Tuesday, June 22, 2021

Lost Spring :: Summary In English

 

Lost Spring :: Summary In English

I. “Sometimes I find a rupee in the garbage’ The author comes across Saheb every morning. Saheb left his home in Dhaka long time ago. He is trying to sponge gold in the heaps of garbage in the neighbourhood. The author asks Saheb why he does that. Saheb mutters that he has nothing else to do. There is no school in his neighbourhood. He is poor and works barefooted.

There are 10,000 other shoeless rag-pickers like Saheb. They live in Seemapuri, on the outer edge of Delhi, in structures of mud, with roofs of tin and tarpaulin but devoid of sewage, drainage or running water. They are squatters who came from Bangladesh back in 1971. They have lived here for more than thirty years without identity cards or permit. They have right to vote. With ration cards they get grains. Food is more important for survival than identity. Wherever they find food, they pitch their tents that become transit homes. Children grow up in them, and become partners in survival. In Seemapuri survival means rag-picking. Through the years rag-picking has acquired the proportions of a fine art. Garbage to them is gold. It is their daily bread and a roof over their heads.

Sometimes Saheb finds a rupee or even a ten-rupee note in the garbage-heap. Then there is hope of finding more. Garbage has a meaning different from what it means to their parents. For children it is wrapped in wonder, for the elders it is a means of survival.

One winter morning the author finds Saheb standing by the fenced gate of a neighbourhood club. He is watching two youngmen playing tennis. They are dressed in white. Saheb likes the game but he is content to watch it standing behind the fence. Saheb is wearing discarded tennis shoes that look strange over his discoloured shirt and shorts. For one who has walked barefoot, even shoes with a hole is a dream come true. But tennis is out of his reach.

This morning Saheb is on his way to the milk booth. In his hand is a steel canister. He works in a tea stall. He is paid 800 rupees and all his meals. Saheb is no longer his master. His face has lost the carefree look. He doesn’t seem happy working at the tea-stall. II. I Want to Drive a Car The author comes across Mukesh in Firozabad. His family is engaged in bangle making, but Mukesh insists on being his own master. “I will be a motor mechanic,” he announces. “I will learn to drive a car,” he says.

Firozabad is famous for its bangles. Every other family in Firozabad is engaged in making bangles. Families have spent generations working around furnaces, welding glass, making bangles for women. None of them know that it is illegal for children like Mukesh to work in the glass furnaces with high temperatures, in dingy cells without air and light. They slog their daylight hours, often losing the brightness of their eyes. If the law is enforced, it could get Mukesh and 20,000 children out of the hot furnaces.

They walk down stinking lanes choked with garbage, past homes that remain hovels with crumbling walls, wobbly doors and no windows. Humans and animals, co-exist there. They enter a half-built shack. One part of it is thatched with dead grass. A frail young woman is cooking evening meal over a firewood stove. She is the wife of Mukesh’s elder brother and already in charge of three men-her husband, Mukesh and their father. The father is a poor bangle maker. Despite long years of hard labour, first as a tailor and then as a bangle maker, he has failed to renovate a house and send his two sons to school. All he has managed to do is teach them what he knows: the art of making bangles.

Mukesh’s grandmother has watched her own husband go blind with the dust from polishing the glass of bangles. She says that it is his destiny. She implies that god-given lineage can never be broken. They have been born in the caste of bangle makers and have seen nothing but bangles of various colours. Boys and girls sit with fathers and mothers welding pieces of coloured glass into circles of bangles. They work in dark hutments, next to lines of flames of flickering oil lamps. Their eyes are more adjusted to the dark than to the light outside. They often end up losing their eyesight before they become adults.

Savita, a young girl in a drab pink dress, sits along side an elderly woman. She is soldering pieces of glass. Her hands move mechanically like the tongs of a machine. Perhaps she does not know the sanctity of the bangles she helps make. The old woman beside her has not enjoyed even one full meal in her entire life time. Her husband is an old man with flowing beard. He knows nothing except bangles. He has made a house for the family to live in. He has a roof over his head.

Little has moved with time in Firozabad. Families do not have enough to eat. They do not have money to do anything except carry on the business of making bangles. The youngmen echo the lament of their elders. They have fallen into the vicious circle of middlemen who trapped their fathers and forefathers. Years of mind-numbing toil have killed all initiative and the ability to dream. They are unwilling to get organised into a cooperative. They fear that they will be hauled up by the police, beaten and dragged to jail for doing something illegal. There is no leader among them. No one helps them to see things differently. All of them appear tired. They talk of poverty, apathy, greed and injustice.

Two distinct worlds are visibleone, families caught in poverty and burdened with the stigma of caste in which they are born; the other, a vicious circle of money-lenders, the middlemen, the policemen, the keepers of law and politicians. Together they have imposed the baggage on the child that he cannot put it down. He accepts it as naturally as his father. To do anything else would mean to dare. And daring is not part of his growing up. The author is cheered when she senses a flash of it in Mukesh who wants to be a motor mechanic.

Lost Spring Summary In Hindi

I. “कभी-कभी मुझे कूड़े के ढेर में एक रुपया मिल जाता है।
लेखिका की प्रतिदिन साहेब से भेंट होती है। वर्षों पहले साहेब बांग्लादेश में अपना घर छोड़कर आ गया था। वह पड़ोस में कूड़े के ढेरों से सोना खंगालने का प्रयास कर रहा होता है। लेखिका साहेब से पूछती है कि वह ऐसा क्यों करता है। साहेब बड़बड़ाता है कि उसके पास करने को और कुछ नहीं है। उसके पड़ोस में कोई विद्यालय नहीं है। वह निर्धन है तथा नंगे पैर काम करता है।

साहेब जैसे अन्य 10,000 जूतेविहीन कूड़े के ढेर में से कबाड़ उठाने वाले हैं। ये लोग दिल्ली के बाहरी किनारे पर सीमापुरी में रहते हैं- मिट्टी के घरौंदों में, जिन पर टीन या तिरपाल की छत है किन्तु वे मल-निकास, गन्दे पानी की नालियों अथवा पेयजल से वंचित हैं। ये अनधिकृत रूप से भूमि पर कब्जा करने वाले वे बांग्लादेशी हैं जो 1971 में यहाँ आये थे। वे पिछले 30 वर्ष से बिना किसी पहचानपत्र या आज्ञा-पत्र के रह रहे हैं।

वे मतदान के पात्र हैं। राशन कार्ड की सहायता से उन्हें अनाज मिल जाता है। जीवित रहने के लिए भोजन पहचान-पत्र से कहीं अधिक आवश्यक है। उन्हें जहाँ कहीं भोजन मिल जाता है, वहीं अपने तम्बू लगा लेते हैं जो उनके गमन-भवन बन जाते हैं। उनमें बच्चे बड़े होते हैं, तथा जीवित रहने में भागीदार बन जाते हैं। सीमापुरी में जीवित रहने का अर्थ है कूड़े-करकट को खंगालना। वर्ष बीतने के साथ कूड़ा-करकट में से मूल्यवान वस्तुएँ ढूंढना एक कला का रूप धारण कर लिया है। उनके लिये कूड़ा तो सोना है। यह उनकी दैनिक रोटी है तथा सिर के ऊपर की छत ।

