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CBSE Sample Papers for Class 10 Science in Hindi Medium Paper 4 is part of CBSE Sample Papers for Class 10 Science in Hindi Medium . Here we have given CBSE Sample Papers for Class 10 Science in Hindi Medium Paper 4

CBSE Sample Papers for Class 10 Science in Hindi Medium Paper 4

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Papers for Class 10 Science in Hindi Medium is given below with free PDF download solutions.

समय : 3 घण्टे
पूर्णांक : 80

सामान्य निर्देश :

• इस प्रश्न पत्र के दो भाग, A व B हैं। आप को दोनों भाग करने हैं।
• सभी प्रश्न अनिवार्य हैं।
• भाग A के सभी प्रश्न एक साथ करने हैं तथा भाग B के सभी प्रश्न एक साथ करने हैं।
• भाग A के प्रश्न सं० 1 व 2 एक अंक के हैं। इनका उत्तर एक शब्द अथवा एक वाक्य में लिखना है।
• प्रश्न सं० 3 से 5 तक दो अंक के हैं। इनका उत्तर 30 शब्दों में (प्रत्येक प्रश्न के लिए) लिखिए।
• प्रश्न सं० 6 से 15 तक तीन अंक के हैं। इनमें से प्रत्येक प्रश्न का उत्तर 50 शब्दों में लिखिए।
• प्रश्न सं० 16 से 21 तक पाँच अंक के हैं। इनमें से प्रत्येक प्रश्न का उत्तर 70 शब्दों में लिखिए।
• भाग B के प्रश्न सं० 22 से 27 तक प्रयोगात्मक कौशल पर आधारित हैं। प्रत्येक प्रश्न दो अंक का है।

SECTION A

प्र०1.
किसी विद्युत उपकरण की विद्युत शक्ति (W), उसमें से प्रवाहित होने वाली विद्युत धारा (amp) तथा विभवांतर (Volt) में संबंध स्थापित कीजिए।

प्र०2.
जीन क्या है?

प्र०3.
मानवों के शुक्राणुओं और अण्डों के बीच प्रमुख अंतर क्या है? इस अंतरे का महत्त्वं लिखिए।

प्र०4.
निम्नलिखित का उत्तर कारण सहित लिखिए :
(a) सोडियम धातु को मिट्टी के तेल में क्यों रखा जाता है?
(b) जब नीले रंग के कॉपर सल्फेट के विलयन में जब कुछ मात्रा में एल्युमिनियम पाउडर डालते हैं। तो वह रंगहीन क्यों हो जाता है?

प्र०5.
घरेलू विद्युत परिपथों में विद्युत उपकरणों को समान्तर क्रम में क्यों लगाया जाता है? कोई दो कारण लिखिए।

प्र०6.
एक बर्तन में ठोस केल्शियम ऑक्साइड लेकर उसमें धीरे से जल डाला गया तो
(i) इस प्रयोग के दो प्रेक्षण लिखिए।
(ii) इस अभिक्रिया का संतुलित रासायनिक समीकरण लिखिए।
अथवा
निम्नलिखित अभिक्रियाओं के लिए संतुलित समीकरण लिखिए :
(a) तनु H₂SO₄ अम्ल की एल्युमिनियम पाउडर से अभिक्रिया।
(b) सोडियम कार्बोनेट की तनु HCl अम्ल के साथ अभिक्रिया।
(c) चूने के पानी में से CO₂ गैस को गुज़ारना।

प्र०7.
किसी वृत्ताकार चालक कुण्डली में विद्युत प्रवाहित होने पर उसके चुबंकीय क्षेत्र के किसी बिंदु पर चुबंकीय प्रबलता का क्या प्रभाव होगा जब :
(i) उसमें प्रवाहित होने वाली विद्युत धारा की मात्रा बढ़ा दी जाए।
(ii) यदि कुण्डली से उस बिंदु की दूरी बढ़ा दी जाए।
(iii) यदि कुण्डली में चालक तार के लपेटे बढ़ा दिए जाएं।

प्र०8.
एक विद्युत बल्ब तथा एक चालक जिसका प्रतिरोध 4 ओम है, एक 6 वोल्ट की बैटरी के साथ श्रेणी क्रम में जुड़े हैं। यदि बल्ब में से 0.25A की विद्युत धारा प्रवाहित हो रही है तो उस बल्ब का प्रतिरोध ज्ञात कीजिए।

प्र०9.
निम्नलिखित की परिभाषा लिखिए।
(a) प्रतिवर्ती क्रिया
(b) अंतग्रंथन (सिनेप्स)
(c) प्रकाशानुवर्तन

प्र०10.
गांव में रामू की खेती की जमीन है। वह उस पर एक ईंटों का भट्ट्ठा लगाना चाहता है। उसका दोस्त श्याम उसे कहता है कि वह भट्ठा लगाने का अपना इरादा बदल दे क्योंकि भट्ठा लगाने से उसकी चिमनी से निकले धुएं से गांववालों के स्वास्थ्य पर बुरा प्रभाव पड़ेगा तथा इसका प्रभाव बाकी लोगों के खेतों से उत्पादित होने वाली फसलों पर भी पड़ेगा। रामू उसकी सलाह मान लेता है तथा अपना इरादा बदल लेता है।
(a) श्याम किस प्रकार की सोच रखता है?
(b) व्याख्या करें कि श्याम की इस सोच ने किस प्रकार गांववालों को नुकसान होने से बचाया।

प्र०11.
कार्बन यौगिकों की समजातीय श्रेणियाँ क्या हैं? ऐल्डिहाइडों की समजातीय श्रेणी के दो क्रमागत सदस्यों के आणिवक-सूत्र लिखिए। उल्लेख कीजिए कि इन यौगिकों का कौन-सा भाग इनके
(i) भौतिक और
(ii) रासायनिक गुणधर्मों का निर्धारण करता है।

प्र०12.
निम्नलिखित सारणी का अध्ययन कीजिए जिसमें छह तत्वों A, B, C, D, E और F की स्थितियाँ उनकी आधुनिक आवर्त सारणी में स्थितियों के अनुसार दर्शायी गई हैं

उपर्युक्त सारणी के आधार पर निम्नलिखित प्रश्नों के उत्तर दीजिए :
(i) उस तत्व का नाम लिखिए जो केवल सहसंयोजी यौगिक बनाता है।
(ii) उस तत्व का नाम लिखिए जिसकी संयोजकता तीन है और वह धातु है।
(iii) उस तत्व का नाम लिखिए जिसकी संयोजकता तीन है और वह अधातु है।
(iv) D और E में से किस तत्व का आकार बड़ा है और क्यों?
(v) तत्व C और F जिस परिवार के सदस्य हैं उनका सामान्य नाम क्या है?

प्र०13.
उपयुक्त उदाहरणों की सहायता से व्याख्या कीजिए कि कुछ लक्षण अगली पीढ़ी को क्यों नहीं दिए जा सकते हैं। इन लक्षणों को क्या कहते हैं?
अथवा
“ऐसा हो सकता है कि कोई लक्षण वंशानुगत हो जाए परन्तु व्यक्त न हो।” एक उपयुक्त उदाहरण की सहायता से इस कथन की पुष्टि कीजिए।

प्र०14.
कोई छात्र किसी दर्पण के ध्रुव से मोमबत्ती की ज्वाला को 15 सेमी दूरी पर रखकर उसके प्रतिबिम्ब को 60 सेमी दूरी पर स्थित पर्दे पर प्रक्षेपित करना चाहता है।
(i) दर्पण के उस प्रकार का नाम लिखिए जिसे उसे उपयोग करना चाहिए।
(ii) बनने वाले प्रतिबिम्ब का रैखिक आवर्धन ज्ञात कीजिए।
(iii) बिम्ब और उसके प्रतिबिम्ब के बीच की दूरी कितनी है?
(iv) इस प्रकरण में प्रतिबिम्ब बनना दर्शाने के लिए किरण आरेख खींचिए।

प्र०15.
श्वेत प्रकाश का एक पतला किरण-पुंज, PQ चित्र में दर्शाए अनुसार, काँच के किसी प्रिज्म ABC में से गुजरता है?
इस चित्र को अपनी उत्तर पुस्तिका पर खींचिए और पर्दे DE पर निर्गत पुंज जैसा दिखाई देगा उसका पथ खींचकर दर्शाइए।
(i) प्रेक्षण की जाने वाली परिघटना का नाम और उसका कारण लिखिए।
(ii) इसके अतिरिक्त प्रकृति में इसे परिघटना को आप और 3 कहाँ प्रेक्षण करते हैं?
(iii) इस प्रेक्षण के आधार पर श्वेत प्रकाश के अवयवों के बारे में जो निष्कर्ष निकाला जा सकता है उसका उल्लेख कीजिए।

प्र०16.
आपको एल्युमिनियम का एक बर्तन दिया है। आपको तनु HCl, तनु HNO₃, ZnCl₂ विलयन तथा जल (H₂O) भी दिये गये हैं। इनमें से कौन-कौन से विलयन एल्युमिनियम के बर्तन में रखे जा सकते हैं? प्रत्येक में होने वाली अभिक्रिया के विषय में भी लिखिए।

प्र०17.
(a) दो एक जैसे 10 Ω के प्रतिरोधकों को 6 वोल्ट की बैटरी के साथ बारी-बारी से पहले श्रेणी क्रम में तथा फिर समान्तर क्रम में लगाया गया। दोनों स्थितियों में शक्ति का अनुपात ज्ञात करें।
(b) एक किलोवाट घण्टा (kWh) तथा ऊर्जा की S.I. इकाई के बीच संबंध स्थापित करें।

प्र०18.
(a) शाकाहारी तथा मांसाहारी प्राणियों की छोटी आंत की लंबाई की तुलना कीजिए।
(b) छोटी आंत की दो विशेषताएँ लिखिए जिसके कारण यह पचे हुए भोजन को अवशोषित कर पाती

प्र०19.
(a) मानव मादा जनन-तंत्र के काट दृश्य का आरेख खींचिए। उस भाग का नामांकन कीजिए जहाँ|
(i) अण्ड विकसित होते हैं।
(ii) निषेचन होता है;
(iii) निषेचित अण्ड आरोपित होता है।
(b) संक्षेप में उन परिवर्तनों का वर्णन कीजिए जो गर्भाशय में उस स्थिति में होते हैं
(i) जब वह युग्मनज ग्रहण करता है।
(ii) जब युग्मनज नहीं बनता।

प्र०20.
(a) गोलीय लेंसों से संबंधित निम्नलिखित पदों की व्याख्या कीजिए :
(i) प्रकाशिक केन्द्र
(ii) वक्रता-केन्द्र
(iii) मुख्य अक्ष
(iv) द्वारक
(v) मुख्य फोकस
(vi) फोकस दूरी
(b) किसी अभिसारी लेंस की फोकस दूरी 12 सेमी है। वह बिम्ब दूरी परिकलित कीजिए जिस पर किसी बिम्ब को रखने पर उसका प्रतिबिम्ब लेंस के दूसरी ओर लेंस से 48 सेमी दूर बनता है।

प्र०21.
(a) हम तत्वों का वर्गीकरण क्यों करते हैं?
(b) आवर्त सारणी का निर्माण करते समय मेंडेलीफ द्वारा अपनाए गए दो मापदंड क्या थे?
(c) मेंडेलीफ ने अपनी आवर्त सारणी में कुछ रिक्त स्थान क्यों छोड़ दिए थे?
(d) मेंडेलीफ की आवर्त सारणी में हीलियम, नियॉन तथा आर्गन जैसी उत्कृष्ट गैसों का उल्लेख क्यों नहीं किया गया था?
(e) क्या आप क्लोरीन के दो समस्थानिकों CI-35 तथा CI-37 को उनके परमाणु द्रव्यमान भिन्न होने के कारण भिन्न-भिन्न स्थानों पर रखना पसंद करेंगे अथवा रासायनिक गुण समान होने के कारण एक ही स्थान पर रखेंगे? अपने उत्तर की पुष्टि कीजिए।

SECTION B

प्र०22.
एक विद्यार्थी ने फेरस सल्फेट को एक परखनली में गर्म किया। गर्म करने पर उसमें से एक तीखी गंध वाली गैस उत्पन्न हुई तथा फेरस सल्फेट का रंग सफेद होने लगा।
(i) तीखी गंध वाली गैस का रासायनिक सूत्र लिखिए।
(ii) FeSO₄ के क्रिस्टलों का रंग सफेद क्यों हो गया?
(iii) इसमें होने वाली अभिक्रिया किस प्रकार की है?

प्र०23.
एक विद्यार्थी ने पत्ती की झिल्ली की अस्थाई स्लाईड बनाकर देखा कि उसकी कोशिकाओं कुछ हरे रंग के बिन्दु हैं।
(i) इन हरे रंग के बिंदुओं का नाम लिखिए तथा उनमें उपस्थित रंगद्रव्यों का नाम लिखिए।
(ii) इन रंगीन कणों का क्या कार्य है?

प्र०24.
किसी विद्युत सैल में किसका मान कम होता है-उसके सिरों के बीच PD(विभवांतर) या उसकी वैद्युत गति बल (e.m.f)?

प्र०25.
एक परखनली में Na₂CO₃ के जलीय विलयन में जब एसिटिक अम्ल डालते हैं तो क्या होता है? निकलने वाली गैस की पहचान के लिए प्रयोग का समीकरण लिखें।

प्र०26.
जब एक प्रकाश किरण, एक कांच के स्लैब से गुजरती है, तो वह अपने पथ से कितनी बार विचलित होती है और क्यों?

प्र०27.
जब किशमिश के दानों को पानी में डालते हैं तो वे पानी को अवशोषित करते हैं क्यों? इस प्रक्रम को समझाएँ।

Answers

उत्तर 1-
शक्ति को विद्युत धारा तथा विभवांतर के गुणनफल के रूप में प्राप्त किया जा सकता है।
शक्ति (P) = विभवांतर (V) x विद्युतधारा (I)
P = V x I

उत्तर 2-
जीन डी०एन०ए० (DNA) की वह सूक्ष्मतम आनुवंशिक इकाई है जो गुणसूत्रों पर स्थित है। प्रत्येक जीन एक विशेष गुण को जनक से संतति तक ले जाता है।

उत्तर 3- शुक्राणु का आकार, अण्डे की तुलना में काफी छोटा होता है, तथा शुक्राण गतिशील होता है, जबकि अंडा (अण्ड कोशिका) गतिशील नहीं होता है। अण्डे में भ्रूण की प्रारम्भिक अवस्थाओं की वृद्धि के लिए भोजन भरा होता है, इसलिए इसका आकार शुक्राणु की तुलना में बड़ा होता है।
महत्त्व-शुक्राणु इसलिए गतिशील होते हैं कि उन्हें नर के शरीर से मादा के शरीर में निषेचन के लिए जाना होता है।

उत्तर 4-
(a) सोडियम धातु एक अति क्रियाशील धातु है। इसे खुली हवा में रखने पर यह वायु की ऑक्सीजन से क्रिया करके सोडियम ऑक्साइड में बदल जाती है। सोडियम धातु को इस अभिक्रिया से बचाने के लिए इसे | मिट्टी के तेल में रखा जाता है।
(b) जब एल्युमिनियम पाउडर को कॉपर सल्फेट विलयन में मिलाया जाता है तब एल्युमिनियम सल्फेट के रंगहीन विलयन के बनने के कारण कॉपर सल्फेट का नीला रंग धीरे-धीरे रंगहीन हो जाता है।

एल्युमिनियम (Al), कॉपर (Cu) से अधिक अभिक्रियाशील है इसलिए यह CuSO₄ विलयन में से Cu विस्थापित कर देता है तथा रंगहीन Al₂(SO₄)₃ बना देता है।

उत्तर 5-
घरेलू विद्युत परिपथों में विद्युत उपकरणों को समान्तर क्रम में लगाने के दो कारण :
(i) इस क्रम में यदि कोई उपकरण खराब होने के कारण, कार्य करना बंद कर देता है, तो बाकी उपकरण ठीक से चलते रहते हैं।
(ii) प्रत्येक उपकरण को इस क्रम में अपने प्रतिरोध के अनुसार विद्युत धारा प्राप्त होती रहती है।

उत्तर 6-
(i) इस प्रक्रम में काफी मात्रा में ऊष्मा उत्पन्न होती है जिससे बर्तन गर्म हो जाता है। बुदबुदाहट की तेज़ आवाज़ उत्पन्न होती है।
(ii)

अथवा

उत्तर 7-
(i) उस बिन्दु पर चुबंकीय क्षेत्र की प्रबलता अधिक हो जायेगी क्योंकि चुबंकीय क्षेत्र की प्रबलता विद्युतधारा के समानुपाती होती है।
(ii) बिंदु की दूरी बढ़ा देने पर चुबंकीय क्षेत्र की प्रबलता कम हो जायेगी क्योंकि चुबंकीय क्षेत्र की प्रबलता दूरी के व्युत्क्रमानुपाती होती है।
(iii) जब कुण्डली में लपेटे बढ़ा देते हैं तो उसमें से प्रवाहित होने वाली विद्युत धारा की मात्रा बढ़ जाती है। जिसके कारण चुबंकीय क्षेत्र की प्रबलता भी अधिक हो जायेगी।

