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Kerala State Syllabus 10th Standard Hindi Solutions Unit 4 Chapter 1 बसंत मेरे गाँव का (लेख)

बसंत मेरे गाँव का Text Book Questions and Answers

बसंत मेरे गाँव का विश्लेषणात्मक प्रश्न

Basanth Mere Gav Ka Summary In Hindi Kerala Syllabus प्रश्ना 1.
उत्तराखंड के हिमालयी अंचल में फूलदेई को बच्चों का सबसे बड़ा त्योहार मानते हैं। क्यों?

उत्तर:
फूलदेई के त्योहार में बच्चों की अहम भूमिका रहती है। इसमें बड़ों की भूमिका केवल सलाह देने तक सीमित रहती है। फूलदेई का त्योहार मनाने के लिए पहले बच्चे फूल चुनते हैं। टोकरियों में रखते हैं। सुबह बच्चे गाँव भर घूमकर घरों की देहरियों को फूलों से सजाते हैं। घरवाले बच्चों को चावल, गुड़, आदि दक्षिणा में देते हैं। इक्कीस दिन तक इसी प्रकार सामग्रियाँ इकट्ठी की जाती हैं। इन्हीं चीज़ों से अंतिम दिन सामूहिक भोज बनाते हैं। ये सारे काम बच्चे ही करते हैं। इसलिए इसे बच्चों का सबसे बड़ा त्योहार मानते हैं।

Hindi Notes 10th Class Kerala Syllabus प्रश्ना 2.
‘आपसी विश्वास के दम पर वर्षों से यहाँ यह लेन-देन चल रहा है।’ यहाँ गाँववालों की कौन सी विशेषता प्रकट होती है?

उत्तर:
यहाँ गाँववाले और पशुचारकों के बीच लेन-देन चल रहा है। वे ईमानदार और निस्वार्थ हैं। मनाफे की कोई प्रतीक्षा के बिना आपसी विश्वास से यह व्यापार चलता रहता है।

Prakrithi Varnana In Malayalam Kerala Syllabus प्रश्ना 3.
‘जब तक हिमालय रहेगा, ऋतुओं के बदलने का उल्लास बना रहेगा।’ इसका क्या तात्पर्य है?

उत्तर:
हिमालय की भौगोलिक स्थिति और इन ऋतुओं के बदलने में घना संबंध है। हिमालय अंचल की संस्कृति, कला, जीवन रीति आदि पर इन ऋतुओं का बड़ा असर है। इसलिए ऐसा कहा . गया है। – प्रकृतिवर्णन से युक्त वाक्य लेख से चुनकर लिखें।

बसंत मेरे गाँव का Text Book Activities and Answers

बसंत मेरे गाँव का अभ्यास के प्रश्न

10th Class Hindi Notes Kerala Syllabus प्रश्ना 1.
प्रकृति वर्णन से युक्त वाक्य लेख से चुनकर लिखें। जैसे,
i. मकर संक्रांति के बाद सूरज पंचाचूली के शिखरों से चौखंभा पर्वत की तरफ खिसकना शुरु कर देता है।
ii. बसंत की गुनगुनी धूप जब दोपहरी में तपाने लगती है तब ऊँचे हिमालय शिखरों पर बुराँस चटकने लगते हैं।
रेखांकित अंशों पर ध्यान दें, और उनका विशेष अर्थ समझें। पाठभाग से ऐसे वाक्यों का चयन करें।
उत्तर:

• सूरज पंचाचूली से खिसककर जब नंदा पर्वत तक पहुँचता है तो पहाडों में फमूली के पीले । फूल खिलने लगते हैं।
• पहाड़ी के ढलानों पर खूबसूरती से कटे सीढ़ीनुमा खेतों में गेहूँ की हरियाली के बीच सरसों की पीलाई पसर जाती है।
• बुराँस के फूल पहाडों पर शानदार लालिमा बिछा देते है।
• गाँव की जड़ में बहती गंगा से सटकर बनी सीली सड़क में हलचल बढ़ जाती है।

10th Standard Hindi Notes Kerala Syllabus प्रश्ना 2.
पदत्त कार्य
मौसम के आधार पर भारत की सामाजिक गतिविधियाँ बदलती रहती हैं। यहाँ की ऋतुओं की बातें तो बिलकुल अनोखी हैं। हर ऋतु के साथ कई त्योहार भी जुड़े हुए हैं। सोचें और इस तालिका की पूर्ति करें:

उपरोक्त तालिका के आधार पर मौसम और त्योहार विषय पर निबंध तैयार करें।
उत्तर:

बसंत मेरे गाँव का बदलते त्योहार बदलते मौसम

हमारे त्योहार और मौसम के बीच घना संबंध है। वेदों में प्रकृति को ईश्वर का साक्षात् रूप मानकर उसके हर रूप की वंदना की गई है। इसके अलावा आसमान के तारों और आकाश मंडल की स्तुति कर उनसे रोग और शोक को मिटाने की प्रार्थना की गई है।

ऋतुएँ छह बताई गई हैं – वसंत, ग्रीष्म, वर्षा, शरद्, हेमंत एवं शिशिर । ऋषियों ने ऐसे प्रत्येक ऋतु में प्रत्येक त्योहार और नियम बनाए जिनका पालन करने से व्यक्ति सुखमय जीवन प्यतीत कर सके। वसंत ऋतु में होली, रंग-पंचमी, बसंत पंचमी, नवरात्रि, रामनवमि, हनुमान जयंती और गुरुपूर्णिमा उत्सव मनाए जाते हैं। ग्रीष्म ऋतु में निर्जला एकादशी वट सावित्री व्रत, शीतलाष्टमी, देवशयनी, एकादशी, और गुरुपूर्णिमा त्योहार आते हैं। श्रावण और भाद्रपद वर्षा ऋतु के मास हैं। वर्षा नया जीवन लेकर आती है।

यह माह जुलाई-सितंबर में पडता है। इस ऋतु के तीज, रक्षाबंधन और कृष्णजन्माष्टमी सबसे बड़े त्योहार हैं। शरद् ऋतु वातावरण में स्वाच्छता का प्रसार दिखाई पड़ता है। यह ऋतु अकतूबर से नवंबर के बीच रहती है। इस ऋतु के त्योहार हैं – श्राध्द पक्ष, नवरात्रि, दशहरा करवा चौथ । हेमंत ऋतु हिंदु माह के मार्गशीर्ष और पौष मास के बीच रहती है। इस ऋतु में शरीर प्रायः स्वस्थ रहता है। करवा चौथ, धनतेरस, रूप चतुर्दशी दीपावली, गोवर्धन पूजा, भाई दूज आदि त्योहार पडेंगे। शिशिर ऋतु माघ और फाल्गुन के महीने अर्थात् पतझड़ माह में आती है।

इस ऋतु में प्रकृति पर बुढ़ापा छा जाता है। इस ऋतु से ऋतुचक्र के पूर्ण होने का संकेत मिलता है। यह 15 जनवरी से पूरे फरवरी माह तक रहती है। इस ऋतु में मकर संक्रांति का त्योहार आता है। इसी ऋतु में फाल्गुन मास कृष्ण चतुर्दशी को महाशिवरात्रि का महापर्व मनाया जाता है।

Sslc Hindi Chapter 1 Notes Kerala Syllabus प्रश्ना 3.
i. बसंत फूलदेई का त्योहार लेकर आता है।
ii. देर शाम तक बच्चे फूल चुनते हैं।
iii. सुबह पौ फटते ही बच्चों की टोलियाँ गाँव भर में घूमती हैं।
रेखांकित शब्द क्रिया के किस समय को सूचित करता है? चर्चा करें।

उत्तर:
यहाँ आता है, चुनते हैं और घूमती हैं क्रियाएँ क्रिया के वर्तमान काल में होने की सूचना देती हैं। वर्तामान काल में क्रिया के आठ रूप बनते हैं। नीचे की तालिका से उन्हें समझें।

10th Std Hindi Notes Kerala Syllabus प्रश्ना 4.
ये वाक्य पढ़ें।
( इन फूलों को रिंगाल से बनी खास तरह की टोकरियों में रखा जाता है।)
यहाँ फूलों को रिंगाल से बनी खास तरह की टोकरियों में किसके द्वारा रखा जाता है?
अब यह वाक्य पढ़ें,
(बच्चे इन फूलों को रिंगाल से बनी खास तरह की टोकरियों में रखते हैं।)
इन दोनों वाक्यों पर चर्चा करें।

उत्तर:
पहले वाक्य में कर्म की प्रमुखता है। इस प्रकार के वाक्य को ‘कर्म वाच्य’ कहते हैं। इसमें कर्म ‘फूल’ है। कर्म वाच्य के वाक्यों की क्रियाएँ कर्म के लिंग-वचन के अनुसार बदलती हैं। लेकिन यहाँ कर्म के साथ ‘को’ परर्सा का प्रयोग हुआ है। इसलिए क्रिया का रूप पुल्लिंग एकवचन में है।

दूसरे वाक्य में कर्ता की प्रमुखता है। इसमें कर्ता ‘बच्चे’ है। इस प्रकार के वाक्य को कर्तृवाच्य कहते हैं। कर्तृवाच्य के वाक्यों की क्रियाएँ कर्ता के लिंग-वचन के अनुसार बदलती हैं।

10th Hindi Notes Pdf Kerala Syllabus प्रश्ना 5.
इस प्रकार के अन्य वाक्य ढूँढकर लिखें।
उत्तर:
1. पिछली शाम चुने गए फूल घरों की देहरियों पर सजाए जाते हैं।
2. दक्षिणा में मिली यह सामग्री पूरे इक्कीस दिन तक इकट्ठी की जाती है।
3. अंतिम दिन इकट्ठी की गई सामग्री से सामूहिक भोज बनाया जाता है।
चर्चा करें:
उपर्युक्त तीन वाक्य कर्मवाच्य के वाक्य हैं। कर्मवाच्य में कर्म की प्रधानता होती है। कर्ता के साथ ‘से’ प्रत्यय जाडेकर क्रिया का भूतकाल रूप के साथ ‘करना’ क्रिया के उचित रूप का प्रयोग होता है।

Hss Live Guru 10th Hindi Kerala Syllabus प्रश्ना 6.
हिमालयी अंचल के लोग प्राकृति से तालमेल रखकर जीवन बिताते हैं। यदि हम प्रकृति के साथ विनाशकारी हस्तक्षेप करें तो क्या-क्या मुसीबतें होंगी?

उत्तर:
मानव के अनियंत्रित हस्तक्षेप के कारण प्रकृतिक संसाधनों पर ज़्यादा असर होता है। प्रकृति के शोषण एवं स्रोतों के नशीकरण से प्राकृतिक संसाधनों पर बुरा असर पड़ता है। ऋतुचक्र में बदलाव आ जाता है। तेज़ गर्मी एवं ठंड इसीका परिणाम है। जनजीवन असहनीय बन जाता है। पेडों के काटने से जंगल नहीं के बराबर होते हैं। जीवजंतुएँ जंगल से निकलकर गाँव की ओर आ जाते हैं। भूकंप, हिम झंझावत, बाढ़, सूखा आदि प्रकृति शोषण के फलस्वरूप होते हैं।

बसंत मेरे गाँव का Orakkum Questions and Answers

Kerala Syllabus 10th Standard Hindi Textbook गतिविधि -1

सूचना : संबंध पहचानें और सही मिलान करें।

उत्तर:

गतिविधि -2

सूचना : बसंत मेरे गाँव का’ लेख का यह अंश पढ़े और अनुबद्ध प्रश्नो के उत्तर लिखें।
बंसत फूलदेई का त्योहार लेकर आता हैं। देर शाम तक बच्चे फूल चुनते हैं। इन फूलों को रिंगाल से बनी खास तरह की टोकरियों में रखा जाता हैं। टोकरियों को रात भर पानी से भरी गागरों के ऊपर रख जाता है ताकि वो सुबह तक मुरझा न पाएँ।

Hindi Notes Class 10 State Syllabus प्रश्ना 1.
फूलदेई को बच्चों का त्योहार क्यों कहा गया है?
उत्तर:
इस त्योहार में सारे काम बच्चे करते हैं। बडों की भूमिका केवल सलाह देना है। इसलिए फूलदेई बच्चों का त्योहार कहा गया है।

10 Std Hindi Notes Kerala Syllabus प्रश्ना 2.
फूलदेई के अवसर पर उत्तराखंड पर्यटन विभाग इसका प्रचार करते हैं। फूलदेई त्योहार से संबंधित पोस्टर तैयार करें।
उत्तर:

गतिविधि – 3
सूचना : ‘बसंत मेरे गाँव का’ लेख का यह अंश पढ़ें और अनुबद्ध प्रश्नों के उत्तर लिखें।
वे जानवरों के साथ-साथ कीड़ाजड़ी, करण और च्यूर जौसी दुर्लभ हिमालयी जड़ी व औषधियाँ भी बेचते हैं। इन गांवों से इनका सादियों का रिश्ता हैं, इसलिए उसी वक्त पूरी कीमत चुकाना ज़रूरी नहीं होता। बर्फीले मौसम में निचले इलाकों की ओर जाते वक्त पुरानी वसूली की जाती है। मज़े की बात हैं कि नकत-उधार के आंकड़े कहीं दर्ज नहीं होते।