कई बार साहेब कूड़े के ढेर में एक रुपया अथवा दस रुपये का नोट पा लेता है। तब अधिक पाने की आशा होती है। कूड़े-करकट का उनके लिए उनके माता-पिता की समझ से अलग अर्थ है। बच्चों के लिए यह आश्चर्य से लिपटा हुआ है, बड़ों के लिए यह जीवित रहने का साधन है।

सर्दी में एक प्रात:काल लेखिका साहेब को पड़ोस के एक क्लब के कांटेदार बाड़ लगे द्वार के पास खड़ा पाती है। वह दो नवयुवकों को टेनिस खेलते हुए देख रहा है। वे सफेद वस्त्र पहने हुए हैं। साहेब को यह खेल अच्छा लगता है, किन्तु वह इस बाड़ के पीछे खड़े रहकर देखने से ही सन्तुष्ट है। साहेब किसी के त्यागे (फॅके) हुए टेनिस के जूते पहने हुए है जो उसकी रंग उड़ी हुई कमीज तथा निकर पर अजीब लगते हैं। किसी ऐसे व्यक्ति के लिए जो नंगे पैर चला हो, छेद वाला जूता भी एक स्वप्न के सत्य होने जैसा है। किन्तु निस उसकी पहुँच से बाहर है।

इस प्रात:काल साहेब दूध की दुकान की ओर जा रहा है। उसके हाथ में एक स्टील का डिब्बा है। वह एक चाय की दुकान पर काम करता है। उसे 800 रुपये तथा उसके तीने समय का भोजन मिलता है। साहेब अब अपनी मर्जी का मालिक नहीं है। उसके चेहरे से चिन्तामुक्त दिखना लुप्त (गायब) हो गया है। चाय की दुकान में काम करके वह प्रसन्न प्रतीत नहीं होता।

II. मैं कार चलाना चाहता हूँ।”
फिरोजाबाद में लेखिका की मुकेश से भेंट होती है। उसका परिवार चूड़ियाँ बनाने में लीन है किन्तु मुकेश स्वयं अपना स्वामी बनने की जिद्द पर डटा हुआ है। वह घोषण करता है, “मैं एक मोटर-मैकेनिक बनूंगा।” वह कहता है, “मैं कार चलाना सीखेंगा”

फिरोजाबाद अपनी चूड़ियों के लिए प्रसिद्ध है। प्रत्येक दूसरा परिवार चूड़ियाँ बनाने के काम में व्यस्त है। परिवारों ने भट्ठियों के सामने काम करते हुए, शीशे को जोड़ लगाते हुए, स्त्रियों के लिए चूड़ियाँ बनाते हुए कई पीढ़ियाँ बिता दी हैं। उनमें से कोई भी यह नहीं जानता कि मुकेश जैसे छोटे बालक के लिए उच्च तापमान वाली शीशे की भट्ठी पर वायु एवं प्रकाश रहित तंग कोठरी में काम करना अवैध (गैर-कानूनी) है। वे दिन के प्रकाश के पूरे समय कठोर परिश्रम करते रहते हैं, प्रायः अपनी आँखों की चमक खो बैठते हैं। यदि कानून को कठोरता से लागू किया जाये, तो यह मुकेश तथा उस जैसे 20,000 बच्चों को गर्म भट्ठियों से मुक्त कर देगा।

वे बदबूदार तंग गलियों से जो कूड़े-करकट से भरी पड़ी हैं, उन घरों के समीप से गुजरते हुए जाते हैं जो ढहती हुई दीवारों, अस्थिर लटकते हुए दरवाजों एवं खिड़की रहित तंग कोठरियाँ मात्र हैं। यहाँ मानव तथा पशु एक साथ निवास करते हैं। वे आधी निर्मित एक फूहड़ झोपड़ी में पहुँचते हैं। इसके एक भाग में सूखी घास की छत लगी है। एक कमजोर नवयुवती लकड़ी के चूल्हे पर शाम का भोजन बना रही है। वह मुकेश के बड़े भाई की पत्नी है तथा तीन पुरुषों की देखभाल करने वाली है उसका पति, मुकेश तथा उनका पिता। पिता एक निर्धन चूड़ियाँ बनाने वाला है। वर्षों तक कठोर परिश्रम करने के बावजूद, पहले एक दर्जी के रूप में तथा फिर चूड़ियाँ बनाने वाले के रूप में, वह एक मकान को पुनः बनाने तथा अपने दोनों बालकों को विद्यालय भेजने में असमर्थ रहा है। जो कुछ वह उन्हें सिखा पाया है वह वही है जो वह जानता है- चूड़ियाँ बनाने की कला।।

मुकेश की दादी ने चूड़ियों के शीशों की पालिश करने से उड़ी धूल से अपने पति को अन्धा होते हुए देखा है। वह कहती है कि यह उसका भाग्य है। उसका निहित अर्थ है कि प्रभु प्रदत्त कुटुम्ब रेखा नहीं तोड़ी जा सकती। वे चूड़ी निर्माताओं की जाति में उत्पन्न हुये हैं और उन्होंने विभिन्न रंग की चूड़ियों के अतिरिक्त कुछ अन्य नहीं देखा है। लड़के तथा लड़कियाँ अपने माता-पिता के साथ बैठकर रंगीन शीशे के टुकड़ों को जोड़कर चूड़ियों के वृत्त बनाते हैं। वे अंधेरी झोंपड़ियों में तेल के दीयों की टिमटिमाती हुए लौ की पंक्तियों के आगे काम करते हैं। उनकी आँखें बाहर के प्रकाश की अपेक्षा अंधेरे में अधिक अभ्यस्त हैं। वयस्क होने से पहले ही प्राय: वे कई बार अपनी आँखों की ज्योति खो देते हैं।

फीकी गुलाबी पोशाक पहने हुए एक युवा लड़की सविता एक बुजुर्ग महिला के साथ बैठी है। वह शीशे के टुकड़ों को टांके लगा रही है। उसके हाथ किसी मशीन के चिमटों की भाँति मशीनी रूप से चलते हैं। शायद वह उन चूड़ियों की पवित्रता के विषय में नहीं जानती जिनको बनाने में वह सहायता करती है। उसके पास बैठी स्त्री ने जीवनपर्यन्त एक बार भी भरपेट भोजन का आनन्द नहीं लिया है। उसका पति लहराती हुई दाढ़ी वाला वृद्ध व्यक्ति है। वह चूड़ियों के अतिरिक्त कुछ नहीं जानता। उसने परिवार के निवास हेतु एक मकान बनाया है। उसके सिर पर छत है।

फिरोजाबाद में समय के साथ बहुत कम बदलाव हुआ है। परिवारों के पास खाने को पर्याप्त भोजन नहीं है। उनके पास इतना धन नहीं है कि चूड़ियाँ बनाने के धन्धे को जारी रखने के अतिरिक्त कोई अन्य काम कर सकें। वे उन बिचौलियों के कुचक्र में फैंस गए हैं। जिन्होंने उनके पिता तथा दादा-परदादा को जाल में फँसाया था। वर्षों तक मस्तिष्क को सुन्न कर देने वाले परिश्रम ने उनके पहल करने की सभी भावनाओं तथा स्वप्न देखने की सामर्थ्य को समाप्त कर दिया है। वह किसी सहकारी संस्था में संगठित होने के अनिच्छुक हैं। उन्हें भय है कि पुलिस द्वारा उनको ही अवैध कार्य करने के लिए पकड़ा जायेगा, पीटा जायेगा तथा कारागार में डाल दिया जायेगा। उनके मध्य कोई नेता नहीं है। कोई भी उन्हें वस्तुओं को पृथक रूप से देखने में सहायता नहीं करता। वे सब थके हुए प्रतीत होते हैं। वे गरीबी (निर्धनता), उदासीनता, लालच तथा अन्याय की बातें करते हैं।

