Maharashtra Board Class 10 Marathi Kumarbharti Solutions Chapter 3 आजी : कुटुंबाचं आगळ

Balbharti Maharashtra State Board Class 10 Marathi Solutions Kumarbharti Chapter 3 आजी : कुटुंबाचं आगळ Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Marathi Kumarbharti Chapter 3 आजी : कुटुंबाचं आगळ

Marathi Kumarbharti Std 10 Digest Chapter 3 आजी : कुटुंबाचं आगळ Textbook Questions and Answers

प्रश्न. पुढील उतारा वाचा आणि दिलेल्या सूचनांनुसार कृती करा :

कृती १ : (आकलन)

प्रश्न 1.
विधाने पूर्ण करा :
(i) स्वत:च्या कपाळावरचे गोंदण दिसू नये, म्हणून आजी —————
उत्तर :
स्वत:च्या कपाळावरचे गोंदण दिसू नये, म्हणून आजी कपाळावर बुक्का लावत असे.

(ii) वर्ष-दीड वर्षाने जन्मणाऱ्या वासरांमध्ये एकाआड एक खोंड नक्की असे, म्हणून —————–
उत्तर :
वर्ष-दीड वर्षाने जन्मणाऱ्या वासरांमध्ये एकाआड एक खोंड नक्की असे, म्हणून दावणीला कायम कपिलीचे बैल असत.

(iii) सुनांनी चहा करून पिऊ नये, म्हणून —————-
उत्तर :
सुनांनी चहा करून पिऊ नये, म्हणून ढाळजेतून सोप्यात येऊन आजी सक्त पहारा करायची.

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

प्रश्न 2.
पुढील मुद्द्यांच्या आधारे आजीचे शब्दचित्र रेखाटा :

(i) आजीचे दिसणे : ——————–
उत्तर :
आजीचे दिसणे : आजीला साडेपाच फूट उंची लाभली होती. तिचा वर्ण गोरा होता. उन्हापावसामुळे तिची त्वचा काळपटली होती. आजीच्या वयाची माणसे कमरेत वाकतात. त्यामुळे चालताना काठी घ्यावी लागते. पण आजी अजूनही ताठ कण्याने चालत होती. अजूनही तिचे सगळे दात शाबूत होते. डोक्यावरचे सगळे केस पिकले होते.

(ii) आजीची शिस्त : ——————–
उत्तर :
आजीची शिस्त : आजीची शिस्त कडक होती. सगळ्यांना सगळी कामे करता आली पाहिजेत, असा तिचा कटाक्ष होता. तिने कामांची वाटणी केली होती. ती कामे आजी सर्व सुनांना आलटूनपालटून करायला लावी. दुपारच्या कामांचेही तिने नियोजन केलेले असे. सुनांनी मुलांच्या बाबतीत आपपरभाव करू नये म्हणून मुलांना खाऊपिऊ घालताना आजीचा सक्त पहारा असे. गल्लीतल्या बायका दुपारी गप्पागोष्टींना येत असत. त्या वेळी ती बायकांनी सांगितलेल्या गोष्टींची शहानिशा करीत असे.

(iii) आजीचे सौंदर्य : —————-
उत्तर :
आजीचे सौंदर्य : आजीचे वय आता सत्तर वर्षांचे होते. उन्हापावसामुळे आजीची त्वचा रापली होती. पण तिचा मूळ गोरा वर्ण लपत नव्हता. तिचे दात मोत्यांसारखे चमकत होते. विशाल कान व धारदार नाक यांनी आजीच्या सौंदर्यात भर पडत होती. चेहऱ्यावर सुरकुत्या पडल्या तरी तिच्या सौंदर्यात उणेपणा आला नव्हता.

(iv) आजीचे राहणीमान : ————-
उत्तर :
आजीचे राहणीमान : त्या काळात इरकली लुगडे उच्च राहणीमानाचे लक्षण होते. हिरव्या व लाल रंगांची नऊवारी इरकली लुगडी व अंगात चोळी हा तिचा पेहराव असे. कपाळावरचं गोंदण दिसू नये म्हणून त्यावर ती बुक्का लावी. ती नेहमी नाली ठोकलेल्या जुन्या वळणाच्या वहाणा वापरत असे.

प्रश्न 3.
चूक की बरोबर सांगा :
(i) राहिलेली अर्धी चरवी घरात आली, की म्हातारी ढाळजंतनं सोप्यात. अवतरायची.
उत्तर :
बरोबर

(ii) आजीच्या डोक्यावरील सर्व केस पांढरे होते……………….
उत्तर :
बरोबर

(iii) सुनांच्या कामाबाबत आजी फारशी काटेकोर नसायची.
उत्तर :
चूक.

कृती २ : (आकलन)

प्रश्न 1.
विधाने पूर्ण करा :
(i) मुलांनी भरपूर खावे-प्यावे व त्यांची आबाळ होऊ नये, म्हणून —————-
उत्तर :
मुलांनी भरपूर खावे-प्यावे व त्यांची आबाळ होऊ नये, म्हणून आजी त्यांना धपाटे घालून घालून खायला घाली.

(ii) प्रत्येक सुनेला प्रत्येक काम आलेच पाहिजे, असा आजीचा आग्रह होता, म्हणून ती ——–
उत्तर :
प्रत्येक सुनेला प्रत्येक काम आलेच पाहिजे, असा आजीचा, आग्रह होता, म्हणून ती रोटेशनप्रमाणे काम बदलत जाई.

(iii) मुलांना दूध प्यायला देण्याबाबत सुना आपपरभाव करतील, अशी भीती आजीला वाटे, म्हणून ————–
उत्तर :
मुलांना दूध प्यायला देण्याबाबतं सुना आपपरभाव करतील, अशी भीती आजीला वाटे, म्हणून ती मुलांना गोठ्यातच दूध प्यायला लावी.

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

प्रश्न 2.
पुढीलपैकी चुकीची वाक्ये दुरुस्त करून बरोबर वाक्ये व दुरुस्त केलेली वाक्ये पुन्हा लिहा :
(i) दुपारची कामे आटोपून आजी ढाळजेत यायची.
उत्तर :
दुपारच्या कामांचे नियोजन करून आजी ढाळजेत यायची.

(ii) गाईने पान्हा सोडला की वासराला सोडायचे.
उत्तर :
गाईने पान्हा सोडला की वासराला धरून ठेवायचे.

(iii) आजीच्या घरी एक गावरान गाय होती.
उत्तर :
आमच्या घरी एक गावरान गाय होती.

कृती ३ : (व्याकरण)

प्रश्न 1.
पुढील वाक्यांतील वाक्प्रचार शोधून अधोरेखित करा :
(i) आजीच्या छत्रछायेखाली आमचे सर्व कुटुंब गुण्यागोविंदाने नांदत होते.
उत्तर :
आजीच्या छत्रछायेखाली आमचे सर्व कुटुंब गुण्यागोविंदाने नांदत होते.

(ii) सत्तरीनंतरही आजीच्या हातात काठी आली नव्हती.
उत्तर :
सत्तरीनंतरही आजीच्या हातात काठी आली नव्हती.

प्रश्न 2.
पुढील अधोरेखित शब्दांच्या जाती ओळखा आणि त्यांचे उपप्रकार लिहा :
(i) चार घरच्या चार सुना नांदायला आल्या.
उत्तर :
चार : विशेषण. संख्यावाचक विशेषण.
सुना : नाम. सामान्य नाम.
आल्या : क्रियापद. अकर्मक क्रियापद.

(ii) प्रत्येकीला काम आलंच पाहिजे.
उत्तर :
काम : नाम. भाववाचक नाम.

(iii) आजीला एकाही सुनेचा भरवसा नव्हता.
उत्तर :
ही : शब्दयोगी अव्यय.
भरवसा : नाम. भाववाचक नाम.

(iv) आमच्या घरी एक गावरान गाय होती.
उत्तर :
एक : विशेषण. संख्यावाचक विशेषण.
गावरान : विशेषण. गुणवाचक विशेषण.
गाय : नाम. सामान्य नाम.

कृती ४ : (स्वमत / अभिव्यक्ती)

प्रश्न 1.
तुम्हांला आठवत असलेला किंवा तुमच्या आईबाबांनी सांगितलेला तुमच्या लहानपणचा एखादा प्रसंग लिहा.
उत्तर :
मी लहान होतो तेव्हा सकाळी उठलो की, दात घासून, तोंड धुऊन मी तडक रानातल्या आमच्या वाड्याकडे धावत सुटायचो. कोकणात गुरांच्या गोठ्याला वाडा म्हणतात. तेथे माझे आजोबा माझी वाटच बघत असत. दरदिवशी ते माझ्यासाठी एक तांब्याएवढे मडके भरून दूध गरम करून ठेवत. मी गेलो की, घटाघटा ते दूध पीत असे. त्यानंतर त्या मडक्याला चिकटलेली साय खरवडून खाणे हा माझा मोठा आनंदाचा भाग असे. माझ्या या अखंडित दिनक्रमामुळे कोणत्या गाई-म्हशीचे दूध कोणत्या चवीचे आहे, हे मी सहज ओळखू शकतो. दुधाला एक स्वत:चे माधुर्य असते. दूध पिताना लोक दुधात साखर का घालतात, हे मला अजून कळलेले नाही.

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

Marathi Kumarbharti Class 10 Textbook Solutions Chapter 3 आजी : कुटुंबाचं आगळ Additional Important Questions and Answers

प्रश्न. पुढील उतारा वाचा आणि दिलेल्या सूचनांनुसार कृती करा :

कृती १ : (आकलन)

प्रश्न 1.
एका शब्दात उत्तर लिहा : (मार्च १९)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 1
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 2

प्रश्न 2.
आकृती पूर्ण करा :
(i)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 3
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 4

(ii)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 5
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 6

(iii)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 7
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 8

(iv)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 9
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 10

कृती २ : (आकलन)

प्रश्न 1.
दुपारच्या वेळी रूढ खेळांखेरीज मुले ज्या गमतीजमती करीत, त्या लिहा.
उत्तर :

  • वडाच्या पारंब्यांना लटकत, लोंबत राहायचे आणि पारंब्यांच्या टोकाला फुटलेली पिवळी पालवी खात बसायचे.
  • देवळातली घंटा वाजवाजवून झोपलेल्यांची झोपमोड करायची.
  • विहिरीत मनसोक्त पोहायचे.
  • शिवणापानी खेळायचे.
  • हातपाय पोटाशी आवळून घेऊन शरीराचे मुटके करून विहिरीत धडादिशी उड्या घ्यायची.
  • ओल्या अंगाने मातीत लोळायचे आणि पुन्हा पाण्यात इंबायचे आणि शेवटी थकून भागून घरी जायचे.

[टीप : परीक्षेत कोणतीही दोन किंवा चार नावे लिहायला सांगितली जाऊ शकतात.]

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

प्रश्न 2.
बालपणी लाभलेल्या रानमेव्यांची नावे सांगा. (कोणतीही चार नावे लिहा.)
उत्तर :

  • गाभोळ्या चिंचा, मिठाचे खडे व कच्च्या कैऱ्या, बोरे, चिंचेची कोवळी पाने.
  • उंबर, ढाळे, भाजलेली कणसे, हुरडा.
  • कच्ची वांगी, गवार, टोमॅटो, शहाळे.
  • कलिंगड, शिंगाडे (चिबूड), करडीची भाजी, ज्वारीचे कणीस.
  • कवठ, तुरी-मटकीच्या शेंगा, उकडलेल्या शेंगा, कुळथाचे पिठले.

प्रश्न 3.
आकृत्या पूर्ण करा :
(i)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 11
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 12

(ii)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 13
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 14

(iii)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 15
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 16

प्रश्न 4.
पुढील घटना उताऱ्याच्या आधारे क्रमाने लिहा : (मार्च ‘१९)
(i) म्हातारीची ढाळज सुटायची
(ii) वाडा शांत व्हायचा
(iii) कडुसं पडायच्या आधी मैफिल मोडायची
(iv) माणसं ढाळजंत बसायची
उत्तर :
(iii) कडुसं पडायच्या आधी मैफिल मोडायची
(i) म्हातारीची ढाळज सुटायची
(iv) माणसं ढाळजंत बसायची
(ii) वाडा शांत व्हायचा

कृती ३ : (व्याकरण)

प्रश्न 1.
सहसंबंध लक्षात घेऊन उत्तरे लिहा :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 17
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 18

प्रश्न 2.
पुढील बोलीभाषेतील शब्दांना प्रचलित प्रमाण मराठी भाषेतील शब्द शोधून लिहा :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 19
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 20

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

प्रश्न 3.
तक्ता पूर्ण करा :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 21
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 22

प्रश्न 4.
पुढील वाक्यांतील अधोरेखित शब्दाचे लिंग बदलून वाक्य पुन्हा लिहा :
आजी त्याची संपादक होती. (सराव कृतिपत्रिका-१)
उत्तर :
आजोबा त्याचे संपादक होते.

कृती ४ : (स्वमत / अभिव्यक्ती)

प्रश्न 1.
‘आमची ढाळज म्हणजे गावाचं वर्तमानपत्र होतं,’ या वाक्याचा तुम्हांला कळलेला अर्थ लिहा.
उत्तर :
‘आगळ’ या कादंबरीच्या नायकाची आजी ही त्याच्या घराची सत्ताधीश होती. घरात तिचा वचक होता. घराबाहेरही तिच्या शब्दाला मान होता. दुपारपर्यंतच्या सर्व कामांचे नियोजन करून आजी ढाळजेत यायची. गल्लीतल्या बायकाही जमा व्हायच्या. निवडटिपण करता करता गप्पाटप्पा व्हायच्या. अनेक बातम्या, गुपिते उघड व्हायची. सगळ्याजणी बातम्यांवर चर्चा करीत. त्यातून बातम्यांची शहानिशा व्हायची.

वर्तमानपत्राचे वार्ताहर गावांतून बातम्या आणतात. संपादक या बातम्यांची शहानिशा करतात. मगच त्या बातम्या वर्तमानपत्रात छापल्या जातात. आजीच्या घराची ढाळज वर्तमानपत्रासारखीच होती. तिथे आलेल्या बातम्यांची शहानिशा झाल्यावरच बायका त्या बातम्या गावभर सांगायला मोकळ्या होत.

प्रश्न 2.
तुलना करा / साम्य लिहा :
आगळ : वाड्याचे संरक्षक कवच
आजी : कुटुंबाचे संरक्षक कवच किंवा
‘आजी म्हणजे घराचा आधार’ हे विधान सोदाहरण पटवून दया. (मार्च ‘१९)
उत्तर :
आगळ म्हणजे उंची-रुंदीला नऊ इंच आणि लांबीला सहा फूट असा भक्कम सागवानी वासा. एकदा आगळ लावली की चोऱ्यामाऱ्या होणे किंवा दरोडा पडणे शक्यच नसे. त्यामुळेच ही आगळ म्हणजे वाड्याला संरक्षणाचे एक भरभक्कम कवच लाभले होते.

प्रस्तुत उताऱ्यात आजीची भूमिकाही अगदी याच प्रकारची आहे. आजीमुळे कुटुंबात सुव्यवस्था नांदत होती. सुना आपापसात हेवेदावे करू शकत नव्हत्या. आपली कामे दुसरीवर ढकलू शकत नव्हत्या. सर्व कामे प्रत्येकीला करावी लागत. या वातावरणामुळे कोणावर अन्याय होत नव्हता. कोणालाही तक्रार करायला वावच राहत नसे. आजीमुळे प्रत्येकीला किंवा प्रत्येकाला भरभक्कम संरक्षण मिळाले होते. हे संरक्षण आगळेइतकेच भक्कम होते.

प्रश्न 3.
आगळ लावण्याची/टाकण्याची पद्धत समजावून सांगा.
उत्तर :
आगळ म्हणजे एक सागवानी अवजड वासा होता. त्याच्या एका टोकाला वाघाचा मुखवटा बसवला होता. वाघाच्या जबड्यात एक भक्कम कडी बसवलेली होती. त्या कडीला धरून आगळ ओढायची किंवा ढकलायची असते. दरवाज्याच्या दोन बाजूंना आगळ अडकवण्यासाठी भिंतींत दोन कोनाडे केलेले असतात. त्यांपैकी एक कोनाडा आगळ पूर्ण सामावली जाईल इतका खोल असतो. कडीला धरून आगळ कोनाड्यात पूर्ण ढकलली की दरवाजा उघडता येतो.

रात्रीच्या वेळी, दरवाजा बंद करून कोनाड्यात ढकलून ठेवलेली आगळ कडीला धरून ओढून बाहेर काढली जाते आणि ते टोक दुसऱ्या भिंतीच्या कोनाड्यात अडकवले जाते. अशा त-हेने आगळ बसवली की दरवाजा कोणीही उघडू शकत नाही.

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

प्रश्न 4.
दुपारच्या वेळी मुलांकडून केल्या जाणाऱ्या कल्पक कृती सांगा.
उत्तर :
कधी कधी बैठ्या खेळांऐवजी मुले वेगवेगळ्या कल्पक कृती करीत असत. एखादया वेळी सरपणातली लाकडे काढून विटीदांडू किंवा भोवरे तयार करीत बसत. भिंगऱ्या तयार करण्यासाठी घेतलेला लाकडाचा तुकडा दगडावर घासून घासून त्याला गोल आकार देण्याचा मुले प्रयत्न करीत. हे मोठे कष्टाचे काम असे. चुलीची काजळी लागून लागून काळ्या कुळकुळीत बनलेल्या खापऱ्या पाटीवर घासून घासून पाठीला काळा कुळकुळीत रंग आणण्याचा प्रयत्न करीत बसत.