Hss Live Guru 10 Hindi Kerala Syllabus प्रश्ना 1.
पशुचारकों की पुरानी वसूली कब की जाती हैं?
उत्तर:
बर्फीले मौसम में निचले इलाकों की ओर जाते समय पुरानी वसूली की जाती हैं।

Class 10 Hindi Chapter 1 Kerala Syllabus प्रश्ना 2.
‘इनका’ – में निहित सर्वनाम कौन – सा हैं?
उत्तर:
ये

प्रश्ना 3.
‘नकद-उधार के आँकड़े कहीं दर्ज नहीं होते’ आपसी विश्वास के आधार पर सादियों से चली आ रही इस सिलसिले पर एक रपट तैयार करें।
उत्तर:
यहाँ नकद – उधार के आँकड़े कहीं दर्ज नहीं होता। उत्तराखड़ मकर संक्रांति से सूरज तपने के कारण गंगा में पानी की धारा तेज़ हो जाती हैं। उस समय यहाँ के हिमालय अंचल में ठंड के मौसम में बर्फीले इलाकों से निचले इलाकों में उतरे पशुचारक वापस घरों को लौटने लगते हैं। रास्ते में आनेवाले गाँवों से उनका लेन-देन भी होता हैं। गाँवों से इनका सादियों का रिश्ता है, इसलिए उसी वक्त पूरी कीमत चुकाना ज़रूरी ही। बर्फीले मौसम में निचले इलाकों की ओर जाते वक्त पुरानी वसूली की जाती है। मज़े की बात है कि नकद-उधार के आंकडे कहीं दर्ज नहीं होते। आपसी विश्वास के दम पर वर्षों से यहाँ ये लेन-देन चल रहा है।

गतिविधि – 4

सूचनाः ‘बसंत मेरे गाँव का’ लेख का यह अंश पढ़ें और अनुबद्ध प्रश्नों के उत्तर लिखें।
सूरज जब चौखंभा पर्वत के पीछे से उदय होने लगता है तब जेठ शुरू हो जाता है। पहाड़ की सड़कें गाड़ियों से भर जाती हैं। अपने घर की छत से जब में बद्रीनाध यात्रा-मार्ग की भीड़ देखता हूँ तो भूल जाता हूँ कि महज दो महीने पहले यह घाटी एकदम शांत थी।

प्रश्ना 1.
संबंध पहचानें और सही मिलान करें।

उत्तर:

प्रश्ना 2.
नमूने के अनुसार वाक्य बदलकर लिखें।
जेठ शुरू हो जाता हैं।
जेठ शुरू हो जाएगा।
सड़कें गाड़ियों से भर जाती हैं। …………………..
उत्तर:

गतिविधि – 5
सूचनाः ‘बसंत मेरे गाँव का’ पाठ के ये वाक्य पढ़ें।
1. सूरज अब नंदा पर्वत से चौखंभा पर्वत की ओर बढ़ने लगता है।
2. पशुचारक वापस घरों को लौटने लगते हैं।
3. बुरांस के फूल पहाड़ों पर शानदार लालिमा बिछा देते हैं।
4. गंगा में पानी की धारा तेज़ हो जाती है।
5. सीली सडक में हलचल बढ़ जाती है।
6. पहाड़ की सड़कें गाड़ियों से भर जाती हैं।

प्रश्ना 1.
प्रत्येक वाक्य के रेखांकित क्रिया रूपों का सीध संबंध वाक्य के किस शब्द से हैं?
उत्तर:
वर्तमान काल के वाक्यों में कर्ता के लिंग, वचन के अनसार क्रिया रूप बनता हैं। इसके लिए ता है, ते है, ती है, ती हैं, ता हूँ, ती हूँ, ते हो, ती हो आदि क्रिया पूरकों का प्रयोग करता है।

बसंत मेरे गाँव का Additional Questions and Answers

बसंत मेरे गाँव का आशयग्रहण के प्रश्न

प्रश्ना 1.
फूलदेई के त्योहार के सिलसिले में बच्चे पहले क्या किया करते हैं?

उत्तर:
बच्चे देर शाम तक फूल चुनते हैं और इन फूलों को रिंगाल से बनी खास तरह की टोकरियों में रखते हैं।

प्रश्ना 2.
फूलों को ताज़ा रखने के लिए क्या करते हैं?

उत्तर:
फूलों को रिंगाल की टोकरियों में रखा जाता है। टोकरियों को रात पानी से भरी गागरों के ऊपर रखा जाता है।

प्रश्ना 3.
फूलदेई त्योहार के दिन सुबह से बच्चे क्या करते हैं?

उत्तर:
बच्चे टोलियाँ बनकर गाँव-भर घूमते हैं। पिछली शाम चुने फूलों को घरों की देहरियों पर सजाते हैं।

प्रश्ना 4.
घरवाले बच्चों को क्या देते हैं?

उत्तर:
घरवाले बच्चों को चावल, गुड़, दाल आदि देते हैं।

प्रश्ना 5.
दक्षिणा में मिली चीज़ों से क्या करते हैं?

उत्तर:
दक्षिणा में मिली चीज़ों को पूरे इक्कीस दिन तक इकट्ठा की जाती है। अंतिम दिन इकट्ठी की गई सामग्री से सामूहिक भोज बनाया जाता है।

प्रश्ना 6.
चैती गीत क्या है? उसकी विशेषता क्या है?

उत्तर:
चैती उत्तराखंड का चैत माह पर केंद्रित लोकगीत है। इन गीतों में पांडवों की हिमालय यात्रा के किस्से होते हैं और पहाड़ के वीरों की शौर्य गाथाएँ भी शामिल होती है।

प्रश्ना 7.
बदलती ऋतुएँ और पशुचारकों के ज़िंदगी में क्या संबंध है?

उत्तर:
ठंड के मौसम में पशुचारक बर्फीले इलाकों से निचले इलाकों में उतर आते हैं। महीनों तक फैले चराहगाहों, घने जंगलों और अनजान बस्तियों में भटकते हैं। पशुचारकों के साथ उनके पालतू जानवर भी होते हैं। गर्मियों के दिन के आने पर वे वापस घरों को लौटने लगते हैं। यह खुशी उत्सव का माहौल रचती है। वे गीत गाते हैं नाचते हैं।

प्रश्ना 8.
रास्ते के गाँववालों का पशुचारकों के जीवन में क्या स्थान है?

उत्तर:
गाँवालों से पशुचारकों का लेन-देन होता रहता है। वे जानवर, दुर्लभ हिमालय जड़ी व औषधियाँ बेचते हैं। इन गाँववालों से उनका सदियों का रिश्ता है। इसलिए उधार में भी व्यापार चलता है। अगले साल ठंड के मौसम में इसकी वसूली की जाती है।

प्रश्ना 9.
जेठ की शुरूआत पर गाँव में कौन-सा परिवर्तन आता है?

उत्तर:
बद्रीनाथ, केदारनाथ, गंगोत्री और यमुनोत्री की ओर आनेवाले यात्रियों की गाड़ियों से सड़कें भर जाती हैं। घाटी की शांत वातावरण बदल जाता है।

बसंत मेरे गाँव का Summary in Malayalam and Translation

बसंत मेरे गाँव का शब्दार्थ

Students can Download Kerala SSLC Biology Previous Year Question Paper March 2019 Malayalam Medium Pdf, Kerala SSLC Biology Model Question Papers helps you to revise the complete Kerala State Board New Syllabus and score more marks in your examinations.

Kerala SSLC Biology Previous Year Question Paper March 2019 Malayalam Medium

Instructions:

• The first 15 minutes is the cool-off time.
• You may use the time to read the questions and plan your answers.
• Answer only on the basis of instructions and questions given.
• Consider score and time while answering.

Time: 1½ Hours
Total Score: 40 Marks

Students can Download Physics Part 2 Chapter 1 Work, Energy and Power Questions and Answers, Summary, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 9th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power in Malayalam Medium

Work, Energy and Power Textual Questions and Answers in Malayalam

You can Download Windows of Knowledge Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Biology Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Biology Solutions Chapter 2 Windows of Knowledge

Windows of Knowledge Questions and Answers

Windows Of Knowledge Kerala Syllabus 10th Chapter 2 Question 1.
Sense organs how helpful us?
Answer:

• For enjoyment.
• For escaping from accidents.
• For predation.
• For communication.

Sslc Biology Chapter 2 Kerala Syllabus Question 2.
Receptors help sense organs in receiving various stimuli from the surroundings. List various sense organs and the stimuli they receive.
Answer:

eye

Windows Of Knowledge Class 10 Notes Kerala Syllabus Chapter 2 Question 3.
How are our eyes protected?
Answer:
Eye socket orbit : Depressions in the skull protects the eyeball.
External eye muscles : Fixes the eyeball in the orbit.
Eyebrow : Protects the eye from dust and sweat.
Eyelashes : Protects the eye from dust.
Eyelids : Protects the eye from dust and external shock.
Conjunctiva : Secretes mucus which pro¬tects the anterior portion of the eyeball from being dry.
Tears : Clean and lubricate the anterior part of the eyeball. Lysozyme, the enzyme present in tears, destroys germs that enter the eyes.
Structure of Eye:

The Eye Parts and Function:

Conjunctiva – The layer which protects the sclera except the cornea.
Cornea – Slightly projected transparent.
anterior part of the sclera which refracts light rays to focus on the retina.
Iris – Part of the choroid seen behind the cornea. Presence of melanin gives the iris a dark colour.
Lens – Elastic transparent convex lens, connected to ciliary muscles by ligaments:
Ciliary muscles – Circular muscles seen around the lens. The contraction and relaxation of these muscles alter the curvature of lens.
Yellow spot – The part of the retina where plenty of photoreceptors are present. It is the point of maximum visual clarity.
Blindspot – The part of the retina from where the optic nerve begins. Photoreceptors are absent.
Optic nerve – Transmits impulses from photoreceptors to the visual center in the brain.

Fluids present in the eye and their functions:

• Aqueous humor – A watery fluid seen in the aqueous chamber (between cornea and lens), oozes from the blood. This fluid supplies nutrients and oxygen to cornea and lens.
• Vitreous humor – A jelly-like fluid filled within the vitreous chamber (between lens and retina), helps to maintain the shape of eyeball.

Nutrition of tissues in the eyes:
The blood and aqueous humor nourishes the tissue in eye.

Regulation Of Light In The Eye

The pupil dilates in dim light and constricts in bright light.
The size of the pupil is regulated by the action of circular muscles and radial muscles. When the radial muscles contract in dim light.

The size of the pupil increases. When the circular muscles contract in intense light, the size of the pupil decreases. Thus the amount of light falling on the lens is regulated according to the intensity of light.

Formation Of Image

Sslc Biology Chapter 2 Notes Kerala Syllabus Question 4.
We have a convex lens inside the eye. What are the peculiarities of image formed by this type of lens? List them.
Answer:

1. Real.
2. Small.
3. Inverted.

Sslc Biology Chapter 2 Questions Kerala Syllabus Question 5.
Observe and analyze table and figure. Write your inferences in the science diary.

Answer:

• While viewing near objects curvature of lens increases. So the focal length increases.
• While viewing near objects the ciliary muscle contracts and ligaments relax. So the focal length increases.
• While viewing distant objects the ciliary muscles relax and the ligament stretch. So the focal length decreases.
• The capacity of the eye to change the curvature of lens depending on the distance between the eye and the object by adjusting the focal length.

10 Biology Chapter 2 Kerala Syllabus Question 6.
Define the power of accommodation of the eye.
Answer:
The capacity of the eye to change the curvature of lens depending on the distance between the eye and the object by adjusting the focal length is called the power of accommodation of the eye.

Retina and The Photoreceptors

Receptor region of the rod cells is rod shaped and contain the pigment rhodopsin, which will be stimulated under dim light. Receptor region of the cone cells is cone shaped and contain the pigment photopsin (iodopsin) which will be stimulated under intense light.

Under dim light, rhodopsin dissociates to form retinal and opsin to produce impulses from rod cells. Under intense light, photopsin (iodopsin) dissociates to form retinal and opsin to produce impulses from cone cells.
Diversities in the cone cells due to the change in amino acid in the opsin molecules.

Chemistry of Vision

Sslc Biology Chapter 2 Questions And Answers Pdf Question 7.
Observe the chemical reaction given below. Discuss based on the indicators and write the inferences in your science diary?

Function of cone cells:
When light rays fall on pigment present in cone cells the photopsin dissociate into retinal and opsin.

Function of rod cells:
When light rays fall on pigments present in rod cells the rhodopsin dissociate into retinal and opsin.

The mechanism of vision:
The chemical change of pigments leads to the formation of impulses. These impulses are transmitted to cerebrum through the optic nerves and this enables vision.

Hss Live Biology 10 Kerala Syllabus Chapter 2 Question 8.
Complete the flowchart showing the process of vision.
Answer:
Light → Cornea → Aqueous humour → Pupil → Lens → Vitreous humour → Retina → Stimulus → Optic nerve → Cerebrum → Sense of sight
Binocular vision:
The images from two sides of the same object are formed in the left and right eye. These two images are combined-together in the visual area of our brain to form a three-dimensional image of the object. This is called binocular vision.