दो स्पष्ट संसार दिखाई देते हैं-एक, गरीबी में फँसे परिवार, जो कि बोझा ढो रहे हैं उसे कलंक का, जिस जाति में उन्होंने जन्म लिया है; दूसरे, महाजनों, बिचौलियों, पुलिसवालों, कानून के रखवालों तथा राजनीतिज्ञों का दुष्चक्र। उन्होंने एक साथ मिलकर बच्चे पर इतना भार (सामान) लाद दिया है कि वह इसे नीचे भी नहीं रख सकता वह इसे उतने ही स्वाभाविक रूप से स्वीकार कर लेता है, जैसे कि उसके पिता ने किया था। कोई अन्य काम करने का अर्थ होगा-साहस करना तथा साहस करने का उनके बड़े होने में कोई हिस्सा नहीं है। लेखिका को तब प्रसन्नता होती है जब वह मुकेश में इसकी चमक देखती है जोकि मोटर-मैकेनिक (मिस्त्री) बनना चाहता है।


Chapter: 2

LOST SPRING

TEXTBOOK EXERCISES

1. What is Saheb looking for in the garbage dumps? Where is he and where has he come from?

Answer: Saheb is a rag picker. Garbage is wrapped in wonder for him. He is looking for “gold” in the garbage dumps. Sometimes he finds a rupee, even a ten rupee note. If luck favours, he can find a silver coin too. There is always hope of finding something more. Saheb has come from Dhaka in Bangladesh. Now he is living in Seemapuri. It is a settlement of rag pickers at the outskirts of Delhi.

2. What explanations does the author offer for the children not wearing footwear?

Answer: Travelling across the country the author has seen poor children walking barefoot, without shoes. One explanation is that it has become a tradition for them to stay barefoot. But the author doubts it. The lack of money is the most valid explanation. Children like Saheb can’t afford shoes. When Saheb gets a pair of shoes he does wear them.

3. Is Saheb happy working at the tea-stall? Explain

Answer: Saheb doesn’t seem to be happy working  at the tea- stall. Now he feels bound and burdened. The steel canister he holds now is very heavy. The plastic bag he used to carry on his shoulder earlier was very light. The bag was his own. The canister belongs to the master. Saheb is no longer his own master.

4. What makes the city of Firozabad famous?

Answer: Firozabad is famous for its bangles. Every other family in Firozabad is engaged in making bangles. It is the centre of India’s glass- blowing industry. Families have spent generations making bangles for all the women in India.

5. Mention the hazards of working in the glass bangles industry?

Answer: Workers in the glass bangles industry have to work in sub-human conditions. They have to face many health hazards. They go blind with the dust from polishing the glass of bangles. They work in dark hutments. Moreover, the temperature around the furnaces remains unbearably high.

6. How is  Mukesh’s  attitude to his situation different from that of his family?

Answer:  Mukesh belongs to a family of bangle makers. But he has no fascination for bangle- making. He insists on being his own master. He wants to become a motor mechanic. He wants to go to a garage and get the required training for the job.


UNDERSTANDING THE TEXT

1. What could be some of the reasons for the migration of people from villages to cities?

Answer: More and more people are migrating to cities. It has become a general trend. People migrate from villages to cities. The reasons for migration are many. First of all, pressure on land is increasing. Land can’t provide job opportunities for all. Over-population and lack of job opportunities have made the people turn to cities. The second reason of migration is the mechanization of farming. Hence, the landless labourers don’t get work at the farms. They are compelled to move to cities for working in industries. The third reason is the destruction of traditional arts and crafts in the villages. Artisans don’t have any market for their goods and crafts in the villages. They need bigger markets for their products. All these factors lead to the migration of people from villages to cities.

2. Would you agree that promises made to poor children are rarely kept? Why do you think this happens in the incidents narrated in the text?

Answer: Yes, the promises made to the poor children are rarely kept. We live in a hypocritical world. We organise seminars to eliminate child labour in the country. India has the dubious distinction of having maximum numbers of child-workers in the world. The more hazardous the industry, the more child workers it will employ. Take for an example, the firework industry of Sivakashi in Tamil Nadu. Every year scores of children die due to blasts in the factories. The administration sleeps over all these incidents.

                 Anees Jung presents a genuine analysis of poor children employed in rag picking and bangle-making industry. The children of 10,000 rag pickers of Seemapuri expose the hollow claims of the authorities. The worst part is that it happens just on the outskirts of New Delhi. Rag pickers of Seemapuri and the child workers in the glass industries of Firozabad have never been to schools. They don’t have even shoes. They have no dreams, no initiatives. They are the softest targets for exploitation.

3. What forces conspire to keep the workers in the bangle industry of Firozabad in poverty?

Answer: There are certain vested interests and forces that conspire to keep the workers in the bangle industry of Firozabad in poverty. Anees Jung rightly analyses that there are two distinct worlds operating in Firozabad. The First world consists of families engaged in the business of making bangles. They are exploited and are caught in a web of poverty. The other world consists of “ sahukars” or moneylenders, the middlemen and the policemen. Together under their eyes 20,000 children work illegally in glass furnaces with high temperatures. Even the young fall into the vicious circle of middlemen. These agents trapped their fathers and forefathers as well. If the young get themselves organised they are “hauled up” by the police. “ Years of mind-numbing toil” has killed all initiative in them. They can’t think of organising themselves into a cooperative. Powerful people keep the workers in bangle industry helpless and poor.

TALKING ABOUT THE TEXT

1. How in your opinion, can Mukesh realise his dream?

Answer: Mukesh belongs to a family of bangle-makers. The bangle makers of Firozabad are condemned to lead a life of poverty, misery and exploitation. But Mukesh seems to be an exception. He has not let poverty kill his dreams. He doesn’t want to follow the traditional job of making bangles. He thinks and acts differently than the other members of his family. He dreams to be a motor mechanic. He wants to drive a car one day. Mukesh seems to be determined. It is said that fortune favours the brave. Mukesh’s dream can be converted into a reality. Only he will have to find out a garage where he can be admitted as an apprentice. Within no time he will graduate himself to be a good mechanic. If he wants to become a taxi –driver, first of all he will have to learn how to drive a car. He will get a licence only when he clears the driving test. After that he can join any travel agency as a driver.

2. Mention the hazards of working in the glass bangles industry?

Answer: Working in the glass bangles industry is quite hazardous. It employs about 20,000 children of tender ages. None of them knows that it is illegal for children to work in such industries. But nobody cares for the law in Firozabad. The bangle-makers work in glass furnaces with high temperature. They work in dingy cells without air and light. Their eyes are more adjusted to the dark than to the light outside. That is why they often end up losing their eye sights before they become adults. Mukesh’s grandfather became blind with the dust from polishing the glass of bangles. Year of mind-numbing toil have killed all their initiative and the ability to dream. Thousands of boys and girls sit with their fathers and mothers in dark hutments. They shape pieces of coloured glass into circles of bangles.

