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ICSE Class 10 Hindi Question Paper 2016 Solved

Section – A (40 Marks)
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Question 1.
Write a short composition in Hindi of approximately 250 words on any one of the following topics : [15]
निम्नलिखित विषयों में से किसी एक विषय पर हिंदी में लगभग 250 शब्दों में संक्षिप्त लेख लिखिए:
(i) पश्चिमी सभ्यता के प्रभाव से फैशन एवं प्रदर्शन की प्रवृत्ति बढ़ती जा रही है, जिसके कारण अनेक प्रकार की समस्याओं का जन्म हो रहा है तथा नैतिक मूल्यों का ह्रास हो रहा है। अपने विचारों द्वारा स्पष्ट कीजिए ।
(ii) ” सादगी भी लोगों के दिलों में अमिट छाप छोड़ सकती है” कथन को ध्यान में रखते हुए अपने देश के किसी ऐसे व्यक्तित्व के विषय में लिखिये जिन्होंने ‘सादा जीवन उच्च विचार’ को आधार मानकर अपना जीवन बिताया, आपके ऊपर उस व्यक्ति का प्रभाव किस प्रकार का रहा यह भी स्पष्ट कीजिए ।
(iii) योग के माध्यम से हम शरीर तथा मन दोनों को स्वस्थ कर सकते हैं, जीवन में योग की अनिवार्यता तथा उससे मिलने वाले लाभों का वर्णन करते हुए अपने विचार लिखिए।
(iv) एक कहानी लिखिए जिसका आधार निम्नलिखित उक्ति हो :-
“मजहब नहीं सिखाता आपस में बैर रखना ”
(v) नीचे दिए गए चित्र को ध्यान से देखिए और चित्र को आधार बनाकर उसका परिचय देते हुए कहानी अथवा लेख लिखिए, जिसका सीधा संबंध चित्र से होना चाहिए ।

Answer :
(i) पश्चिमी सभ्यता से प्रभावित युवा वर्ग की मानसिकता
पश्चिमी सभ्यता के प्रभाव से फैशन और प्रदर्शन की प्रवृत्ति बढ़ती जा रही है । इस कथन में लेशमात्र भी मिथ्या नहीं है । पाश्चात्य सभ्यता इतनी प्रभावशाली हो गई है कि लोगों की विवेकबुद्धि कुंठित हो गई है। क्या सही है, क्या गलत है, यह सोचना ही बंद कर दिया है । कभी-कभी तो यह फैशन इतना हास्यास्पद प्रतीत होता है कि लोगों की बुद्धि पर दया आती है। फैशन में भी तो एक संतुलन होना चाहिए। फैशन के नाम पर भारतीय संस्कृति को तिलांजलि देकर पाश्चात्य संस्कृति को अपनाना अपनी ही जड़ों पर कुठाराघात किया जाने के समान है |

नवीनता लाना या परिवर्तन लाना बुरी बात नहीं है, लेकिन उसमें शालीनता भी होनी चाहिए। फैशन के नाम पर नग्नता, अश्लीलता, मर्यादाहीनता आदि, सराहनीय नहीं है। इसमें समाज में विकृति उत्पन्न होती है। ऐसा प्रतीत होता है कि पश्चिमी देशों की नकल करके हम लोग फूहड़ और लज्जाहीन हो गये हैं । पाश्चात्य संस्कृति को आत्मसात करके हम अपने रीति-रिवाजों और श्रेष्ठ परंपराओं को नष्ट करते जा रहे हैं । फलस्वरूप समाज में अपराध, अशिष्टता, अश्लीलता, पारिवारिक विघटन उत्पन्न हो रहे हैं । आज लोगों को वेलेन्टाइन डे, अप्रैल फूल, रोज डे, मदर्स डे, फादर्स डे आदि तो याद रहते हैं । परंतु माता-पिता, गुरु आदि के प्रति आदर भाव याद नहीं रहता है। पाश्चात्य सभ्यता का ही परिणाम है कि शादी के बाद बहुत शीघ्र संबंध विच्छेद हो जाते हैं ।

हमारे पूर्वजों ने एक लंबे और गहन अध्ययन के बाद मनुष्य के लिए एक सभ्य समाज की संरचना हेतु हर क्षेत्र में कुछ आदर्श, कुछ मूल्य, कुछ सीमाएँ निर्धारित की थीं। जिनके अनुपालन से भारतीय संस्कृति विश्व में पूजनीय बनी। परंतु आज उन्हीं आदर्शों की अवहेलना से हमारे नैतिक मूल्यों का निरंतर पतन हो रहा है। यह पाश्चात्य संस्कृति का ही प्रभाव है कि पति- पत्नी के संबंधों के बीच तीसरे की उपस्थिति न्यायोचित ठहराई जा रही है। शर्म की बात तो यह है कि भारतीय नारी जिसकी सुंदरता तथा शृंगार उसकी लज्जा हुआ करती थी’ आज वह पूर्णतया लज्जारहित हो गई है। युवक-युवतियों पर माता-पिता का कोई अंकुश नहीं रहा ।

‘गर्लफ्रेंड’, ‘बॉय फ्रेंड’ आधुनिक फैशन है। देर रात तक घर से बाहर रहना, क्लब में पार्टी, ड्रिंक व डांस करना यह सब पाश्चात्य संस्कृति का ही प्रभाव है । ‘प्रेम विवाह’ इसी संस्कृति की देन है । पाश्चात्य सभ्यता ने माँ को ‘मौम’ व पिता को ‘डैड’ कर दिया। गुरु-शिष्य तथा भाई- बहन आदि के अतिरिक्त समाज के अहम व महत्त्वपूर्ण संबंध भी इतने दूषित हो गये हैं जिन्हें सुनकर लोग शर्मसार हो जाते हैं ।

इतिहास साक्षी है कि जब कभी भी किसी देश, समाज, धर्म या समुदाय को हानि पहुँची तो उसका कारण उसकी सभ्यता व संस्कृति तथा उसके आदर्शों पर कुठाराघात है। मार्क्स ने कहा था कि इस पाश्चात्य संस्कृति का भारत पर हावी होने का कारण अर्थ अथवा पूँजी है। यह पूँजी ही आधुनिक परिवेश में समाज के लिए स्वादिष्ट विष परोस रही है जो संपूर्ण मानव जाति को पतन की ओर ले जा रहा है ।

(ii) ‘सादा जीवन उच्च विचार’
‘सादा जीवन उच्च विचार’ बड़ी सारगर्भित उक्ति है । भले ही आज के आकर्षण युक्त परिवेश में इस उक्ति का महत्त्व न हो परंतु युगों-युगों से यह कथन महत्त्वपूर्ण रहा है क्योंकि अनेक महापुरुषों ने इसके महत्त्व को द्विगुणित किया है। कौन नहीं जानता महान विभूति, लाल बहादुर शास्त्री को, जो गुदड़ी के लाल कहे जाते थे । वह सादगी में पले बढ़े तथा जीवन में संघर्षों से निरंतर लड़ते रहे । प्रतिदिन नदी पार करके पाठशाला जाना और उसी तरह वापस लौटना । उन्होंने अपनी सच्चाई, ईमानदारी व परिश्रम के बल पर हर कठिनाई को सहज बनाया तथा सदैव जीवन में स्वावलंबी बने रहे। जीवन कदाचित् सादा था, परंतु विचारों में अप्रत्याशित उच्चता थी । अपने इन्हीं गुणों के फलस्वरूप जनता ने उन्हें भारत के प्रधानमंत्री पद पर आसीन कर दिया ।

ऐसी ही एक महान विभूति ईश्वरचन्द्र विद्यासागर जो सादगी की प्रतिमूर्ति थे । एक महान आयोजन में मुख्य अतिथि के सम्मान हेतु आमंत्रित थे । मार्ग में उन्होंने देखा एक भारी ‘बीम’ को कुछ अशक्त लोग उठाने के प्रयास कर रहे थे, परंतु भारी होने के कारण नहीं उठा पा रहे थे। वहीं उनका स्वामी जो सहारा देकर उठवा सकता था, किंतु अपने बड़प्पन के कारण ‘उठाओ’, ‘ धक्का दो’ आदि शब्द तो कह रहा था, परंतु उनकी सहायता नहीं कर रहा था । ईश्वरचन्द्र विद्यासागर ने जब देखा तो आगे बढ़कर जैसे ही हाथ से जोर लगाया काम हो गया । आयोजन में वह व्यक्ति जो आदेश दे रहा था, उसने उन्हें मुख्य अतिथि के रूप में देखा तो शर्म से पानी-पानी हो गया। इसी तरह सादगी अपनी अमिट छाप छोड़ जाती है ।

डॉ. राजेन्द्र प्रसाद सन 1950 में स्वतंत्र भारत के प्रथम राष्ट्रपति बने । उनके आवास की व्यवस्था ‘तत्कालीन वायसराय के भवन में की गई। वहाँ उन्हें बहुत से कमरे दिखाये गये । उन्होंने स्पष्ट कहा उन्हें अधिक कमरों की आवश्यकता नहीं। वह अपनी आवश्यकतानुसार ही कमरों में रहेंगे। अधिकारी उन्हें शयन कक्ष में ले गये । वहाँ का स्प्रिंगदार कीमती शाही पलंग देखकर उन्होंने उसे हटाने को कहा। अधिकारियों के यह कहे जाने पर कि आपका पद गरिमावाला है और वायसराय के समकक्ष है ।

अतः पद के अनुरूप आपको इसका उपयोग करना चाहिए । तब उन्होंने उत्तर दिया वह वायसराय के अनुरूप नहीं है । शाही पलंग उनके लिए काँटों की सेज होगी। उन्होंने कहा “मेरे रहन-सहन के कुछ उसूल हैं, कुछ आचार-विचार हैं, मैं उनके विरुद्ध नहीं जा सकता अन्यथा मेरे आदर्शों को आघात पहुँचेगा।” भारत एक कृषि प्रधान देश है। यहाँ का किसान परिश्रमी तो है, परंतु उसकी अवस्था बड़ी दीन-हीन है। ऐसे देशवासियों का प्रतिनिधि ऐशो-आराम की जिंदगी कैसे बिता सकता है ?

उनके इन शब्दों को सुनकर अधिकारी अचंभित हुए बिना न रह सके। उन्होंने समझ लिया कि यह सच्चे जनसेवक सादगी पसंद तथा सिद्धान्तवादी महापुरुष हैं। उनके हृदय में अपने नेता के प्रति हार्दिक श्रद्धा की भावना का प्रादुर्भाव हुआ ।

(iii) योग साधना का महत्त्व
हमारे जीवन में योग का महत्त्वपूर्ण स्थान है। यह अक्षरशः सत्य है कि योग के माध्यम से हम अपने मन व शरीर को पूर्णरूपेण स्वस्थ बना सकते हैं । यदि वास्तव में हम अपने जीवन को सुखी व समृद्ध बनाना चाहते हैं तथा दुःखों से छुटकारा पाना चाहते हैं तो हमें नियमित रूप से योग करना चाहिए। यूँ तो जीवन में सुख-दुःख आते रहते हैं, लेकिन मन को स्वस्थ रखने के लिए हमें शरीर को चुस्त-दुरुस्त रखना अति आवश्यक है ।

कर्म साधना के लिए शरीर का स्वस्थ होना आवश्यक है, लेकिन मन की स्वस्थता ही कर्म करने की प्रेरणा देती है । अत: नियमित योग के माध्यम से ही तन व मन दोनों स्वस्थ रह सकते हैं । नियमित योगासनों को करने से शरीर में शक्ति तो आती ही है साथ ही रूप में लावण्य आता है। खुली हवा में योग करना फेफड़ों में ऑक्सीजन की वृद्धि करता है । शरीर का अंग- प्रत्यंग क्रियाशील होता है तथा शरीर में लचीलापन आता है । योग के द्वारा अनेक रोगों से भी रक्षा होती है तथा यह रक्त को शुद्ध करता है। हृदय को शुद्ध रक्त प्राप्त होता है । शरीर का आलस्य दूर होता है तथा क्रियाशीलता बढ़ती है। योग करने का सबसे श्रेष्ठ समय प्रातःकाल होता है। सूर्य नमस्कार करने से भयंकर से भयंकर रोगों से निवृत्ति मिल जाती है। योग करने से खूब भूख लगती है तथा पाचन शक्ति दुरुस्त रहती है। आत्मविश्वास, एकाग्रता एवं मस्तिष्क में बल वृद्धि होती है । शरीर से पसीना निकलता है जिससे शरीर की गंदगी दूर होती है।

योग न करने से शरीर आलसी, रोगी व अनमना – सा रहता है। काम करने में मन नहीं लगता । भोजन में आसक्ति नहीं रहती । हमेशा प्रतीत होता है कि जैसे हम बीमार हैं। चिकित्सक के पास जाकर भी संतुष्टि नहीं होती । योग बिना पैसे का उपचार है। यह सब योग का ही चमत्कार है कि हमारे ऋषि-मुनियों ने योग के बल पर निराहार रहकर तीन-तीन हजार वर्षों तक एक पैर पर खड़े होकर तपस्याएँ कीं। यही कारण है आज भी ऋषि- मुनियों ने व्यायाम और योगासन पर बहुत बल दिया है। बाबा रामदेव का योग आज के युग का स्पष्ट उदाहरण है। उन्होंने योग के क्षेत्र में क्रांति ला दी है। उन्होंने योग के माध्यम से लोगों
लाइलाज रोगों को भी दूर किया है। उन्होंने प्राणायाम और योग के अनेक आसन बताये हैं । इन योगासनों का ज्ञान बाबा रामदेव मौखिक और लिखित रूप में टेलीविजन के आस्था और संस्कार चैनलों पर नियमित देते हैं ।

योग का सबसे बड़ा लाभ तो यह है कि योगी व्यक्ति को बहुत जल्दी बुढ़ापा नहीं आता । उसके शरीर का हर अंग नियमित व्यायाम व योग से क्रियाशील बना रहता है । आज की दुनिया में बी.पी., डाइबिटीज, आर्थराइटिस, थायरॉइड जैसे अनेक रोग लगे हुए हैं, परंतु इसमें संदेह नहीं है कि जिन्होंने बचपन से योग व व्यायाम को अपनी नियमित दिनचर्या बनाई होगी, उसे कोई बीमारी छू नहीं सकती।
हमारे देश के प्रधानमन्त्री श्री नरेन्द्र मोदी जी तो व्यायाम व योग के प्रबल समर्थक है । उन्होंने प्रत्येक के लिए योग की आवश्यकता पर जोर दिया है। इसके महत्त्व को दर्शाने के लिए उन्होंने “21 जून योग दिवस” का भी आयोजन किया था । वह स्वयं योग करते हैं । कहने का अभिप्राय यह है कि योग करने वाले की कार्यक्षमता एवं मस्तिष्क क्षमता में अत्यधिक वृद्धि होती है । इसलिए मन, बुद्धि एवं शरीर को स्वस्थ रखने के लिए योग करना चाहिए ।

(iv) “मजहब नहीं सिखाता आपस में बैर रखना।” इस वाक्य का भाव है कि मजहब अर्थात् कोई धर्म परस्पर दुश्मनी करने की शिक्षा नहीं देता । हम अपने प्राचीन इतिहास को उठाकर देखें तो पता चलता है कि भिन्न-भिन्न धर्मों और संप्रदायों के धर्म-गुरुओं ने आपस में प्रेम, बंधुत्व और सद्भाव काही पाठ पढ़ाया है। मिल-जुलकर सभी समस्याओं का समाधान संभव है। किंतु धर्म न संप्रदायों का आश्रय लेकर उनसे जुड़ी मानसिक संकीर्णताओं में फंसकर लड़ना-झगड़ना केवल वैमनस्य और अराजकता को ही जन्म देता है।

जब तक हम एक-दूसरे की भावनाओं को नहीं समझेंगे, परस्पर एक-दूसरे का आदर नहीं करेंगे, एक-दूसरे के हित का चिंतन नहीं करेंगे तब तक आपसी सद्भाव का अभाव ही रहेगा। हमें अपने आपको, अपनी जबान को और अपनी नस्ल को झूठी शान-शौकत और श्रेष्ठता की चकाचौंध से बचाना होगा। हमारे मंदिरों, मस्जिदों, गिरजाघरों और गुरुद्वारों के उत्थान में हमारी संस्कृति और मेहनत के दर्शन होते हैं। इनमें हमारे पूर्वजों के परिश्रम व प्यार की मिसालें हैं हमें उनका आदर करना चाहिए । हम सब एक ही परमात्मा की संतानें हैं। हिंदू, मुस्लिम, सिक्ख, ईसाई से भी बढ़कर एक धर्म है और वह है मानव धर्म । हमें उसकी अवहेलना नहीं करनी चाहिए। यदि इस भाव को हम हृदय में धारण करें तो बहुत सहज है कि हम एक-दूसरे से भाई – बहन जैसा प्यार कर सकेंगे, फिर बैर कहाँ होगा ?