Food habits and health of the eyes:
The pigment present in photoreceptors is derived from vitamin A. The deficiency of vitamin A results in the low production of retinal. This inturn creates the deficiency of rhodopsin in rod cells. The resynthesis of rhodopsin also get blocked. This causes night blindness.

Due to prolonged deficiency of vitamin A, the conjunctiva and cornea will become dry and opaque. This causes xerophthalmia and leads ultimately to blindness. Hence it is necessary to include food containing vitamin Ain our diet.

Eye – Defects and Diseases

Hsslive Guru 10th Biology Kerala Syllabus Chapter 2 Question 9.
Prepare a table to show reason of various disorders and diseases that affect on our eyes. Suggest remedial measures also.
Answer:

10th Standard Biology Notes Kerala Syllabus Chapter 2 Question 10.
Prepare posters for an awareness program on the protection of eyes and donation of eyes.
Answer:

Kerala Syllabus 10th Standard Biology Chapter 2 Question 11.
Mention a few points to a discussion related with the health of our eyes.
Answer:

• Avoid falling of bright source of light directly to the eyes. Don’t look at the sun.
• Avoid reading under dim light.
• Do not watch TV or other screens continuously.
• Frequently wash our eyes.
• Include vitamin A contained items in our daily food.

Ear

External Ear: Ear pinna, Ear canal, Eardrum /Tympanum
Middle Ear: Malleus, Incus, Stapes, Eustachian tube.
Internal Ear: Cochlea, Vestibule (Sacule, Utricle), Semicircular canal, Auditory nerve
External Ear:

Biology Class 10 Chapter 2 Kerala Syllabus Question 12.
What are the parts of the external ear.
Answer:
Pinna and auditory canal, tympanum.

Hsslive Biology 10th Kerala Syllabus Chapter 2 Question 13.
How does the pinna help in hearing.
Answer:
Carries sound waves to the auditory canal.

Biology Class 10 Kerala Syllabus Chapter 2 Question 14.
What is the importance of ear wax?
Answer:
Prevent dust and foreign particles from entering the ear.
Middle ear:

Middle ear is a small chamber between the external ear and the internal ear. Ear ossicles namely malleus, incus and stapes are the important parts seen inside the middle ear. A bony wall separates the middle ear from the internal ear. In this bony wall, there are two openings which are covered by membranes.

The upper opening is oval window and the lower one is round window. The ear ossicles connect the tympanum the internal ear through the oval window. Eustachian tube that connects the middle ear to the pharynx.
Internal ear:

The internal ear, seen inside the bony labyrinth, as membranous labyrinth. A coiled tube-like cochlea and the vestibular apparatus (vestibule & three semicircular canals) are the parts of inner ear.

The membranous labyrinth is filled with a fluid, named endolymph and the space between the bony and membranous labyrinths is filled with another fluid, named perilymph.
Cochlea of the internal ear functions in hearing, while the vestibular apparatus helps to maintain body balance through transmitting impulses to the cerebellum.

Ear and Hearing

Cochlea is a coiled tube like a snail shell. It consists of three chambers. Specialized sensory hair cells which are present in basilar membrane that separates the middle and lower chambers, function as auditory receptors. The basilar membrane and sensory hair cells together constitute the organ of Corti. The sound waves which passes through the external ear vibrates the tympanum.

The vibration of the tympanum is transmitted to the ear ossicles which causes the vibration of the membrane in the oval window. Thus vibration causes the movement inside the cohlea. As a result, the sensory hair cells seen in the organ of Corti the cochlea are stimulated and impulses are generated. These impulses reach the cerebrum through the auditory nerve and hearing is affected.

Question 15.
Observe the figure given below. Prepare notes on process of hearing in the science diary.

Answer:
Structure of Cochlea:
Cochlea is a coiled tube like a snail shell. It consists of three chambers. Perilymph and endolymphs are filled inside this.

Auditory receptors in cochlea:
Specialized sensory hair cells which are present in basilar membrane that separates the middle and lower chambers, function as auditory receptors.

The process which sound waves becomes impulses:
The vibration of the tympanum is transmitted to the ear ossicles which causes the vibration of the membrane in the oval window. This vibration causes the movement of fluid inside the cochlea. As a result of the sensory hair cells seen in the organ of Corti of the cochlea are stimulated and impulses are generated.

Sense of hearing:
The impulses reach the cerebrum through the auditory nerve and hearing is affected.

Question 16.
Complete the flow chart of hearing?
Answer:

Ear and Body Balancing

Question 17.
Parts of the internal ear which help in body balance.
Answer:
The vestibular apparatus is constituted of structure namely utricle, saccule and three semicircular canals.

Question 18.
Peculiarities of receptors in the vestibule.
Answer:
Cluster of receptors in the form of hair cells, seen inside the vestibule and semicircular canals help in balancing of the body.

Question 19.
How does the cerebellum act in accordance with the variation in the balance of the body?
Answer:
Receptors (hair cells) seen inside the ampulla of semicircular canals, saccule and utricle are stimulated according to the movement of head. The impulses formed thus will be transmitted to the cerebellum through the vestibular nerve. Cerebellum functions so as to maintain the equilibrium of body.
To detect taste:

Chemoreceptors seen inside the mouth and tongue help us to detect taste. The chemoreceptors seen inside the papillae of the tongue are called as taste buds. Smell also influences taste.

Taste buds are stimulated by four types of taste namely sweet, salt, sour, bitter umami. Substances which dissolves in saliva, stimulate the taste buds and generate impulses. These impulses reach the brain and we experience taste.

Food particles dissolve in the saliva → Taste receptors (taste buds) are stimulated → Impulses form and transmit through nerves → Brain → Experience of taste.
To detect smell:

When particles enter the nose and disperse in the mucus, the olfactory receptors in the mucous membrane get stimulate and the impulses reach the brain through the olfactory nerve. Brain helps in the perception of smell.

Aromatic particles diffuse in the air and enter the nostrils → These aromatic particles dissolve in the mucus inside the nostrils → Impulses form and transmit through nerves → Brain → Experience of taste.

Sense Through Skin

Question 20.
What all receptors are seen in the skin to receive stimuli? Observe figure and write inference in your science diary.

Answer:
Heat, cold, touch, pressure and pain are felt by our skin. When these receptors are stimulated, impulses from the reach in the cerebrum for its perception.

Question 21.
Are these receptors uniformly distributed all over the skin?
Answer:
The fingertip contains more touch receptors.

Question 22.
Observe the illustration on the various receptors of certain organisms around us and their peculiarities. Write your inferences in your science diary.

Answer:

Windows of Knowledge Let Us Assess

Question 1.
When we view nearby objects.
a) Ciliary muscles relax.
b) Curvature of lens decreases.
c) Ciliary muscles contract.
d) Focal length increases.
Answer:
c) Ciliary muscles contract.

Question 2.
Identify the odd one and write down the common feature of others.
Malleus, Eustachian tube, Stapes, Incus.
Answer:
Eustachian tube. Others are ear ossicles.

Question 3.
Copy the figure. Identify the parts according to the hints and label them.

Hints :
a) The part where the muscles that regulate the size of the pupil are seen.
b) The chamber which is filled with vitreous humor.
c) The layer of eye where photoreceptors are seen.
Answer:
a) Iris.
b) Vitreous chamber.
c) Retina (label correctly).

Question 4.
Figure of Ear is given. Redraw it and name and label the parts mentioned.

a) Part that receives the vibrations of the eardrum.
b) The tube that connects the pharynx.
c) Part where the auditory receptors are seen.
Answer:
a) A thin circular membrane.
b) Eustachian tube.
c) Cochlea.

Windows of Knowledge Extended Activities

• Conduct a class seminar on the topic “Eye diseases and protection of the eyes”.
• Prepare notes on the peculiarities of sense organs of different organisms by collecting more information.
• Take the section of a bull’s or buffalo’s eye and observe the parts. Write your observation and inferences in the science diary.

Windows Of Knowledge More Questions and Answers

Question 1.
Find the odd one and write the peculiarities of others.
a) Iris, malleus, lens, retina.
b) Utricle, sacule, vestibular nerve, cochlea.
Answer:
a) malleus, others are the parts of eyes.
b) cochlea: others are related to body balancing.

Question 2.
Find the relation and fill in the blanks.
a) Cochlea : Auditory nerve.
Vestibule : ………
b) Rod cells : Rhodopsin.
Cone cells : ……..
c) Glaucoma : Laser surgery.
Cataract :
Answer:
a) Cochlea : Auditory nerve.
Vestibule : Vestibular nerve.
b) Rod cells : Rhodopsin.
Cone cells : Photopsin.
c) Glaucoma : Laser surgery.
Cataract : Replacing the lens through surgery.

Question 3.
Correct the sentence if it is wrong.
a) Conjunction is the outer layer which gives firmness the eye.
b) Aqueous humour nourishes the tissues of the eye.
c) Round window enables the vibration of conchleae fluid through the vibration of stapes.
d) Hair cells in cochlea are help in balancing the body.
Answer:
a) False conjunctiva is the layer which protects the sclera except the cornea.
b) Hair cells in the cochlea are help in hearing.

Question 4.
Slightly projected transparent anterior part of the sclera?
Answer:
Cornea.

Question 5.
The transparent membrane which protects the sclera, except the cornea?
Answer:
Conjunctiva.

Question 6.
The middle layer of eye, …… which contains a large number of blood vessels.
Answer:
Choroid.

Question 7.
The dark colored anterior part of choroid is,…… which contains the pigment melanin.
Answer:
Iris.

Question 8.
The aperture at the centre of iris?
Pupil.

Question 9.
The antagonistic muscles in the iris that regulate the size of eye pupil?
Answer:
Radial muscles (pupillary dilator) and Circular muscles (pupillary constrictor).

Question 10.
Vitamin A helps us for better vision. Give reason.
Answer:
Retinal, the visual pigment found in the photo receptors, is formed from vitamin A.

Question 11.
The bird, owl has no vision in daytime. Why?
Answer:
Owl’s retina is devoid of cone cells and hence no vision in daytime.

Question 12.
Analyze the statement that animals like cat and owl have more vision at night.
Answer:
Cat and owl have more rod cells in their retina so that they have more vision at night.

Question 13.
A kite can locate its prey even from high altitude. How is this possible?
Answer:
The eyes of kite are closer to each other and contain a large number of cone cells. Hence it has high power of vision.

Question 14.
“When the first chamber of the eye is cut open, its fluid spreads out but the second chamber is cut open, its fluid doesn’t spread out”. These words are taken from Rahul’s Practical Observation Record.
a) Name the fluids?
b) What are the functions of those fluids?
c) Name the eye-fluid related with Glaucoma?
Answer:
a) Aqueous humor and Vitreous humor
b) Aqueous humor supplies nutrients and oxygen to cornea and lens. Vitreous humor maintains the shape of eyeball.
c) Aqueous humor.

Question 15.
Analyse the illustration of image formation in retina and make a flowchart.

Light rays from the object → cornea → Aqueous humor/pupil → lens → vitreous humor → image on retina.

Question 16.
Though images of object are formed in both eyes, we can see only one object. Give reason.
Answer:
Cerebrum coordinates the two images and hence get a three-dimensional view of the object.

Question 17.
Carefully observe the above picture and make any two inferences we get from this picture.

Answer:
The process of seeing object through the coordination of impulses by the cerebrum. Binocular vision.

Question 18.
What is the need of closing one eye while shooting an object?
Answer:
Binocular vision help us to get a three-dimensional image of the object and to calculate the correct distance, depth, height and width of the object. But in the case of aiming an object, we need to get correct line instead of common focus through binocular vision.

Question 19.
The sight of one of the eyes of a participant in the hurdles competition has lost. Is the participant is apt for participating hurdles competition? Justify your answer.
Answer:
Since the participant lost one of his eyes’ sight, he has no binocular vision and he cannot calculate the correct distance, depth, height and width of the hurdles.

Question 20.
A patient having following symptoms consulted a doctor.

•  Dry conjunctiva.
• Dry and opaque cornea.
• Vision problem.
• Name his defect. Should the patient take any change in his food habit?

Answer:
Xerophthalmia. He should include food containing more vitamin A in his diet.

Question 21.
Select correct answer from the box.
Hyper metropia, Myopia, Night blindness, Colour blindness, Cataract, Glaucoma, Conjunctivitis, Xerophthalmia.
a) Disorder related to the eye lens?
b) Disorder related to the aqueous humour?
c) Disorders due to the difference in the size of eyeball?
d) Disorder that can be rectified by convex lens?
e) An infection?
1) Genetic disorder?
g) Disorder due to the deficiency of vitamin A?
h) Disorder due to the prolonged deficiency of vitamin A?
Answer:
a) Cataract.
b) Glaucoma.
c) Hyper metropia and myopia.
d) Hyper metropia.
e) Conjunctivitis.
f) Colourblindness.
g) Night blindness.
h) Xerophthalmia.

Question 22.
The tube that connects the middle ear to the pharynx? What is its function?
Answer:
Eustachian tube. It helps to regulate the pressure inside the middle ear.