3. Why should child labour be eliminated and how?

Answer: It is crying shame that India has the maximum number of child-workers in the world. It is a stigma that puts our heads in shame. Childhood is the most tender age. A child needs love and care. It is quite unfortunate that all the major industries employ a large number of child-workers. About 20,000 children work in the furnaces with high temperature in bangles industry in Firozabad. Carpet industry of Mirzapur employs thousands of children. The worst criminals are the firework-factories in Sivakasi. Employment of children in such hazardous industries is illegal. It is banned by the law.

                  But the laws against child-labour don’t have teeth in them. Those who employ children must be punished. And those who employ them in hazardous industries must be sent behind the bars. Only exemplary punishment can put an end to this shameful practice.

Important Question for CBSE Class 12 English Flamingo Book Chapter 1 The Last Lesson

 

Important Question for CBSE Class 12 English Flamingo Book Chapter 1 The Last Lesson

 

Video of The Last Lesson Important Questions

 

The Last Lesson 2 Marks Important Questions (30 to 40 words)

Q.- Why did Franz not want to go to school that day?

A. M Hamel had asked the class to revise the grammar topic of Participles for a test. Franz did not know participles and feared the scolding. So, he did not want to go to school.

 

Q.- How is the mother tongue important to a person? What does M Hamel, the teacher say about it?

A. Mother tongue is the common factor which unites the countrymen. M Hamel made the villagers realize the importance of the mother tongue. He spoke about the beauty of their mother tongue – the French language. He asked the class  to guard it because it was the key to their freedom.

 

See Video of The Last Lesson Important Questions

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Q1.- i. Why did the elders of the village attend the Last Lesson?
ii. Comment on the significance of the villagers sitting at the back in M Hamel’s classroom.

A. Berlin had ordered that French language would no longer be taught across schools in Alsace and Lorraine. The village elders were present in the class which was the last class of the French language. They were there to pay respect to the teacher, M Hamel who had taught there for forty years. They regretted not having attended school in their childhood days.

 

Q2.- What words did M Hamel write on the blackboard before dismissing the class? What did they mean?

A. Before dismissing the class, M Hamel wrote the following words on the blackboard – “Vive la France”. “Vive la France” means ‘Long live France’. It was a way of showing his love and support towards his mother tongue and his country.

 

Q3.- What changes did the order from Berlin bring about on the day of the last lesson?

A. The order from Berlin brought all the routine hustle-bustle of the school life to a stand- still. The teacher, M. Hamel, was kind towards his students and taught with more patience. The students became more attentive and concerned about education.

 

Q4.- How was Mr Hamel dressed differently that day? Why?

A. M Hamel was dressed in his special dress that he wore on a few occasions. It consisted of his beautiful green coat, frilled shirt and a little black silk cap, all embroidered. He wore the special dress because it was the last lesson that he would deliver in the school where he had been teaching for the last forty years.

 

Q5.- After sitting down at his desk, what unusual thing did Franz observe about M Hamel?

A. Franz observed that M. Hamel was wearing his special dress that he wore on selected occasions only. He was not holding the ruler in his hands. He was calm and kind towards the students. The village elders were present in the class too.

 

Q6.- Why was Franz not scolded for reaching school late that day?

A. M Hamel was very kind and calm that day because it was the last lesson in French. Berlin had ordered that instead of French, Germany would be taught in the schools of Alsace and Lorraine. So, M Hamel did not scold Franz for being late.

 

Q7.- How did M Hamel make his last lesson a special one? What did he emphasize on in it?

A. M Hamel made the last lesson by wearing his special dress to the class. He got new copies for the students which had the words “France, Alsace” written beautifully on them.

 

Q8.- How did Franz react to the declaration that it was their last French lesson?

A. Franz was shocked to know that he could not learn French anymore. He repented that he had not been serious before. He wished he had revised participles and would be able to answer M Hamel’s questions. Franz gained a sudden liking for his teacher and did not want him to leave.

 

Class 12 English Important Links

 

The Last Lesson 5 Marks Important Questions (120 to 150 words)

Q1. How different from usual was the atmosphere at school on the day of the last lesson?           

A. There was a lot of sadness due to the order from Berlin that only the German language would be taught in the schools of Alsace and Lorraine. The school was unusually still and quiet as was on a Sunday morning. There was no sound of the opening and closing of desks and lessons being repeated in unison.

The French teacher, M. Hamel, was wearing his special dress which he wore only on special occasions like inspection and prize days. There was also a profound change in his behavior and he was more polite than usual.

The back benches of the classroom were occupied by the village elders who were attending the class as a mark of respect for their mother-tongue.

 

Q2. What shows Mr Hamel’s love for the French language?

A. M Hamel was a linguistic chauvinist. He was proud of French, his mother tongue. He said that French was the most beautiful, clearest and logical language. He wrote the words “Vive la France” meaning ‘long live France’ on the blackboard. He added that people of a nation had their mother tongue as the key to freedom from slavery.

M Hamel added that Frenchmen had not been serious in learning their mother tongue and now, the enemy soldiers would mock at them. The French would be enslaved easily. They had been procrastinating but now the opportunity had come to an end. The village elders had also not studied during their childhood and felt sorry for themselves. The teacher, students and their parents – all were to be blamed for the sorry state of affairs.

The Last Lesson : Lesson Summary

 

The Last Lesson :: Lesson Summary In English

Franz started for school very late that morning. He was afraid of being scolded because M. Hamel was to question them on participles, and he did not know the first word about them. He thought of running away and spending the day out of doors. The warm bright day, the chirping birds, and the Prussian soldiers drilling in the open field back of the sawmill were tempting. But he resisted the temptation and hurried off to school.

There was a crowd in front of the bulletin-board near the town-hall. Wachter, the blacksmith asked Franz not to go so fast. He assured the boy that he would get to his school in plenty of time. Usually there was a great bustle when the school began but that day everything was as quiet as Sunday morning.

Through the window Franz saw his classmates, already in their places and M. Hamel walking up and down with his terrible iron ruler under his arm. Franz opened the door and went in. He blushed and was frightened. M. Hamel very kindly asked him to go to his place.

Franz noticed that their teacher had put on his beautiful green coat, his frilled shirt, and the little black silk cap, all embroidered. He wore these only on inspection and prize days. The village people were sitting quietly on the usually empty back benches. Everybody looked sad; and Hauser had brought an old primer.

M. Hamel said that it was the last lesson he would give them. Henceforth, only German was to be taught in the schools of Alsace and Lorraine. The new master would come the next day. This was their last lesson of French. He wanted them to be very attentive.

Franz felt sorry that he had not learnt his lessons properly. The idea that M. Hamel was going away made the narrator forget all about his ruler and how cranky he was. Now Franz understood why M. Hamel had put on his fine Sunday clothes and why the old men of the village were sitting there. They had come to thank the master for his forty years’ faithful service and to show their respect for the country that was theirs no more.

M. Hamel asked Franz to recite, but he stood there silent. The teacher did not scold him. He confessed that his parents and he (the teacher) were at fault. Then he talked of the French language-the most beautiful language in the world—the clearest, the most logical. He asked them to guard it among them and never forget it. Their language was the key to their prison.

Then they had lesson in grammar and writing. The pigeons cooed very low on the roof. Franz thought if they would make even the pigeons sing in German. All the while M. Hamel was sitting motionless in his chair and gazing at one thing or the other. His sister was packing their trunks in the room above as they had to leave the country next day.

