CBSE - Page 8 of 12 - Vidhyarthi Darpan

Hindi 10th Previous Year Question Paper 2019 SET-III (CBSE)

हिन्दी

खण्ड ‘क’

प्रश्न 1. निम्नलिखित गद्यांश को ध्यानपूर्वक पढ़िए और पूछे गए प्रश्नों का उत्तर लिखिए।

आजकल दूरदर्शन पर आने वाले धारावाहिक देखने का प्रचलन बढ़ गया है बाल्यावस्था में ही शौक हानिकारक है दूरदर्शन पर दिखाए जाने वाले धारावाहिक निम्न स्तर के होते हैं। उन में अश्लीलता, अन आस्था, फैशन तथा नैतिक बुराइयां ही अधिक देखने को मिलती है। छोटे बालक मानसिक रुप से परिपक्व नहीं होते। इसमें भेजो भी देखते हैं उनका प्रभाव उनके दिमाग पर अंकित हो जाता है। बुरी आदतों को वैसी ही अपना लेते हैं समाज शास्त्रियों के एक वर्ग का मानना है कि समाज के चारों ओर फैली बुराइयों का एक बड़ा कारण है। दूरदर्शन से आत्मसीमितता, जड़ता, पंगुता अकेलापन आदि दोष बढ़े है । बिना समय की पाबंदी के घटकों दूरदर्शनबिल्कुल गलत है। कैसे मान सकता विकास रुक जाता है, नजर कमजोर हो सकती है और तनाव बढ़ सकता है।

(क) आजकल दूरदर्शन के धारावाहिकों का स्तर कैसा है?

(ख) दूरदर्शन का दुष्प्रभाव केंद्र अधिक पड़ता है और क्यों?

(ग) दूरदर्शन के क्या-क्या दुष्प्रभाव है?

(घ) ‘ बाल्यावस्था’ शब्द का संधि विच्छेद कीजिए।

(ड) उपन्यास के लिए उपयुक्त शीर्षक लिखिए।

उत्तर: (क) आजकल दूरदर्शनओं के धारावाहिकों का स्तर का घटता जा रहा है उसमें दर्शकों को अश्लीलता, अनास्था, फैशन लता नैतिक बुराइयां ही अधिक देखने को मिलती है।

(ख) दूरदर्शन का दुष्प्रभाव सबसे अधिक छोटे बालकों पर पड़ता है क्योंकि वह मानसिक रूप से परिपक्व होता है। वे जो इस छोटी उम्र में देखते हैं उसका प्रभाव उन पर अधिक पड़ता है।

(ग) दूरदर्शन के कई दुष्प्रभाव है जैसे इससे समाज में फैली बुराइयों को बढ़ावा मिलता है। इससे आत्मसीमितता, जड़ता, पंगुता अकेलापन आदि दोष बढ़े है। नजर कमजोर पड़ती है घंटो दूरदर्शन देखने से समय की पाबंदी कट जाती है।

(घ) ‘बाल्यावस्था’ का संधि विच्छेद बाल +अवस्था है

(ड़) उपयुक्त का शीर्षक दूरदर्शन के प्रभाव।

प्रश्न 2. निम्नलिखित काव्यांश को ध्यानपूर्वक पढ़िए और पूछे गए प्रश्नों के उत्तर लिखिए

कोलाहल हो,

या सन्नाटा, कविता सदा सृजन करती है,

जब भी आँसू

हुआ पराजित, कविता सदा जंग लड़ती है।

जब भी कर्ता हुआ अकर्ता,

कविता ने जीना सिखलाया

यात्राएँ जब मौन हो गईं

कविता ने चलना सिखलाया

जब भी तम का

जुल्म चढ़ा है, कविता नया सूर्य गढ़ती है,

जब गीतों की फसलें लुटती

शीलहरण होता कलियों का,

शब्दहीन जब हुई चेतना

तब-तब चैन लुटा गलियों का

जब कुर्सी का

कंस गरजता, कविता स्वयं कृष्ण बनती है।

अपने भी हो गए पराए

यूँ झूठे अनुबंध हो गए

घर में ही वनवास हो रहा

यूँ गूंगे संबंध हो गए।

(क) कविता कैसी परिस्थितियों में सूजन करती है? स्पष्ट कीजिए।

(ख) भाव समझाइए जब भी तम का जुल्म बढ़ा है, कविता नया सूर्य गढ़ती है।’

(ग) गलियों का चैन कब लुटता है?

(घ) “परस्पर संबंधों में दरियाँ बढ़ने लर्गी-यह भाव किस पंक्ति में आया है?

(ङ) कविता जीना कब सिखाती है?

अथवा

जो बीत गई सो बात गई

जीवन में एक सितारा था,

माना, वह बेहद प्यारा था,

वह डूब गया तो डूब गया।

अंबर के आनन को देखो,

कितने इसके तारे टूटे

कितने इसके प्यारे छूटे,

जो छूट गए फिर कहाँ मिले;

पर बोलो टूटे तारों पर,

कब अंबर शोक मनाता है?

जो बीत गई सो बात गई।

जीवन में वह था एक कुसुम,

थे उस पर नित्य निछावर तुम,

वह सूख गया तो सूख गया;

मधुबन की छाती को देखो,

सूखी कितनी इसकी कलियाँ,

मुरझाई कितनी बल्लरियाँ,

जो मुरझाई फिर कहाँ खिल,

पर बोलो सूखे फूलों पर,

कब मधुबन शोर मचाता है?

जो बीत गई तो बात गई।

(क) जो बीत गई सो बात गई से क्या तात्पर्य है। स्पष्ट कीजिए।

(ख) आकाश की ओर कब देखना चाहिए, और क्यों?

(ग) “सूखे फूल’ और ‘मधुबन के प्रतीकार्य स्पष्ट कीजिए।

(घ) टूटे तारों का शोक कौन नहीं मनाता है?

(ङ) आपके विचार से जीवन में एक सितारा किसे माना होगा?

उत्तर: (क) कविता हमेशा ही कठिन परिस्थितियों को हमारे अनुकूल कर नए पथ का सृजन करती है। जब आँसू पराजित हो जाते हैं तो कविता अपनी लेखनी द्वारा प्रतिकूल परिस्थितियों से जंग लड़ती है। जब कर्ता हताश हो जाता है तो उसमें नई उमंग भरती है। कविता एक ऐसे नए सूरज को निर्माण करती है जो नया सवेरा लाता है।

(ख) “जब भी तम का जुल्म बढ़ा है, कविता नया सूर्य गढ़ती है-को भाव यह है कि जब-जब अंधेरा अर्थात् प्रतिकूल परिस्थितियाँ अपने चरम पर होती हैं तो कविता अपनी लेखनी द्वारा इन प्रतिकूल अंधकारमय परिस्थितियों को अपने पथ प्रदर्शक शब्दों द्वारा नया सूर्य दिखाकर उन्हें प्रतिकूल बनाती है।

(ग) गलियों का चैन शब्दहीन निर्दय चेतना द्वारा लुटता है।

(घ) परस्पर संबंधों में दूरियाँ बढ़ने लगी :- यह भाव अपने भी हो गए पराए में आया है।

(ङ) जब कर्ता अकर्ता हो जाता है अर्थात् प्रतिकूल परिस्थितियों के समक्ष हार जाता है तो कविता उसे जीना सिखाती है।

अथवा

(क) जो बीत गई सो बात गई से तात्पर्य है कि जो बीत गया वो हमारा कल था और वह दोबारा नहीं आएगा। अतीत के दुःखों को याद कर रोने से कोई लाभ नहीं।

(ख) अंबर की ओर रात्रि में देखना चाहिए, जब उसमें अनगिनत तारे होते हैं क्योंकि तारे प्रतिदिन टूटते हैं पर अंबर हमेशा ही वैसा का वैसा रहता है। चाहे नए तारे आए या पुराने टूटे।

(ग) मधुबन का अर्थ बगीचा एवं सूखे फूल मधुबन में मुरझाए फूल।।

(घ) टूटे तारों का शोक अंबर नहीं मनाता है।

(ङ) आपके विचार से जीवन में एक सितारा हमारे जीवन का महत्वपूर्ण व्यक्ति होता है जिसके इर्द-गिर्द हमारी दुनियाँ घुमती है।

खण्ड ‘ख‘

प्रश्न 3. निर्देशानुसार किन्हीं तीन के उत्तर लिखिए-

(क) मैंने उस व्यक्ति को देखा जो पीड़ा से कराह रहा था। (संयुक्त वाक्य में बदलिए)

(ख) जो व्यक्ति परिश्रमी होता है, वह अवश्य सफल होता है। (सरल वाक्य में बदलिए)

(ग) वह कौन-सी पुस्तक है जो आपको बहुत याद है। रेखांकित उपवाक्य का भेद बदलिए)

(घ) कश्मीरी गेट के निकल्सन कब्रगाह में उनका ताबूत उतारा गया। मिश्र वाक्य में बदलिए)

उत्तर: (क) संयुक्त वाक्य मैंने उस व्यक्ति को देखा पर वह पीड़ा से कराह रहा था।

(ख) सरल वाक्य-परिश्रमी व्यक्ति अवश्य सफल होता है।

(ग) वह कौन-सी पुस्तक है’-प्रधान उपवाक्य

(ग) वह कौन-सी पुस्तक है-प्रधान उपवाक्य

(घ) जहाँ कश्मीरी गेटका निकाल्सन कब्रगाह तथा वहाँ उनका ताबूत उतारा गया।

अथवा

(घ) उनका ताबूत उतारा गया जहाँ कश्मीरी गेट का निकत्सन कब्रगाह था।

प्रश्न 3.निम्नलिखित वाक्यों में से किन्हीं चार वाक्यों का निर्देशानुसार वाच्य परिवर्तन कीजिए-

(क) बालगोबिन भगत प्रभातियाँ गाते थे। (कर्मवाच्य में बदलिए)

(ख) बीमारी के कारण वह यहाँ न आ सका।। (भाववाच्य में बदलिए)

(ग) माँ के द्वारा बचपन में ही चोषित कर दिया गया था। (कर्तवाच्य में बदलिए)

(घ) अवनि चाय बना रही है। (कर्मवाच्य में बदलिए)

(ङ) घायत हंस उड़ न पाया। भाववाच्य में बदलिए।

उत्तर. (क) कर्मवाच्य- बालगोबिन भगत द्वारा प्रभातियाँ गाई जाती थी।

(ख) भाव वाच्य- बीमारी के कारण वह यहाँ नहीं आ सकता।

(ग) कर्तृवाच्य-माँ ने बचपन में ही घोषित कर दिया था।

(घ) अवनि के द्वारा चाय बनायी गयी।

(ङ) भाववाच्य घायल हंस उड़न सका।

प्रश्न5. निम्नलिखित वाक्यों में से किन्हीं चार रेखांकित पदों का पद-परिचय लिखिए

(क) दादी जी प्रतिदिन समाचार पत्र पढ़ती हैं।

(ख) रोहन यहाँ नहीं आया था।

(ग) वे मुंबई जा चुके हैं।

(घ) परिश्रमी अंकिता अपना काम समय में पूरा कर लेती है।

(ङ) रवि रोज सवेरे दौड़ता है।

उत्तर: (क) पढ़ती है- एकवचन, क्रिया, स्त्रीलिंग

(ख) यहाँ- सर्वनाम, स्थानवाचक क्रिया विशेषण

(ग) वे- बहुवचन, सर्वनाम (पुरुषवाचक), कर्ता कारक।

(घ) परिश्रमी- जातिवाचक संज्ञा, एकवचन, विशेषता विशेषता स्पष्ट करता है।

(ङ) रवि- व्यक्ति वाचक संज्ञा, एकवचन, पुल्लिंग कर्ताकारक।

प्रश्न 6. निम्नलिखित में से किन्हीं चार प्रश्नों के उत्तर दीजिए

(क) करुण रस का एक उदाहरण लिखिए।

(ख) निम्नलिखित काव्य पंक्तियों में निहित रस पहचान कर लिखिए

तंबूरा ले मंच पर बैठे प्रेमप्रताप,

साज मिले पंद्रह मिनट, घंटा भर आलाप।

घंटा भर आलाप राग में मारा गोता.

धीरे-धीरे खिसक चुके थे सारे श्रोता।

(क) उत्साह किस रस का स्थायी भाव है?

(ख) वात्सल्य रस का स्थायी क्या है?

(ग) शृंगार रस के कौनसे दो भेद हैं।

उत्तरः करुण रस का उदाहरण उभरी नसों वाले हाथ घिसे नाखूनों वाले हाथ पीपल के पत्ते से नए-नए हाथ जूही की डाल से खुशबूदार हाथ गंदे कटे-पिटे हाथ जख्म से फटे हुए हाथ खुशबू रचते हैं

(क) हाथा

(ख) हास्यरस

(ग) उत्साह वीर रस का स्थायी भाव है।

(घ) वात्सल्य रस का स्थायी भाव वत्सल है।

(ङ) श्रृंगार रस के दो भेद संयोग श्रृंगार और वियोग श्रृंगार

खण्ड ‘ग’

प्रश्न 7. निम्नलिखित गद्यांश को ध्यानपूर्वक पढ़कर पूछे गए प्रश्नों के उत्तर लिखिए

किसी दिन एक शिष्या ने डरते-डरते खाँ साहब को टोका, बाबा। आप यह क्या करते हैं, इतनी प्रतिष्ठा है आपकी। अब तो आपको भारतरत्न भी मिल चुका है, यह फटी तहमद न पहना करें। अच्छा नहीं लगता, जब भी कोई आता है आप इसी फटी तहमद में सबसे मिलते हैं। खाँ साहब मुसकराए। लाड़ से भरकर बोले, “धत् ! पगली, ई भारतरत्न हमको शहनईया पे मिला है. लुंगिया पे नाहीं तुम लोगों की तरह बनाव-सिंगार देखते रहते, तो उमर ही बीत जाती, हो चुकती शहनाई। तब क्या रियाज हो पाता?”