Question 23.
Why certain person feel ear pain as they goes higher altitudes?
Answer:
When we goes higher altitudes, the atmospheric pressure decreases. This creates low pressure on one side of the eardrum to cause pain.

Question 24.
Recognize the figure A and B.

Answer:
A. Taste bud.
B. Olfactory receptors.

Question 25.
Convert the following chart of sensing smell to a flow chart of feeling taste.
Olfactory particles dissolve in mucus → Stimulation in olfactory receptors → Formation of impulse → Transmission of impulses through nerve → Olfactory center in the cerebrum → Experiencing smell
Answer:
Food particles dissolve in the saliva → Taste receptors (taste buds) are stimulated → Impulses form and transmit through nerves → Brain → Experience of taste.

Question 26.
We are unable to recognize smell when we have cold. Can you say the reason?
Answer:
Smell influences tasting. During cold, excess mucus will be produced in the nasal cavity. This will obstruct the normal stimulation of olfactory receptors and hence, we are unable to recognize smell and feel the food tasteless.

Question 27.
The ability of shark to sense smell is sharp. Why?
Answer:
Shark has highly sensitive olfactory receptors.

Question 28.
Give example for animals in which the taste receptors are seen on its legs.
Answer:
Housefly.

Question 29.
Find out the odd one. Write down the common feature of other Cornea, Cochlea, Choroid, Sclera. (Model 2016)
Answer:
Cochlea. Others are part of eye.

Question 30.
‘Giddiness is felt when we turn round and round’. Analyze the statement. (Model 2016)
Answer:
This statement is true. When we turn round continuously, the endolymph inside the vestibular apparatus also move and there will be continuous stimulation of the receptors and passing of the impulses to the cerebellum. Thus our muscular coordination cannot be maintained to feel giddiness.

Question 31.
Redraw the diagram and label the parts that perform the following functions: (March 2016)

a) The tissues in the eye receive nutrients and oxygen.
b) Regular the amount of light entering the eye.
Answer:
a) Choroid layer
b) Iris/pupil.

Question 32.
Observe the diagram.

a) Redraw and label the diagram.
b) Complete the table appropriately. (Model 2015)

Answer:

Question 33.
Find out the relationship between the pair of words and fill up the blanks.
Rod cells : Rhodopsin
Cone cells : …….. (March 2015)
Answer:
Photopsin.

Question 34.
“Receptors are modified neurons”. Justify the statement with examples of receptors in different sense organs. (March 2015)
Answer:
Receptors are modified neurons. Rods and cones in the eyes, auditory receptors in the ear, taste receptors in the tongue are examples.

Question 35.
Name the process represented in following illustration? Write the peculiarity of image formed in labelled sense organ? (March 2015)

Answer:
Formation of image in the brain. Peculiarity of the image formed: Real, small.

Question 36.
The diagram given below shows the structure of ear:

a) Redraw the diagram.
b) Label the parts A, B, C, D.
c) Write down the name and function of E and F. (Model 2014)
Answer:
a)

b) A – tympannum (eardrum)
B – Auditory nerve.
C – Ear ossicles.
D – Eustachian tube-labelling.
c) E – Semicircular canal- balancing.
F – Cochlea – hearing.

Question 37.
Make a flow chart showing the sequential stages to feel the smell. (March 2015)
Answer:
Olfactory particles enter the nose through air → Olfactory particles dissolve in the mucus and reach the olfactory receptors → Olfactory receptors are stimulated and impulses are formed Olfactory center of brain → Sense of smell.

Question 38.
Find out ‘word pair relation’ and fill in the blanks.
a) Dim light : Rod cells.
Bright light : ……
b) Euglena : ……
Insects : Ommatidia.
c) Frog : Urea
Tadpole : …….. (March 2014)
Answer:
a) Cone cells.
b) Eye spot.
c) Ammonia.

Question 39.
Correct the mistakes and rearrange the given flow chart of hearing process.
Ear canal → Cochlea → Eardrum → Earossicles → Oval window → Auditory nerve. (March 2014)
Answer:
Ear canal → Eardrum → Ear ossicles → Oval window → Cochlea → Auditory nerve.

Question 40.
Analyse the word pair relationship and fill up the blanks.
Photopsin : Colourblindness.
………. : Night blindness. (Model 2013)
Answer:
Rhodopsin.

Question 41.
Redraw the picture.
a) Label the parts A, B and C.
b) Write down the functions of B, C & D.

Answer:
a) A – Lens.
B – Aqueous chamber .
C – Layers of the eye.

b) B – Aqueous chamber:
This is the chamber between lens and the cornea. It is filled with a watery fluid called aqueous humour which supplies nutrients arid oxygen to the cells of the cornea and the lens.

c – Layer of the eye:
Retina, the innermost layer of the eyeball it imparts the firmness to the eye ball.

b – Vitreous chamber:
This is the large chamber seen between the lens and the retina which is filled with the jelly like vitreous humor. This helps to maintain the shape of the eye ball.

Question 42.
The flow chart sound waves passing through the ear is shown below. Complete it.
a) Ear Canal.
b) ………
c) Mallius.
d) ……….
e) Stapes.
f) ……….
g) Cochlea.
h) Perilymph.
i) …………
j) Auditory receptors. (Model 2012)
Answer:
b) Tympanum / Eardrum.
d) Incus.
f) Oval window.
i) Endolymph.

Question 43.
Copy the figure. Label the name of following parts on the figure.

a) Part on which image form.
b) Circular anterior part of the choroid layer.
c) The main part which regulate the focal length of light rays. (Model 2012)
Answer:

Question 44.
Find the odd one, note down common features of the others.
Thyroxin, Prolactin, Ethylene, Oxytocin. (March 2012)
Answer:
Ethylene is a plant hormone, others are animal hormones.

Question 45.
Suku feels difficulty of vision in dimlight.
a) Identify the disease, from among the following.
i) Haemophilia.
ii) Night blindness.
iii) Glaucoma.
b) What is the reason for this disease? (March 2012)
Answer:
Night blindness, caused due to the deficiency of vitamin A.

Question 46.
“Allow anyone to take my eyes after my death”. These are the words of Raju, who was admitted to the hospital after a serious accident.
a) How will you react to this statement?
b) Is it difficult to donate eyes after death?
c) Is it possible to donate an organ while alive? (March 2012)
Answer:
a) Raju is a noble and kind hearted person because Eye donation is a noble deed, as it brings happiness into the dark life of a blind person.
b) No, Eyes can be removed within 6 hours and transplanted to the patient by surgery.
c) Certain organs like, kidney, liver can be donated by a live donor.

Windows of Knowledge SCERT Questions and Answers

Question 1.
Complete the flowchart filling the blanks: (Question Pool-2017)

Answer:
A. Pinna.
B. Ear ossicles.
C. Oval window.
D. Cochlea.
E. Hair cells.
F. Cerebrum.

Question 2.
Given below is the portion of a poster.
Don’t take high eye pressure easy.
It may lead to blindness in future.
a) Which is the eye disease mentioned in the poster?
b) What is the reason for this condition?
c) What is the remedy for this disease? (Question Pool -2017)
Answer:
a) Glaucoma
b) 1) Blockage of reabsorption of aqueous humour.
2) Increase in the pressure inside the eyes
D) Laser surgery.

Question 3.
Observe the figure given below and answer the question.

a) Identify A, B
b) Which is the pigment in A?
c) Which is the eye disorder related to B? (Question Pool-2017)
Answer:
a) A – Rod cells.
B – Cone cells.
b) Rhodopsin.
c) Colourblindness.

Question 4.
Processes related to maintaining the equilibrium of the body are given below. Analyse and arrange them in the correct order.
a) Impulses are formed.
b) Muscular movements are coordinated.
c) Maintains the equilibrium of the body.
d) Sensory hair cells of the vestibular apparatus are stimulated.
e) Impulses reach the cerebellum.
f) Body movements create the movement of fluid inside the vestibule and semicircular canal. (Question Pool – 2017)
Answer:

Question 5.
Analyse the illustration related to the power of accommodation of the eye and answer the questions.

a) Which figure indicates the change in the lens while viewing distant objects?
b) Give the reason for the change in the curvature of the lens in figure II than in figure 1. (Question Pool -2017)
Answer:
a) Figure 1
b) 1. While viewing nearby objects.
2. Ciliary muscles contract.
3. Ligaments relax.

Question 6.
Statements related to sense organs are given below:
Choose the correct ones.
a) Taste buds are the chemoreceptors seen in the papilla?
b) Receptors are uniformly distributed all over the skin.
c) Impulses from the olfactory receptors reach the cerebrum through the olfactory nerve.
d) We experience taste when impulses from the taste buds reach the cerebellum. (Question Pool-2017)
Answer:
a) Taste buds are the chemoreceptors seen in the papilla?
c) Impulses from the olfactory receptors reach the cerebrum through the olfactory nerve.

Question 7.
Complete the table suitably in accordance with the given model. (Question Pool -2017)

Answer:
a) Planaria.
b) Jacobson’s organ.
c) Detects smell.
d) Receptors in the lateral line.

Question 8.
Observe the figure and answer the questions.

a) What does the figure indicate?
b) Identify X,Y.
c) How do X and Y differ in function? (Question Pool-2017)
Answer:
a) Internal ear
b) 1. X-Vestibular nerve
2. Y – Auditory nerve
c)1. X- Transmits impulses form the vestibular apparatus to cerebellum.
2. Y-Transmits Impulses from the cochlea to the cerebrum.

Question 9.
Light rays which reflect from the object are focussed on the retina and an image is formed.
a) Write the peculiarities of this image.
b) How do the images formed in the two eyes combine? What is its advantage? (Question Pool 2017)
Answer:
a) Small, inverted, real.
b) 1. Three-dimensional image of the object is formed due to the activity of the brain.
2. Binocular vision becomes possible.

Question 10.
“The focal length of the lens in the eye can be adjusted to view distant and nearby objects.” Evaluate this statement and arjswerthe questions given below.
a) When does historical length of the lens decrease?
b) What is the change in the focal length of the lens while viewing distant objects? How does it become possible? (Question Pool-201?)
Answer:
a) While viewing nearby objects.
b) Focal length increases, while viewing distant object the ciliary muscles relax ligaments stretch and the curvature of lens decreases. So the focal length decreases.

Question 11.
Vision is enabled when the impulse from the retina reaches the cerebrum through the optic nerve.
a) Draw a flow chart showing the pathway of light from cornea to retina.
b) There is no vision at the point where the optic nerve starts. Why? (Question Pool 2017)
Answer:
a) Light → Cornea → Aqueous humor → Pupil → Lens → Vitreous humor → Retina.
b) Rod cells and cone cells are absent in the part from where the optic nerve starts photoreceptors are absent. So there is no vision.

Question 12.
Examine the terms given below and choose the parts related to the maintenance of equilibrium of the body.
Cerebrum, Eustachian tube, Sacule, Cochlea, Utricle, Vestibular nerve, Oval window, Cerebellum, Semi circular canals. (Question Pool 2017)
Answer:
Sacule, Utricle, Vestibular nerve, Cerebellum.

Question 13.
Redraw the diagram given below. Identify the parts and label it.

a) Connects the middle ear with the pharynx.
b) Auditory receptor hair cells are present.
c) Separates the external ear from the middle ear. (Question Pool 2017)
Answer:

a) Eustachian tube.
b) Cochlea.
c) Tympanum.

Question 14.
Analyse the statements given below and write reasons.
a) Tears have antiseptic properties.
b) We can see objects in three-dimension. (Question Pool 2017)
Answer:
a) Lysozyme the enzyme present in tears destroys germs that enter the eyes.
b) Binocular vision.

Question 15.

a) How is this reaction related to vision?
b) How does the deficiency of vitamin A cause poor vision in dim light? (Question Pool 2017)
Answer:
a) 1. When light rays fall rhodopsin dissociate into retinal and opsin.
2. Generates impulses.
3. When impulses reach the cerebrum vision is experienced.
b) Retinal, the visual pigment found in the photoreceptors, is formed from vitamin A. The deficiency of vitamin A results in the low production of retinal. This, in turn, creates the deficiency of rhodopsin in rod cells. The resynthesis of rhodopsin also gets blocked. This causes a condition called night blindness.

Question 16.
Prepare two placards to be used in a rally organized by the school Science Club to propagate the importance of eye donation. (Question Pool 2017)
Answer:
Appropriate idea
Example : Let the eyes glow even after death!
Eye donation great donation.

Question 17.
Redraw the diagram and label the parts given below.

a) Transparent anterior part of the sclera.
b) Fluid that nourishes the tissues of the eye.
c) The layer that has photoreceptors. (Question Pool 2017)
Answer:

a) Comea.
b) Aqueous humor.
c) Retina.

Question 18.
Observe the figure given below and answer the questions:

a) Which is the receptor seen in the figure?
b) Which sense organ is this receptor seen in?
c) What is the function of this receptor? (Question Pool 2017)
Answer:
a) Olfactory receptor
b) Nose
c) Gets stimulated by aromatic particles and generate impulses

Question 19.
A fluid is formed in the eye just as the cerebrospinal fluid is formed in the brain.
a) Name the fluid.
b) What is its function? (Question Pool 2017)
Answer:
a) Aqueous humor.
b) Nourishes the tissues of the eye.