After writing, they had a lesson in history, and then the babies chanted their ba, be, bi, bo, bu. Even old Hauser was crying. All at once the church-clock struck twelve and then the midday prayers. At the same moment the trumpets of the Prussians, returning from drill, sounded under the windows. M. Hamel stood up. He wanted to speak but something choked him.

Then he took a piece of chalk and wrote on the blackboard as large as he could “Vive La France!” After this he stopped and leaned his head against the wall. Without a word, he made a gesture with his hand to indicate that the school was dismissed and they might go.

The Last Lesson Summary In Hindi

उस प्रात:काल फ्रेन्ज विद्यालय के लिए बहुत देर से चला। उसे धमकाये जाने का भय था क्योंकि मि० हैमल उनसे Participles के विषय में प्रश्न पूछने वाले थे, जबकि वह उसके विषय में पहला अक्षर भी नहीं जानता था। उसने भाग जाने तथा खुले में दिन व्यतीत करने की सोची। गर्म, चमकीला दिन, चहचहाते हुए पक्षी तथा आरा मशीन के पिछवाड़े खुले खेतों में अभ्यास करते हुए प्रशिया के सैनिक, प्रलोभित करने वाले थे। किन्तु उसने प्रलोभन को दबाया तथा शीघ्रता से विद्यालय की ओर चला गया।

टाउन हॉल के समीप सूचना-पट्ट के सामने भीड़ थी। लोहार वैचर ने फ्रेन्ज को इतनी तीव्रता से न जाने को कहा। उसने बालक को विश्वास दिलाया कि वह अपने विद्यालय में समय से पहले पहुँच जाएगा। प्रायः विद्यालय आरम्भ होने के समय काफी कोलाहल होता था किन्तु उस दिन प्रत्येक चीज़ ऐसे शान्त थी जैसे कि रविवार को प्रात:काल हो।

खिड़की से फ्रेन्ज ने देखा कि उसके सहठी पहले ही अपने-अपने स्थान पर थे तथा मि० हैमल अपने बाजू के नीचे लोहे का अपना भयंकर पैमाना (लकीर खींचने की पटरी) रखे इधर-उधर घूम रहे थे। फ्रेन्ज ने द्वार खोला तथा भीतर चला गया। शर्म से उसके चेहरे पर लालिमा आ रही थी तथा वह भयभीत था। मि० हैमल ने अत्यन्त दया भाव से उसे अपने स्थान पर जाने को कहा।

फ्रेन्ज ने ध्यान से देखा कि उनके अध्यापक ने अपना सुन्दर हरा कोट, झालरदार कमीज तथा छोटी काली रेशमी टोपी पहनी हुई थी। इन सबपर कसीदा कढ़ा हुआ था। वह इन्हें निरीक्षण अथवा पारितोषिक वितरण के दिन पहनते थे। ग्रामीण लोग चुपचाप प्रायः खाली रहने वाली पीछे की सीटों पर बैठे हुए थे। प्रत्येक व्यक्ति दु:खी दिखाई देता था। वृद्ध हॉसर तो एक पुरानी प्रवेशिका (वर्णमाला की बच्चों की पुस्तिका) भी ले आया था।

मि० हैमल ने कहा कि यह अन्तिम पाठ था जो वह उन्हें पढ़ायेगा। इसके पश्चात Alsace तथा Lorrain के विद्यालयों में केवल जर्मन भाषा ही पढ़ाई जायेगी। नया अध्यापक अगले दिन आ जायेगा। यह उनका फ्रांसीसी भाषा का अन्तिम पाठ था। वह चाहता था कि वे अत्यन्त ध्यानपूर्वक रहें।

फ्रन्ज को अफसोस (खेद) हुआ कि उसने अपना पाठ उचित ढंग से याद नहीं किया था। मि० हैमल के जाने की खबर से वर्णनकर्ता यह भूल गया कि उनका पैमाना (पट्टी) और वे (मिस्टर हैमल) कितने क्रूर हैं। अब फ्रेन्ज की समझ में आया कि मि० हैमल ने अपनी सर्वश्रेष्ठ (रविवासरीय) पोशाक क्यों पहनी हुई थी और गाँव के वृद्ध लोग वहाँ क्यों बैठे हुए थे। वे अध्यापक की चालीस वर्ष की स्वामीभक्ति पूर्ण सेवा के लिए उनका धन्यवाद करने तथा उस देश के प्रति सम्मान प्रकट करने आए थे जो अब उनका नहीं था।

मि० हैमल ने फ्रेन्ज को सस्वर गाने को कहा, किन्तु वह वहाँ चुपचाप खड़ा रहा। अध्यापक ने उसे नहीं धमकाया। उसने स्वीकार किया कि स्वयं वह (अध्यापक) तथा फ्रेन्ज के माता-पिता दोषी थे। फिर उसने फ्रांसीसी भाषा के विषय में बातें की कि यह संसार की सबसे सुन्दर भाषा, सबसे साफ एवं अत्यधिक तर्कपूर्ण है। उसने हमें कहा कि अपने बीच इसकी रक्षा करें तथा कभी न भूलें । उनकी भाषा उनके कारावास की कुंजी थी।

फिर उन्हें व्याकरण तथा लेखन के पाठ मिले। छत पर कबूतर बहुत धीमे गुटरगूं कर रहे थे। फ्रेन्ज सोचने लगा कि क्या वे कबूतरों को भी जर्मन भाषा में गाना गवायेंगे। इस पूरे समय मि० हैमल कुर्सी पर गतिहीन बैठे हुए किसी न किसी वस्तु को देखते रहे। उनकी बहन ऊपर वाले कमरे में ट्रंकों में अपना सामान लगा रही थी क्योंकि उन्हें अगले दिन देश छोड़कर जाना था।

लेखन के पश्चात, उनका इतिहास का पाठ था, तथा फिर बच्चों ने मंत्र की भाँति उच्चारित किया—बा, बे, बी, बो, बु। वृद्ध हॉसर भी रो रहा था। अचानक गिरजाघर के घंटे ने बारह बजाये तब दोपहर की प्रार्थना हुई। उस क्षण प्रशिया के सैनिकों के सैन्य अभ्यास से लौटने की तुरही की ध्वनि खिड़कियों के नीचे से आई। मि० हैमल खड़ा हो गया। वह बोलना चाहता था किन्तु किसी चीज़ से उसका गला सँध गया।

फिर उन्होंने चॉक का एक टुकड़ा उठाया तथा, जितने बड़े अक्षरों में वह लिख सकता था, उसने श्यामपट पर लिखा “फ्रांस अमर रहे।” इसके पश्चात् वह रुक गया तथा दीवार के साथ अपना सिर झुका लिया। बिना कोई शब्द बोले उसने हाथ से ही यह इंगित करने की मुद्रा बनाई कि विद्यालय की छुट्टी हो गई थी तथा वे जा सकते थे।

Thursday, June 17, 2021

Class X : Advance Mathematics Solutions

 

LESSON: SET

                                                   Exercise 1.1


1.    For the set A= { x : x Є N and x 10} and ф, find the followings –
(a)   n(A)  and n(ф)             (b) n(Aꓴф) and n(A ꓵ ф)

Solution:

Given, A=    { x : x Є N and x 10}

(a)   n(A)=10           and      n(ф)=0

(b)   n(Aꓴф)= n(A) + n(ф)

