 24ab²c² = 2 × 2 × 2 × 3 × a × b × b × c × c
(ii) 36 x³y²z = 2 × 2 × 3 × 3 × x × x × x × y × y × z
(iii) 56 mn²p² = 2 × 2 × 2 × 7 × m × n × n × p × p

Question 4.
Using the identity (x + a)(x + b) – x² + x(a + b) + ab, find the following product.
(i) (x + 3) (x + 7)
(ii) (6a + 9) (6a – 5)
(iii) (4x + 3y) (4x + 5y)
(iv) (8 + pq) (pq + 7)
Solution:
(i) (x + 3) (x + 7)
Let a = 3; b = 7, then
(x + 3) (x + 7) is of the form x² + x (a + b) + ab
(x + 3) (x + 7) = x² + x (3 + 7) + (3 × 7) = x² + 10x + 21

(ii) (6a + 9) (6a – 5)
Substituting x = 6a ; a = 9 and b = -5
In (x + a) (x + b) = x² + x (a + b) + ab, we get
(6a + 9)(6a – 5) = (6a)² + 6a (9 + (-5)) + (9 × (-5))
6² a² + 6a (4) + (-45) = 36a² + 24a – 45
(6a + 9) (6a – 5) = 36a² + 24a – 45

(iii) (4x + 3y) (4x + 5y)
Substituting x = 4x ; a = 3y and b = 5y in
(x + a) (x + b) = x² + x (a + b) + ab, we get
(4x + 3y) (4x – 5y) = (4x)² + 4x (3y + 5y) + (3y) (5y)
= 4² x² + 4x (8y) + 15y² = 16x² + 32xy + 15y²
(4x + 3y) (4x + 5y) = 16x² + 32xy + 15y²

(iv) (8 + pq) (pq + 7)
Substituting x = pq ; a = 8 and b = 7 in
(x + a) (x + b) = x² + x (a + b) + ab, we get
(pq + 8) (pq + 7) = (pq)² + pq (8 + 7) + (8) (7)
= p² q² + pq (15) + 56
(8 + pq) (pq + 7) = p² q² + 15pq + 56

Question 5.
Expand the following squares, using suitable identities.
(i) (2x + 5)²
(ii) (b – 7)²
(iii) (mn + 3p)²
(iv) (xyz – 1)²
Solution:
(i) (2x + 5)²
Comparing (2x + 5)² with (a + b)² we have a = 2x and b = 5
a = 2x and b = 5,
(a + b)² = a² + 2ab + b²
(2x + 5)² = (2x)² + 2(2x) (5) + 5² = 2² x² + 20x + 25
= 2² x² + 20x + 25
(2x + 5)² = 4x² + 20x + 25

(ii) (b – 7)²
Comparing (b – 7)² with (a – b)² we have a = b and b = 7
(a – b)² = a² – 2ab + b²
(b – 7)² = b² – 2(b) (7) + 7²
(b – 7)² = b² – 14b + 49

(iii) (mn + 3p)²
Comparing (mn + 3p)² with (a + b)² we have
(a + b)² = a² + 2ab + b²
(mn + 3p)² = (mn)² + 2(mn) (3p) + (3p)²
(mn + 3p)² = m² n² + 6mnp + 9p²

(iv) (xyz – 1)²
Comparing (xyz – 1)² with (a – b)² we have = a + xyz and b = 1
a = xyz and b = 1
(a – b)² = a² – 2ab + b²
(xyz – 1)² = (xyz)² – 2 (xyz) (1) + 1²
(xyz -1)² = x² y² z² – 2 xyz + 1

Question 6.
Using the identity (a + b)(a – b) = a² – b², find the following product.
(i) (p + 2) (p – 2)
(ii) (1 + 3b) (3b – 1)
(iii) (4 – mn) (mn + 4)
(iv) (6x + 7y) (6x – 7y)
Solution:
(i) (p + 2) (p – 2)
Substituting a = p ; b = 2 in the identity (a + b) (a – b) = a² – b², we get
(p + 2) (p – 2) = p² – 2²

(ii) (1 + 3b)(3b – 1)
(1 + 3b) (3b -1) can be written as (3b + 1) (3b – 1)
Substituting a = 36 and b = 1 in the identity
(a + b) (a – b) = a² – b², we get
(3b + 1)(3b – 1) = (3b)² – 1² = 3² × b² – 1²
(3b + 1) (3b – 1) = 9b² – 1²