बैलगाड्या बनवण्यासाठी ज्वारीची ताटे वापरीत. लाल माती आणून बैल बनवत बसत. गोल आकाराचे गोटे जमवून ते सर्व बाजूंनी दगडावर घासून घासून त्याने छान गोल आकार दयायचा प्रयत्न करीत बसत. अशा अनेक कल्पक कृती करण्यात मुले दंग होत.

प्रश्न 5.
पाठात चित्रित झालेल्या एकत्र कुटुंबपद्धतीबाबतचे तुमचे विचार स्पष्ट करा.
उत्तर :
या पाठात ग्रामीण भागातील मागील पिढीचे चित्रण आले आहे. त्या काळातील हे एक एकत्र कुटुंब होते. आजी ही कुटुंबप्रमुख होती. कुटुंबाच्या सर्व बाबींवर, सर्व व्यक्तींवर आजीचीच सत्ता चालत असे. कोणी कोणकोणती कामे कधी व कशी करावीत, हे आजी ठरवत असे.

या पद्धतीमुळे कुटुंबातील सर्व व्यवहारांना एकसूत्रीपणा येतो आणि कामे सुरळीतपणे पार पडतात; याचा कुटुंबालाच फायदा होतो, हे खरे आहे. पण या पद्धतीमध्ये कोणालाही स्वातंत्र्य राहत नाही. सुनांना साधा चहासुद्धा करून पिण्याची मोकळीक नव्हती. म्हणजे कोणालाही जरासुद्धा हौसमौज करण्याची परवानगी नव्हती. याचाच अर्थ कुटुंबातल्या सदस्यांना जीवनातील लहानसहान आनंदसुद्धा घेता येत नव्हते. त्यातही स्त्रियांना तर पूर्ण पारतंत्र्यातच राहावे लागे. ही चांगली स्थिती अजिबात नाही. आधुनिक काळात म्हणूनच एकत्र कुटुंबपद्धत टिकली नाही.

प्रश्न 6.
पाठाच्या शीर्षकाची समर्पकता थोडक्यात स्पष्ट करा.
उत्तर :
ग्रामीण भागात पूर्वी घराभोवती एक भलीमोठी, मजबूत भिंत बांधली जाई. भिंतीत एक मजबूत दार असे. त्याला कड्याकोयंडे असतच; शिवाय एक भलीभक्कम आगळ बसवलेली असे. एकदा ही आगळ लावली की घर पूर्णपणे बंद होत असे. घरातील कोणीही बाहेर जाऊ शकत नसे किंवा कोणीही बाहेरून आत येऊ शकत नसे. घरावर कोणाचाही हल्ला होणे शक्य नसे. यामुळे घर पूर्णपणे सुरक्षित होई. म्हणून ग्रामीण जीवनात या आगळीला एक महत्त्वपूर्ण स्थान लाभले होते.

पाठाच्या शीर्षकावरून असे दिसते की, त्या कुटुंबातील आजी ही त्या कुटुंबाची एक प्रकारे आगळच होती. तिच्या दराऱ्यामुळे कुटुंबाचे सर्व व्यवहार सुरळीत चालत असत. कुटुंबाला आपोआपच पूर्ण संरक्षण लाभायचे. घराची आगळ लावल्यावर आपल्या माणसांना बाहेर जाता येत नसे. म्हणजेच त्यांच्यावर बंधने येत. त्याचप्रमाणे आजीच्या नियंत्रणामुळे कुटुंबातील व्यक्तींवर बंधने येत. या बंधनांचा एक चांगला फायदा होई. कुटुंबातील कोणीही गैरवर्तन करू शकत नसे. त्यामुळे कुटुंबाचे व्यवहार कोलमडून पडत नसत. कुटुंबाला अंतर्गत व बाह्य असे दोन्ही अंगांनी संरक्षण मिळे. म्हणून ‘आजी : कुटुंबाचं आगळ’ हे शीर्षक अत्यंत समर्पक आहे.

प्रश्न 7.
आजच्या काळात एकत्र कुटुंब पद्धतीची आवश्यकता वाटते का? तुमचे मत सोदाहरण लिहा. (सराव कृतिपत्रिका-१)
उत्तर :
एकत्र कुटुंब पद्धतीचे काही फायदे आहेत; तसे काही तोटेही आहेत. तोटे काढून टाकले, तर एकत्र कुटुंब पद्धत आजच्या काळात खूप उपयोगी होऊ शकते. एकत्र कुटुंब पद्धतीत व्यक्तीला स्वत:चा वैयक्तिक विकास करून घेण्याची संधी खूप कमी प्रमाणात मिळते. व्यक्ती कुटुंबाशी बांधली जाते. कुटुंबाच्या अडीअडचणी, कुटुंबाची कामे, जबाबदाऱ्या यांच्यात ती गुरफटली जाते. कुटुंब एका व्यक्तीच्या नियंत्रणात राहते.

एकत्र कुटुंब पद्धतीमुळे कुटुंबात सर्व वयोगटातील माणसे असतात. वयस्क माणसांची कार्यक्षमता खुप कमी झालेली असते. त्यांना स्वत:च्या गरज भागवणे जिकिरीचे बनते. त्या वयात त्यांना इतरांच्या मदतीची खूप गरज असते. त्यांना एकत्र कुटुंब पद्धतीमध्ये आधार मिळतो. त्याचप्रमाणे लहान मुलांनाही एकत्र कुटुंब पद्धतीत खूप आधार मिळतो. आजच्या काळात आई-बाबा दोघेही नोकरी करतात. त्यामुळे घरी मुलांची काळजी घेणारे कोणीही नसते. मुलांची आबाळ होते. त्यांचे खाणेपिणे, त्यांचा अभ्यास वगैरे बाबींकडे लक्ष देणारे कोणी नसते. अशा स्थितीत घरी आजी-आजोबा असतील, तर ते मुलांकडे लक्ष देऊ शकतात. आजीआजोबांना समाधानही मिळते. पगारी माणसे घरातल्या माणसांप्रमाणे काळजी घेऊ शकत नाहीत. म्हणून आजच्या काळात एकत्र कुटुंब पद्धतीचा फायदा होऊ शकतो.

व्याकरण व भाषाभ्यास

कृतिपत्रिकेतील प्रश्न ४ (अ) आणि (आ) यांसाठी…

अ. व्याकरण घटकांवर आधारित कृती : .

१. समास :

प्रश्न 1.
पुढील विग्रहांवरून सामासिक शब्द ओळखा :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 23
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 24
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 25

प्रश्न 2.
तक्ता पूर्ण करा : (मार्च ‘१९)
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 26
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 27

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

२. अलंकार :

प्रश्न 1.
पुढील उदाहरण वाचून तक्ता पूर्ण करा :
देवाहुनही महान आहे माझी आई
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 28
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 29

३. शब्दसिद्धी :

प्रश्न 1.
‘बे’ हा उपसर्ग लागून तयार होणारे चार शब्द लिहा.
उत्तर :

  • बेबंद
  • बेजबाबदार
  • बेहिशेब
  • बेबनाव.

प्रश्न 2.
पुढील शब्दांना ‘अनीय’ हा प्रत्यय लावून शब्द तयार करा :
(i) श्रवण – ……………
उत्तर :
श्रवणीय

(ii) वाचन – ………….
उत्तर :
वाचनीय.

प्रश्न 4.
दोन अभ्यस्त शब्द लिहा.
उत्तर :

  • लालेलाल
  • गारेगार.

४. सामान्यरूप :

प्रश्न 1.
पुढील तक्ता पूर्ण करा :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 30
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 31

५. वाक्प्रचार :

(१) कंसातील वाक्प्रचारांचा त्या खालील वाक्यांत योग्य ठिकाणी वापर करून वाक्ये पुन्हा लिहा :
(शाबूत असणे, कानोसा घेणे, कडुसं पडणे, शहानिशा होणे, गुण्यागोविंदाने नांदणे)
प्रश्न 1.
शाळेतून घरी आलो की बाबा नेहमी अभ्यासाची चौकशी करायचे.
उत्तर :
शाळेतून घरी आलो की बाबा नेहमी अभ्यासाची शहानिशा करायचे.

प्रश्न 2.
मावळतीला सूर्य गेला नि अंधार पडला.
उत्तर :
मावळतीला सूर्य गेला नि कडुसं पडले.

प्रश्न 3.
घरातील दोन्ही जावा अगदी खेळीमेळीने राहत.
उत्तर :
घरातील दोन्ही जावा अगदी गुण्यागोविंदाने नांदत होत्या.

प्रश्न 4.
पाऊस पडणार आहे की नाही, याचा आमचा मोती कान टवकारून अंदाज घेतो.
उत्तर :
पाऊस पडणार आहे की नाही, याचा आमचा मोती कान टवकारून कानोसा घेतो.

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

(२) पुढील वाक्यांतील वाक्प्रचार शोधून अधोरेखित करा :

प्रश्न 1.
सहलीच्या वेळी शिस्तभंग होऊ नये याकडे शिक्षकांचा कटाक्ष असतो.
उत्तर :
सहलीच्या वेळी शिस्तभंग होऊ नये याकडे शिक्षकांचा कटाक्ष असतो.

प्रश्न 2.
दोन व्यक्तींतील संवादाचा तिसऱ्या व्यक्तीने कानोसा घेणे अयोग्यच.
उत्तर :
दोन व्यक्तींतील संवादाचा तिसऱ्या व्यक्तीने कानोसा घेणे अयोग्यच.

प्रश्न 3.
कारण नसताना हुकमत गाजवणाऱ्या व्यक्ती इतरांच्या नजरेतून उतरतात.
उत्तर :
कारण नसताना हुकमत गाजवणाऱ्या व्यक्ती इतरांच्या नजरेतून उतरतात.

Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी : कुटुंबाचं आगळ

आ. (भाषिक घटकांवर आधारित कृतीः

१. शब्दसंपत्ती :

प्रश्न 1.
विरुद्धार्थी शब्दांच्या योग्य जोड्या जुळवा :
Maharashtra Board Class 10 Marathi Solutions Chapter 3 आजी कुटुंबाचं आगळ 32
उत्तरे :
(i) आळस × उत्साह
(ii) आदर × अनादर
(iii) आस्था × अनास्था
(iv) आपुलकी × दुरावा

आजी : कुटुंबाचं आगळ Summary in Marathi

पाठाचा आशय या कथानकातील आजी खमकी आहे. तिच्या देहाची ठेवण, तिचे दिसणे, तिचा पेहराव इत्यादी बाबींमध्ये तिचा खमकेपणा दिसून येतो.

‘आगळ ‘मधील नायकाच्या घरी एक गाय होती. ती वरवर्षी व्यायची. त्यामुळे घरामध्ये दुधाची खूप रेलचेल होती. मुलांना भरपूर दुध प्यायला मिळे. आजीला चार सुना होत्या. आजीचा दरारा असल्यामुळे मुलांना खायलाप्यायला देताना सुना आपपरभाव करू शकत नव्हत्या. कामचुकारपणा करू शकत नव्हत्या. आपली कामे दुसरीवर टाकू शकत नव्हत्या. सर्व कामे प्रत्येकीला करावी लागत. या वातावरणामुळे कोणावर अन्याय होत नव्हता. कोणालाही तक्रार करायला वावच नसे. आजीमुळे प्रत्येकीला किंवा प्रत्येकाला भरभक्कम संरक्षण मिळाले होते. हे संरक्षण आगळइतकेच भक्कम होते.

दुपारच्या वेळी मुलांचे बैठे खेळ किंवा क्वचितप्रसंगी मैदानी खेळ खेळले जात. अशा प्रकारे मुलांचे बालपण तर निसर्गामध्ये सहजगत्या घडत होते. या सगळ्याला आजीच्या मायेच्या सावलीचा आधार होता.

दुपारपर्यंतची कामे आटोपल्यावर थोडीशी विश्रांती घेऊन आजी ढाळजेत येऊन बसायची. गल्लीतल्या बायका जमायच्या. निवडटिपण असली कामे करता करता गप्पा होत. गावभरच्या बातम्या कळत. आजीच्या समोरच बातम्यांची शहानिशा होई. ही ढाळज म्हणजे एक प्रकारे गावाचे वर्तमानपत्रच होती. रात्री आठ वाजता वाड्याचा दरवाजा बंद होई आणि आगळ बसवली जाई. आगळ बसवली की संरक्षणाची हमी मिळे.

आजी : कुटुंबाचं आगळ शब्दार्थ

  • रापणे – त्वचेवर काळपटलेली छटा येणे.
  • गोंदण – विशिष्ट प्रकारच्या सुईने त्वचेवर टोचून टोचून नक्षी काढणे.
  • गावरान – गावठी. वेत – वासराला जन्म देणे.
  • धार काढणे – गाई-म्हशीचे दूध काढणे.
  • चरवी – दूध काढण्याचे भांडे.
  • धारोष्ण – उष्णपणा निवला नाही असे ताजे दूध.
  • सरपण – इंधन (विशेषतः लाकडांचे).
  • ढाळज – मोठा वाडा वगैरेंसारख्या घराच्या मुख्य दरवाजाजवळची आतल्या बाजूची
  • जागा, पडवी, वाकळ – गोधडी.
  • शहानिशा – खातरजमा,
  • कडुसं – काळोख होण्याची वेळ,
  • आगळ – अडसर (येथे अर्थ-भक्कम आधार.).
  • देवळी – भिंतीतला कोनाडा.
  • चिंचोके – चिंचेच्या बिया.
  • गजगे – सागरगोटे,
  • जिबल्या – अर्धा कापलेला जोड्यांचा तळ.
  • मुटके – हातपाय पोटाशी आवळून घेऊन आजूबाजूला खूप पाणी उडेल अशा रितीने पाण्यात मारलेली उडी.

आजी : कुटुंबाचं आगळ वाक्प्रचार व त्यांचे अर्थ

  • हुकूमत गाजवणे : अधिकार गाजवणे.
  • हातात काठी येणे : म्हातारपण येणे; तोल सांभाळता न येणे.
  • गुण्यागोविंदाने नांदणे : समंजसपणे व आनंदाने राहणे.
  • धार काढणे : गायी-म्हशीचे दूध काढणे.
  • कटाक्ष असणे : खास लक्ष देणे.
  • धान्य निवडणे : धान्यातले गोटे इत्यादी वेचून बाहेर काढणे.
  • हुक्की येणे : लहर येणे.

Maharashtra Board Class 10 Marathi Kumarbharti Solutions Chapter 4 उत्तमलक्षण

Balbharti Maharashtra State Board Class 10 Marathi Solutions Kumarbharti Chapter 4 उत्तमलक्षण (संतकाव्य) Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Marathi Kumarbharti Chapter 4 उत्तमलक्षण (संतकाव्य)

Marathi Kumarbharti Std 10 Digest Chapter 4 उत्तमलक्षण Textbook Questions and Answers

प्रश्न. पुढील कवितेच्या आधारे दिलेल्या सूचनांनुसार कृती करा :

कृती १ : (आकलन)

प्रश्न 1.
आकृत्या पूर्ण करा :
(i)
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 1
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 2

(ii)
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 3
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 4

Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य)

प्रश्न 2.
चौकटी पूर्ण करा :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 5
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 6

प्रश्न 3.
पुढील गोष्टींबाबत संत रामदास कोणती दक्षता घ्यायला सांगतात. (मार्च ‘१९)
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 7
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 8

कृती २ : (आकलन)

प्रश्न 1.
शब्दजाल पूर्ण करा :
(i)
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 9
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 10

(ii)
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 11
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 12
(iii)
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 13
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 14

Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य)

प्रश्न 2.
पुढील व्यक्तींशी कसे वागावे, असे संत रामदास म्हणतात :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 15
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 16

प्रश्न 3.
पुढील गोष्टींबाबत कोणती दक्षता घ्यावी, ते लिहा :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 17
उत्तर :
(i) तोंडाळासी भांडू नये.
(ii) संतसंग खंडू नये.
(iii) सत्यमार्ग सोडू नये.

प्रश्न 4.
असत्य विधान ओळखा :
(i) संतसंग सोडू नये.
(ii) अपकार घेऊ नये.
(iii) व्यापकपण सांडू नये.
(iv) खोटेपणाच्या पंथाला जाऊ नये.
उत्तर :
असत्य विधान – अपकार घेऊ नये.

प्रश्न 5.
अचूक विधान ओळखा : (मार्च ‘१९)
(i) पैज, होड लावावी.
(ii) सत्याची वाट धरावी.
(iii) पापद्रव्य सहज जोडावे.
(iv) नेहमी अभिमानाने वागावे.
उत्तर :
अचूक विधान – सत्याची वाट धरावी.

प्रश्न 6.
तुमच्यातील प्रत्येकी तीन गुण व तीन दोष शोधून लिहा.
उत्तर :
नमुना उत्तर :
गुण : (i) मी रोज व्यायाम करतो.
(ii) मी खोटे बोलत नाही.
(iii) मी आईला कामात मदत करतो.