        =10 + 0

        = 10

And      n(A ꓵ ф)= n(A) + n(B) – n(A ꓴ ф)

                                        = 10 + 0 – 10


            2.      Let A and B be two sets and U be their Universal set. If n(U)= 120, n(A)= 42, n(B)= 50 and             n(AꓵB)= 21 then find,

(i)                 n (AUB), n(A – B), n(B – A) and n(AˊꓵBˊ)

(ii)               n(Bˊ), n(Aˊ),  n(AꓴB)ˊ

(iii)             n(PUQ) and n(PꓵQ) where P= A – B and Q= AꓵB

(iv)              How many elements are there in the set U – (AUB)

Solution:

Given, n(U)=120,  n(A)= 42,  n(B)= 50,  n(AꓵB)= 21

i)                    We know that

n (AUB)= n(A)+n(B) – n(AꓵB)

                =42+50 – 21

                =71

            Again, n (A  ̶  B)= n(A) ̶  n(AꓵB)

                                       = 42 – 21

                                       = 21

                        n (B ̶  A)= n(B) ̶   n(AꓵB)

                                       = 50 – 21

                                       = 29

And                  n(AˊꓵBˊ)= n(AꓴB)ˊ

                                                   = n(U) – n(AUB)

                                                   = 120 – 71

                                                   = 49

(ii)             n(Bˊ)= n(U) – n(B)

                                    = 120 – 50

                                    = 70

                            n (Aˊ)= n(U) – n(A)

                                    = 120 – 42

                                    = 78

And      n(AUB)ˊ= n(U) – n(AUB)

                                                = 120 – 71

                                                = 49

(iii)              n(PUQ)= n [(A  ̶  B) U (AꓵB)]

                                    = n[(AꓵBˊ) U (AꓵB)]

                                    = n[Aꓵ(BˊUB)]             (distribution law)

                                    = n [AꓵU]

                                    = n(A)

                                    = 42

And      n(PꓵQ)= n [(A  ̶  B) ꓵ (AꓵB)]

                                    = n[(AꓵBˊ) ꓵ (AꓵB)]

                                    = n[ A ꓵ (BꓵBˊ)]

                                    = n [A ꓵ ф]

                                    = n(ф)

                                    = 0

(iv)       n[U – (AUB)] = n(U) – n(AUB)

                                                = 120 – 71

                                                = 49

 

 

3. If n(AꓵB)= 36, n(A – B)= 25, n(B – A)= 20, Then find n(AUB), n(A) and  n(B).

Solution:

We know that,

                                    n(AUB)=  n(A – B) + n(B – A) + n(AꓵB)

                                                = 25 + 20 + 36

                                                = 81

                                    n(A)= n(A – B) + n(AꓵB)

                                            = 25 + 36

                                            = 61

                                    n(B)= n(B – A) + n(AꓵB)

                                            = 20 + 36

                                             = 56

4. Draw a Venn diagram to demarcate the regions AꓵB, A – B and B – A with respect to the question (3) above and hence verify the results obtained already.

Solution:

5. In a class test in mathematics and English, it was found that 55 students have passed in Mathematics, 46 have passed in English and 35 passed in both the subjects. If the number of students who appeared in the test is 100, the find

                                (i) the percentage of unsuccessful students in both subjects.

                                (ii) the percentage of students who have passed in Mathematics only.

                                (iii) the percentage of students who have passed in English only.

Solution:

Let, set of students who have passed in Mathematics be M

Set of students who have passed in English be E

                                Given, n(U)=100,  n(M)= 55,       n(E)= 46,             and  n(MꓵE)= 35

                                We know that,

                                    n(MUE)= n(M) + n(E) – n(MꓵE)

                                                = 55 + 46 – 35

                                                = 66

(i)                 Number of students unsuccessful both subjects= n(MꓵE)ˊ

                                                                                                     = n(U) - n(MUE)

                                                                                               = 100 – 66

                                                                                               = 34

                          Therefore, the percentage of unsuccessful students in both subjects= 34%

(ii)               Number of students who have passed in Mathematics= n(M – E)

                                                                                                         = n(M) - n(MꓵE)

                                                                                                         = 55 – 35

                                                                                                         = 20

          Therefore, the percentage of students who have passed in Mathematics= 20%

(iii)             Number of students who have passed in English = n(E – M)

                                                                                                       = n(E) - n(MꓵE)

                                                                                                       = 46 – 35

                                                                                                       = 11

                                Therefore, the percentage of students who have passed in English= 11%

6.  In a survey of 550 students in a school, it was found that 175 students drink milk, 300 students drink tea and 110 students drink both milk and tea. Find the number of students who drinks neither milk nor tea.

Solution:

Let, the set of students drink milk be M

The set of students drink tea be T

                                Given,  n(M)= 175, n(T)= 300 and n(MꓵT)= 110

                                We know that,

                                    n(MUT)= n(M) + n(T) - n(MꓵT)

                                                = 175 + 300 – 110

                                                = 365

                   Therefore, the number of students who drink neither milk nor tea= 550 – 365

                                                                                                                             = 185

7. In a survey among a group of employees of a central Govt. office in Assam, it was found that 80 of them can speak the Assamese, 70 of them can speak English and 50 of them can speak both Assamese and English. If each of the employees who took part in the survey can speak either Assamese or English or both of the two languages, then find

(i) the total number of employees who participated in the survey?

(ii) how many of them can speak Assamese only?

(iii)     how many of them can speak English only?

Solution:

 Let the set of employees can speak Assamese be A and the set of employees can speak English be E

                                Given, n(A)= 80, n(E)= 70 and n(AꓵE)= 50

(i)                 We know that, n(AUE)= n(A) + n(E) – n(AꓵE)

                                                      = 80 + 70 – 50

                                                      = 100

                          Therefore, the total number of employees= 100

(ii)               the number of employees who can speak Assamese only = n(AUE) – n(E)

                                                                                                              = 100 – 70

                                                                                                              = 30

(iii)             the number of employees who can speak English only = n(AUE) – n(A)

                                                                                                               = 100 – 80

                                                                                                               = 20

8.  In a club of 250 members, it is found that 130 of them drink tea and 85 of them drink tea but not coffee. If each of the members of the club drinks at least one of the items between tea and coffee, the find-

                                (i) how many members drink coffee

                                (ii) how many members drink coffee, but not tea?

               Solution:

                                Let the set of members drink tea be T and

                                The set of members drink coffee be C

                                Given, n(T)= 130, n(TUC)= 250, n(T – C)= 85

(i)                 The number of members who drink coffee, n(C)= n(TUC) – n(T – C)

                                                                                          = 250 – 85

                                                                                          = 165

(ii)                       n(TꓵC)= n(T) + n(C) – n(TUC)

                                          = 130 + 165 – 250

                                          =45

               Therefore, number of members drink coffee, but not tea, n(C – T)= n(C) – n(TꓵC)

                                                                                                                        = 165 – 45

                                                                                                                         = 120

9. In a class of 90 students, 60 students play volleyball, 53 students play badminton and 35 students play both of the games. Find the number of students

                          (i)  who do not play any one of the two games

                          (ii) who play badminton only, but not volleyball

(iii)             who play volleyball only, but not badminton

(iv)              who play atleast one of two games?