(iii) (4 – mn) (mn + 4)
(4 – mn) (mn + 4) can be written as (4 – mn) (4 + mn) = (4 + mn) (4 – mn)
Substituting a = 4 and b = mn is
(a + b) (a – b) = a² – b², we get
(4 + mn) (4 – mn) = 4² – (mn)² = 16 – m² n²

(iv) (6x + 7y) (6x – 7y)
Substituting a = 6x and b = 7y in
(a + b) (a – b) = a² – b², We get
(6x + 7y) (6x – 7y) = (6x)² – (7y)² = 6²x² – 7²y²
(6x + 7y) (6x – 7y) = (6x)² – (7y)² = 6²x² – 7²y²
(6x + 7y) (6x – 7y) = 36x² – 49y²

Question 7.
Evaluate the following, using suitable identity.
(i) 51²
(ii) 103²
(iii) 998²
(iv) 47²
(v) 297 × 303
(vi) 990 × 1010
(vii) 51 × 52
Solution:
51²
= (50 + 1)²
Taking a = 50 and b = 1 we get
(a + b)² = a² + 2ab + b²
(50 + 1)² = 50² + 2 (50) (1) + 1² = 2500 + 100 + 1
51² = 2601

(ii) 103²
103² = (100 + 3)²
Taking a = 100 and b = 3
(a + b)² = a² + 2ab + b² becomes
(100 + 3)² = 100² + 2 (100) (3) + 3² = 10000 + 600 + 9
103² = 10609

(iii) 998²
998² = (1000 – 2)²
Taking a = 1000 and b = 2
(a – b)² = a² + 2ab + b² becomes
(1000 – 2)² = 1000² – 2 (1000) (2) + 2²
= 1000000 – 4000 + 4
998² = 10,04,004

(iv) 47²
47² = (50 – 3)²
Taking a = 50 and b = 3
(a – b)² = a² – 2ab + b² becomes
(50 – 3)² = 50² – 2 (50) (3) + 3²
= 2500 – 300 + 9 = 2200 + 9
47² = 2209

(v) 297 × 303
297 × 303 = (300 – 3) (300 + 3)
Taking a = 300 and b = 3, then
(a + b) (a – b) = a² – b² becomes
(300 + 3) (300 – 3) = 300² – 3²
303 × 297 = 90000 – 9
297 × 303 = 89,991

(vi) 990 × 1010
990 × 1010 = (1000 – 10) (1000 + 10)
Taking a = 1000 and b = 10, then
(a – b) (a + b) = a² – b² becomes
(1000 – 10) (1000 + 10) = 1000² – 10²
990 × 1010 = 1000000 – 100
990 × 1010 = 999900

(vii) 51 × 52
= (50 + 1) (50 + 1)
Taking x = 50, a = 1 and b = 2
then (x + a) (x + b) = x² + (a + b) x + ab becomes
(50 + 1) (50 + 2) = 50² + (1 + 2) 50 + (1 × 2)
2500 + (3) 50 + 2 = 2500 + 150 + 2
51 × 52 = 2652

Question 8.
Simplify: (a + b)² – 4ab
Solution:
(a + b)² – 4ab = a² + b² + 2ab – 4ab = a² + b² – 2ab = (a – b)²

Question 9.
Show that (m – n)² + (m + n)² = 2(m² + n²)
Solution:
Taking the LHS = (m – n)² + (m + n)²

Question 10.
If a + b = 10 , and ab = 18, find the value of a² + b².
Solution:
We have (a + b)² = a² + 2ab + b²
(a + b)² = a² + b² + 2ab
given a + b = 0 and ab = 18
10² = = a² + b² + 2(18)
100 = = a² + b² + 36
100 – 36 = a² + b²
a² + b² = 64

Question 11.
Factorise the following algebraic expressions by using the identity a² – b² = (a + b)(a – b).
(i) z² – 16
(ii) 9 – 4y²
(iii) 25a² – 49b²
(iv) x⁴ – y⁴
Solution:
(i) z² – 16
z² – 16 = z² – 4²
We have a² – b² = (a + b) (a – b)
let a = z and b = 4,
z² – 4² = (z + 4) (z – 4)

(ii) 9 – 4y²
9 – 4y² = 3² – 2² y² = 3² – (2y)²
let a = 3 and b = 2y, then
a² – b² = (a + b) (a – b)
∴ 3² – (2y)² = (3 + 2y) (3 – 2y)
9 – 4y² = (3 + 2y) (3 – 2y)