(क) एक दिन एक शिष्या ने खाँ साहब को क्या कहा? क्यों?

(ख) खाँ साहब ने शिष्या को क्या समझाया।

(ग) इससे खाँ साहब के स्वभाव के बारे में क्या पता चलता

उत्तर: (क) एक दिन एक शिष्या ने खाँ साहब से कहा कि बाबा अब तो आपको बहुत प्रतिष्ठा व सम्मान मिल चुका है, फिर भी आप यह फटी हुई तहमद (लुंगी) क्यों पहनते हो? उस (शिष्या) ने ऐसा इसलिए कहा क्योंकि खाँ साहब इसी तहमद में ही सभी से मिलते थे और उसे यह अच्छा नहीं लगता था।

(ख) खाँ साहब ने नम्रतापूर्वक अपनी शिष्या को समझाते हुए कहा कि मुझे भारतरत्न’ शहनाई बजाने पर मिला है, न कि लुंगी पर। मैंने बनाव सिंगार पर ध्यान न देकर अपनी साधना शहनाई पर ध्यान दिया है।

(ग) इससे खाँ साहब के स्वभाव का पता चलता है कि वे सादा जीवन उच्च विचार के प्रबल समर्थक थे। वे सादगी पसंद और उन्होंने अपना सारा जीवन अपनी साधना में समर्पित कर दिया। वे सच्चे अर्थों में सच्चे कलाकार थे।

प्रश्न 8. निम्नलिखित में से किन्हीं चार प्रश्नों के उत्तर संक्षेप में लिखिए :

(क) दो उदाहरण दीजिए जिनसे आपको लग हो कि बालगोबिन भगत सामाजिक रूढ़ियों से न बँध कर प्रगतिशील विचारों का परिचय देते हैं।

(ख) नवाब साहब ने खीरा खाने की पूरी तैयारी की और उसके बाद उसे बिना खाए फेंक दिया। इस नवाबी व्यवहार पर टिप्पणी कीजिए।

(ग) फादर कामिल बुल्के के हिन्दी के प्रति लगाव के दो उदाहरण पाठ के आधार पर दीजिए।

(घ) मन्नू भण्डारी के व्यक्तित्व पर किन-किन व्यक्तियों का प्रभाव किस रूप में पड़ा?

(ङ) कैसे कह सकते हैं कि “काशी संस्कृति की प्रयोगशाला” है? ‘नौबतखाने में इबादत’ पाठ के आधार पर लिखिए।

उत्तर:(क) जब बालगोबिन भगत के बेटे की मृत्यु हुई तो उन्होंने अपनी पतोहू को दूसरा विवाह करने के लिए कहा। उनके विचार से पति की मृत्यु के बाद एक स्त्री के लिए अकेले जीवन बिताना बहुत ही दुखपूर्ण और चुनौती भरा कार्य है। उनका यह व्यवहार सामाजिक रूढ़ियों से न बंधकर ऊपर उठाता है।

भगत गृहस्थ होकर भी एक साधु की परिभाषा पर खरे उतरते थे खेतीबाड़ी करते थे और जो भी खेत में अन्न उपजता उसको सबसे पहले कबीर के मठ में ले जाते और जो भी प्रसाद रूप में मिलता उसी से गुजर बसर करते।

(ख) नवाब साहब अपनी नवाबी का दिखावा कर रहे थे। उन्होंने खीरे को पहले धोया, सुखाया, छीला और फिर फॉकों में काटकर सँघकर खिड़की से बाहर फेंक दिया इससे उनके दिखावटी पूर्ण जीवन का पता चलता है। कि वे खीरे को अपदार्थ और तुच्छ समझते हैं।

(ग) फादर कामिल बुल्के हिंदी वालों के द्वारा हिन्दी भाषा की उपेक्षा से बहुत दुखी होते थे। हर मंच से वे हिन्दी को राष्ट्रभाषा बनाये जाने की बात करते और हिन्दी को राष्ट्रभाषा बनाने के लिए कार्य करते।

(घ)मन्नू भंडारी पर उनके पिता और हिन्दी की प्राध्यापिका शीला अग्रवाल का प्रभाव पड़ा। पिता की तरह वह भी देशभक्त और आजादी के आंदोलन में भागीदारी देने वाली देशभक्त थीं। पिता की तरह ही शक्की स्वभाव और तमाम गुण मन्नू में समाहित थे।

शीला अग्रवाल ने उन्हें साहित्य के क्षेत्र में आगे बढ़ने को कार्य किया। मन्नू को चुनिंदा उच्च साहित्यकारों की पुस्तकें पढ़ने के लिए प्रेरित किया। साथ ही घर की चारदीवारी से बाहर निकालकर उनको आंदोलन के लिए प्रेरित किया।

(ङ) काशी संस्कृति की पाठशाला है क्योंकि काशी में संगीत की एक अद्भुत परंपरा रही है। बड़े-बड़े रसिक कण्ठे महाराज ने भी यहीं सबको संस्कृति का पाठ पढ़ाया। काशी में बाबा विश्वनाथ हैं, संकटमोचक हनुमान का मंदिर है। काशी में गंगा जमुनी संस्कृति है। इसको शास्त्रों में आनंद कानन के नाम से भी जाना जाता है।

प्रश्न 9. निम्नलिखित काव्यांश को ध्यानपूर्वकपढ़कर पूछे गए प्रश्नों के उत्तर लिखिए

यश है या न वैभव है, मान है न सरमायाः

जितनी ही दौड़ातू उतना ही भरमाया।

प्रभुता का शरण-बिंब केवल मृगतृष्णा है,

हर चंद्रिका में छिपी एक रात कृष्णा है।

जो है यथार्थ कठिन उसका तू कर पूजन

छाया मत छूना मन,

होगा दुख दूना।

(क) हर चंद्रिका में छिपी एक रात कृष्णा है- इस पंक्ति से कवि किस तथ्य से अवगत करवाना चाहता है ?

(ख) कवि ने यथार्थ के पूजन की बात क्यों कही है?

(ग) मृगतृष्णा का प्रतीकात्मक अर्थ लिखिए।

उत्तर (क) हर चंद्रिका में छिपी एक रात कृष्णा है- इस पंक्ति में कवि यह तथ्य अवगत कराना चाहते हैं कि मनुष्य को इस यथार्थको स्वीकार कर लेना चाहिए कि जीवन में सुख-दुख का चोली दामन का साथ होता है। जीवन में केवल सुख रूपी चाँदनी रातें ही नहीं अपितु दुख रूपी अमावस्या भी आती है।

(ख) कवि ने यथार्थ पूजन की बात इसलिए कही है क्योंकि यथार्थ ही जीवन की वास्तविकता है, इसका सामना हर किसी को करना पड़ता है। भविष्य को सुंदर बनाने के लिए वर्तमान में परिश्रम करना पड़ता है।

(ग) मृगतृष्णा का शाब्दिक अर्थ है-धोखा या भ्रम रेगिस्तान में रेत के टीलों पर चिलचिलाती धूप को पानी समझकर हिरण प्यास बुझाने दौड़ता है। इसी को मृगतृष्णा कहते हैं। इसका प्रतीकात्मक अर्थ भ्रमक चीजों से है जो सुख का भ्रम पैदा करती है। जो न होकर भी होने का आभास कराती है वही मृगतृष्णा है।

प्रश्न 10. निम्नलिखित में से किन्हीं चीर प्रश्नों के उत्तर संक्षेप में लिखिए :

(क) ‘उत्साह’ कविता में कवि बादल से क्या अनुरोध करता है?

(ख) भाव स्पष्ट कीजिए :“छू गया तुमसे कि झरने लग पड़े शेफालिका के फूल बाँस था कि बबूल?”

(ग) “छाया मत छूना मन’ में ‘छाया’ किसका प्रतीक है? उसे छूने को मना क्यों किया गया है?

(घ) ‘कन्यादान’ कविता का संदेश संक्षेप में लिखिए।

(ङ) मॅजे हुए प्रतिष्ठित संगीतकार को भी अच्छे संगतकार की आवश्यकता क्यों होती है?

उत्तर: (क) उत्साह कविता में कवि बादल से अनुरोध करता है। कि हे बादलो तुम गगन को चारों ओर से घेर लो, घोर अंधकार कर लो और क्रांति करो। अनंत दिशा से आकर घरघोर गर्जना करके बरसों और तपती हुई धरा को शीतल कर दो। क्रांति के द्वारा परिवर्तन ले आओ।

(ख) कवि अपने शिशु की दंतुरित मुस्कान को देखकर कहता है कि तुम्हारे नए दाँतों के मुस्कान में एक आकर्षण है। जैसे ही मैंने तुम्हें छुआ ऐसे लगा कि शेफालिका के सफेद फूल झड़ रहे हैं। तुम्हारी मुस्कान देखकर बांस और बबूल में भी शेफालिका के जैसे फूल खिलने लगेंगे।

(ग) छाया मत छूना अतीत की स्मृतियों की प्रतीक है। कवि अतीत को याद करने से मना करता है क्योंकि अतीत को याद करके वर्तमान का दुःख दुगुना हो जाता है। हम आज के सुख को भी खो देते हैं। अतः हमें अतीत को भूल जाना चाहिए और वर्तमान में जीना चाहिए और आने वाले समय को सुखी बनाने के लिए कार्य करना चाहिए।

(घ) कन्यादान कविता का सन्देश यह है कि हमारे समाज में स्त्रियों के लिए कुछ प्रतिमान स्थापित कर दिए जाते हैं। समाज उनको कमजोर समझता है और अत्याचार करता है। अपने संचित अनुभव के आधार पर माँ कन्यादान के समय अपनी बेटी को शिक्षित कर रही है। ताकि समाज में वह एक उच्च सुखी जीवन जी सके और समाज की मानसिकता से वह परिचित हो सके। विवाह पश्चात् लड़की परिवार की केन्द्र बिन्दु होती है। अतः लड़की को उसके कर्तव्यों से परिचित करा रहा।

(ङ) जब कभी मँजा हुआ संगीतकार अपने सुरों के जंगल में भटक जाता है अनहद में चला जाता है तब संगतकार ही उसके सुरों को सँभालने का कार्य करता है। संगीतकार ही मुख्य संगीतकार या गायक का अस्तित्व बचाता है और स्वयं हमेशा मुख्य गायक के पीछे रहता।

प्रश्न 11. ‘माता का अंचल’ पाठ की दो बातों का उल्लेख कीजिए जो आपको अच्छी लगी हों। इनसे आपको क्या प्रेरणा मिली?

अथवा

प्रश्न 11. ‘जॉर्ज पंचम की नाक’ पाठ में निहित व्यंग्य को स्पष्ट कीजिए।

उत्तर:‘माता का अंचल’ पाठ ग्रामीण संस्कृति के बच्चों के बचपन की एक जीवंत झलक है। इस पाठ में बच्चों के स्वच्छंद बचपन का वर्णन है कि किस प्रकार वे अपने हमजोलियों के बीच मिट्टी में ही बिना खेल खिलौनों के अपना जीवन बिताते हैं। पिताजी का भोलेनाथ के हर खेल में शामिल होना हर खेल पर अपनी बच्चों सी टिप्पणी देना बहुत अच्छा लगा। जब चूहे के बिल में से सांप निकल आया और दशहत में आकर संकट के समय भोलेनाथ का माँ के आँचल में जाकर छुप जाना बहुत अच्छा लगा। इस पाठ में गुदगुदाने वाले प्रसंग भी अनेक हैं। पिता का इस प्रकार बच्चा बन जाना बहुत सुखद अनुभव है जो सभी पाठकों को गुदगुदा देता है।

अथवा

उत्तर:‘जॉर्ज पंचम की नाक’ पाठ एक सटीक व्यंग्य है हमारे शाही तंत्र की गुलामी की मानसिकता पर जब रानी एलिजाबेथ द्वितीय भारत दौरे पर आ रही थी तो बड़े-बड़े हुक्कामों ने दिल्ली का काया पलट कर दिया। वे भूल चुके हैं कि इसी महारानी के देश ने ही उन्हें कभी गुलाम बनाया था। इस निबंध में सरकारी कार्यप्रणाली पर भी व्यंग्य है। नाजनीनों की तरह दिल्ली को सजाया संवारा गया । जॉर्ज पंचम की मूर्ति से गायब नाक के लिए मूर्तिकार को नाक लगाने का आदेश दे दिया गया। मूर्तिकार हिंदुस्तान के कोने कोने में गया किन्तु मूर्ति की नाप की नाक ढूँढने में असफल रहा।

अंत में जिंदा नाक लगाकर कार्य पूरा किया गया। यह एक जीता जागता उदाहरण है हमारे शाही तंत्र की मानसिकता पर कि किस प्रकार अपनी नाक बचाने के लिए जनता की नाक तक काट देते हैं।

खण्ड ‘घ’

प्रश्न 12. निम्नलिखित में से किसी एक विषय पर दिए गए संकेत बिन्दुओं के आधार पर लगभग 200 से 250 शब्दों में निबंध लिखिए

(क) कमरतोड़ महँगाई

महँगाई के कारण
समाज पर प्रभाव
व्यावहारिक समाधान

(ख) स्वच्छ भारत अभियान

विकास में स्वच्छता का योगदान
अस्वच्छता से हानिया
रोकने के उपाय

(ग) बदलती जीवन शैल

जीवन शैली का आशय
बदलाव कैसा
परिणाम

उत्तर: निबंध

कमरतोड़ महँगाई महँगाई का अर्थ होता है वस्तुओं की कीमत में वृद्धि होना। इस महँगाई पर ही पूरे देश की अर्थव्यवस्था टिकी होती है। मॅहगाई मनुष्य के जीवन शैली को प्रभावित करती है। आप समाज की यह प्रमुख समस्या है जिसने उच्च, मध्यम व निम्न सभी वर्गों की कमर तोड़ रखी