Question 20.
Justify the statements given below:
a) Smell can be detected only in the presence of mucus.
b) Persons with colour blindness cannot distinguish. between green and red colours. (Question Pool 2017)
Answer:
a) Aromatic particles dissolve in the mucus and stimulate olfactory receptors.
b) Reduction/damage of cone cells that help to distinguish red and green colors.

Question 21.
Given below is a table related to eye diseases. Fill up suitably. (Question Pool -2017)

Answer:
a) Lens in the eye becomes opaque.
b) Glaucoma.
c) Laser surgery.
d) Xerophthalmia.

Question 22.
It is because of its taste that we like food. Given below are the different stages of experiencing taste. Analyze and arrange them in the correct order.
a) Experience of taste.
b) Causes impulses.
c) Food particles dissolve in saliva.
d) Reaches taste buds.
e) Impulses reach the brain.
f) Chemoreceptors get stimulated. (Question Pool 2017)
Answer:
c.
d.
f.
b.
e.
a.

Question 23.
Identify the odd one. Write the common feature of the others.
a) Cold, temperature, Pressure, Taste
b) Tympanum, Oval window, Cochlea, Papilla
c) Ommatidia, Pupil, Iris, Conjunctiva. (Question Pool 2017)
Answer:
a) 1. Taste.
2. Others senses through skin.
b) 1. Papilla.
2. Others are parts of the ear.
c) 1. Ommatidia.
2. Others are parts of eye.

Question 24.
Identify the word pair relationship and fill in the blanks.
a) Retina : The inner layer which has photoreceptors.
………. : The transparent anterior part of the sclera.
b) Blind spot : The part from where the optic nerve begins.
………. : The part where the image has maximum clarity.
c) ……… : Nourishes the tissues of the eye.
Vitreous humor : Shape of the eye. (Question Pool 2017)
Answer:
a) Cornea.
b) Yellow spot.
c) Aqueous humour.

Question 25.
Choose the correct statements from the following ones.
a) Cochlea helps in maintaining the equilibrium of the body.
b) The oval window helps in the movement of fluid inside the cochlea.
c) We detect smell when impulses from the olfactory receptors reach the cerebellum.
d) Glaucoma is a condition in which the lens of the eyes become opaque resulting in blindness. (Question Pool 2017)
Answer:
b) The oval window helps in the movement of fluid inside the cochlea.

Question 26.
Choose the correct pairs related to the eye defects from the pairs given felow.
a) Cataract – Reabsorption of aqueous humor does not occur.
b) Glaucoma – Rectified by Replacing the lens.
c) Colourblindness – Infection of conjunctiva.
d) Cataract – Lens of the eye becomes opaque.
e) Conjunctivitis – Rectified by Laser surgery.
f) Glaucoma – Rectified by Laser surgery.
g) Colorblindness-Inability to distinguish colours. (Orukkam 2017)
Answer:
Correct pairs:
d) Cataract – Lens of the eye becomes opaque.
f) Glaucoma – Rectified by Laser surgery.
g) Colour Blindness – Inability to distinguish colours.

Question 27.
Redraw the diagram and label the parts based on the functions given below.

a) The part which is adjusted with the intensity of light.
b) The part where photoreceptors are present.
c) The transparent anterior part of the sclera.
d) The muscles which help to alter the curvature of the lens.
e) Transmits impulses from photoreceptors to the visual center in the brain.
f) The chamber which is filled with a jelly-like substance.
g) The layer made up of connective tissues which to the eye. (Orukkam 2017)
Answer:

Question 28.
Rearrange the flowchart related to sense of vision.
Light → Lens → Pupil → Vitreous humor → Aqueous humor → Optic nerve → Retina → Cerebrum → Cornea → Impufse → Sense of vision. (Orukkam 2017)
Answer:
Light → Cornea → Aqueous humor → Pupil → Lens → Vitreous humor → Retina → Impulse → Optic nerve → Cerebrum → Sense of vision.

Question 29.
Select suitable words from the box and complete the flowchart given below related to hearing. (Orukkam 2017)
Lens, Cerebrum, Cochlea, Retina, Auditory canal, Oval window, Ear ossicles, Auditory nerve, Hair cells, Optic nerve, impulse, Cornea, Pinna.

Answer:

Question 30.
Stages related to the maintenance of body balance are given below. Rearrange these statements in order.
a) Generates impulses.
b) Body movements create movement of fluid inside the vestibule and semicircular canals.
c) Impulses are transmitted by the vestibular nerve to the cerebellum.
d) Creates movements of the sensory hair cells.
e) Maintains equilibrium of the body.
f) Cerebellum coordinates muscular activities. (Orukkam 2017)
Answer:
b) Body movements create movement of fluid inside the vestibule and semicircular canals.
d) Creates movements of the sensory hair cells, a) Generates impulses.
c) Impulses are transmitted by the vestibular nerve to the cerebellum.
f) Cerebellum coordinates muscular activities.

Question 31.
Prepare a poster for creating awareness about the importance of eye donation. (Orukkam 2017)
Answer:
– Eye donations noble deed.
– No magician can cure blindness, but you donate your eyes.
– May your eyes see more after your life too.

Question 32.
Explain the necessity of including Vitamin A rich food items in daily diet for the health our eyes? (Orukkam 2017)
Answer:
Retinal, the pigment presented in the photoreceptors, is formed from vitamin A.

Question 33.
Complete the flowchart related to the sense of smell. (Orukkam 2017)

Answer:

Students can Download Kerala SSLC Maths Model Question Paper 1 English Medium, Kerala SSLC Maths Model Question Papers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala SSLC Maths Model Question Paper 1 English Medium

Time: 2½ Hours
Total Score: 80 Marks

Instructions

• Read each question carefully before writing the answer.
• Give explanations wherever necessary.
• First 15 minutes is Cool-off time. You may use the time to read the questions and plan your answers.
• No need to simplify irrationals like √2, √3, π etc., using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Wandoor Ganitham Sslc Model Question Paper Question 1.
Consider the arithmetic sequence 13, 23, 33, …..
a) What is its common difference?
b) What is the first three-digit term of this sequence?
Answer:
(a) 10
(b) 103

Wandoor Ganitham Model Question Paper Question 2.
In the figure, O is the centre of the circle ∠A = 60°.

a) ∠BOD = __________
b) ∠C = _________
Answer:
(a) ∠BOD = 120°
(b) ∠C = 120°

Wandoor Ganitham Answer Key Polynomials Question 3.
If x – 1 is a factor of the polynomial 5x³ – 4x² + x – k, what number is k?
Answer:
P(x) = 5x³ – 4x² + x – k
P(1) = 5 × 1³ – 4 × 1² + 1 – k = 0
5 – 4 + 1 – k = 0
k = 2

Wandoor Ganitham Question Paper Question 4.
A circle is drawn with the origin as centre. It passes through the point (3, 3).
a) What is the radius of the circle?
b) Write the coordinates of a point where the circle meets the x-axis.
Answer:
(a) Radius = 3√2 cm
(b) (-3√2, 0)

Answer any five questions from 5 to 11. Each question carries 3 scores. (5 × 3 = 15)

Wandoor Ganitham Answer Key 2021 Question 5.
a) If a square is inscribed in a circle of diameter 4cm, what will be the length of a side of the square?
b) What is the base edge of a square pyramid of maximum size that can be carved out from a hemisphere of radius 5 cm?
Answer:
(a) Base edge of the square pyramid = 2√2 cm
(b) Base edge = \(\frac{10}{\sqrt{2}}=\frac{5 \times 2}{\sqrt{2}}\)
\(\begin{aligned}
&=\frac{5 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2}}\\
&=5 \sqrt{2} \mathrm{cm}
\end{aligned}\)

Wandoor Ganitham Model Question Paper 2021 Question 6.
Draw the co-ordinate axes and mark the point (4, 0). Draw an isosceles right angled triangle with this point as one of its vertices.
Answer:

Wandoor Ganitham Answer Key Arithmetic Sequence Question 7.
Draw a circle of radius 3.5 centimetres. Draw a triangle of angles 50°, 60°, 70°with its vertices as points on the circle.
Answer:

Wandoor Ganitham Model Question Paper Answer Key Question 8.
The algebraic form of an arithmetic sequence is 5n + 5
(a) What is its first term?
(b) What is the difference pf its 10th and 20th terms?
(c) Can the difference between any two terms of this sequence be 368? Justify.
Answer:
(a) x₁ = 5 × 1 + 4 = 9
(b) Difference: x₂₀ – x₁₀ = 10d = 10 × 5 = 50
(c) The difference between any two terms of an arjthmetic sequence is divisible by the common difference.
368 is not divisible by 5, the common difference that is, 368 is not the difference btween two terms.

Sslc Model Question Paper Maths English Medium Question 9.
The length of a rectangle is 6 centrimetres more than its breadth. Its area is 1216 square centimetres. Find its length.
Answer:
Breadth = x
Length = x + 6
x(x + 6) = 1216
⇒ x² + 6x = 1216
⇒ x² + 6x + 9 = 1216 + 9 = 1225
⇒ (x + 3)² = 1225
⇒ x + 3 = √1225 = ±35
⇒ x + 3 = 35 or x + 3 = -35
⇒ x = 35 – 3 = 32 or x = -35 – 3 = -38
i.e Breadth = 32
Length = 32 + 6 = 38

Sslc Maths Question Paper 2020 Kerala English Medium Question 10.
In the figure, C, D are points on the circle AD is a diameter of the circle. ∠C = 30°, AB = 4 centrimetres.

a) ∠D = ________
b) ∠ABD = _________
c) What is the length of the diameter?
Answer:
(a) ∠D = 30° (Angle in the same arc)
(b) ∠ABD = 90° (Angle in the semicircle is 90°)
(c) 8 cm

Sslc Maths Model Question Paper English Medium Question 11.

In the figure C is the centre of the circle, PA and PB are tangents. PC = 5 centimetres and radius of the circle is 3 centimetres.
a) Find the length of PA
b) What is the area of the quadrilateral PACB?
Answer:
(a) PA = 4 cm
(b) Area of PAC = \(\frac { 1 }{ 2 }\) × 4 × 3 = 6 cm²
Area of PACB = 2 × 6 = 12 cm²

Answer any 7 questions from 12 to 21. Each question carries 4 scores. (7 × 4 = 28)

Wandoor Ganitham Sslc Model Question Paper Answer Key Question 12.
The coordinates of two opposite vertices of a rect-angle are (7, 8) and (1, 3).
a) Without drawing coordinate axes, mark these points as the vertices of a rectangle with left-right, top-bottom positions correct.
b) Find the co-ordinates of other two vertices.
c) What is the length of its diagonals?
Answer:

Wandoor Ganitham Sslc Model Answer Key 2021 Question 13.
Consider the polynomial P(x) = ax³ – x² – bx – 1
a) Find P(1).
b) What is the relation between a and b if x – 1 is a factor of P(x)?
c) What is the relation between a and b if x + 1 is a factor of P(x)?
d) Will, P(x) have both (x + 1) and (x – 1) as factors for any numbers a and b? Justify.
Answer:
P(x) = ax³ – x² – bx – 1
(a) P(1) = a × 1³ – 1² – b × 1 – 1 = a – 1 – b – 1 = a – b – 2
(b) If (x – 1) is a factor, P(1) = 0
ie. a – b – 2 = 0
ie. a – b = 2
(c) If x + 1 is a factor, P(-1) = 0
ie. a (-1)³ – (-1)² – b(-1) – 1 =0
ie. -a – 1 + b – 1 = 0
ie. b – a = 2
(d) It is not possible to find a and b such that a – b = b – a = 2
Therefore we cannot find a and b.