(iii) 25a² – 49b²
25a2 – 49b2 = 52 – a2 – 72 = (5a)2 – (7b)2
let A = 5a and B = 7b
A²  B²
(5a)² – (7b)² = (5a + 7b) (5a – 7b)

(iv) x⁴ – y⁴
Let x⁴ – y⁴ = (x²)² – (y²)²
We have a² – b² = (a + b) (a – b)
(x²)² – (y²)² = (x² + y²) (x² – y²)
x⁴ – y⁴ = (x² + y²) (x² – y²)
Again we have x² – y² = (x + y) (x – y)
∴ x⁴ – y⁴ = (x² + y²) (x + y) (x – y)

Question 12.
Factorise the following using suitable identity.
(i) x² – 8x + 16
(ii) y² + 20y + 100
(iii) 36m² + 60m + 25
(iv) 64x² – 112xy + 49y²
(v) a² + 6ab + 9b² – c²
Solution:
(i) x² – 8x + 16
x² – 8x + 16 = x² – (2 × 4 × x) + 4²
This expression is in the form of identity
a² – 2ab + b² = (a – b)²
x² – 2 × 4 × x + 4² = (x – 4)²
∴ x² – 8x + 16 = (x – 4) (x – 4)

(ii) y² + 20y + 100
y² + 20y + 100 = y² + (2 × (10)) y + (10 × 10)
= y² + (2 × 10 × y) + 10²
This is of the form of identity
a² + 2 ab + b² = (a + b)²
y² + (2 × 10 × y) + 10² = (y + 10)²
y² + 20y + 100 = (y + 10)²
y² + 20y + 100 = (y + 10) (y + 10)

(iii) 36m² + 60m + 25
36m² + 60m + 25 = 62 m² + 2 × 6m × 5 + 5²
This expression is of the form of identity
a² + 2ab + b² = {a + b)²
(6m)² + (2 × 6m × 5) + 5²
= (6m + 5)²
36m² + 60m + 25 = (6m + 5) (6m + 5)

(iv) 64x² – 112xy + 49y²
64x² – 112xy + 49y² = 82 x² – (2 × 8x × 7y) + 7²y²
This expression is of the form of identity
a² – 2ab + b² = (a- b)²
(8x)² – (2 × 8x × 7y) + (7y)² = (8x – 7y)²
64x² – 112xy + 49y² = (8x – 7y) (8x – 7y)

(v) a² + 6ab + 9b² – c²
a² + 6ab + 9b² – c² = a² + 2 × a × 3b + 3² b² – c²
= a² + (2 × a × 3b) + (3b)² – c²
This expression is of the form of identity
[a² + 2ab + b²] – c² = (a + b)² – c²
a² + (2 × a × 36) + (3b)² – c² = (a + 3b)² – c²
Again this RHS is of the form of identity
a² – b² = (a + b) (a – b)
(a + 3b)² – c² = [(a + 3b) + c] [(a + 3b) – c]
a² + 6ab + 9b² – c² = (a + 3b + c) (a + 3b – c)

Objective Type Questions

Question 1.
If a + b = 5 and a² + b² = 13, then ab = ?
(i) 12
(ii) 6
(iii) 5
(iv) 13
Answer:
(ii) 6
Hint: (a + b)² = 25
13 + 2ab = 25
2ab = 12
ab = 6

Question 2.
(5 + 20)(-20 – 5) = ?
(i) -425
(ii) 375
(iii) -625
(iv) 0
Answer:
(iii) -625
Hint: (50 + 20) (-20 – 5) = -(5 + 20)² = – (25)² = – 625

Question 3.
The factors of x² – 6x + 9 are
(i) (x – 3)(x – 3)
(ii) (x – 3)(x + 3)
(iii) (x + 3)(x + 3)
(iv) (x – 6)(x + 9)
Answer:
(i) (x – 3)(x – 3)
Hint: x² – 6x + 9 = x² – 2(x) (3) + 3²
a² – 2ab + b² – (a- b)² = (x – 3)² = (x – 3) (x – 3)

Question 4.
The common factors of the algebraic expression ax²y, bxy² and cxyz is
(i) x²y
(ii) xy²
(iii) xyz
(iv) x
Ans :
(iv) xy
Hint: ax²y = a × x × x × y
bxy² = b × x × y × y
cxyz = C × x × y × z
Common factor = xy