महँगाई की समस्या न केवल भारत बल्कि पूरे विश्व की गंभीर समस्या बन गई है। इस समस्या के कारण बहुत से देशों का आर्थिक स्तर घटता है। हमारा देश, भारत जनसंख्या की दृष्टि से दूसरा बड़ा देश है। पर उस तरह की पैदावार नहीं हो पा रही है जिससे आए दिन सामानों के दाम बढ़ते हैं। आजादी के बाद भारत में तीन चीजें हमेशा बढ़ती रही हैं भ्रष्टाचार, असमानता और महँगाई। ये तीनों सगी बहनें हैं। ये एक साथ बढ़ती हैं। भ्रष्टाचार बढ़ता है तो धनवान और धनी होते जाते हैं और गरीब बिल्कुल लगोंटी यारी हो जाते हैं। काले धन के कारण कालाबाजारी बढ़ती है। उससे मँहगाई बढ़ती है।

हमारे देश के अमीर लोग इस मॅहगाई के सबसे अधिक जिम्मेदार होते हैं। आपात काल के शुरू-शुरू में वस्तुओं की कीमत कम रखने की परंपरा चली लेकिन व्यापरी अपनी मनमानी करते हैं, सामान्य जनता पर महँगाई का सबसे अधिक प्रभाव पड़ता है। उसके पास सीमित पूँजी होती है और उसकी खरीदने की शक्ति कमजोर हो जाती है। अफसरशाही नेता, व्यापारी ये सभी मँहगाई को बढ़ाने के लिए बहुत अधिक जिम्मेदार हैं। इससे समाज पर दुष्प्रभाव पड़ता है।

यदि सरकार मुद्रास्फीति पर रोक लगाए तो शायद महँगाई पर कुछ तो लगाम लग सकेगी। सरकार को अधिक पैमाने पर गांवों का विकास कर उन्हें जागरूक करना चाहिए जिससे वे आधुनिक संसाधनों का प्रयोग करे और पैदावार बढ़ाए।

हमारी अधिकांश समस्याओं का मूल कारण जनसंख्या है। जब तक जनसंख्या को वश में नहीं किया जाएगा मॅहगाई वश में नहीं आयेगी। महँगाई को वश में करने के लिए उचित राष्ट्र नीति जरूरी है। मॅहगाई को रोकने के लिए लोगों को अपनी जमाखोरी की प्रवृत्ति छोड़नी होगी। उपभोक्ता को भी उतनी ही वस्तुएं खरीदनी चाहिए जितनी कि उसे आवश्यकता हो । दोबारा जरूरत पड़ने पर उपभोक्ता को तभी सामान लेना चाहिए। इस तरह से महँगाई पर काबू रखा जा सकता है।

(ख) स्वच्छ भारत अभियान

एक कदम स्वच्छता की ओर स्वच्छता आज केवल घर या मुहल्ले तक नहीं बल्कि इसका दायरा काफी बड़ा बन गया है। देश और राष्ट्र की स्वच्छता से ही वास्तविक विकास हो सकता है। जिस देश का नागरिक स्वच्छता के प्रति सजग होगा उस देश का विकास अबाध गति से होगा। प्रत्येक नागरिक का कर्तव्य है कि वह अपने देश की स्वच्छता में अपना सहयोग दें। इसी को ध्यान में रखते हुए हमारी सरकार ने स्वच्छ भारत अभियान चलाया है जिसमें यह प्रण लिया गया है कि प्रत्येक गली, मुहल्ला, सड़कों से लेकर शौचालय का निर्माण कराना और बुनियादी ढांचे को बदलना एवं स्वच्छता का संदेश फैलाना।

अस्वच्छता से विभिन्न प्रकार की हनियाँ हैं। स्वच्छ वातावरण से पर्यावरण का विकास संभव है। हम सभी जानते हैं कि हमारा वातावरण ग्लोबल वार्मिंग की समस्या से ग्रस्त है पर इससे कुछ अंशों में युक्त। का एकमात्र साधन स्वच्छता है। जिस तरह का स्वच्छ परिवेश से वातावरण में कोई रोग-बीमारी का खतरा नहीं होगा। पहले गांवों में लोग खुले में शौच जाते थे पर इस अभियान द्वारा जनता जागरूक हो गई एवं उन्हें शौचालय का महत्व समझ में आया है। सड़कों, गलियों को स्वच्छ रखना जिससे वहाँ रहने वाले लोग स्वस्थ रहेंगे।

हमारे पूज्यनीय गांधी जी स्वच्छता के प्रति अत्यंत जागरूक थे उन्हीं के सिद्धांतों को आगे बढ़ाते हुए हमारे प्रधानमंत्री श्री नरेन्द्र मोदी जी ने 2 अक्टूबर 2014 को इस आंदोलन से जोड़ने की मुहिम चलाई है। इस आंदोलन को जन-जन तक पहुँचाने के लिए हमारे अभिनेता-अभिनेत्री सब स्कूल, कॉलेजों वे सरकारी कार्यालयों ने अहम भूमिका निभाई है। अब वह समय दूर नहीं जब हम गांधी जी के सपने को साकार कर पाएंगे।

इसलिए हमें ‘स्वच्छ भारत अभियान में बढ़चद कर हिस्सा लेना चाहिए और कुछ नहीं तो हमें कम से कम रोज हमारी गली को साफ करना चाहिए। शिक्षा के प्रसार प्रचार को बढ़ावा देकर जनता को स्वच्छता के प्रति जागरूक करना चाहिए। स्वच्छ भारत अभियान में आप भी भागीदार बने लोगों को स्वच्छता के प्रति जागरूक बनाएं।

(ग) बदलती जीवन शैली स्वस्थ जीवन शैली एक अच्छे जीवन की नींव है। हालांकि इस जीवन शैली को हासिल करने में ज्यादा मेहनत नहीं लगती बल्कि कई लोग व्यावसायिक प्रतिबद्धताओं दृढ़ संकल्प की कमी और अन्य कारणों द्वारा इसका पालन नहीं कर पाते स्वस्थ रहने के लिए किस प्रकार की शैली अपनाना है यह जानना ज्यादा जरूरी हैं।

आजकल की पीढ़ी कम्यूटर मोबाइल, बर्गर, पिज्जा और देर रात की पार्टियों पर आधारित है-मूल रूप से ये सब अस्वास्थ्यकर है। पेशेवर प्रतिबद्धताओं और व्यक्तिगत मुद्दों ने सभी को जकड़ लिया है। और इन सभी आवश्यकताओं के बीच वे अपना स्वास्थ्य खो रहे हैं। इन दिनों लोग अपने दैनिक जीवन में इतने व्यस्त हो गए हैं कि वे भूल गए हैं कि एक स्वस्थ जीवन जीने के क्या मायने हैं। हम स्वस्थकर आदतें अपनाकर अपनी जीवन शैली में सुधारा जा सकता है यदि हम प्रातः भ्रमण, योगा व मेडिटेशन का जीवन में समावेश करें तो हमारा स्वास्थ्य अच्छा हो पाएगा। शरीर में अनेक शक्तियाँ निहित हैं, यदि हम उन शक्तियों को पहचान लेंगे तो हम निरोग रहेंगे।

प्रश्न 13. ‘पड़ोस में आग लगने की दुर्घटना की खबर तुरंत दिए जाने पर भी दमकल अधिकारी और पुलिस देर से पहुँचे जिससे आग ने भीषण रूप ले लिया। इसके बारे में विवरण सहित एक शिकायती पत्र अपने जिला अधिकारी को लिखिए।

अथवा

प्रश्न 13. पढ़ाई छोड़कर घर बैठे छोटे भाई को समझाते हुए पत्र लिखिए कि पढ़ना क्यों आवश्यक है। पत्र ऐसा हो कि उसमें नई उमंग का संचार हो सके।

उत्तर:

परीक्षा भवन,

क, ख, ग, आगरा।

दिनांक 25 मार्च 20XX

सेवा में,

जिलाधिकारी,

अ, ब, स

आगरा ।

विषय – दमकल अधिकारी और पुलिस की लापरवाही की शिकायत हेतु पत्र।

महोदय,

सविनय निवेदन है कि मैं च. छ. ज. क्षेत्र का निवासी इस पत्र के माध्यम से आपको दमकल विभाग और पुलिस की लापरवाही की शिकायत करना चाहता हूँ। कल मेरे पड़ोसी के यहाँ अचानक आग लग गयी, जिस कारण पड़ौसियों का घर पूरी तरह से जल कर स्वाहा हो गया।

आग लगने की घटना की जानकारी तुरंत दमकल विभाग को दी गयी परन्तु कई घण्टे तक भी दमकल की कोई गाड़ी नहीं आयी और न ही पुलिस ने आकर घटना की जानकारी ली। इतनी देर में आग की लपटें दूसरे पड़ौसियों के घर तक आकर भी फैल गयी।

दमकल विभाग के कर्मचारी समय पर आते तो इतने बड़े नुकसान को बचाया जा सकता था। आपसे अनुरोध है कि दमकल विभाग और पुलिसकर्मियों के प्रति सख्त कार्यवाही करें जो सूचना देने के बाद भी घटनास्थल पर नहीं आए।

सधन्यवाद।

भवदीय

प. फ. ब.

अथवा

उत्तर:

परीक्षा भवन

क, ख, ग

आगरा

दिनांक 22.20.20XX

प्रिय अनुज,

शुभाशीष। इस पत्र के द्वारा मैं तुम्हें पढ़ाई का महत्व समझाना चाहता हूँ मुझे मालूम हुआ है कि तुम पढ़ाई छोड़कर घर पर बैठे हो और विद्यालय भी नहीं जा रहे हो। यदि तुम विद्यालय नहीं जाओगे तो घर पर बैठकर तुम्हारा अर्जित ज्ञान भी धूमिल हो जाएगा और खाली दिमाग शैतान का घर होता है। बिना पढ़ाई के जीवन व्यर्थ है तुम अपना भविष्य बिना पढ़ाई के कैसे बना सकते हो। यदि तुम पढ़ोगे तो आगे चलकर अपने पैरों पर खड़े हो सकते हो और आत्मसम्मान आत्मविश्वास प्राप्त करोगे और जो चाहो जीवन में हासिल कर पाओगे।

आशा है कि तुम मेरी बात समझ गए होंगे और कल से नियमित रूप से विद्यालय जाओगे तथा परिश्रम करोगे। घर में माता पिताजी को मेरा दंडवत् प्रणाम देना और छोटी बहन को प्यार।

तुम्हारा अग्रज,

च.छ.ज.

प्रश्न 14. अतिवृष्टि के कारण कुछ शहर बाढ़ ग्रस्त हैं। वहाँ के निवासियों की सहायतार्थ सामग्री एकत्र करने हेतु एक विज्ञापन लगभग 50 शब्दों में तैयार कीजिए।

अथवा

प्रश्न 14. बॉल पेनों की एक कंपनी सफल’ नाम से बाजार में आई है। उसके लिए एक विज्ञापन लगभग 50 शब्दों में तैयार कीजिए।

उत्तर: दान एक पुण्य

आइए पुण्य कमाइए

रोटरी क्लब की ओर से केरल के बाढ़पीड़ितों के लिए सहायता आपकी सहायता किसी की जिंदगी को सुरक्षित कर सकती है। तो देर किस बात की गाँधी मैदान में आकर अपना सहयोग दें.

(सहयोग राशि कुछ भी हो सकती है) शिविर 7,8,9 अप्रैल 2019 तक ही।

अन्य जानकारी हेतु संपर्क करें 8979645321, 011-254639

रोटरी क्लब ऑफिस गाँधी मैदान,

जनकपुरी नई दिल्ली।

अथवा

उत्तर:

Maths 10th Previous Year Question Paper 2019 SET-II (CBSE)

Maths

Section -A

Q.1. If HCF (336,54) = 6, find LCM (336, 54),

Answer : Given, HCF (336, 54) = 6

We know HCF × LCM = one number × other number

6 × LCM = 336 × 54

LCM = 336×54/6 = 336 × 9 = 3024

Q.2. Find the nature of roots of the quadratic equation 2x² – 4x + 3 = 0.

Answer: Given, 2x² – 4x + 3 = 0

Comparing it with quadratic equation ax² + bx + c = 0

Here, a = 2, b = -4 and c = 3

D = b² – 4ac = (-4)² – 4 × (2)(3) = 16 – 24 = -8 < 0

Hence, D < 0 this shows that roots will be imaginary.

Q.3. Find the common difference of the Arithmetic Progression (A.P.).

Answer :

Q.4. Evaluate: sin² 60° + 2 tan 45° – cos² 30°

Q.4. If sin A = ¾, Calculate sec A.

Answer :

We know,

Q.5. Write the coordinates of a point P on the x-axis which is equidistant from point A(-2, 0) and B(6, 0).

Answer: Let coordinates of P on x-axis is (x, 0)

Given, A(-2, 0) and B(6, 0)

Here, PA = PB

On squaring both sides, we get

(x + 2)² = (x – 6)²

⇒ x² + 4 + 4x = x² + 36 – 12x

⇒ 4 + 4x = 36 – 12x

⇒ 16x = 32

⇒ x = 2

Coordinates of P are (2, 0)

Q.6. Find the 21st term of the A.P. -4½, – 3, -1½ ………

Answer :

Section- B

Q.7. For what value of k, will the following pair of equations have infinitely many Answers:

2x + 3y = 7 and (k + 2)x – 3(1 – k)y = 5k + 1 [2]

Answer: Given, The system of equations is

2x + 3y = 7 and (k + 2) x – 3 (1 – k) y = 5k +1

These equations are of the form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

where, a₁ = 2, b₁ = 3, c₁ = -7

a₂ = (k + 2), b₂ = -3(1 – k), c₂ = -(5k + 1)

Since, the given system of equations have infinitely many Answers.

Hence, the given system of equations has infinitely many Answers when k = 4.

Q.8. Find a relation between x and y if the points A(x, y), B(-4, 6) and C(-2, 3) are collinear.

Q.8. Find the area of a triangle whose vertices are given as (1, -1) (-4, 6) and (-3, -5).

Answer: Given, A(x, y), B(-4, 6), C(-2, 3)

x¹ = x, y¹ = y, x² = -4, y² = 6, x³ = -2, y³ = 3

If these points are collinear, then area of triangle made by these points is 0.