हमारे धर्मों के पैगंबर और गुरुओं ने लोगों को समानता की शिक्षा दी। जैन गुरु महावीर स्वामी के समीपस्थ वातावरण में तो हिंसक जंगली पशु भी हिंसा त्यागकर प्रेम से रहते थे । बौद्ध धर्म के अनुयायी महात्मा बुद्ध का कहना था कि ” लोगों के दिलों में मोहब्बत के फूल खिलाना हजारों तीर्थों से बेहतर है । ” भगवान श्रीकृष्ण ने प्रेम के वशीभूत होकर अपने मित्र सुदामा के कंटक भरे चरणों को अपने अश्रुजल से ही धो डाला। महान् बलिदानी प्रभु ईसा मसीह मानव जाति की भलाई के लिए सूली पर चढ़ गये। ममतामयी माँ मदर टेरेसा ने बिना किसी धर्म व संप्रदाय का विचार किये हर पीड़ित, निराश्रित व दुःखी को हृदय से लगाया। हम यदि इतने बड़े कार्य न भी करें तो क्या मज़हब को बीच में न लाकर प्यार से नहीं रह सकते ? यह तो बहुत छोटी-सी बात है जो हर प्राणी को प्रसन्नता दे सकती है। इसी संदर्भ में मुझे एक कहानी याद आ गई ।

राम और रहीम दो मित्र थे। दोनों बड़ी घनिष्ठता थी । राम के घर में कोई उत्सव होता या कोई परेशानी होती तो रहीम जी जान से राम के उस कार्य में शरीक होता । रहीम के घर में ईद की सिंवइयों का भरपूर आनन्द राम का पूरा परिवार लेता था । एक बार हिन्दू मुसलमानों में झगड़ा छिड़ गया । राम का घर रहीम की बस्ती में था। सारे मुसलमानों ने राम के घर पर धावा बोल दिया। रहीम को इस बात का अंदेशा था । वह अपनी कटार लेकर राम के दरवाजे पर बैठ गया । उसने चेतावनी दी कि मेरा कत्ल करने के बाद ही कोई राम के घर में घुसेगा । लोग दो दिन तक सिर पीटते रहे। कोई परिणाम न निकला ।

वह केवल एक पंक्ति गुनगुनाता रहा “ईश्वर अल्लाह तेरो नाम सबको सन्मति दे भगवान ।” अंत में लोगों को इस बात का ज्ञान हुआ कि परमात्मा ही ईश्वर है, वही अल्लाह है, वही यीसू मसीह और वही वाहे गुरु है और हम सभी उसी एक की संतानें हैं। आपस में भाई-बहन हैं। फिर एक-दूसरे के दुश्मन कैसे हो सकते हैं ? यह बात सबको समझ आ गई। सभी ने रहीम से माफी माँगी और राम को गले लगाया । सब आपस में मित्र बन गये । बात सद्भाव की है अगर मन में सभी प्राणियों के प्रति सद्भाव होगा तो बैर अपने आप समाप्त हो जायेगा ।

(v) चित्र प्रस्ताव
प्रस्तुत चित्र में दो 12-13 वर्ष की बालिकाएँ दर्शायी गई हैं। वह दोनों एक नल के पास बैठी हुई हैं। नल में बहुत तेज पानी आ रहा है जिसे देखकर वे दोनों अप्रत्याशित रूप से प्रसन्न हो रही हैं। ऐसा प्रतीत हो रहा है जैसे बहुत दिनों के बाद उन्हें नल में पानी की प्राप्ति हुई हो। उनकी हँसी ऐसी है जैसे बहुत समय के पश्चात् जब बादल बरसते हैं तो लोग खुशी से झूम उठते हैं । जल ही जीवन है । जल के बिना जीवन का अस्तित्व नहीं है । रहीम कवि ने जल के महत्त्व को दर्शाते हुए लिखा है-
” रहिमन पानी राखिये बिन पानी सब सून ।
पानी गये न ऊबरे मोती मानुष चून” ।।

प्राणिमात्र के जीवन का आधार जल है । जल के बिना जीवन की संभावना कदाचित् असंभव है। हमारे देश में इस समय जल संकट बहुत अधिक है । आपने अक्सर नलों पर प्रातः काल से ही बर्तनों की भीड़ देखी होगी। पानी प्राप्त करने के लिए लोगों को झगड़ते भी देखा होगा । हमारी सरकार प्रकृति प्रदत्त जल सबको प्राप्त कराने में सर्वथा अशक्त है । इस चित्र के माध्यम से इसी भाव की अभिव्यक्ति हो रही है ।

यह एक कस्बे का चित्र है जहाँ पर वहाँ के निवासियों ने बहुत दिनों से जल संस्थान अधिकारियों से कस्बे में जल की व्यवस्था के लिए अनेक प्रार्थना-पत्र दिये। बड़े समूह में एकत्रित होकर बार-बार जल संकट की शिकायत की। उन्होंने यह भी बताया कि कूएँ सूख गये हैं न नलकूप व्यर्थ हो गये हैं। यहाँ के निवासी प्यास से व्याकुल हैं तथा पशुओं के लिए भी पानी की व्यवस्था नहीं है। बार-बार प्रार्थना किए जाने पर शायद अधिकारियों को दया आ गई, और उन्होंने इस क्षेत्र में पाइप लाईन बिछवा दी। अब पाइप लाइन तो बिछ गई परंतु पाइप में पानी नहीं था । पुनः सभी क्षेत्रवासियों ने प्रदेश की सरकार से अपनी कठिनाई बताई । उसमें भी एक-दो महीने लग गये। कई बार तो जल संस्थान से पानी का टैंकर भेजकर लोगों की कठिनाई को दूर किया गया, लेकिन यह स्थायी समाधान नहीं था । पुनः कुछ लोग सीधे मुख्य

मंत्री के पास पहुँचे और अपनी दीन-दशा का वर्णन किया। मंत्री जी ने लोगों के कष्ट को समझा और परंतु अपने अधीनस्थ अधिकारियों को कड़े आदेश देकर तुरंत कार्यवाही करने के लिए कहा। परिणामस्वरूप वहाँ के निवासियों की मेहनत रंग लाई और आज नलों में पानी आने लगा जिसे देखकर यह दोनों बालिकाएँ अत्यंत प्रसन्न हैं तथा संपूर्ण कस्बे के निवासियों की समस्या का भी समाधान हो गया ।
यह कहानी बताती है कि सही दिशा में परिश्रम करने से सफलता अवश्य मिलती है।

Question 2.
Write a letter in Hindi in approximately 120 words on any one of the topics given below: [7]

निम्नलिखित में से किसी एक विषय पर हिंदी में लगभग 120 शब्दों में पत्र लिखिए :
(i) आपके टेलीविजन द्वारा विभिन्न चैनलों पर अंधविश्वास तथा तंत्र-मंत्र से संबंधित कार्यक्रम दिखाकर जनता को भ्रमित किया जा रहा है। भारत सरकार के सूचना एवं प्रसारण मंत्री को इसकी जानकारी देते हुए ऐसे कार्यक्रमों पर रोक लगाने का अनुरोध कीजिए ।
(ii) आपकी चचेरी बहन जो गाँव में रहती है, उसकी दसवीं के बाद शिक्षा रोक दी गई है अतः उसकी आगे की शिक्षा जारी रखने का निवेदन करते हुए अपने चाचा जी को पत्र लिखिए जिसमें नारी शिक्षा की आवश्यकता और उसके लाभों की भी चर्चा कीजिए ।
Answer:
(i) प्रतिष्ठा में,
माननीय सूचना एवं प्रसारण मंत्री,
भारत सरकार,
दिल्ली ।
विषय – टेलीविजन पर प्रसारित अंधविश्वास एवं तंत्र-मंत्र से संबंधित कार्यक्रमों पर प्रतिबंध लगाने हेतु ।
माननीय महोदय,

आज के इस वैज्ञानिक युग में टेलीविजन के चैनलों पर अनेक प्रकार के अंधविश्वासों एवं तंत्र-मंत्र मिश्रित कार्यक्रमों की भरमार है । आश्चर्य की बात तो यह है कि इनके प्रभाव से कमिश्नर, कलेक्टर, वकील जैसे प्रबुद्ध लोग भी अछूते नहीं रह सके हैं। सुख व दुःख मानव जीवन के दो अंग हैं। कभी सुख तो कभी दुःख आते ही रहते हैं । इसी प्रकार से मानव शरीर मानसिक व शारीरिक रोग से ग्रसित होता रहता है । उसका उपाय ओझा, ज्योतिषी या फिर झाड़-फूंक करने वाले तांत्रिक नहीं कर सकते हैं । उसका उपचार तो चिकित्सकों व वैद्यों के पास होता है, न कि अंधविश्वासी धन ऐंठने वाले या भ्रमित करने वाले पाखण्डी पण्डितों या तांत्रिकों के पास हैं ।

महोदय, कभी गणेशजी की पत्थर की मूर्ति को दूध पिला रहे हैं। कभी भूत-प्रेत का बहाना कर एक निर्दोष पुरुष या स्त्री को कोड़ों से मार रहे हैं। दूध उफन गया तो अपशकुन हो गया। बिल्ली रास्ता काट गई तो अशुभ है। लंबी बीमारी हो गई तो ग्रह शांति के लिए यज्ञ, दान आदि करना- ये सारे कृत्य भोले-भाले लोगों को पथभ्रष्ट करके धन अर्जित करने के साधन हैं ।

टेलीविजन का कोई भी चैनल खोलकर देखिये हर चैनल पर एक नया ज्योतिषी दिखाई देगा। जिसका कार्य ऐसे कृत्य कर भोली-भाली जनता से धन ऐंठना है। सच तो यह है कि मनुष्य पुरुषार्थ करना भूल गया है। यही पर्याप्त नहीं है, कई पुरोहित तो लड़कियों के इलाज के बहाने व्यभिचार करते हैं ।

महोदय मेरा आपसे विनम्र निवेदन है कि जनता की ईश्वर में आस्था बनी रहने देने के लिए और लोगों की सोच सकारात्मक बनाने के लिए अतिशीघ्र इन ढोंगी व पाखण्डियों के कार्यक्रमों पर प्रतिबंध लगाने का कष्ट करें ताकि व्यक्ति, समाज एवं देश का उद्धार हो सके। आशा है आप मेरे इस प्रस्ताव पर गंभीरता विचार करेंगे।
सधन्यवाद ।
प्रार्थी,
रोहित सहगल
इंदौर |
दिनांक : 25.02.20….

(ii) 40 / 28 गौतम नगर
दिल्ली ।
दिनांक : 15.04.20…
आदरणीय चाचाजी,
सादर चरण स्पर्श ।
कुशलपूर्वक रहकर आपकी सपरिवार कुशलता की कामना करता हूँ। कल ही मुझे निशा का पत्र मिला और ज्ञात हुआ कि आपने उसकी कक्षा दस की परीक्षा उत्तीर्ण करने के पश्चात् उसे आगे अध्ययन करने की अनुमति नहीं दी है । यह जानकर बहुत कष्ट हुआ। चाचाजी आप एक अध्यापक हैं। लड़कियों की शिक्षा कितनी महत्त्वपूर्ण है, इस विषय में आपसे अधिक कौन जान सकता है ? शिक्षा लड़कियों का वह शस्त्र है जिसके सहारे वे अपने जीवन को मजबूत आधार दे सकती हैं। क्या आप जानते नहीं कि समाज में कितना नारी उत्पीड़न हो रहा हैं ? जीवन में कभी यदि कोई विषमता आ जाए तो शिक्षित नारी अपने जीवन की समस्याओं का समाधान करने योग्य तो होती है। आज के इस विकसित परिप्रेक्ष्य में भी आप मेरी बहन को केवल चूल्हे चाकी तक ही सिमटा देना चाहते हैं । आप नहीं चाहते कि वह भी अच्छी पढ़ाई करके अपने पैरों पर खड़ी हो जाये और खुशहाल जिंदगी जी सके।

चाचा जी मेरी आपसे सानुरोध प्रार्थना है कि आप निशा की आगे की पढ़ाई आरंभ कराइये तथा वह जो सपने हृदय में संजोये है, उन्हें पूरे करने में उसकी सहायता कीजिए ।
आशा है आप मेरे इस सुझाव का आदर करते हुए निशा की भावना को समझेंगे। चाचा जी को सादर प्रणाम । घर में सबको
यथा योग्य।
आपके उत्तर पत्र की प्रतीक्षा में ।
आपका भतीजा
करन

Question 3.
Read the passage given below and answer in Hindi the questions that follow, using your own words as far as possible :

निम्नलिखित गद्यांश को ध्यान से पढ़िए तथा उसके नीचे लिखे प्रश्नों के उत्तर हिंदी में लिखिए। उत्तर यथासंभव आपके अपने शब्दों में होने चाहिए :-

बहुत समय पहले एक गाँव में हरिहर नाम का एक दयालु और सीधा- सच्चा किसान रहता था । वह खेती-बाड़ी का काम करता था । वह पूरा दिन अपने खेत में जी-तोड़ मेहनत करता था और शाम का समय ईश्वर की प्रार्थना में बिताता था। जीवन में मात्र उसकी एक इच्छा थी । वह उडुपि के मंदिर में भगवान श्रीकृष्ण के दर्शन करना चाहता था । वह उडुपि दक्षिण कर्नाटक का प्रमुख तीर्थस्थान है । वह अपनी गरीबी के कारण तीर्थयात्रा की इच्छा पूरी नहीं कर पाता था । इसी तरह कुछ वर्ष बीत गए। समय के साथ-साथ हरिहर की आर्थिक स्थिति भी सुधरती गई। अब उसने तीर्थयात्रा की योजना बनाई । उसकी पत्नी ने उसके लिए पर्याप्त भोजन बाँध दिया।

हरिहर तीर्थयात्रियों के एक दल के साथ उडुपि की ओर चल दिया । मार्ग में उसे एक स्थान पर एक बूढ़ा आदमी मिला। उसकी दशा बहुत दी दयनीय थी । वह कई दिनों से भूखा-प्यासा था और पीड़ा के कारण कराह रहा था । जैसे ही हरिहर की नजर उस पर पड़ी, उसका हृदय करुणा से भर गया। उसने बूढ़े के पास जाकर पूछा, “बाबा, क्या तुम भी तीर्थयात्रा करने उडुपि जा रहे हो ?” बूढ़े आदमी ने उत्तर दिया, ‘मेरा एक बेटा बीमार है और दूसरे बेटे ने भी तीन दिनों से कुछ नहीं खाया। फिर मैं तीर्थ यात्रा कैसे करूँ ।”

हरिहर समझता था कि दीन-दुखियों की सेवा ही ईश्वर की सबसे बड़ी सेवा है इसलिए उसने उडुपि जाने से पहले उस बूढ़े के घर पहले जाने का निश्चय किया। उसके साथियों ने उसे बहुत समझाया “बहुत मुश्किल से तुमने धन एकत्र किया है। अगर यह नष्ट हो गया तो फिर तुम कभी तीर्थयात्रा नहीं कर पाओगे ।” हरिहर पर उनकी बातों का कोई प्रभाव नहीं पड़ा। वह बूढ़े के घर जा पहुँचा । उसने सबसे पहले घर के सभी व्यक्तियों को भरपेट भोजन कराया। फिर वह बीमार बच्चे के लिए दवा ले आया। उसने बूढ़े को खेत में बोने के लिए बीज भी ला दिये। वह कुछ दिन वहाँ रुका। उसने बूढ़े आदमी के बेटे की सेवा की, जिससे वह कुछ दिनों में स्वस्थ हो गया लेकिन इन सारे कार्यों में उसके सारे पैसे खर्च हो गये ।

अब उसने अपनी तीर्थयात्रा बीच में ही छोड़कर वापस घर लौटने का निश्चय किया । उसे उडुपि न जा पाने का बिल्कुल भी दुःख न था । क्योंकि वह जानता था कि उसने अपना सारा धन दीन-दुखियों की सेवा में खर्च किया था । घर पहुँचकर उसने अपनी पत्नी को सारी बातें बता दीं । पत्नी भी इस पर प्रसन्न हुई क्योंकि वह भी धार्मिक स्वभाव की महिला थी । उस रात हरिहर ने सपने में भगवान श्रीकृष्ण को देखा, जो उससे कह रहे थे, ” हरिहर तुम मेरे सच्चे भक्त हो। तुमने उस बुढ़े आदमी की सहायता की और अपनी इच्छा का बलिदान कर दिया । वह बूढ़ा आदमी कोई और नहीं मैं ही था । तुम्हारी परीक्षा के लिए ही मैं उस बूढ़े आदमी का वेश धारण कर वहाँ आया था। तुम मेरे सच्चे सेवक हो ।” इस तरह हरिहर बगैर तीर्थयात्रा पर गए पुण्य का भागीदार बना ।