Hss Live Guru Model Question Paper Question 14.
The radius and height of a cone are 12 centimetres and 6 centimetres respectively.
a) What is its volume?
b) If this cone is cut parallel to its base, along the midpoint of its height, what is the radius of the small cone obtained?
c) What is the volume of the small cone?
d) Find the ratio of the volumes of the small cone and the first cone.
Answer:
r = 12 cm, h = 6 cm
(a) Volume = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × π × 12² × 6 = 288π cm³
(b) R = 12 cm, H = 6 cm, h = 3 cm

\(\begin{aligned}
&\frac{r}{R}=\frac{h}{H}\\
&\frac{r}{12}=3
\end{aligned}\)
r = 6 cm
(c) Volume = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × π × 6² × 3 = 36π cm³
(d) Ratio = 36π : 288π = 1 : 8

Wandoor Ganitham Sslc Unit Test 2021 Answer Key Question 15.
In the figure ABCD is a parallelogram. ∠E = 90°. A(3, 5) B(8, 5) are two vertices BE = 3 units, CE = 4 units.

a) Write the coordinates of C.
b) What are the coordinates of D?
c) Find the coordinates of meeting point of the di-agonals of the parallelogram.
Answer:
(a) B(8, 5), BE = 2
E(11, 5), C (11, 9)
(b) D (6, 9)

(c) The point of intersection of the diagonals is the midpoint of a diagonal.
The mid point of AC is \(=\left(\frac{11+3}{2}, \frac{9+5}{2}\right)\) = (7, 7)

Wandoor Ganitham Pre Model Answer Key Question 16.
Draw a circle of radius 3 centrimetres. mark a point 7 centimetres away from its centre. Draw the tan¬gents to the circle from that point.
Answer:

Hsslive Guru Question Paper Question 17.
In triangle PQR, ∠Q = 90°, ∠R = x°. Lengths of the sides PQ, QR and PR are a, b, c respectively.

a) Which among the following is tan x°?
\(\left(\frac{a}{c}, \frac{b}{a}, \frac{a}{b}, \frac{b}{c}\right)\)
b) Similarly write sin x° and cos x° from this triangle.
c) Prove that \(\frac{\sin x^{0}}{\cos x^{0}}=\tan x^{0}\)
Answer:

Question 18.
Consider two fractions having numerator 1. The denominator of one fraction is 2 more than the denominator of the other. Sum of these fractions is \(\frac{5}{12}\).
a) Write the above fact as an algebraic equation.
b) Find the fractions.
Answer:
Let x be the denominator of the first fraction.
The denominator of the other fraction is x + 2

Question 19.
A bag contains some red and green balls. If we take a ball from it, without looking, the probability of getting a red ball is \(\frac{1}{4}\)
a) What is the total number of balls, if there are 8 red balls?
b) What is the probability that a ball taken is green?
c) Find the sum of both the probabilities.
d) From a box containing some red balls and some blue balls th& probability of getting a red ball is \(\frac{a}{b}\). What is the probability of getting a blue ball?
Answer:

Question 20.
In the figure, ∠P = 90°. Sides of triangle APC are extended to B and D.

a) If a circle is drawn with AC as its diameter, where will be the position of P with respect to that circle?
b) What about the position of P, if the circle is drawn with AD as diameter?
c) Prove that, the circles drawn with the sides of a quadrilateral with perpendicular diagonals, will meet at a common point.
Answer:
(a) On the circle
(b) On the circle
(c) Since AB and CD are perpendicular to each other, the circle with diameter AB, diameter AD, diameters DB and diameter GB passes through P.

Question 21.
a) Write the sequence of odd numbers greater than 1.
b) What is the algebraic form of this sequence?
c) What is the algebraic form of the arithmetic sequence \(\frac{3}{6}, \frac{5}{6}, \frac{7}{6} \ldots .\)
d) Prove that this sequence does not contain any natural number.
Answer:
(a) 3, 5, 7, …
(b) x_n = 2n + 1
(c) x_n = \(\frac{2 n+1}{6}\)
(d) To get integer terms, the numerator should be a multiple of 6.
2n + 1 is odd. It is not a multiple of 6 for carry n.
The sequence does, not contain integers.

Answer any five questions from 22 to 28. Each question carries 5 score. (5 × 5 = 25)

Question 22.
The table below shows the students of a class sorted according to their scores in a test

a) If the students are arranged in the increasing order of their scores, the score of the student at what position is taken as the median?
b) What may be assumed as the score of the 15th student in the arrangement?
c) Find the median score.
Answer:

(a) n = 41 (odd number)
\(\frac{41+1}{2}=\frac{42}{2}\) = 21st score comes in the middle.
It is in between 20 and 30
(b) When 10 scores we equally divided among 10 children, each one’s share is \(\frac { 10 }{ 10 }\) = 1
Score of 15th child = 20 + \(\frac { 1 }{ 2 }\) = 20.5
(c) Consider the arithmetic sequence with first term 20.5 and common difference 1
7th term will be the median Median = 20.5 + 6 × 1 = 20.5 + 6 = 26.5

Question 23.
A solid is made by fixing a hemisphere of same radius on the flat face of a cone. The height of the cone is 12 centimetres and its slant height is 13 centimetres.

a) What is the radius of the cone?
b) What is the curved surface area of the hemisphere?
c) What is the total surface area of the solid?
Answer:
l = 13 cm, h = 12 cm

Question 24.
Draw a triangle of sides 5 dentimetres, 6 centimetres and 7 centimetres. Draw its incircle.
Answer:
Draw triangle, Draw angle bisectors. They meet at a point. That point will be the incenter.

Question 25.
A boy is standing between two buildings of equal height. The boy and the buildings are in a straight line. He sees the tops of those buildings at elevations of 45° and 30°. The nearest building is 20 metres away from him.
a) Drawa rough figure.
b) Find the height of the buildings and distance between the buildings.
Answer:
(a)

(b) BC = 20m, ∠CBD = 45°, CD = 20 m
Height of the building = 20 m
AE = 20 m
ie. AB = 20√3 m
Distance between the building 20 + 20√3 = 20(1 + √3)m

Question 26.
In the figure chords AB and CD intersect at P. PA = 8 centrimetres, PB = 6 centimetres, PC = 4 centimetres, BC = 4 centimetres.

a) Which angle is equal to ∠A?
b) Write one more pair of equal angles.
c) Find the length of PD.
d) What is the length of AD?
Answer:
(a) ∠A = ∠C
(b) ∠D = ∠B
(c) PD × PC = PA × PB
PD × 4 = 8 × 6
PD = 12 cm
(d) ΔAPD , ΔCPB are similar
\(\begin{aligned}
&\frac{A P}{P C}=\frac{P D}{P B}=\frac{A D}{B C}\\
&\frac{8}{4}=\frac{A D}{4}
\end{aligned}\)
AD = 8 cm

Question 27.
The sum of first 9 terms of an arithmetic sequence is 45 and the sum of first 18 terms is 171.
a) What is the sum of its 10th to 18th terms?
b) What is tis 5th term?
c) Find its 14th term.
d) Find the sum of 5th to 14th terms.
Answer:
S₉ = 45, S₁₈ = 171
(a) If S_n denotes the sum of first n terms = S₁₈ – S₉ = 171 – 45 = 126
(b) \(x_{5}=\frac{S_{9}}{9}=\frac{45}{9}=5\)
(c) \(x_{14}=\frac{126}{9}=14\)
(d) Sum of 5th to 14th terms = \(\frac { 10 }{ 2 }\) (X₅ + X₁₄)
= 5(5 + 14)
= 5 × 19
= 95

Question 28.
In the figure ABCD is a square. The axes are drawn throgh the midpoints of its sides. The length of its side is 6 units.

a) Write the coordinates of its vertices.
b) Write the equation of its diagonal BD.
Answer:
(a) A (-3, -3) B(3, -3) C (3, 3) D (-3, 3)
(b) Slope of BD = \(\frac{Y_{2}-Y_{1}}{X_{2}-X_{1}}=\frac{3–3}{-3-3}=\frac{6}{-6}=-1\)
Let (x, y) be a point on BD
\(\frac{y-3}{x-3}=-1\)
y + 3 = -1(x – 3)
y + 3 = -x + 3
x + y = 0

Read the following. Understand the Mathematical Idea expressed in it and answer the questions that follow. (6 × 1 = 6)

Question 29.
Consider the number 2751. The sum of its digits is 2 + 7 + 5 + 1 = 15. Adding the digits of 15 we get 1 + 5 = 6. This number 6 is called the ‘digital root’ of the number 2751. That is, to find the digital root of a number, find the sum of its digits (Dont forget to find the sum of the digits again if the first sum has more than one digit)
Let us see one more example.
The sum of the digits of the number 679412 is 6 + 7 + 9 + 4 + 1 + 2 = 29
Sum of digits of 29 = 2 + 9 = 11
Sum of digits of 11 = 1 + 1 = 2
Therefore the digital root of 679412 is 2.
Digital roots have an interesting property.
To see this, consider the product 43 × 27 = 1161.
The digital roots of the numbers 43 and 27 are 4 + 3 = 7 and 2 + 7 = 9.
Product of the digital roots = 7 × 9 = 63.
Digita root of 63 = 6 + 3 = 9.
The digital root of 1161 is also 9 (1 + 1 + 6 + 1 = 9)
That is the digital root of 1161 = The digital root fo 63, where 63 is the product of the digital roots of 43 and 27.
This property is true for all other natural numbers.
a) What is the digital root of 345?
b) What is the digital root of 927?
c) What is the digital root fo 345 × 927?
d) The digital root of the number 63 □ 5 is 8 (□ represents a missing digit). Find the missing digit.
e) 121 × 92 = 11□32. Find the missing digit.
f) If the digital root of a is 5 and the digital root of b is 2. Then what is the digital root of ab?
Answer:
(a) 3 + 4 + 5 = 12 = 1 + 2 = 3
(b) 9 + 2 + 7 = 18 = 1 + 8 = 9
(c) 9 × 3 = 27 = 2 + 7 = 9
(d) 6 + 3 + x + 5 = 8
14 + x = 8
1 + 4 + x = 8
5 + x = 8
x = 8 – 5 = 3
(e) (1 + 2 + 1) (9 + 2) = 1 + 1 + X + 3 + 2
4(1 + 1) = 7 + x
4 × 2 = 7 + x
8 = 7 + x
x = 1
(f) Digital root f a × b = 5 × digital root of 2
5 × 2 = 10
Digital root of 10 is 1.

You can Download Equal Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus 8th Standard Maths Solutions Chapter 8 Area of Quadrilaterals

Area of Quadrilaterals Text Book Questions and Answers

Textbook Page No 149

Area Of Quadrilateral Class 8 Scert Chapter 8 Question 1.
Draw a parallelogram of sides 5 centimetres, 6 centimetres and area 25 square centimetres.
Solution:

• Area of parallelogram = one side × distance to the opposite side.
• Since 5x distance to the opposite side is 25 sq.cm. Distance to the opposite side is 5 cm.
• Draw a square ABCD of side 5cm.
• Draw an arc of radius 6cm with centre A to cut DC at E.
• Length of radius 6cm with centre B and arc of radius 5 cm with centre E meet at F.

Class 8 Mathematics Area Of Quadrilaterals Kerala Scert Solutions Question 2.
Draw a parallelogram of area 25 square centimetres and perimeter 24 centimetres.
Solution:
Area of the parallelogram to be 25 cm². The product of one side and distance to the opposite side to be 25 cm².
One side and distance to the opposite side can be 5 cm each.
The other side is = 12 – 5 = 7 cm.
Draw a square with side cm and draw an arc of radius 7 cm with A as centre such that it intersects the side CD. Mark the point E and join AE. Draw an arc of radius 7 cm with B as centre and draw another arc of radius 5 cm with E as centre, Two arcs meet at the point
P. JoinBF and CF.

Area Of Quadrilaterals Class 8 Kerala Syllabus Question 3.
In the figure, the two bottom corners of a parallelogram are joined to a point on the top side.

Solution:
Area of dark triangle = \(\frac{1}{2} b h\) cm²
Area of parallelogram = bh
= 2 × area of triangle 10 cm²

Class 8 Mathematics Construction Of Quadrilaterals Kerala Scert Solutions Question 4.
The picture below shows the parallelogram formed by the intersection of two pairs of parallel lines?

What is the area of this parallelogram? And the perimeter?
Solution:
One side of the parallelogram = 4 cm
Distance to the opposite side = 3 cm
Area of the parallelogram = 4 × 3 = 12 sq.cm
BC × DE = 12
BC × 2 = 12
BC = 6, AD = 6

Perimeter = 4 + 6 + 4 + 6 = 20 cm.

Area Of Quadrilateral Questions For Class 8 Kerala Syllabus Question 5.
Compute the area of the parallelogram below:

Solution:

Textbook Page No 153

Hss Live Guru 8 Maths Kerala Syllabus Chapter 8 Question 6.
Draw a square of area 4½ square centimetres.
Solution:


Draw a circle of radius 3 cm. Draw the diameter AC and construct the perpendicular bisector of AC which meets the circle at B and D. Join AC, AD, CD and BD to get the square ABCD.

Area Of Quadrilateral Formula Class 8 Kerala Syllabus  Question 7.
Draw a non-square rhombus of area 9 square centimetres.
Solution:

• Draw AC = 1 cm
• Find midpoint of AC by drawing per perpendicular bisector
• Find B and D, such that OB = OD = 9 cm

Similarly we can draw the other two rhombus
AC = 9 cm,
OB = OD = 1 cm
AC = 3 cm, OD = OB = 3 cm

8th Standard Maths Notes Kerala Syllabus Chapter 8 Question 8.
The area of the dark triangle in the figure is 5 square centimetres. What is the area of the parallelogram?
Solution:

iii. Perimeter = 4 × 15 = 60 m
iv. Rhombus is a parallelogram. So,
Area of parallelogram = one side × distance to the opposite side
15 × h = 216
h = \(\frac{216}{15}=14.4 \mathrm{cm}\)

8th Standard Maths Notes PdfKerala Syllabus Chapter 8 Question 9.
A 68 centimetre long rope is used to make a rhombus on the ground. The distance between a pair of opposite corners is 16 metres.
i. What is the distance between the other two corners?
ii. What is the area of the ground bounded by the rope?
Solution:

Length of one side of a parallelogram

Diagonals are perpendicular bisector

Kerala Syllabus 8th Standard Maths Notes Pdf Chapter 8 Question 10.
In the figure, the midpoints of the diagonals of a rhombus are joined to form a small quadrilateral:

i. Prove that this quadrilateral is a rhombus.
ii.The area of the small rhombus is 3 square centimetres. What is the area of the large rhombus?
Solution:

The diagonals of the rhombus intersect at O and they bisect each other at right angles.
OA = OC
\(\frac{\mathrm{OA}}{2}=\frac{\mathrm{OC}}{2}\)
OQ = OS
Similarly, since OB = OD
ie., OR = OP
The diagonals of PQRS bisect each other. (Since OQ = OS and OR = OP)
Since the diagonal AC is perpendicular to BD, the diagonals of PQRS are also mutually perpendicular bisectors.
therefore PQRS is a rhombus

ii. Area of quadrilateral PQRS = 3 sq. cm.