दोष : (i) मला चटकन राग येतो.
(ii) मी ताटात अन्न टाकतो.
(iii) माझे अक्षर चांगले नाही.

कृती ३ : (काव्यसौंदर्य)

प्रश्न 1.
‘सभेमध्ये लाजों नये। बाष्कळपणे बोलों नये।’ या ओळीतील विचार स्पष्ट करा.
उत्तर :
‘उत्तमलक्षण’ या कवितेमध्ये संत रामदास यांनी आदर्श गुणसंपन्न व्यक्तीची लक्षणे समजावून सांगितली आहेत. त्यांपैकी एक लक्षण उपरोक्त चरणात सूचित केले आहे.

मनुष्य हा समाजप्रिय प्राणी आहे. माणसांमध्ये तो नित्य वावरत असतो. समूहामध्ये आदर्श व्यक्तीचे वर्तन कसे असावे, हे सांगताना संत रामदास म्हणतात – सभेमध्ये वावरताना, आपले मत मांडताना कधीही लाजू नये. स्पष्टपणे आपले म्हणणे मांडावे; परंतु त्याच वेळी बालिशपणे बोलून आपले हसे करून घेऊ नये. निरर्थक असे वक्तव्य करू नये. बाष्कळपणे बोलू नये. उत्तम पुरुषाचे एक मर्मग्राही लक्षण या ओवीतून मांडले आहे.

प्रश्न 2.
‘आळसें सुख मानूं नये,’ या ओळीचा तुम्हांला समजलेला अर्थ स्पष्ट करा.
उत्तर :
‘उत्तमलक्षण’ या ओव्यांमध्ये संत रामदास यांनी उत्तम व्यक्तीची लक्षणे विशद करताना आळस हा माणसाचा शत्रू आहे, असे ठासून प्रतिपादिले आहे.

‘आळस’ हा माणसाच्या अंगी असलेला दुर्गुण आहे. आळसामुळे कार्य करायला उत्साह राहत नाही व त्यामुळे बरीच कामे खोळंबून राहतात. ‘आळसे कार्यभाग नासतो!’ या समर्थ रामदासांच्या उक्तीमध्ये हेच तत्त्व सांगितले आहे. माणसाच्या मनाला जे षड्विकार जडतात, त्यात ‘आळस’ हा एक विकार आहे. दैनंदिन कामांमध्ये आळसाला स्थान देऊ नये. आळसामुळे प्रगती खुंटते, भविष्य अंधारते. आळसामुळे मनाला जडत्व प्राप्त होते व माणूस नाकर्ता होतो. आळशी माणसाला समाजात मान मिळत नाही. म्हणून आळसात सुख मानू नये, समाधान मानू नये, असे समर्थ रामदास सांगतात.

Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य)

Marathi Kumarbharti Class 10 Textbook Solutions Chapter 4 उत्तमलक्षण Additional Important Questions and Answers

प्रश्न. पुढील कवितेसंबंधी त्याखाली दिलेल्या मुद्द्यांच्या आधारे कृती सोडवा :

प्रश्न 1.
कविता-उत्तमलक्षण.
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 20
उत्तर : उत्तमलक्षण.
(१) प्रस्तुत कवितेचे कवी : संत रामदास.
(२) कवितेचा रचनाप्रकार : ओवी.
(३) कवितेचा काव्यसंग्रह : श्रीदासबोध.
(४) कवितेचा विषय : उत्तम माणसाची लक्षणे.
(५) कवितेतून व्यक्त होणारा (स्थायी) भाव : आदर्श माणसे घडवण्याचा ध्यास.

(६) कवितेच्या कवींची लेखनवैशिष्ट्ये : प्रस्तुत कवितेची रचना ओवी या छंदात केलेली आहे. ओवी हा उच्चारणाला सुलभ असा अत्यंत लवचीक रचनाप्रकार आहे. त्यामुळे या कवितेतील भाषा ओवी या छंदाला साजेशी सुबोध व सर्वसामान्यांना सहज समजेल अशी आहे. साहजिक कवितेची आवाहन शक्ती वाढली आहे. चुकीचे वर्तन व चांगले वर्तन या दोन्ही बाबी समर्थांनी स्पष्ट शब्दांत कोणतीही संदिग्धता न ठेवता सांगितल्या आहेत. समर्थांच्या या रचनेत प्रासादिकता आढळते.

(७) कवितेची मध्यवर्ती कल्पना : प्रस्तुत कवितेत संत रामदासांनी उत्तम, गुणसंपन्न, आदर्श व्यक्तीची लक्षणे सांगितली आहेत. समाजात वावरताना कसे वागावे, काय करावे व काय टाळावे यांचे सुस्पष्ट शब्दांत निवेदन केले आहे. समाज नीतिमान व कर्तबगार व्हावा, ही तळमळ या पदयपाठातून स्पष्टपणे जाणवते.

(८) कवितेतून व्यक्त होणारा विचार : प्रस्तुत कवितेत संत रामदासांनी, व्यक्तीने समाजात कसे वागावे याचे मार्गदर्शन केले आहे. नेहमीच सावध मनाने वागावे. इतरांना सहकार्य करावे. कोणाशीही कपटाने वागू नये. तोंडाळ, वाचाळ माणसांना टाळावे. आळस झटकून टाकावा. पूर्ण विचार करून वागावे. उपकाराची परतफेड करावी. नेहमी उदारपणाने वागावे. मनाचा मोठेपणा बाळगावा. परावलंबी होऊ नये. नेहमी सत्याने वागावे. कुप्रसिद्धी टाळावी इत्यादी अनेक गुणांचे आचरण करण्यास या कवितेत संत रामदासांनी सांगितले आहे.

(९) कवितेतील आवडलेलो ओळ :
अपकीर्ति ते सांडावी। सत्कीर्ति वाडवावी।।
विवेके दृढ धरावी । वाट सत्याची ।।

(१०) कविता आवडण्याची वा न आवडण्याची कारणे : मला ही कविता खूप आवडली. एकतर हे विचार पटकन पटण्यासारखे आहेत. ही कविता उच्चारताना, वाचताना आनंद होतो. उच्च, उदात्त विचार मनात घोळवल्याने मनही आनंदित होते. सर्व गुण-अवगुण रामदासांनी अत्यंत स्पष्टपणे, परखडपणे सांगितले आहेत. कुठेही शब्दांचा बोजडपणा नाही. प्रत्येक शब्दागणीक अर्थ स्पष्ट होत जातो. मुख्य म्हणजे दैनंदिन जीवनात वागताना आवश्यक असलेल्या गुणांचे मार्गदर्शन घडत असल्याने कविता आपली, स्वतःची, स्वतःसाठी असलेली वाटत राहते.

(११) कवितेतून मिलणाग संदेश : प्रत्येक व्यक्तीने चांगले वागण्याचाच प्रयत्न केला पाहिजे. त्यासाठी वाईट गुण कोणते व चांगले गुण कोणते हे स्पष्टपणे समजून घ्यावे. प्रत्येकाने उत्तम माणूस बनण्याचा प्रयत्न करावा. यातूनच चांगला व समर्थ समाज निर्माण होतो.

कृतिपत्रिकेतील प्रश्न २ (इ) साठी…

प्रश्न. पुढील ओळींचे रसग्रहण तुमच्या शब्दांत लिहा :

प्रश्न 1.
‘जनी आर्जव तोडूं नये । पापद्रव्य जोडूं नये ।
पुण्यमार्ग सोडूं नये । कदाकाळीं ।।’ (मार्च ‘१९)
उत्तर :
आशयसौंदर्य : ‘उत्तमलक्षण’ या श्रीदासबोधातील एका समासामध्ये समर्थ रामदासांनी गुणसंपन्न आदर्श व्यक्तिमत्त्वाची महत्त्वाची लक्षणे सांगितली आहेत. त्यांपैकी उपरोक्त ओवीमध्ये तीन लक्षणांचा ऊहापोह केला आहे.

काव्यसौंदर्य : समाजात वावरताना व्यक्तीने कोणते आचरण करावे हे सांगताना संत रामदास म्हणतात – लोकांचे मन मोडू नये. लोकांनी केलेली विनंती धुडकावू नये. उलट जनभावनांचा आदर . करावा. तसेच वाईट मार्गाने संपत्ती साठवू नये. अशी संपत्ती हे पापाचे धन असते. म्हणून सत्शील मार्गाने जीवन व्यतीत करावे. पुण्यमार्ग आचारावा. कधीही पुण्यमार्गाने जाण्याचे सोडू नये.

भाषिक वैशिष्ट्ये : वरील ओवीमध्ये जनांसाठी खूप सुगम निरूपण केले आहे. ‘तोडू नये, जोडू नये, सोडू नये’ अशा सोप्या यमकांद्वारे संदेशामध्ये आवाहकता आली आहे. ओवीछंदाला साजेशी सुबोध भाषा वापरल्यामुळे जनमानसावर तत्त्व ठसवणे सुलभ झाले आहे. पापद्रव्य व पुण्यमार्ग यांतील विरोधाभास ठळकपणे उठून दिसतो. ओवीमध्ये प्रासादिकता हा गुण आढळतो.

Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य)

प्रश्न 2.
‘अपकीर्ति ते सांडावी। सत्कीर्ति वाडवावी।
विवेकें दृढ धरावी। वाट सत्याची।।’ (मार्च ‘१९)
उत्तर :
आशयसौंदर्य : संत रामदासांनी ‘उत्तमलक्षण’ या कवितेत आदर्श गुणवान व्यक्तीची वैशिष्ट्ये सांगताना या ओळींमधून सद्वर्तन कशा प्रकारे करावे, याची शिकवण दिली आहे.

काव्यसौंदर्य : संत रामदास म्हणतात – लोक आपल्याला दूषणे देतील व निंदा करतील असे वर्तन कदापिही करू नये. ज्या वागण्याने आपली अपकीर्ती होईल, असे वागणे टाळावे. उलट आपल्या व्यक्तिमत्त्वाची कीर्ती पसरेल, अशी वागणूक करायला हवी. स्वतः चांगले वागून सत्कीर्ती वाढवायला हवी. त्यासाठी बुद्धीचा विवेक महत्त्वाचा ठरतो. सद्विचाराने, विवेकाने सत्याचा मार्ग ठामपणे आचरावा. विवेकबुद्धी ठोस असणे गरजेचे आहे.

भाषिक वैशिष्ट्ये : सन्मार्गाचे लक्षण सांगताना जनसामान्यांना समजतील असे तीन मुद्दे या ओळीत सहजपणे सांगितले आहेत. अपकीर्ती व सत्कीर्ती तसेच सांडावी व वाढवावी या विरोधी शब्दांमुळे ओवीची खुमारी वाढली आहे. दृढ धरणे हा वाक्प्रचार चपखलपणे उपयोगात आणला आहे. जनमानसावर तत्त्व ठसवण्याची समर्थांची हातोटी समर्थपणे व्यक्त झाली आहे.

व्याकरण व भाषाभ्यास

कृतिपत्रिकेतील प्रश्न ४ (अ) आणि (आ) यांसाठी…

अ. व्याकरण घटकांवर आधारित कृती:

१. समास :

प्रश्न 1.
पुढील तक्ता पूर्ण करा : (ठळक अक्षरांत उत्तरे दिली आहेत.)
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 21

२. अलंकार :
पुढील ओळी वाचून तक्ता पूर्ण करा :
प्रश्न 1.
‘ऊठ पुरुषोत्तमा । वाट पाही रमा ।
दावि मुखचंद्रमा । सकळिकांसी।।’

प्रश्न 2.
‘नयनकमल’ हे उघडित हलके जागी हो जानकी.
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 22
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 23

Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य)

३. वृत्त :
पुढील ओळींचे लगक्रम लिहा :

प्रश्न 1.
ऐकू न ये तुज पिकस्वर मंजुळे का?
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 24

प्रश्न 2.
मना सज्जना भक्तिपंथेचि जावे.
उत्तर :
Maharashtra Board Class 10 Marathi Solutions Chapter 4 उत्तमलक्षण (संतकाव्य) 25

उत्तमलक्षण Summary in Marathi

कवितेचा भावार्थ :
उत्तम पुरुषाची (आदर्श व्यक्तीची) लक्षणे सांगताना संत रामदास श्रोत्यांना म्हणतात – श्रोतेहो, तुम्हांला आता मी उत्तम, गुणवान व्यक्तीची लक्षणे सांगतो, ती सावध मनाने ऐकावीत. ही लक्षणे ऐकून तुम्ही सर्व गोष्टी जाणण्याची खूण अंगी बाणवावी. ।।१।।

पूर्ण माहिती असल्याशिवाय कुठल्याही रस्त्याने जाऊ नये. वाटेमधील अडथळे जाणून घेतल्याशिवाय प्रवास करू नये. फळाचे अंतरंग (गुण) ओळखल्याशिवाय ते खाऊ नये. रस्त्यात पडलेली वस्तू अचानक पटकन उचलू नये. ।।२।।

लोकांनी केलेली विनंती लक्षात घ्यावी. तिला अमान्य करू नये. पाप किंवा कपट करून मिळवलेल्या संपत्तीचा साठा करू नये. कपट करून पैसा मिळवू नये. पुण्याचा मार्ग म्हणजे सच्छील मार्ग, सद्वर्तन कधीही सोडू नये. ।।३।।

जी तोंडाळ, भांडकुदळ व्यक्ती असेल, त्या व्यक्तीशी कधीही भांडू नये. व्यर्थ बडबड करणाऱ्या वाचाळ माणसाशी तंटा करू नये. सज्जनाची संगत कधी मध्येच सोडू नये, (हे कायम मनात ठेवा.) मनापासून झालेली संतांची मैत्री सोडू नये. ।।४।।

काम न करता ऐदीपणे, आळस करून आनंद घेऊ नये. आळसात सुख नसते. कुणाबद्दल उखाळ्यापाखाळ्या करून चहाडी करू नये. दुसऱ्याबद्दल खोटेनाटे बोलणे मनातही आणू नये. संपूर्ण शोध घेतल्याशिवाय कोणतेही काम करू नये. (काम अर्धवट सोडू नये.) ।।५।।

सभेमध्ये, समूहाच्या बैठकीमध्ये लाजू नये, मुखदुर्बळ राहू नये. मोकळ्या मनाने बोलावे परंतु बालिशपणे बोलू नये. बाष्कळ बडबड करू नये, कोणत्याही प्रकारची पैज, शर्यत लावू नये. कुणाशीही स्पर्धा करू नये. ।।६।।

कोणाचे उपकार सहसा घेऊ नयेत आणि घेतलेच तर त्यांची लगेच परतफेड करावी, उपकारातून लवकर उतराई व्हावे. दुसऱ्यांना दुःख देऊ नये, त्यांना व्यथित करू नये. कुणाचा विश्वासघात किंवा बेइमानी मुळीच करू नये. ।।७।।

स्वत:च्या मनाचा मोठेपणा कधी सोडून देऊ नये. मन कोते करू नये, ते व्यापक ठेवावे. परावलंबी होऊ नये. कुणावरी आपल्या आयुष्याचे ओझे लादू नये. ।।८।।

सत्याचा मार्ग कधी सोडू नये. खरेपणाने वागावे, असत्याच्या, खोटेपणाच्या मार्गाने जाऊ नये. खोटारडेपणा करून वागू नये. खोटेपणाचा कधीही वृथा अभिमान, व्यर्थ गर्व करू नये. ।।९।।

अपकीर्तीला बळी पडू नये. कुप्रसिद्धी टाळावी. चांगली कीर्ती वाढवावी. चांगल्या प्रकारे प्रसिद्ध पावावे. सारासार विचाराने, विवेकाने वर्तन करून सत्यमार्ग पत्करावा. ।।१०।।

उत्तमलक्षण शब्दार्थ

  • उत्तम – आदर्श, गुणसंपन्न,
  • लक्षण – गुणवैशिष्ट्य.
  • श्रोतीं – ऐकणाऱ्या लोकांनो.
  • सावधान – सावध होणे, सजग होणे.
  • उत्तम – उत्कृष्ट, जेणें – ज्याने.
  • बाणे – बाणणे, अंगिकारणे, सवय लागणे,
  • सर्वज्ञ – सारे जाणणारा.
  • पुसल्याविण – विचारल्याशिवाय,
  • येकायेकी – एकदम, पटकन.
  • जनीं – लोकांचे. आर्जव – विनंती.
  • पापद्रव्य – पापाने (कपटाने) मिळवलेली संपत्ती.
  • जोडू नये – साठवू नये.
  • पुण्यमार्ग – चांगला रस्ता, सद्वर्तन,
  • कदाकाळी – कोणत्याही वेळी.
  • तोंडाळ – वाटेल ते बोलणारा, भांडकुदळ.
  • वाचाळ – व्यर्थ बडबड करणारा.
  • तंडो नये – तंटा (भांडण) करू नये.
  • संतसंग – सज्जन माणसाची संगत.
  • खंडू नये – तोडू नये.
  • अंतर्यामी – मनातून, हृदयातून.
  • आळस – काम न करणे.
  • चाहाडी – एखादयाबद्दल वाईट सांगणे, आगलावेपणा.
  • कार्य – काम,
  • सभा – समूहाची बैठक.
  • बाष्कळपणा – बालिशपणा.
  • पैज – स्पर्धा, शर्यत.
  • होड – पैज,
  • परपीडा – दुसऱ्याला छळणे, दुःख देणे.
  • विश्वासघात – बेइमानी.
  • व्यापकपण – (मनाचा) मोठेपणा.
  • पराधेन – परावलंबी, दुसऱ्यावर विसंबणे.
  • वोझें – (स्वत:चा) भार,
  • कोणीयेकासी – कोणावरही.
  • सत्यमार्ग – खऱ्याचा मार्ग, सद्वर्तन.
  • असत्य – खोटेपणा.
  • पंथे – मार्गाने, वाटेने.
  • कदा – कधीही.
  • अभिमान – व्यर्थ गवं.
  • अपकीर्ति – बेअब्रू, वाईट प्रसिद्धी,
  • सत्कीर्ति – चांगली प्रसिद्धी.
  • वाडवावी – वाढवावी.
  • विवेके – चांगल्या विचाराने.
  • दृढ – ठाम, मजबूत, ठोस.