              Solution:

                                 Let the set students play volleyball be V and

                                The set of students play badminton be B and set of universal be U

                                Given, n(U)= 90, n(V)= 60, n(B)= 53 and n(VꓵB)= 35

                                Therefore, n(VUB)= n(V) + n(B) - n(VꓵB)

                                                            = 60 + 53 – 35

                                                            = 78

(i)                 number of students who do not play any one of the two games,=n(U)-n(VUB)

                                                                                                                  = 90 – 78

                                                                                                                  = 12

(ii)               the number of students who play badminton only, but not volleyball,

                              n(B – V)= n(B) - n(VꓵB)

                                          = 53 – 35

                                          = 18

(iii)             the number of students who play volleyball only, but not badminton,

                  n(V – B)= n(V) - n(VꓵB)

                              = 60 – 35

                              = 25

(iv)              the number of students who play atleast one of two games,

                                 n(VUB)= n(V) + n(B) - n(VꓵB)

                                = 60 + 53 – 35

                                = 78

10. In a survey among 1500 families of a town it was found that 1263 families have TV, 639 families have radio and 197 families have neither TV nor Radio.

                                (i) how many families in that town have both radio and TV?

                                (ii) how many families have TV only, but no radio?

                                ( iii) how many families have radio only, but no TV.

               Solution:

                                Let total set of families be U

                                The set of families have TV be T

                                The set of families have radio be R

                                Given, n(U)= 1500, n(T)= 1263, n(R)=639 and n(TˊꓵRˊ)= 197

                  Now,        n(TˊꓵRˊ)= n(U) – n(TUR)

ð  197 = 1500 - n(TUR)

ð  n(TUR)= 1500 – 197

ð  n(TUR)= 1306

(i)                 the number of families in that town have both radio and TV,

                          n(TꓵR) = n(T) + n(R) – n(TUR)

                                          = 1263 + 639 – 1303

                                          = 599

(ii)               the number of families have TV only, but no radio,

                          n(T – R)= n(T) - n(TꓵR)

                                          = 1263 – 599

                                          = 664

(iii)             the number of families have radio only, but no TV,

              n(R – T)= n(R) - n(TꓵR)

                              = 639 – 599

                              = 40

 

11. Out of 180 students of a class, 76 of them study Mathematics, 81 of them study Physics and 80 of them study Chemistry. Also, 34 of them study Mathematics and Physics, 30 of them study Mathematics and Chemistry and 33 of them study Physics and Chemistry. If 18 students study all the three subjects then find –

                                (i) the number of students who study only Physics

                                (ii) the number of students who study only Chemistry

                                (iii )the number of students who study only Mathematics

                    (iv ) the number of students who study mathematics and Physics, but not Chemistry

                    ( v) the number of students who study Physics and Chemistry, but not Mathematics

                    (vi ) the number of students who study chemistry and Mathematics, but not Physics

                                (vii ) the number of students who study none of the three subjects.

Solution:

Let the set of total students are U

Set of students study Mathematics be M

Set of students study Physics be P

Set of students study Chemistry be C

Given, n(U)= 180, n(M)= 76, n(P)= 81, n(C)= 80,

n(MꓵP)= 34, n(MꓵC)= 30, n(PꓵC)= 33 and n(MꓵPꓵC)= 18



 (i)                  the number of students who study only Physics= n(P) – n(MꓵP)–n(PꓵC)+ n(MꓵPꓵC)

                                                                              = 81 – 34 – 33 + 18

                                                                              = 32

 

(ii)               the number of students who study only Chemistry=n(C)–n(MꓵC)- n(PꓵC)+n(MꓵPꓵC)

                                                                                 = 80 – 30 – 33 + 18

                                                                                 = 35

 

(iii)             the number of students who study only Mathematics

                                                                  = n(M)–n(MꓵP)–n(MꓵC)+n(MꓵPꓵC)

                                                                  = 76 – 34 – 30 + 18

                                                                  = 30

 

(iv)              the number of students who study mathematics and Physics, but not Chemistry

                                                                  = n(MꓵP) -  n(MꓵPꓵC)

                                                                  = 34 – 18

                                                                  = 16

 

(v)                the number of students who study Physics and Chemistry, but not Mathematics

                                                                  = n(PꓵC) - n(MꓵPꓵC)

                                                                  = 33 – 18

                                                                  = 15

 

(vi)              the number of students who study chemistry and Mathematics, but not Physics

                                                                  = n(CꓵM) - n(MꓵPꓵC)

                                                                  = 30 – 18

                                                                  = 12

 

(vii)            the number of students who study none of the three subjects

= n(U) – n(MUPUC)

= n(U) – [n(M) + n(P) + n(C) – n(MꓵP) – n(PꓵC) – n(MꓵC) + n(MꓵPꓵC)]

= 180 – [ 76 + 81 + 80 – 34 – 33 – 30 + 18]

= 180 – 158

= 22

 

 

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Chapter 1: Set

Lesson: Set

Exercise 1.2

1.       If A = {1, 3, 5} and B= {2, 4, -1} then find A×B and B×A . Find the number of elements in A×B and B×A.

Solution: Given, A = {1, 3, 5} and B= {2, 4, -1}

A×B = {(1, 2), (1, 4), (1, -1), (3, 2), (3, 4), (3, -1), (5, 2), (5, 4), (5, -1)}

B×A= {(2, 1), (2, 3), (2, 5), (4, 1), (4,3), (4, 5), (-1, 1), (-1, 3), (-1, 5)}

Therefore, n(A×B) = 9  and n(B×A)= 9

2.       (a) If A = {-1, 3, 6} and B= {-3, 5} then find A×B and B×A. Also draw the graph of the Cartesian products.

Solution:  (a) Given, A = {-1, 3, 6} and B= {-3, 5}

Now, A×B= {(-1, -3), (-1, 5), (3, -3), (3, 5), (6, -3), (6, 5)}

And B×A= {(-3, -1), (-3, 3), (-3, 6), (5, -1), (5, -3) (5, 6)}

 

 

 (b) If A= {2, 4}, B= {-1, 3, 7}, C={1, 0} then with the help of tree diagram find A×B, B×A and A×B×C.

Solution:  Given A= {2, 4}, B= {-1, 3, 7}, C= {1, 0}

                                  

                                       


A×B= {(2, 1), (2, 3), (2, 7), (4, -1), (4, 3), (4, 7)}


B×A = {(-1, 2), (-1, 4), (3, 2), (3, 4), (7, 2), (7, 4)}

 

 

 A×B×C = {(2, -1, 1), (2, -1, 0), (2, 3, 1), (2, 3, 0), (2, 7, 1), (2, 7, 0), (4, -1, 1), (4, -1, 0), (4, 3, 1), (4, 3, 0), (4, 7, 1), (4, 7, 0)}


3.        If A= {x, y, u, v} and B= {a, b, l, m} then find A×B and represent the Cartesian product by a coordination diagram.

Solution:  Given, A= {x, y, u, v} and B= {a, b, l, m}

A×B= {(x, a), (x, b), (x, l), (x, m), (y, a), (y, b), (y, l), (y, m), (u, a), (u, b), (u, l), (u, m), (v, a), (v, b), (v, l), (v, m)}

 

 

4.       If A= {3, 5}, B= {1, 2, 4}, C= {3, 4, 6} then show that

(i)                  A×(BUC) = (A×B) U (A×C)

(ii)                A×(BꓵC) = (A×B) ꓵ (A×C)

Solution: Given, A= {3, 5}, B= {1, 2, 4}, C= {3, 4, 6}

Now, BUC= {1, 2, 3, 4, 6}

            BꓵC = {4}

           A×B = {(3, 1), (3, 2), (3, 4), (5, 1)


   (A×C)= {(3, 3), (3, 4), (3, 6), (5, 3), (5, 4), (5, 6)}

(i)                  LHS= A×(BUC)

= {3, 5} × {1, 2, 3, 4, 6}

= {(3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (5,1), (5, 2), (5, 3), (5, 4), (5,6)}

                       RHS= (A×B) U (A×C)

                                = {(3, 1), (3, 2), (3, 4), (5, 1), (5, 2), (5, 4)} U {(3, 3), (3, 4), (3, 6), (5, 3), (5, 4), (5, 6)}

                                = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (5,1), (5, 2), (5, 3), (5, 4), (5,6)}

                Therefore, LHS = RHS

(ii)                LHS= A×(BꓵC)

= {3, 5} × {4}

= {(3, 4), (5, 4)}

                        RHS= (A×B) ꓵ (A×C)

                                = {(3, 1), (3, 2), (3, 4), (5, 1), (5, 2), (5, 4)} ꓵ {(3, 3), (3, 4), (3, 6), (5, 3), (5, 4), (5, 6)}

                                = {(3, 4), (5, 4)}

                Therefore, LHS = RHS

Showed

5.                 5. For the set A= {x:x Є I and -2 < x < 4} and B= {y:y is a root of y2 – 9=0} draw the graph of A×B and find n(A×B)

Solution: Given, A= {x: x Є I and -2 < x < 4}

                                    = {-1, 0, 1, 2, 3}

                        B= {y: y is a root of y2 – 9=0}

                        = {-3, 3}

A×B= {(-1, -3), (-1, 3), (0, -3), (0, 3), (1, -3), (1, 3), (2, -3), (2, 3), (3, -3), (3, 3)}

n (A×B)= 10

 

 

 6.        Draw the graphs of A×B and B×A on two separate coordinate planes for the set A={-3, 0, 3} and B= {-1, 1}

Solution: Given, A={-3, 0, 3} and B= {-1, 1}

A×B= {(-3, -1), (-3, 1), (0, -1), (0, 1), (3, -1), (3, 1)}

B×A={(-1,-3), (-1, 0), (-1, 3), (1, -3), (1, 0), (1, 3)}


 

 7. (i) If A=ф and B={-1, 1} then determine the power set of A×B, B×A and B2

(ii ) If A= {0} and B= {1} then determine the A×B and B×A. Also determine P(A×B) and P(B×A)

Solution: (i) Given, A=ф and B={-1, 1}

Now, A×B= ф

                        B×A= ф

Therefore, P(A×B) ={ ф}

                        P(B×A)= { ф}

Again,    B2 = B × B

                        = {-1, 1} × {-1, 1}

                        = {(-1, -1), (-1, 1), (1, -1), (1, 1)}

Therefore,

 P( B2)= { ф, {(-1, -1)},{ (-1, 1)}, {(1, -1)}, {(1, 1)}, {(-1, -1), (-1, 1)}, {(-1, -1),(1, -1)}, {(-1, -1), (1, 1)}, {(-1, 1), (1, -1)}, {(-1, 1), (1, 1)}, {(1, -1), (1, 1)},  {(-1, -1), (-1, 1), (1, -1)}, {(-1, -1), (-1, 1), (1, 1)}, {(-1, -1), (1, -1), (1, 1)}, {(-1, 1), (1, -1), (1, 1)}, {(-1, -1), (-1, 1), (1, -1), (1, 1)}}

(ii ) Given, A= {0} and B= {1}

Therefore, A×B = {(0, 1)}

                        B×A = {(1, 0)}

        P (A×B) = {ф, {(0, 1)}}

        P (B×A) = {ф, {(1, 0)}}


8.       If A= {2, 4} and B = {4, 2}, Can we write A×B = A2 and B×A= B2 ?

Solution: Given, A= {2, 4} and B = {4, 2}

Now, A×B   = {2, 4} × {4, 2}

                        = {(2, 4), (2, 2), (4, 4), (4, 2)}

          A2 = {2, 4} × {2, 4}

                        = {(2, 2), (2, 4), (4, 2), (4, 4)}

And, B×A = {4, 2} × {2, 4}

                        = {(4, 2), (4, 4), (2, 2), (2, 4)}

                        B2 = {4, 2} × {4, 2}B

                            = {(4, 4), (4, 2), (2, 4), (2, 2)}

Therefore, A×B = A2 and B×A= B2

9.    If n(A) = 3 and two of the elements of A × A be (a, a), (b, c) then determine the sets A and A × A.

Solution: Given, n(A) = 3

                        A = {a, b, c}

        Now, A × A = {a, b, c} ×{a, b, c}

                             = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}

10      If A   B and C ⊆ D then show that A×C  B×D

Solution: let (a, b) be one element of A×C

                                           (a, b) Є A×C

ð  a Є A and b Є C

ð  a Є B and b Є D   (since, A   B and C ⊆ D)

ð  (a, b) Є B×D

                Therefore, A ×C  B × D

11.        If A×B= {(m, 1), (n, 3), (m, 3), (n, 1), (m, 2), (n, 2)}, then find A and B.

Solution:  Given,

                        A×B= {(m, 1), (n, 3), (m, 3), (n, 1), (m, 2), (n, 2)}

                                = {(m, 1), (m, 2), (m, 3), (n, 1), (n, 2), (n, 3)}

Therefore, A= {m, n}, B= {1, 2, 3}

12.     If A and B be non-empty sets and A × B= A × C then show that B=C.

Solution: Let y be an arbitrary element of B

Therefore, y Є B

ð  (x,y) Є A×B (ꓯ x Є A)

ð  (x,y) Є A×C ( A×B = A×C)

ð  y Є C      ………………………….(1)

Again, Let x be any arbitary element of C

Therefore, x Є C

ð  (y, x) Є A×C (ꓯ y Є A)

ð  (y, x) Є A ×B (A×C= A ×B)

ð  X Є B      ……………………………(2)

From (1) and (2), we get

                        B=C

13.     A= {1, 2}, B= {1, 3, 2} and C= {4, 2, 1} then show that A2 = B2ꓵC2

Solution: Given, A= {1, 2}, B= {1, 3, 2} and C= {4, 2, 1}

                        A2 = A×A

                            = {1, 2} × {1, 2}

                            = {(1, 1), (1, 2), (2, 1), (2, 2)}

                        B2= B×B

                            = {1, 3, 2} × {1, 3, 2}

                            = {(1, 1), (1, 3), (1, 2), (3, 1), (3, 3), (3,2}, (2, 1), (2, 3), (2, 2)}

                        C2 = {4, 2, 1} × {4, 2, 1}

                             = {(4, 4), (4, 2),(4, 1), (2, 4), (2, 2), (2,1), (1, 4), (1, 2), (1, 1)}

        B2ꓵC2= {(1, 1), (1, 2), (2, 1), (2, 2)}

Therefore, A2 = B2ꓵC2                  (Showed)

14.     If A={0, 1} then find A×A×A

Solution: Given, A= {0, 1}

Now, A × A × A= {0, 1} × {0, 1} × {0, 1}

                              = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)}

 

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