उत्तर 8-

बल्ब :
प्रतिरोध, R₁ = ?
प्रवाहित विद्युत धारा, I₁ = 0.25A
अन्य चालक :
प्रतिरोध, R₂ = 4Ω
प्रवाहित विद्युत धारा, I₂ = ?
परिणामी प्रतिरोध, R = R₁ + R₂
∴विभवांतर, V = 6V
परिपथ में प्रवाहित विद्युत धारा, I = 0.25 A
∵I = I₁ = I₂ …[श्रृंखला संयोजन में
∴V = IR …[ओम के नियमानुसार

अब R₁ + R₂ = R
R₁ + 4 = 24
∴R₁ = 24 – 4 = 20 Ω

उत्तर 9-
(a) प्रतिवर्ती क्रिया-किसी बाहरी उद्दीपन के प्रति यदि अनुक्रिया तेजी से तथा बिना सोचे समझे होती है। तथा जो मस्तिष्क के एच्छिक नियंत्रण में नहीं होती, प्रतिवर्ती क्रिया कहलाती है। ये क्रियाएँ अधिकतर मेरुरज्जु द्वार नियंत्रित होती हैं। उदाहरण : गर्म प्लेट के छूने पर तत्काल हाथ का हट जाना या नुकीली वस्तु पर पैर पड़ जाने पर पैर का हट जाना।
(b) अंतर्ग्रथन (सिनेप्स)-निकटवर्ती तंत्रिका कोशिकाओं की जोड़ी के बीच अति सूक्ष्म रिक्त स्थान, जिसके पार तंत्रिका आवेगों को एक तंत्रिका कोशिका से अगली तंत्रिका कोशिका तक पहुँचाया जाता है, अंतर्ग्रथन कहलाता है।
(c) प्रकाशानुवर्तन-पौधे के प्ररोह (तने तथा पत्तियाँ) का प्रकाश की दिशा की ओर वृद्धि करना प्रकाशानुवर्तन कहलाता है। यह क्रिया ऑक्सिन नामक पादप हार्मोन द्वारा नियंत्रित होती है।

उत्तर 10-
(a) श्याम एक समझदार व सुलझा हुआ व्यक्ति है। वह पर्यावरण संरक्षण तथा अपने गांववालों के स्वास्थ्य व भलाई की सोच रखता है।
(b) (i) ईंटों के भट्ठे लगाने से धरती की उपजाऊ परत नष्ट हो जाएगी तथा वह ज़मीन गहरे बंजर खड्डों में परिवर्तित हो जायेगी।
(ii) भट्ठे में ईंटों को पकाने के लिए जीवाश्मी ईधन को जलाया जाता है जिसके धुएं से वातावरण दूषित होता है तथा लोगों को कई प्रकार की बीमारियाँ होने लगती हैं।
(iii) ईंटों के भट्ठे से बहुत अधिक मात्रा में राख निकलती है तथा यह राख वायु में मिलकर वायु प्रदूषण का कारण बनती है।
इन सभी कारणों के फलस्वरूप रामू श्याम की सलाह मानकर अपना इरादा बदल देता है।

उत्तर 11-
एक जैसे कार्बनिक यौगिकों का समूह, जिनके रासायनिक गुण काफी मिलते हैं, समजातीय श्रेणी कहलाते हैं।
एल्डिहाइडों के क्रमागत सदस्यों के आणिवक-सूत्र :
H–CHO
CH₃-CHO
C₂H₅-CHO
(i) एल्डिहाइड समूह से जुड़ी कार्बन श्रृंखला, जैसे H-, CH₃-, C₂H₅– इनके भौतिक गुणधर्मों का निर्धारण करती है।
(ii) प्रकार्यात्मक समूह से जुड़े सभी एल्डिहाइड, जैसे -CHO इनके रासायनिक गुणधर्मों का निर्धारण करते हैं।

उत्तर 12-
(i) E
(ii) D
(iii) B
(iv) तत्व ‘D’ का आकार बड़ा है क्योंकि किसी आवर्त में बाएँ से दाएँ जाने पर तत्वों के परमाणुओं का आकार छोटा होता जाता है। नाभिक में आवेश के बढ़ने से यह इलेक्ट्रॉनों को नाभिक की ओर खींचता है जिससे परमाणु का आकार घटता जाता है।
(v) उत्कृष्ट गैस समूह।

उत्तर 13-
वे लक्षण, जो जनन कोशिकाओं के जीनों में परिवर्तन नहीं ला पाते हैं, वे लक्षण अगली पीढ़ी में नहीं जा सकते हैं। ऐसे लक्षणों को उपार्जित लक्षण कहते हैं।
उदाहरण के लिए :
(i) यदि किसी झींगुर को काफी दिनों से पर्याप्त भोजन न मिल पाए और इस कारण उसका वज़न घट जाए तो वज़न का घटना एक उपार्जित लक्षण है।
(ii) चूहे की कटी हुई पूंछ एक उपार्जित लक्षण है।
अथवा
एक पीढ़ी से दूसरी पीढ़ी तक प्रभावी लग DNA द्वारा स्थानांतरित होते हैं। उपार्जित लक्षण DNA में नहीं आते हैं अतः ये अगली पीढ़ी में वंशानुगत नहीं होते।
उदाहरण-जब मटर के एक शुद्ध लम्बे पौधे को एक बौने पौधे से संकरण करवाते हैं तो F, पीढ़ी के पौधे केवल लम्बे दिखाई देते हैं जबकि बौनेपन का लक्षण भी वंशानुगत होता है। जब F, पीढ़ी के पौधों का पुनः संकरण करवाते हैं, तो F, पीढ़ी में बौने पौधे भी दिखाई देते हैं। इससे पता चलता है कि बौनेपन का लक्षण भी वंशानुगत हुआ है।
अतः कहा जा सकता है, “ऐसा हो सकता है कि कोई लक्षण वंशानुगत हो जाए परन्तु व्यक्त न हो।”

उत्तर 14-
(i) उसे अवतल दर्पण का उपयोग करना चाहिए।
(ii) u = -15 cm (u सदैव ऋणात्मक)
υ = -60 cm (वास्तविक स्थिति के लिए ν ऋणात्मक)
m = \(\frac { -v }{ u }\) => -4
(iii) बिम्ब तथा उसके प्रतिबिम्ब के बीच की दूरी = 45 सेमी
(iv) किरण आरेख :

उत्तर 15-

(i) प्रेक्षण की जाने वाली परिघटना का नाम प्रिज्म द्वारा श्वेत प्रकाश का विक्षेपण कहलाता है। इसका कारण यह है कि जब एक श्वेत प्रकाश किरणे एक प्रिज्म से गुजरती है तो यह सात अवयवी वर्गों में विक्षेपित हो जाती है।
(ii) इन्द्रधनुष का बनना प्रकृति में श्वेत प्रकाश के विक्षेपण का एक उदाहरण है।
(iii) इस प्रेक्षण के आधार पर श्वेत प्रकाश के अवयवों के विषय में यह निष्कर्ष निकलता है कि श्वेत प्रकाश सात रंगों की किरणों से मिलकर बनता है। ये रंग हैं-बैंगनी, जामुनी, नीला, हरा, पीला, नारंगी तथा लाल।

उत्तर 16-
(i) तनु HCl अम्ल को एल्युमिनियम के बर्तन में नहीं रखा जा सकता क्योंकि यह इस धातु से क्रिया कर एल्युमिनियम क्लोराइड बनाएगा जिससे हाइड्रोजन गैस निष्कासित होगी।

(ii) तनु HNO₃ अम्ल Al धातु से अभिक्रिया करके उस पर Al₂O₃ की रक्षा परत बना देता है इसलिए तनु HNO₃ अम्ल को एल्युमिनियम के बर्तन में रखा जा सकता है।

(iii) ZnCl₂ को भी एल्युमिनियम के बर्तन में नहीं रखा जा सकता क्योंकि Al धातु Zn से अधिक क्रियाशील है इसलिए अधिक क्रियाशील Al कम क्रियाशील Zn को इस विलयन में से विस्थापित कर देगा।

(iv) Al धातु H₂O से अभिक्रिया नहीं करता इसीलिए जल को एल्युमिनियम के बर्तन में रखा जा सकता है।

उत्तर 17-

(a) श्रेणी क्रम में, R = R₁ + R₂ = 10 + 10 = 20 Ω

उत्तर 18-
(a) छोटी आंत की लंबाई विभिन्न प्राणियों में उनके द्वारा खाए जाने वाले भोजन पर निर्भर करती है। उदाहरण के लिए, सेलुलोज़ एक कार्बोहाइड्रेट भोजन होता है जो शरीर में बहुत कठिनाई से पचता है। इसकी तुलना में मांस शरीर में आसानी से पच जाता है। शाकाहारी जीवों की छोटी आंत अधिक लंबी होती है क्योंकि उन्हें हरे पत्तियों के सेलुलोज़ को पचाने में अधिक समय लगता है। इनकी तुलना में मांसाहारी जीवों की
छोटी आंत कम लंबी होती है क्योंकि मांस को पचाने में उन्हें कम समय लगता है।
(b) छोटी आंत की दो विशेषताएँ ।
(i) छोटी आंत की भीतरी दीवार पर बहुत अधिक ऊंगलीनुमा संरचनाएँ होती है जिन्हें दीर्घरोम कहते हैं।
(ii) ये संरचनाएँ छोटी आंत का सतही क्षेत्र बढ़ा देती हैं जिसके कारण पचे हुए भोजन का अवशोषण जल्दी हो जाता है।

उत्तर 19-
(a) मानव मादा जनन-तंत्र के काट दृश्य का आरेख : अंडवाहिका

(i) अण्ड विकसित होते हैं-अण्डाशय।
(ii) निषेचन होता है-अण्डवाहिका।
(iii) निषेचित अण्ड आरोपित होता है-गर्भाशय।
(b) (i) जब गर्भाशय युग्मनज ग्रहण करता है तो इसकी परत मोटी होती जाती है भ्रूण के विकसित होने के लिए इसकी भित्ति पर एक संरचना बनती है जिसे प्लेसेंटा कहते हैं। इसी के माध्यम से भ्रूण
को विकसित होने के लिए ऑक्सीजन तथा भोजन प्राप्त होता है।
(ii) जब युग्मनज नहीं बनता तो गर्भाशय की परत टूटने लगती है तथा अण्डकोशिका ऋतुस्राव के समय गर्भाशय की टूटी परत के साथ योनिमार्ग से शरीर से बाहर निकल जाती है।

उत्तर 20-
(a) (i) प्रकाशिक केन्द्र-लेंस के केंद्रीय बिंदु को लेंस का प्रकाशिक केन्द्र कहते हैं।
(ii) वक्रता केन्द्र-लेंस के दो गोलीय पृष्ठ होते हैं। इनमें से प्रत्येक पृष्ठ एक गोले का भाग होता है। इन गोलों के केन्द्र लेंस के वक्रता केन्द्र कहलाते हैं।
(iii) मुख्य अक्ष-किसी लेंस के दोनों वक्रता केंद्रों से गुजरने वाली एक काल्पनिक सीधी रेखा लेंस का मुख्य अक्ष कहलाती है।
(iv) द्वारक-गोलीय लेंस की वृत्ताकार रूपरेखा का प्रभावी व्यास इसकी द्वारक कहलाता है।
(v) मुख्य फोकस-किसी लेंस के मुख्य अक्ष के समांतर जाने वाले प्रकाश किरणें लेंस से अपवर्तन के पश्चात जिस बिंदु पर मिलती है अथवा मिलती हुई प्रतीत होती है उस बिंदु को लेंस का मुख्य फोकस कहते हैं।
(vi) फोकस दूरी-किसी लेंस के मुख्य फोकस की प्रकाशिक केंद्र से दूरी लेंस की फोकस दूरी कहलाती है।

∴ लैंस से बिम्ब 16 सेमी दूरी पर रखा गया है।

उत्तर 21-
(a) हम तत्वों का वर्गीकरण उनके गुणों का अध्ययन आसानी से तथा एक क्रमबद्ध तरीके से करने के लिए करते हैं। इन तत्वों को खोजने के पश्चात् उनकी संपूर्ण जानकारी तथा गुण प्राप्त करने में कई मुश्किलों को सामना करना पड़ता था इसीलिए वैज्ञानिकों ने तत्वों के अध्ययन को सुगम बनाने के लिए तत्वों को वर्गीकरण किया।
(b) मेंडेलीफ द्वारा अपनाए गए दो मापदंड :
(i) परमाणु द्रव्यमान तथा
(ii) रासायनिक गुणधर्मों में समानता।
(c) मेंडेलीफ ने ये रिक्त स्थान इसलिए छोड़ दिए थे ताकि जब नए तत्व खोजे जाएँ तो वे अपने द्रव्यमान तथा गुणों के अनुसार ये रिक्त स्थान प्राप्त कर सकें।
(d) उत्कृष्ट गैसें, जैसे हीलियम, नियॉन तथा आर्गन, की खोज उस समय तक नहीं की गई थी क्योंकि ये अक्रिय गैस हैं तथा वायुमण्डल में इनकी मात्रा बहुत अल्प है। इनकी खोज के बाद इन अक्रिय गैसों को मेंडेलीफ की सारणी में बिना किसी परिवर्तन किए नए समूहों में रखा गया।
(e) क्लोरीन के समस्थानिक Cl-35 तथा Cl-37 को मेंडेलीफ की आवर्त सारणी में उनके परमाणु द्रव्यमान भिन्न होने के कारण अलग-अलग स्थान पर रखेंगे परंतु इनका रासायनिक गुणधर्म समान होने के कारण उन्हें एक ही स्थान पर रखेंगे। ये एक-दूसरे के प्रतिवादी हैं। अतः क्लोरीन के इन समस्थानिकों को मेंडेलीफ ने अपनी आवर्त सारणी में एक स्थान नहीं दिया।

उत्तर 22-
(i) तीखी गंध वाली गैसों का रासायनिक सूत्र :
SO₂ (सल्फर डाइऑक्साइड) व SO₃ (सल्फर ट्राईऑक्साइड)।
(ii) गर्म करने पर FeSO₄ के क्रिस्टलीय जल अणु वाष्पित हो जाते हैं तथा सफेद रंग का चूर्ण में परिवर्तित हो जाते है।
(ii) यह एक तापीय विघटन अभिक्रिया है क्योंकि इस अभिक्रिया में एक पदार्थ को गर्म करने पर वह तीन | पदार्थों में विभाजित हो जाता है।

उत्तर 23-
(i) ये हरे रंग के बिंदु हरित लवक है तथा उनमें उपस्थित हरे रंग के रंगद्रव्य क्लोरोफिल हैं।
(ii) हरे रंग के क्लोरोफिल कण प्रकाश संश्लेषण क्रिया में सौर ऊर्जा को अवशोषित करते हैं।

उत्तर 24-
इन दोनों में से विद्युत सैल का PD (विभवांतर) का मान कम होता है क्योंकि उस सैल के e.m.f. का कुछ भाग उसके आंतरिक प्रतिरोध को समाप्त करने में प्रयोग हो जाता है।

उत्तर 25-
इस प्रक्रिया में CO₂ गैस बुदबुदाहट के साथ उत्पन्न होती है।

उत्तर 26-
जब एक प्रकाश किरण एक कांच के स्लैब से गुजरती है तो वह अपने पथ से दो बार विचलित होती है
• पहली बार वायु से कांच में जाने पर वह विरल माध्यम से सघन माध्यम में जाती है और अभिलम्ब की ओर मुड़ जाती है।
• दूसरी बार जब वह कांच से वायु में वापिस आती है। यह सघन माध्यम से विरल माध्यम में जाती है। तो वह अभिलम्ब से दूर हो जाती है।

उत्तर 27-
किशमिश के दाने परसरण प्रक्रम के कारण जल को अवशोषित करते हैं। जल के अणु किशमिश के दानों की अर्धपरागम्य झिल्ली से बाहर से भीतर की ओर प्रवाहित होते हैं।

We hope the CBSE Sample Papers for Class 10 Science in Hindi Medium Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 10 Science in Hindi Medium Paper 4, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 9 Maths Paper 4 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 4

CBSE Sample Papers for Class 9 Maths Paper 4

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
Simplify: 3^2/3 x 7^2/3.

Question 2.
When P(x) = x³ – ax² + x is divided by (x – a). Find the remainder.

Question 3.
On which axis does the point A(0, 4) lie?

Question 4.
Euclid divided his book “Elements” into how many chapters?

Question 5.
A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting head?

Question 6.
What is the total surface area (T.S.A.) of a cone whose radius is \(\frac { r }{ 2 }\) and slant height is 2l?