Space Missions Class 10 Questions And Answers Maharashtra Board

Class 10 Science Part 1 Chapter 10

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions Notes, Textbook Exercise Important Questions and Answers.

Std 10 Science Part 1 Chapter 10 Space Missions Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 10 Space Missions Question Answer Maharashtra Board

Question 1.
Fill in the blanks and explain the statements with reasoning:
a. If the height of the orbit of a satellite from the earth’s surface is increased, the tangential velocity of the satellite will ………………
Answer:
If the height of the orbit of a satellite from the earth’s surface, is increased, the tangential velocity of the satellite will decrease.
Explanation: The gravitational force (F) exerted by the earth on the satellite will decrease if the height of the orbit of the satellite from the earth’s surface is increased. Hence, the tangential velocity of the satellite will decrease.
The formula
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 1
shows that υc decreases with increasing h.

b. The initial velocity (during launching) of the Mangalyaan must be greater than ………… from the earth.
Answer:
The initial velocity (during launching) of the Mangalyaan must be greater than the escape velocity from the earth.
Explanation: If a satellite is to travel beyond the gravitational pull of the earth, its velocity must be more than the escape velocity from the earth.
[Note: The velocity must be atleast equal to the escape velocity. Refer the definition of escape velocity.]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 2.
State with reasons whether the following statements are true or false.
a. If a spacecraft has to be sent away from the influence of the earth’s gravitational field, its velocity must be less than the escape velocity.
Answer:
False.
Explanation: The escape velocity of a body is the minimum velocity with which it should be projected from the earth’s surface, so that it can escape the influence of the earth’s gravitational field. This clearly shows that the given statement is false.

b. The escape velocity on the moon is less than that on the earth.
Answer:
True.
Explanation: Escape velocity of an object from the earth,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 2

c. A satellite needs a specific velocity to revolve in a specific orbit.
Answer:
True.
Explanation:
Centripetal force on the satellite \(\frac{m v_{c}^{2}}{R+h}\) = gravitational force exerted by the earth on the satellite \(\frac{G M m}{(R+h)^{2}}\)
where,
m: mass of the satellite
υc: critical velocity of the satellite
h: height of the satellite from the surface of the earth
M: mass of the earth
R: radius of the earth
G: gravitational constant
∴ \(v_{\mathrm{c}}^{2}=\frac{G M}{R+h}\)
∴ \(v_{\mathrm{c}}=\sqrt{\frac{G M}{R+h}}\)
Thus, if the value of h changes, the value of υc also changes. It means a satellite needs to be given a specific velocity (in the tangential direction) to keep it revolving in a specific orbit.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 3

d. If the height of the orbit of a satellite increases, its velocity must also increase.
Answer:
False.
Explanation:
Centripetal force on the satellite \(\frac{m v_{c}^{2}}{R+h}\) = gravitational force exerted by the earth on the satellite \(\frac{G M m}{(R+h)^{2}}\)
where,
m : mass of the satellite
υc : critical velocity of the satellite
h : height of the satellite from the surface of the earth
M : mass of the earth
R : radius of the earth
G : gravitational constant
∴ \(v_{\mathrm{c}}^{2}=\frac{G M}{R+h}\)
∴ \(v_{\mathrm{c}}=\sqrt{\frac{G M}{R+h}}\)
Thus, if the value of h changes, the value of υc also changes. It means a satellite needs to be given a specific velocity (in the tangential direction) to keep it revolving in a specific orbit.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 4
As per the formula υc = \(\sqrt{\frac{G M}{R+h}}\) , if the value of h increases, the value of υc decreases. Hence, if the height of the satellite from the surface of the earth increases, its velocity decreases.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 3.
Answer the following questions:
a. What is meant by an artificial satellite? How are the satellites classified based on their functions?
(OR)
Write the importance of artificial satellites in your words. (Practice Activity Sheet – 3)
Answer:
A manmade object orbiting the earth or any other planet is called an artificial satellite. Satellites work on solar energy and hence photovoltaic panels are attached on both sides of the satellite, which look like wings. Satellites are also installed with various transmitters and other equipment to receive and transmit signals between the earth and the satellites.

Classification of satellites depending on their functions:
(1) Weather satellites: weather satellites collect the information regarding weather conditions of the region. It records temperature, air pressure, wind direction, humidity, cloud cover, etc. this information is sent to the space research station on the earth and then with this information weather forecast is made.

(2) Communication satellites: In order to establish communication between different places on the earth through mobile phones or computer assisted internet, communication satellites are used. Many artificial satellites placed at various locations in the earth’s orbit are well interconnected and help us to have communication with any place, from anywhere, at any time and in any form including voicemail, email, photographs, audio mail, etc.

(3) Broadcasting satellites: Broadcasting satel¬lites are used to transmit various radio and television programs and even live programs from any place on the earth to any other place. As a result, one can have access to information about current incidents, events, programs, sports and other events right from his drawing room with these satellites.

(4) Navigational satellites: Navigational satel¬lites assist the surface, water and air transportation and coordinate their busy schedule. These satellites also assist the user with current live maps as well as real time traffic conditions.

(5) Military satellites: Every sovereign nation needs to keep the real time information about the borders. Satellites help to monitor all movements of neighboring countries or enemy countries. Military satellites also help to guide the missiles effectively.

(6) Earth observation satellites: These satellites observe and provide the real time information about the earth. These satellites also help us to collect information about the resources, their management, continuous observation about a natural phenomenon and the changes within it.

(7) Other satellites: Apart from these various satellites, certain satellites for specific works or purposes are also sent in the space. E.g. India has sent EDUSAT for educational purpose; CARTOSAT for surveys and map making. Similarly, satellites with telescopes, like Hubble telescope or a satellite like International Space Station help to explore the universe. In fact, ISS (International Space Station) provides a temporary residence where astronauts can stay for a certain short or long period and can undertake the research and study space activities.
The various functions listed above show the importance of artificial satellites.

b. What is meant by the orbit of a satellite? On what basis and how are the orbits of artificial satellites classified?
Answer:
Orbit of a satellite is its path around the earth.
Orbits of artificial satellites can be classified on various basis.
(1) On the basis of the angle of the orbital plane: Orbital plane of a satellite can be the equatorial plane of the earth or it can be at an angle to it.
(2) On the basis of the nature of the orbit: Orbital plane can be circular or elliptical in shape.
(3) On the basis of the height of the satellite: Orbit of a satellite can be HEO, MEO or LEO.

(i) High Earth Orbit (HEO) satellite: A satellite orbiting at a height equal to or greater than 35780 km above the earth’s surface is called a High Earth Orbit satellite. The critical velocity (υc) of a satellite revolving in an orbit at 35780 km above the earth surface is 3.08 km/s. Such a satellite will take about 23 hours 54 minutes to complete one revolution around the earth. The earth completes one rotation about its axis in the same time. The orbital plane of such a satellite is the equatorial plane of the earth. The satellite’s relative position appears stationary with respect to a place on the earth. This satellite is, therefore, called a geostationary satellite or geosynchronous satellite.

(ii) Medium Earth Orbit (MEO) satellite: A satellite orbiting at a height between 2000 km and 35780 km above the earth’s surface is called a Medium Earth Orbit satellite. The orbital path of such a satellite is normally elliptical and passes through the North and the South polar regions. These satellites take about 12 hours to complete one revolution around the earth.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 5

(iii) Low Earth Orbit (LEO) satellite:
A satellite orbiting at a height between 180 km and 2000 km above the earth’s surface is called a Low Earth Orbit satellite. Normally, these satellites take 90 minutes to complete one revolution around the earth. Weather satellites, space telescopes and International Space Station are Low Earth Orbit satellites.

c. Why are geostationary satellites not useful for studies of polar regions? (Practice Activity Sheet – 4)
(OR)
Explain the following statement. A geostationary satellite is not useful in the study of polar regions. (Practice Activity Sheet – 1)
Answer:
Geostationary satellites have two distinct characteristics:
(1) Geostationary satellites are HEO satellites and are placed at 35780 km above the earth’s surface.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 6

(2) A geostationary satellite revolves in the equatorial plane of the earth, and thus, it can never fly above the polar regions.
Hence, geostationary satellites are not useful for studies of polar regions.

d. What is meant by a satellite launch vehicle? Explain the satellite launch vehicle developed by ISRO with the help of a schematic diagram.
Answer:
A rocket used to carry an artificial satellite to a desired height above the earth’s surface and then project it with a proper velocity so that the satellite orbits the earth in the desired orbit is called a launch vehicle. A satellite launch vehicle needs a specific velocity as well as a thrust to reach the desired height above the earth’s surface. The velocity and the thrust of a satellite launch vehicle depend on the weight and orbital height of the satellite.

Accordingly, the structure of the launch vehicle is decided and designed. The weight of the fuel also contributes a major portion in the total weight of the launch vehicle. This also influences the structure of the launch vehicle. In order to use the fuel optimally, multiple stage launch vehicles are now designed and used.

The Polar Satellite Launch Vehicle (PSLV) developed by ISRO is shown below in a schematic diagram.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 7

e. Why is it beneficial to use a satellite j launch vehicle made up of more than one stage?
Answer:
Earlier Satellite Launch Vehicles (SLV) used to be of a single stage vehicles. Such SLVs used to be very heavy as well as expensive in terms of its fuel consumption. As a result, SLVs with multiple stages were developed.

In multistage SLVs, as the journey of the launch vehicle progresses and the vehicle achieves a specific velocity and a certain height, the fuel of the first stage is exhausted and the empty fuel tank gets detached from the main body of the launch vehicle and falls back into a sea or on unpopulated land. As the fuel in the first stage is exhausted, the engine in the second stage is Ignited. However, the weight of the launch vehicle is now less than what it was earlier and hence it can move with higher velocity, Thus, it saves fuel consumption. Hence, it is beneficial to use a multistage satellite launch vehicle.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 4.
Complete the following table:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 8
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 9

Question 5.
Solve the following problems:
a. If the mass of a planet is eight times the mass of the earth and its radius is twice the radius of the earth, what will be the escape velocity for that planet?
Answer:
Given:
(1) The mass of the planet (M) is eight times the mass of the earth, i.e., 8 × 6 × 1024 kg
(2) The radius of the planet (R) is twice the radius of the earth, i.e., 2 × 6.4 × 106 km
(3) G = 6.67 × 10-11 N·m2/kg2
Escape velocity for that planet
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 10
= 2.237 × 104 m/s
= 22.37 km/s

b. How much time would a satellite in an orbit at a height of 35780 km above the earth’s surface take to complete one revolution around the earth, if the mass of the earth were four times its original mass?
Answer:
Given: R (Earth) = 6400 km = 6.4 × 106 m,
M (Earth) = 6 × 1024 kg
∴ M’ = 4M = 4 × 6 × 1024 kg
h = 35780 km = 3.578 × 107 m = 35.78 × 106 m,
G = 6.67 × 10-11 N·m2/kg2, T = ?
The time that the satellite would take to complete one revolution around the earth,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 11
= Approx 4.303 × 104 s
= Approx 11.95 h
or 11 hours 57 minutes 10 seconds.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

c. If the height of a satellite completing one revolution around the earth in T seconds is h1 meters, then what would be the height of a satellite taking 2\(\sqrt{2}\) T seconds for one revolution?
Answer:
Given:
(1) Time: T seconds
(2) Height: h1
Let us assume the height of the satellite completing one revolution in 2\(\sqrt{2}\) T seconds as h2.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 12
∴ R + h2 = 2R + 2h1
∴ h2 = R + 2h1

Project:

Project 1.
Collect information about the space missions undertaken by Sunita Williams.
Hints:
The following sources can be used to get the information on the above topic:
(1) Google Search Engine
(2) YouTube
(3) E-books on Sunita Williams
(4) English and other regional language books on Sunita Williams available in your library
(5) Newspaper clippings

Based on the information you have collected from the above sources, complete the project in about 5 pages. You can do value addition to your project with the help of suitable photos, clippings, charts, graphs and sketches.

Project 2.
Assume that you are interviewing Sunita Williams. Prepare a questionnaire and also the answers.
Answer:
Points to make a list of a questionnaire for the interview of Sunita Williams :
(1) Primary and higher education
(2) The source of inspiration to become an astronaut
(3) Information about her mentor
(4) General and specific training
(5) Initial experience of being an astronaut
(6) First space mission, its nature, duration and experience
(7) Natureofresearchcarriedoutinspace
(8) Some special memories
(9) Future plans
(10) Tips and guidance for the younger generation.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Can you recall? (Text Book Page No. 135)

Question 1.
What is the difference between space and sky?
Answer:

  1. The visible portion of the atmosphere and outer space seen by simple eyes, without any equipment from the earth, is known as the sky.
  2. The infinite three-dimensional expanse in which the Solar system, stars, celestial bodies, galaxies and the endless Universe exist is known as space.
  3. Both sky and space lack a definite boundary. However, the sky is a very tiny part of space.

Question 2.
What are different objects in the Solar system?
Answer:

  1. Our Solar system is a very tiny part of a huge Galaxy-Milky Way.
  2. The Sun is at the centre of the Solar system. Sun is a star.
  3. Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune are planets in our Solar system. These planets revolve around the Sun. Some of these planets have their own natural satellites.
  4. Besides, there are asteroids, meteoroids, comets and meteors in the Solar system.

Question 3.
What is meant by a satellite?
Answer:

  1. An astronomical object orbiting any planet of our Solar system is called a satellite.
  2. Mercury and Venus have no satellites.
  3. Some planets have more than one satellite. E.g. Jupiter has 69 satellites.

Question 4.
How many natural satellites does the earth have?
Answer:
The earth has one natural satellite called the moon.

Question 5.
Which type of telescopes are orbiting around the earth? Why is it necessary to put them in space?
Answer:
(1) The following three types of telescopes are orbiting around the earth:

  • Optical Refracting Telescope.
  • Optical Reflecting Telescope.
  • Radio Telescope.

(2) Visible light and radio waves emitted by celestial bodies in space pass through the atmosphere before reaching the earth’s surface. During this journey, some light is absorbed by the atmosphere. Hence, the intensity of the light reaching the earth’s surface decreases. Besides, temperature and air pressure cause the atmospheric turbulence. Hence, light rays change their path, resulting in a change in the position of the image of a celestial body.

City lights during night, and bright sunlight during day also put limitations on usage of optical telescopes on the earth. To minimize these problems, optical telescopes are situated on mountain top, away from inhabited places. However, limitations caused by the atmosphere still persist.

To get rid of these problems scientists have successfully launched telescopes in space. Images obtained by these telescopes are brighter and clearer than those obtained by the telescopes located on the earth’s surface.

Can you recall? (Text Book Page No. 135)

Question 1.
Where does the signal in your cellphone come from?
Answer:
In nearby area of our residence, many mobile towers are installed at various places. Cellphones receive signals from one of these mobile towers.

Question 2.
Where from does it come to mobile towers?
Answer:
All mobile towers are connected to satellites. Cellphone signal reaching the nearest mobile tower in our vicinity is first transmitted to the satellite. The satellite transmits the signal to the mobile tower near the destination.

Question 3.
Where does the signal to your TV set come from?
Answer:
(1) Television Centre or Studio transmits the TV program which first reaches the satellite. The dish antenna of the cable operator in our area receives these signals. The TV programs reach our TV set through a cable connected between the cable operator’s receiving station and our TV set.

(2) Alternatively, a small portable dish antenna fixed on the rooftop is also used to receive the TV signals directly from the satellites. Finally, a cable connected to the dish antenna and TV set brings the programme to our TV set.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 4.
You may have seen photographs showing the position of monsoon clouds over the country in the newspaper. How are these images obtained?
Answer:
Weather satellites take photographs of the sky above the earth’s surface at regular intervals. Some satellites, capable of receiving radio signals, also collect the information of weather conditions and finally images of the sky are built with computers. Territorial boundaries of the states and the country are drawn later on these images. Such satellite images with imposed boundaries are printed in media or shown on the television.

Fill in the blanks:

Question 1.
A man-made object revolving around the earth in a fixed orbit is called …………..
Answer:
A man-made object revolving around the earth in a fixed orbit is called an artificial satellite.

Question 2.
Chandrayaan-I discovered the presence of ………….. on the moon.
Answer:
Chandrayaan-I discovered the presence of water on the moon.