8th Area Of Quadrilateral Formula Class 8 Kerala Syllabus Question 11.
What is the area of the largest rhombus that can be drawn inside a rectangle of sides 6 centimetres and 4 centimetres?
Solution:
If we join the midpoints of the opposite sides of rectangle, we get the longest diagonal of the rhombus.

Textbook Page No 156

Hsslive 8th Maths Kerala Syllabus Chapter 8 Question 12.
Draw a rectangle of sides 7 centimetres and 4 centimetres. Draw isosceles trapeziums of the same area, with the following specifications.
i. Lengths of parallel sides 9 centimetres, 5 centimetres.
ii. Lengths of non-parallel sides 5 centimetres.
Solution:
i. Draw a rectangle of length 7 cm and breadth 4 cm.

Extent AB either side by 1 cm and mark E and F. Also mark the points G and H on CD at a distance 1 cm from C and D respectively. Draw the rhombus EFGH.

ii. Draw the rectangle ABCD.The arc of radius 5cm drawn with A as centre intersects DC at P. The arc of 5cm drawn with C as the centre intersect extented AB at Q.

Std 8 Maths Kerala Syllabus Kerala Syllabus Chapter 8 Question 13.
Calculate the area of the isosceles trapezium drawn below:

Solution:

8th Std Maths Notes Kerala Syllabus Chapter 8 Question 14.
The parallel sides of an isosceles trapezium are 8 centimetres and 4 centimetres long; and non – parallel sides are 5 centimetre long. What is its area?
Solution:
Divide the isosceles trapezium into a rectangle and two triangles.

Textbook Page No 158

Area Of Quadrilateral Class 8 Kerala Syllabus Chapter 8 Question 15.
The lengths of the parallel sides of a trapezium are 30 centimetres, 10 centimetres and the distance between them is 20 centimetres. What is its area?
Solution:

Question 16.
Compute the area of the trapezium shown below:

Solution:
Consider the ∆ PQS, it is a right angled triangle

Textbook Page No 159

Question 17.
Compute the area of the hexagon below.

Solution:

Total area = 108 + 108 + 168 = 384 cm²

Question 18.
This is a picture drawn in the lesson Construction of Quadrilaterals.

What is the area of the large trapezium made up of the four smaller ones?
Solution:

Question 19.
What is the area of the quadrilateral shown below?

Solution:

Question 20.
Prove that for any quadrilateral with diagonals perpendicular, the area is half the product of the diagonals.
Solution:

Question 21.
Compute the area of the quadrilateral shown below:

Solution:

Question 22.
Compute the area of the parallelogram shown below:

Solution:
The diagonal divides the parallelogram into two equal triangles.

Area of the parallelogram = 2 × area of triangle = 2 × 96 = 192 cm²

Question 23.
The three blue lines in the picture below are parallel:

Prove that the areas of the quadrilaterals ABCD and PQRS are in the ratio of the lengths of the diagonals AC and PR.
i. How should the diagonals be related for the quadrilaterals to have equal area?
ii. Draw two quadrilaterals, neither parallelograms nor trapeziums, of area 15 square centimetres
Solution:


i. For the quadrilaterals to have equal area, the diagonals AC and PR should be equal.
ii. Draw three parallel lines, distance between them is 5 cm each. Also draw two quadrilaterals with diagonals PQ and RS 6 cm each.

Additional Questions And Answers

Question 1.
Which of the following has greater area; a parallelogram with one side 12 cm and distance between parallel sides is 6 cm or a square having diagonals 12 cm each?
Solution:

Question 2.
Area of a trapezium is 128 sq. cm and the distance between its parallel sides is 8 cm. Length of one parallel side is 28 cm. Find the length of the other parallel side.
Solution:

Length of the other Parallel side = 4 cm

Question 3.
Find the area of a rhombus of one side 6 cm and one diagonal 6 cm.
Solution:

Question 4.
It was decided to lay tiles in the room of Hari’s house with tiles in the shape of trapezium. The parallel sides of a tile is 40 cm and 20 cm and the distance between them is 10 cm. If 200 tiles are needed to lay tiles in the hall, find the area of the hall in square metres.
Solution:

Number of tiles required = 200
Area of the hall = 200 × 300
= 60000 cm²
= 6 m²

Question 5.
In the pictures given below, which is has more area?

Solution:
Rectangle has the maximum area among the parallelograms with same sides.

Question 6.
The ratio of the two adjacent sides of a parallelogram is 3 : 2. Distance between longer sides is 10 cm. If area is 900 cm², find the sides of the parallelogram?
Solution:
Let the sides be 3x and 2x

Question 7.
If one side of a parallelogram is ‘a’ and height of that side is h, prove that area = ah.
Solution:
In the figure, ABCD is a parallelogram. AB = a

The perpendicular distance from D to AB = h
By drawing the diagonal BD, we can divide the parallelogram into two equal triangles. ∆ ABD and ∆ BCD are equal triangles. ∴ their areas are equal, ie; area of the parallelogram is two times of the area of ∆ ABD.

Question 8.
PQRS is a rhombus. If the diagonals are 8 cm and 9 cm each, compute the area.

Solution:
Let the diagonals of the rhombus are d₁ and d₂

Question 9.
The perimeter of a rhombus is 40 cm. If the length of one diagonal is 16 cm. What is the length of the other diagonal. Find the area?
Solution:
Perimeter = 40 cm
One side = 10 cm .
∆ POQ is a right angled triangle Since d₁ = 16 cm, OP = 8 cm
∴ 8² + OQ² = 10²

Question 10.
Draw a square of area 12.5 cm² and write the geometric principles of the construction.
Solution:

Draw a line of length 5 cm and its perpendicular bisector.
Draw a circle with the midpoint of this line as centre and the distance to one end as radius. This line will be the diameter of the circle. Mark the point of intersection of the perpendicular bisector as the circle and complete the square.

Question 11.
Compute the area of the trapezium shown below.

Solution:

Question 12.
Compute the area of the quadrilateral ABCD

Solution:

Question 13.
In the figure AB || CD. AD = BC = 13 cm
Distance between the parallel side is 12 cm. If CD = 20 cm. Find the area of ABCD.

Solution:
To compute AB

Question 14.
Draw a square and mark the mid-points. How is the area of the first square with the second square.
Solution:
Draw the square ABCD.
The perpendicular bisector of the side AB bisects CD also at right angles.

Similarly draw the perpendicular bisectors of AD and BC. Which intersect the sides. Join the points of intersection to complete the square.

We see, 8 triangles in the figure. The square PQRS is formed by joining 4 among these. The area of the square PQRS is half of the area of the square ABCD.

Question 15.
What is the maximum area of parallelogram of sides 8 cm and 5 cm? What is the speciality of the parallelogram of maximum area?
Solution:
The area will be maximum for a rect¬angle and the maximum area is 40 cm².

Question 16.
See the quadrilateral Ravi drew. In it AC = 12 cm and perpendiculars to AC are 6 cm and 12 cm.

a. Find the area of the quadrilateral.
b. Raju has to draw a parallelogram of this area. If its base is 12 cm what should be its height?
Solution:

b. If the height of the parallelogram is h its area is bh = 12 × h
This is 54 sq. cm

Question 17.
The area of a rhombus is 112 sq. cm and one of its diagonal is 16 cm long. Find the length of the other diagonal.
Solution:
Area of the rhombus \(=\frac{1}{2} \mathrm{d}_{1} \times \mathrm{d}_{2}\)
= 112 sq. cm

Question 18.
Draw a parallelogram of sides 6 cm, 4 cm and area 18 cm².
Solution:
Draw a line AB of length 6 cm. Draw a circle with centre A and radius 4 cm

In the circle (as shown in the picture) mark a point D and join AD. Draw an arc of radius 4 cm with centre B as centre and radius 4 cm and draw another arc with D as centre and radius 6 cm. Two arcs meet at C. Join CD and BC.

ABCD will be a parallelogram.

Question 19.
In the figure ABCD, AB parallel to CD and the distance between them is 8 cm. ?

AB = 12 cm, CD = 10 cm. Compute the area of quadrilateral (trapezium) ABCD?
Solution:

Question 20.
Compute the area of the quadrilateral ABCD in the figure.

Solution:

You can Download सुख-दुख Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख (कविता)

सुख-दुख पाठ्यपुस्तक के प्रश्न और उत्तर

सुख दुख कविता का सारांश Kerala Syllabus 8th प्रश्ना 1.
‘सुख-दुख की खेल मिचौनी खोले जीवन अपना मुख।’ इन पंक्तियों का आशय क्या है?

उत्तर:
सुख और दुख एक सिक्के के दो पहलू हैं। जीवन में सुख और दुख का होना स्वाभाविक है। जीवन का असली मुख सुख-दुख के मिलन में खुलता है।

Sukh Dukh Poem Summary In Hindi Kerala Syllabus 8th प्रश्ना 2.
‘फिर घन में ओझल हो शशि फिर शशि में ओझल हो घन।’ यहाँ घन और शशि किन-किन के प्रतीक हैं?

उत्तर:
यहाँ घन दुख का प्रतीक है और शशि सुख का प्रतीक है।

सुख-दुख कविता का सारांश Kerala Syllabus 8th प्रश्ना 3.
‘अविरत सुख भी उत्पीड़न’ क्या आप इससे सहमत है? क्यों?

उत्तर:
मैं इस कथन से सहमत हूँ। क्योंकि जीवन का असली मुख सुख-दुख पूर्ण है। हमेशा सुख हो तो जीवन की असलियत की पहचान नहीं होगी। सदा सुखी रहनेवालों के जीवन में दुख के आने पर वे उसके सामना करने में असफल हो जाते हैं।

सुख-दुख Textbook Activities

Sukh Dukh Kavita Ka Saransh Kerala Syllabus 8th प्रश्ना 1.
कविता से शब्द युग्मों का चयन करके लिखें।

उत्तर:
1. सुख-दुख
2. निशा-दिवा
3. सोता-जागता
4. साँझ-उषा
5. विरह-मिलन
6. हास-अश्रु

Sukh Dukh Poem Kerala Syllabus 8th प्रश्ना 2.
वर्गपहेली की पूर्ति करें।


दाईं ओर:

1. ‘मुख’ का समानार्थी शब्द।
5. इसका अर्थ है- ‘आकाश’।
6. सुख-दुख के मिलन से यह परिपूर्ण हो जाता है।
9. यह ‘निरंतर’ का समानार्थी है।

नीचे की ओर:

2. सफलता पाने के लिए इसकी ज़रूरत है।
3. ‘मेघ’ के लिए इस कविता में प्रयुक्त शब्द।
4. कविता में ‘संसार’ का प्रतीकात्मक शब्द।
7. ‘उषा’ का अर्थ।
8. ‘सुख-दुख’ किस विधा की रचना है?
10. यह कभी नहीं बोलना चाहिए।
उत्तर:
दाईं ओर:

1. आनन,
5. गगन,
6. जीवन,
9. अविरत
नीचे की ओर:

2. मेहनत,
3. घन,
4. जग,
7. प्रभात,
8. कविता,
10. झूठ

सुख दुख कविता का आशय Kerala Syllabus 8th प्रश्ना 3.
सुख-दुख कविता का आशय लिखें।

उत्तर:
सुख-दुख कविता श्री. सुमित्रानंदन पंत की है। इसमें कवि जीवन की असलियत को समझाने का प्रयास किया है। वे कहते हैं कि जीवन सुख-दुखपूर्ण है। सुख-दुख की खेलमिचौनी से जीवन अपना वास्तविक मुख खोल देता है। सुख-दुख के मधुर मिलन से जीवन पूर्ण हो जाता है। जीवन के दुख कभी सुख में बदलता है तो कभी सुख दुख में परिणित हो जाता है। संसार अति सुख और दुख से पीड़ित है। सुख-दुख मानव जीवन में बराबर रहें। निरंतर दुख और निरंतर सुख दोनों पीड़ा देनेवाले हैं। सुख-दुख रूपी दिन-रात में संसार का जीवन सोता और जागता है। इस संध्या और उषा के आँगन में विरह और मिलन का आलिंगन हो रहा है। मानव जीवन का मुख सदा हँसी और आँसू से भरा है।

सुख-दुख Summary in Malayalam and Translation

सुख-दुख शब्दार्थ Word meanings

You can Download Equations Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 2 Equations