SECTION-B

Question 7.
Factorize: a(a – 1) – b(b – 1).

Question 8.
The angles of a triangle are in the ratio 2:3:7. Find the measure of each angle of triangle.

Question 9.
Find the angle at which the bisectors of any two adjacent angles of a parallelogram intersect.

Question 10.
Show that a median of a triangle divides it into two triangles of equal area.

Question 11.
Find the area of isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.

Question 12.
The mean of 40 numbers was found to be 38. Later on, it was detected that a number 56 was misread as 36. Find the correct mean of the (data) given numbers.
OR
Following data gives a number of children in 40 families.
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3,4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2,4, 3, 2, 1, 0, 5, 1, 2,4, 3, 4, 1, 6, 2, 2.
Represent it in the form of frequency distribution taking classes 0-2, 1-4, etc.

SECTION-C

Question 13.
Simplify

Question 14.
Find the product: (a-b-c) (a² + b² + c² + ab + ac – bc).
OR
The polynomial f(x) = x⁴ – 2x³ + 3x² – ax + b when divided by (x – 1) and (x + 1) leaves the remainder 5 and 19 respectively. Find the value of a and b. Hence find the remainder when fix) is divided by (x – 2).

Question 15.
Fill in the blanks:
(i) For the line 4x + 3y = 12, x-intercept =___ and y-intercept = ___
(ii) If x = 4, y = 3, is a solution of 2x + ky = 14, then k =___
(iii) If the point P(P, 4) lies on the line 3x + y = 10, then P =___

Question 16.
In the adjoining figure, ABCD is a parallelogram in which E and F are the mid points of the sides AB and CD respectively. Prove that the line segment CE and AF trisect the diagonal BD.

Question 17.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that
ar(∆AOD) = ar(∆BOC)

Question 18.
Construct an isosceles triangle whose base is 5 cm and whose vertical angle is 72°

Question 19.
Plot the points A(-5, 2), B(3, 2), C(-4, -3) and D(6, 0), E(0, 2), F(4, -4) on the graph paper. Write the name of quadrant also.

Question 20.
The total surface area of a cylinder is 462 cm². Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Question 21.
A tent is in the form of a right circular cylinder, surmounted by a cone. The diameter of a cylinder is 24 m. The height of the cylindrical portion is 11m, while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.

Question 22.
Three coins are tossed simultaneously 150 times and it is found that 3 tails appeared 24 times; 2 tails appeared 45 times 1 tail appeared 72 times and no tail appear 9 times. If three coins are tossed simultaneously at random, find the probability of getting
(i) 3 tails,
(ii) 2 tails,
(iii) 1 tail.

SECTION-D

Question 23.

Question 24.
If the polynomials (2x³ + ax² + 3x – 5) and (x³ + x² – 2x +a) leave the same remainder when divided by (x – 2). Find the value of a. Also, find the remainder in each case.

Question 25.
Two men starts from point A and B respectively, 42 km apart. One walks from A to B at 4 km/hr and another walks from B to A at a certain uniform speed. They meet each other after 6 hours. Find the speed of the second man.

Question 26.
In a ∆ABC, ∠B > ∠C. If AM is the bisector of ∠BAC and AN ⊥ BC, prove that ∠MAN = \(\frac { 1 }{ 2 }\) (∠B – ∠C).

Question 27.
In the given figure, ABCD is a square, M is the mid-point of AB and PQ ⊥ CM meets AD at P and CB produced at Q. Prove that
(i) PA = BQ,
(ii) CP = AB + PA

Question 28.
Prove that the sum of either pair of the opposite angles of a cyclic quadrilateral is 180°.
OR
Prove that the opposite angles of a cyclic quadrilateral are supplementary.

Question 29.
A copper wire of diameter 6 mm is evenly wrapped on a cylinder of length 15 cm and diameter 49 cm to cover its whole surface. Find the length and volume of the wire. If the specific gravity of copper be 9 g per cubic cm, find the weight of the wire.

Question 30.
15 students of Govt. School spend the following number of hours in a month for doing cleaning in their street 25, 15, 20, 20, 9, 20, 25, 15, 7, 13, 20, 12, 10, 15, 8.
(i) Find mean, median and mode from above data.
(ii) Which value is depicted from above data?

Solutions

Solution 1.
3^2/3 x 7^2/3 = (3 x 7)^2/3 = (21)^2/3 

Solution 2.
P(x) = x³ – ax² + x, x – a = 0 ⇒ x = a
When P(x) is divided by (x – a), we get
Remainder = P(+a) = a³ – a(+a)² + (+a) = a³ – a³ + a = a
Remainder = P(a) = a
⇒ Remainder = a

Solution 3.
y-axis ⇒ coordinates of y-axis is (0, b).

Solutions 4.
13 chapters.

Solution 5.

Solution 6.

Solution 7.
a(a – 1) – b(b – 1) = a² – a – b² + b = a² – b² – (a – b) = (a – b)(a + b) – (a – b)
= (a – b) (a + b – 1)

Solution 8.
Let the angles of the given triangles measure (2x)°, (3x)° and (7x)°.
2x + 3x + 7x = 180° [By angle sum property of A]
12x = 180° ⇒ x = 15°
Hence angles are
2x = 30°
3x = 45°
7x =105°

Solution 9.

∠A + ∠B = 180°
⇒ \(\frac { 1 }{ 2 }\)∠A + \(\frac { 1 }{ 2 }\)∠B = 90°
In ∆AOB
\(\frac { 1 }{ 2 }\)∠A + \(\frac { 1 }{ 2 }\)∠B + ∠AOB = 180°
⇒ 90° + ∠AOB = 180°
⇒ ∠AOB = 90°

Solution 10.

Given: A ∆ABC in which AD is median.
To Prove: ar(∆ABD) = ar(∆ADC)
Construction: Draw AL ⊥ BC
Proof: BD = DC
(AD is median so D is midpoint of BC)

=> ar(∆ABD) = ar(∆ADC) [∵ ar(∆) = \(\frac { 1 }{ 2 }\) x base x height]
Hence, a median of a triangle divides it into two triangles of equal area.

Solution 11.
a = 13,b = 13,c = 24

Hence, area of the triangle (∆) = 60 cm²

Solution 12.
Calculated mean of 40 numbers = 38
∴ Calculated sum of these numbers = 38 x 40 = 1520
Correct sum of these numbers = [1520 – (wrong item) + correct item]
= 150 – 36 + 56 = 1540
The correct mean = \(\frac { 1540 }{ 40 }\) = 38.5
Hence, correct mean = 38.5

Lowest data = 0, highest data = 6

Solution 13.

Solution 14.
(a – b – c) (a² + b² + c² + ab + ac – bc)
= [a + (-b) + (-c)] [a² + (-b)² + (-c)² – a(-b) – (-b) (-c) – (-c)a]
= a³ + (-b)³ + (-c)³ – 3a (-b)(-c)
= a³ – b³ – c³ – 3abc (a – b – c) (a² + b² + c² + ab + ac – bc)
= a³ – b³ – c³ – 3abc

OR
f(x) = x⁴ – 2x³ + 3x² – ax + b
If f(x), divided by (x – 1), let remainder be R₁ = 5.
x – 1 = 0, x = 1

Solution 15.
(i) For the line 4x + 3y = 12, x-intercept = 3 and y-intercept = 4, i.e., Ans = 3, 4
(ii) If x = 4, y = 3 is a solution of 2x + ky = 14, then k = 2.
=> Putting x = 4 and y = 3 in eqn. 2x + ky = 14
2 x 4 + k x 3 = 14 =>3k = 14 – 8 = 6
3k = 6 => k = 2
(iii) If the point P(P, 4) lies on the line 3x +y = 10, then P = 2
=> Putting x = P and y = 4 in eqn. 3x + y = 10
3P + 4 = 10 =>3P = 6 => P = 2.

Solution 16.

Let BD be intersected by CE and AF at P and Q respectively. AB || DC and AB = DC [opposite sides of || gm]
=> AE = FC and \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) DC => AE || FC and AE = FC
=> AECF is a || gm => AF || CE => EP || AQ and FQ || CP.
In ∆BAQ, E is the mid point of AB.
and EP || AQ so, P is the mid point of BQ
BP = PQ [By converse of midpoint theorem]
Again in ∆DPC, F is mid point of DC and FQ || CP. So Q is the mid point of DP.
=> PQ = QD [By converse of mid point theorem]
∴ BP = PQ = QD
Hence, CE and AF trisect the diagonal BD.

Solution 17.

∆ACD and ∆BCD being on the same base DC and between the same parallels DC and AB, then
ar(∆ACD) = ar(∆BCD)
Subtracting ar(∆DOC) on both sides
ar (∆ACD) – ar(∆DOC) = ar(∆BCD) – ar(∆DOC)
ar(∆AOD) = ar(∆BOC)

Solution 18.

Steps of Construction:
1. Draw a line segment BC = 5 cm
2. Make ∠CBX = 72°, below the line segment
3. Make ∠XBY = 90°
4. Draw the right bisector PQ of BC, intersecting BY at O.
5. With O as centre and radius OB, draw a circle, intersecting PQ at A.
6. Join AB and AC.
Then ∆ABC is the required isosceles triangle in which AB = AC.

Solution 19.
Point A(-5, 2) lies in => II quadrant
B(3, -2) lies in => I quadrant
C(-4, -3) lies in => III quadrant
D(6, 0) lies in => x-axis (abscissa)
E(0, 2) lies in => y-axis (ordinate)
F(4, -4) lies in => IV quadrant

Solution 20.
C.S.A. of cylinder = \(\frac { 1 }{ 3 }\) x (T.S.A. of cylinder)
= \(\frac { 1 }{ 3 }\) x 462 = 154 cm²
(T.S.A. of cylinder – C.S.A. of cylinder) = 462 – 154 = 308 cm²
2πr(h + r) – 2πrh = 308

Solution 21.

Radius of the cylinder = \(\frac { 24 }{ 2 }\) = 12
m = R
Height, H = 11 m
Curved surface area of the cylindrical portion
= 2πRH sq. units
= (2π x 12 x 11) m²
= (264 π) m²
Radius of the cone = r = 12m,
height of the cone = (16 – 11) = 5m

Hence, the area of the canvas required for tent = 1320 m².

Solution 22.
Total number of trials = 150
Number of times 3 tails appeared = 24
Number of times 2 tails appeared = 45
Number of times 1 tail appeared = 72
Number of times 0 tail appeared = 9

Solution 23.

Solution 24.
Let f(x) = 2x³ + ax² + 3x – 5 and g(x) = x³ + x² – 2x + a
When f(x) is divided by (x – 2), then remainder = f(2) [x – 2 = 0 => x = 2]
f(2) = 2(2)³ + a(2)² + 3(2) – 5 = 16 + 4a – 5 + 6 = 17 + 4a
When g(x) is divided by (x – 2), remainder = g(2) [x – 2 = 0 => x = 2]
g(2) = (2)³ + (2)² -2 x 2 + a = 8 + a
But f(2) = g(2)
17 + 4a = 8 + a
3a = -9 => a = -3
Remainder in each case = 8 + (-3) = 8 – 3 = 5

Solution 25.
Let the speed of another man be x km/h.
Average speed of both men = (x + 4) km/h.

=> 6x = 18 =>x = \(\frac { 18 }{ 6 }\) = 3
x = 3 km/h

Solution 26.
Given: In ∆ABC, in which ∠B > ∠C, AN ⊥ BC and AM is the bisector of ∠A. Also find the angle MAN if ∠B = 65° and ∠C = 30°.
To Prove: ∠MAN = \(\frac { 1 }{ 2 }\) (∠B – ∠C)
Proof: Since AM is the bisector of ∠A

Solution 27.

In ∆PAM and ∆QBM
AM = BM (M is the mid point)
∠PAM = ∠QBM (each 90°)
∠PMA = ∠BMQ (V.O.A)
∴ ∆AMP ≅ ∆BMQ (ASA congruency)
∆AMP ≅ ∠BMQ (CPCT)
PA = BQ …(i)
and MP = MQ
Now join PC. Again in ∆CMP and ∆CMQ
PM = MQ (Proved)
∠CMP = ∠CMQ (each 90°)
CM = CM (Common)
∆CMP = ∆CMQ (RHS congruency)
CP = CQ (CPCT)
CP = CQ = BC + BQ = AB + PA[BC = AB, PA = BQ]
CP = AB + PA

Solution 28.
Given: A cyclic quadrilateral ABCD
To Prove: ∠A + ∠C = 180°
∠B + ∠D = 180°
Construction: Join AC and BD

Solution 29.
Length of the cylinder = 15 cm
Radius = 24.5 cm, diameter of the wire = 0.6 cm

Solution 30.
Data in ascending order 7, 8, 9, 10, 12, 13, 15, 15, 15, 20, 20, 20, 20, 25, 25 => n = 15.

Median =15
Mode = Maximum number of observations = 20 (4 times)
(ii) Value => Social work.

We hope the CBSE Sample Papers for Class 9 Maths Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 4, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 9 Maths Paper 3 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 3

CBSE Sample Papers for Class 9 Maths Paper 3

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 3 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
Write a rational number between √2 and √3

Question 2.
Find the remainder when P(x) = x³ – 3x² + 4x + 32 is divided by (x + 2).

Question 3.
What is the equation of y-axis?

Question 4.
If 3∠A = 4∠B = 6∠C, then ∠A : ∠B : ∠C = ?

Question 5.
In which quadrant point (x, y) lies if x < 0, y > 0?

Question 6.
The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio of 5:3, then find the ratio of their volumes.

SECTION-B

Question 7.
Factorize: x(x – y)³ + 3x²y (x -y).

Question 8.
The base of an Isosceles triangle measure 24 cm and its area is 192 cm². Find its perimeter.

Question 9.
The runs scored by two batsmen in a cricket match is 164. Write a linear equation in two variables x and y. Also write a solution of this equation.

Question 10.
12 packets of salts, each marked 2 kg, actually contained the following weights (in kg) of salt.
1.950, 2.020, 2.060, 1.980, 2.030, 1.970
2.040, 1.990, 1.985, 2.025, 2.000, 1.980
Out of these packets one packet is chosen at random. What is the probability that the chosen packet contains more than 2 kg of salt?

Question 11.
The percentage of marks obtained by a student in six unit test are given below.

The unit test is selected at random. What is the probability that the student gets more than 60% marks in the test?

Question 12.
Evaluate (104)³ using suitable identity.

SECTION-C

Question 13.
If x = 7 + 4√3 , then find the value of \(\left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) \)
OR
Find the value of √4.3 geometrically and represent √4.3 on a given real line.

Question 14.
a³ – b³ + 1 + 3ab

Question 15.
In the figure, AB || CD. Prove that ∠p + ∠q – ∠r = 180°.

Question 16.
In the given figure, ABC is a triangle in which AB = AC. Side BA is produced to D, such that AB = AD. Prove that ∠BCD = 90°.

Question 17.
Mid Point Theorem: Prove that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.
OR
Prove that parallelograms on the same base and between same parallels are equal in area.

Question 18.
In the given figure O is the centre of circle, prove that ∠x + ∠y = ∠z

Question 19.
Construct a ∆ABC in which BC = 6 cm, ∠B = 60° and sum of other two sides is 9 cm [i.e., AB + AC = 9 cm].

Question 20.
Plot the points A(3, 3), B(2, 4), C(5, 5), D(0, 2), E(3, -3) and F(-5, -5) on a graph paper. Which of these points are the mirror images in
(i) x-axis,
(ii) y-axis?

Question 21.
A cylindrical metallic pipe is 14 cm long. The difference between the outside and inside surface is 44 cm². If the pipe is made up of 99 cubic cm of metal, find the outer and inner radii of the pipe.

Question 22.
The total surface area of a solid cylinder is 231 cm² and its curved surface area is \(\frac { 2 }{ 3 }\) of the total surface area. Find the volume of the cylinder.

SECTION-D

Question 23.
Express \(0.6+0.\overline { 7 } +0.\overline { 47 } \) in the form \(\frac { p }{ q }\), wherep and q are integers and q ≠ 0.

Question 24.
If both (x – 2) and \(x-\frac { 1 }{ 2 }\) are factors of polynomial Px² + 5x + r show that \(\frac { P }{ r }\) = 1.

Question 25.
A pharmacist needs to strengthen a 15% alcoholic solution to one of 32% alcohol. How much pure alcohol should be added to 800 ml of 15% solution?

Question 26.
In the figure, POQ is a line, Ray OR is perpendicular to line PQ, OS is another ray lying between rays OP and OR. Prove that
∠ROS = \(\frac { 1 }{ 2 }\)(∠QOS – ∠POS)

Question 27.
In the given figure, C is the mid point of AB. If ∠DCA = ∠ECB and ∠DBC = ∠EAC, prove that
(i) ∆DBC ≅ ∆EAC
(ii) DC = EC

Question 28.
In the adjoining figure, D, E, F are mid points of the sides BC, CA and AB of ∆ABC. If BE and DF intersect at X while CF and DE intersect at Y, prove that XY = \(\frac { 1 }{ 4 }\)BC

Question 29.
Draw a histogram to represent the following frequency distribution:

Question 30.
Kartikeya wanted to make a temporary shelter for street dogs by making a box like structure with tarpaulin that covers all the four sides and the top of the house. How much tarpaulin would be referred to make the shelter of height 2.5 m with base dimensions 4m x 3m. Assuming stitching margin is negligible. Which values are depicted in this question?

Solutions

Solution 1.
√2 = 1.414, √3 = 1.732
Rational number between √2 and √3 = \(\frac { \sqrt { 2 } +\sqrt { 3 } }{ 2 } \) = 1.573 ~ 1.6 upto one place of decimal = \(\frac { 8 }{ 5 }\)

Solution 2.
P(x) = x³ – 3x² + 4x + 32,x + 2 = 0 => x = -2
When P(x) is divided by (x + 2), we get
Remainder = P(-2) = [(-2)³ – 3(-2)² + 4(-2) + 32] = -8 – 12 – 8 + 32 = 4
Remainder = 4

Solution 3.
x = 0.

Solution 4.
Let 3∠A = 4∠B = 6∠C = k°

Solution 5.
If x < 0, y > 0 => x = -ve and y = +ve => P(x, y) lies in quadrant II because it is (-, +).

Solution 6.
Let their radius be 2x cm and 3x cm and their height be 5y cm and 3y cm respectively and volume of cylinder = πr²h

Solution 7.
x(x – y)³ + 3x²y (x – y) = x(x – y) [(x – y)² + 3xy] = x(x – y) [(x² + y² – 2xy + 3xy)]
= x (x – y) (x² + y² + xy).

Solution 8.
AL ⊥ BC

Hence perimeter of Isosceles triangle = 20 + 20 + 24 = 64 cm.

Solution 9.
Both Batsmen make runs = x + y = 164
So, linear equation = x + y = 164
If x = 100, then y = 164 – 100 = 64
One solution is

Solution 10.
Total number of packets of salts = 12
Number of packets containing more than 2 kg of salt = 5.

Solution 11.
Number of test in which student gets more than 60% marks = 2.
Total number of test = 6

Solution 12.
(104)³ = (100 + 4)³ = (100)³ + (4)³ + 3(100)(4)(100 + 4)
= 1000000 + 64 + 1200 (100 + 4)
= 1000000 + 64 + 120000 + 4800 = 1124864

Solution 13.

Steps of Construction:
Draw a line segment AB = 4.3 units and extend it to C such that BC = 1 unit.
Find the mid point O of AC.
Draw a semicircle with O as centre and OA as radius.
Draw BD ⊥ AC intersecting the semicircle at D.
Then BD = √4.3 units
with B as centre and BD as radius, draw an arc meeting with AC produced at E. Then BE = BD = √4.3 units.

Solution 14.
a³ – b³ + 1 + 3ab = a³ + (-b)³ + 1 + 3ab = a³ + (-b)³ + (1)³ – 3 x a x (-b) x 1
a³ – b³ + 1 + 3ab = (a – b + 1) (a² + b² + 1 + ab + b – a)
= (a – b + 1) (a² + b² + ab – a + b + 1)
a³ – b³ + 1 + 3 ab = (a – b + 1) (a² + b² + ab – a + b + 1)

Solution 15.

AB || CD.
Through F, draw PFQ || AB || CD
So, q = ∠1 + ∠2
AB || PF and EF is transversal line
∠p° + ∠1 = 180° …(i)
Again
PF || CD and GF is transversal line
∠2 = ∠r [Alternate Interior angles] …(ii)

Adding (i) and (ii)
∠p + ∠1 + ∠2 = 180° + ∠r
∠p + (∠1 + ∠2) = 180° + ∠r
∠p + ∠q = 180° + ∠r
∠p + ∠q – ∠r = 180°

Solution 16.

AB = AC => ∠ABC = ∠ACB
AC = AD => ∠ADC = ∠ACD
∴ ∠ABC + ∠ADC = ∠ACB + ∠ACD = ∠BCD
∠DBC + ∠BDC = ∠BCD
[∵ ∠ABC = ∠DBC, ∠ADC = ∠BDC]
Adding ∠BCD on both sides
∠DBC + ∠BDC + ∠BCD = 2∠BCD
180° = 2 ∠BCD[Angle sum property of A]
=> ∠BCD = 90°
Hence, ∠BCD = 90°.

Solution 17.

Given: A ∆ABC in which D and E are the mid points of AB and AC respectively, DE is joined.
To prove: DE || BC and DE = \(\frac { 1 }{ 2 }\)BC
Construction: Draw CF || BA, meeting DE produced in F.
Proof: In ∆AED and ∆CEF
AE = CF (given)
∠AED = ∠CED (V.O.A.)
∠DAE = ∠FCE (alternate interior angles) BD || CF
∴ ∆AED ≅ ∆CEF (By ASA congruency)
=> AD = CF (CPCT)
But AD = BD
=> BD = CF and BD || CF (by construction)
∴ BCFD is a parallelogram.∴DF || BC and DF = BC
=> DE || BC [∵ DF || BC]
DE = \(\frac { 1 }{ 2 }\) DF = \(\frac { 1 }{ 2 }\) BC [∵ DF = BC]
DE || BC and DE = \(\frac { 1 }{ 2 }\) BC
OR
Given: Two parallelograms ABCD and ABEF on the same base AB and between the same parallel lines AB and FC.

Solution 18.
In ∆ACF, side CF is produced to B
∴ ∠y = ∠1 + ∠3 …(i)
[exterior angle = sum of interior opposite angles]
In ∆AED, side ED is produced to B
∠1 + ∠x = ∠4 …(ii)
Adding eq. (i) and (ii)
∠1 + ∠x + ∠y = ∠1 + ∠3 + ∠4
⇒∠x + ∠y = ∠3 + ∠4 [∵∠4 = ∠3 angles in the same segment ]
⇒∠x + ∠y = 2∠3 [∵∠AOB = 2∠ACB]
∠x + ∠y = ∠z
Hence ∠x + ∠y = ∠z

Solution 19.
Steps of Construction:
(i) Draw BC = 6 cm
(ii) Construct ∠CBX = 60°
(iii) Along BX, set off BP = 9 cm
(iv) Join CP
(v) Draw the perpendicular bisector of CP to intersect BP at A.
(vi) Join AC, then ∆ABC is the required triangle.

Solution 20.
(i) (3, -3) is the mirror image of (3, 3) in x-axis.
(ii) No mirror image in y-axis.

Solution 21.
Let the outer and inner radii be R cm and r cm respectively.
h = 14 cm, outer surface area = 2πRh = 2 x \(\frac { 22 }{ 7 }\) x R x 14 = (88R) cm²

Solution 22.
TSA of cylinder = 2πrh + 2πr² = 231 …(1)
CSA of cylinder = 2πrh = \(\frac { 2 }{ 3 }\) of (TSA of cylinder)

Solution 23.

Solution 24.
Let f(x) = Px² + 5x + r
If (x – 2) is a factor of P(x), then remainder = 0
x – 2 = 0 => x = 2
Remainder = f(2) = P(2)² + 5x² + r = 0
4P + 10 + r = 0
4P + r = -10 ..(i)

Solution 25.
Let x ml of the pure alcohol be added to be 15% solution to get 32% alcoholic solution. New volume of alcoholic solution = (800 + x) ml
Quantity of pure alcohol in (800 + x) ml of 32% solution = Quantity of pure alcohol in 800 ml of 15% solution + x ml of pure alcohol

Solution 26.
∠QOS = ∠ROQ + ∠ROS …(i)
∠POS = ∠POR – ∠ROS …(ii)
Subtracting eqn. (i) by (ii)
∠QOS – ∠POS = ∠ROQ – ∠POR + 2∠ROS
∠QOS – ∠POS = (∠ROQ – ∠ROQ) + 2∠ROS (∠ROQ = ∠POR = 90°)
=> 2∠ROS = ∠QOS – ∠POS
=> ∠ROS = \(\frac { 1 }{ 2 }\) [∠QOS – ∠POS]

Solution 27.

(i) ∠DCA = ∠ECB
=> ∠DCA + ∠DCE = ∠ECB + ∠DCE
=> ∠ACE = ∠BCD
Now in AAEC and ABCD
∠DBC = ∠EAC (given)
AC = BC (C is the mid point)
∠ACE = ∠BCD (Proved)
∴ ADBC ≅ AEAC (ASA congruency)
(ii) ADBC ≅ AEAC
DC = EC (CPCT)

Solution 28.

In ∆ABC, F and E are the mid-points of AB and AC respectively.
∴FE || BC and FE = \(\frac { 1 }{ 2 }\) BC = BD
∴FE || BD and FE = BD [∵FE||BC => FE||BD]
=> BDEF is a parallelogram whose diagonals BE and DF intersect each other at X.
∴X is the midpoint of DF.
Similarly, Y is the mid point of DE.
Thus, in ∆DEF, X and Y are the midpoints of DF and DE respectively.
So, XY || FE

Solution 29.
In the given frequency distribution, class sizes are different. So, to calculate the adjusted frequency for each class, Minimum class size = 5

Solution 30.
Required tarpaulin = Area of shelter = Curved surface area of cuboid + Top area
= 2 (l + b) x h + lb
= 2(4 + 3) x 2.5 + 4 x 3
= 35 + 12
= 47 m²
Values: (i) Care for animal, (ii) Kindness.

We hope the CBSE Sample Papers for Class 9 Maths Paper 3 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 3, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 9 Maths Paper 2 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 2

CBSE Sample Papers for Class 9 Maths Paper 1

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 2 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
If √2 = 1.414 , then \(\sqrt { \frac { (\sqrt { 2 } -1) }{ (\sqrt { 2 } +1) } } =\) ?

Question 2.
If \(\frac { x }{ y } +\frac { y }{ x } =-1\), where x ≠ 0,y ≠ 0, then find the value of x³ – y³.

Question 3.
Who is known as the “Father of Geometry”?

Question 4.
Find the area of the ∆OAB with 0(0, 0). A(4, 0) and B(0, 6).

Question 5.
Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface area.

Question 6.
In 50 tosses of a coin, tail appears 32 times. If a coin is tossed at random, what is the probability of getting a head?

SECTION-B

Question 7.
If a + b + c = 0, then \(\left( \frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { c^{ 2 } }{ ab } \right) =?\)

Question 8.
Find the measure of an angle, if six times its complement is 12° less than twice its supplement.

Question 9.
If an angle of a parallelogram is two third of its adjacent angle, find the smallest angle of the parallelogram.

Question 10.
In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that \(\frac { ar(\Delta ABD) }{ ar(\Delta ADC) } =\frac { m }{ n } \)

Question 11.
The sides of a triangle are in the ratio 5:12:13 and its perimeter is 150 cm. Find the area of triangle.

Question 12.
Find the mean of the first six prime numbers.

SECTION-C

Question 13.
Rationalise the denominator of \(\frac { \left( \sqrt { 3 } +\sqrt { 2 } \right) }{ \left( \sqrt { 3 } -\sqrt { 2 } \right) } \)
OR
If a and b are rational numbers and \(\frac { \left( \sqrt { 11 } -\sqrt { 7 } \right) }{ \left( \sqrt { 11 } +\sqrt { 7 } \right) } =a-b\sqrt { 77 } \)
Find the value of a and b.

Question 14.
Factorize (p – q)³ + (q – r)³ + (r – p)³

Question 15.
Draw the graph of the equation 2x + 3y = 11. From your graph, find the value of y, when x = -2.

Question 16.
In a quadrilateral ABCD the line segment bisecting ∠C and ∠D meet at E.
Prove that ∠A + ∠B = 2∠CED.

Question 17.
In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN, then prove that AC bisect BD.

Question 18.
Without plotting the given points on a graph paper, indicate the quadrants in which they lie
(a) ordinate = 6, abscissa = -3
(b) ordinate = -6, abscissa = 4
(c) abscissa = -5, ordinate = -7
(d) ordinate = 3, abscissa = 5
(e) (-6, 5)
(f) (2,-9)

Question 19.
A well of inner diameter 14 m is dug to a depth of 15 m. Earth taken out of it has been evenly spread all around it to a width of 7 m to form an embankment. Find the height of embankment so formed.

Question 20.
Find the number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 5 cm and diameter of 4.5 cm.

Question 21.
The table given below shows the ages of 75 teachers in a school.

Here 18-29 means from 18 to 29 including both. A teacher from the school is chosen at random. What is the probability that the teacher chosen is
(i) 40 years or more than 40 years old?
(ii) 49 years or less than 49 years old?
(iii) 60 years or more than 60 years old?

Question 22.
Construct a ∆ABC in which BC = 5.6 cm, ∠B = 30° and the difference between the other two sides is 3 cm.

SECTION D

Question 23.
If a and b are rational numbers and \(\frac { 4+3\sqrt { 5 } }{ 4-3\sqrt { 5 } } =a+b\sqrt { 5 } \) Find the value of a and b.

Question 24.
If x + y + z = 0, prove that x³ + y³ + z³ = 3xyz. Without actual calculation find the value of (-12)³ + 7³ + 5³ using above identity.

Question 25.
A taxi charges RS 20 for the first km and @ RS 12 per km for subsequent distance covered. Taking ‘the distance covered as x km and total fare RS y, write a linear equation depicting the relation in ,x and y
(i) Draw the graph between x and y.
(ii) From your graph find the taxi charges for covering 14 km.

Question 26.
In the given figure, AB || CD. Find the value of x.

Question 27.
In a ∆ABC, the bisectors of ∠B and ∠C intersect each other at a point O. Prove that ∠BOC = 90° + \(\frac { 1 }{ 2 }\)∠A

Question 28.
Prove that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Question 29.
The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be \(\frac { 1 }{ 27 }\) of the volume of the given cone, at what height above the base, the section has been made?

Question 30.
A doctor suggests two ways for treatment of a particular disease one is by taking medicine only and other is by doing meditation and yoga.

(i) Draw frequency polygon for the given data on the same graph.
(ii) What is the importance of yoga and meditation in our life?

Solutions

Solution 1.

Solution 2.
\(\frac { x }{ y } +\frac { y }{ x } =-1\) ⇒ x² + y² = -xy
⇒ x² + y² + xy = 0
⇒ x³ – y³ = (x – y)(x² + y² + xy) = (x – y) x 0 = 0
⇒ x³ – y³ = 0

Solution 3.
Mathematician Euclid is known as “Father of Geometry’”.

Solution 4.
OA = 4 units, OB = 6 units
ar(∆OAB) = \(\frac { 1 }{ 2 }\) x OA x OB = \(\frac { 1 }{ 2 }\) x 4 x 6 = 12 sq. unit

Solution 5.
Volume of cube = (side)³, surface area of cube = 6(side)²
Let the volume of first cube = a³; and second cube = b³

Solution 6.
Probability of getting head

Solution 7.
a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc

Solution 8.
Let the measure of the required angle = x°
Measure of its complement = (90° – x).
Measure of its supplement = (180° – x)
6(90° – x) = 2(180° – x) – 12
⇒ 540° – 6x = 360° – 2x – 12 ⇒ 4x = 192°
⇒ x = 48°
Hence the measure of required angle = 48°.

Solution 9.
Let one angle of the parallelogram be x°.
We know that sum of two adjacent angles of a parallelogram is equal to 180°.

Solution 10.

Solution 11.
Let the sides of the triangle be 5x, 12x and 13x as a, b and c. Then
5x + 12x + 13x = 150 => 30x =150 => x = 5
∴ a = 25 cm, b = 60 cm and c = 65 cm
given 2s = 150 cm => s = 75 cm
(s – a) = 50, (s – b)= 15, (s – c) = 10

Solution 12.
The first six prime numbers are 2, 3, 5, 7, 11 and 13.

Solution 13.

Solution 14.
Putting p – q = x,q – r = y and r – p = z
x + y + z = p – q + q – r + r – p = O if x + y + z = 0 then x³ + y³ + z³ = 3xyz
∴ (p – q)³ + (q – r)³ + (r – p)³ = 3(p – q)(q – r)(r – p)

Solution 15.
2x + 3y = 11


Line AB is the required graph of 2x + 3y= 11.
Reading the Graph: Given x = -2. Take a point M on the x-axis such that OM = -2.
Draw PM, parallel to they-axis, meeting the line AB at P, it is PM = 5 = y.
So, when x = -2 then y = 5.

Solution 16.
Let CE and DE be the bisector of ∠C and ∠D respectively.

Solution 17.
We want to prove that AC bisect BD i.e., OB = OD
Let AC and BD intersect at O and BM = DN
Now ∴ ∆OND ≅ ∆OMB (By ASA)
=> OD = OB (CPCT)
[ ∵ ∠OND = ∠OMB (each 90°), DN = BM (given), ∠DON = ∠BOM (V.O.A.)]
=> AC bisect BD.

Solution 18.
(a) A(-3, 6) lies in quadrant II
(b) B(4, -6) lies in quadrant IV
(c) C(-5, -7) lies in quadrant III
(d) D(5, 3) lies in quadrant I
(e) E(-6, 5) lies in quadrant II
(f) F(2, -9) lies in quadrant IV

Solution 19.
For well: r = 7m, h = 15m = depth
Volume of the earth dug out = Volume of the well
= Volume of cylinder = πr²h

Solution 20.
Each coin is cylindrical in shape.

= 45 x 5 = 225
The number of coins required = 225

Solution 21.
Total number of teachers = 75
(i) Number of teachers who are 40 years or more than 40 years old = 35 + 10 = 45

(ii) Number of teachers who are 49 years or less than 49 years old = 5 + 25 + 35 = 65

(iii) Number of teachers who are 60 years or more = 0

Solution 22.
Steps of Construction:
(i) Draw BC = 5.6 cm
(ii) Construct ∠CBX = 30°
(iii) Set off BP = 3 cm
(iv) Join PC
(v) Draw the right bisector of PC, meeting BP produced at A.
(vi) Join AC. Then ∆ABC is the required triangle.

Solution 23.

Solution 24.
x + y + z = 0 => x + y = – z => (x+y)³ = (-z)³
=> x³ + y³ + 3xy (x + y) = -z³
=> x³ + y³ + 3xy (-z) = -z³
=> x³ + y³ – 3xyz = -z³
=> x³ + y³ + z³ – 3xyz = 0
=> x³ + y³ + z³ = 3xyz
Putting x = -12, y = 7, z = 5
We get x + y + z = -12 + 7 + 5 = -12 + 12 = 0 then x³ + y³ + z³ = 3xyz
=> (-12)³ + (7)³ + (5)³ = 3 x (-12) x 7 x 5
= -3 x 7 x 60
= -21 x 60
= -1260
(-12)³ + 7³ + 5³ = -1260.

Solution 25.
y = 20 + 12 (x – 1) = 20 + 12x – 12
=> y = 12x + 8

On the graph paper take distance along x-axis and fare (in Rs) along y-axis. We plot the points A(6, 80) and B(11, 140) on the graph paper. Join AB and produce it on both sides to obtain the required graph.
From the graph, when x = 14, we find y = Rs 176.

Solution 26.

Through E draw a line GEH || AB || CD
Now GE || AB and EA is transversal
∴ ∠GEA = ∠EAB = 50°
[Alternate Interior Angles]
Again EH || CD and EC is a transversal
∠HEC + ∠ECD = 180°
[Cointerior angles of the same side of transversal]
=> ∠HEC + 100° = 180°
=> ∠HEC = 80°
Now GEH is a straight angle
∠GEA + ∠AEC + ∠HEC = 180°
=> 50° + x° + 80° =180° => x = 50°

Solution 27.
In ∆ABC we have
∠A + ∠B + ∠C = 180°

Solution 28.
Given: A circle C(O, r) in which arc AB subtends ∠AOB at the centre and ∠ACB at any point C on the remaining part of the circle.
To prove: (i) and (ii) ∠AOB = 2 ∠ACB, when AB is a minor arc or a semicircle.
(iii) Reflex ∠AOB = 2∠ACB, when AB is a major arc.
Construction: Join AB and CO. Produce CO to a point D outside the circle.

We know that when one side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles.
∴ ∠AOD = ∠OAC + ∠OCA
∠BOD = ∠OBC + ∠OCB
[∵ OC = OA = r ,OC = OB = r]
∠AOD = 2∠OCA and ∠BOD = 2∠OCB
In Figure (i) and Figure (ii) i.e., Case (i) and Case (ii)
∠AOD + ∠BOD = 2∠OCA + 2∠OCB => ∠AOB = 2(∠OCA + ∠OCB)
=> ∠AOB = 2∠ACB
In Figure (iii) i.e., Case (iii)
∠AOD + ∠BOD = 2∠OCA + 2∠OCB
=> Reflex ∠AOB = 2 (∠OCA + ∠OCB) => Reflex ∠AOB = 2∠ACB

Solution 29.
Let the smaller cone have radius r and height h cm and let the radius of the given original cone be R cm.

Solution 30.
(i) We take two imagined classes, one at the beginning, namely (10-20) and the other at the end, namely (70-80) each with frequency zero.
With these two classes, we have the following frequency table:

Importance of Yoga and Meditation:
1. Meditation gives you an experience, an inner state, where what is you and what is your is separated.
2. By meditation life seems to be flowing without conflict and just feel more cohesive.
3. Yoga is important for us to achieve:
(i) Good Health
(ii) Success
(iii) Overall well being
(iv) For peace, joy and love
(v) For inner exploration.

We hope the CBSE Sample Papers for Class 9 Maths Paper 2 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 2, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 9 Maths Paper 5 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 5

CBSE Sample Papers for Class 9 Maths Paper 5

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 5 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
Write the value of \(\frac { \sqrt { 32 } +\sqrt { 48 } }{ \sqrt { 8 } +\sqrt { 12 } } \)
OR
Find \(\sqrt [ 4 ]{ \sqrt [ 3 ]{ { 2 }^{ 2 } } } \)

Question 2.
For what value of a, x – 3 is a factor of x³ + x² – 17x + a?

Question 3.
In figure below, ∆PQR is an isosceles right triangle right angled at Q. Find angle P (∠P).

Question 4.
In a parallelogram ABCD, E and F are any two points on the sides AB and BC respectively. If ar (∆DCE) is 12 cm², then find ar (∆ADF).

Question 5.
What is the volume of the hollow right circular cylinder?

Question 6.
In a class consisting of 18 boys and 22 girls, one student is absent. Find the probability that the absent student is boy.

SECTION-B

Question 7.
If \(\frac { { x }^{ 2 }+1 }{ x } =7\), then find the value of \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \)

Question 8.
Find the angle which is two times its supplementary angle.

Question 9.
Which of the following points lie on the x-axis? Show the position of other points also.
A (1, 1), B (1, 0), C (0, 1), D (0, 0), E (-1, 0), F (0, -1) G (4, 0), H (0, 7)

Question 10.
Answer the following:
(i) A point lies on y-axis, then which coordinate is zero?
(ii) If a point is at a distance of 2 units from y-axis and 3 units from x-axis. Write the coordinates.

Question 11.
Into a circular drum of radius 4.2 m and height 3.5 m, how many full bags of wheat can be emptied if the space required for wheat on each bag is 2.1 cubic m? (Take π = 3.14)

Question 12.
Find the median of first 11 multiples of 3.

Question 13.
Simplify

OR
If 5^2x-1 – (25)^x-1 = 2500, then find the value of x.

Question 14.
If 3x + y + z = 0, show that 27x³ + y³ + z³ = 9xyz.
OR
If a + b = c, then show that b² + ac = c² – ab.

Question 15.
In the given figure, ∠ABC = 60°, ∠BCE = 25°, ∠DCE = 35° and ∠CEF = 145°. Prove that AB || EF.

Question 16.
Find the value of x and y, if (x + 4, 3y – 2) = (7, -5).

Question 17.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
OR
Show that the diagonals of a square are equal and bisect each other at right angle.

Question 18.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

OR
In the following figure, AB is equal to the radius of the circle. Find the value of ∠AMB, if ‘O’ is the centre of the circle.

Question 19.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Question 20.
An isosceles triangle with base 48 cm has area 240 sq cm. Find the remaining two sides of the triangle.

Question 21.
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square meter, find the
(i) Inside surface area of the dome
(ii) Volume of the air inside the dome.

Question 22.
A jar contains 3000 white, black and red beads. The beads are thoroughly mixed and a sample of 60 is taken. The sample is found to contain 17 white beads, 32 black beads and 11 red beads. Estimate the number of beads of each colour in the jar.

SECTION-D

Question 23.
If 2^a = 3^b = 6^c, then show that \(c=\frac { ab }{ a+b }\)

Question 24.
If both (x – 2) and \(x-\frac { 1 }{ 2 }\) are factors px² + 5x + 8, show that \(\frac { p }{ r }=1\)

Question 25.
In figure below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Question 26.
Bisectors of angle A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.Prove that the angles of a triangle DEF are 90° – \(\frac { 1 }{ 2 }\)A, 90° – \(\frac { 1 }{ 2 }\)B, and 90° – \(\frac { 1 }{ 2 }\)C

Question 27.
A cylinder is within the cube touching all the vertical faces. A cone is inside the cylinder. If their heights are same with the same base. Find the ratio of their volumes.

Question 28.
Construct a combined histogram and frequency polygon for the following frequency distribution.

Question 29.
ABC is a triangle right angled at C. A line through the mid¬point M of hypotenuse AB and parallel to BC intersect AC at D, show that:
(i) D is the mid-point of AC.
(ii) MD ⊥ AC
(iii) CM = MA = \(\frac { 1 }{ 2 }\)AB

Question 30.
Kartikeya and Pallavi of class IX decided to collect Rs 25 for class cleanliness. Write it in linear equations in two variables. Also draw the graph. What values of both the students are depicted here?

Solutions

Solution 1.

Solution 2.
x – 3 = 0 => x = 3, p(x) = x³ + x² – 17x + a
For factor x – 3, remainder p(3) = 0
P(3) = (3)³ + (3)² – 17 x 3 + a = 0 => 27 + 9 – 51 + a = 0
a = 51 – 36 = 15 => a = 15

Solution 3.
∠P = ∠R (Isosceles A), ∠Q = 90°
∠P + ∠Q + ∠R = 180° => ∠P + ∠P + 90° = 180°
2 ∠P = 90° =>∠P = 45°

Solution 4.
ar (∆ADF) = \(\frac { 1 }{ 2 }\) ar (||gm ABCD) = ar (∆DCE)
ar (∆ADF) = ar (∆DCE)
ar (∆ADF) = 12 cm²
[In same base and between same parallels area of triangle is half the area of parallelogram.

Solution 5.
Volume = πh (R² – r²) cubic unit.

Solution 6.
Total students = 18 + 22 = 40, Boys = 18, girls = 22
P (Absent boys) = p (E) = \(\frac { 18 }{ 40 }\) = \(\frac { 9 }{ 20 }\)

Solution 7.

Solution 8.
Let angle be x. It supplementary angle = (180 – x)°
=> x = 2(180 – x)
=> x = 360 – 2x =>3x = 360° => x = 120°

Solution 9.
(i) Points lie on x-axis may be (a, 0) => B (1, 0), E (-1, 0), G (4, 0).
(ii) Points lie on y-axis may be (0, b) => C (0, 1), F (0, -1), H (0, 7)
(in) Point lies on origin => D(0, 0)
(iv) Point lies of I quadrant may be (x, y) = (1, 1).

Solution 10.
(i) If a point lies on y-axis, then its x-coordinate will be zero, i.e., it is represented by (0, b).
(ii) Coordinates of points be (2, 3).

Solution 11.
r = 4.2m
h = 3.5m

Complete or full bags of wheat = 92 bags.

Solution 12.
First 11 multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33
Median = Middle term of the observations
Total terms = n = 11

Median = 18 (6th term)

Solution 13.

Solution 14.
If a + b + c = 0 then a³ + b³ + c³ = 3abc
If a = 3x, b = y, c = z
then a + b + c = 3x + y + z = 0
and, a³ + b³ + c³ = 3 abc
(3x)³ + (y³) + (z)³ = 3 x (3x) x y x z
27 x³ + y³ + z³ = 9 xyz
OR
Since a + b = c
On squaring both sides (a + b)² = c²
a² + b² + 2 ab = c²
a² + b² + ab + ab = c²
(a² + ab) + b² = c² – ab
a (a + b) + b² = c² – ab
ac + b² = c² – ab [since a + b = c]
b² + ac = c² – ab

Solution 15.
∠BCD = ∠BCE + ∠DCE
= 25° + 35° = 60°

Solution 16.
[(x + 4), (3y – 2)] = (7, -5)
On comparing
x + 4 = 7 ⇒ x = 7 – 4 = 3 ⇒ x = 3
3y – 2 = -5 ⇒ 3y = -5 + 2 = -3
⇒ y = \(\frac { -3 }{ 3 }\) = -1
x = 3,y = -1

Solution 17.

Given: In quadrilateral ABCD
OA= OC and ∠AOB = ∠BOC = ∠COD
OB = OD = ∠DOA = 90°
To prove: ABCD is a rhombus
i.e. ABCD is a parallelogram having all sides equal.
Proof: In ∆AOD and ∆COD
OA = OC(Diagonals bisect each other)
∠AOD = ∠COD (Each 90°)
OD = OD (Common)
∴ ∆AOD ≅ ∆COD (By S.A.S)
=> AD = CD (By CPCT)
Similarly, AD = AB, and CD = BC
So, AB = BC = CD = DA
Since opposite sides of a quadrilateral ABCD are equal, so it is a parallelogram.
Here, all sides of the parallelogram are equal. So it is rhombus.
OR

Given: ABCD is a square and diagonals AC and BD intersect each other at O.
To Prove: AC = BD, OA = OC, OB = OD and ∠AOB = 90°
Proof: In ∆ABC and ∆DCB
AB = DC
∠ABC = ∠DCB
BC = BC
∴ ∆ABC ≅ ∆DCB
AC = DB
Hence the diagonals of a square are equal in length.
In ∆AOB and ∆COD ,
∠AOB = ∠COD
∠ABO = ∠CDO
AB = CD
∴ ∆AOB ≅ ∆COD
=> AO = CO and OB = OD
Hence diagonals of a square bisect each other.
Now in ∆AOB and ∆COB
OA = OC; AB = BC and BO = BO => ∴ ∆AOB ≅ ∆COB(SSS Congruence)
∠AOB = ∠COB (CPCT)
∠AOB = ∠COB
But ∠AOB + ∠COB = 180° (Linear pair)
2 ∠AOB = 180°
∠AOB = \(\frac { 180 }{ 2 }\) = 90°
∠AOB = 90°
Hence, the diagonals of a square bisect each other at right angle.

Solution 18.

Given: In trapezium ABCD, AB || DC and BC = AD
To Prove: ABCD is a cyclic quadrilateral.
Construction: Draw A ⊥ DC and BN ⊥ DC
Proof: In ∆AMD and ∆BNC
AD = BC (Given)
∠AMD = ∠BNC (Each 90°)
AM = BM
(Perpendicular distance between two parallel lines are equal)
∴ ∆AMD ≅ ∆BNC (By RHS congruency)
∴∠ADC = ∠BCD …(1) (CPCT)
∵ ∠BAD and ∠ADC are on the same side of transversal AD.
∠BAD + ∠ADC = 180° …(2)
=> ∠BAD + ∠BCD = 180° (∵ ∠ADC = ∠BCD)
=> This shows that opposite angles are supplementary.
=> ABCD is a cyclic quadrilateral.
OR

Given: O is the centre of circle and AB is equal to the radius.
Construction: Join OA, OB and OC.
To Find: ∠AMB
In ∆OAB, OA = OB = OC (Each equal to radius)
=> ∆OAB is an equilateral triangle.
=> ∠AOB = 60°
=> ∠AOB = \(\frac { 180 }{ 2 }\)= 90°
∠ACB = 30°
(Angle subtended by an arc at the centre is double that subtended at any part of the circle)
But ∠DAC = 90° (Angle in the semi-circle is 90°)
Now ∠CAM = 180°- 90° = 90° (Linear pair)
Now in ∆ (CAM)
∠A + ∠C + ∠M= 180°
90° + 30° + ∠M = 180°
∠M = 180° – 90° – 30° = 60°
∠AMB = 60°

Solution 19.
Steps of Construction:
1. Draw a line segment AB of 11 cm (XY + YZ + ZX =11 cm).
2. Construct an angle ∠PAB = 30° at point A and an angle ∠OBA of 90° at point B.
3. Bisect ∠PAB and ∠QBA. Let these bisectors intersect each other at point X.
4. Draw perpendicular bisector ST of AX and UV of BX.
5. Let ST intersects AB at Y and UV intersects AB at Z. Join XY, XZ.
Thus, ∆XYZ is the required triangle.

Solution 20.
Let two sides of Isosceles triangle = x cm
Area = 240 cm², Here, a = x, b = x, c = 48 cm

Hence two sides of Isosceles triangle = 26 cm.

Solution 21.
(i) Cost occured in white-washing the dome from inside = Rs 498.96
Cost of white-washing 1 m² = Rs 2

V = 523.908 m³ = 523.9 m³ approximately.
Hence, volume of air inside the dome is 523.9 m³.

Solution 22.
Sample sum = 17 + 32 + 11 = 60

Hence number of beads
(i) White = 850
(ii) Black = 1600
(iii) Red = 550

Solution 23.
2^a = 3^b ⇒ 2 = 3^b/a ….(1)
6^c = 3^b ⇒ 6 = 3^b/c …..(2)

Solution 24.
Let f(x) = px² + 5x + r. If (x – 2) and \(x-\frac { 1 }{ 2 }\) are factors of polynomial f(x), then remainder must be zero.
If x – 2 = 0, => x = 2
Remainder = f(2) = p(2)² + 5(2) + r = 0
4p + 10 + r = 0
4p + r = – 10 …(1)

Solution 25.
∠BAD = ∠EAC (Given)
Adding ∠DAC on both sides
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE …(1)
In ∆BAC and ∆ADE
AC = AE (Given)
∠BAC = ∠DAE (Proved in Eqn. (1)]
=> AB = AD (Given)
∴ ∆ABC ≅ ∆DAE (By SAS congruency)
BC = DE (CPCT)
Hence, BC = DE .

Solution 26.
It is given that BF is the bisector of ∠B

Solution 27.
Let the length of each edge of the cube be ‘a’ units.
V1 = Volume of cube = a³ cubic units …(1)
Since a cylinder is within the cube and touches all the faces of the cube.
r = Radius of the base of the cylinder = \(\frac { a }{ 2 }\)
H = Height of the cylinder = a

Solution 28.
For frequency polygon 2 more class intervals (-10 – 0) and (50 – 60) are taken and classmarks of all class intervals are taken.
The class marks are -5, 5, 15, 25, 35,45, 55.
The graph is plotted on graph paper. Coordinates on both axes are given here in the graph paper.

Solution 29.
(i) In ∆ABC, M is the mid-point of AB and MD || BC so, D is the mid-point of AC.
(Converse .of mid-point theorem)
(ii) As DM || CB and AC is a transversal line for them
∠MDC + ∠DCB = 180° (Co-interior angles)
∠MDC + 90° = 180°
∠MDC = 90°
MD⊥AC
(iii) Join MC
In ∆AMD and ∆CMD AD = CD (D is mid-point of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (Common)

Solution 30.
Let Kartikaya collected = Rs x
Pallavi collected = Rs y
ATQ, Linear Equation in two variables
x + y = 25
(i) If x = 10,y = 25 – x = 25 – 10 = 15
(ii) If x = 15, y = 25 – 15 = 10


The graph is plotted.
Values: (i) Co-operation
(ii) Awareness about cleanliness,
(iii) Responsibility.

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CBSE Sample Papers for Class 9 Maths Paper 1 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 1

CBSE Sample Papers for Class 9 Maths Paper 1

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 1 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.

Question 2.
What is the degree of zero polynomial?

Question 3.
In ∆ABC, BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then ∠A = ?

Question 4.
Write the signs of abscissa and ordinate of a point in quadrant (II).

Question 5.
The total surface area of a cube is 216 cm². Find its volume.

Question 6.
Find the slope of the line 2x + 3y – 4 = 0.

SECTION-B

Question 7.
If a + b + c = 0, then a³ + b³ + c³ = ?

Question 8.
Find four different solutions of 2x + y = 6

Question 9.
The base of an isosceles triangle is 16 cm and its area is 48 cm². Find the perimeter of a triangle.

Question 10.
If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 13. Find the value of x.

Question 11.
1500 families with 2 children each were selected randomly and the following data were recorded.

Out of these families, one family is selected at random. What is the probability that the selected family has
(i) 2 girls
(ii) no girl.

Question 12.
Two coins are tossed 1000 times and the outcomes are recorded as under.

A coin is thrown at random. What is the probability of getting
(i) at most one head ?
(ii) atleast one head ?

SECTION-C

Question 13.
If x = 3 + √8 , find the value of \(\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) \)

Question 14.
If p = 2 – a, prove that a³ + 6ap + p³ – 8 = 0.

Question 15.
In the given figure, prove that x = α + β + γ.

Question 16.
If the bisector of the vertical angle of a triangle bisect the base, prove that the triangle is Isosceles.

Question 17.
The sides BA and DC of quad. ABCD are produced as shown in the given figure. Prove that x° + y° = a° + b°.

Question 18.
In an equilateral triangle, prove that the centroid and the circumcentre coincide.

Question 19.
Construct a Δ ABC whose perimeter is 14 cm and sides are in the ratio 2:3:4.

Question 20.
Plot the points A(-5, 2), B(-4, -3), C(3, -2) and D(6, 0) on a graph paper. Write the name of figure formed by joining them.

Question 21.
The diameter of a roller, 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground. Find the cost of levelling it at 75 paise per square metre.

Question 22.
The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost of painting the vessel all over at 14 paise per cm².

SECTION-D

Question 23.
Show that

Question 24.
Prove that x³ + y³ + z³ – 3xyz =(x + y + z) (x² + y² + z² – xy – yz – zx). Hence find the value of x³ + y³ + z³ if x + y + z = 0.

Question 25.
In a ∆ABC, the side AB and AC are produced to P and Q respectively. The bisector of ∠PBC and ∠QCB intersect at a point O. Prove that ∠BOC = 90° – \(\frac { 1 }{ 2 }\) ∠A.

Question 26.
In each of the following figures, AB || CD. Find the value of x in each case.

Question 27.
Draw the graph of the equation 3x + 2y = 12. At what points does the graph cut the x-axis and the y-axis.

Question 28.
A river 2m deep and 45 m wide is flowing at the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.
Now answer these questions.
1. Why we should preserve and not waste the water and other natural resources? Write one method to restore the water at your home.
2. Why it is necessary to clean all rivers in the country?

Question 29.
The mean of 25 observations is 36. If the mean of first 13 observations is 32 and that of last 13 observations is 39, find the 13th observation.

Question 30.
Plot the points A(2, 5), B(-2, 2) and C(4, 2) on a graph paper. Join AB, BC and AC. Calculate the area of ∆ABC.

Solutions

Solution 1.

Solution 2.
Degree of zero polynomial is not defined.

Solution 3.

∠A + ∠B = ∠ACD
⇒ ∠A + 40° = 120°
⇒ ∠ A = 80°

Solution 4.
The points in quadrant II are in the form (-, +).

Solution 5.
The Total Surface Area of cube (T.S.A.) = 6a² = 216
=> a² = 36 => a = √36 = 6 cm
Volume of cube = (a)³ = (6)³ = 216 cm³
Volume of cube = 216 cm³

Solution 6.
2x + 3y – 4 = 0 ⇒ 3y = – 2x + 4

Solution 7.
a + b + c = 0 ⇒ a + b = – c
⇒ (a + b)³ = (- c)³ ⇒ a³ + b³ + 3ab (a + b) = -c³
⇒ a³ + b³ + 3ab (- c) = -c³ ⇒ a³ + b³ + c³ = 3abc

Solution 8.
y = 6 – 2x
(i) If x = 1,y = 6 – 2 = 4
(ii) If x = 2,y = 6 – 4 = 2
(iii) If x = 3,y = 6 – 6 = 0
(iv) If x = 4,y = 6 – 8 = -2

Solution 9.
b =16 cm, area = 48 cm²

Perimeter of Isosceles triangle = 2a + b = 20 + 16 = 36 cm.

Solution 10.
Mean of the given observations

Solution 11.
Total number of families = 1500

Solution 12.
Total number of tosses = 266 + 540 + 194 = 1000
(i) At most one head = Number of times one head or no head appears
= 540+ 194 = 734
P(E1) = P(at most one head) = \(\frac { 734 }{ 1000 }\) =0.734
(ii) Atleast one head = Number of times one head or two head appears
= 540 + 266 = 806
P(E2) = P(atleast one head) = \(\frac { 806 }{ 1000 }\) = 0.806

Solution 13.
x + 3√8

Solution 14.
We have p = 2 – a => a + p + (-2) = 0
=> Putting a = x,p = y and (-2) = z.
we get a + p + (-2) = 0 =>x + y + z = 0
⇒a³ + p³ + (-2)³ = 3 × a × p × (-2) [If x + y + z = 0 then x³ + y³ + z³ = 3xyz]
⇒a³ + p³ + (-2)³ = -6ap
⇒a³ + 6ap + p³ – 8 = 0

Solution 15.

Join B and D and produce BD to E.
p + q = β and s + t = x
Side BD of A ABD is produced to E
∴ p + α = s ..(1)
Side BD of A CBD is produced to E.
∴ q + γ = t ..(2)
Adding (1) and (2)
(p + α) + (q + γ) = s + 1
(p + q) + (α) + (γ) = x
β + α + γ = x
⇒ x = α + β + γ

Solution 16.

Given: ∆ AABC in which AD is the bisector of ∠ A, which meets BC in D such that BD = DC.
To prove: AB = AC
Construction: Produce AD to E such that AD = DE and join EC.
Proof: In ∆ABD and ∆ECD
BD = DC (given)
AD = DE (by construction)
∠ADB = ∠EDC (V.O.A)
=> ∴ ∆ABD ≅ ∆ECD
∴ ∆ABC ≅ ∆ECD
=> AB = EC and ∠1 = ∠3 …(2) (CPCT)
Also ∠1 = ∠2 …(3) [∵AD bisects ∠A]
∴ ∠2 = ∠3 [using (2)and(3)]
=> EC = AC …(4) [side opposite to equal angles]
=> AB = AC [using (1)and(4)]
=> ∆ABC is Isosceles.

Solution 17.
We have ∠ A + b° = 180° (Linear pair)
=> ∠A = 180° – b° …(1)
Also ∠C + a° = 180° (Linear pair)
∠C = 180° – a° …(2)

Solution 18.

Let ∆ ABC be the given equilateral triangle and let its median AD,
BE and CF intersect at G.
Then G is the centroid of ∆ ABC
In ∆ BCE and ∆ CBF
BC = CB (common)
∠B = ∠C [each 60°]

This shows that G is the circumcentre of ∆ ABC
=> G is the centroid as well as circumcentre of ∆ ABC
Note: Centroid: The point of intersection of the medians of a triangle is called centroid.
The centroid of the triangle is the point located at \(\frac { 2 }{ 3 }\) of the distance from a vertex along the median.
Circumcentre: The centre of the circumcircle of the triangle is called the circumcentre.

Solution 19.
Steps of construction:
(i) Draw a line segment PQ =14 cm. Draw a ray PX making an acute angle with PQ and draw in the downward direction.
(ii) From P, mark set of (2 + 3 + 4) = 9 equal distance along PX.
(iii) Mark points L, M, N on PX such that PL = 2 cm, LM = 3 cm and MN = 4 cm.
(iv) Join NQ. Through L and M draw LB || NQ and MC || NQ. Cutting PQ at B and C respectively.
(v) With B as centre and radius BP draw an arc. With C as centre and radius CQ draw another arc, cutting the previous arc at A.
(vi) Join AB and AC. ∆ ABC is the required triangle.

Solution 20.
Points A(-5, 2), B(-4, -3), C (3, -2) and D(6, 0) are shown on graph paper.

The figure formed by joining them is called quadrilateral.

Solution 21.
Radius of roller = r = 42 cm, Length = h = 120 cm
Area covered by the roller in 1 revolution = C.S.A. of the roller = 2πrh sq. unit.

Solution 22.
Outer radius of the vessel R = 14 cm
Inner radius of the vessel = r = 10 cm
Area of the outer surface = 2πR² cm²
= (2π x 14 x 14) cm² = 392 π cm²
Area of the inner surface = 2πt² cm²
= (2π x 10 x 10) cm² = 200 π cm²
Area of the ring (shaded) at the top = π(R² – r²)
= π[( 14)² – (10)²] = π(14 + 10) (14 – 10)
= 96 π cm²
Total area to be painted = 392 π + 200 π + 96 π
= 688 π cm²

Solution 23.

Solution 24.
x³ + y³ + z³ – 3xyz = (x³ + y³) + z³ – 3xyz
x³ + y³ + z³ – 3xyz = [(x + y)³ – 3xy(x + y)] + z³ – 3xyz [let x + y = u]
= u³ – 3xyu + z³ – 3xyz
= u³ + z³ – 3xy (u + z) = (u³ + z³) – 3xy (u + z)
= (u + z) (u² – uz + z²) – 3xy (u + z) = (u + z) [u² + z² – uz – 3xy]
= (x + y + z) [(x + y)² + z² – (x + y)z – 3xy] [u = x + y]
= (x + y + z) [x² + y² + z² + 2xy – xz -yz – 3xy]
x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)
if x + y + z = 0
x³ + y³ + z³ – 3xyz = 0 × (x² + y² + z² – xy – yz – zx)
x³ + y³ + z³ – 3xyz = 0
x³ + y³ + z³ = 3xyz

Solution 25.
We have ∠B + ∠CBP = 180° (Linear pair)

Solution 26.
(i) Draw EF || AB || CD

=> ∠1 + ∠2 = x°
Now AB || EF and AE is transversal.
∠1 + ∠BAE = ∠1 + 104° = 180°
(co interior angles)
∠1 = 180°- 104° = 76°
Again EF || CD and EC is the transversal
∠2 + ∠ECD = ∠2 + 116° = 180°
=> ∠2 = 64°
Hence x = ∠1 + ∠2 = 76° + 64° = 140°
=> x = 140°
(ii) Draw EO || AB || CD

x° = ∠1 + ∠2
Now EO || AB and OB is transversal
∠1 + ∠ABO = 180° [Co-interior angles]
∠1 + 40°= 180°
=> ∠1 = 180°- 40° = 140° => ∠1 = 140°
Again EO || CD and DO is transversal
∠2 + ∠CDO = 180°[Co-interior angles]
∠2 + 35° = 180°
=> ∠2 = 180° – 35° = 145°
=> ∠2 = 145°
∠1 + ∠2 = (140° + 145°) = 285°
=> x = ∠1 + ∠2 = 285°
=> x = 285°

Solution 27.
Linear equation 3x + 2y = 12

The point A(4, 0) cuts on the x-axis and point B(0, 6) cuts on the y-axis.

Solution 28.

Volume of the water running into the sea per minute
= Volume of cuboid
= l x bx h
= 50 x 45 x 2
= 4500 m³
(i) We should preserve the water and other natural resources for ourself and for our next generation because they can not be polluted or finished very fast.
(ii) Method of restoring fresh water is by ‘rain water harvesting’ in home.
(iii) Our rivers are very polluted and causing the problems in ecosystem and becoming the shortage of fresh water at large scale. So, it is necessary to clean all rivers of our country.

Solution 29.
Mean of first 13 observations = 32
Sum of the first 13 observations = 32 x 13 = 416
Mean of last 13 observations =39
Sum of last 13 observations = 39 x 13 = 507
Mean of 25 observations = 36
Sum of all the 25 observations = 36 x 25 = 900
∴ the 13th observation = (416 + 507 – 900) = 23
Hence the 13th observation = 23

Solution 30.

From Graph:
BC = CE + BE = (4 – 0) + (0 – (-2))
Base = BC = 4 – (-2) = 4 + 2 = 6 units
Height = AD = OF – OE = 5 – 2 = 3 units
ar (∆ ABC) = \(\frac { 1 }{ 2 }\) x Base x height
= \(\frac { 1 }{ 2 }\) x 6 x 3 = \(\frac { 18 }{ 2 }\) = 9 sq units
ar (∆ ABC) = 9 sq. units.

We hope the CBSE Sample Papers for Class 9 Maths Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 1, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Accountancy. Here we have given CBSE Sample Papers for Class 12 Accountancy Paper 7

CBSE Sample Papers for Class 12 Accountancy Paper 7

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 7 of Solved CBSE Sample Papers for Class 12 Accountancy is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

(i) Please check that this paper contains 23 questions.
(ii) The paper contains two parts A and B.
(iii) Part A is compulsory for all.
(iv) Part B has two options—Option-1 Analysis of Financial Statements and Option-II Computerized Accounting.
(v) Attempt only one option of Part B.
(vi) All parts of a question should be attempted at one place.

PART – A
Partnership Firms and Company Accounts

Question 1.
A, B and C are partners decided that no interest on drawings is to be charged to any partner. But, after one year, ‘C’ wants that interest on drawings should be charged to every partner. State how C can do this?

Question 2.
State the liability of partners in case of dissolution of a firm.

Question 3.
When a liability is to be discharged by a partner, why is his capital account credited?

Question 4.
In case of dissolution of a firm, which items on the liabilities side are to be paid last?

Question 5.
State the steps other than rejecting applications that a company can take in case of over subscription.

Question 6.
What is meant by capital reserve?

Question 7.
Sharma and Verma were partners in a firm sharing profits in the ratio of 4 :1. Their capitals on 1-4-2012 were : Sharma Rs 5,00,000 and Verma Rs 1,00,000. The partnership deed provided that Sharma will get a commission of 10% on the net profit after allowing a salary of Rs 5,000 per month to Verma. The profit of the firm for the year ended 31st march was Rs 2,80,000. Prepare profit and loss appropriation account of Sharma and Verma for the year ended 31st March.

Question 8.
Raghav Ltd. purchased a running business from Krishna Ltd. for a sum of Rs 15,00,000, payable Rs 3,00,000 by cheque and for the balance issued 9% debentures of Rs 100 each at par.
The assets and liabilities consisted of the following:
Plant and machinery Rs 4,00,000
Buildings Rs 6,00,000
Stock Rs 5,00,000
Sundry debtors Rs 3,00,000
Sundry creditors Rs 2,00,000
Record necessary journal entries in the books of Raghav Ltd.

Question 9.
G. Ltd. had a balance of Rs 22,00,000 in its statement of profit and loss. Instead of declaring a dividend, it decided to redeem its Rs 20,00,000,9% debentures at a premium of 10% out of profits on 31st March, 2015. The company invested the required amount in fixed deposit in a bank on 30th April, 2014 earning Interest @ 6% P.a Tax was deducted on interest earned @ 10% P.a by the bank. Pass the necessary journal entries in the books of the company for the redemption of debentures.

Question 10.
Shauiya Ltd. offered 4,00,000 shares to public for subscription. Applications were received for 7,00,000 shares and prorata allotment was made to the applicants. Sakshi applied for 3,500 shares and Vandana was alloted 1,600 shares.
On the basis of above information, calculate:
(i) How many applications have been rejected altogether?
(ii) What is pro-rata ratio?
(iii) How many shares were alloted to Sakshi?
(iv) How many share were applied by Vandana?

Question 11.
P, Q and R are partners sharing profits and losses in the ratio 5:3:2. From 1st Januaiy 2006, they dedded to share profits and losses in equal proportion. The partnership deed provides that in the event of any change in ratio, the goodwill should be valued at three year’s purchase of the average of the five year’s profit The profit and loss of proceeding five years are:
Profits 2001 – Rs 60,000 2002 Rs 1,50,000; 2003 – Rs 1,70,000;
2004 – Rs 190,000
Loss: 2005 – Rs 70,000
Give necessary journal entry to record the change.

Question 12.
A and B are partners sharing profit in ratio of 5 : 3, they admit C into firm for 3/10th profit which he takes 2/10th from A and 1/10th from B and brings a part of his share of premium for goodwill in cash. Goodwill account does not appear in books of A and B. Fill the missing information in following journals entries and compute the new ratio of A, B and C.

Question 13.
(a) Vishwakarma Ltd. issued 15,000 Preference shares of Rs 100 each at a premium of 50%. Payments were to be made as — Rs 25 on application; Rs 35 on allotment (including premium) and Rs 45 on first and final call.
The applications for 14,000 shares were received and all are accepted. All the money was duly received except the first and final call on 600 shares.
Show the share capital in the balance sheet of the company, also prepare notes to accounts for the same.
(b) Ratan steel Ltd. decided that 5% jobs in the company will be reserved for the children of company’s employees. State the value involved in such decision.

Question 14.
X and Y are partners sharing profits and losses in the ratio of 3:2. They agree to take Z into partnership for 1/9 share. For this purpose, goodwill is to be valued at two years’ purchase of the average profit of last four years which were as follows,
Year ending on 31March 2009 50,000 (Profit)
Year ending on 31March 2010 120,000 (Profit)
Year ending on 31March 2011 180,000 (Profit)
Year ending on 31March 2012 70,000 (loss)
On 1 april, 2011 a motor bike costing Rs 50000 was purchased and debited to travelling expense account on which depreciation is charged @ 20%. Calculate the value of goodwill.

Question 15.
Babul and Vinay were partners. The partnership deed provides for
(a) Profit to be divided as 1/2 vinay -1/3 Babul and 1/6 to be transfer as reserve.
(b) The accounts are closed on 31 March each year.
(c) In the event of death of a partner, the executors will be entitled to the following:
(i) Capital to the credit on date of death.
(ii) Interest on capital 12% pa.
(iii) Proportion of profit to the date of death based on average profit credited for last 3 years.
(iv) Share of goodwill on the date of death based on average profit credited for the last 3 years.
The following information is provided to you:
Babul’s capital Rs 90,000; Vinay’s capital Rs 60,000; Reserves 30,000, cash Rs 1,10,000, Investment Rs 70,000.
Prepare Vinay’s A/c to be presented to his executors; as he died on April 30th 2007. The profits for the three preceeding years were Rs 48,000, Rs 42,000 and Rs 45,000.

Question 16.
Bhamashah Company Limited made an issue of 1,00,000 equity shares of Rs 10 each at a premium of 20% payable as follows:
On application Rs 2.50 per share
On allotment Rs 4.50 per share
On first and final call – balance.
Applications were received for 2,00,000 equity shares and directors made pro rata allotment Ramu, who had applied for 800 shares did not pay allotment and final call money with result his shares were forfeited. Later on, 80% of forfeited shares were reissued Rs 8 per share fully paid up.
Pass necessary journal entries for the above mentioned transactions to the books of the company.
OR
X Ltd. issued 40,000 equity shares of Rs 10 each at a premium of Rs 2.50 per share. The amount was payable as follows:
On application — Rs 2 per share
On allotment — Rs 4.50 per share (including premium)
and on call — 6 per share.
Owing to heavy subscription, the allotment was made on prorata basis.
(a) Applicants for 20,000 shares were alloted 10,000 shares.
(b) Applicants for 56,000 shares were alloted 14,000 shares.
(c) Applicants for 48,000 shares were alloted 16,000 shares.
It was decided that excess amount received on applications would be utilized on allotment and the surplus will be refunded.
Ram, to whom 1,000 shares were allotted, who belong to category (a) failed to pay allotment money. His shares were forfeited after final call. Pass necessary journal entries in the books of X Ltd. for above transactions.

Question 17.
X, Y and Z were partners sharing profit in the ratio of 2:2:1. The balance sheet on 31 March 2010, when they dissolved the firm was as follows.

It was agreed that:
(i) X to take over furniture at Rs 8,000 and debtors Rs 1,20,000 and the creditors of Rs 16,000 were to be paid by him at figure.
(ii) Y to take all stock Rs 17,000 and Sundry assets at Rs 72,000 (being 10% less than the book value).
(iii) Z to take over remaining assets at 80% of book value and assume responsibility of discharge of loan together with accrued interest of Rs 2,300.
(iv) The expenses of realisation were 27,000. The remaining debtors were sold to a debt collecting agency at 50% of the value. Prepare necessary accounts to close books of firm.
OR
S and K were partners in a firm sharing profit as S – 75% and K – 25%. On 1 April 2008, their position was.

R is now joins the partnership. He agrees to pay partners Rs 20000 by way of goodwill and introduces 1/2 of combined capital of two existing partners after depreciation plant stock at 20% and 10% respectively and raising a provision of 10% against sundry debtors. The new partner is allowed 1/4th profit of firm.
You are required to record the transaction in the books of firm and give the new balance sheet of the new firm.

PART -B
‘Analysis of Financial Statements’

Question 18.
A mutual fund company received a dividend of Rs 20 lakhs on its investments in another company’s share, where will it appear in cash flow statement?

Question 19.
Mention the net amount of source or use when discount of Rs 2,000 is received on making payment to a creditor of Rs 27,000.

Question 20.
List any four items which are included under the head reserves and surplus of the company’s balance sheet, as per schedule III to companies Act 2013.

Question 21.
From the following information, prepare comparative balance sheet of SMC Ltd.

Question 22.
Inventory turnover ratio: 4 times, inventory at the end was Rs 20,000 more than inventory , in the beginning. Revenue from operations – Rs 3,00,000, Gross profit ratio – 25%. Current liabilities – Rs 40,000, Quick ratio – 0.75 :1. Calculate current ratio.

Question 23.
X Ltd. made a profit of Rs 1,00,000 after charging depredation of Rs 20,000 on assets and transfer to general reserve of Rs 3,000. The goodwill amortised was Rs 7,000 and gain on sale of machinery was Rs 3,000. The other information available to you (changes in value of current assets and current liabilities) is as follows.
At the end of year, Debtor showed an increase of Rs 6,000; creditors an increase of Rs 10,000, prepaid expenses an increase of Rs 200; bills receivable a decrease of Rs 3,000; Bills payable a decrease of Rs 4,000 and outstanding expenses a decrease of Rs 2,000. Ascertain the cash flow from operating activities.

Answers

Answer 1.
C can do this only if A and B, both agree to his proposal.

Answer 2.
(The liabilities of partners are to pay the external debts from realised amount and to give public notice in general).

Answer 3.
This liability actually belongs to the firm but if a partner pays it, then he is entitled to receive it from the firm, therefore, his capital account is credited.

Answer 4.
Capital accounts or capital of partners are paid at last.

Answer 5.
Excess money to be adjusted in allotment and calls.

Answer 6.
Reserve created out of capital profits, not generally available for dividend distribution, e.g. profit prior to incorporation.

Answer 7.
Profit and Loss Appropriation A/c

Answer 8.
Journal

Answer 9.
Books of G. Ltd.
Journal

Answer 10.
(i) Number of applications rejected altogether since Pro-rata allotment was made to all the applicants.
(ii) Pro rata ratio = No. of shares allotted: No. of shares applied
4,0, 000: 7,00,000 = 4:7
It mans that 4 shares will be allotted to an applicant who has applied for 7 shares.
(iii) No. of shares allotted to Sakshi = 3,500 x \(\frac { 4 }{ 7 }\) = 2,000 shares.

Answer 11.

Answer 12.
Journal
Calculation of new profit sharing ratio:

Answer 13.
(a) Extract of balance sheet of Vishwakarma Ltd.
as at…..

Answer 14.
Calculation of average profit:

Note:
(i) Cost of motor bike was wrongly debited to travelling expense account. After rectification, the loss of 2012 will be reduced by Rs 50,000.
(ii) Depreciation on motor bike Rs 10,000 (20% of 50,000) was not charged to profit and loss account of 2012 which will increase the loss of 2012 by Rs 10,000.
Hence, the final loss will be Rs 70,000 – 50,000 + 10,000 = 30,000

Answer 15.
Dr. Vinay’s Capital A/c Cr.

Answer 16.
In the books of Bhamashah Ltd.
Journal

Answer 17.
Realisation A/c

Answer 18.
It will be shown under cash inflow from operating activities. The reason being that a mutual fund company is a finance company and it has been received dividend from its primary revenue generating activities.

Answer 19.
Use Rs 25,000.

Answer 20.
(i) Capital reserve,
(ii) Capital redemption reserve,
(iii) Security premium reserve,
(iv) Debenture redemption reserve,
(v) Revaluation reserve,
(vi) Surplus i.e. balance in statement of profit and loss.

Answer 21.
Comparative balance sheet
as at 31st March 2014 and 2015

Answer 22.

Answer 23.
Calculation of net cash flow from operating activities

We hope the CBSE Sample Papers for Class 12 Accountancy Paper 7 help you. If you have any query regarding CBSE Sample Papers for Class 12 Accountancy Paper 7, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Accountancy. Here we have given CBSE Sample Papers for Class 12 Accountancy Paper 6

CBSE Sample Papers for Class 12 Accountancy Paper 6

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 6 of Solved CBSE Sample Papers for Class 12 Accountancy is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

(i) Please check that this paper contains 23 questions.
(ii) The paper contains two parts A and B.
(iii) Part A is compulsory for all.
(iv) Part B has two options—Option-1 Analysis of Financial Statements and Option-II Computerized Accounting.
(v) Attempt only one option of Part B.
(vi) All parts of a question should be attempted at one place.

PART – A
Partnership Firms and Company Accounts

Question 1.
State any two items from accounting point of view that may be included in a partnership deed?

Question 2.
Give one reason in favour of having partnership deed?

Question 3.
The firm XYZ earned a profit of Rs 2,75,000 during the year ending on 31 March, 2009.10% of the profit was to be transferred to general reserve. Pass necessary journal entry for the same.

Question 4.
Why a retiring/deceased partner entitled to a share of goodwill of the firm?

Question 5.
Give any two alternatives to a company for the allotment of shares in case of over-subscription?

Question 6.
DN Ltd. issued 50,000 shares of Rs 10 each at a discount of 10% payable at Rs 2 per share on application, Rs 3 per share on allotment and Rs 2 on each first and final call. Applications were received for 70,000 shares. It was decided that
(a) refuse allotment to the applicants of 10,000 shares.
(b) allot 20000 shares to Mohan who had applied for similar number and
(c) allot the remaining shares on pro data basis. Mohan failed to pay the allotment money and Sohan who belonged the category C and was alloted 3000 shares paid both the calls with allotment Calculate the amount received on allotment.

Question 7.
X, Y and Z are partners sharing protits and losses in the ratio 3:2:1. After final accounts have prepared, it was discovered that interest on drawings @ 5% p.a had not been taken into consideration. The drawings of the partners were X Rs 15,000, Y Rs 12,600 Z Rs 12,000. Give necessary adjusting journal entry.

Question 8.
X Ltd. redeemed 100, 6% debentures of Rs 100 each by converting them into equity shares of Rs 100 each. The 6% debentures were redeemable at premium of 10% for which the equity shares were issued at 25% premium. Pass the necessary journal entries for the redemption of the above mentioned debentures in the books of X Ltd.

Question 9.
Adit Taneja applied for 4,000 shares of Rs 100 each issued at a premium of Rs 2 per share. After having paid Rs 2 per share on application, he did not pay allotment money on Rs 5 per share (including premium) and on his subsequent failure to pay the first call, his shares were forfeited-Fill in the following journal entries.

Question 10.
(a) Q Ltd. forfeited 1000 equity shares of Rs 10 each issued at Rs 14 per share for non payment of final call of Rs 7 (including premium) per share. Pass Journal entry.
(b) Tata chemicals Ltd. provided Rs 20 lac for a solar water purification, plant set up by a Non Govt organisation. People from economically backward class living in the vicinity will be given five litres of clean portable water daily for each family member. State the values involved in such decision.

Question 11.
(a) P, Q and R sharing profits and losses in the ratio of 3 :2 1 decide to share future profits and losses in the ratio of 4:3:2 with effect from 1 April, 2012. Following is an extract of their balance sheet as at 31 March, 2012:

Show the accounting treatment if a claim account of workmen’s compensation is estimated at Rs 1,20,000.
(b) They also decided that 5% of profits of firm will be devoted for providing school uniforms to the students belonging to low income group admitted to private schools as per the provisions of Right to Education Act 2009. Identify any two values involved in making such a decision.

Question 12.
L and M were partners in a firm sharing profits 4:3 ratio. They admitted O as a new partner. The new profit sharing ratio of L, M and O will be 3 : 3 : 4. O brought Rs 2,00,000 for his capital. The goodwill of the firm on O’s admission was valued at Rs 70,000. O brought his share of goodwill in cash. Calculate the sacrificing ratio of L and M and pass necessary journal entries for above transactions on O’s admission

Question 13.
(a) Sundram Ltd. purchased furniture for Rs 3,00,000 from Ravindram Ltd, Rs 1,00,000 were paid by drawing a promissory note in favour of Ravindram Ltd. The balance was paid by issue of equity shares of Rs 10 each at a premium of 25%. Pass necessary journal entries in books of Sundram Ltd.
(b) SSP Ltd forfeited 300 shares of Rs 10 each issued at a premium of Rs 2 per share for the non payment of allotment money of Rs 4 per share (including premium). The first and final call of Rs 3 per share has not been made yet. 50% of forfeited shares were re-issued at Rs 8 per share fully paid up. Pass necessary journal entries for forfeiture and reissue of shares.

Question 14.
Lalan and Balan were partners in a firm sharing profit in the ratio of 3 : 2. Their fixed capitals on 1-4-2010 were:
Lalan Rs 1,00,000 and Balan Rs 2,00,000. They agreed to allow interest on capital @ 12% P.a and to charge on drawings @ 15% P.a. The firm earned a profit, before all above adjustments were Rs 30,000 for the year ended 31-3-2011. The drawings of Lalan and Balan during the year were Rs 3,000 and Rs 5,000 respectively. Showing your calculations clearly, prepare profit and loss appropriation A/c of Lalan and Balan. The interest on capital will be allowed even if the firm incurs a loss.

Question 15.
A and B shares profits and losses in the ratio of 3 : 2. They have decided to dissolve the firm. Assets and external liabilities have been transferred to realisation A/c. Pass the journal entries to effect the following:
(a) Bank loan of Rs 12,000 is paid off.
(b) A was to bear all expenses of realisation for which he is given a commission of Rs 400.
(c) Deferred Advertisement expenditure A/c appeared in the books at Rs 28/100.
(d) Stock worth Rs 1,600 was taken over by B at Rs 1,200.
(e) An unrecorded computer realised Rs 7,000.
(f) There was an outstanding bill for repairs for Rs 2,000, which was paid off.

Question 16.
Metallic Ltd. invited applications for 40,000 equity shares of Rs 50 each issued at a premium of Rs 10 per share. The amount was payable as follows:
On application and allotment — Rs 20 per share. Balance (including premium) on first and final call.
Applications for 70,000 shares were received. Applications for 20,000 shares were rejected and pro-rata allotment was made to the remaining applicants. First and final call was made and duly received except on 400 shares allotted to Nitesh and his shares were forfeited.
Journalise the above transactions.
OR
Arti Ltd. invited applications for issuing 80,000 shares of Rs 10 each at a premium of Rs 4 per share. The amount was payable as follows: .
On application — Rs 5 per share.
On allotment — Rs 9 per share (including premium)
Applications were received for 1,40,000 shares.
Allotment was made on the following basis:
(i) To applicants for 80,000 shares — 60,000 shares
(ii) To applicants for 60,000 shares — 20,000 shares
Rajiv belonging to category (i), had applied for 1,200 shares failed to pay his dues and his shares were forfeited. Pass journal entries in the books of Arti Ltd. to record the above transactions.

Question 17.
Following is the balance sheet of X, Y and Z as on 31-3-2012. They shared profits in the ratio of 3:2:2.

On 1st April 2012, Y decided to retire from the firm on the following terms:
(a) Stock to be depreciated by Rs 12,000.
(b) Advertisement suspense A/c to be written off.
(c) Provision for bad and doubtful debts to be increased to Rs 6,000.
(d) Fixed assets to be appreciated by 10%.
(e) Goodwill of the firm, valued at Rs 80,000 and the amount due to the retiring partner be adjusted in X’s and Z’s capital accounts.
Prepare revaluation account, partners’ capital accounts and balance sheet of the firm.
OR
X and Y were partners in a firm sharing profits in 5:3 ratio. They admitted Z as a new partner for 1/3 share in profit Z was to contribute Rs 20,000 as his capital. The balance sheet of x and y on 1-4-2013, the date of Z’s admission was as follows:

Other terms agreed upon were:
(i) Goodwill of the firm was valued at Rs 12,000.
(ii) Land and building were to be valued at Rs 35,000 and plant and machinery at Rs 25,000.
(iii) The provision for doubtful debts was found to be in excess by Rs 400.
(iv) A liability for Rs 1,000 included in Sundry creditors was not likely to arise.
(v) The capitals of the partners be adjusted on the basis of Z’s contribution of capital in the firm.
(vi) Excess or shortfall, if any, to be transferred to current account.
Prepare revaluation account, partners’ capital accounts and the balance sheet of the new firm.

PART – B
‘Analysis of Financial Statements’

Question 18.
What is meant by financial statements?

Question 19.
Dividend paid by a finance company is classified under which kind of activity while preparing a cash flow statement?

Question 20.
List any four items which are included under the head current assets of the company’s balance sheet, as per the schedule III to the companies act 2013.

Question 21.
Prepare common size statement of profit and loss from the following information:

Question 22.
Net profit after interest but before tax Rs 1,40,000, 15% long term borrowings Rs 4,00,000, shareholders funds Rs 2,40,000, tax rate 50%. Calculate return on capital employed.

Question 23.

From the following information. Prepare a cash flow statement:

Answers

Answer 1.
(i) Interest on capital
(ii) Profit sharing ratio.

Answer 2.
To avoid litigation in future.

Answer 3.
Journal

Answer 4.
Retiring/deceased partner losses his share in profit, to compensate this, he becomes entitled to a share of goodwill of the firm.

Answer 5.
(i) Some applications are accepted in full and excess applications are rejected and their application money is refunded.
(ii) Some application may be alloted shares in fixed proportion called Pro-rata allotment.

Answer 6.
Total amount due on allotment = 50,000 x 3 = 1,50,000
Calls in arrears = 2,0000 x 3 = 60,000
Calls in advance = 3000 x 4 = 12000
Already received = Rs 20,000
Calculation of allotment money received
Total amount = 1,50,000

Answer 7.
Journal

Answer 8.
Journal

Answer 9.
Journal


(iii) Since share capital A/c is debited with Rs 24,000 (i.e. Rs 8 per share), the amount of first call must be Rs 3 per share.

Answer 10.
Q. Ltd.
Journal

(b) (i) Use of solar power will be helpful in environmental protection.
(ii) It will help to reduce the health hazards and improve the quality of life of poor sections of the society.

Answer 11.

(b) Two values involved are:
(i) Assisting the implementation of Right to Education Act 2009.
(it) Helping the low income group segment of the society.

Answer 12.
Journal

Answer 13.
(a) In the books of Sundram Ltd.
Journal

Answer 14.

Answer 15.
Journal

Answer 16.
Journal

Answer 17.

Answer 18.
The financial statements are historical documents which show the organised summary of detailed information about the financial performance of an accounting entity for an accounting period and financial position at the end of an accounting period.

Answer 19.
Financing activity.

Answer 20.
(i) Current investments,
(ii) Inventories,
(iii) Trade receivables,
(iv) Cash and cash equivalents,
(v) Short term loans and advances,
(vi) Other current Assets.

Answer 21.
Common size Statement of Profit and Loss
for the year ended 31st March 2014 and 2015

Answer 22.

Answer 23.
Cash flow statement for the year ended 31st March 2015

We hope the CBSE Sample Papers for Class 12 Accountancy Paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 12 Accountancy Paper 6, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Accountancy. Here we have given CBSE Sample Papers for Class 12 Accountancy Paper 4

CBSE Sample Papers for Class 12 Accountancy Paper 4

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Papers for Class 12 Accountancy is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

(i) Please check that this paper contains 23 questions.
(ii) The paper contains two parts A and B.
(iii) Part A is compulsory for all.
(iv) Part B has two options—Option-1 Analysis of Financial Statements and Option-II Computerized Accounting.
(v) Attempt only one option of Part B.
(vi) All parts of a question should be attempted at one place.

PART – A
Partnership Firms and Company Accounts

Question 1.
When the partners’ capitals are fixed, where the drawings made by a partner will be recorded?

Question 2.
State the ratio in which the partners share profits or losses on revalution of assets and liabilities, when there is a change in profit sharing ratio amongst existing partners?

Question 3.
Name the account which is opened to credit the share of profit of the deceased patner, till the time of his death to his capital account.

Question 4.
Give the journal entry to distribute workmen compensation reserve of Rs 60,000 at the time of retirement of Sajjan, when there is no claim against it The firm has three partners- Rajat, Sajjan and Kavita.

Question 5.
What is meant by securities Premium?

Question 6.
What is meant by issue of debentures as a collateral security?

Question 7.
X, Y and Z were sharing profits and losses in the ratio of 5 : 3 : 2. They decided to share future profits and losses in the ratio of 2:3:5 with effect from 1-4-2010. They decided to record the effect of the following, without affecting their book values:
(i) Profit and loss account Rs 24,000
(ii) Advertisement suspense account Rs 12,000
Pass the necessary adjusting entry.

Question 8.
Mohit Ltd. took over assets of Rs 8,40,000 and liabilities of Rs 80,000 of Ram Ltd. at agreed value of Rs 7,20,000. Mohit Ltd. Paid to Ram Ltd. by issuing of 9% debentures of Rs 100 each at a premium of 20%.
Pass the necessary journal entries to record the above transactions in the books of Mohit Ltd.

Question 9.
Rise Ltd. had issued 40,000, 8% debentures of Rs 100 each redeemable on 31st March 2015. It was decided to invest 15% of the face value of debentures to be redeemed towards debenture redemption investment on 30th April 2014. Investments were encashed and debentures were redeemed on due date. Record necessary journal entries for redemption of debentures.

Question 10.
A and B were partners in a firm sharing profits in 4 :3 ratio. They admitted C into partnership who has won a gold medal in world chess championship for 20% share in the profits. C acquired his share of profits in the ratio of 1:2 from A and B.
You are required to:
(i) Identify the value in admitting C into partnership.
(ii) Calculate the new profit sharing ratio of the partners.

Question 11.
A, B, C and D are partners sharing profits in the ratio of 3:3:2:2 respectively. D retires and A, B and C decide to share the future profits in the ratio of 3 : 2:1. Goodwill of the firm is valued at Rs 6,00,000. Goodwill already appears in the books at Rs 4,50,000. The profit for the 1st year after D’s retirement amount to Rs 12,00,000. Give the necessary journal entries to record goodwill and to distribute the profits. Show your calculations clearly.

Question 12.
Kavita Ltd. issued 2000,7% debentures of Rs 500 each at a discount of 6% on April, 2013 redeemable after 6 years. On 1st October 2016, the debentures were redeemed by conversion into equity shares of Rs 100 each at a premium of Rs 25 per share at the option of debenture holders. Record necessary entries for issue and redemption of debentures.

Question 13.
(a) Green Valley Ltd. offered 5,00,000 shares to public for subscription. Applications were received for 7,50,000 shares and pro-rata allotment was made to the applicants of 6,00,000 shares. Arushi applied for 4,800 shares and Navya was alloted 3,000 shares.
From the above information, calculate:
(i) How many applications have been rejected altogether?
(ii) What is the prorata ratio?
(iii) How many shares were allotted to Arushi?
(iv) How many shares were applied by Navya?
(b) Which value has been affected by rejecting the 150000 applications. Suggest a better alternative for the same.

Question 14.
Ravi and Mohan were partners in a firm sharing profits in the ratio of 7 : 5. Their respective fixed capitals were Ravi Rs 10,00,000 and Mohan Rs 7,00,000. The partnership provided for the following:
(i) Interest on capital @ 12% P.a
(ii) Ravi’s salary Rs 6,000 per month and Mohan’s salary Rs 60,000 per year.
The profit for the year ended 31-3-2012 was Rs 5,04,000 which was distributed equally, without providing for the above. Pass an adjustment entry.

Question 15.
Priya and Riya were partners in a firm sharing profits equally. In spite of repeated reminders by the authorities, they kept evading the taxes. The Court ordered for the dissolution of their partnership firm on 31st March 2015. Priya was deputed to realise the assets and to pay the liabilities. She was paid Rs 1,000 as commission for her services. They were having Rs 8,000 in Profit and loss A/c on the date of dissolution. From the information given below, complete realisation A/c, partners’ capital A/c and cash A/c

Question 16.
Dinesh Ltd. invited applications for issuing 10,000 euity shares of Rs 10 each. The amount was payable as follows:
On application Rs 1
On allotment Rs 2
On first call Rs 3
On second and final call-balance.
The issue was fully subscribed. Ram to whom 100 shares were allotted, failed to pay the allotment money and his shares were forfeited immediately after allotment. Shyam to whom 150 shares were allotted, Failed to pay the first call. His shares were also forfeited after the first call. Afterwards, the second and final call was made. Mohan to whom 50 shares were alloted failed to pay the second and final call. His shares were also forfeited. All the forfeited shares were re-issued at Rs 9 per share fully paid up. Pass necessary journal entries in the books of Dinesh Ltd.
OR
Som Ltd. invited applications for issuing 60,000 equity shares of Rs 100 each at a premium of Rs 50 per share. The amount was payable as follows:
On application Rs 75 per share (including premium Rs 25) on allotment Rs 50 per share (including premium Rs 25) on first and final call – the balance amount. Applications for 55,000 shares were received. Allotment was made to all the applicants and the company received all money due on allotment except K, who was allotted 500 shares and his shares were immediately forfeited afterwards the first and final call was made. L to whom 300 shares were allotted failed to pay the first and final call. His shares were also forfeited. 300 shares of K and 200 shares of L were re-issued for Rs 75,000 fully paid up.
Pass necessary journal entries in the books of Som Ltd. for the above transactions.

Question 17.
On 31st march 2012, the balance sheet of Ram and Shyam who were sharing profits in the ratio of 3 :1 was as follows:

They decided to admit, Mohan on April 1st 2012 for 1/5th share on the following terms:
(i) Mohan shall bring Rs 6,000 as his share of premium.
(ii) That unaccounted accrued income of Rs 100 be provided for.
(iii) The market value of investments was Rs 4,500
(iv) A debtor whose dues of Rs 500 was written off as bad debts paid Rs 400 in full settlement.
(v) Mohan to bring in capital to the extent of 1/5th of the total capital of the new firm. Prepare revaluation A/c, partners’ capital A/c and the balance sheet of the new firm.
OR
A, B and C were partners sharing profits in the ratio of 3:1:1. Their balance as on March 312011, the date on which they dissolve their firm, was as follows:

It was agreed that:
(a) A to take over bills receivables at Rs 800, debtors amounting to Rs 20,000 at Rs 17,200 and the creditors of Rs 6,000 were to be paid by him at this figure.
(b) B is to take over all stock for Rs 7,000 and some sundry assets at Rs 7,200 (being 10% less than the book value).
(c) C to take over remaining Sundry assets at 90% of the book value and assume the responsibility of discharge of loan together with accrued interest of Rs 300.
(d) The expenses of realisation were Rs 270. The remaining debtors were sold to a debt collecting agency at 50% of the book value. Prepare realisation A/c, partners capital A/c and cash A/c.

PART -B
‘Analysis of Financial Statements’

Question 18.
Sale of marketable securities at par would result in inflow, outflow or No flow of cash.

Question 19.
Where will a manufacturing company record dividend received while preparing cash flow statement?

Question 20.
Give major heading and sub heading under which following items will be shown in a company’s balance sheet as per schedule III, part I of the companies Act 2013:
(a) Preliminary expenses,
(b) Goodwill,
(c) Long term investments,
(d) Securities premium reserve,
(e) Unclaimed dividend,
(f) Encashment of employees earned leave payable on retirement,
(g) Provision for doubtful debts,
(h) Interest on calls in arrears.

Question 21.
From the following information, calculate the working capital turnover ratio.
Cost of revenue from operations Rs 5,00,000 Gross profit ratio 20%
Fixed assets Rs 5,00,000
Capital employed Rs 7,50,000

Question 22.
(a) What is the meaning and purpose of analysis of financial statements?
(b) Name any two tools of analysis of financial statements.

Question 23.
Following is the balance sheet of Aran Ltd. as on 31st March 2015 and 2014.


Prepare cash flow statement after taking into account the following adjustments: Tax paid during the year amounted to Rs 98,000.

Answers

Answer 1.
Drawing (except permanent drawings) are recorded on the debit side of the partners’ current accounts when there partners capitals are fixed.

Answer 2.
Partners share profits or losses in the old profit sharing ratio, in the event of revaluation of assets and liabilities.

Answer 3.
Profit and loss suspense account is opened to credit the share of profit of the deceased partner, till the time of his/her death to his/her capital account.

Answer 4.
In case of no claim against workmen compensation Reserve, the following, entry needs to be passed:

Answer 5.
Securities premium is the excess amount, charged by a company for issuing securities (Shares, debentures etc.) over its face value.

Answer 6.
Issue of debentures as a collateral security means securing underlying asset or loan taken,by issuing debentures as additional or secondary security over and above the prime or principal security.

Answer 7.

Answer 8.
Journal

Answer 9.
Books of Rise Ltd.
Journal

Note: Interest on investment is not calculated because rate of interest is not given.

Answer 10.
(i) Value involved in admitting C is promoting sports culture in the country.
(ii) C acquired his share of profit (20% or \(\frac { 1 }{ 5 }\)) from A and B in the ratio of 1 : 2, this means

Answer 11.
Journal

Answer 12.
Books of Kavita Ltd.

Answer 13.
(a) (i) Application for 1,50,000 shares have been rejected altogether since allotment was made to the applicants of 6,00,000 shares.
(ii) Pro rata ratio = No of shares allotted : No. of shares applied = 5,00,000 : 6,00,000 = 5:6
(iii) No. of shares allotted to Arushi = 4,800 x \(\frac { 5 }{ 6 }\) = 4000 shares
(iv) No. of shares applied by Navya = 3,000 x = \(\frac { 6 }{ 5 }\) = 3600 shares
(b) (i) Value of equality has been affected by rejecting the 1,50,000 applications.
(ii) Better alternative – Pro-rata allotment to all applicants so that the rejected applicants may not be demotivated from applying for shares in future.

Answer 14.

Answer 15.

Answer 16.

Answer 17.

Answer 18.
Sale of marketable securities at par would result no flow of cash.

Answer 19.
Dividend and interest received by a manufacturing company will be recorded under investing activities.

Answer 20.

Answer 21.
Working capital = Capital employed – Fixed assets = 7,50,000 – 5,00,000 = 2,50,000
Calculation of revenue from operations:
When sales is = 100, Gross profit is = 20
Cost is 100 – 20 = 80

Answer 22.
(a) Analysis of financial statements is a systematic process of identifying the financial strengths and weaknesses of the firm by establishing relationship between the items of the balance sheet and income statement.
Objectives of financial statements analysis are:
(i) Measure the short term solvency of the enterprise.
(ii) Measure the long term solvency of the enterprise.
(iii) Measure the operating efficiency and profitability of the enterprise.
(iv) Compare intra-firm position in the industry.
(b) The tools of analysis of financial statements are:
(i) Common size statement
(ii) Ratio analysis
(iii) Cash flow statement.

Answer 23.
Cash Flow Statement
for the year ended March 31, 2015.

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