Question 3.
Apart from launching a satellite around the earth, India has been able to launch a satellite around ……………
Answer:
Apart from launching a satellite around the earth, India has been able to launch a satellite around Mars.

Question 4.
All satellites work on …………… energy.
Answer:
All satellites work on solar energy.

Question 5.
……………. are used to carry and place a satellite in a specific orbit.
Answer:
Satellite launchers are used to carry and place a satellite in a specific orbit.

Question 6.
USA has developed ……………. as an alternative to space launch vehicles.
Answer:
USA has developed space shuttles as an alternative to space launch vehicles.

Question 7.
Hubble telescope is a ………….. satellite.
Answer:
Hubble telescope is a Low Earth Orbit (LEO) satellite.

Question 8.
……………. executed the first ever mission to the moon in the world.
Answer:
Russia executed the first ever mission to the moon in the world.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 9.
………… executed the first manned mission to the moon in the world.
Answer:
USA executed the first manned mission to the moon in the world.

Select the appropriate answer from given options:

Question 1.
Which one of the following is a Low Earth Orbit (LEO) satellite?
(a) Navigational satellite
(b) Geostationary satellite
(c) International Space Station
(d) All of the above
Answer:
(c) International Space Station

Question 2.
Which of the following satellite launchers is developed by India?
(a) INSAT
(b) IRNSS
(c) EDUSAT
(d) PSLV
Answer:
(d) PSLV

Question 3.
Which of the following astronauts travelled through space shuttle ‘Discovery’ first time? (Practice Activity Sheet – 4)
(a) Kalpana Chawla
(b) Rakesh Sharma
(c) Sunita Williams
(d) Neil Armstrong
Answer:
(c) Sunita Williams

Considering the correlation between the words of the first pair, pair the third word accordingly with proper answer. (OR) Considering the first correlation, complete the second.

Question 1.
IRNSS : Direction showing satellite :: INSAT :………… (Practice Activity Sheet – 1)
Answer:
IRNSS : Direction showing satellite :: INSAT : Weather satellite

Question 2.
Hubble telescope : 569 km high from the earth’s surface :: Revolving orbit of Hubble telescope :………. (Practice Activity Sheet – 2; March 2019)
Answer:
Hubble telescope : 569 km high from the earth’s surface :: Revolving orbit of Hubble telescope : Low Earth Orbit.

Match the column:

Question 1.

Column A Column B
(1) Clouds over India (a) Low Earth Orbit
(2) Global communication (b) PSLV
(3) Launch vehicle made by ISRO (c) Communication satellite
(4) International Space Station (d) EDUSAT
(5) Navigational satellite (e) Weather satellite
(f) Medium Earth Orbit

Answer:
(1) Clouds over India – Weather satellite
(2) Global communication – Communication satellite
(3) Launch vehicle made by ISRO – PSLV
(4) International Space Station – Low Earth Orbit
(5) Navigational satellite – Medium Earth Orbit.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Answer the following questions in one sentence each:

Question 1.
What do you mean by the orbit of a satellite?
Answer:
Orbit of a satellite is its path around the earth.

Question 2.
Which factor decides the orbit of a satellite?
Answer:
The function of a satellite decides the orbit of the satellite.

Question 3.
What is a High Earth Orbit satellite?
Answer:
A satellite orbiting at a height equal to or greater than 35780 km above the earth’s surface is called a High Earth Orbit satellite.

Question 4.
Give two examples of Low Earth Orbit satellites.
Answer:
Weather satellite and International Space Station are Low Earth Orbit satellites.

Question 5.
What is a launch vehicle?
Answer:
A rocket used to carry an artificial satellite to a desired height above the earth’s surface and then project it with a proper velocity so that the satellite orbits the earth in the desired orbit is called a launch vehicle.

Question 6.
Name the launch vehicle developed by India.
Answer:
The launch vehicle developed by India is known as PSLV, i.e., Polar Satellite Launch Vehicle.

Answer the following questions:

Question 1.
Write the proper name of the orbits of satellites shown in the following figure with their height from the earth’s surface. (Practice Activity Sheet – 4)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 13
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 14
(a) Low earth orbits: height above the earth’s surface: 180 km to 2000 km
(b) Medium earth orbits: height above the earth’s surface: 2000 km to 35780 km
(c) High earth orbits: height from the earth’s surface > 35780 km

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 2.
Explain the need and importance of space missions.
Answer:
Man has always been curious about the sun, moon, stars and the world beyond the earth. Initially, man tried to observe space with the help of telescopes. However, later he dreamt to fly into space and finally succeeded to reach into space.

Space missions are now essential to understand the origin and evolution of our solar system as well as to study the Universe beyond the Solar system.

Space missions have given us many benefits and made our life simpler. It is because of space missions that the real-time immediate communication and exchange of information across the globe is now possible. We can receive the abundant information at the desk at our home or office. We also get information about any topic at any time and anywhere at fingertips through the Internet. Besides, the advanced alerts about some natural calamities like cyclones or storms are received through satellites sent as a part of space missions. Satellites have also helped us in entertainment. Programmes, sports events, etc., can be telecast live and can reach millions at a time throughout the world.

Satellite surveillance of the enemy, exploring the reserves of various minerals resources, access to various activities like trade, tourism and navigation, and easy global reach to make world a global village is all possible due to the space missions. Thus, space missions are extremely important in defence, communication, weather forecast, observation, direction determination, etc.

Question 3.
What are space expeditions? Explain their need and importance in your words. (Practice Activity Sheet – 2)
Answer:
A mission planned (i) for establishing artificial satellites in the earth’s orbit, using them for research or for the benefit of life, or (ii) for sending a spacecraft to the various components of the solar system or outside is called a space expedition.

Man has always been curious about the sun, moon, stars and the world beyond the earth. Initially, man tried to observe space with the help of telescopes. However, later he dreamt to fly into space and finally succeeded to reach into space. Space missions are now essential to understand the origin and evolution of our solar system as well as to study the Universe beyond the Solar system.

Space missions have given us many benefits and made our life simpler. It is because of space missions that the real-time immediate communication and exchange of information across the globe is now possible. We can receive the abundant information at the desk at our home or office. We also get information about any topic at any time and anywhere at fingertips through the Internet. Besides, the advanced alerts about some natural calamities like cyclones or storms are received through satellites sent as a part of space missions. Satellites have also helped us in entertainment. Programmes, sports events, etc., can be telecast live and can reach millions at a time throughout the world.

Satellite surveillance of the enemy, exploring the reserves of various minerals resources, access to various activities like trade, tourism and navigation, and easy global reach to make world a global village is all possible due to the space missions. Thus, space missions are extremely important in defence, communication, weather forecast, observation, direction determination, etc.

Question 4.
What are the objectives of the space mission?
Answer:
Man initially tried to satisfy his curiosity to know the world and universe beyond the earth with the help of telescopes. However, it has some obvious limitations and to overcome these limitations, man later ventured into space missions.

Space missions carried out by man were aimed at four specific objectives:

  1. To launch artificial satellites in the earth’s orbit for study and research.
  2. To launch artificial satellites in the earth’s orbit for various purposes like telecommunication, weather forecast, radio and TV programme transmission, etc.
  3. To send artificial satellites beyond the earth’s orbit to observe, study and collect the information from other planets, meteors, meteoroids, asteroids and comets.
  4. To sense and understand space beyond the solar system.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 5.
Write on significant space missions carried out by man.
Answer:
Man has carried out many space missions within and beyond the earth’s orbit. Significant space missions are as follows:
(1) Space missions within the earth’s orbit: Man has so far sent many artificial satellites of various types in the earth’s orbit. These satellites have made the life of man simpler. Besides, it has also helped us in resource management, communication, disaster management, etc.

(2) Moon missions : Moon is the natural satellite , of the earth and it is the nearest celestial body to us. Naturally, our initial space missions were directed to the moon. As of now, only Russia, USA, European Union, China, Japan and India have successfully undertaken . moon missions. Russia executed 15 moon missions between 1959 and 1976. Of these, last 4 missions brought the stone samples from the moon for study and analysis. However all these missions were unmanned. USA executed moon missions between 1962 and 1972. Some of these missions were unmanned.

However, the historic moon mission took place on 20th July, 1969, when American astronaut Neil Armstrong became the first human to step on the moon. India has undertaken the moon mission. Indian Space Research Organisation (ISRO) successfully launched Chandrayaan-I and placed it in orbit of the moon. It sent useful information to the earth for about a year. The most important discovery made during the mission was the presence of water on moon’s surface. India was the first country to discover this.

(3) Mars mission: The second nearest celestial object to the earth is Mars and many nations sent spacecraft towards it. But only few of them have been successful. However, the performance of Mangalyaan, the Indian spacecraft sent by ISRO towards Mars, was remarkable. Mangalyaan was launched in November 2013 and was placed in the orbit of Mars successfully in September 2014. It has obtained useful information about the surface and atmosphere of Mars.

(4) Space missions to other planets: Other than moon and Mars missions, many other space missions were undertaken for studying other planets. Some spacecraft orbited the planets, some landed on some planets, and some just observed the planets, passed near them and went further to study other celestial bodies. Some spacecraft were sent specifically to study asteroids and comets. Some spacecraft’s have brought dust and stone samples from asteroids for the study.

All these space missions are very useful in getting information and helping us in clarifying our concepts about the origin of the earth and the Solar system.

Question 6.
Bring out the contribution of India’s space missions.
Answer:
Successful space missions as well as scientific and technological accomplishments by India in space technology have made a significant contribution in the national and social development of our country.
India has indigenously built various launchers and these launchers can put the satellites having the mass up to 2500 kg in orbit.

Indian Space Research Organisation (ISRO) has designed and built two important launchers: Polar Satellite Launch Vehicle (PSLV) and Geosynchronous Satellite Launch Vehicle (GSLV).

Many satellites in INS AT and GSAT series are active in telecommunication, television broadcasting, meteorological services, disaster management and in monitoring and management of natural resources. EDUSAT is used specifically for education while satellites in IRNSS series are used for navigation. Thumba, Sriharikota and Chandipur are Indian satellite launch centers.

Vikram Sarabhai Space Centre at Thiruvananthapuram, Satish Dhawan Space Research Centre at Sriharikota and Space Application Centre at Ahmedabad are space research organizations of India.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 7.
What is meant by space debris? Why is there need to manage the debris? (March 2019)
Answer:
In a space nonessential objects such as the parts of launchers and satellites, revolving around the earth are called the debris in space.

The debris can be harmful to the artificial satellites. It can collide with the satellite or spacecrafts and damage them. Therefore the future of artificial satellites or spacecrafts are in danger.
Hence, it is necessary to manage the debris.

Solve the following examples/numerical problems:
[Note: See the textbook for the relevant data.]

Problem 1.
If the mass of a planet is 8 times that of the earth and its radius is twice the radius of the earth, what will be the escape velocity for that planet? (Escape velocity for the earth = 11.2 km/s) (Practice Activity Sheet – 2)
Answer:
Given:
Mass of the planet = 8ME radius of the planet, Rp = 2RE,
escape velocity for the earth, υescE = 11.2 km/s
escape velocity for the planet, υescP = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 15

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Problem 2.
Calculate the critical velocity (υc) of the satellite to be located at 35780 km above the surface of the earth.
Answer:
Given:
G: 6.67 × 10-11 N·m2/kg2,
M(Earth): 6 × 1024 kg,
R(Earth): 6.4 × 106 m,
h: 35780 km = 35780 × 103 m,
υc = ?
Critical velocity of the satellite
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 16
= 3.08 × 103 m/s
= 3.08 Km/s.

Problem 3.
In the above example (2) how much time will the satellite take to complete one revolution around the earth?
Answer:
Given:
R: 6400mkm = 6.4 × 106 m
h: 35780 km = 3.5780 × 107 m
v: 3.08 km/s = 3.08 × 103 m/s
T = ?
The time required for the satellite to complete one revolution around the earth,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 17
= Approx 86060 s
= 23 hours 54 minutes 20 seconds

Problem 4.
Calculate the critical velocity (υc) of the satellite to be located at 2000 km above the surface of the earth.
Answer:
Refer to the example (2) above.
Here,h = 2 × 106 m
υc = 6902 m/s

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Problem 5.
In the above example (4), how much time will the satellite take to complete one revolution around the earth?
Answer:
Refer to example (3) above.
Approx 7647 s
= 2 hours 7 minutes 27 seconds.
[Note: For more solved problems and problems for practice, refer Chapter 1 (Gravitation)]

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9

Question 1.
A man buys a house for ₹ 10 lakh and rents it. He puts 10% of the annual rent aside for repairs, pays ₹ 1,000 as annual taxes, and realizes 8% on his investment thereafter. Find the annual rent of the house.
Solution:
Let ₹ ‘x’ be the annual rent of the house.
The man keeps 10% of the annual rent aside for repairs.
i.e., \(\frac{10}{100}\) × x or ₹ \(\frac{x}{10}\) aside tor repairs.
In addition, he pays ₹ 1000 as annual taxes.
After incurring these expenses he is left with an amount which is 8% of his investment for the house.
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q1
∴ The annual rent of the house is ₹ 90,000.

Question 2.
Rose got 30% of the maximum marks in an examination and failed by 10 marks. However, Lily who appeared for the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?
Solution:
Let maximum marks be x
Rose scored 30% of maximum marks
i.e. Rose scored \(\frac{30}{100}\)x
Rose failed by 10 marks
∴ passing marks = \(\frac{30}{100}\)x + 10 …..(i)
Lily scored 40% of maximum marks
i.e. Lily scored \(\frac{40}{100}\)x
Lily scored 15 marks more than passing marks
∴ passing marks = \(\frac{40}{100}\)x – 15 ……(ii)
equating (i) and (ii),
\(\frac{30x}{100}\) + 10 = \(\frac{40x}{100}\) – 15
∴ 10 + 15 = \(\frac{40 x-30 x}{100}\)
∴ 10x = (25)(100)
∴ x = 250
From (i), passing marks = \(\frac{30}{100}\)(250) + 10
= 75 + 10
= 85
∴ Passing marks for the examination were 85.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9

Question 3.
Ankita’s Salary was reduced by 50%. Again the reduced salary was increased by 50%. Find loss in terms of percentage.
Solution:
Let Ankita’s initial salary be ₹ ‘x’.
Her salary was reduced by 50%.
∴ Ankita’s salary after reduction = x(1 – \(\frac{50}{100}\))
= x(1 – \(\frac{1}{2}\))
= \(\frac{x}{2}\)
Ankita’s reduced salary was then increased by 50%
∴ Ankita’s final salary after the increase
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q3
∴ Loss in Ankita’s salary after the decrease and increase = x – \(\frac{3 x}{4}\) = \(\frac{x}{4}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q3.1
∴ Ankita lost 25% of her salary.

Question 4.
By selling 300 lunch boxes, a shopkeeper gains the selling price of 100 lunch boxes. Find his gain percent.
Solution:
Let ₹ x be the selling price (S.P.) of one lunch box.
∴ S.P. of 300 lunch boxes = 300x
and S.P. of 100 lunch boxes = 100x
Gain = 100x ……[given]
C.P. of 300 lunch boxes = S.P. – Gain
= 300x – 100x
= 200x
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q4
∴ The shopkeeper’s gain percentage is 50%.

Question 5.
A salesman sold an article at a loss of 10%. If the selling price has been increased by ₹ 80, there would have been a gain of 10%. What was the cost of the article?
Solution:
Let ₹ x be the cost price of the article.
S.P. of the article = x – \(\frac{10}{100}\)x = \(\frac{9x}{100}\) …….(i)
Given that, S.P. increased by ₹ 80 would have given 10% gain
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q5
∴ The cost price of the article is ₹ 400

Question 6.
Find the single discount equivalent to a series discount of 10%, 20%, and 15%.
Solution:
Let the marked price be ₹ 100
After 1st discount the price = 100(1 – \(\frac{10}{100}\)) = 90
After 2nd discount the price = 90(1 – \(\frac{20}{100}\)) = 72
After 3rd discount the price = 72(1 – \(\frac{15}{100}\)) = 61.2
∴ The selling price after 3 discounts is ₹ 61.2.
∴ Single equivalent discount = marked price – selling price
= 100 – 61.2
= ₹ 38.8
∴ The single equivalent discount is ₹ 38.8 on ₹ 100.
i.e. The single equivalent discount is 38.8%.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9

Question 7.
Reshma put an amount at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, she would have received ₹ 360 more. Find the sum.
Solution:
Let P and R represent the principal amount and rate of interest p.a. respectively.
Given duration = T = 3 years
Simple interest = \(\frac{\mathrm{PRT}}{100}=\frac{3 \mathrm{PR}}{100}\)
Given that, had the amount been kept at 2% more, then the gain would have been ₹ 360 more.
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q7
∴ The sum of money is ₹ 6,000.

Question 8.
The compound interest on ₹ 30000 at 7% p.a. is ₹ 4347. What is the period in years?
Solution:
Given that,
Principal (P) = ₹ 30,000
Rate of interest (R) = 7% p.a.
Compound interest = ₹ 4,347
Amount after compound interest
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q8
= \(\left(\frac{107}{100}\right)^{2}\)
= (1.07)2
∴ T = 2
∴ Amount is invested for 2 years.