(i) हरिहर क्या काम करता था ? उसकी एकमात्र इच्छा क्या थी ? [2]
(ii) हरिहर को तीर्थयात्रा के मार्ग में कौन मिला ? उसकी क्या स्थिति थी ? [2]
(iii) हरिहर ने बूढ़े व्यक्ति की कैसे सहायता की ? [2]
(iv) हरिहर ने घर लौटने का निश्चय क्यों किया ? वहाँ लौटने पर रिहर ने क्या स्वप्न देखा ? [2]
(v) इस गद्यांश से आपको क्या शिक्षा मिलती है ? [2]
Answer :
(i) सीधे-सादे स्वभाव का व्यक्ति हरिहर एक कृषक था । वह पूरे दिन अपने खेत में जी-तोड़ मेहनत करता था । संध्या समय वह अपने प्रभु श्रीकृष्ण की आराधना में व्यतीत करता था । दक्षिण कर्नाटक में उडुपि नाम का एक प्रमुख तीर्थस्थान है। उसमें उसकी घनिष्ठ आस्था थी । उसके दर्शन करने की उसकी प्रबल तथा एकमात्र अभिलाषा थी ।

(ii) हरिहर ने कड़ी मेहनत करके कुछ धन एकत्रित किया और वह तीर्थयात्रियों के एक समूह के साथ उडुपि नामक तीर्थस्थान की
ओर चल दिया। अभी वह थोड़ी ही दूर चला, तभी मार्ग में उसे एक वृद्ध व्यक्ति मिला जिसकी दशा अत्यंत दयनीय थी। वह कई दिनों से भूख-प्यास से पीड़ित और दर्द से कराह रहा था । उसकी दयनीय अवस्था से हरिहर दया द्रवित हो गया । उस बूढ़े से बातचीत करके हरिहर को ज्ञात हुआ कि उसका एक पुत्र बीमार है, दूसरा तीन दिन से भूखा है हरिहर समझ रहा था कि शायद इस भीड़ में मिलने वाला वह बूढ़ा भी तीर्थयात्री है, परंतु जब उसे उसका सच पता चला तो उसका हृदय करुणा से भर गया।

(iii) हरिहर दीन-दु:खियों की सेवा को परमात्मा की सच्ची भक्ति समझता था। उसने उडुपि जाने का निश्चय त्याग दिया। उसके साथियों के बार-बार समझाने के उपरांत भी उसने दीन- दुःखयों की सेवा को ही अपना प्रथम कर्तव्य समझा । हरिहर की पत्नी ने तीर्थयात्रा पर चलते समय बहुत – सा भोजन रखा था। सर्वप्रथम हरिहर ने बूढ़े के घर जाकर उस भोजन से उसके घर के भूखे व्यक्तियों को भरपेट भोजन करवाकर उन्हें तृप्त किया। तत्पश्चात् वह बीमार बच्चे के लिए दवा लेकर आया। यही नहीं उसने बूढ़े के खेत में अनाज बोने के लिए बीज लाकर दिये । वह कुछ दिन वहाँ ठहरा। उसने उसके बीमार बच्चे की सेवा – सुश्रुषा की जिससे वह स्वस्थ हो गया।’

(iv) उस बूढ़े व्यक्ति के परिवार की सेवा न सहायता करते हुए उसके कई दिन व्यतीत हो गये साथ ही उडुपि नामक तीर्थस्थान के दर्शन हेतु जो उसने धन एकत्रित किया था, वह भी वहाँ समाप्त हो गया। उसने अपने इस कर्म को हरि इच्छा समझा और प्रसन्न मन से अपने घर के लिए लौट पड़ा। घर पहुँचकर उसने सारी घटना पत्नी को बताई क्योंकि पत्नी भी धार्मिक स्वभाव वाली व सेवाभावी थी। अतः उसने भी पति के इस कृत्य की सराहना की। उसी रात हरिहर ने स्वप्न में भगवान श्रीकृष्ण के दर्शन किये जो कह रहे थे ” हरिहर तुम मेरे सच्चे भक्त हो, वह बूढ़ा कोई और नहीं, मैं ही था, मैं तो तुम्हारी परीक्षा ले रहा था। तुम परीक्षा में सफल हुए और तुम्हें तीर्थ का पुण्य भी प्राप्त हुआ । ”

(v) इस गद्यांश से हमें यह शिक्षा मिलती है कि जो सुख व संतुष्टि दीन-दुःखी की सेवा और पीड़ा दूर करने में है वह तीर्थ आदि के दर्शन करने में नहीं है । तीर्थ में स्थित परमात्मा हर मनुष्य के हृदय में स्थित है। मनुष्य की पीड़ा हरने से उसके अंदर विद्यमान परमात्मा प्रसन्न होता है । इसलिए पीड़ितों की सेवा परमात्मा की सच्ची सेवा है।

Question 4.
Answer the following questions according to the instructions given :

निम्नलिखित प्रश्नों के उत्तर निर्देशानुसार लिखिए:-

(i) निम्न शब्दों के विशेषण बनाइए लोभ, इतिहास | [1]
(ii) निम्न शब्दों में से किसी एक शब्द के दो पर्यायवाची शब्द लिखिए-
बादल, स्वतंत्र । [1]
(iii) निम्नलिखित शब्दों में से किन्हीं दो शब्दों के विपरीतार्थक शब्द लिखिए-
उपकार, कोमल, नूतन, स्वामी । [1]
(iv) निम्नलिखित मुहावरों में से किसी एक की सहायता से वाक्य बनाइए-
अपने पैर पर आप कुल्हाड़ी मारना, बाल-बाल बचना।
(v) भाववाचक संज्ञा बनाइए- [1]
अधिक, भक्त ।
(vi) कोष्ठक में दिए गए निर्देशानुसार वाक्यों में परिवर्तित कीजिए-
(a) महाराणा प्रताप के साहस की तुलना नहीं की जा सकती है। [1]
(रेखांकित शब्दों के स्थान पर एक शब्द का प्रयोग कीजिये)
(b) मेरे घर में जो नौकर काम करता है वह भाग गया है। [1] (सरल वाक्य बनाइये )
(c) राजा का सेवक बहुत बुद्धिमान था । (लिंग बदलकर वाक्य दोबारा बनाइये )
Answer :
(i) लोभ – लोभी
इतिहास – ऐतिहासिक

(ii) बादल – मेघ, वारिद
स्वतंत्र – स्वच्छन्द, स्वेच्छाचारी ।

(iii) उपकार – अपकार
कोमल – कठोर
नूतन – पुरातन
स्वामी – सेवक ।

(iv) अपने पैर पर आप कुल्हाड़ी मारना-
मैंने यह भेद उसे बताकर स्वयं ही अपने पाँव पर कुल्हाड़ी मार ली ।
बाल-बाल बचना वह कार के नीचे आने से बाल-बाल बच गया ।

(v) अधिक – अधिकता
भक्त – भक्ति ।

(vi) (a) महाराणा प्रताप का साहस अतुलनीय था ।
(b) मेरे घर में काम करने वाला नौकर भाग गया है।
(c) रानी की सेविका बहुत बुद्धिमती थी ।

Note: Section B is not given due to change in present syllabus

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**Answer is not provided due to change in the present syllabus.

Section — A (Drama)
The Merchant of Venice: Shakespeare

Question 1.
Read the extract given below and answer the questions that follow:

Portia : As from her lord, her governor, her king.
Myself and what is mine to you and yours
Is now converted : but now I was the lord
Of this fair mansion, master of my servants,
Queen o’er myself; and even now, but now,
This house, these servants, and this same myself,
Are yours, my lord :

(i) Where are Portia and Bassanio ? What has just taken place which makes Portia to speak these words ? What, was the inscription given in the lead casket ? [3]
(ii) What does Bassanio say in praise of Portia’s portrait ? [3]
(iii) What news saddens Bassanio on this happy occasion ? What does Portia ask him to do ? [3]
(iv) Who is Balthazar ? What was the work assigned to him by Portia ? [3]
(v) Where does Portia really plan to go ? What similarity do we find between Portia and Antonio ? What does this scene reveal about the character of Portia ? Give a reason to justify your answer. [4]
Answers:
(i) Portia and Bassanio are in a room in Portia’s house. Bassanio has just chosen the leaden casket which contained Portia’s portrait. Portia is oveijoyed at this.
The leaden casket had this inscription on it:
“Who chooseth me must give and hazard all he hath.”

(ii) Looking at the portrait of Portia, Bassanio says that some semi-divine artist has made this portrait. It is so lifelike. The eyes seem to move and the lips seem to part to say something. It is really excellent.

(iii) On this happy occasion, Bassanio receives from Salerio Antonio’s letter. This letter informs him that Antonio’s ships have met with a disaster. His creditors have become cruel, as his means have fallen low. Antonio’s bond to Shylock has been forfeited.
When Portia, learns about Antonio’s plight, she asks Bassanio to expedite all his business immediately go to Venice to help Antonio.

(iv) Balthazar is the trusted servant of Portia. She asks him to go to Padua immediately and deliver a letter to her cousin Dr. Bellario. Moreover, whatever papers or garments he might give to him, he must bring them to her at a very fast speed.

(v) Portia plans to appear in the court disguised as a representative of Dr. Bellario.
Both Portia and Antonio are very sincere. Both of them love Bassanio and are ready to do anything for him. This scene reveals Portia’s essential wisdom. As soon as she sends Bassanio, she decides to send Balthazar to Dr. Bellario to seek legal opinion and help from him in the case of Antonio.

Question 2.
Read the extract given below and answer the questions that follow:

But mercy is above this sceptred sway;
It is enthroned in the hearts of kings,
It is an attribute to God himself;
And earthly power doth then show likest God’s
When mercy seasons justice.

(i) Name the speaker. Why did the speaker appeal to the Jew for mercy ? Earlier who else in the play appealed for mercy ? [3]
(ii) What are the three qualities of mercy which the speaker has statedjust before the extract ? [3]
(iii) Give the meaning of But mercy is above this sceptred sway ’. How does Shylock turn down Portia s plea for mercy ? What does he insist on ? [3]
(iv) What is Bassanio ready to do for Antonio in the court ? Why is Bassanio snubbed immediately by the disguised Portia ? [3]
(v) Mention two prominent character traits of Shylock as highlighted through the scene from which the extract has been taken. Substantiate your answer with examples from the text. [4]
Answers:

(i) The speaker is Portia dressed as a doctor of laws. The speaker appealed to the Jew for mercy to diminish the force of his demand for strict justice. She added that if he still insisted on his demand of a pound of flesh, the strict court of Venice would have to pronounce a judgement against Antonio.
Earlier the Duke himself had appealed to Shylock to show mercy.

(ii) The speaker has stated that mercy is shown by human beings as naturally and spontaneously as the gentle drops of rain fall from the sky above down to the ground below. Secondly, mercy is the most powerful feeling in those who have the highest authority. It suits a king more than his crown. Mercy is greater than the worldly authority which is symbolized by the sceptre which he holds. Thirdly, mercy is the essential quality of God who is the supreme power of the universe. Earthly power acquires a divine quality when a man softens the rigours of the demands of justice by showing mercy.

(iii) The line “But mercy is above this sceptred sway” means that the king’s sceptre is the emblem of his worldly power. It is the symbol of fear which kingship inspires in others. But mercy is higher and greater than the worldly authority which is symbolized by rod. Shylock turns down Portia’s plea for mercy asking her to let him face the consequences of his actions. He takes full responsibility for his act. He persists in his demand. He asks Portia to enforce the law and let him have the penality specified in the bond which Antonio has forfeited.

(iv) Bassanio is ready to pay twice the borrowed money. If that is not sufficient, he is ready to pay ten times that amount. If even this is not enough, he is ready to forfeit his hands and his heart.
Bassanio asks Portia (disguised as doctor of laws) to bend the law once and use the authority of the court to do what is morally right, to save the life of Antonio. But Portia snubs him saying that there is no power that can change the established law of Venice.

(v) The trial scene shows that Shylock is a revengeful and bloodthirsty man. He turns down the appeals of the Duke and Portia for mercy and insists on having a pound of flesh. He is offered ten times the amount of the borrowed money but he refuses it. He wants only revenge. Secondly, the scene shows that Shylock is firm in his resolve. He bluntly tells Portia that there is no power in the tongue of man to alter his resolve to take a pound of flesh. He says, “My deeds upon my head”. Thus he invokes the law to let him have the penality as mentioned in the bond.

Loyalties : John Galsworthy

Question 3.**
Read the extract given below and answer the questions that follow:

Inspector : [Sharply] Are you sure there was nobody in the room already ?
De Levis : [Taken aback] I don’t know. I never thought. I didn’t look under the bed, if you mean that.
Inspector : [Jotting] Did not look under bed. Did you look under it after the theft ?
De Levis : No. I didn’t.
Inspector : Ah ! Now, what did you do after you came back from your bath ? Just give us that precisely.
(i) What reply did De Levis give to the inspector’s last question in the extract? [3]
(ii) What made De Levis check the contents of his pocket book ? What did he find there ? Whom did he go to upon discovering the theft ? [3]
(iii) Who was Robert ? Where was Roberts room ? At what time did he take De Levis ’clothes and boots ? [3]
(iv) What is the Inspector s final theory of the theft? [3]
(v) Whom did De Levis accuse of stealing his money ? What were his reasons for making this accusation ? [4]

Question 4.**
Read the extract given below and answer the questions that follow:

Mabel: [ Utterly surprised] Ronny! Do they want me in Court?
Dancy: No.
Mabel: What is it, then ? Why are you back ?
Dancy: Spun.
Mabel: [Blank] Spun ? What do you mean ? What’s spun ?
Dancy : The case. They’ve found out through those notes.
Mabel: Oh ! [Staring at his face.] Who ?
(i) Where are Mabel and Dancy at this time? What was Mabel doing just before this conversation ? [3]
(ii) Why did Mabel say, “Do they want me in court? ” Explain the meaning of ‘spun ’in the extract ? [3]
(iii) What ‘notes ’ is Dancy talking about now ? How does Mabel react immediately after the extract? [3]
(iv) Dancy leaves a note for his best friend towards the end of the play. What is the name of his best friend ? What is written in the note ? [3]
(v) What does Dancy do at the end ? Why does he do that ? What is your opinion of Mabel and Dancy ? [4]

Section – B (Poetry)
A Collection of Poems

Question 5.**
Read the extract given below and answer the questions that follow:

The woods are lovely, dark and deep.
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
(Stopping by Woods on a Snowy Evening: Robert Frost)

(i) Who is ‘I’referred to in the extract ? Which season of the year is it ? What evidence is there in the poem to support your answer ? [3]
(ii) Who has made him aware of his mistake ? How does it make the speaker aware of his mistake? What does it seem to say ? [3]
(iii) What are the three sounds heard ? [3]
(iv) What has been said earlier by the poet about the owner of the woods ? [3]
(v) What does lovely, dark and deep suggest ? What is the underlying significance in the repetition of the last two lines of the extract ? Mention the moral tag that the poet attaches to the poem. [4]

Question 6.**
Read the extract given below and answer the questions that follow:

The morning stretched calm, beautiful, and warm.
Sprawling half clad, I gazed out at the form
Of shimmering leaves and shadows. Suddenly
A strong flash, then another, startled me.
I saw the old stone lantern brightly lit.

(i) In which country did the incident described in the poem occur ? What had caused the flashes? What happened to his home soon after ? [3]
(ii) Describe the injuries suffered by the Doctor. [3]
(iii) What was his wife s name ? What happened to a house standing before them ? [3]
(iv) How does he describe the people he met on the way ?[3]
(v) Why were the people walking with their hands away from their bodies ? What was common to all of them ? What message does the poem convey ? [4]

Section – C (Prose)
Collection of Short Stories

Question 7.**
Read the extract given below and answer the questions that follow:

“That only leaves October, November and December,” said the Queen. “And after that we shall have to begin all over again.”
“No, we shan’t,” said the King, “because I think twelve daughters are enough for any man and after the birth of dear little December I shall be reluctantly compelled to cut off your head.”
He cried bitterly when he said this, for he was extremely fond of the Queen.

(i) Why did the king change the names of his daughters so many times ? [3]
(ii) In what way was Princess September different from her sisters ? What reason does the author give for this difference
in their temperaments ? [3]
(iii) Which unusual birthday tradition did the King of Siam observe ? Mention some of the gifts that he gave. [3]
(iv) Why did Princess September put the Nightingale in a cage? What reasons did she give to the bird for putting it in a cage and then keeping it there ? [3]
(v) How did the bird behave upon being locked in a cage ? What is the message of the story ? [4]

Question 8.**
The Last Leaf explores the theme of Friendship and Self-sacrifice. Discuss this with close reference to O’Henry’s ‘The Last Leaf. ’ [16]

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Part – I (30 Marks)
Attempt all questions from this Part

Question 1.
(a) Who administers the oath of office to the Council of Ministers ? [1]
Answer:
The President

(b) Why is the Rajya Sabha called a ‘Permanent house ? [1]
Answer:
The Rajya Sabha is a Permanent House, hence it cannot be dissolved. Each member is elected for a term of six years. l/3rd of its total members retire at the end of every two years, and the equal number of new members are elected to fill the vacancies caused by the retirement of 1/3rd members.

(c) State the minimum number of times the Lok Sabha must meet in a year [1]
Answer:
3 Sessions.