Q.9. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is ⅕. The probability of selecting a black marble at random from the same jar is ¼. If the jar contains 11 green marbles, find the total number of marbles in the jar.

Answer: Let the probability of selecting a blue marble, black marble and green marbles are P(x), P(y), P(z) respectively.

P(x) = ⅕ , P (y) = ¼ Given

We Know,

P(x) + P(y) + P(z) = 1

Q.10. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique Answer.

Answer: Given, x + 2y = 5, 3x + ky + 15 = 0

Comparing above equations with

a¹x + b¹y + c¹ = 0 and a²x + b²y + c² = 0,

We get,

a¹ = 1, b¹ = 2, c¹ = -5

a² = 3, b² = k, c² = 15

Condition for the pair of equations to have unique Answer is

k can have any value except 6.

Q.11. The larger of the two supplementary angles exceeds the smaller by 18°. Find the angles.

Q.11. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?

Answer: Let two angles A and B are supplementary.

A + B = 180° …(i)

Given, A = B + 18°

On putting A = B + 18° in equation (i),

we get B + 18° + B = 180°

⇒ 2B + 18° = 180°

⇒ 2B = 162°

⇒ B = 81°

A = B + 18°

⇒ A = 99°

Answer: Let age of Sumit be x years and age of his son be y years.Then, according to question we have, x = 3y …… (i)

Five years later, x + 5 = 2½ (y + 5)……….(ii)

On putting x = 3y in equation (ii)

Q.12. Find the mode of the following frequency distribution:

Answer :

Here, the maximum frequency is 50.

So, 35 – 40 will be the modal class.

l = 35, f₀ = 34, f₁ = 50, f₂ = 42 and h = 5

Section – C

Q.13. Point A lies on the line segment XY joining X(6, -6) and Y (-4, -1) in such a way that XA/XY = ⅖.

If Point A also lies on the line 3x + k (y + 1) = 0, find the value of k.

Answer: Given,

Since, point A(2, -4) lies on line 3x + k (y + 1) = 0.

Therefore it will satisfy the equation.

On putting x = 2 and y = -4 in the equation, we get

3 × 2 + k(-4 + 1) = 0

⇒ 6 – 3k = 0

⇒ 3k = 6

⇒ k = 2

Q.14. Solve for x: x² + 5x – (a² + a – 6) = 0 [3]

Answer: Taking (a² + a – 6)

= a² + 3a – 2a – 6

= a(a + 3) – 2 (a + 3)

= (a + 3) (a – 2)

x² + 5x – (a + 3) (a – 2) = 0

⇒ x² + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0

⇒ x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0

⇒ (x – a + 2)(x + a + 3) = 0

Hence, x – a + 2 = 0 and x + a + 3 = 0

x = a – 2 and x = -(a + 3)

Required values of x are (a – 2), -(a + 3).

Q.15. Find A and B if sin (A + 2B) = √3/2 and cos (A + 4B) = 0, where A and B are acute angles.

Answer: Given

Sin (A + 2B) = √3/2 and cos (A + 4B) = 0

⇒ sin (A + 2B) = 60° (∵ sin 60° = √3/2)

A + 2B =60 …(i)

cos (A + 4B) = cos 90° (∵ cos 90° = 0)

⇒ A + 4B = 90° …(ii)

On solving equation (i) and (ii), we get

B = 15° and A = 30°

Q.16. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection.

Answer: Let the required ratio be k : 1

By section formula, we have

Q.17. Evaluate:

Answer :

Q.18. In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use π = 3.14)

Q.18. In Figure 5, ABCD is a square with side 2√2 cm and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14)

Answer: Given, OABC is a square with OA = 15 cm

OB = radius = r

Let side of square be a then,

a² + a² = r²

⇒ 2a² = r²

⇒ r = √2 a

⇒ r = 15√2 cm (∵ a = 15 cm)

Area of square = Side × Side = 15 × 15 = 225 cm²

Area of shaded region = Area of quadrant OPBQ – Area of square

= 353.25 – 225 = 128.25 cm

Answer: Given, ABCD is a square with side 2√2 cm

BD = 2r

In ∆BDC,

BD² = DC² + BC²

⇒ 4r² = 2(DC)² (∵ DC = CB = Side = 2√2 )

⇒ 4r² = 2 × 2√2 × 2√2

⇒ 4r² = 8 × 2

⇒ 4r² = 16

⇒ r²= 4

⇒ r = 2 cm

Area of square BCDA = Side × Side = DC × BC = 2√2 × 2√2 = 8 cm²

Area of circle = πr² = 3.14 × 2 × 2 = 12.56 cm²

Area of shaded region = Area of circle – Area of square. = 12.56 – 8 = 4.56 cm²

Q.19. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use π = 22/7).

Answer: ABCD is a cylinder and BFC and AED are two hemispheres which has radius (r) = 7/2 cm

Total volume of solid = Volume of two hemisphere + Volume of cylinder

= 179.67 + 500.5 = 680.17 cm

Q.20. The marks obtained by 100 students in an examination are given below:

Find the mean marks of the students.

Answer :

Q.21. For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5?

Q.21. Find the zeroes of the quadratic polynomial 7y2 – 11/2y – ⅔ and verify the relationship between the zeroes and the coefficients.

Answer: Given, f(x) = 3x⁴ – 9x³ + x² + 15x + k

It is completely divisible by 3x² – 5

Q.22. Write all the values of p for which the quadratic equation x² + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.

Answer: Given, equation is x² + px + 16 = 0

This is of the form ax² + bx + c = 0

where, a = 1, b = p and c = 16

D = b² – 4ac = p² – 4 × 1 × 16 = p² – 64

for equal roots, we have D = 0

p² – 64 = 0

⇒ p² = 64

⇒ p = ±8 Putting p = 8 in given equation we have,

x2 + 8x + 16 = 0

⇒ (x + 4)² = 0

⇒ x + 4 = 0⇒ x = -4

Now, putting p = -8 in the given equation, we get

x² > – 8x + 16 = 0

⇒ (x – 4)² = 0

⇒ x = 4

Required roots are -4 and -4 or 4 and 4.

Section – D

Q.23. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.

Answer: Given, ΔABC ~ ΔDEF

Q.24. Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of a pole is 60° and the angle of depression from the top of the other pole of point P is 30°. Find the heights of the poles and the distance of the point P from the poles.

Answer: Let AC is the road of 80 m width. P is the point on road AC and height of poles AB and CD is h m.

h———–(i)

Equating the values of h from equation (i) and (ii) we get

⇒ x√3 =

⇒ 3x = 80 – x

⇒ 4x = 80

⇒ x = 20m

On putting x = 20 in equation (i), we get

h = √3 × 20 = 20√3

h = 20√3 m

Thus, height of poles is 20√3 m and point P is at a distance of 20 m from left pole and (80 – 20) i.e., 60 m from right pole.

Q.25. The total cost of a certain length of a piece of cloth is ₹ 200. If the piece was 5 m longer and each metre of cloth costs ₹ 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre?

Answer: Let the original length of the piece of cloth is x m and rate of cloth is ₹ y per metre.

Then according to question, we have

x × y = 200 …(i)

and if length be 5 m longer and each meter of cloth be ₹ 2 less than

(x + 5) (y – 2) = 200

⇒ (x + 5) (y – 2) = 200

⇒ xy – 2x + 5y – 10 = 200 …(ii)

On equating equation (i) and (ii), we have

xy = xy – 2x + 5y – 10

⇒ 2x – 5y = -10 …… (iii)

⇒ y = 200/x from equation (i)

⇒ 2x – 5 × 200/x = -10

= 2x – 1000/x -10

⇒ 2x² – 1000 = -10x

⇒ 2x² + 10x – 1000 = 0

⇒ x² + 5x – 500 = 0

⇒ x² + 25x – 20x – 500 = 0

⇒ x(x + 25) – 20 (x + 25) = 0

⇒ (x + 25) (x – 20) = 0

⇒ x = 20 (x ≠ -25 length of cloth can never be negative)

∴ x × y = 200

20 × y = 200

y = 10

Thus, the length of the piece of cloth is 20 m and original price per metre is ₹ 10.

Q.26. Construct an equilateral ∆ABC with each side 5 cm. Then construct another triangle whose sides are ⅔ times the corresponding sides. Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA.

Answer: Steps for construction are as follows:

Draw a line segment BC = 5 cm
At B and C construct ∠CBX = 60° and ∠BCX = 60°
With B as centre and radius 5 cm, draw an arc cutting ray BX at A. On graph paper, we take the scale.
Join AC. Thus an equilateral ∆ABC is obtained.

Below BC, make an acute angle ∠CBY
Along BY, mark off 3 points B₁, B₂, B₃ Such that BB₁, B₁B₂, B₂B₃ are equal.
Join B³C
From B₂ draw B₂D || B₃C, meeting BC at D
From D, draw DE || CA, meeting AB at E.

Then ∆EBD is the required triangle, each of whose sides is ⅔ of the corresponding side of ∆ABC.

Q.27. Change the following data into ‘less than type’ distribution and draw its olive:

Answer :

Q.28. Prove that:

Answer :

Q.29. Which term of the Arithmetic Progression -7, -12, -17, -22,…..will be -82? Is -100 any term of the A.P.? Give a reason for your answer.

Q.29. How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180? Explain the double answer.

Answer:-7, -12, -17, -22, …….

Here a = -7, d = -12 – (-7) = -12 + 7 = -5

Let T_n = -82T_n = a + (n – 1) d

⇒ -82 = -7 + (n – 1) (-5)

⇒ -82 = -7 – 5n + 5

⇒ -82 = -2 – 5n

⇒ -82 + 2 = -5n

⇒ -80 = -5n

⇒ n = 16

Therefore, 16th term will be -82.

Let T_n = -100

Again, T_n = a + (n -1) d

⇒ -100 = -7 + (n – 1) (-5)

⇒ -100 = -7 – 5n + 5

⇒ -100 = – 2 – 5n

⇒ -100 + 2 = -5n

⇒ -98 = -5n

⇒ n = 98/5

But the number of terms can not be in fraction.

So, -100 can not be the term of this A.P.

Answer: 45, 39, 33, …..

Here a = 45, d = 39 – 45 = -6

Let S_n = 180

⇒ n/2 [ 2a + (n – 1) d] = 180

⇒ n/2 [2 × 45 + (n – 1) (-6)] = 180

⇒ n/2 [90 – 6n + 6] = 180

⇒ n/2 [96 – 6n] = 180

⇒ n(96 – 6n) = 360

⇒ 96n – 6n² = 360

⇒ 6n² – 96n + 360 = 0

On dividing the above equation by 6

⇒ n² – 16n + 60 = 0

⇒ n² – 10n – 6n + 60 = 0

⇒ n(n – 10) – 6 (n – 10) = 0

⇒ (n – 10) (n – 6) = 0

⇒ n = 10, 6

Sum of first 10 terms = Sum of first 6 terms = 180

This means that the sum of all terms from 7th to 10th is zero.

Q.30. In a class test, the sum of Aran’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.

Answer: Let Aran marks in Hindi be x and marks in English be y.

Then, according to question, we have

x + y = 30 …(i)(x + 2)(y – 3) = 210 …(ii)

from equation (i) put x = 30 – y in equation (ii)

(30 – y + 2) (y – 3) = 210

⇒ (32 – y) (y – 3) = 210

⇒ 32y – 96 – y² + 3y = 210

⇒ y² – 35y + 306 = 0

⇒ y² – 18y – 17y + 306 = 0

⇒ y(y – 18) – 17(y – 18) = 0

⇒ (y – 18) (y – 17) = o

⇒ y = 18, 17

Put y = 18 and 17 in equation (i), we get x = 12, 13

Hence his marks in hindi can be 12 and 13 and in english his marks can be 18 and 17.

Maths 10th Previous Year Question Paper 2019 SET-III (CBSE)

Maths

Section -A

Q.1. If HCF (336,54) = 6, find LCM (336, 54),

Answer : Given, HCF (336, 54) = 6

We know HCF × LCM = one number × other number

6 × LCM = 336 × 54

LCM = 336×54/6 = 336 × 9 = 3024

Q.2. Find the nature of roots of the quadratic equation 2x² – 4x + 3 = 0.

Answer: Given, 2x² – 4x + 3 = 0

Comparing it with quadratic equation ax² + bx + c = 0

Here, a = 2, b = -4 and c = 3

D = b² – 4ac = (-4)² – 4 × (2)(3) = 16 – 24 = -8 < 0

Hence, D < 0 this shows that roots will be imaginary.

Q.3. Find the common difference of the Arithmetic Progression (A.P.).

Answer :

Q.4. Evaluate: sin² 60° + 2 tan 45° – cos² 30°

Q.4. If sin A = ¾, Calculate sec A.

Answer :

We know,

Q.5. Write the coordinates of a point P on the x-axis which is equidistant from point A(-2, 0) and B(6, 0).

Answer: Let coordinates of P on x-axis is (x, 0)

Given, A(-2, 0) and B(6, 0)

Here, PA = PB

On squaring both sides, we get

(x + 2)² = (x – 6)²

⇒ x² + 4 + 4x = x² + 36 – 12x

⇒ 4 + 4x = 36 – 12x

⇒ 16x = 32

⇒ x = 2

Coordinates of P are (2, 0)

Q.6. In Figure 1, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB.

Q.6. In Figure 2, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.

Answer:

Given, ∠C = 90° and AC = 4 cm, AB = ?

∆ABC is an isosceles triangle so, BC = AC = 4 cm

On applying Pythagoras theorem, we have

AB² = AC² + BC²

⇒ AB² = AC² + AC² (∵ BC = AC)

⇒ AB² = 42 + 42 = 16 + 16 = 32

⇒ AB = √32 = 4√2 cm

Answer: Given, DE || BCOn applying, Thales theorem, we have

Section – B

Q.7. A die is thrown twice. Find the probability that

(i) 5 will come up at least once.