Question 9.
The value of the machine depreciates at the rate of 15% p.a. It was purchased 2 years ago. Its present value is ₹ 7,225. What was the purchase price of the machine?
Solution:
Given,
Rate of depreciation = r = 15%
Number of years = n = 2 years
Present value of machine = P.V. = ₹ 7,225
The purchase price (V) of the machine can be found using
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q9
∴ The purchase price of the machine was ₹ 10,000/-.

Question 10.
A tree increases annually by \(\frac{1}{8}\) of its height. By how much will it increase after 2\(\frac{1}{2}\) years. If its length today is 8 m?
Solution:
The height of the tree today is 8m.
The height of the tree increases by \(\frac{1}{8}\)th of its height every year.
At the end of 1st year, height of the tree will be = 8 + \(\frac{1}{8}\) × 8 = 9 m
And, at the end of the 2nd year, height of the tree will be = 9 + \(\frac{1}{8}\) × 9
= 9(1 + \(\frac{1}{8}\))
= 9 × \(\frac{9}{8}\)
= \(\frac{81}{8}\)
After six more months, the height of the tree will be
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q10
∴ Increase in the height of the tree after 2\(\frac{1}{2}\) years = 10.75 – 8 = 2.75 m.

Question 11.
A building worth ₹ 1,21,000 is constructed on land worth ₹ 81,000. After how many years will the value of both be the same if land appreciates at 10% p.a and buildings depreciate at 10% p.a.
Solution:
Given,
Value of the building = V.B. = ₹ 1,21,000
Value of land = V.L. = ₹ 81,000/-
Rate of appreciation of land = rate of depreciation of building = r = 10%.
For the value of building and land to be the same.
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q11
∴ n = 2 years.
∴ After two years value of the building and land will be the same.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9

Question 12.
Varun invested 25%, 30%, and 20% of his savings in buying shares of three different companies, ‘A’, ‘B’, and ‘C’ which declared dividends, 10%, 12%, and 15% respectively. If his total income on account of dividends is ₹ 6,370/-, find the amount he invested in buying shares of company ‘B’.
Solution:
Let ‘T’ be Varan’s total savings.
∴ Investment of Varan in:
Company A = 25% of T = \(\frac{25}{100}\) × T = \(\frac{T}{4}\),
Company B = 30% of T = \(\frac{30}{100}\) × T = \(\frac{3T}{10}\),
Company C = 20% of T = \(\frac{20}{100}\) × T = \(\frac{T}{5}\)
Company A, B and C declared dividends 10%, 12% and 15% respectively.
∴ Dividend from company A = 10% of \(\frac{T}{4}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q12
∴ Varan invested ₹ 21,000 in company B.

Question 13.
Find the annual dividend received from ₹ 25,000, 8% stock at ₹ 108.
Solution:
Amount invested = ₹ 25,000
Dividend = 8%
Assuming face value F.V. as ₹ 100
Annual income per share = \(\frac{\text { Dividend }}{100} \times \text { Face value }\)
= \(\frac{8}{100}\) × 100
= ₹ 8
Market value of the share M.V. = ₹ 108
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Q13
Annual dividend on amount invested = Rate of return × amount invested
= \(\frac{7.4}{100}\) × 25,000
= ₹ 1850
∴ Annual dividend of ₹ 1,850 is received from 8% stock at ₹ 108.
Alternate approach
Assuming ₹ 25,000 as the total face value of all the shares.
Since the dividend is 8%,
Annual dividend = \(\frac{8}{100}\) × 25,000 = ₹ 2,000

Question 14.
A, B, and C enter into a partnership. A invests 3 times as much as B invests and B invests two-thirds of what ‘C’ invests. At the end of the year, the profit earned is ₹ 8,800. What is the share of ‘B’?
Solution:
Let ‘a’, ‘b’ and ‘c’ be the amounts invested by A, B and C respectively.
Given that, A invests 3 times as much as B and B invests two third of what ‘C’ invests.
∴ a = 3b and b = \(\frac{2}{3}\)c
∴ \(\frac{a}{b}=\frac{3}{1}\) and \(\frac{b}{c}=\frac{2}{3}\)
or \(\frac{a}{b}=\frac{6}{2}\) and \(\frac{b}{c}=\frac{2}{3}\)
∴ a : b = 6 : 2 and b : c = 2 : 3
∴ a : b : c = 6 : 2 : 3
Given that profit earned = ₹ 8800
∴ Share of ‘B’ in profit = \(\frac{2}{11}\) × 8800 = ₹ 1600
∴ B’share in profit is ₹ 1600.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9

Question 15.
The ratio of investment of two partners Santa and Banta is 11 : 12 and the ratio of their profits is 2 : 3. If Santa invested the money for 8 months, then for how much time did Banta his money?
Solution:
Let ‘x’ be the time in months for which Banta invested his money
Santa and Banta invested their money in the ratio 11 : 12.
Santa invested his money for 8 months and the ratio of their profits is 2 : 3.
∴ 11 × 8 : 12 × x = 2 : 3
∴ \(\frac{88}{12 x}=\frac{2}{3}\)
∴ x = \(\frac{88 \times 3}{2 \times 12}\)
∴ x = 11
∴ Banta invested his money for 11 months.

Question 16.
Akash, Sameer, and Sid took a house on rent for one year for ₹ 16,236. They stayed together for 4 months and then Sid left the house. After 5 more months, Sameer also left the house. How much rent should each pay?
Solution:
Let ‘R’ be the rent per month to be paid to the landlord.
Given that, Sid left the house after 4 months
∴ Rent paid by Sid = \(\frac{R}{3}\) × 4 = \(\frac{4R}{3}\)
Sameer left the house after another 5 months,
∴ Rent paid by Sameer = \(\frac{R}{2}\) × 5 + \(\frac{R}{3}\) × 4
= R(\(\frac{5}{2}+\frac{4}{3}\))
= \(\frac{23R}{6}\)
Akash stayed in the house for the entire year.
∴ Rent paid by Akash = 3R + \(\frac{R}{2}\) × 5 + \(\frac{R}{3}\) × 4
= R(3 + \(\frac{5}{2}+\frac{4}{3}\))
= \(\frac{41R}{6}\)
∴ The rent paid by the three of them, over that period of one year must be in the proportion.
\(\frac{41 \mathrm{R}}{6}: \frac{23 \mathrm{R}}{6}: \frac{4 \mathrm{R}}{3}\)
i.e. in the proportion
41 : 23 : 8 …..(multiplying throughout by \(\frac{6}{R}\))
Let x be the constant of proportionality.
Rent to be paid by Akash = ₹ 41x
Rent to be paid by Sameer = ₹ 23x
and rent to be paid by Sid = ₹ 8x
The total rent for the house was ₹ 16236.
∴ 41x + 23x + 8x = ₹ 16236
∴ 72x = 16236
∴ x = 225.5
∴ Akash should pay 41x = 41 × 225.5 = ₹ 9245.5
Sameer should pay 23x = 23 × 225.5 = ₹ 5186.5
and Sid should pay 8x = 8 × 225.5 = ₹ 1804

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Miscellaneous Exercise 9

Question 17.
Ashwin Auto Automobiles sold 10 motorcycles. Total sales amount was ₹ 6,80,000. 18% GST is applicable. Calculate how much CGST and SGST the firm has to pay.
Solution:
Given, total sales amount for Ashwin Automobiles was ₹ 6,80,000.
18% GST is applicable.
∴ GST payable = 18% of 6,80,000
= \(\frac{18}{100}\) × 6,80,000
= ₹ 1,22,400
Now CGST = SGST = 9%
= \(\frac{\text { GST payable }}{2}\)
= \(\frac{1,22,400}{2}\)
= ₹ 61,200
∴ CGST = SGST = ₹ 61,200

Question 18.
‘Sweet 16’ A ready made garments shop for Women’s garments, purchased stock for ₹ 4,00,000 and sold that stock for ₹ 5,50,000 (12% GST is applicable) Find,
(i) Input Tax Credit
(ii) CGST and SGST paid by the firm.
Solution:
Given that, stock purchased by ‘Sweet 16’ was worth ₹ 4,00,000
GST applicable is 12%.
∴ Input tax = 12% of 4,00,000
= \(\frac{12}{100}\) × 4,00,000
= ₹ 48,000
∴ Input tax Credit (ITC) = ₹ 48,000
The garment stock was sold for ₹ 5,50,000
Output tax = 12% of 5,50,000
= \(\frac{12}{100}\) × 5,50,000
= ₹ 66,000
∴ GST payable = output tax – ITC
= 66,000 – 48,000
= ₹ 18,000
∴ CGST = SGST = \(\frac{\text { GST payable }}{2}\) = ₹ 9,000

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.7 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Question 1.
Shantanu has a choice to invest in ₹ 10 shares of two firms at ₹ 13 or at ₹ 16. If the first firm pays a 5% dividend and the second firm pays a 6% dividend per annum, find:
(i) Which firm is paying better?
(ii) If Shantanu invests equally in both the firms and the difference between the return from them is ₹ 30. Find how much, in all, does he invest.
Solution:
(i) For firm 1:
Face value of the share (F.V.) = ₹ 10
Market value of the share (M.V.) = ₹ 13
Dividend = 5%
∴ Annual income from the share = \(\frac{5}{100}\) × 10 = ₹ 0.5
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q1
For firm 2:
Face value of the share (F.V.) = ₹ 10
Market value of the share (M.V.) = ₹ 16
Dividend = 5%
∴ Annual income from the share = \(\frac{6}{100}\) × 10 = ₹ 0.6
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q1.1
Since, the profit percentage from firm 1 > profit percentage from firm 2, the first firm is paying better.

(ii) Let ‘X’ be the amount Shantanu invests in each of the firms.
Given that difference between the return from them is ₹ 30, we have
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q1.2
In all, Shantanu invests = 2X
= 2 × 31,200
= ₹ 62,400/-

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Question 2.
A dividend of 9% was declared on ₹ 100 shares selling at a certain price in the stock market. If the rate of return is 7.5% calculate
(i) The market price of each share, and
(ii) The amount to be invested to obtain an annual dividend of ₹ 630.
Solution:
(i) Given that,
Face value of the share (F.V) = ₹ 100
Dividend = 9%
Rate of return = 7.5%
Annual income from the share = \(\frac{9}{100}\) × 100 = ₹ 9
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q2
∴ The market price of the share is ₹ 120.

(ii) Let ‘X’ be the amount to be invested to obtain an annual dividend of ₹ 630.
∴ 7.5% of X is ₹ 630
∴ \(\frac{7.5}{100}\) × X = 630
∴ X = \(\frac{630 \times 100}{7.5}\)
∴ X = 8400
∴ ₹ 8400 need to be invested to obtain an annual dividend of ₹ 630.

Question 3.
Nilesh has the option of investing his money in 8% ₹ 10 shares at a premium of ₹ 3.50 or 7% ₹ 100 shares at a premium of 20%. Which of the two investments will be more profitable for him?
Solution:
For share 1:
Face value of the share (F.V.) = ₹ 10
Premium = ₹ 3.5
∴ Market value of the share (M.V.) = 10 + 3.5 = ₹ 13.5
Dividend = 8 %
∴ Annual income from the share = \(\frac{8}{100}\) × 10 = ₹ 0.8
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q3
= \(\frac{800}{135}\)
= 5.93%

For share 2:
Face value of the share (F.V.) = ₹ 100
Premium = 20%
∴ Market value of the share (M.V.) = 100 + (\(\frac{20}{100}\) × 100) = ₹ 120
Dividend = 7%
Annual income from the share = \(\frac{7}{100}\) × 100 = ₹ 7
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q3.1
Since, profit percentage from share 1 > profit percentage from share 2, investing in the first kind of shares will be more profitable for Nilesh.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Question 4.
Sudhakar invests ₹ 1344 in buying shares of face value ₹ 24 selling at a 12% premium. The dividend on the shares is 15% per annum. Calculate
(i) The number of shares Sudhakar buys, and
(ii) The dividend he receives annually.
Solution:
Given that,
Face value of the share (F.V.) = ₹ 24
Premium = 12%
∴ Market value of the share (M.V.) = 24 + (\(\frac{12}{100}\) × 24) = ₹ 26.88
(i) Sudhakar invests ₹ 1344 in the shares
∴ Number of shares purchased by Sudhakar = \(\frac{1344}{26.88}\) = 50
∴ Sudhakar buys 50 shares.
(ii) Dividend on the share = 15%
Annual income on one share = \(\frac{15}{100}\) × 24 = ₹ 3.6
∴ The total dividend he receives annually = 50 × 3.6 = ₹ 180
∴ Sudhakar receives ₹ 180 as his annual dividend.

Question 5.
Sameer invests ₹ 5625 in a company paying 7% per annum when the share of ₹ 10 stands for ₹ 12.50. Find Sameer’s income from this investment. If he sells 60% of these shares of ₹ 10 each, find his gain or loss in this transaction.
Solution:
Given:
Face value of the share (F.V.) = ₹ 10
Market value of the share (M.V.) = ₹ 12.5
Amount invested in shares = ₹ 5625
∴ Number of shares purchased by Sameer = \(\frac{5625}{12.5}\) = 450
Dividend = 7%
Annual income from one share = \(\frac{7}{100}\) × 10 = ₹ 0.7
∴ Sameer’s income from this investment = number of shares × annual income from one share
= 450 × 0.7
= ₹ 315
Sameer sells 60 % of these shares = \(\frac{60}{100}\) × 450 = 270 shares
Sameer purchased these shares at ₹ 12.5 per share.
∴ Purchase price for these shares = 270 × 12.5 = ₹ 3375
If he sells these shares at ₹ 10 per share, he would receive 270 × 10 = ₹ 2700
∴ In this transaction, Sameer would incur a loss of 3375 – 2700 = ₹ 675

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Question 6.
Geeta buys ₹ 100 shares of a company that pays a 15% dividend. She buys the shares at a price from the market that gives her a 10% return on her investment. At what price did she buy each share?
Solution:
Given that,
Face value of the share (F.V.) = ₹ 100
Dividend = 15%
∴ Annual income from the share = \(\frac{15}{100}\) × 100 = ₹ 15
Rate of return on investment = 10%
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q6
∴ Geeta bought each share from the market at ₹ 150.

Question 7.
Tejas invests in 9% ₹ 100 shares at ₹ 145 but Shail invests in 7% ₹ 100 shares at ₹ 116. Whose investment is better?
Solution:
Investment of Tejas:
Given that, the Face value of the share (F.V.) = ₹ 100
The market value of the share (M.V.) = ₹ 145
Dividend = 9%
Annual income from the share = \(\frac{9}{100}\) × 100 = ₹ 9
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q7

Investment of Shail:
Face value of the share (F.V.) = ₹ 100
Market value of the share (M.V.) = ₹ 116
Dividend = 7%
Annual income from the share = \(\frac{7}{100}\) × 100 = ₹ 7
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q7.1
Since the rate of return for Tejas’s investment is greater than that for Shail’s, Tejas’s investment is better.

Question 8.
A 6% share yields 8%. Find the market value of a ₹ 100 share.
Solution:
Given that,
Face value of the share = ₹ 100
Dividend = 6%
Yield = 8%
Annual income on the share = \(\frac{6}{100}\) × 100 = ₹ 6
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q8
∴ The market value of the share = ₹ 75

Question 9.
Ashwini bought ₹ 40 shares at a premium of 40%. Find the income, if Ashwini invests ₹ 14,000 in these shares and receives a dividend at the rate of 8% on the nominal value of the shares.
Solution:
Given,
Face value of the shares (F.V.) = ₹ 40
Premium = 40%
Market value of the shares (M.V.) = 40 + (40 × \(\frac{40}{100}\))
= 40 + 16
= ₹ 56
Ashwini invests ₹ 14000 in these shares
∴ Number of shares bought by Ashwini = \(\frac{Amount Invested}{Market value of one share}\)
= \(\frac{14000}{56}\)
= 250
Dividend = 8%
∴ Annual income on one share = \(\frac{8}{100}\) × 40 = ₹ 3.2
∴ Income of Ashwini on 250 shares = 250 × 3.2 = ₹ 800
∴ Ashwini earns ₹ 800 on her investment.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Question 10.
Mr. Rutvik invests ₹ 30,000 in buying shares of a company that pays a 12% dividend annually on ₹ 100 shares selling at a premium of ₹ 50. Find
(i) The number of shares bought Mr. Rutvik and
(ii) His annual income from the shares.
Solution:
Given that,
Face value of a share (F.V.) = ₹ 100
Premium = ₹ 50
∴ Market value of a share (M.V.) = 100 + 50 = ₹ 150
Dividend =12%
Mr. Rutvik invests ₹ 30,000 in the shares.
(i) Number of shares bought by Mr. Rutvik = \(\frac{Amount invested}{Market value}\)
= \(\frac{30000}{150}\)
= 200

(ii) Dividend on the share = 12%
∴ Annual income from one share = \(\frac{12}{100}\) × 100 = ₹ 12
∴ His annual income from shares = number of shares × income from one share
= 200 × 12
= ₹ 2400

Question 11.
Rasika bought ₹ 40 shares at a discount of 40%. Find the income, if she invests ₹ 12,000 in these shares and receives a dividend at the rate of 11% on the nominal value of the shares.
Solution:
Given,
Face value of the shares (F.V.) = ₹ 40
Discount = 40%
∴ Market value of the shares (M.V.) = 40 – (40 × \(\frac{40}{100}\))
= 40 – 16
= ₹ 24
Rasika invests ₹ 12,000 in these shares.
∴ Number of shares bought by Rasika = \(\frac{Amount invested}{Market value of one share}\)
= \(\frac{12000}{24}\)
= 500
Dividend = 11%
∴ Annual income on one share = \(\frac{11}{100}\) × 40 = ₹ 4.4
∴ Rasika’s income on 200 such shares = 500 × 4.4 = ₹ 2200
∴ Rasika earns ₹ 2200 from her investment.