1. Summer Session (Feb-May)
2. Monsoon Session (July-Sept.)
3. Winter Session (Nov.-Dee.)

(d) State the composition of the Electoral College in the election of the President ofIndia. [ 1 ]
Answer:
The joint body of elected M.Ps and M.L.As (i.e. elected members of both the Houses of the Parliament and elected members of the Legislative Assemblies of States) which elects the President, is known as the Electoral College. It is a temporary body which is dissolved immediately after the election of the President is over. Nominated M.Ps and nominated M.L.As have no hand in electing the President.

(e) What is an ‘Ordinance’? When can it be passed? [1]
Answer:
An Ordinance is a Presidential decree having authority of law.
The President can issue an Ordinance:

• When the President is satisfied that immmediate necessary legislative action is required.
• When the Parliament is not in session.

(f) Mention one way by which the authority of the Prime Minister can be checked? [1]
Answer:
India has a well established democratic system for the past 67 years. The presence of the opposition in the Parliament, with its leader keeps him under check.

(g) What is understood by the term ‘Individual Responsibility’ in a Parliamentary Democracy? [1]
Answer:
If a minister has taken a departmental decision independently in the best interest of the department, then he is individually responsible to the President, practically to the Prime Minister who can ask him to resign from his office. The concerned Minister acts accordingly.

(h) What is meant by a ‘Single Integrated Judicial System ’ as provided in the Indian Constitution? [1]
Answer:

• All the courts enforce the state as well as laws made by the Union Parliament.
• There are no separate sets of law and a single civil and criminal system operates through out the country,

(i) What is the ‘Appellate Jurisdiction’ of the High Court [1]
Answer:
Appellate Jurisdiction means that the High Court has the power to accept appeals against the decision of District Court in civil as well as criminal matters.

(j) State one point of distinction between the District Judge and the Session Judge. [1]
Answer:

Question 2.
(a) State any two political causes responsible for the First War of Independence. [2]
Answer:

• Lord Dalhousie’s policy of annexation and the Doctrine of Lapse made the Indian rulers angry and insecure. The prominent states which fell victim to the Doctrine were : Satara, Jhansi and Nagpur.
• Lord Canning declared that Bahadur Shah’s successor would not be allowed to use the imperial title, i.e. the ‘King’.

(b) What was the role of the press in promoting nationalistic sentiments amongst the Indians? [2]
Answer:

• It played an important role in the campaigns for social reforms and to influence activities of the state.
• It played an important role in the growth of National Movement.
• Press criticized the unjust policies of the British government in India. They spread the message of patriotism and liberal ideals of liberty, equality, etc. There were other dailies which voiced.

(c) Mention two important contributions of Lala Lajpat Rai. [2]
Answer:

• Lala Lajpat Rai emphasised the importance of education in moulding the character of people.
• Through his writings he preached radical nationalism, inspired the Indian youth and kindled the fire of patriotism in them.

(d) What were the two basic reasons responsible for the Surat Split in 1907 between the Early Nationalists and the Assertive Nationalists ? [2]
Answer:
To placate the Assertive Nationalists the Calcutta session of the Congress in 1906 A.D. adopted “Swaraj ” as its goal and passed resolutions supporting Swadeshi and Boycott and National education. But the moderate section in the Congress did not like this development. The Radicals wanted to extend the Swadeshi and Boycott Movement to rest of India but the Moderates wanted to confine it to Bengal.

(e) Mention any one provision each of the Gandhi-Irwin Pact signed in 1931. [2]
Answer:

• Withdrawal of all ordinances issued by the British Government imposing curbs on the activities of the Indian National Congress.
• Release of prisoners arrested for participating in the Civil Disobedience Movement.

(f) State two important objectives of the Indian National Army. [2]
Answer:

• The aim of the army was to overthrow the British Raj in colonial India, with Japanese assistance.
• To establish an Interim Government.

(g) Name the three members of the Cabinet Mission. [2]
Answer:

• Lord Pethick- Lawrence,
• Sir Stafford Cripps
• Mr. A.V Alexander.

(h) Mention any two terms of the Treaty of Versailles signed on June 28, 1919. [2]
Answer:

• Germany was declared guilty of war and was compelled to pay 33 billion dollars as war indemnity to the victor nations.
• Germany had to evacuate the places she had captured during the war.

(i) State two factors which were responsible for the failure of the League of Nations. [2]
Answer:

• Actually, the League of Nations had no power to act on its own initative to preserve peace in the world.
• The USA did not become its member, because the USA Senate did not ratify the Covenant of the League of Nations.

(j) Mention two functions of the General Assembly. [2]
Answer:

• It considers and approves the budget of the United Nations.
• It appoints the Secretary-General upon the recommendation of the Security Council.

Question 3.
With reference to the Indian Parliament, explain the following:
(a) The tenure of the members ofthe Lok Sabha and the Rajya Sabha. [2]
Answer:
Lok Sabha: The term of the Lok Sabha is five years. But it can be dissolved earlier by the President if a motion expressing the No-Confidence in the Government is moved and passed by the opposition.

Rajya Sabha: The Rajya Sabha is a permanent House. It cannot be dissolved. Each member has a term of six years. One third of its total number of members retire at the end of every two years, and the equal number of new members are elected or nominated to fill the vacancies caused by the retirement of l/3rd members. Thus, the Rajya Sabha continues to exist, though its members retire when they complete their six years tenure. They can also be re-elected or renominated.

(b) The composition of the Lok Sabha and the Rajya Sabha. [4]
Answer:
Lok Sabha:
The Lok Sabha is composed of representatives of the people chosen by direct election on the basis of adult suffrage. The maximum strength of the House envisaged by the Constitution is about 552,530 members to represent the States, 20 members to represent the Union Territories and not more than two members of the Anglo-Indian Community to be nominated by the President, if, in his opinion, that community is not adequately represented in the House. The total elected membership is distributed among the States roughly on population basis.

Rajya Sabha:
The strength of the Rajya Sabha is fixed at 250 members. There are two categories of members: First, the elected members; and second, the nominated members.

Elected Members:
There are 238 elected members, who represent the 28 States and the Union Territories, including the National Capital Territory, Delhi. Seats are allotted to them on the basis of their population. For example, Uttar Pradesh, with a population of nearly 16 crores is represented in the Rajya Sabha by 31 members, which is the largest. Goa, Manipur, Meghalaya, Mizoram, Nagaland, Sikkim, Tripura send only one member each to the Rajya Sabha.

Nominated Members:
The President nominates 12 members to the Rajya Sabha. These are eminent persons from the fields of Literature, Science, Art and Social Service. They are nominated for six years, with a provision that one-third (i.e., 4 out of 12) shall retire after eveiy two years. These nominations are made by the President on the advice and recommendations of the Prime Minister.

(c) Its powers to make laws on subjects mentioned in the [4]
(i) Union List
(ii) Concurrent List.
Answer:
(i) Union List: The Parliament has exclusive powers to make laws on the 97 subjects in the Union List, including important subjects like Defence, Foreign Policy, Communications, etc.

(ii) Concurrent List: The Legislative Assembly and Parliament can make laws on 47 subjects listed in the Concurrent List. If there is any conflict between the Union Parliament and the State Legislature on any law over this list, the Union Law will prevail.

Question 4.
The Cabinet holds a pivotal position in the working of the Indian Parliamentary Government In this context discuss the following:
(a) The formation of the Cabinet. [2]
Answer:

• Cabinet is the most important part of the government.
• Cabinet is consisted of the most senior leaders of the ruling party.
• Cabinet ministers are appointed by the Prime Minister. They hold the charge of the most important portfolios like Defence, Home, Finance, Foreign affairs, etc.

(b) Any two administrative powers of the Cabinet. [4]
Answer:

• The whole administration of the state is run by the ministers.
• Every minister is incharge of one particular department and is also responsible to the Lok Sabha for running the administration of that department. He has to answer the questions of the members of Parliament relating to his department.
• The Ministers also enforce properly the laws made by the Parliament.
• The President appoints the Attorney General of India, Governor of a state, Chief Election Commissioner, Election Commissioners, Judges of Supreme Court, High Courts and other high dignitaries of the state on the advice of the Cabinet.

(c) Any two legislative powers of the Cabinet. [4]
Answer:

• The sessions of the Parliament are summoned by the President on the advice of the Cabinet. The initiative in this matter is taken by the Ministry of Parliamentary Affairs. The Cabinet decides dates for holding sessions and for proroguing the Parliament.
• The special address that the President delivers to the Parliament is prepared by the Cabinet and the President simply reads it.
• The Cabinet gets the bills drafted. The members of the Council of Ministers introduce, explain and defend the bills in the Parliament and get them passed.
• The President issues ordinances at the instance of the Cabinet.

Question5
With reference to the Supreme Court, explain its functions stated below:
(a) Original Jurisdiction. [3]
Answer:
Original Jurisdiction means the authority and power of the Supreme Court to hear a case in the first instance directly. The original and exclusive jurisdiction of the Supreme Court extends to the following cases:
(i) Interpretation of the Constitution in case of a dispute between the Union Government and one or more States.

(ii) Interpretation of the Constitution in case of dispute between two or more States interests, relating to some justiciable right.

(iii) The Supreme Court has the exclusive jurisdiction to determine the Constitutional Validity of the Union laws. Such cases are debated by a majority of the full Constitutional Bench of seven Judges.

(iv) The Supreme Court has a special jurisdiction and responsibility for the enforcement of Fundamental Rights.
It can issue writs of Habeas Corpus, Mandamus, Prohibition, Quo-warranto and Certiorari for this and other purposes. As such a citizen can resort directly for such relief to the Supreme Court for the enforcement of his Fundamental Rights, without first resorting to a High Court.

(v) The Supreme Court has also the exclusive and original jurisdiction to inquire into and decide all doubts and disputes in connection with the election of the President and Vice-President of India.

(b) Advisory Function. [3]
Answer:
The President may seek Supreme Court’s opinion or advice on a question of law or fact, which is of public importance in the opinion of the President, whether it has actually arisen or is likely to arise. The Supreme Court may give its opinion, after such hearing as it may think fit. It is done by a Bench of five judges, by a majority vote. The opinion of the Supreme Court is pronounced in the open court. But the opinion of the Supreme Court is riot binding on the President, as it is not a judicial decision or pronouncement. It is only a consultative view. It must be noted that the Supreme Court is also not bound to give its advisory opinion when a reference is made to it by the President. The Supreme Court can decline to give its opinion. Usually, its opinion is rejected by all courts and tribunals.

(c) As a guardian of Fundamental Rights. [4]
Answer:
The Supreme Court is the guardian of the Fundamental Rights. A citizen, whose fundamental right has been violated by any individual, government or institution, may move the Supreme Court for the protection of his fundamental right, A writ is a legal instrument to enforce obedience to the orders of a court. It is a form of written command or order from a judicial authority asking a particular person or official to act or not to act in a particular way. The Constitution empowers the Supreme Coin! and High Courts to issue the following writs of Habeas Corpus, Mandamus, Prohibition and Certiorari. Out of all these writ of Habeas Corpus is very important.

Writ of Habeas Corpus : Habeas Corpus (Latin: means that you have the body) is a writ through which a person can seek relief from unlawful detention (i.e., held in police custody or prison) of himself or another person.

Section-B
Attempt any three questions from this Section

Question 6
The establishment of the Indian National Congress led to the development of the National Movement in India. In this context answer the following:
(a) When was the Indian National Congress established ?
Who presided over its first session? [2]
Answer:
1885. W.C. Bannerjee.

(b) What were the four aims of the Congress ? [4]
Answer:

• To enable national workers from all parts of India to become personally known to each other.
• To end all racial, religious and provincial prejudices and to promote a feeling of national unity among all the patriots of the country. To promote friendly relations between the Hindus and the Muslims.
• To formulate popular opinion on vital Indian problems and to present these before the government.
• To create, train and organize a strong body of public opinion in the country.

(c) Mention four basic beliefs of the Early Nationalists. [4]
Answer:
1. The English were just and freedom loving people. They had an abiding faith in the British sense of justice and fairplay and believed in loyalty to the British crown.

2. So they believed that simply by presenting and pleading their case before the British Parliament and the British people, justice would be done and their grievances would be redressed.

3. The Congress was proud of its British connections and looked to England for inspiration and guidance, (z’v) They recognised the benefits of the English rule, especially the English language and modem means of transport and communication. They also tried to purge the society of its evils like sati, child marriage, untouchability, etc.

Question 7.
In the Nagpur session, 1920, the Congress ratified the resolution to launch the Non-Cooperation movement under the leadership of Gandhiji. In this context:
(a) What do you understand by the term Non-Cooperation? [2]
Answer:
It was a movement launched by Gandhiji in 1920 over the question of the injustice done to Turkey, the repression in Punjab and for Swaraj. It is called the Non-Cooperation Movement because of the methods adopted in the movement.

(b) What were the objectives which the movement sought to achieve ? [3]
Answer:

• Redressal of the wrongs committed in Punjab that entailed the Jallianwala Bagh massacre and the atrocities related to the martial laws.
• Restoring the old status of the Sultan of Turkey.
• Accomplishment of Swaraj.
  In September, 1920, a special session of the Congress, presided over by Lala Lajpat Rai was convened at Calcutta that sought to approve the scheme.

(c) Explain the impact of the Non-Cooperation movement in India s struggle for freedom. [5]
Answer:
1. New Direction to Freedom Movement: It can be said without any doubt that the Non-Cooperation Movement and the role played by Gandhi in it took the Indian freedom movement to new heights. It was responsible for creating a new political fervour among the people and taught the Indian fearlessness. Under the leadership of Mahatma Gandhi, Swarajya found popular expression and a patriotic zeal gripped the entire nation. The weapon of passive resistance or satyagraha, that Gandhi gave to the nation, emerged as the greatest asset of the Indians.

An important programme of the Non-Cooperation Movement was the removal of untouchability, poverty and the promotion of Khadi. Under the guidance of Gandhi, charkha and Indian handloom products gained back their glory. Many weavers were employed. The freedom struggle assumed an all India character under his leadership. It gave a sense of unity – like common language (Hindi), national uniform (Khadi) and national activity (spinning). It continued with the constructive programme. It undermined the power and prestige of the British government.

2. Mass Movement: The Movement gave rise to the sentiments of nationalism on a wider scale. When thousands and thousands of people walked together shoulder to shoulder and resisted the British for months, the National Movement naturally became quite wide in its scope. It was no longer limited to only an educated urban people. All sections of society- peasants, workers, students, women and oppressed people participated in the movement. Many local revolts like the revolt against taluqdars in Awadh, and Akali Dal movement merged into it and provided the movement a wide base.

3. Hindu-Muslim Unity: The Movement fostered Hindu-Muslim unity which could be seen in the merger of the Khilafat Movement with this Movement. It provided an opportunity to the Congress to bring the urban Muslims into the National Movement.

Question 8
With reference to the picture given above answer the following:

(a) Identify the Viceroy in the picture. [1]
Answer:
Lord Mountbatten.

(b) Why was he sent to India? [2]
Answer:
The British Prime Minister, announced on February 20, 1947, in the House of Commons that Britain would withdraw from India and transfer power to reponsible Indians by 30th June 1948. Even if Indian leaders failed to come to an agreement by that time, the British would withdraw and hand over power to some form of Central government or in some areas to the existing Provincial governments. It would not accept the forcing of unwilling provinces into a United Indian government.

For the purpose of taking necessary steps in that direction Lord Wavell was recalled and Lord Mountbatten was appointed the new Viceroy. Lord Mountbatten assumed office on March 24, 1947.

(c) How did he plan to solve the communal problem existing in India ? [2]
Answer:
Lord Mountbatten lost no time in holding negotiations with the party leaders. Nehru, Patel, Azad, Liaqat Ali, Baldev Singh and others. In the course of his talks with them, Mountbatten was convinced that there was absolutely no possibility of an agreed solution on the basis of the Cabinet Mission Plan and that the partition of India was inevitable. He was successful in convincing both Patel and Nehru and other Congress leaders to the same view.

(d) Why did the Congress accept the Plan? State three reasons to justify its acceptance. [5]
Answer:
(i) Non-Cooperative attitude of the League: The League had joined the interim government to obstruct the working of the Congress and not to cooperate with it.

(ii) Communal Riots: The large scale communal riots that engulfed, the whole country convinced all that the only solution to the communal problem lay in the partition of India. These riots were the outcome of “Direct Action” Day by the League.

(iii) To get rid of Undemocratic Procedures: The leaders felt that the Constitution of separate electorate, British policy of Divide and Rule and other undemocratic procedures would weaken the democracy. So to get rid of these, partition was a must.

Question 9.
With reference to the causes of the Second World War answer the following:
(a) (i) Explain how the ideologies of Fascism and Nazism led to the Second World War. [3]
(ii) How did the Japanese invasion of China create conditions for the outbreak of the war ? [3]
Answer:
(i) Both Nazism and Fascism believed in the principle of the ‘survival of the fittest ’which is the law of life. The strongest alone have the right to survive and rule. Both believed in the principle of expansion. Hitler believed that nation became great only by defeating other nations in war. Both Hitler and Mussolini pursued an expansionist policy abroad. In May, 1936 Italian troops captured Addis Ababa, capital of Ethiopia. Hitler also captured Austria, Denmark, Czechoslovakia, Norway, Luxemberg. Holland and Belgium and Paris too, fell to the Nazis.