(ii) 5 will not come up either time.

Answer: When two dice are thrown simultaneously, all possible outcomes are

(1.1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total number of outcomes = 36Total outcomes where 5 comes up at least once = 11

Q.8. Find a relation between x and y if the points A(x, y), B(-4, 6) and C(-2, 3) are collinear.

Q.8. Find the area of a triangle whose vertices are given as (1, -1) (-4, 6) and (-3, -5).

Answer: Given, A(x, y), B(-4, 6), C(-2, 3)

x¹ = x, y¹ = y, x² = -4, y² = 6, x³ = -2, y³ = 3

If these points are collinear, then area of triangle made by these points is 0.

Answer: Let the probability of selecting a blue marble, black marble and green marbles are P(x), P(y), P(z) respectively.

P(x) = ⅕ , P (y) = ¼ Given

We Know,

P(x) + P(y) + P(z) = 1

Q.10. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique Answer.

Answer: Given, x + 2y = 5, 3x + ky + 15 = 0

Comparing above equations with

a¹x + b¹y + c¹ = 0 and a²x + b²y + c² = 0,

We get,

a¹ = 1, b¹ = 2, c¹ = -5

a² = 3, b² = k, c² = 15

Condition for the pair of equations to have unique Answer is

k can have any value except 6.

Q.11. The larger of the two supplementary angles exceeds the smaller by 18°. Find the angles.

Q.11. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?

Answer: Let two angles A and B are supplementary.

A + B = 180° …(i)

Given, A = B + 18°

On putting A = B + 18° in equation (i),

we get B + 18° + B = 180°

⇒ 2B + 18° = 180°

⇒ 2B = 162°

⇒ B = 81°

A = B + 18°

⇒ A = 99°

Answer: Let age of Sumit be x years and age of his son be y years.Then, according to question we have, x = 3y …… (i)

Five years later, x + 5 = 2½ (y + 5)……….(ii)

On putting x = 3y in equation (ii)

Q.12. Find the mode of the following frequency distribution:

Answer :

Here, the maximum frequency is 50.

So, 35 – 40 will be the modal class.

l = 35, f₀ = 34, f₁ = 50, f₂ = 42 and h = 5

Section – C

Q.13. Find the ratio in which the y-axis divides the line segment joining the points (-1, -4) and (5, -6). Also, find the coordinates of the point of intersection.

Answer: Let the y-axis cut the line joining point A(-1, -4) and point B(5, -6) in the ratio k : 1 at the point P(0, y)

Then, by section formula, we have

Q.14. Find the value of:

Answer :

Q.15. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.

Answer: Given, a radius of small sphere be r = 3 cm

Both spheres are made by the same metal, then their densities will be same.

Let radius of bigger sphere = r’ then,

Then according to question, we have,

Volume of bigger sphere + Volume of smaller sphere = Volume of new sphere.

4/3(r)³ + 4/3 (r)³= 4/3 (R)³

⇒ r’³ + r³ = R³

⇒ 189 + 27 = R³

⇒ 216 = R³

⇒ R = 6

D = 6 × 2 = 12

Radius of new sphere is 6 cm.

So, the diameter is 12 cm.

Q.16. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection.

Answer: Let the required ratio be k : 1

By section formula, we have

Q.17. Evaluate:

Answer :

Q.18. In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use π = 3.14)

Q.18. In Figure 5, ABCD is a square with side 2√2 cm and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14)

Answer: Given, OABC is a square with OA = 15 cm

OB = radius = r

Let side of square be a then,

a² + a² = r²

⇒ 2a² = r²

⇒ r = √2 a

⇒ r = 15√2 cm (∵ a = 15 cm)

Area of square = Side × Side = 15 × 15 = 225 cm²

Area of shaded region = Area of quadrant OPBQ – Area of square

= 353.25 – 225 = 128.25 cm

Answer: Given, ABCD is a square with side 2√2 cm

BD = 2r

In ∆BDC,

BD² = DC² + BC²

⇒ 4r² = 2(DC)² (∵ DC = CB = Side = 2√2 )

⇒ 4r² = 2 × 2√2 × 2√2

⇒ 4r² = 8 × 2

⇒ 4r² = 16

⇒ r²= 4

⇒ r = 2 cm

Area of square BCDA = Side × Side = DC × BC = 2√2 × 2√2 = 8 cm²

Area of circle = πr² = 3.14 × 2 × 2 = 12.56 cm²

Area of shaded region = Area of circle – Area of square. = 12.56 – 8 = 4.56 cm²

Answer: ABCD is a cylinder and BFC and AED are two hemispheres which has radius (r) = 7/2 cm

Total volume of solid = Volume of two hemisphere + Volume of cylinder

= 179.67 + 500.5 = 680.17 cm

Q.20. The marks obtained by 100 students in an examination are given below:

Find the mean marks of the students.

Answer :

Q.21. For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5?

Q.21. Find the zeroes of the quadratic polynomial 7y2 – 11/2y – ⅔ and verify the relationship between the zeroes and the coefficients.

Answer: Given, f(x) = 3x⁴ – 9x³ + x² + 15x + k

It is completely divisible by 3x² – 5

Q.22. Write all the values of p for which the quadratic equation x² + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.

Answer: Given, equation is x² + px + 16 = 0

This is of the form ax² + bx + c = 0

where, a = 1, b = p and c = 16

D = b² – 4ac = p² – 4 × 1 × 16 = p² – 64

for equal roots, we have D = 0

p² – 64 = 0

⇒ p² = 64

⇒ p = ±8 Putting p = 8 in given equation we have,

x2 + 8x + 16 = 0

⇒ (x + 4)² = 0

⇒ x + 4 = 0⇒ x = -4

Now, putting p = -8 in the given equation, we get

x² > – 8x + 16 = 0

⇒ (x – 4)² = 0

⇒ x = 4

Required roots are -4 and -4 or 4 and 4.

Section – D

Q.23. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite the first side is a right angle.

Answer: Given, ∆ABC in which AC² = AB² + BC²

To prove: ∠B = 90°Construction : Draw a ∆DEF such that

DE = AB, EF = BC and ∠E = 90°.

Proof: In ∆DEF we have ∠E = 90°

So, by Pythagoras theorem, we have

DF² = DE² + EF²

⇒ DF² = AB² + BC² …(i)

(∵ DE = AB and EF = BC)

AC² = AB² + BC² …(ii) (Given)

From equation (i) and (ii), we get

AC² = DF² ⇒ AC = DF.

Now, in ∆ABC and ∆DEF, we have

AB = DE, BC = EF and AC = DF.

∆ABC = ∆DEF.

Hence, ∠B = ∠E = 90°.

Hence Proved.

Q.24. From a point P on the ground, the angle of elevation of the top of the tower is 30° and that of the top of the flag-staff fixed on the top of the tower is 45°. If the length of the flag-staff is 5 m, find the height of the tower. (Use √3 = 1.732) [4]

Answer: Let AB be the tower and BC be the flag-staff.

Let P be a point on the ground such that

∠APB = 30° and ∠APC = 45°, BC = 5 m

Let AB = h m and PA = x metres

From right ∆PAB, we have

Hence, the height of the tower is 6.83 m

Q.25. A right cylindrical container of radius 6 cm and height 15 cm is full of ice-cream, which has to be distributed to 10 children in equal cones having a hemispherical shape on the top. If the height of the conical portion is four times its base radius, find the radius of the ice-cream cone.

Answer: Let R and H be the radius and height of the cylinder.

Given, R = 6 cm, H = 15 cm.

Volume of ice-cream in the cylinder = πR²H = π × 36 × 15 = 540π cm³

Let the radius of cone be r cm

Height of the cone (h) = 4r

Radius of hemispherical portion = r cm.

Volume of ice-cream in cone = Volume of cone + Volume of the hemisphere

Number of ice cream cones distributed to the children = 10

⇒ 10 × Volume of ice-cream in each cone = Volume of ice-cream in cylindrical container

⇒ 10 × 2πr =540π

⇒ 20r³ = 540

⇒ r³ = 27

⇒ r = 3

Thus, the radius of the ice-cream cone is 3 cm.

Answer: Steps for construction are as follows:

Draw a line segment BC = 5 cm
At B and C construct ∠CBX = 60° and ∠BCX = 60°
With B as centre and radius 5 cm, draw an arc cutting ray BX at A. On graph paper, we take the scale.
Join AC. Thus an equilateral ∆ABC is obtained.

Below BC, make an acute angle ∠CBY
Along BY, mark off 3 points B₁, B₂, B₃ Such that BB₁, B₁B₂, B₂B₃ are equal.
Join B³C
From B₂ draw B₂D || B₃C, meeting BC at D
From D, draw DE || CA, meeting AB at E.

Then ∆EBD is the required triangle, each of whose sides is ⅔ of the corresponding side of ∆ABC.

Q.27. Change the following data into ‘less than type’ distribution and draw its olive:

Answer :

Q.28. Prove that:

Answer :

Q.29. Which term of the Arithmetic Progression -7, -12, -17, -22,…..will be -82? Is -100 any term of the A.P.? Give a reason for your answer.

Q.29. How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180? Explain the double answer.

Answer:-7, -12, -17, -22, …….

Here a = -7, d = -12 – (-7) = -12 + 7 = -5

Let T_n = -82T_n = a + (n – 1) d

⇒ -82 = -7 + (n – 1) (-5)

⇒ -82 = -7 – 5n + 5

⇒ -82 = -2 – 5n

⇒ -82 + 2 = -5n

⇒ -80 = -5n

⇒ n = 16

Therefore, 16th term will be -82.

Let T_n = -100

Again, T_n = a + (n -1) d

⇒ -100 = -7 + (n – 1) (-5)

⇒ -100 = -7 – 5n + 5

⇒ -100 = – 2 – 5n

⇒ -100 + 2 = -5n

⇒ -98 = -5n

⇒ n = 98/5

But the number of terms can not be in fraction.

So, -100 can not be the term of this A.P.

Answer: 45, 39, 33, …..

Here a = 45, d = 39 – 45 = -6

Let S_n = 180

⇒ n/2 [ 2a + (n – 1) d] = 180

⇒ n/2 [2 × 45 + (n – 1) (-6)] = 180

⇒ n/2 [90 – 6n + 6] = 180

⇒ n/2 [96 – 6n] = 180

⇒ n(96 – 6n) = 360

⇒ 96n – 6n² = 360

⇒ 6n² – 96n + 360 = 0

On dividing the above equation by 6

⇒ n² – 16n + 60 = 0

⇒ n² – 10n – 6n + 60 = 0

⇒ n(n – 10) – 6 (n – 10) = 0

⇒ (n – 10) (n – 6) = 0

⇒ n = 10, 6

Sum of first 10 terms = Sum of first 6 terms = 180

This means that the sum of all terms from 7th to 10th is zero.

Answer: Let Aran marks in Hindi be x and marks in English be y.

Then, according to question, we have

x + y = 30 …(i)(x + 2)(y – 3) = 210 …(ii)

from equation (i) put x = 30 – y in equation (ii)

(30 – y + 2) (y – 3) = 210

⇒ (32 – y) (y – 3) = 210

⇒ 32y – 96 – y² + 3y = 210

⇒ y² – 35y + 306 = 0

⇒ y² – 18y – 17y + 306 = 0

⇒ y(y – 18) – 17(y – 18) = 0

⇒ (y – 18) (y – 17) = o

⇒ y = 18, 17

Put y = 18 and 17 in equation (i), we get x = 12, 13

Hence his marks in hindi can be 12 and 13 and in english his marks can be 18 and 17.

Maths 10th Previous Year Question Paper 2019 SET-I (CBSE)

Maths

Section -A

Q.1. If HCF (336,54) = 6, find LCM (336, 54),

Answer : Given, HCF (336, 54) = 6

We know HCF × LCM = one number × other number

6 × LCM = 336 × 54

LCM = 336×54/6 = 336 × 9 = 3024

Q.2. Find the nature of roots of the quadratic equation 2x² – 4x + 3 = 0.

Answer: Given, 2x² – 4x + 3 = 0

Comparing it with quadratic equation ax² + bx + c = 0

Here, a = 2, b = -4 and c = 3

D = b² – 4ac = (-4)² – 4 × (2)(3) = 16 – 24 = -8 < 0

Hence, D < 0 this shows that roots will be imaginary.

Q.3. Find the common difference of the Arithmetic Progression (A.P.).

Answer :

Q.4. Evaluate: sin² 60° + 2 tan 45° – cos² 30°

Q.4. If sin A = ¾, Calculate sec A.

Answer :

We know,

Answer:

Q.5. Write the coordinates of a point P on the x-axis which is equidistant from point A(-2, 0) and B(6, 0).

Answer: Let coordinates of P on x-axis is (x, 0)

Given, A(-2, 0) and B(6, 0)

Here, PA = PB

On squaring both sides, we get

(x + 2)² = (x – 6)²

⇒ x² + 4 + 4x = x² + 36 – 12x

⇒ 4 + 4x = 36 – 12x

⇒ 16x = 32

⇒ x = 2

Coordinates of P are (2, 0)

Q.6. In Figure 1, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB.

Q.6. In Figure 2, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.

Answer:

Given, ∠C = 90° and AC = 4 cm, AB = ?

∆ABC is an isosceles triangle so, BC = AC = 4 cm

On applying Pythagoras theorem, we have

AB² = AC² + BC²

⇒ AB² = AC² + AC² (∵ BC = AC)

⇒ AB² = 42 + 42 = 16 + 16 = 32

⇒ AB = √32 = 4√2 cm

Answer: Given, DE || BCOn applying, Thales theorem, we have

Section – B

Q.7. Write the smallest number which is divisible by both 306 and 657.

Answer: Smallest number which is divisible by 306 and 657 is,

LCM (657, 306)

657 = 3 × 3 × 73

306 = 3 × 3 × 2 × 17

LCM =3 × 3 × 73 × 2 × 17 = 22338

Q.8. Find a relation between x and y if the points A(x, y), B(-4, 6) and C(-2, 3) are collinear.

Q.8. Find the area of a triangle whose vertices are given as (1, -1) (-4, 6) and (-3, -5).

Answer: Given, A(x, y), B(-4, 6), C(-2, 3)

x¹ = x, y¹ = y, x² = -4, y² = 6, x³ = -2, y³ = 3

If these points are collinear, then area of triangle made by these points is 0.

Answer: Let the probability of selecting a blue marble, black marble and green marbles are P(x), P(y), P(z) respectively.

P(x) = ⅕ , P (y) = ¼ Given

We Know,

P(x) + P(y) + P(z) = 1

Q.10. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique Answer.

Answer: Given, x + 2y = 5, 3x + ky + 15 = 0

Comparing above equations with

a¹x + b¹y + c¹ = 0 and a²x + b²y + c² = 0,

We get,

a¹ = 1, b¹ = 2, c¹ = -5

a² = 3, b² = k, c² = 15

Condition for the pair of equations to have unique Answer is

k can have any value except 6.

Q.11. The larger of the two supplementary angles exceeds the smaller by 18°. Find the angles.

Q.11. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?

Answer: Let two angles A and B are supplementary.

A + B = 180° …(i)

Given, A = B + 18°

On putting A = B + 18° in equation (i),

we get B + 18° + B = 180°

⇒ 2B + 18° = 180°

⇒ 2B = 162°

⇒ B = 81°

A = B + 18°

⇒ A = 99°

Answer: Let age of Sumit be x years and age of his son be y years.Then, according to question we have, x = 3y …… (i)

Five years later, x + 5 = 2½ (y + 5)……….(ii)

On putting x = 3y in equation (ii)

Q.12. Find the mode of the following frequency distribution:

Answer :

Here, the maximum frequency is 50.

So, 35 – 40 will be the modal class.

l = 35, f₀ = 34, f₁ = 50, f₂ = 42 and h = 5

Section – C

Q.13. Prove that 2 + 5√3 is an irrational number, given that √3 is an irrational number.

Q.13. Using Euclid’s Algorithm, find the HCF of 2048 and 960.

Answer: Let 2 + 5√3 = r, where, r is rational.

⇒ (2 + 5√3)2 = r2

⇒ 4 + 75 + 20√3 = r2

⇒ 79 + 20√3 = r2

⇒ 20√3 = r2 – 79

⇒ √3 = r² – 79/20

Now, r² – 79/20 is a rational number. So, √3 must also be a rational number.

But √3 is an irrational number (Given).

So, our assumption is wrong.

2 + 5√3 is an irrational number.

Hence Proved.

Answer: Step I: Here 2048 > 960 so, On applying Euclid’s algorithm, we get 2048 = 960 × 2 + 128

Step II: Because remainder 128 ≠ 0, so, On applying Euclid’s algorithm between 960 and 128, we get

960 = 128 × 7 + 64

Step III: Again remainder 64 ≠ 0, so

128 = 64 × 2 + 0

Here the remainder is 0. So, the process ends here. And the dividend is 64 so, required HCF is 64.

Q.14. Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × DP.

Q.14. Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of triangles POQ and ROS.

Answer: Given, ∆ABC, ∆DBC are right-angle triangles, right-angled at A and D, on the same side of BC.

AC & BD intersect at P.

In ∆APB and ∆PDC,

∠A = ∠D = 90°

∠APB = ∠DPC (Vertically opposite)

∆APB ~ ∆PDC (By AA Similarity)

AP/BP = PD/PC (by c.s.s.t.)

⇒ AP × PC = BP × PD.

Hence Proved.

Answer: Given, PQRS is a trapezium where PQ || RS and diagonals intersect at O and PQ = 3RS

In ∆POQ and ∆ROS, we have

∠ROS = ∠POQ (vertically opposite angles)

∠OQP = ∠OSR (alternate angles)

Hence, ∆POQ ~ ∆ROS by AA similarity then,

If two triangles are similar, then the ratio of areas is equal to the ratio of square of its corresponding sides. Then,

Q.15. In Figure 3, PQ and RS are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting PQ at A and RS at B. Prove that ∠AOB = 90°.

Answer: Given, PQ || RS

To prove: ∠AOB = 90°

Construction: Join O and C, D and E

In ∆ODA and ∆OCA

OD = OC (radii of circle)

OA = OA (common)

AD = AC (tangent drawn from the same point)

By SSS congruence

∆ODA = ∆OCA

Then, ∠DOA = ∠AOC …(i)

Similarly, in ∆EOB and ∆BOC, we have

∆EOB = ∆BOC∠EOB = ∠BOC …(ii)

EOD is a diameter of the circle, therefore it is a straight line.

Hence, ∠DOA + ∠AOC + ∠EOB + ∠BOC = 180°

⇒ 2(∠AOC) + 2(∠BOC) = 180°

⇒ ∠AOC + ∠BOC = 90°

⇒ ∠AOB = 90°.Hence Proved.

Q.16. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection.

Answer: Let the required ratio be k : 1

By section formula, we have

Q.17. Evaluate:

Answer :

Q.18. In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use π = 3.14)

Q.18. In Figure 5, ABCD is a square with side 2√2 cm and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14)

Answer: Given, OABC is a square with OA = 15 cm

OB = radius = r

Let side of square be a then,

a² + a² = r²

⇒ 2a² = r²

⇒ r = √2 a

⇒ r = 15√2 cm (∵ a = 15 cm)

Area of square = Side × Side = 15 × 15 = 225 cm²

Area of shaded region = Area of quadrant OPBQ – Area of square

= 353.25 – 225 = 128.25 cm

Answer: Given, ABCD is a square with side 2√2 cm

BD = 2r

In ∆BDC,

BD² = DC² + BC²

⇒ 4r² = 2(DC)² (∵ DC = CB = Side = 2√2 )

⇒ 4r² = 2 × 2√2 × 2√2

⇒ 4r² = 8 × 2

⇒ 4r² = 16

⇒ r²= 4

⇒ r = 2 cm

Area of square BCDA = Side × Side = DC × BC = 2√2 × 2√2 = 8 cm²

Area of circle = πr² = 3.14 × 2 × 2 = 12.56 cm²

Area of shaded region = Area of circle – Area of square. = 12.56 – 8 = 4.56 cm²

Answer: ABCD is a cylinder and BFC and AED are two hemispheres which has radius (r) = 7/2 cm

Total volume of solid = Volume of two hemisphere + Volume of cylinder

= 179.67 + 500.5 = 680.17 cm

Q.20. The marks obtained by 100 students in an examination are given below:

Find the mean marks of the students.

Answer :

Q.21. For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5?

Q.21. Find the zeroes of the quadratic polynomial 7y2 – 11/2y – ⅔ and verify the relationship between the zeroes and the coefficients.

Answer: Given, f(x) = 3x⁴ – 9x³ + x² + 15x + k

It is completely divisible by 3x² – 5

Q.22. Write all the values of p for which the quadratic equation x² + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.

Answer: Given, equation is x² + px + 16 = 0

This is of the form ax² + bx + c = 0

where, a = 1, b = p and c = 16

D = b² – 4ac = p² – 4 × 1 × 16 = p² – 64

for equal roots, we have D = 0

p² – 64 = 0

⇒ p² = 64

⇒ p = ±8 Putting p = 8 in given equation we have,

x2 + 8x + 16 = 0

⇒ (x + 4)² = 0

⇒ x + 4 = 0⇒ x = -4

Now, putting p = -8 in the given equation, we get

x² > – 8x + 16 = 0

⇒ (x – 4)² = 0

⇒ x = 4

Required roots are -4 and -4 or 4 and 4.

Section – D

Q.23. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer: Given, A ∆ABC in which DE || BC and DE intersect AB and AC at D and E respectively.

To prove: AD/DB = AE/EC

Construction: Join BE and CD

Draw EL ⊥ AB and DM ⊥ AC

Proof: we have

area (∆ADE) = ½ × AD × EL

and area (∆DBE) = ½ × DB × EL (∵ ∆ = ½ × b × h)

Now, ∆DBE and ∆ECD, being on same base DE and between the same parallels DE and BC, We have

area (∆DBE) = area (∆ECD) …..(iii)

from equations (i), (ii) and (iii), we have

AD/DC = AE/EC

Hence Proved.

Q.24. Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an elevation of 30°. Deepak standing on the roof of a 50 m high building, finds the angle of elevation of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.

Answer: Let Amit be at C point and the bird is at A point. Such that ∠ACB = 30°. AB is the height of bird from point B on ground and Deepak is at D point, DE is the building of height 50 m.

Hence, the distance of bird from Deepak is 50√2 m.

Q.25. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 gm mass. (Use π = 3.14)

Answer: Let AB be the iron pole of height 220 cm with base radius 12 cm and there is the other cylinder CD of height 60 cm whose base radius is 8 cm.

Volume of AB pole = πr¹h¹ = 3.14 × 12 × 12 × 220 = 99475.2 cm³

Volume of CD pole = πr²h² = 3.14 × 8 × 8 × 60 = 12057.6 cm³

Total volume of the poles = 99475.2 + 12057.6 = 111532.8 cm³

It is given that,

Mass of 1 cm³ of iron = 8 gm

Then mass of 111532.8 cm³ of iron = 111532.8 × 8 gm

Then the total mass of the pole is = 111532.8 × 8 gm = 892262.4 gm = 892.2624 kg

Answer: Steps for construction are as follows:

Draw a line segment BC = 5 cm
At B and C construct ∠CBX = 60° and ∠BCX = 60°
With B as centre and radius 5 cm, draw an arc cutting ray BX at A. On graph paper, we take the scale.
Join AC. Thus an equilateral ∆ABC is obtained.

Below BC, make an acute angle ∠CBY
Along BY, mark off 3 points B₁, B₂, B₃ Such that BB₁, B₁B₂, B₂B₃ are equal.
Join B³C
From B₂ draw B₂D || B₃C, meeting BC at D
From D, draw DE || CA, meeting AB at E.

Then ∆EBD is the required triangle, each of whose sides is ⅔ of the corresponding side of ∆ABC.

Q.27. Change the following data into ‘less than type’ distribution and draw its olive:

Answer :

Q.28. Prove that:

Answer :

Q.29. Which term of the Arithmetic Progression -7, -12, -17, -22,…..will be -82? Is -100 any term of the A.P.? Give a reason for your answer.

Q.29. How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180? Explain the double answer.

Answer:-7, -12, -17, -22, …….

Here a = -7, d = -12 – (-7) = -12 + 7 = -5

Let T_n = -82T_n = a + (n – 1) d

⇒ -82 = -7 + (n – 1) (-5)

⇒ -82 = -7 – 5n + 5

⇒ -82 = -2 – 5n

⇒ -82 + 2 = -5n

⇒ -80 = -5n

⇒ n = 16

Therefore, 16th term will be -82.

Let T_n = -100

Again, T_n = a + (n -1) d

⇒ -100 = -7 + (n – 1) (-5)

⇒ -100 = -7 – 5n + 5

⇒ -100 = – 2 – 5n

⇒ -100 + 2 = -5n

⇒ -98 = -5n

⇒ n = 98/5

But the number of terms can not be in fraction.

So, -100 can not be the term of this A.P.

Answer: 45, 39, 33, …..

Here a = 45, d = 39 – 45 = -6

Let S_n = 180

⇒ n/2 [ 2a + (n – 1) d] = 180

⇒ n/2 [2 × 45 + (n – 1) (-6)] = 180

⇒ n/2 [90 – 6n + 6] = 180

⇒ n/2 [96 – 6n] = 180

⇒ n(96 – 6n) = 360

⇒ 96n – 6n² = 360

⇒ 6n² – 96n + 360 = 0

On dividing the above equation by 6

⇒ n² – 16n + 60 = 0

⇒ n² – 10n – 6n + 60 = 0

⇒ n(n – 10) – 6 (n – 10) = 0

⇒ (n – 10) (n – 6) = 0

⇒ n = 10, 6

Sum of first 10 terms = Sum of first 6 terms = 180

This means that the sum of all terms from 7th to 10th is zero.

Answer: Let Aran marks in Hindi be x and marks in English be y.

Then, according to question, we have

x + y = 30 …(i)(x + 2)(y – 3) = 210 …(ii)

from equation (i) put x = 30 – y in equation (ii)

(30 – y + 2) (y – 3) = 210

⇒ (32 – y) (y – 3) = 210

⇒ 32y – 96 – y² + 3y = 210

⇒ y² – 35y + 306 = 0

⇒ y² – 18y – 17y + 306 = 0

⇒ y(y – 18) – 17(y – 18) = 0

⇒ (y – 18) (y – 17) = o

⇒ y = 18, 17

Put y = 18 and 17 in equation (i), we get x = 12, 13

Hence his marks in hindi can be 12 and 13 and in english his marks can be 18 and 17.

Science 10th Previous Year Question Paper 2019 SET-III (CBSE)

Science

Section – A

Q.1. Why does the cord of an electric oven not glow while it’s heating element does?

Answer: The resistance of the heating element of an electric oven is very high. As the current flows through the heating element it becomes too hot and glows red. On the other hand the cord of an electric oven low resistance hence it does not become red during the flow of current.

Q.2. Although coal and petroleum are produced by the degradation of biomass, yet we need to conserve these resources. Why?

Answer: We need to conserve coal and petroleum because they are formed in millions of years and are non-renewable also and their rate of consumption is very high. So these are limited and cannot last for longer time.

Section – B

Q.3. What is atmospheric refraction? List two phenomena which can be explained on the basis of atmospheric refraction.

Answer: In atmosphere, there are layers of different densities and refractive indices, when the light ray is passed through these layers refraction of light takes place which is called atmospheric refraction.

Two phenomenon that can be explained on the basis of atmospheric refraction are:

Twinkling of stars.
Early sunset and delayed sunrise.

Q.4. Name a metal of medium reactivity and write three main steps in the extraction of this metal from its sulphide ore.

Answer: Zinc

The steps involved in the extraction of zinc from zinc sulphide are:

1. Roasting of sulphide ore in the presence of air to convert it into metal oxide.

2. Reduction of metal oxide with carbon to get free metal.

3. Refining of impure metal to get pure metal.

Q.5. List two chemical properties on the basis of which ethanol and ethanoic acid may be differentiated and explain how.

Q.5. Unsaturated hydrocarbons contain multiple bonds between two carbon atoms and these compounds show addition reactions. Out of saturated and unsaturated carbon compounds, which compounds are more reactive? Write a test to distinguish ethane from ethene.

Answer: 1. Ethanol does not react with sodium bicarbonate but ethanoic acid reacts with sodium bicarbonate releasing CO₂ gas.

2. Ethanol does not change the colour of blue litmus paper but ethanoic acid changes the colour of blue litmus to red due to presence of carboxylic acid group.

Answer: Unsaturated carbon compounds are more reactive than saturated carbon compounds. Bromine water is decolourized by ethene but there is no change of colour of bromine water with ethane.

Section – C

Q.6. What happens to a beam of white light when it gets refracted through a glass prism? Which colour deviates the most and the least after refraction through a prism? What is likely to happen if a second identical prism is placed in an inverted position with respect to the first prism. Justify your answer.

Q.6. A student needs spectacles of power -0.5 D for the correction of his vision.

(i) Name the defect in vision the student is suffering from.

(ii) Find the nature and focal length of the corrective lens.

(iii) List two causes of this defect.

Answer: When white light is refracted through a glass prism, it gets split into its constituting colours at different angles. This phenomenon is called Dispersion of Light.

Forming a rainbow,

Least deviated colour is red whereas most deviated colour is violet.

When second identical prism is placed in an inverted position with respect to first prism, recombination of the spectrum will take place and white light will be obtained.

Answer: (i) Myopia.

(ii) Concave lens with the focal length of 200 cm

Given, P = -0.5 D

We have,

P = 1/f

f=1/P

f= 1/(-0.5)

f= -2m = -200cm

(iii) Two causes of Myopia are:

Elongation of eye ball.
High converging power of eyeless.

Q.7. Define a food chain. Design a terrestrial food chain of four trophic levels. If a pollutant enters at the producer level, the organisms of which trophic level will have the maximum concentration of the pollutant in their bodies? What is this phenomenon called?

Answer: It is the sequence of arrangement of a living organism in a community in which one organism consumes another organism to transfer food energy.

Grass → Insect → Frog → Bird

The organism at a higher tropic level will have a maximum concentration of pollutants. This phenomenon is called biological magnification.

Q.8. What are amphoteric oxides? Give an example. Write balanced chemical equations to justify your answer.

Answer: Those oxides which behave both acidic and basic oxides are called amphoteric oxides. Example: Al₂O₃ (Alumina)

Q.9. During the reaction of some metals with dilute hydrochloric acid, the following observations were made by a student:

(a) Silver does not show any change.

(b) Some bubbles of gas are seen when a lead is reacted with the acid.

(d) The temperature of the reaction mixture rises when aluminium is added to the acid.

Explain these observations giving an appropriate reason.

Answer:

(a) Silver is covered with a thin layer of silver chloride, so it does not react with dilute hydrochloric acid.

(b) Bubbles of hydrogen gas are evolved when lead is reacted with the acid.

(d) The temperature of the reaction mixture rises when aluminium is added to the acid because the reaction is highly exothermic in nature.

Q.10. Given below are the steps for the extraction of copper from its ore. Write the chemical equation of the reactions involved in each case.

(i) Roasting of copper (I) sulphide.

(ii) Reduction of copper (I) oxide from copper (I) sulphide

(iii) Electrolytic refining.

Answer:

Q.11. What is transpiration? List its two functions.

Q.11. (a) What is translocation? Why is it essential for plants?

(b) Where do the substances in plants reach as a result of translocation?

Answer: The evaporation of water from the leaves of a plant is called transpiration. Functions of transpiration

1. It helps in the upward movement of water and minerals from the root to the leaves through the stem.

2. Helps in cooling the plant surface.

Answer: (a) The transport of food from leaves to other parts of the plant is called translocation. Translocation is essential for plants because without it food prepared by the leaves cannot reach other parts of the plant for their growth and development.

(b) The substances in plants reach other tissues in plants from the leaves as a result of translocation.

Q.12 What is carpel? Write the function of its various parts.

Answer: The flask-shaped organ in the centre of a flower is called carpel. It is also called a female reproductive organ of the plant. It is made up of three parts:

1. Stigma

2. Style

3. Ovary

1. Stigma is the top part of carpel and is sticky. So, it receives the pollen from the anther of stamen.

2. Style connects stigma to ovary.

3. Ovary contains female gametes of the plant and helps in reproduction.

Q.13 A student holding a mirror in his hand directed the reflecting surface of the mirror towards the Sun. He then directed the reflected light on to a sheet of paper held close to the mirror.

(a) What should he do to bum the paper?

(b) Which type of mirror does he have?

(c) Will he be able to determine the approximate value of the focal length of this mirror from this activity ? Give reason and draw ray diagram to justify your answer in this case.

Q.13. A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. The distance of the object from the lens is 18 cm. Find the nature, position and size of the image formed.

Answer: (a) He should place the sheet of paper at the focus of the mirror to burn the paper.

(b) He has a concave mirror.

(c) Yes, the sheet of paper will start burning at the focus of the mirror which will give an approximate value of focal length, i.e., the distance between the mirror and the point where the sheet of paper starts burning.

Answer: A concave mirror forms a real image of the sun,

Given: Height of object, h₁ = +10 cm. Focal length, f = +12 cm.

Object distance, u = -18 cm. From the lens formula,

The position of image formed is at distance of 36 cm from convex lens. Since the value of the magnification is more than 1 (it is 2), the image formed is larger than object. The minus sign of magnification shows that image is formed below the principal axis. Hence, the image formed is real and inverted,

Q.14 What are solar cells? Explain the structure of solar panel. List two principal advantages associated with solar cells.

Answer: Solar cells are the devices which convert solar energy into electricity. A simple solar cell is made up of a sandwich of a silicon-boron layer and a silicon-arsenic layer. Boron and arsenic are present in a very small amount. A piece of wire is soldered into the top of the upper layer of cell and another piece of wire is soldered at the bottom of the lower layer to pass on the current. The solar cell is covered with a glass cover for protection. Advantages:

• Solar cells have no moving parts.

• It requires no maintenance.

Q.15. (a) Budding, fragmentation and regeneration, all are considered as an asexual mode of reproduction. Why?

(b) With the help of neat diagrams, explain the process of regeneration in Planaria.

Answer: (a) Budding, fragmentation and regeneration are considered as an asexual mode of reproduction because only one parent is involved no sex cells are involved.

(b) Regeneration in planaria.

The process of getting back a full organism from its body parts is called regeneration. Planaria reproduces by this method in which if the body of Planaria somehow gets cut into a number of pieces, then each body piece can regenerate into a complete Planaria by growing all the missing parts.

Section – D

Q.16. A 6 cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 30 cm. The distance of the object from the mirror is 45 cm. Use mirror formula to determine the position, nature and size of the image formed. Also, draw labelled ray diagram to show the image formation in this case.

Q.16. An object 6 cm in size is placed at 50 cm in front of a convex lens of focal length 30 cm. At what distance from the lens should a screen be placed in order to obtain a sharp image of the object? Find the nature and size of the image. Also, draw a labelled ray diagram to show the image formation in this case.

Answer: Given, Height of the object = 6 cm

Focal length, f = -30 cm

Object distance, u = -45 cm

Image distance, v = ?

Height of image, h_i = ?

We have,

1/f = 1/v + 1/u

Hence, the image formed is virtual, erect and magnified.

Q.17. (a) Why is the use of iodised salt advisable? Name the disease caused due to deficiency of iodine in our diet and state its one symptom.

(b) How do nerve impulses travel in the body? Explain.

Q.17. What is hydrotropism? Design an experiment to demonstrate this phenomenon.

Answer: (a) lodised salt is advisable because iodine is necessary for the formation of thyroxine hormone by the thyroid gland Goitre is the disease caused due to its deficiency. Symptom: The neck of the person appears to be swollen due to the enlargement of thyroid gland.

(b) Two neurons are not joined to one another completely. There is a small gap between a pair of neuron. This gap is called synapse. The nerve impulse are carried out to this gap by the help of neurotransmitter (chemical substance). The conduction of nerve impulse through the synapse takes place in the form of electrical nerve impulse. When a stimulus acts on the receptor an electrical impulse is produced with the help of chemical reaction. This electrical impulse passes through the synapse and then to the other neuron. Thus, in this way nerve impulses travel in the body.

Answer: The movement of root of plants towards water is called hydrotropism. Take two glass troughs A and B fill each one of them two-thirds with soil. In trough A plant a tiny seedling figure

(a). In trough B plant a similar seedling and also place a small clay pot inside the soil figure

(b). Water the soil in trough A daily and uniformly. Do not water the soil in trough B but put some water in the clay pot buried in the soil. Leave both the troughs for a few days.

Now, dig up the seedlings carefully from both the trough without damaging their roots. We will find that the root of seedling in through A is straight. On the other hand, the root of seedling in trough B is found to be bent to the right side (towards the clay pot containing water) figure (b). This can be explained as follows.

In through A, the root of seedling gets water from both sides (because the soil is watered uniformly) in trough B, the roots gets water oozing out from the clay pot which is kept on the right side. So, the root of seedling in trough B grows and bends towards the source of water to the right side. The experiment shows that the root of a plant grows towards water. In other words, the root of a plant is positively hydrotropic.

Q.18. (a) What are homologous structures ? Give an example.

(b) “The sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it.” justify this statement with the help of a flow chart showing sex-determination in human beings.

Answer: (a) The structures which have the same basic design but different functions are called homologous structures or homologous organs. Example: Forelimbs of a man, a lizard, a frog they have same basic design of bones but perform different functions. (b) The sex of a newborn depends on what happens at the time of fertilization.

1. If a sperm carrying X chromosome fertilizes the ovum carrying X chromosome, then the girl child will be born and the child will have XX combination of sex chromosomes.

2. If a sperm carrying Y chromosome fertilizes the ovum carrying X chromosome, then the child born will be.

The above presentation clearly shows that it is matter of chance whether the newborn will be boy or girl and none of the parents may be considered responsible for it.

Q.19. When do we consider a person to be myopic or hypermetropic ?

List two causes of -hypermetropia.

Explain using ray diagrams how the defect associated with hypermetropic eye can be corrected.

Answer: Myopia is the defect in vision in which a person cannot see the distant objects clearly whereas in hypermetropia is the defect in which a person cannot see nearby objects clearly. Hypermetropia is caused due to:

1. Decrease in converging power of eye-lens.

2. Too short eye ball.

In a hypermetropic eye, the image of near by object lying at normal near point N (at 25 cm) is formed behind the retina.

Hypermetropic eye can be corrected using convex lenses. When a convex lens of suitable power is placed in front of hypermetropic eye, then the diverging rays of light from the object are converged first by the convex lens used. This form a virtual image of the object at another near point N’. Now, the rays can be easily focused by the eye lens to form an image on retina.

Q.20. What is sexual reproduction?

Explain how this mode of reproduction gives rise to more viable variations than asexual reproduction.

How does this affect evolution?

Answer: The production of new organisms from two parents by making use of their sex cells is called sexual reproduction.

In sexual reproduction, more viable variations are observed than asexual reproduction as two parents is involved. In the case of asexual reproduction single parent organism gives rise to offsprings which are genetically identical to their parents. This limits the scope of variation in characters of offsprings. Due to sexual reproduction, the extent of variation is much large and therefore choices of evolution is also high.

The variations in the characters of the obtained offsprings from sexual reproduction enable them to adapt according to the environment and struggle for their existence. Over a period of time, the characters accumulate within the species and the formed species give rise to new species and this process goes on.

Q.21. Draw the paitern of magnetic field lines produced around a current-carrying straight conductor passing perpendicularly through horizontal cardboard. State and apply the right-hand thumb rule to mark the direction of the field lines. How will the strength of the magnetic field change when the point where magnetic field is to be determined is moved away from the straight conductor ?

Give reason to justify your answer.

Answer: Maxwell’s Right Hand Thumb rule states that if the current-carrying wire is imagined to be held in the right hand so that thumb points in the direction of current, then the direction in which fingers encircle the wire will give the direction of magnetic field lines around the wire. If we hold the current-carrying straight wire so that thumbs Magnetic field pattern due to a straight current-carrying wire in upward direction points the direction of current, the direction of magnetic field lines will be anticlockwise. The strength of the magnetic field is inversely proportional to the distance of the point of observation from the wire So, as we move away from the wire the strength of magnetic decreases Current (upwards).

Section – E

Q.22. A teacher provided acetic acid, water, lemon juice, aqueous solution of sodium hydrogen carbonate and sodium hydroxide to students in the school laboratory to determine the pH values of these substances using pH papers. One of the students reported the pH values of the given substances as 3, 12, 4, 8 and 14 respectively. Which one of these values is not correct? Write its correct value stating the reason.

Q.22. What would a student report nearly after 30 minutes of placing duly cleaned strips of aluminium, copper, iron and zinc in freshly prepared iron sulphate solution taken in four beakers?

Answer: The value of pH for water is not correct. The correct value of pH of water is 7 because it has almost equal concentration of H⁺and OH^–, due to which it is neutral.

Answer: Aluminium displaces the iron from iron sulphate and the colour of two solution changes from green to brown. No change takes place when copper strip is dipped in iron sulphate solution. No cfiange will be observed when iron strips are dipped in iron sulphate solution. The colour of the solution changes from green to colourless when zinc is added to iron sulphate solution.

Q.23 What is observed when a pinch of sodium hydrogen carbonate is added to 2 mL of acetic acid taken in a test tube? Write chemical equation for the reaction involved in this case.

Answer: CO2 gas is evolved with brisk effervescence when sodium hydrogen carbonate is added to acetic acid.

Q.24. List in proper sequence four steps of obtaining germinating dicot seeds.

Q.24. After examining a prepared slide under the high power of a compound microscope, a student concludes that the given slide shows the various stages of binary fission in a unicellular organism. Write two observations on the basis of which such a conclusion may be drawn.

Answer: 1. The root is formed when radicle of seed grows.

2. The root grows downward into the soil and absorbs water and minerals from the soil.

3. The shoot is formed from the upward growth of plumule.

4. The green leaves are developed when the shoot comes above the ground.

Answer: 1. A single parent divides to form two daughter cells.

2. The nucleus of mature cell seems elongated and a grove is formed in cell which divides the nucleus.

Q.25. List four precautions which a student should observe while preparing a temporary mount of a leaf peel to show stomata in his school laboratory.

Answer: 1. Freshly plucked leaf should be taken for an epidermal peel.

2. Hold the slide by its edges.

3. Peel should be cut to a proper size.

4. The peel should be allowed to dry.

Q.26. Draw the path of a ray of light when it enters one of the faces of a glass slab at an angle of nearly 45°. Label

(i) angle of refraction

(ii) angle of emergence and

(ii) lateral displacement.

Q.26. A student traces the path of a ray of light through a glass prism as shown in the diagram, but leaves it incomplete and unlabelled. Redraw and complete the diagram. Also label on it zi, ze, zr, and ZD.

Answer:

Q.27. The current flowing through a resistor connected in a circuit and the potential difference developed across its ends are as shown in the diagram by milliammeter and voltmeter readings respectively:

(a) What are the least counts of these meters?

(b) What is the resistance of the resistor?

Answer: (a) 10 mA and 0.1 V

(b) V = 2.4 volt, 1 = 250 mA = 0.25 A

From Ohm’s law.

R=v/I = 2.4/0.25 = 9.6Ω

Science 10th Previous Year Question Paper 2019 SET-II (CBSE)

Science

Section – A

Q.1. Write two advantages associated with water harvesting at the community level.

Answer:

The exploitation of water resources will be reduced.
It helps to recharge natural wells.

Q.2. Should the resistance of a voltmeter be low or high? Give reason.?

Answer:The resistance of a voltmeter should be high because voltmeter is connected parallel to the component of a circuit and it also takes negligible current from the circuit in order to measure the potential difference accurately.

Section – B

Q.3. Draw electron dot structure of carbon dioxide and write the nature of bonding between carbon and oxygen in its molecule.

Q.3. List two properties of carbon which lead to the huge number of carbon compounds we see around us, giving the reason for each.

Answer:

Covalent bond (double bond) is present in between C and O.

Answer: Two properties of carbon which lead to the huge number of carbon compounds are :

Catenation: It is the ability of carbon to form bonds with other atoms of carbon.
Tetravalency: With the valency of four, carbon is capable of bonding with 4 other atoms. This forms huge number of compounds.

Q.4. Given reasons:

(a) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

(b) Aluminium is a highly reactive metal; still, it is widely used in making cooking utensils. [2]

Answer:(a) Carbonate and sulphide ores are usually converted into oxides during the process of extraction because obtaining a metal from its metal oxide is much easier than from metal carbonates and sulphides.

(b) Aluminium is highly reactive metal still it is widely used in making cooking utensils because it reacts with 02 present in air to from aluminium oxide that forms a protective layer and protects the metal from corrosion.

Q.5. The power of a lens in +5 diopters. What is the nature and focal length of this lens ? At what distance from this lens should an object be placed so as to get its inverted image of the same size?

Given, P = + 5D

We have

Power, P = 1/ f(in meter)

+5= 1/f

f= ⅕ m {1 meter = 100cm}

f= 100/5 = 20cm

Focal length, f = 20 cm (or +20 cm).

Since the focal length of the lens is positive. Therefore, the nature of the lens is convex.

Same size and inverted image is formed when

Magnification, m = -1

Also, m = v/u

v = -u

From the lens formula,

Section – C

Q.6. List two types of transport system in human beings and write the functions of any one of these.

Answer:

Lymphatic system and blood circulatory system are two types of transport system in human beings.

Functions of blood circulatory system:

It carries nutrients and oxygen to all cells in the body.
Removes CO₂ from the body cells.
It carries digested food from the small intestine to other parts of the body.
It carries hormones from endocrine glands to different organs fo the body.

Q.7. Distinguish between pollination and fertilisation. Mention the site and the product of fertilisation in a flower. ?

Answer:

The transfer of pollen grains from anther of a stamen to the stigma of a carpel is called pollination whereas fertilisation is the process when the male gamete present in pollen grain joins the female gamete present in ovule.
Pollination is an external mechanism whereas fertilisation is an internal mechanism which takes place inside the flower.

Site of fertilisation in flower is ovary. Product of fertilisation in flower is zygote.

Q.8. What are amphoteric oxides? Give an example. Write balanced chemical equations to justify your answer.

Answer: Those oxides which behave both acidic and basic oxides are called amphoteric oxides. Example: Al₂O₃ (Alumina)

Q.9. What is a homologous series of carbon compounds ? Give an example and list its three characteristics.

Answer: A homologous series is a group of organic compounds having similar structures and similar chemical properties in which the successive compounds differ by CH₂ group. Example: Alkanes with general formula C_nH_2n+2 Characteristics:

• All the members of a homologous series can be represented by the same general formula.

• Any two adjacent homologues differ by 1 carbon atom and 2 hydrogen atoms in their molecular formulae.

• The difference in the molecular masses of any two adjacent homologues is 14 u.

Q.10. List three environmental consequences of using fossil fuels. Suggest three steps to minimise the pollution caused, by various energy sources.

Answer: The combustion of fossil fuels releases different harmful products. Three environmental consequences of using fossil fuels are:

It releases CO₂ which is a greenhouse gas which traps the solar energy falling on earth and it leads to global warming.
Carbon monoxide is poisonous gas which when enters in the bloodstream stops the functioning of red blood cells to carrying oxygen from lungs to other parts of the body. It also causes death.
Sulphur dioxide released during the burning of fossil fuels is harmful for lungs and causes bronchitis and other diseases.

Steps to minimise the pollution caused by various energy sources are:

Solar cookers should be used to cook food wherever possible.
Use of Biogas as domestic fuel should be encouraged in rural areas.
Three R’s strategy-Reduce, Reuse and Recycle should be practised

Q.11. What is transpiration? List its two functions.

Q.11. (a) What is translocation? Why is it essential for plants?

(b) Where do the substances in plants reach as a result of translocation?

Answer: The evaporation of water from the leaves of a plant is called transpiration. Functions of transpiration

1. It helps in the upward movement of water and minerals from the root to the leaves through the stem.

2. Helps in cooling the plant surface.

(b) The substances in plants reach other tissues in plants from the leaves as a result of translocation.

Q.12. What is carpel? Write the function of its various parts.

Answer: The flask-shaped organ in the centre of a flower is called carpel. It is also called a female reproductive organ of the plant. It is made up of three parts:

1. Stigma

2. Style

3. Ovary

1. Stigma is the top part of carpel and is sticky. So, it receives the pollen from the anther of stamen.

2. Style connects stigma to ovary.

3. Ovary contains female gametes of the plant and helps in reproduction.

Q.13. A student holding a mirror in his hand directed the reflecting surface of the mirror towards the Sun. He then directed the reflected light on to a sheet of paper held close to the mirror.

(a) What should he do to bum the paper?

(b) Which type of mirror does he have?

(c) Will he be able to determine the approximate value of the focal length of this mirror from this activity ? Give reason and draw ray diagram to justify your answer in this case.

Answer: (a) He should place the sheet of paper at the focus of the mirror to burn the paper.

(b) He has a concave mirror.

Answer: A concave mirror forms a real image of the sun,

Given: Height of object, h₁ = +10 cm. Focal length, f = +12 cm.

Object distance, u = -18 cm. From the lens formula,

Q.14. Which compounds are called

(i) alkanes,

(ii) alkenes and

(iii) alkynes?

C₄H₁₀ belongs to which of these? Draw two structural isomers of this compound.

Answer: (i) The hydrocarbons in which carbon atoms are connected by only single covalent bonds are called alkanes.

(ii) The hydrocarbons in which carbon atoms are connected by the double bond are called alkene.

(iii) The hydrocarbons in which carbon atoms are connected by the triple bond are called alkynes.

C₄H₁₀ belongs to alkane

Q.15. Write the essential function performed by ozone at the higher levels of the Earth’s atmosphere? How is it produced ? Name the synthetic chemicals mainly responsible for the drop of amount of ozone in the atmosphere. How can the use of these chemicals be reduced?

Answer: Ozone layer absorbs most of the harmful ultraviolet radiations from the sun to the earth. It is formed high up in the atmosphere by the action of ultraviolet radiation on oxygen gas. Chlorofluorocarbons are the synthetic chemicals responsible for the drop of amount of ozone in the atmosphere.

The use of these chemicals can be reduced by:

• Replacement of chlorofluorocarbons with hydrochlorofluorocarbons because it breaks down more quickly.

• Safe disposal of old appliances such as refrigerators and freezers.

Section – D

Q.16. (a) What are the dominant and recessive traits?

(b) “Is it possible that a trait is inherited but may not be expressed in the next generation ?” Give a suitable example to justify this statement.

Answer: (a) The trait which can express its effect over contrasting trait is called dominant trait whereas the trait which cannot express its effect over contrasting trait or which gets suppressed by the contrasting trait is called recessive trait. The inherited trait which is not expressed will be a recessive trait.

(b) In Mendel’s experiment, when pure tall pea plants were crossed with pure dwarf pea plants, only tall pea plants were obtained in F₁ generation. On selfing, the pea plants of F₁ generation both tall and dwarf pea plants were obtained in F₂ generation. The reappearance of the dwarf pea plants in F₂ generation proves that the dwarf trait was inherited but not expressed in F₁ generation. The recessive trait does not express itself in the presence of the dominant trait. So, it is possible that one trait may be inherited but may not be expressed in an organism.

Q.17. (a) Why is the use of iodised salt advisable? Name the disease caused due to deficiency of iodine in our diet and state its one symptom.

(b) How do nerve impulses travel in the body? Explain.

Q.17. What is hydrotropism? Design an experiment to demonstrate this phenomenon.

Answer: The movement of root of plants towards water is called hydrotropism. Take two glass troughs A and B fill each one of them two-thirds with soil. In trough A plant a tiny seedling figure

(a). In trough B plant a similar seedling and also place a small clay pot inside the soil figure

(b). Water the soil in trough A daily and uniformly. Do not water the soil in trough B but put some water in the clay pot buried in the soil. Leave both the troughs for a few days.

Q.18. (a) What are homologous structures ? Give an example. (b) “The sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it.” justify this statement with the help of a flow chart showing sex-determination in human beings.

1. If a sperm carrying X chromosome fertilizes the ovum carrying X chromosome, then the girl child will be born and the child will have XX combination of sex chromosomes.

2. If a sperm carrying Y chromosome fertilizes the ovum carrying X chromosome, then the child born will be.

The above presentation clearly shows that it is matter of chance whether the newborn will be boy or girl and none of the parents may be considered responsible for it.

Q.19. When do we consider a person to be myopic or hypermetropic ?

List two causes of -hypermetropia.

Explain using ray diagrams how the defect associated with hypermetropic eye can be corrected.

1. Decrease in converging power of eye-lens.

2. Too short eye ball.

In a hypermetropic eye, the image of near by object lying at normal near point N (at 25 cm) is formed behind the retina.

Q.20 (a) What is the scattering of light? Explain how the colour of the scattered light depends on the size of the scattering particles.

(b) Explain the reddish appearance of the Sun at sunrise or sunset. Why does it not appear red at noon?

Answer: (a) Scattering of light is the phenomenon in which a part of the incident light is dispersed in different directions.

Dependence of colour and scattered light on the size of particles:

When the particles like dust and water droplets present in the atmosphere are large in size, the scattered light appears white.
When the particles are extremely minute in size, they will scatter blue light present in the white sunlight.

(b) The reddish appearance of the sun at sunrise ‘ and sunset is due to the scattering of blue colour present in the sunlight away from our line of sight and leaves behind the mainly red colour of the direct sunlight which reaches the human eye.

The reason for Sun not appearing red at the noon is that the light has to travel a relatively shorter distance through the atmosphere to reach us and therefore, only a litte of blue colour of the white light is scattered.

Section – E

Q.22. What would a student report nearly after 30 minutes of placing duly cleaned strips of aluminium, copper, iron and zinc in freshly prepared iron sulphate solution taken in four beakers?

Answer: The value of pH for water is not correct. The correct value of pH of water is 7 because it has almost equal concentration of H⁺and OH^–, due to which it is neutral.

Q.23. What is observed when a pinch of sodium hydrogen carbonate is added to 2 mL of acetic acid taken in a test tube? Write chemical equation for the reaction involved in this case.

Answer: CO₂ gas is evolved with brisk effervescence when sodium hydrogen carbonate is added to acetic acid.

Q.24. List in proper sequence four steps of obtaining germinating dicot seeds. OR After examining a prepared slide under the high power of a compound microscope, a student concludes that the given slide shows the various stages of binary fission in a unicellular organism. Write two observations on the basis of which such a conclusion may be drawn.

Answer: 1. The root is formed when radicle of seed grows.

2. The root grows downward into the soil and absorbs water and minerals from the soil.

3. The shoot is formed from the upward growth of plumule.

4. The green leaves are developed when the shoot comes above the ground.

Q.24. 1. A single parent divides to form two daughter cells.

2. The nucleus of mature cell seems elongated and a grove is formed in cell which divides the nucleus.

Q.25. List four precautions which a student should observe while preparing a temporary mount of a leaf peel to show stomata in his school laboratory.

Answer: 1. Freshly plucked leaf should be taken for an epidermal peel.

2. Hold the slide by its edges.

3. Peel should be cut to a proper size.

4. The peel should be allowed to dry.

Q.26. Draw the path of a ray of light when it enters one of the faces of a glass slab at an angle of nearly 45°. Label ( angle of refraction (ii) angle of emergence and (ii) lateral displacement.