Question 12.
Nisha invests ₹ 15,840 in buying shares of nominal value ₹ 24 selling at a premium of 10%. The company pays a 15% dividend annually. Find
(i) The dividend she receives annually, and
(ii) The rate of return from her investment.
Solution:
Given that,
Face value of the share (F.V.) = ₹ 24
Premium = 10%
∴ Market value of the share (M.V.) = 24 + (24 × \(\frac{10}{100}\))
= 24 + 2.4
= ₹ 26.4
Dividend = 15%
∴ Annual income on the share = \(\frac{15}{100}\) × 24 = ₹ 3.6
Nisha invests ₹ 15,840 in these shares.
∴ Number of shares bought by Nisha
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q12
(i) Annual dividend received by Nisha = Number of shares × annual income from one share
= 600 × 3.6
= ₹ 2160

(ii) Rate of return from the investment
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q12.1

Question 13.
Ashutosh buys 80, ₹ 100 shares at a discount of 20% and receives a return of 12% on his money. Calculate
(i) The amount invested by Ashutosh.
(ii) The rate of dividend paid by the company.
Solution:
Given
Face value of the shares (F.V.) = ₹ 100
Discount = 20%
∴ Market value of the shares (M.V.) = 100 – (100 × \(\frac{20}{100}\)) = ₹ 80
(i) Amount invested by Ashutosh = number of shares × market value of the shares
= 80 × 80
= ₹ 6400

(ii) Ashutosh receives a return of 12% on his money.
∴ Ashutosh’s income from shares = \(\frac{12}{100}\) × 6400 = ₹ 768
∴ Ashutosh’s annual income from one share = \(\frac{768}{80}\) = ₹ 9.6
Annual income from one share = \(\frac{\text { Dividend }}{100} \times \text { Face value }\)
∴ 9 6 = \(\frac{\text { Dividend }}{100} \times 100\)
∴ Rate of dividend = 9.6%

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Question 14.
Vaishnavi bought 1000, ₹ 100 shares from the stock market carrying 8% dividend quoted at ₹ 130. A few days later the market value of the shares went up by 10%. Vaishnavi sold all her shares. What was her total income from this transaction?
Solution:
Given that,
Face value of the shares (F.V.) = ₹ 100
The market value of the shares (M.V.) = ₹ 130
Dividend = 8%
Income from the each share = \(\frac{8}{100}\) × 100 = ₹ 8
Number of shares bought by Vaishnavi = 1000
∴ Vaishnavi’s income from dividend = 1000 × 8 = ₹ 8000
The price of the shares went up by 10%
New market value of the shares = 130 + (130 × \(\frac{10}{100}\)) = ₹ 143
Vaishnavi sold the shares at ₹ 143 which she bought at ₹ 130 each.
∴ Vaishnavi’s profit on one share =143 – 130 = ₹ 13
∴ Vaishnavi’s profit after selling all her shares =1000 × 13 = ₹ 13,000
Vaishnavi’s total income from this transaction = Income from dividend + income from sale of shares
= 8,000 + 13,000
= ₹ 21,000
∴ Vaishnavi’s total income from this transaction was ₹ 21,000.

Question 15.
Mr. Dinesh invests ₹ 20,800 in 6% ₹ 100 shares at ₹ 104, and ₹ 14,300 in 10.5% ₹ 100 shares at ₹ 143. What will be his annual income from the shares?
Solution:
For 1st kind of shares,
Face value of shares (F.V.) = ₹ 100
Dividend = 6%
∴ Annual income from one share = \(\frac{6}{100}\) × 100 = ₹ 6
Market value of the share (M.V.) = ₹ 104
Total amount invested = ₹ 20,800
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q15
∴ Total income from 1st kind of shares = 200 × 6 = ₹ 1200
For 2nd kind of shares,
Face value of shares (F.V.) = ₹ 100
Dividend = 10.5%
∴ Annual income from one share = \(\frac{10.5}{100}\) × 100 = ₹ 10.5
Market value of the share (M.V.) = ₹ 143
Total amount invested = ₹ 14300
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q15.1
∴ Total income from 2nd kind of shares = 100 × 10.5 = ₹ 1050
∴ Total annual income of Dinesh from both these shares = 1200 + 1050 = ₹ 2250

Question 16.
A company declares a semi-annual dividend of 5%. Daniel has 400 shares of the company. If Daniel’s annual income from the shares is ₹ 1,000, find the face value of each share.
Solution:
Given that,
Semi-annual dividend = 5%
∴ Annual dividend = 10%
Number of shares with Daniel = 400
Daniel’s annual income from the shares = ₹ 1000
∴ Annual income from one share = \(\frac{1000}{400}\) = ₹ 2.5
But annual income from one share = \(\frac{\text { Annualdividend }}{100} \times \text { Face value }\)
∴ 2.5 = \(\frac{10}{100}\) × Face value of the share
∴ Face value of the share = ₹ 25

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Question 17.
Bhargav buys 400, ₹ 20 shares at a premium of ₹ 4 each and receives a dividend of 12%. Find
(i) The amount invested by Bhargav.
(ii) His total income from the shares.
(iii) Percentage return on his money.
Solution:
Given that,
Face value of the shares (F.V.) = ₹ 20
Premium = ₹ 4
∴ Market value of the shares (M.V.) = ₹ 24
Dividend = 12%
∴ Annual income from the share = \(\frac{12}{100}\) × 20 = ₹ 2.4
Bhargav buys 400 shares
(i) The amount invested by Bhargav = number of shares × market value
= 400 × 24
= ₹ 9600

(ii) Bhargav’s income from the shares = number of shares × annual income from one share
= 400 × 2.4
= ₹ 960

(iii) Percentage return on Bhargav’s money
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q17
∴ Bhargav gets 10% as the rate of return on his money.

Question 18.
Anil buys 350 ₹ 100 shares of a company at a premium of 20% from the market. The company pays 12% dividend annually. Find
(i) the investment made by Anil,
(ii) his annual income from the shares, and
(iii) the rate of return from the shares.
Solution:
Given that,
Face value of shares (F.V.) = ₹ 100
Premium = 20%
∴ Market value of shares (M.V.) = 100 + (\(\frac{20}{100}\) × 100) = ₹ 120
Dividend = 12%
∴ Annual income from one share = \(\frac{12}{100}\) × 100 = ₹ 12
Anil buys 350 shares.
(i) Amount invested by Anil = number of shares × market value
= 350 × 120
= ₹ 42,000

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

(ii) Anil’s annual income from the shares = number of shares × annual income from one share
= 350 × 12
= ₹ 4200

(iii) Rate of return from shares
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7 Q18
∴ The rate of return from Anil’s shares is 10%.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.6 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6

Question 1.
M/s Janaseva sweet mart sold sweets of ₹ 3,86,000. What CGST and SGST he will pay if the rate of GST is 5%?
Solution:
Given that M/s Janaseva sweet mart sold sweets of ₹ 3,86,000
∴ Bill amount = ₹ 3,86,000
GST payable at the rate 5%
∴ CGST and SGST applicable is 2.5% each
∴ CGST on the bill = \(\frac{2.5}{100}\) × 3,86,000 = ₹ 9650
and SGST on the bill = \(\frac{2.5}{100}\) × 3,86,000 = ₹ 9650

Question 2.
Janhavi Gas Agency purchased some gas cylinders for ₹ 5,00,000 and sold them to the customers for ₹ 5,90,000. Find the amount of GST payable and the amount of ITC. 5% GST is applicable.
Solution:
Given that, Janhavi Gas Agency purchased some gas cylinders for ₹ 5,00,000 and GST applicable is 5%.
∴ Input tax (ITC) = 5% of 5,00,000
= \(\frac{5}{100}\) × 5,00,000
= ₹ 25,000
Janhavi Gas Agency sold the gas cylinders for ₹ 5,90,000
∴ Output tax for Janhavi Gas Agency = 5% of 5,90,000
= \(\frac{5}{100}\) × 5,90,000
= ₹ 29,500
GST payable = Output tax – Input tax (ITC)
= 29,500 – 25,000
= ₹ 4500
∴ GST payable for Janhavi Gas Agency is ₹ 4,500 and ITC is ₹ 25,000.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6

Question 3.
A company dealing in mobile phones purchased mobile phones worth ₹ 5,00,000 and sold the same to customers at ₹ 6,00,000. Find the amount of ITC and amount of GST if the rate of GST is 12%.
Solution:
Given that the rate of GST applicable is 12%.
The company purchased mobile phones worth ₹ 5,00,000.
∴ Input tax (ITC) = 12% of 5,00,000
= \(\frac{12}{100}\) × 5,00,000
= ₹ 60,000
The company dealing in mobile phones sold the same to customers at ₹ 6,00,000.
∴ Output tax of the company = 12% of 6,00,000
= \(\frac{12}{100}\) × 6,00,000
= ₹ 72,000
GST payable for the company = Output tax – Input tax (ITC)
= 72,000 – 60,000
= ₹ 12,000
∴ The ITC for the company is ₹ 60,000 and GST payable is ₹ 12,000.

Question 4.
Prepare business to customers (B2C) tax invoice using given information. Write the name of supplier, address, state, Date, Invoice Number, GSTIN etc. as per your choice
Supplier: ___________
Address: ___________
State: ___________
Date: ___________
Invoice No: ___________
GSTIN: ___________
Particular: Rate of Sarees – ₹ 2750
Rate of GST 5% HSN 5407 – 2 pcs
Rate of Kurta – ₹ 750
Rate of GST 12% HSN 5408
Solution:
Supplier: M/s Swaglife Fashions
Address: 143, Shivaji Rasta, Mumbai 400001
Mobile No. 9263692111
Email: abc@gmail.com
State: Maharashtra
Date: 31/08/19
Invoice No: GST/110
GSTIN: 27ABCDE1234HIZS
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6 Q4
∴ Rate of 1 saree = ₹ 2750
∴ Rate of 2 sarees = 2 x 2750 = ₹ 5500
∴ GST on sarees = 12% of 5500
= \(\frac{12}{100}\) × 5500
= ₹ 660
∴ CGST = SGST = ₹ 330
∴ Rate of 1 Kurta = ₹ 750
∴ GST on Kurta = 12% of 750
= \(\frac{12}{100}\) × 750
= ₹ 90
∴ CGST = SGST = ₹ 45

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6

Question 5.
Heena Enterprise sold cosmetics worth ₹ 25,000 to Leena traders, a retailer. Leena Traders sold it further to Meena Beauty Products for ₹ 30,000. Meena Beauty Product sold it further to the customers for ₹ 40,000. The rate of GST is 18%. Find
(i) GST Payable by each party
(ii) CGST and SGST
Solution:
The trading chain,
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6 Q5
∴ Output tax for Heena Enterprises = 18% of 25,000
= \(\frac{18}{100}\) × 25,000
= ₹ 4,500
∴ GST payable by Heena Enterprises
Now output tax for Leena traders = 18% of 30,000
= \(\frac{18}{100}\) × 30,000
= ₹ 5,400
∴ GST payable by Leena traders = Output tax – Input tax
= 5,400 – 4,500
= ₹ 900
∴ Output tax for Meena beauty products = 18% of 40,000
= \(\frac{18}{100}\) × 40,000
= ₹ 7,200
∴ GST payable by Meena beauty products = Output tax – Input tax
= 7,200 – 5,400
= ₹ 1,800

(ii) Now, CGST = SGST = \(\frac{\text { GST }}{2}\) = 9%
∴ Statement of GST payable at each stage can be tabulated as:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6 Q5.1

Question 6.
‘Chitra furnishings’ purchased tapestry (curtain cloth) for ₹ 28,00,000 and sold for ₹ 44,80,000. Rate of GST is 5%. Find
(i) Input Tax
(ii) Output Tax
(iii) ITC
(iv) CGST and SGST
Solution:
Given, that ‘Chitra furnishings’ purchased tapestry (curtain cloth) for ₹ 28,00,000 and rate of GST is 5%
(i) Input tax = 5% of 28,00,000
= \(\frac{5}{100}\) × 28,00,000
= ₹ 1,40,000
The tapestry was sold at ₹ 44,80,000

(ii) Output tax = 5% of 44,80,000
= \(\frac{5}{100}\) × 44,80,000
= ₹ 2,24,000

(iii) Now ITC = Input tax = ₹ 1,40,000
GST payable = Output tax – ITC
= 2,24,000 – 1,40,000
= ₹ 84,000

(iv) CGST = SGST = \(\frac{\text { GST Payable }}{2}\)
= \(\frac{84,000}{2}\)
= ₹ 42,000
∴ CGST = SGST = ₹ 42,000

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6

Question 7.
Two friends ‘Aditi’ and ‘Vaishali’ went to a restaurant. They ordered 2 Masala Dosa costing ₹ 90 each 2 coffee costing ₹ 60 each and 1 sandwich costing ₹ 80. GST is charged at 5%. Find the Total amount of the bill including GST.
Solution:
Aditi and Vaishali ordered for 2 Masala Dosas, 2 Coffees and 1 Sandwich
∴ Total price of their order = 2 × 90 + 2 × 60 + 80 = ₹ 380
GST is charged at 5%
∴ GST on the total order = 5% × 380
= \(\frac{5}{100}\) × 380
= ₹ 19
∴ Total bill amount including GST = 380 + 19 = ₹ 399

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5

Question 1.
Three partners shared the profit in a business in the ratio 5 : 6 : 7. They had partnered for 12 months, 10 months, and 8 months respectively. What was the ratio of their investments?
Solution:
Let the ratio of investments of the three partners be p : q : r.
They partnered for 12 months, 10 months, and 8 months respectively.
∴ The profit shared by the partners will be in proportion to the product of capital invested and their respective time periods.
∴ 12 × p : 10 × q : 8 × r = 5 : 6 : 7
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5 Q1
From (i) & (ii), we have
p : q : r = 50 : 72 : 105
∴ The ratio of their investments was 50 : 72 : 105.

Question 2.
Kamala, Vimala and Pramila enter into a partnership. They invest ₹ 40,000, ₹ 80,000 and ₹ 1,20,000 respectively. At the end of the first year, Vimala withdraws ₹ 40,000, while at the end of the second year, Pramila withdraws ₹ 80,000. In what ratio will the profit be shared at the end of 3 years?
Solution:
Given that, Kamala, Vimala, and Pramila invest ₹ 40,000, ₹ 80,000, and ₹ 1,20,000 respectively.
The ratio of profits is to be calculated at the end of 3 years.
Vimala withdraws ₹ 40,000 at the end of the first year.
∴ Vimala invested ₹ 80,000 for one year and 40,000 for 2 years.
Pramila withdraws ₹ 80,000 at the end of the second year.
∴ Pramila invested ₹ 1,20,000 for two years and 40,000 for one year.
Kamala invested ₹ 40,000 for all the 3 years.
∴ The ratio of profits to be shared at the end of 3 years will be
= 40,000 × 3 : 80,000 × 1 + 40,000 × 2 : 1,20,000 × 2 + 40,000 × 1
= 1,20,000 : 1,60,000 : 2,80,000
= 12 : 16 : 28
= 3 : 4 : 7

Alternate Method:
Given that, Kamala, Vimala and Pramila invest ₹ 40,000, ₹ 80,000 & ₹ 1,20,000 respectively.
Given, information can be tabulated as:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5 Q2
∴ The profits to be shared at the end of 3 years will be
= 1,20,000 : 1,60,000 : 2,80,000
= 12 : 16 : 28
= 3 : 4 : 7

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5

Question 3.
Sanjeev started a business investing ₹ 25,000 in 1999. In 2000, he invested an additional amount of ₹ 10,000 and Rajeev joined him with an amount of ₹ 35,000. In 2001, Sanjeev invested another additional amount of ₹ 10,000 and Pawan joined them with an amount of ₹ 35,000. What will be Rajeev’s share in the profit of ₹ 1,50,000 earned at the end of 3rd year from the start of the business in 1999?
Solution:
The given information can be tabulated as:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5 Q3
∴ The ratio of profits to be shared at the end of 3 years will be 1,05,000 : 70,000 : 35,000
i.e. in the proportion 3 : 2 : 1
Given, profit earned ₹ 1,50,000/-
∴ Rajeev’s share in the profit = \(\frac{2}{6}\) × 1,50,000 = ₹ 50,000/-

Question 4.
Teena, Leena, and Meena invest in a partnership in the ratio: 7/2, 4/3, 6/5. After 4 months, Teena increases her share by 50%. If the total profit at the end of one year is ₹ 21,600, then what is Leena’s share in the profit?
Solution:
Investment of Teena, Leena and Meena are in the ratio \(\frac{7}{2}: \frac{4}{3}: \frac{6}{5}\)
After 4 months, Teena’s share increases by 50%.
i.e. \(\frac{7}{2}+\left(\frac{7}{2} \times \frac{50}{100}\right)=\frac{7}{2}+\frac{7}{4}\)
i.e. \(\frac{21}{4}\)
The profit will be shared in the proportion of product of capitals and respective time periods in months.
i.e. \(\frac{7}{2} \times 4+\frac{21}{4} \times 8: \frac{4}{3} \times 12: \frac{6}{5} \times 12\)
i.e. 56 : 16 : \(\frac{72}{5}\)
i.e. 7 : 2 : \(\frac{9}{5}\)
i.e. in the proportion 35 : 10 : 9 …..[Multiplying throughout by 5]
Given that profit at the end of one year = ₹ 21,600/-
∴ Leena’s share in the profit = \(\frac{10}{54}\) × 21,600
= 5 × 800
= 4000
∴ Leena’s share in the profit is ₹ 4000/-.