(ii) In Asia, Japan harboured expansionist desires. It attacked China in 1931 and annexed Manchuria. China appealed to the League of Nations to declare sanctions against Japan, Britain and France leading, members of the League ignored the appeal. In 1933, Japan left the League of Nations and started occupying the British and American properties in China. Britain and France followed the policy of appeasement under which they ignored aggressive policy of Japan, thinking that Japanese could be used to weaken China. The Japanese were determined to dominate the Far East. America warned Japan of the dire consequences but Japan ignored. She joined “Rome-Berlin-Tokyo” Axis. She also attacked the American naval base at Pearl Harbour.
The League of Nations failed to check these nations in their aggressive policies of territorial annexation.

(b) Explain the consequences of the war with reference to the formation of the United Nations. [4]
Answer:
Mankind looked for a better, a happier and a peaceful world after it had witnessed the nuclear holocaust and the widespread destruction and death in the Second World War. To save the coming generations from the scourge of War, the United Nations Organization (U.N.O.) was established on October 24,1945, with its headquarters at New York (U.S.A.). Peace, human dignity, justice, human brotherhood, cooperation, etc. are the main objectives of the U.N.O. It is the largest world organisation, and has 192 member-States, and it has done a commendable job in almost all spheres of life. We can say that the U.N.O. is the great gift of the World War II to save and serve mankind.

Question 10
With reference to the Non-Aligned Movement, explain the
following:
(a) ‘Non-Alignment’ [2]
Answer:
Non-Alignment is the international policy of a sovereign state according to which it does not align itself with any of the power blocks and at the same time actively participates in the world affairs to promote international peace, harmony and cooperation.

(b) Two factors responsible for its formation. [4]
Answer:
(i) Global tension caused by Cold War: Most of the newly independent countries of Asia and Africa realised that the division of the world into two power blocs was not in their larger interest and this might endanger world peace. These nations felt that by maintaining distance from both the super powers they would put off the danger of another war.

(ii) Struggle against imperialism: The newly independent nations opted for non-alignment because most of these countries remained under big powers for a long period of time. They wanted to enjoy their newly acquired freedom and the power that had come with it.

(c) Role of Pt. Jawaharlal Nehru. [4]
Answer:
(i) Asian-Relations Conference, 1947: India with Nehru as her Interim P.M. convened the Asian Relations Conference in New Delhi in March, 1947. He laid down the basic principle of Non-Alignment, when he said, “We have no design against any body ours is the great design of promoting peace and progress all over the world”.

(ii) Enunciation of Panchsheel, 1954 : Later in 1954, Nehru along with the Chinese Premier, Chon-Enlai, enunciated the five Principles of Peaceful co-existence, popularly known as Panchsheel. These principles provided the firm basis on which friendly relations between nations could be maintained, as well as economic development and peace in the world secured. Panchsheel became the bed-rock of the NAM, later on. This was Nehru’s singular contribution to international relations.

(iii) Nehru’s Role in the Belgrade Conference, 1961: The first formal NAM Summit/Conference was held, from September 1 to 6 in 1961 at Belgrade in Yugoslavia, in which 25 nations participated. This conference adopted a 27-Points Declaration, in preparing which Nehru is said to have played a crucial role. In the Belgrade Conference, he laid emphasis on the following principles :
(a) negotiation for peace,
(b) disarmament,
(c) ending colonialism and imperialism in all its forms,
(d) condemnation of racial discrimination being practised in South Africa or else where,
(e) stress on socio-economic development of Afro-Asian nations on the basis of mutual benefit and equality. The Belgrade Summit also made an appeal to Super-Powers to help in maintaining peace and security in the world. The Conference also invited other Afro-Asian and Latin American nation to join the NAM. Jawaharlal Nehru, Marshall Tito and Colonel Nasser were the most prominent leaders of the NAM at Belgrade. The Trio carried a great influence.

Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2017 is the best way to boost your preparation for the board exams.

ICSE Class 10 Maths Question Paper 2017 Solved

Section – A (40 Marks)
(Attempt all questions from this Section)

Question 1.
(a) If b is the mean proportion between a and c, show that: [3]
\(\frac{a^4+a^2 b^2+b^4}{b^4+b^2 c^2+c^4}=\frac{a^2}{c^2}\)
Solution:
Here, b is the mean proportion between a and c.
∴ b² = ac

(b) Solve the equation 4x² – 5x – 3 = 0 and give your answer correct to two decimal places. [4]
Solution:
Given equation is : 4x² – 5x – 3 = 0
By using quadratic formula, we obtain

= 1.693 or -0.443
= 1.69 or -0.44

(c) AB and CD are two parallel chords of a circle such that AB = 24 cm and CD = 10 cm. If the radius of the circle is 13 cm, find the distance between the two chords. [3]

Solution:
Here, O is the centre of the given circle of radius 13 cm. AB and CD are two parallel chords, such that AB = 24 cm and CD = 10 cm.
Join OA and OC.
Since ON ⊥ AB and OM ⊥ CD.
∴ M, O and N are collinear and M, N are mid-points of CD and AB.
Now, in rt. ∠ed ∆ANO, we have
ON² = OA² – AN²
= 13² – 12²
= 169 – 144 = 25
∴ ON = \(\sqrt{25}\) = 5 cm

Similarly, in rt. ∠ed ∆CMO, we have
OM² = OC² – CM²
= 13² – 5²
= 169 – 25
= 144
OM = \(\sqrt{144}\) = 12cm
Hence, distance between the two chords
NM = NO + OM
= 5 + 12
= 17 cm

Question 2.
(a) Evaluate without using trigonometric tables, sin²28° + sin²62° + tan²38° – cot² 52° + \(\frac{1}{4}\) sec² 30° [3]
Solution:
sin²28° + sin²62° + tan²38° – cot² 52° + \(\frac{1}{4}\) sec² 30°
= sin² 28° + sin² (90° – 28°) + tan² 38° – cot² (90° – 38°) + \(\frac{1}{4}\left(\frac{2}{\sqrt{3}}\right)^2\)
= sin² 28° + cos² 28° + tan² 38° – tan² 38° + \(\frac{1}{4}\left(\frac{4}{3}\right)\)
[∵ sin (90° – θ) = cos θ, cot (90° – θ) = tan θ]
= 1 + \(\frac{1}{3}\) [sin² θ + cos² θ = 1]
= \(\frac{3+1}{3}\)
= \(\frac{4}{3}\)
= 1 \(\frac{1}{3}\)

(b) If A = \(\left[\begin{array}{ll}
1 & 3 \\
3 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
-2 & 1 \\
-3 & 2
\end{array}\right]\) and A² – 5B² = 5C.
Find matrix C where C is a 2 by 2 matrix. [4]
Solution:

(c) Jay a borrowed ₹ 50000for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹ 33000 at the end of the first year. Find the amount she must pay at the end of the second year to clear her debt. [3]
Solution:
Principal = ₹ 50000
Time = 1 year
Rate = 12%
Amount after one year = P(1 + \(\frac{\mathrm{R}}{\mathrm{100}}\))
= 50000 (1 + \(\frac{12}{100}\))
= 50000 × \(\frac{112}{100}\) = ₹ 56000
Amount paid at the end of first year = ₹ 33000
∴ Principal for second year = ₹ (56000 – 33000)
= ₹ 23000
Time = 1 year
Rate= 15%
Amount due after second year = P(1 + \(\frac{\mathrm{R}}{\mathrm{100}}\))
= ₹ 23000 [1 + \(\frac{15}{100}\))
= ₹ 23000 × \(\frac{115}{100}\)
= ₹ 26450.

Question 3.
(a) The catalogue price of a computer set is ₹ 42000. The shopkeeper gives a discount of 10% on the listed price. He further gives an off-season discount of 5% on the discounted price. However, sales tax at 8% is charged on the remaining price after the two successive discounts.
Find: [3]
(i) the amount of sales tax a customer has to pay.
(ii) the total price to be paid by the customer for the computer set.
Solution:
List price of the computer = ₹ 42000
Rate of discount = 10%
Total discount = ₹ \(\frac{10}{100}\) × 42000
= ₹ 4200
Discounted price = ₹ (42000 – 4200)
= ₹ 37800
Another off-season discount = 5%
Total off-season discount = ₹ \(\frac{5}{100}\) × 37800
= ₹ 1890
Remaining price after two successive discounts = ₹ 35910
Total sales tax at 8% = ₹ \(\frac{8}{100}\) × 35910
= ₹ 2872.80
Total price of the computer set
= ₹ (35910 + 2872.80)
= ₹ 38782.80
Hence, amount of sales tax is ₹ 2872.80 and total price to be paid by the customer for the computer set is ₹ 38782.80.

(b) P(1, -2) is a point on the line segment A (3, -6) and B(x, y) such that AP : PB is equal to 2 : 3. Find the coordinates of B. [4]
Solution:
Here, AP : PB = 2 :3, therefore, P divides AB in the ratio 2 : 3
Thus, coordinates of P are : P \(\left(\frac{2 x+9}{2+3}, \frac{2 y-18}{2+3}\right)\)

Also, coordinates of P are P(1, -2)
∴ \(\frac{2 x+9}{5}\) = 1 and \(\frac{2 y-18}{5}\) = -2
2x = 5 – 9 and 2y = 18 – 10
x = \(\frac{-4}{2}\) and y = \(\frac{8}{2}\)
x = -2 and y = 4
Hence, coordinates of B are B (-2, 4).

(c) The marks of 10 students of a class in an examination arranged in ascending order is as follows : [3]
13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Solution:
Here, number of students are 10 i.e., even number of observations.
∴ Median = \(\frac{1}{2}\)
(\(\frac{10^{\text {th }}}{2}\) observation + \(\left(\frac{10}{2}+1\right)^{\text {th }}\) observation)
= \(\frac{1}{2}\) (5^th observation + 6^th observation)
∴ 48 = \(\frac{1}{2}\) (x + x + 4)
96 = 2x + 4
92 = 2x
⇒ x = 46
Thus, the observations are 13, 35, 43, 46, 46, 50, 55, 61, 71, 80.
Hence, the mode of the given data is 46.

Question 4.
(a) What must be subtracted from 16x³ – 8x² + 4x + 7 so that the resulting expression has 2x + 1 as a factor ? [3]
Solution:
Let p(x) = 16x³ – 8x² + 4x + 7
and g(x) = 2x + 1
Put 2x + 1 = 0 ⇒ x = –\(\frac{1}{2}\)
∴ Remainder = p\(\left(-\frac{1}{2}\right)\)
= 16\(\left(-\frac{1}{2}\right)^3\) – 8\(\left(-\frac{1}{2}\right)^2\) + 4\(\left(-\frac{1}{2}\right)\) + 7
= 16 × \(\left(-\frac{1}{8}\right)\) – 8 × \(\frac{1}{4}\) – 4 × \(\frac{1}{2}\) + 8
= -2 – 2 – 2 + 7
= -6 + 7 = 1
Hence, 1 is subtracted from p(x), so that g(x) is a factor of p(x).

(b) In the given figure ABCD is a rectangle. It consists of a circle and two semi-circles each of which are of radius 5 cm. Find the area of the shaded region. Give your answer correct to three significant figures. [4]

Solution:
Here, radius of a circle and two semi-circles = 5 cm
∴ Length of the rectangle = 5 + 10 + 5
= 20 cm
Breadth of the rectangle = 10 cm
Now, area of the shaded part = Area of rectangle -2 × Area of circle
= 20 × 10 – 2 × \(\frac{22}{7}\) × 5 × 5
= 200 – \(\frac{1100}{7}\) = 200 – 157.14
= 42.86 cm² or 42.9 cm².

(c) Solve the following inequation and represent the solution set on a number line. [3]
-8\(\frac{1}{2}\) < – \(\frac{1}{2}\) – 4x ≤ 7\(\frac{1}{2}\), x ∈ I
Solution:
Given inequation is :
– 8\(\frac{1}{2}\) < – \(\frac{1}{2}\) – 4x ≤ 7\(\frac{1}{2}\), x ∈ I
⇒ –\(\frac{17}{2}\) < –\(\frac{1}{2}\) – 4x ≤ \(\frac{15}{2}\)
⇒ \(\frac{1}{2}\) – \(\frac{17}{2}\) < -4x ≤ \(\frac{1}{2}\) + \(\frac{15}{2}\)
⇒ – 8 < – 4x ≤ 8 ⇒ \(\frac{-8}{-4}\) > \(\frac{-4x}{-4}\) ≥ \(\frac{8}{-4}\)
2 > x ≥ -2
Solution set on number line

Section – B (40 Marks)
Attempt any four questions from this section.

Question 5.
(a) Given matrix B = \(\). Find the matrix X if, X = B² – 4B. [4]
Hence, solve for a and b given X\(\left[\begin{array}{l}
a \\
b
\end{array}\right]=\left[\begin{array}{l}
5 \\
50
\end{array}\right]\)
Solution:

(b) How much should a man invest in ₹ 50 shares selling at ₹ 60 to obtain an income of ₹ 450, if the rate of dividend declared is 10%. Also, find his yield percent, to the nearest whole number. [3]
Solution:
Market value of a share = ₹ 60
Face value of a share = ₹ 50
Rate of dividend = 10%
Total income = ₹ 450
If income is ₹ 5, then investment = ₹ 60
If income is ₹ 1, then investment = ₹ \(\frac{60}{5}\)
= ₹ 12
If income is 1450, then investment = ₹ 12 × 450
= ₹ 5400
Thus, total investment is ₹ 5400
Yield percent = ₹ \(\frac{450}{5400}\) × 100
= 8.33
= 8 (to the nearest whole number)

(c) Sixteen cards are labelled as a, b, c … m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is: [3]
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median ?
Solution:
Total number of cards = 16
(i) Number of vowels = 4 (a, e, i, o)
Probability = \(\frac{4}{16}\) = \(\frac{1}{4}\)

(ii) Number of consonant = 16 – 4 = 12
Probability = \(\frac{12}{16}\) = \(\frac{3}{4}\)

(iii) Probability (none of the letters of the word median)
= \(\frac{10}{16}\) = \(\frac{5}{8}\)

Question 6.
(a) Using a ruler and a compass construct a triangle ABC in which AB = 7 cm, ∠CAB = 60° and AC = 5 cm. Construct the locus of: [4]
(i) points equidistant from AB and AC.
(ii) points equidistant from BA and BC.
Hence, construct a circle touching the three sides of the triangle internally.
Solution:
Steps of Construction :

1. Draw a line segment AB = 7 cm.
2. At A construct an angle of 60° such that AC = 5 cm.
3. Join BC to get ∆ABC.
4. Draw angle bisector of ∠BAC, which is the required locus of the points equidistant from AB and AC.
5. Draw angle bisector of ∠ABC, which is the required locus of the points equidistant from BA and BC.
6. Let the two angle bisectors intersect each other in I.
7. Through I, draw ID ⊥ AB. With 1 as centre and radius = ID draw a circle which touches all the sides of the ∆ABC internally.

(b) A conical tent is to accommodate 77 persons. Each person must have 16 m³ of air to breathe. Given the radius of the tent as 7m, find the height of the tent and also its curved surface area. [3]
Solution:
Total number of persons accommodated = 77
Volume of air required for each person = 16 m³
Volume of the conical tent = 77 × 16
= 1232 m³
Radius of the tent = 7 m
Let h be the height of the conical tent
∴ \(\frac{1}{3}\)πr²h = 1232
∴ \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 7 × h = 1232
⇒ h = \(\frac{1232 \times 7 \times 3}{22 \times 7 \times 7}\)
= 24 m
Now, Slant height (l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{7^2+24^2}\)
= \(\sqrt{49+576}\)
= \(\sqrt{625}\) = 25 m
∴ Curved surface area = πrl
= \(\frac{22}{7}\) × 7 × 25
= 550 m²

(c) If = \(\frac{7 m+2 n}{7 m-2 n}=\frac{5}{3}\), use properties of proportion to find : [3]
(i) m : n
(ii) \(\frac{m^2+n^2}{m^2-n^2}\)
Solution:
\(\frac{7 m+2 n}{7 m-2 n}=\frac{5}{3}\)
Using componendo and dividendo, we have

[Using componendo and dividendo]

Question 7.
(a) A page from a savings bank account passbook is given below: [5]

(i) Calculate the interest for the 6 months from January to June 2016, at 6% per annum.
(ii) If the account is closed on 1st July 2016, find the amount received by the account holder.
Solution:
Minimum balance for the month Jan., 2016 = ₹ 5600
Minimum balance for the month Feb., 2016 = ₹ 4100
Minimum balance for the month Mar., 2016 = ₹ 4100
Minimum balance for the month Apr., 2016 = ₹ 2000
Minimum balance for the month May, 2016 = ₹ 8500
Minimum balance for the month June, 2016 = ₹ 10000
Total = ₹ 34300
Principal = ₹ 34300
Rate = 6% p.a.
Time = \(\frac{1}{12}\) year
∴ I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\frac{34300 \times 6 \times 1}{100 \times 12}\)
= ₹ 171.50
Amount on 1st July 2016 = ₹ (34300 + 171.50)
= ₹ 34471.50

(b) Use a graph paper for this question (Take 2 cm =1 unit on both x and y axis) [5]
(i) Plot the following points :
A(0, 4), B(2, 3), C(1, 1) andD(2, 0).
(ii) Reflect points B, C, D on the y-axis and write down their coordinates. Name the images as B’, C’, D’ respectively.
(iii) Join the points A, B, C, D, D’, C’,B’ and A in order, so as to form a closed figure. Write down the equation of the line of symmetry of the figure formed.
Solution:
(i) On graph
(ii) B'(- 2, 3), C'(-1, 1), D'(-2, 0)
(iii) Equation of the line of symmetry is x = 0

Question 8.
(a) Calculate the mean of the following distribution using step deviation method. [4]

Solution:

Let assumed (a) = 35
∴ Mean = a + \(\frac{\sum f_i u_i}{\sum f_i}\) × h
= 35 – \(\frac{37}{100}\) × 10 = 35 – 3.7 = 31.3

(b) In the given figure PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that: [3]
(i) BD is a diameter of the circle.
(ii) ABC is an isosceles triangle.