Equations Text Book Questions and Answers

Textbook Page No. 34

Kerala Syllabus 8th Standard Maths Solution Chapter 2 Question 1.
“Six more marks and I would’ve got full hundred marks in the maths test.” Rajan was sad. How much mark did he actually get? the num¬ber.
Solution:
The marks Rajan got is six less from 100
∴ 100 – 6 = 94

Kerala Syllabus Class 8 Maths Solutions Chapter 2 Question 2.
Mother gave Rs.6o to Lissy for buying hooks. She gave back the 13 rupees left. For how much money did she buy books?
Solution:
Money Lissy got from mother = Rs. 6o
Amount she returned = Rs. 13
Amount Lissy used to buy books = 60 – 13 = Rs. 47

Kerala Syllabus 8th Standard Maths Notes Chapter 2 Question 3.
Gopalan bought a bunch of bananas. 7 of them were rotten which he threw away. Now there are 46. How many bananas were there in the bench?
Solution:
No. of rotten bananas = 7
No. of bananas left =46
∴ Total number of bananas
= 46 + 7 = 53

Hss Live Guru 8th Maths Chapter 2  Question 4.
Vimala spent 163 rupees shopping and now she has 217 rupees. How much money did she have at first?
Solution:
Amount spent by Vimala for shopping = Rs. 163
Amount left with her after shopping Rs. 217
Total amount = 163 + 217 = Rs. 380

Hsslive Guru 8th Class Maths Chapter 2 Question 5.
264 added to a number makes it 452. What is the number?
Solution:
Number = 452 – 264 = 188

Kerala Syllabus 8th Standard Notes Maths Chapter 2 Question 6.
198 subtracted from a number makes it 163. What is the number.
Solution:
Number = 198 + 163 = 361

Textbook Page No. 35

8th Scert Maths Solutions Chapter 2 Question 1.
In a company the manager’s salary in five times that of a peon. The manager gets Rs, 40,000 a month. How much does a peon get a month?
Solution:
Salary of the manager = Rs.40,000.
\(\frac{1}{5}\) th salary of the manager in the salary of the peon.
∴ Salary of the peon = \(\frac{40000}{5}\)
= 8000

Kerala Syllabus 8th Standard Maths Textbook Solutions Question 2.
The travellers of a picnic split equally the 5200 rupees spent. Each gave 1300 rupees. How many travellers were there?
Solution:
Total amount spent = Rs. 5200
Share of one = Rs.1300
∴ Members in the group = \(\frac{5200}{1300}\) = 4

Kerala Syllabus 8th Maths Solutions Chapter 2 Question 3.
A number multiplied by 12 gives 756. What is the number?
Solution:
Number = \(\frac{756}{12}\) = 63

Class 8 Maths Chapter 2 Kerala Syllabus Question 4.
A number divided by 21 gives 756. What is the number
Solution:
Number 756 × 21 = 15,876

Textbook Page No. 37

8th Standard Maths Guide Kerala Syllabus Chapter 2 Question 1.
Anita and her friends bought pens. For five pens bought toge¬ther, they got a discount of three rupees and it cost them 32 rupees. Had they bought the pens separately, how much would each have to spend?
Solution:
Cost for 5 pens = Rs. 32
discount = Rs.3
Real cost = 32 + 3 = 35
∴ Real cost of 1 pen = \(\frac{35}{5}\) = 7

8th Class Maths Notes Kerala Syllabus Chapter 2 Question 2.
The perimeter of a rectangle is 25 metres and one of its side is 5m. How many metres is the other side?
Solution:
Perimeter of the rectangle = 25m
One side = 5m
Double of the sum of length and breadth is perimeter = \(\frac{25}{2}\) – 5 = 7.5

Maths Class 8 Kerala Syllabus Chapter 2 Question 3.
In each of the problems below, the result of doing some operations on a number is given. Find the number?
(i) Three added to double is 101
(ii) Two added to triple is 101
(iii) Three substracted from double is 101
(iv) Two substracted from triple is 101
Solution:

Hsslive 8th Class Maths Chapter 2 Question 4.
Half a number added to the number gived III. What is the number?
Solution:
1\(\frac{1}{2}\) of the numbers = 111
ie \(\frac{3}{2}\) of the number = 111
number = 111 × \(\frac{2}{3}\) = 74

Hss Live Guru 8 Maths Chapter 2 Question 5.
A piece of folk maths. A child asked a flock of birds, ‘How many are you”? A bird replied
“We and us again
with half of us
And half of that
with one more.
would make hundred” How many birds were there?
Solution:
We and us ⇒ double (2 times)

Textbook Page No. 41

Hsslive Guru Maths 8th Standard Chapter 2 Question 1.
The perimeter of a rectangle is 80 metre and its length is one metre more than twice the breadth. What are its length and breadth?
Solution:
If x be the breadth, Then length
= 2x + 1
2 (x + 2x + 1) = 80
2(3x + 1) = 80
6x + 2 = 80
6x = 80 – 2
x = \(\frac{80-2}{6}\) = 13.
breadth = 13 metre
Length = 2 × 13 + 1 = 27 metre

Question 2.
From a point on a line another line is to be drawn such that the angle on one side is 50° more than the angle on the other side. How much is the smaller angle?
Solution:
Let one angle is x, the second angle = x + 50
Sum of two angles on a line = 180°
ie x + x + 50 = 180°
2x + 50 = 180°, 2x = 180 – 50
x = \(\frac{180-50}{2}\) = 65
The angles are 65°, 115°

Question 3.
The price of a book is 4 rupees more than the price of a pen. The price of a pencil is 2 rupees less than the price of the pen. The total price of 5 books, 2 pens and 3 pencils is 74 rupees. What is the price of each?
Solution:
Let the cost of a pen = x
cost of a book = x + 4
Cost of a pencil = x – 2
ie.5(x + 4) + 2x + 3(x – 2) = 74
5x + 20 + 2x + 3x – 6 = 74
10 x + 14 = 74, 10x = 74 – 14
x = \(\frac{74-14}{10}\) = 6
Cost of pen = Rs. 6
Cost of book = Rs. 10
Cost of pencil = Rs. 4

Question 4.
(i) The sum of three consecutive natural numbers is 36. What are the numbers
(ii) The sum of three consecutive even numbes is 36. What are the numbers?
(iii) Can the sum of three consecutive odd numbers be 36. Why?
(iv) The sum of three consecutive odd numbers is What are the numbers?
(v) The sum of three consecutive natural numbers is 33. What are the numbers?
Solution:
x + x + 1 + x + 2 = 36
3x + 3 = 36
x = \(\frac{36-3}{3}\) = 11
∴ Numbers are 11, 12, 13
x – 1 + x + x + 1 = 36
3x = 36
x = 12
Or
∴ Numbers are 11, 12, 13

(ii) (x – 2) + x + (x + 2) = 36
3x = 36
x = 12
∴ The numbers are 10, 12, 14

(iii) The sum of three odd numbers is odd.
∴ It is not possible to get 36 as the sum of three odd numbers.

(iv) (x – 2) + x + (x + 2) = 33
3x = 33
x = 11
∴ Numbers are 9, 11, 13

(v) (x – 1)+ x + (x + 1) = 33
3x = 33
x = 11
The numbers are 10, 11, 12

Question 5.
(i) In a calender, a square of four numbers is marked. The sum of the numbers is 80. What are the numbers?
(ii) A square of nine numbers is marked in a calendar. The sum of all there numbers is 90. What are the numbers?
Solution:
(i) Let,
x x + 1
x + 7 x + 8
four numbers in the square of 4 numbers.
Sum of the numbers = x + (x + 1) + (x + 7) + (x + 8) = 80
4x + 16 = 80, 4x = 80 – 16

Textbook Page No. 44

Question 1.
Ticket rate for the science exhibition is rupees 10 for a child and 25 rupees for the adult. Rs. 740 was got from 50 persons. How many children among them?
Solution:
Let the numbers of children be x
Then the number of adult = 50 – x
∴ 10 x + 25 (50 – x) = 740
10 x + 1250 – 25 x = 740
1250 – 15 x = 740
15 x = 1250 – 740
x = \(\frac{1250-740}{15}\) = 34
Number of adults = 50 – 34 = 16

Question 2.
A class has the same numbers of girls and boys. Only 8 boys were absent on a particular day and then the number of girls was double the number of boys and girls?
Solution:
Let the number of boys =number of girls = x
2 (x – 8) = x
2x – 16 = x, 2x – x = 16
x – 16 = 0
∴ x = 16
∴ The number of boys = No. of girls = 16

Question 3.
Ajayan is ten years older than Vijayan. Next year Ajayan’s age would be double that of Vijayan. What are their ages now?
Solution:
Let the age of Vijayan = x
Age of Ajayan = x + 10
Age of Vijayan after an year = x + 1
Age of Ajayan after an year = x + 11
2 (x + 1) = x + 11
2x + 2 = x + 11
2x – x = 11 – 2
x = 9
∴ Age of Vijayan = 9
Age of Ajayan = 19

Question 4.
Five times a number is equal to three times the sum of the number and 4. What is the number?
Solution:
Let the number = x
Five times number = 5x
4 more than the number = x + 4
Three times of it = 3(x + 4)
ie. 5 x = 3 (x + 4)
5 x = 3x + 12
5 x – 3 x = 12
2 x = 12
x = 6

Question 5.
In a co-operative society, the number of men is thrice the number of women 29 women and 16 men more joined the society and now the number of men is double the number of women. How many women were there in the society at first?
Solution:
Let the number of women = x
No of men = 3 x
No. of women when 29 more were joined = x + 29
No. of men when 16 more were joined = 3x + 16
3 x + 16 = 2 (x + 29)
3x + 16 = 2x + 58
3x – 2x = 58 – 16
x = 42
No. of women = 42;
No. of men = 3 × 42 = 126

Equations Additional Questions & Answers

Question 1.
“If you give rupees 5 to me both of us have equal amounts with us”. Ajay told to Vineeth. Then Vineeth told to Ajay “ If you give rupees 5 to me I will have 5 times more money than yours. Find out the amount both of them have?
Solution:
Let Ajay has Rs.x with him and when Vineeth gives Rs. 5, both of them have x + 5 rupees with them. If Vineeth gets RS.5 he has 5 times more than Ajay.
x + 5 + 5 = 5x
4 x = 10
x = \(\frac{10}{4}=\frac{5}{2}\) = 2.5 Rupees
Amount with Ajay = Rs. 2.50
Amount with Vineeth = Rs.12.50

Question 2.
The perimeter of a triangle is 49 cm. One side is 7 cm more than the second side and 5 cm less than the third side. Find out the lengths of three sides?
Solution:
Let the second side = xcm
First side = x + 7 cm
Third side = x + 12 cm
Perimeter = x + x + 7 + x + 12 = 49
= 3x = 49 – 19 = 30
3x = 20, x = 10 cm
The sides are = 10 cm, 17 cm, 22 cm

Question 3.
The sum of two numbers is 9, 8 times the smaller number is 2 more than 6 times the bigger number. Write an equation to find the numbers?
Solution:
Let the smaller number = x
Bigger number = 9 – x
Equation is, 8 x = 6 (9 – x) + 2

Question 4.
Anitha and friends bought pens. When they bought a packet of 5 pens they got a discount of rupees 2. They paid rupees 18. Find the cost of each one buys separately?
Solution:
Total cost in including discount = Rs. 20
Cost of 5 pens = Rs.20
Cost of 1 pen = \(\frac{20}{5}\) = 4

Question 5.
The cost of a chair and a table is Rs. 1500. The cost of table is 4 times the cost of chair. Find the expense of each?
Solution:
Let the cost of chair = x
Cost of table = 4x
x + 4 x = 1500
5x = 1500
x = 300
cost of chair = Rs. 300
Cost of table = Rs. 1200

Question 6.
There were 25 questions in the examination written by Jafar. Each correct answer gets 2 marks. There is a loss of mark for each wrong answer. Jafar answered all the questions. He got 35 marks. Find^ut the number of correct answers?
Solution:
Let the number of correct answers = x
Total number of questions = 25
No. of wrong answers = 25 – x
Marks for correct answers = 2x
marks losses for wrong answers = 25
ie. 2x – (25 – x) = 25
2x – 25 + x = 35
3x = 60
x = \(\frac{60}{3}\) = 20, ∴ number of correct answers = 20.

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Instructions:

• The first 15 minutes is the cool-off time.
• You may use the time to read the questions and plan your answers.
• Answer only on the basis of instructions and questions given.
• Consider score and time while answering.

Time: 1½ Hours
Total Score: 40 Marks

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Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 20 Static Electricity in Malayalam

Static Electricity Text Book Questions and Answers

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Kerala SSLC Social Science Model Question Paper 3 Malayalam Medium

Instructions:

• The first 15 minutes is the cool-off time. You may use the time to read the questions and plan your answers.
• Answer all questions in PART – A. Answer any one from the questions given under each question number in PART – B.

Time: 2½ Hours
Total Score: 80 Marks

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General Instructions:

1. The first 15 minutes is the cool off time. You may – use the time to read and plan your answers.
2. Answer the questions only after reading the instructions and questions thoroughly.
3. Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
4. Five questions are given in each section. Answer any four from each section.
5. Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks