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.

 

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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

Class 12 English Flamingo and Vistas Book Notes, Lesson Explanation, Question Answers	Class 12 Flamingo Book Chapter wise Video Explanation
Class 12 Flamingo and vistas book MCQs	Class 12 English Flamingo and Vistas book MCQs Videos
Class 12 English Important Questions	Class 12 English Important Questions Videos

 

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।

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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