Solution:
Here, PQ is a tangent to the circle at A.
∠BAQ = 30°. Since AB is angle bisector of ∠CAQ.
∴ ∠CAB = ∠BAQ = 30°
Again, ∠PAC = 180° – ∠CAQ
= 180° – 30° – 30°
= 120°

Also, AD is angle bisector of ∠PAC
∴ ∠PAD = ∠CAD = 60°
Since angles in the corresponding alternate segment are equal
∴ ∠ADB = ∠BAQ = 30°
and ∠DBA = ∠PAD = 60°
Also, angles in same segment are equal
∴ ∠DCA = ∠DBA = 60°
and ∠ACB = ∠ADB = 30°
Now, ∠DCB = ∠DCA + ∠ACB
= 60° + 30° = 90°
We know that angle in a semi-circle is right angle.
Thus, BD is a diameter of the circle.
In AACB, ∠ACB = ∠CAB = 30°
Hence, ∆ABC is an isosceles triangle.

(c) The printed price of an air conditioner is ₹ 45000/-. The wholesaler allows a discount of 10% to the shopkeeper. The shopkeeper sells the article to the customer at a discount of 5% of the marked price. Sales tax (under VAT) is charged at the rate of 12% at every stage. Find: [3]
(i) VAT paid by the shopkeeper to the government.
(ii) The total amount paid by the customer inclusive of tax. [3]
Solution:
List price of air conditioner = ₹ 45000
Discount = 10%
Total discount = ₹ \(\frac{10}{100}\) × 45000
= ₹ 4500
.’. Selling price = ₹ (45000 – 4500)
= ₹ 40500
VAT paid = ₹ \(\frac{12}{100}\) × 40500
= ₹ 4860
Marked price of air conditioner = ₹ 45000
Discount = 5%
Total discount = ₹ \(\frac{5}{100}\) × 45000
= ₹ 2250
Selling price = ₹ (45000 – 2250)
= ₹ 42750
VAT paid = ₹ \(\frac{12}{100}\) × 42750
= ₹ 5130
Thus, VAT paid by the shopkeeper to the government
= ₹ (5130 – 4860) = ₹ 270
Total amount paid by the customer = ₹ (42750 + 5130)
= ₹ 47880

Question 9.
(a) In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of: [4]
(i) ∠BCD
(ii) ∠BOD
(iii) ∠OBD

Solution:
Here, ∠DAE = 70°
∴ ∠BAD = 180° – ∠DAE [a linear pair]
= 180° – 70° = 110°
ABCD is a cyclic quadrilateral
∴ ∠BCD + ∠BAD = 180°
∠BCD + 110° = 180°
⇒ ∠BCD = 180° – 110°
= 70°
Since angle subtended by an arc at the centre of a circle is twice the angle subtended at the remaining part of the circle.

∴ ∠BOD = 2∠BCD = 2 × 70° = 140°
In ∆OBD, OB = OD = radii of same circle.
∴ ∠OBD = ∠ODB
Thus, ∠OBD = \(\frac{1}{2}\) (180° – ∠BOD)
= \(\frac{1}{2}\) (180° – 140°) = \(\frac{1}{2}\) × 40° = 20°

(b) A(-1, 3), B(4, 2) and C(3, -2) are the vertices of a triangle. [3]
(i) Find the coordinates of the centroid G of the triangle.
(ii) Find the equation of the line through G and parallel to AC.
Solution:
A(- 1, 3), B(4, 2) and C(3, -2) are the vertices of ∆ABC.
∴ Coordinates of the centroid G of the ∆ABC are :
G\(\left(\frac{-1+4+3}{3}, \frac{3+2-2}{3}\right)\)
G(2, 1)
Here, line 7’ is drawn through G(2, 1) and parallel to the line AC.
∴ Slope of the line l = Slope of the line AC
= \(\frac{-2-3}{3+1}\) = \(\frac{-5}{4}\)

Thus, the equation of the required line l is :
y – 1 = \(\left(\frac{-5}{4}\right)\) (x – 2)
4y – 4 = -5x + 10
5x + 4y = 14

(c) Prove that: \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\) = tan θ [3]
Solution:
L.H.S.

Question 10.
(a) The sum ofthe ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages. [4]
Solution:
Let Vivek’s age be x years
∴ Amit’s age = 47 – x
Also, product of their ages = 550
∴ x(47 – x) = 550
47x – x² = 550
⇒ x² – 47x + 550 = 0
⇒ x² – 25x – 22x + 550 = 0
⇒ x(x – 25) – 22(x – 25) = 0
⇒ (x – 25) (x – 22) = 0
⇒ x = 25
or x = 22
Since Vivek is elder brother of Amit.
Hence, age of Vivek is 25 years and age of Amit is 22 years.

(b) The daily wages of 80 workers in a project are given below: [6]

Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = ₹ 50 on x-axis and 2 cm = 10 workers on y-axis. Use your ogive to estimate:
(i) the median wage of the workers.
(ii) the Ipwer quartile wage of workers.
(iii) the number of workers who earn more than ₹ 625 daily.
Solution:
The cumulative frequency distribution for the given data is:

Plot the points (450, 2), (500, 8), (550, 20), (600, 38), (650, 62), (700, 75), (750, 80).
Join them free hand to get the required ogive.
Now, from the graph, we obtain :
(i) median wage of the workers = ₹ 605
(ii) lower quartile wage of workers = ₹ 550
(iii) Number of workers who earn more than ₹ 625 daily
= 80 – 50
= 30.

Question 11.
(a) The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number. [4]
Solution:
Let PQ be the light house of height 60 m, A and B are the two ships on the opposite sides of the light house, such that: ∠PAQ = 60°, ∠PBQ = 45°

In rt.∠ed ∆PQB, we have
\(\frac{\mathrm{PQ}}{\mathrm{QB}}\) = tan 45° = 1
⇒ PQ = QB = 60 m
In rt. ∠ed ∆PQA, we have
\(\frac{\mathrm{PQ}}{\mathrm{AQ}}\) = tan 60° = \(\sqrt{3}\)
AQ = \(\frac{P Q}{\sqrt{3}}=\frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=20 \sqrt{3}\)
= 20(1.73) = 34.6 m
Now, AB = AQ + QB = 60+ 34.6 = 94.6 m
Hence, the distance between the two ships is 95 m (nearest to whole number).

(b) PQR is a triangle. S is a point on the side QR of ∆PQR such that ∠PSR = ∠QPR. [3]
Given QP = 8 cm, PR = 6 cm and SR = 3 cm.

(i) Prove ∆PQR ~ ∆SPR
(ii) Find the length of QR and PS
(iii) \(\frac{\text { area of } \triangle \mathrm{PQR}}{\text { area of } \triangle \mathrm{SPR}}\)
Solution:
Given : QP = 8 cm, PR = 6 cm and SR = 3 cm
In ∆PQR and ∆SPR
∠QPR = ∠PSR (given)
∠QRP = ∠SRP (common)
∴ ∆PQR ~ ∆SPR (by AA similarity rule)
Since APQR-ASPR
∴ \(\frac{\mathrm{QR}}{\mathrm{PR}}\) = \(\frac{\mathrm{PR}}{\mathrm{SR}}\)
⇒ QR = \(\frac{P R \times P R}{S R}=\frac{6 \times 6}{3}\) = 12 cm
And \(\frac{\mathrm{PS}}{\mathrm{QP}}\) = \(\frac{\mathrm{SR}}{\mathrm{PR}}\)
⇒ PS = \(\frac{\mathrm{SR} \times \mathrm{QP}}{\mathrm{PR}}=\frac{3 \times 8}{6}\) = 4 cm
\(\frac{\text { area of } \triangle \mathrm{PQR}}{\text { area of } \triangle \mathrm{SPR}}=\frac{\mathrm{QP}^2}{\mathrm{PS}^2}=\frac{8^2}{4^2}=\frac{64}{16}=\frac{4}{1}\)

(c) Mr. Richard has a recurring deposit account in a bank for 3 years at 7.5% p.a. simple interest. If he gets ₹ 8325 as interest at the time of maturity, find : [3]
(i) The monthly deposit
(ii) The maturity value.
Solution:
Let the monthly deposit be ₹ x
Time = 3 years or 36 months, R = 7.5%, Interest = ₹ 8325.
∴ x × \(\frac{36(36+1)}{2}\) × \(\frac{1}{12}\) × \(\frac{7.5}{100}\) = 8325
x = \(\frac{8325 \times 100 \times 2 \times 12}{36 \times 37 \times 7.5}\)
= ₹ 2000
Thus, the monthly deposit is ₹ 2000
The maturity value = ₹ 36 × 2000 + ₹ 8325
= ₹ 72000 + ₹ 8325
= ₹ 80325.

Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2014 is the best way to boost your preparation for the board exams.

ICSE Class 10 Maths Question Paper 2014 Solved

Time Allowed: 2 1/2 hours
Max. Marks: 80

Answers to this paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper. The time given at the head of this Paper is the time allowed for writing the answers.

Attempt all questions from Section A and any four questions from Section B. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of questions are given in brackets [ ]
Mathematical tables are provided.

Section-A (40 Marks)
(Attempt all questions from this section)

Question 1.
(a) Ranbir borrows ₹ 20,000 at 12% per annum compound interest. If he repays ₹ 8400 at the end of the first year and ₹ 9680 at the end of the second year, find the amount of loan outstanding at the beginning of the third year. [3]
(b) Find the values of x, which satisfy the inequation
-2\(\frac{5}{6}\) < \(\frac{1}{2}\) – \(\frac{2 x}{3}\) ≤ 2, x ∈ W. Graph the solution set on the number line. [3]
(c) A die has 6 faces marked by the given numbers as shown below:

The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than -3.
(iii) the smallest integer.
Solution :
(a) Amount borrowed (Loan) = ₹ 20000
Rate of interest = 12% p.a.
Interest due after one year = ₹ \(\frac{20000 \times 12 \times 1}{100}\)
= ₹ 2400
Amount of loan outstanding after one year
= ₹ (20000 + 2400)
= ₹ 22400
Total payment at the end of first year = ₹ 8400
Principal for second year = ₹ (22400 – 8400)
= ₹ 14000
Interest due after second year = ₹ \(\frac{14000 \times 12 \times 1}{100}\) = ₹ 1680
Amount of loan outstanding after second year
= ₹ (14000 + 1680) = ₹ 15680
Total payment at the end of second year = ₹ 9680
Thus, amount of loan outstanding at the beginning of the third year = ₹ (15680 – 9680) = ₹ 6000.

(b) Given inequation is -2\(\frac{5}{6}\) < \(\frac{1}{2}\) – \(\frac{2 x}{3}\) ≤ 2, x ∈ W
⇒ –\(\frac{17}{6}\) < \(\frac{1}{2}\) – \(\frac{2 x}{3}\) ≤ 2
⇒ –\(\frac{17}{6}\) – \(\frac{1}{2}\) < –\(\frac{2 x}{3}\) ≤ 2 – \(\frac{1}{2}\)
⇒ –\(\frac{20}{6}\) < –\(\frac{2 x}{3}\) ≤ \(\frac{3}{2}\)
⇒ -20 < -4x ≤ 9 ⇒ 5 > x ≥ –\(\frac{9}{4}\)
Hence, the required solution set on number line is :

(c)
(i) Here, we have only three positive integers (1, 2, 3)
∴ Probability (getting a positive integer) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

(ii) Now, we have five integers greater than – 3 i.e., -2, -1, 1, 2, 3.
∴ Probability (getting a integer greater than – 3) = \(\frac{5}{6}\)

(iii) Here, we have only one smallest integer i.e., -3
∴ Probability (getting smallest integer) = \(\frac{1}{6}\)

Question 2.
(a) Find x, y if \(\left[\begin{array}{cc}
-2 & 0 \\
3 & 1
\end{array}\right]\left[\begin{array}{c}
-1 \\
2 x
\end{array}\right]\) + 3\(\left[\begin{array}{c}
-2 \\
1
\end{array}\right]\) = 2\(\left[\begin{array}{l}
y \\
3
\end{array}\right]\) [3]

(b) Shahrukh opened a Recurring Deposit Account in a bank and deposited ₹ 800 per month for 1 1/2 years. If he received ₹ 15,084 at the time of maturity, find the rate of interest per annum. [3]

(c) Calculate the ratio in which the line joining A(-4, 2) and B (3, 6) is divided by point P (x, 3). Also find
(i) x
(ii) Length of AP. [4]
Solution:
(a)

Comparing the corresponding entries of equal matrix, we obtain
2y = -4 and 2x = 6
y = -2 and x = 3

(b)
Monthly deposit = ₹ 800
Period of deposit (n) = 1\(\frac{1}{2}\) years or 18 months
Total deposited amount = ₹ 800 × 18
= ₹ 14400
Let rate of interest be r% p.a.
Equivalent principal for one month
= \(\frac{n(n+1)}{2}\) × monthly deposit
= ₹ \(\left(\frac{18 \times 19}{2} \times 800\right)\) = ₹ 136800
Interest paid by the bank = \(\frac{136800}{100 \times 12}\) = 114r
Now, total deposited amount + interest amount = amount on maturity
14400 + 114r = 15084
⇒ 114 r = 15084 – 14400
⇒ 114r = 684
⇒ r = \(\frac{684}{114}\) = 6%
Hence, the rate of interest per annum is 6%

(c) Let the line joining A(-4, 2) and B (3, 6) is divided by P(x, 3) in the ratio k : 1

Hence, the required ratio is 1 : 3, x = –\(\frac{9}{4}\) and length of AP is \(\frac{\sqrt{65}}{2}\) units.

Question 3.
(a) Without using trigonometric tables, evaluate
sin² 34° + sin² 56° + 2 tan 18° tan 72° – cot² 30°. [3]
(b) Using the Remainder and Factor Theorem, factorise the following polynomial:
x³ + 10x² – 37x + 26. [3]
(c) In the figure given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle.
(Take π = 22/7) [4]

Solution:
(a) sin² 34° + sin² 56° + 2 tan 18° tan 12° – cot² 30°
= sin² 34° + sin² (90° – 34°) + 2 tan 18°. tan (90° – 18°) – (\(\sqrt{3}\))²
= sin² 34° + cos² 34° + 2 tan 18°. cot 18° – 3
[∵ sin (90°- θ) = cos θ, tan (90° – θ) = cot θ]
= 1 + 2 – 3 = 0

(b) Let p(x) = x³ + 10x² – 37x + 26
Now, factors of 26 are ± 2 and ±13
Put x = 2, we obtain p(2) = (2)³ + 10(2)² – 37(2) + 26
= 74 – 74 = 0
⇒ (x – 2) is a factor of p(x)
By long division, we obtain

Thus, x³ + 10x² – 37x + 26 = (x – 2)(x² + 12x – 13)
= (x – 2) (x² + 13x – x – 13)
= (x – 2) (x – 1) (x + 13)

(c) Area of rectangle ABCD = 14 × 7 = 98 cm²
Area of quarter circle BFEC = \(\frac{1}{4}\) × \(\frac{22}{7}\) × 7 × 7
= 38.5 cm²
Area of semicircle DGE = \(\frac{1}{2}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\)
= 19.25 cm²
Area of the remaining piece of the rectangle = 98 – 38.5 – 19.25
= 40.25 cm²

Question 4.
(a) The numbers 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x. [3]

(b) In the figure, ZDBC 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC [3]

(c) Use a graph paper to answer the following questions. (Take 2 cm 1 unit on both axis)
(i) Plot the points A (-4, 2) and B (2, 4)
(ii) A’ is the image of A when reflected in the y-axis. Plot it on the graph paper and write the coordinates of A’.
(iii) B’ is the image of B when reflected in the line AA’. Write the coordinates of B’.
(iv) Write the geometric name of the figure ABA’B’.
(v) Name a line of symmetry of the figure formed. [4]
Solution:
(a) Here, the given observations are 6, 8, 10, 12, 13 and x.
Now, mean = \(\frac{6+8+10+12+13+x}{6}\) = \(\frac{49+x}{6}\)
(6/2)^th observation
Median = \(\frac{+(6 / 2+1)^{\text {th }} \text { observation }}{2}\)
= \(\frac{3^{\text {th }} \text { observation }+4^{\text {th }} \text { Obsrvation }}{2}\)
= \(\frac{10+12}{2}\) = 11
According to given condition, we have \(\frac{49+x}{6}\) = 11
⇒ 49 + x = 66
⇒ x = 66 – 49 = 17

(b) Since BD is a diameter of the circle
∴ ∠BCD = 90°
∠DBC = 58° (given)
∠BDC = 90° – 58° = 32°
Also, BECD is a cyclic quadrilateral.
∴ ∠BEC + ∠BDC = 180°
⇒ ∠BEC + 32° = 180°
⇒ ∠BEC = 180° – 32° = 148°
Again, ∠BAC = ∠BDC = 32
[∠s is the same segment]
Hence, ∠BDC = 32°, ∠BEC = 148° and
∠BAC = 32°.

(c) (i) Points A(-4, 2) and B(2, 4) are plotted in the graph.
(ii) A’ (4, 2)
(iii) B’ (2, 0)
(iv) Kite
(v) AA’ is the line of symmetry of the figure ABA’B’.

Section—B (40 Marks)
(Attempt any four questions from this Section)

Question 5.
(a) A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being ₹ 18,000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax is 8%, find:
(i) the VAT paid by the shopkeeper.
(ii) the total amount that the consumer pays for the washing machine. [3]

(b) If \(\frac{x^2+y^2}{x^2-y^2}\) = \(\frac{17}{8}\) then find the value of:
(i) x : y
(ii) \(\frac{x^3+y^3}{x^3-y^3}\)

(c) In ΔABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ΔACD is similar to ΔBCA.
(ii) Find BC and CD.
(iii) Find area of ΔACD : area of ΔABC. [4]

Solution.
(a) Printed price of washing machine = ₹ 18000
Rate of discount = 20%
Sale price of washing machine of wholesaler
= ₹\(\left(18000 \times \frac{80}{100}\right)\) = ₹ 14400
VAT@8% = ₹\(\left(\frac{8}{100} \times 14400\right)\) = ₹ 1152
Sale price of washing machine for shopkeeper
= ₹\(\left(18000 \times \frac{90}{100}\right)\) = ₹ 16200
VAT@ 8% = ₹ \(\left(\frac{8}{100} \times 16200\right)\) = ₹ 1296
Thus,VAT paid by the shopkeeper
= ₹ (1296 – 1152) = ₹ 144
Total amount that the consumer pays for the washing machine = ₹ (16200 + 1296) = ₹ 17496

(b) Given that: \(\frac{x^2+y^2}{x^2-y^2}\) = \(\frac{17}{8}\)
By componendo and dividendo, we obtain

(c) (i) In ΔACD and ΔBCA, we have
∠ACD = ∠BCA [common]
∠DAC = ∠ABC [given)
⇒ ΔACD \(\sim\) ΔBCA
[by AA similarity axiom]

(ii) Since corresponding sides of similar triangles are proportional

(iii)

Question 6.
(a) Find the value of ‘a’ for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line. [3]

(b) Salman invests a sum of money in ₹ 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is ₹ 600, calculate:
(i) the number of shares he bought.
(ii) his total investment.
(iii) the rate of return on his investment. [3]

(c) The surface area of a solid metallic sphere is 2464 cm². It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate:
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = 22/7) [4]
Solution:
(a) Since the points A(u, 3), B(2,1) and C(5, a) are collinear
∴ ar (Δ ABC) = 0
⇒ \(\frac{1}{2}\)|a(1 – a) + 2(a – 3) + 5(3 – 1)| = 0
⇒ a – a² + 2a – 6 + 15 – 5 = 0
⇒ 3a – a² + 4 = 0
a² – 3a – 4 = 0
(a – 4) (a + 1) = 0
a = 4
a = -1
Thus, points are A(4, 3) , B (2, 1) and C (5, 4) or A(-1, 3), B(2, 1) and C(5, -1).
Now, equation of line AC is
y – 3 = \(\frac{1-3}{2-4}\)(x – 4)
⇒ y – 3 = \(\frac{-2}{-2}\)(x – 4)
⇒ y – 3 = x – 4
⇒ x – y – 1 = 0

(b) Dividend on 1 share of ₹ 50 = 15% of ₹ 50 = ₹ \(\frac{15}{100}\) × 50
= ₹ 7.5
Annual dividend = ₹ 600
∴ Number of shares bought = \(\frac{600}{7.5}\) = 80
His total investment = ₹ (60 × 80)
[∵ a share of ₹ 50 is quoted at 20% premium]
= ₹ 4800
Rate of return on investment = \(\frac{600}{4800}\) × 100
= 12.5%

(c) Surface area of metallic sphere = 2464 cm²
4πr² = 2464
r² = \(\frac{2464 \times 7}{4 \times 22}\) = 196
∴ r = 14 cm
Let the total number of cones recasted be n, then n × volume of a cone = volume of solid metallic sphere
n × \(\frac{1}{3} \pi r^2\) × h = \(\frac{4}{3} \pi r^3\)nri 3 3
n × 3.5 × 3.5 × 7 = 4 × 14 × 14 × 14
n = \(\frac{4 \times 14 \times 14 \times 14}{3.5 \times 3.5 \times 7}\)
= 128
Hence, the radius of the sphere is 14 cm and number of cones recasted is 128.

Question 7.
(a) Calculate the mean of the distribution given below using the short cut method.

[3]
(b) In the figure given below, diameter AB and chord CD of a circle meet at R PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find:
(i) AB.
(ii) the length of tangent PT. [3]

(c) Let A = \(\left[\begin{array}{cc}
2 & 1 \\
0 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & -2
\end{array}\right]\), and C = \(\left[\begin{array}{cc}
-3 & 2 \\
-1 & 4
\end{array}\right]\). Find A² + AC – 5B.
Solution:
(a)

Let a (assumed mean) = 45.5
Mean (\(\overline{\mathrm{X}}\)) = a + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × h
= 45.5 + \(\frac{7}{50}\) × 10
= 45.5 + 1.4 = 46.9

(b) Here, two chords, AB and CD,
when produced, meet at P
∴ PA × PB = PC × PD
= PA × 4 = (7.8 + 5) × 5
⇒ PA × 4 = 64
⇒ PA = 16
⇒ AB + PB = 16
⇒ AB + 4 = 16
⇒ AB = 12 cm

Now, ABP is secant line and PT be a tangent
∴ PT² = PA × PB
⇒ PT² = 16 × 4
⇒ PT = \(\sqrt{16 \times 4}\) = 8 cm

(c)
A² + AC – 5B

Question 8.
(a) The compound interest, calculated yearly, on a certain sum of money for the second year is ₹ 1320 and for the third year is ₹ 1452. Calculate the rate of interest and the original sum of money. [3]
(b) Construct a ΔABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle. [3]
(c) (use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:

Draw a histogram representing the above distribution and estimate the mode from the graph .[4]
Solution:
(a) Compound interest for second year = ₹ 1320
Compound interest for third year = ₹ 1452
S.I. on ₹ 1320 for one year = ₹ (1452 – 1320)
= ₹ 132
Principal = ₹ 1320,
Time = 1 year
∴ R = \(\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}\) = \(\frac{132 \times 100}{1320 \times 1}\) = 10%
Let principal be ₹ 100,Interest for first year
= \(\frac{100 \times 10 \times 1}{100}\) = ₹ 10
Interest for second year = \(\frac{110 \times 10 \times 1}{100}\) = ₹ 11
If interest for second year is ₹ 11, then, principal = ₹ 100
If interest for second year is ₹ 1320, then, principal = ₹ \(\left(\frac{100}{11} \times 1320\right)\) = ₹ 12000
Hence, the rate of interest is 10% p.a. and the original sum of money is ₹ 12000.

(b) Steps of construction :
(i) Construct ΔABC, such that AB = 5.5 cm, BC = 6.5 cm and AC = 5 cm.
(ii) Draw the angle bisectors BO and CO of ∠B and ∠C and let they intersect in O.
(iii) Draw OE ⊥ BC, intersecting BC is E.
(iv) With O as centre and OE as radius draw the required incricle of ΔABC. Radius of incircle = 1.5 cm

(c)Required histogram is as given below :

Clearly, from the graph estimated mode is ₹ 22.

Question 9.
(a) If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x. [3]
(b) Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(x – 1)² – 3x + 4 = 0
(c) A page from the savings bank account of Priyanka is given below:

If the interest earned by Priyanka for the period ending September, 2006 is ₹ 175, find the rate of interest. [4]
Solution:
(a) Duplicate ratio of 4 : 9 is (x – 9): (3 + 6).
∴ \(\frac{4^2}{9^2}\) = \(\frac{x-9}{3 x+6}\)
⇒ \(\frac{16}{81}\) = \(\frac{x-9}{3 x+6}\)
⇒ 48x + 96 = 81x – 729
⇒ 81x – 48x = 729 + 96
⇒ 33x = 825
⇒ x = \(\frac{825}{33}\) = 25

(b) Given quadratic equation is
(x – 1)² – 3x + 4 = 0
x² – 2x + 1 – 3x + 4 = 0
x² – 5x + 5 = 0
⇒ x = \(\) = \(\)
= \(\frac{5 \pm 2.236}{2}\) = \(\frac{7.236}{2}\), \(\frac{2.764}{2}\)
= 3.618, 1.382
x = 3.62,1.38 upto two significant figures

(c) Minimum qualifying balance for April, 2006 = ₹ 6000
May, 2006 = ₹ 7000
June, 2006 = ₹ 10000
July, 2006 = ₹ 6000
August, 2006 = ₹ 6000
September, 2006 = ₹ 7000
Total = ₹ 42000
Interest = ₹ 175, Time = \(\frac{1}{12}\) year
∴ R = \(\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}\) = \(\frac{175 \times 100 \times 12}{42000 \times 1}\) = 5% p.a

Question 10.
(a) A two digit positive number is such that the product of its digits is 6. 1f 9 is added to the number, the digits interchange their places. find the number. [4]
(b) The marks obtained by 100 students in a Mathematics test are given below :

Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm = 10 units on both axis.
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students who did not pass in the test if the pass percentage was 35. [6]
Solution:
(a) Let one’s place digit be x
∴ Ten’s place digit = \(\frac{6}{x}\)
Original number = 10 × \(\frac{6}{x}\) + x = \(\frac{60}{x}\) + x
Number obtained by interchanging their digits
= 10x + \(\frac{6}{x}\)
According to question, we obtain \(\frac{6}{x}\) + x + 9 = 10x + \(\frac{6}{x}\)
⇒ 60 + x² + 9x = 10x² + 6
⇒ 9x² – 9x – 54 = 0
⇒ x² – x – 6 = 0
⇒ (x – 3) (x + 2) = 0
⇒ x = 3 or x = -2
[Rejecting -ve value, ∵ given no. is positive]
Thus, one’s place digit is 3 and ten’s place digit is \(\frac{6}{3}\)
i.e., 2.
Hence, the required number is 23.

(b) Cummulative frequency distribution table is as below:


Plot the points (0, 0), (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85),
(80, 91), (90, 96) and (100, 100). Join them free hand to get the required ogive. Now, from the above graph, we obtain
(i) median = 45
(ii) lower quartile = 32
(iii) number of students who obtained more than 85% marks in the test = 100 – 94 = 6
(iv) number of students who did not pass in the test if the pass percentage was 35 = 30

Question 11.
(a) In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the:
(i) radius of the circle.
(ii) length of chord CD. [3]

(b) Prove the identity : (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ. [3]
(c) An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. [4]
Solution:
(a) We know that perpendicular drawn from the centre to the chord, bisects the chord.
∴ AM = MB = 12 cm and
CN = ND = \(\frac{1}{2}\) CD
OM = 5 cm, ON = 12 cm
Join AO and CO.
In rt. ∠ed ΔAMO, by Pythagoras Theorem, we have

AO² = AM² + OM² = 12² + 5²
= 144 + 25 = 169
AO = 13 cm
Thus, radius of the circle is 13 cm.
Now, in rt. ∠ed ΔCNO, we obtain
CN² = CO² – ON² = 13² – 12² = 169 – 144 = 25 CN = 5 cm
∴ CD = CN + ND
= CN + CN
= 2CN = 2 × 5 = 10 cm

(b) L.H.S. = (sin θ + cos θ) (tan θ + cot θ)
= (sin θ + cos θ) \(\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)\)
= (sin θ + cos θ) \(\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\right)\)
= \(\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}\) = \(\frac{\sin \theta}{\cos \theta \sin \theta}\) + \(\frac{\cos \theta}{\cos \theta \sin \theta}\)
= sec θ + cosec θ
= R.H.S.

(c) Let AB be the altitude of an aeroplane, P and Q are the positions of two boats on either sides of AB, such that AB = 250 m, ∠P = 45°, ∠Q = 60°

In rt. ∠ed ΔPBA, we have
\(\frac{A B}{P B}\) = tan 45°
⇒ \(\frac{250}{\mathrm{~PB}}\) = 1 ⇒ PB = 250 m
In rt. ∠ed ΔQBA, we have
\(\frac{A B}{B Q}\) = tan 60°
⇒ \(\frac{250}{\mathrm{BQ}}\) = \(\sqrt{3}\)
⇒ BQ = \(\frac{250}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{250 \sqrt{3}}{3}\) m

Thus, PQ (width of the river) = PB + BQ
= 250 + \(\frac{250 \sqrt{3}}{3}\)
= \(\left(\frac{750+250 \sqrt{3}}{3}\right) \mathrm{m}\)
= 394.33 m

Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2013 is the best way to boost your preparation for the board exams.

ICSE Class 10 Maths Question Paper 2013 Solved

Time Allowed: 2 1/2 hours
Max. Marks : 80

Answers to this paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper. The time given at the head of this Paper is the time allowed for writing the answers.

Attempt all questions from Section A and any four questions from Section B. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of questions are given in brackets [ ]
Mathematical tables are provided.

Section-A (40 Marks)
(Attempt all questions from this section)

Question 1.
(a) Given A = \(\left[\begin{array}{cc}
2 & -6 \\
2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-3 & 2 \\
4 & 0
\end{array}\right]\), C = \(\left[\begin{array}{ll}
4 & 0 \\
0 & 2
\end{array}\right]\)
Find the matrix X such that A + 2X = 2B + C. [3]
(b) At what rate % p.a. will a sum of ₹ 4000 yield ₹ 1324 as compound interest in 3 years? [3]
Answer:
(c) The median of the following observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean. [4]
Solution:
(a) Given that A + 2X = 2B + C

(b) Here, P = ₹ 4000, C.I. = ₹ 1324, T = 3 years r

⇒ \(\frac{r}{100}\) = \(\frac{1}{10}\)
⇒ r = 10%

(c) Given observations in ascending order are :
11, 12, 14, x – 2, x + 4, x + 9, 32, 38 and 47
Here, number of observations are 9 (odd number)
∴ Median = \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\left(\frac{9+1}{2}\right)^{t h}\) term = 5^th term
⇒ x + 4 = 24
⇒ x = 20 [∵ Median = 24(given)]
Thus, the observations are 11, 12, 14, 18, 24, 29, 32, 38, 47.
Now, mean = \(\frac{11+12+14+18+24+29+32+38+47}{9}\)
= \(\frac{225}{9}\) = 25

Question 2.
(a) What number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional? [3]
(b) If (x – 2) is a factor of the expression 2x³ + ax² + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b. [3]
(c) Draw a histogram from the following frequency distribution and find the mode from the graph : [4]

Solution:
(a) Let x be added to each number 6, 15, 20 and 43 to make these numbers proportional.
∴ (6 + x) : (15 + x) = (20 + x) : (43 + x)
⇒ \(\frac{6+x}{15+x}\) = \(\frac{20+x}{43+x}\)
⇒ (6 + x) (43 + x) = (15 + x) (20 + x)
⇒ 258 + 6x + 43x + x² = 300 + 15x + 20x + x²
⇒ 258 + 49x = 300 + 35x
⇒ 49x – 35x = 300 – 258
⇒ 14x = 42
⇒ x = 3

(b) Since (x – 2) is a factor of 2x³ + ax² + bx – 14.
⇒ 2(2)³ + a(2)² + 6(2) – 14 = 0
⇒ 16 + 4a + 2b – 14 = 0
⇒ 4a + 2b = – 2
⇒ 2a + b = -1 …..(i)
Also, when 2x³ + ax² + bx – 14 is divided by x – 3, it leaves remainder 52
⇒ 2(3)³ + a(3)² + 6(3) – 14 = 52
⇒ 54 + 9a + 36 = 52 + 14
⇒ 9a + 3b = 66 – 54
⇒ 9a + 3b = 12
⇒ 3a + b = 4 …(ii)
Subtracting (i) from (ii), we have
a = 5
From (i), we have
2(5) + b = – 1
⇒ b = -11
Hence, the required values of a and b are a = 5 and 6 = – 11.

(c) Required histogram from the given frequency distribution is:
Now, in the highest rectangle, draw two straight lines AC and BD from the comers of the rectangles on either ‘sides of the highest rectangle to the opposite comers of the highest rectangle. Let they intersect in L. Draw LM ⊥ OX. The abscissa of the point M represents 14. Hence, the required mode is 14.

Question 3.
(a) Without using tables evaluate 3 cos 80°.
Cosec 10° + 2 sin 59° sec 31°. [3]
(b) In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.

[4]
sin 59°. cosec 59°
(c) AB is a diameter of a circle with centre C = (-2, 5). IfA = (3, -7). Find
(i) the length of radius AC
(ii) the coordinates of B.
Solution:
(a) 3 cos 80° cosec 1o° + 2 sin 59° sec 31°
= 3 cos (90° – 10°) cosec 1o° + 2 sin (90° – 31°) sec 31°
= 3 sin 10° cosec 1o° + 2 cos 31° sec 31°
[∵ cos(90° – θ) = sinθ and sin(90° – θ) = cos θ]
= 3 sin 10°.\(\frac{1}{\sin 10^{\circ}}\) + 2 cos 31°.\(\frac{1}{\cos 31^{\circ}}\)
= 3 + 2 = 5

(b) (i) By using angle sum property of triangle, in ΔABD,
we have
∠ADB + ∠ABD + ∠BAD = 180°
∠ADB + 70° + 65° = 180°
∠ADB = 180° – 135°
= 45°
Now, ∠ADC = ∠ADB + ∠BDC
∠ADC = 45° + 45°
[∴ ∠BDC = 45° (given)]
∠ADC = 90°
Thus, AC is a diameter.
[∵ angle in a semicircle is 90°]

(ii) Also, ∠ACB = ∠ADB
[∵ angles of the same segment]
⇒ ∠ACB = ∠ADB = 45°
Thus, ∠ACB = 45°

(c) Let coordinates of B be (x, y)

∴ \(\frac{3+x}{2}\) = -2 and \(\frac{-7+y}{2}\) = 5
⇒ 3 + x = – 4and -7 + y = 10
⇒ x = -7 and y = 17
Thus, the coordinates of B are B (-7, 17)
Now, |AC| = \(\sqrt{(-2-3)^2+(5+7)^2}\)
= \(\sqrt{(-5)^2+(12)^2}\)
= \(\sqrt{25+144}\) = \(\sqrt{169}\) = 13
Hence, length of the radius AC is 13 units.

Question 4.
(a) Solve the following equation and calculate the answer correct to two decimal places:
x² – 5x – 100. [3]
(b) In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ΔABC – ΔDEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm. Calculate CD.
(iii) Find the ratio of the area of ΔABC : area of ΔDEC. [3]

(c) Using a graph paper, plot the points A (6, 4) and B (0, 4).
(i) Reflect A and B in the origin to get the images A’ and B’.
(ii) Write the co-ordinates of A’ and B’.
(iii) State the geometrical name for the figure ABA’B’.
(iv) Find its perimeter. [4]
Solution:
(a) x² – 5x – 10 = 0

(b)
Here, AB ⊥ BC, DE ⊥ BC

(i) Now, in ΔABC and ΔDEC
∠C = ∠C [common]
∠ABC = ∠DEC = 90° [given]
∴ ΔABC \(\sim\) ΔDEC [by AA similarity axiom]

(ii)
\(\frac{\mathrm{CD}}{\mathrm{CA}}\) = \(\frac{\mathrm{DE}}{\mathrm{AB}}\) [∵ sides of the similar triangles are proportional]
⇒ \(\frac{C D}{15}\) = \(\frac{4}{6}\)
⇒ CD = \(\frac{4 \times 15}{6}\) = 10 cm

(iii)
Since ΔABC \(\sim\) ΔDEC

(c)
(i)

(ii) Coordinates of A’ and B’ are A’ (-6, -4) and B’(0, -4).
(iii) ABA’B’ is a parallelogram.
(iv) Perimeter of ||^gm ABA’B’
= AB + BA’ + A’B’ + B’A
= 6 + 10 + 6 + 10
= 32 units.

Section-B (40 Marks)
(Attempt any four questions from this Section)

Question 5.
(a) Solve the following inequation, write the solution set and represent it on the number line:
–\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – 1\(\frac{1}{3}\) < \(\frac{1}{6}\), x ∈ R
(b) Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets ₹ 8088 from the bank after 3 years, find the value of his monthly installment. [3]
(c) Salman buys 50 shares of face value ₹ 100 available at ₹ 132.
(i) What is his investment?
(ii) If the dividend is 7.5%, what will be his annual income?
(iii) If he wants to increase his annual income by 150, how many extra shares should he buy? [4]
Solution:
(a)
–\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – 1\(\frac{1}{3}\) < \(\frac{1}{6}\), x ∈ R
⇒ –\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – \(\frac{3}{4}\) < \(\frac{1}{6}\)
⇒ -2x ≤ 3x – 8 < 1
⇒ -2x ≤ 3x – 8 and 3x – 8 < 1
⇒ 8 ≤ 5x and 3x < 9
⇒ \(\frac{8}{5}\) ≤ x and x < 3
⇒ \(\frac{8}{5}\) ≤ x < 3
⇒ 1\(\frac{3}{5}\) ≤ x < 3

(b) Here, rate of interest (r) = 8% p.a.
Maturity amount = ₹ 8088
Time = 3 years
or n = 3 × 12 = 36
Let the monthly installment be ₹ x
∴ 36x + x × \(\frac{36(36+1)}{2}\) × \(\frac{1}{12}\) × \(\frac{8}{100}\) = 8088
36x + x × \(\frac{111}{25}\) = 8088
\(\frac{36 \times 25 x+111 x}{25}\) = 8088
\(\frac{1011}{25} x\) = 8088
x = 8088 × \(\frac{25}{1011}\)
x = 8 × 25 = 200
Hence, the required monthly installment is ₹ 200

(c) Here, face value of a share = ₹ 100
Market value of a share = ₹ 132
No. of shares = 50
(i) His investment ₹ 132 × 50
= ₹ 6600
(ii) His annual income = ₹ \(\frac{7.5}{100}\) × 50 × 100
= ₹ 375
(iii) Total income required = ₹ (375 + 150)
= ₹ 525
⇒ x = \(\frac{525 \times 100}{7.5}\)
= ₹ 7000
∴ Extra investment = ₹ (7000 – 5000)
= ₹ 2000
Thus, number of extra shares required
= \(\frac{2000}{100}\) = 20

Question 6.
(a) Show that \(\sqrt{\frac{1-\cos A}{1+\cos A}}\) = \(\frac{\sin A}{1+\cos A}\)
(b) In the given circle with centre O, ∠ABC = 1000, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.

(c) Given below are the entries in a Savings Bank A/c pass book:

Calculate the interest for six months from February to July at 6% p.a.
Solution:
(a)

(b)
Here, ABCD is a cyclic quadrilateral.
∴ ∠ADC + ∠ABC = 180°
∠ADC + 100° = 180°
⇒ ∠ADC = 180° – 100° = 80°
Thus, ∠ADC = 80°
Now, in ΔACD, by using angle sum property, we have
∠CAD + ∠ACD + ∠ADC = 180°
∠CAD + 40° + 80° = 180°
∠CAD = 180° – 40° – 80° = 60°

Thus, ∠DCT = ∠CAD = 60°
[∠s in the corresponding alternate segment]

(c)

Minimum balance for the month:
February = ₹ 4500
March = ₹ 4500
April = ₹ 4500
May = ₹ 6730
June = ₹ 1730
July = ₹ 7730
Total principle for one month = ₹ 29690
∴ Interest = \(₹ \frac{29690 \times 6 \times 1}{100 \times 12}\) = ₹ 148.45

Question 7.
(a) In ΔABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A. [3]
(b) A shopkeeper sells an article at the listed price of ₹ 1500 and the rate of VAT is 12% at each stage of sale. If the shopkeeper pays a VAT of ₹ 36 to the Government, what was the price, inclusive of Tax, at which the shopkeeper purchased the articles from the wholesaler? [3]
(c) In the figure given, from the top of a building AB =60 m high, the angles of depression of the top
and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :

(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post. [4]
Solution:
(a) Here, coordinates of the vertices of ΔABC are A(3, 5), B(7, 8) and C(1, -10).
For equation of the median through A, we have D as mid-point of BC and coordinates of D are:
D\(\left(\frac{7+1}{2}, \frac{8-10}{2}\right)\) i.e., D(4, -1)
Now, equation of AD is given by
y – 5 = \(\left(\frac{-1-5}{4-3}\right)\)(x – 3)
y – 5 = -6(x – 3)
y – 5 = -6x + 18
6x + y – 23 = 0

(b)
Let selling price of an article by the wholesaler be ₹ x.
Rate of VAT = 12%
∴ VAT charged = ₹ \(\frac{12}{100}\) × x
= ₹ 180
Selling price of the article by the shopkeeper = ₹ 1500
VAT at the rate of 12% = ₹ \(\frac{12}{100}\) × 1500
= ₹ 180
VAT paid to the Govt, by the shopkeeper = ₹ 36
VAT charged = ₹ (180 – 36) = ₹ 144
Now, \(\frac{12}{100}\) × x = ₹ 144
⇒ x = ₹ \(\frac{14400}{12}\) = ₹ 1200
Thus, price inclusive of tax at which the shopkeeper purchased the article from the wholesaler
= ₹ 1200 + ₹ \(\frac{12}{100}\) × 1200
= ₹ 1200 + ₹ 144
= ₹ 1344

(c)
Here, AB = 60 m, ∠ADE = 30° and ∠ACB = 60°. Let height of vertical lamp post CD be h m.
Now, in rt.∠ed ΔABC, we have
\(\frac{\mathrm{AB}}{\mathrm{BC}}\) = tan 60°
⇒ \(\frac{60}{\mathrm{BC}}\) = \(\sqrt{3}\)
⇒ BC = \(\frac{60}{\sqrt{3}}\) = \(\frac{60 \sqrt{3}}{\sqrt{3} \sqrt{3}}\)
= 20\(\sqrt{3}\) m
AE = AB – EB
= AB – DC [∵ EB = DC]
= 60 – h

Again, in it ∠ed ΔAED, we have
\(\frac{\mathrm{AE}}{\mathrm{ED}}\) = tan 30°
\(\frac{60-h}{20 \sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\) [∵ ED = BC = 20\(\sqrt{3}\)m]
60 – h = 20
⇒ h = 60 – 20 = 40 m
Hence, the horizontal distance between AB and CD is 20\(\sqrt{3}\) m and the height of the lamp post is 40 m.

Question 8.
(a) Find x and y if \(\left[\begin{array}{ll}
x & 3 x \\
y & 4 y
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
5 \\
12
\end{array}\right]\) [3]
(b) A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast. [3]
(c) Without solving the following quadratic equation, find the value of ‘p’ for which the given equation has real and equal roots x² + (p – 3)x + p = 0. [4]
Solution:
(a)

(b)
Let the number of right circular cones recasted be n.
∴ Volume of n right circular cones = Volume of solid sphere n × \(\frac{1}{3}\) × π × 2.5 × 2.5 × 8
= \(\frac{4}{3}\) × π × 15 × 15 × 15
n × 50 = 60 × 225
n = \(\frac{60 \times 225}{50}\) = 270
Hence, the required number of right circular cones recasted is 270.

(c) Here, given quadratic equation is x² + (p – 3)x + p = 0
For real and equal roots, we have
D = b² – 4ac – 0
⇒ b² = 4ac
⇒ (p – 3)² = 4(1)(p)
⇒ p² + 9 – 6p – 4p = 0
⇒ p² – 10p + 9 = 0
⇒ (p – 9)(p – 1) = 0
⇒ Either p – 9 = 0 or p – 1 = 0
⇒ p = 9 or p = 1

Question 9.
(a) In the figure alongside, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of shaded portion. (Take π = \(\frac{22}{7}\))
(b) A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box. [3]
(c) Find the mean of the following distribution by step deviation method: [4]

Solution:
(a) Clearly, required area = area of quadrant – area of right-angled triangle AOD
= \(\frac{1}{4}\)(πr²) – \(\frac{1}{4}\) × Base × Perpendicular
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 3.5 × 3.5 – \(\frac{1}{2}\) × 3.5 × 2
= \(\frac{11}{2}\) × 0.5 × 3.5 – 3.5 = 3.5 (5.5 × 0.5 – 1)
= 3.5 × 1.75 = 6.125 cm²

(b) Let number of black balls in the box be n
Number of white balls = 30
Total number of balls in the box = 30+ n
As per condition of the question, we have
\(\frac{n}{30+n}\) = \(\frac{2}{5} \times \frac{30}{30+n}\)
n = \(\frac{2 \times 30}{5}\) = 12
Hence, the required number of black balls is 12.

(c)

Let assumed mean (a) = 55
Now, mean (\(\bar{x}\)) = a + \(\frac{\sum f_i u_i}{\sum f_i}\) × h
= 55 + \(\frac{-27}{50}\) × 10
= 55 – \(\frac{27}{5}\) = 55 – 5.4
\(\bar{x}\) = 49.6

Question 10.
(a) Using a ruler and compasses only:
(i) Construct a triangle ABC with the following data:
AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure ∠BCP. [4]

(b) The marks obtained by 120 students in a test are given below :

Draw an ogive for the given distribution on a graph sheet.
Use suitable scale for ogive to estimate the following:
(i) The median.
(ii) The number of students who obtained more than 75% marks in the test.
(iii) The number of students who did not pass the test if minimum marks required to pass is 40. [6]
Solution:
(a) (i) Steps of Construction :
1. Draw a line segment BC = 6 cm.
2. Construct ∠XBC = 120° at B.
3. With B as centre and radius 3.5 cm cut off AB = 3.5 cm.
4. Join AC.
ΔABC is the required triangle

(ii) Steps of Construction :
1. Draw perpendicular bisector of BC which cuts BC at point D.
2. With D as centre and radius = CD draw a circle passing through points B and C.
3. Draw angle bisector of ∠ABC which intersects the circle at point P.
4. ∴ P is the point which is equidistant from AB and BC.

(iii) Join CP and measure ∠BCP.
Thus, ∠BCP = 30°.

(b)

(i) Median = 43.08
(ii) 10 students

(iii) 52 students

Question 11.
(a) In the figure given below, the line segment AB meets X-axis at A and Y-axis at B. The point P (-3, 4) on AB divides it in the ratio 2 : 3. Find the coordinates of A and B. [3]

(b) Using the properties of proportion. solve for x, given
\(\frac{x^4+1}{2 x^2}\) = \(\frac{17}{8}\) [3]
(c) A shopkeeper purchases a certain number of books for ₹ 960. If the cost per book was ₹ 8 less, the number of books that could be purchased for ₹ 960 would be 4 more. Write an equation, taking the original cost of each book to be ₹ x, and solve it to find the original cost of the books. [4]
Solution:
(a) Let coordinates of A and B be A(-x, 0) and B(0, y).
Now, P(-3, 4) on AB divides AB in the ratio 2 : 3
∴ \(\left(\frac{2(0)-3 x}{2+3}, \frac{2 y+3(0)}{2+3}\right)\) = (-3, 4)
⇒ \(\left(\frac{2 x}{5}, \frac{2 y}{5}\right)\) = (-3, 4)
⇒ \(\frac{-3 x}{5}\) = -3 and \(\frac{2 y}{5}\) = 4
⇒

Hence, the coordinates of A and B are A(-5, 0) and B(0, 10)

(b)
\(\frac{x^4+1}{2 x^2}\) = \(\frac{17}{8}\)
Using componendo and dividendo, we have

Again, applying componendo and dividendo, we have
\(\frac{x^2+1+x^2-1}{x^2+1-x^2+1}\) = \(\frac{5+3}{5-3}\)
⇒ \(\frac{2 x^2}{2}\) = \(\frac{8}{2}\) ⇒ x² = 4
⇒ x = ±2

(c) Let the original cost price of each book be ₹ x
Total cost price = ₹ 960
∴ Number of books purchased = \(\frac{960}{x}\)
New cost price of each book = ₹ (x – 8)
∴ Number of books purchased = \(\frac{960}{x-8}\)
Now, as per condition of the statement, we have
\(\frac{960}{x-8}\) – \(\frac{960}{x}\) = 4
⇒ 960\(\left(\frac{x-x+8}{(x-8) x}\right)\) = 4
⇒ \(\frac{960 \times 8}{4}\) = x² – 8x
⇒ x² – 8x – 1920 = 0
Which is the required quadratic equation.
⇒ x² – 48x + 40x – 1920 = 0
⇒ x(x – 48) + 40(x – 48) = 0
⇒ (x – 48)(x + 40) = 0
⇒ Either x – 48 = 0 or x + 40 = 0
⇒ x = 48 or x = -40
(Rejecting – ve value, cost per book cannot be negative)
Thus, x = 48
Hence, the original cost price of each book is ₹ 48.