Question 5.
Dilip and Pradeep invested amounts in the ratio 2 : 1, whereas the ratio between amounts invested by Dilip and Sudip was 3 : 2. If ₹ 1,49,500 was their profit, how much amount did Sudip receive?
Solution:
Let the amounts invested by Dilip, Pradeep and Sudip be ₹ ‘d’, ₹ ‘p’ and ₹ ‘s’ respectively.
Given that, d : p = 2 : 1
∴ d : p = 6 : 3 …..(i)
and d : s = 3 : 2
∴ d : s = 6 : 4 …..(ii)
From (i) and (ii),
d : p : s = 6 : 3 : 4
∴ The ratio of profits to be shared among Dilip, Pradeep and Sudip will be 6 : 3 : 4.
Given, profit earned = ₹ 1,49,500/-
∴ Sudip’s share in the profit = \(\frac{4}{13}\) × 1,49,500
= 4 × 11,500
= ₹ 46,000/-

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5

Question 6.
The ratio of investments of two partners Jatin and Lalit is 11 : 12 and the ratio of their profits is 2 : 3. If Jatin invested the money for 8 months, find for how much time Lalit invested his money.
Solution:
Let ‘x’ be the time in months for which Lalit invested his money
Jatin and Lalit invested their money in the ratio 11 : 12.
Jatin invested his money for 8 months and the ratio of their profits is 2 : 3.
∴ 11 × 8 : 12 × x = 2 : 3
∴ \(\frac{88}{12 x}=\frac{2}{3}\)
∴ x = \(\frac{88 \times 3}{2 \times 12}\)
∴ x = 11
∴ Lalit invested his money for 11 months.

Question 7.
Three friends had dinner at a restaurant. When the bill was received, Alpana paid \(\frac{2}{3}\) as much as Beena paid and Beena paid \(\frac{1}{2}\) as much as Catherin paid. What fraction of the bill did Beena pay?
Solution:
Let ‘T’ be the total bill amount at the restaurant and ‘a’, ‘b’, and ‘c’ be the share of Alpana, Beena, and Catherin respectively.
Given, that Alpana paid \(\frac{2}{3}\) as much as Beena paid
∴ a = \(\frac{2}{3}\) b …..(i)
Also, Beena paid \(\frac{1}{2}\) as much as Catherin paid.
∴ b = \(\frac{1}{2}\) c
∴ c = 2b …….(ii)
∴ Three friends paid the total bill amount.
∴ a + b + c = T …..(iii)
Using (i) and (ii) in (iii), we get
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5 Q7
Thus, Beena paid \(\left(\frac{3}{11}\right)^{\text {th }}\) fraction of the total bill amount.

Question 8.
Roy starts a business with ₹ 10,000, Shikha joins him after 2 months with 20% more investment than Roy, after 2 months Tariq joins him with 40% less than Shikha. If the profit earned by them at the end of the year is equal to twice the difference between the investment of Roy and ten times the investment of Tariq. Find the profit of Roy?
Solution:
Given that, Roy starts the business with ₹ 10,000.
Shikha joins him after 2 months with 20% more investment than Roy.
∴ Shikha’s investment = 10,000 + (10,000 × \(\frac{20}{100}\)) = ₹ 12,000
Tariq joins after two more months with an investment 40% less than Shikha.
∴ Tariq’s investment = 12,000 – (12,000 × \(\frac{40}{100}\)) = ₹ 7,200
Now, the profit will be shared in the proportion of product of capitals and respective periods in months.
i.e. 10,000 × 12 : 12,000 × 10 : 7,200 × 8
i.e. in the proportion, 25 : 25 : 12 …..(i) [Dividing throughout by 4,800]
Given that, profit at the end of the year = twice of the difference between investment of Roy and ten times the investment of Tariq.
∴ Profit = 2 [(10 × 7,200) – 10,000]
= 2[72,000 – 10,000]
= 2 × 62,000
= ₹ 1,24,000
∴ Roy’s share of profit = \(\frac{25}{62}\) × 1,24,000 …..[From (i)]
= ₹ 50,000/-

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5

Question 9.
If 4(P’s Capital) = 6(Q’s Capital) = 10 (R’s Capital), then out of the total profit of ₹ 5,580, what is R’s share?
Solution:
Let ‘p’, ‘q’ and ‘r’ be P, Q and R’s Capital for the business respectively.
∴ 4p = 6q = 10r
L.C.M of 4, 6, 10 = 60
∴ We take 4p = 6q = 10r = 60x
∴ p = 15x, q = 10x, r = 6x
∴ p : q : r = 15 : 10 : 6
Given that total profit = ₹ 5580
R’s share in the profit = \(\frac{6}{31}\) × 5580 = ₹ 1080/-

Question 10.
A and B start a business, with A investing the total capital of ₹ 50,000, on the condition that B pays interest at the rate of 10% per annum on his half of the capital. A is a working partner and receives ₹ 1,500 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year?
Solution:
Let ‘x’ and ‘y’ be the profits earned by A and B respectively and let ‘z’ be the total profit for the year.
A is the working partner and receives ₹ 1500 per month from the total profit.
i.e. 12 × 1500 = ₹ 18,000 at the end of the year.
The remaining profit is shared between A and B equally.
∴ y = \(\frac{z-18000}{2}\) …..(i)
Thus, profit earned by A at the end of that year is given by
x = 18000 + \(\left(\frac{z-18000}{2}\right)\)
∴ x = \(\frac{z+18000}{2}\) ……(ii)
A invests the entire capital on the condition that B pays A interest at the rate of 10% per annum on his half of the capital.
∴ At the end of the first year,
A will receive \(\frac{10}{100}\) × 25,000 i.e. ₹ 2500/- over and above his share of profit.
∴ A’s income = Profit of A + 2500 = x + 2500
Given that,
income of A = twice the income of B
∴ x + 2500 = 2y …..(iii)
Using (i) and (ii) in (iii), we get
\(\frac{z+18000}{2}\) + 2500 = 2\(\left(\frac{z-18000}{2}\right)\)
z + 18000 + 5000 = 2(z – 18000)
z + 23000 = 2z – 36000
∴ z = 59,000
∴ The total profit for the year = ₹ 59,000/-

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4

Question 1.
Kanchan purchased a Maruti car for ₹ 2,45,000/- and the rate of depreciation is 14\(\frac{2}{7}\)% per annum. Find the value of the car after two years?
Solution:
Given, purchase price of the car = V = ₹ 2,45,000
Rate of depreciation per annum = r
= 14\(\frac{2}{7}\)%
= \(\frac{100}{7}\)%
∴ Value of the car after two years = \(\mathrm{V}\left(1-\frac{\mathrm{r}}{100}\right)^{\mathrm{n}}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4 Q1
∴ The value of the car after two years is ₹ 1,80,000.

Question 2.
The value of a machine depreciates from ₹ 32,768 to ₹ 21,952/- in three years. What is the rate of depreciation?
Solution:
Given, initial value of machine = V = ₹ 32,768/-
Depreciated value of the machine = D.V. = ₹ 21,952/-
Numher of years = n = 3
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4 Q2
∴ r = 12.5%
∴ Rate of depreciation is 12.5% per annum.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4

Question 3.
The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. Its present value is ₹ 2,18,700/-. What was the purchase price of the machine?
Solution:
Given, the rate of depreciation per annum = r = 10%
Number of years = n = 3
Present value of the machine = P.V. = ₹ 2,18,700/-
∴ Purchase price of the machine
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4 Q3
∴ The purchase price of the machine is ₹ 3,00,000.

Question 4.
Mr. Manish purchased a motorcycle at ₹ 70,000/-. After some years he sold his motorcycle at its exact depreciated value of it that is ₹ 51,030/-. The rate of depreciation was taken as 10%. Find out how many years he sold his motorcycle.
Solution:
Given, purchase price of the motorcycle = V = ₹ 70,000/-
Depreciated value of the motorcycle = D.V. = ₹ 51,030/-
∴ Rate of depreciation = r = 10%
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4 Q4
∴ n = 3
∴ Manish sold his motorcycle after 3 years.

Question 5.
Mr. Chetan purchased furniture for his home at ₹ 5,12,000/-. Considering the rate of depreciation as 12.5%, what will be the value of furniture after 3 years.
Solution:
Given, purchase price of furniture = V = ₹ 5,12,000/-
Rate of depreciation = r = 12.5%
Number of years = n = 3 years
∴ Value of furniture after 3 years = \(\mathrm{V}\left(1-\frac{\mathrm{r}}{100}\right)^{\mathrm{n}}\)
= 5,12,000 \(\left(1-\frac{12.5}{100}\right)^{3}\)
= 5,12,000 \(\left(1-\frac{1}{8}\right)^{3}\)
= 5,12,000 \(\left(\frac{7}{8}\right)^{3}\)
= 5,12,000 × \(\frac{343}{512}\)
= 3,43,000
∴ The value of furniture will be ₹ 3,43,000/-

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4

Question 6.
Grace Fashion Boutique purchased a sewing machine at ₹ 25,000/-. After 3 years machine was sold at depreciated value of ₹ 18,225/-. Find the rate of depreciation.
Solution:
Given, purchase price of sewing machine = V = ₹ 25,000/-
Selling price of machine = D.V. = ₹ 18,225/-
Number of years = n = 3 years
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4 Q6
∴ 100 – r = 90
∴ r = 10%
∴ Rate of depreciation is 10% per annum.

Question 7.
Mr. Pritesh reduced the value of his assets by 5% each year, which were purchased for ₹ 50,00,000/-. Find the value of assets after 2 years.
Solution:
Given, initial value of assets = V = ₹ 50,00,000/-
Rate of depreciation per annum = r = 5%
Number of years = n = 2 years
∴ Value of assets aftertwo years = \(V\left(1-\frac{r}{100}\right)^{n}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4 Q7
= 12,500 × 361
= 45,12,500
∴ The value of assets after two years is ₹ 45,12,500/-.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4

Question 8.
A manufacturing company is allowed to charge 10% depreciation on its stock. The initial value of the stock was ₹ 60,000/-. After how many years value of the stock will be ₹ 39366?
Solution:
Given, rate of depreciation = r = 10%
Initial value of stock = V = ₹ 60,000
Depreciated value of stock = D.V. = ₹ 39,366/-
By using,
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4 Q8
∴ n = 4
∴ The value of the stock will be ₹ 39,366/- after 4 years.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3

Question 1.
What would be the simple interest on an amount of ₹ 9,600 at the rate of 6% per annum after 3 years?
Solution:
Given Principal P = ₹ 9600
Rate of interest R = 6% p.a.
Number of years = T = 3
Simple Interest I = \(\frac{\text { PRT }}{100}\)
= \(\frac{9600 \times 3 \times 6}{100}\)
= 96 × 18
= 1728
∴ Simple interest after 3 years would be ₹ 1728

Question 2.
What would be the simple interest at the rate of 9\(\frac{1}{2}\)% per annum on ₹ 6,000 for 2\(\frac{1}{2}\) years?
Solution:
Rate of interest per annum R = 9\(\frac{1}{2}\)% = \(\frac{19}{2}\)%
Principal P = ₹ 6000
Duration T = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) years
∴ Simple Interest, I = \(\frac{\text { PRT }}{100}\)
= 6000 × \(\frac{19}{2} \times \frac{5}{2} \times \frac{1}{100}\)
= 15 × 19 × 5
= 1425
∴ Simple interest would be ₹ 1425.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3

Question 3.
What would be the simple interest on ₹ 8,400 in 9 months at the rate of 8.25 percent per annum?
Solution:
Principal P = ₹ 8400
Rate of interest R = 8.25%
Duration T = 9 months = \(\frac{3}{4}\) years
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3 Q3
∴ Simple interest would be ₹ 519.75.

Question 4.
What would be the compound interest on ₹ 4200 for 18 months at 10% per annum compounded half yearly?
Solution:
Principal P = ₹ 4200
Rate of interest R = 10%
Duration T = 18 months = 1.5 years
compounding is done half yearly
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3 Q4
= \(\frac{4200 \times 9261}{2000}\)
= 4862.025
I = A – P
= 4862.025 – 4200
= 662.025
∴ Compound interest would be ₹ 662.025.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3

Question 5.
Find compound interest on ₹ 10,000 for 2 years at 8% per annum compounded half yearly.
Solution:
Principal P = ₹ 10,000
Rate of interest R = 8% p.a. compounded half yearly
Duration T = 2 years
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3 Q5
I = A – P
= 11648.58 – 10000
= 1698.58
∴ Compound interest is ₹ 1698.58.

Question 6.
In how many years ₹ 1,00,000 will become ₹ 1,33,100 at compound interest rate of 10% per annum?
Solution:
Principal P = ₹ 1,00,000
Amount A = ₹ 1,33,100
Rate of interest R = 10% p.a.
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3 Q6
∴ ₹ 1,00,000 will become ₹ 1,33,100 after 3 years.

Question 7.
A certain sum of money becomes three times of itself in 20 years at simple interest. In how many does it become double of itself at the same rate of simple interest?
Solution:
Given that, sum of money triples itself in 20 years
∴ P + I = 3P
∴ I = 2P
and T = 20 years
Now simple interest I = \(\frac{\text { PRT }}{100}\)
∴ 2P = \(\frac{\mathrm{P} \times \mathrm{R} \times 20}{100}\)
∴ R = 10
∴ Rate of interest = 10% per annum
The time period is to be calculated for the condition that the sum doubles itself i.e. for the condition
P + I = 2P
i.e. I = P
\(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\) = P
∴ \(\frac{10 \times T}{100}\) = 1
∴ T = 10
∴ The sum will become double of itself in 10 years.

Question 8.
A person borrows 10,000 for 2 year at 4% p.a. simple interest he immediately lends it to another person at 6.5% p.a. for 2 years. Find his total gain in the transaction.
Solution:
Person borrows money at 4% per annum and lends it at 6.5% per annum.
∴ His gain is (6.5 – 4) = 2.5% on ₹ 10000 for 2 years
i.e. gain = \(\frac{1000 \times 2.5 \times 2}{100}\)
= 100 × 5
= ₹ 500
∴ The person will gain ₹ 500 in this transaction.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3

Question 9.
A man deposits X 200 at the end of each year in recurring account at 5% compound interest. How much will it become at the end of 3 years?
Solution:
At end of 1st year, 2nd year and 3rd year ₹ 200 were deposited.
Rate of interest R = 5% p.a.
At end of 3 years, amount
= 200 + \(200\left[1+\frac{5}{100}\right]+200\left[1+\frac{5}{100}\right]^{2}\)
= 200 [1 + 1.05 + (1.05)2]
= 200 [2.05 + 1.1025]
= 200 [3.1525]
= 630.5
At end of 3 years, the account will have a balance of ₹ 630.5.

Question 10.
A man gets a simple interest of ₹ 2,000 on a certain principal at the rate of 5% p.a. in 4 years. What compound interest will the man get on twice the principal in 2 years at the same rate.
Solution:
Let Principal amount = P
Simple Interest I = ₹ 2000
Rate of interest R = 5% p.a.
Time duration T = 4 years
I = \(\frac{\text { PRT }}{100}\)
∴ 2000 = \(\frac{\mathrm{P} \times 5 \times 4}{100}\)
∴ P = 10000
Twice the principal was invested for compound interest with the same rate of interest for 2 years.
Here, P = 2 × 10,000 = ₹ 20,000
∴ Amount received,
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3 Q10
I = A – P = 22050 – 20000 = 2050
The man will receive ₹ 2050 as compound interest.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3

Question 11.
The difference between simple interest and compound interest on a certain sum of money is ₹ 32 at 8% per annum for 2 years. Find the amount.
Solution:
Compound Interest = A – P = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-\mathrm{P}\)
Simple interest = \(\frac{\text { PRT }}{100}\)
Given R = 8%, T = 2 years and
compound interest – simple interest = ₹ 32
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3 Q11
∴ The man will receive a compound interest of ₹ 5000.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Question 1.
Mr. Sarad purchased a laptop for ₹ 24,000 and sold it for ₹ 30,000. What was the profit percentage?
Solution:
Cost price (C.P.) = ₹ 24000
Selling price (S. P.) = ₹ 30,000
Profit = S.P. – C.P.
= 30,000 – 24,000
= 6,000
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q1
∴ Profit Percentage = 25%

Question 2.
Shraddha purchased a mobile phone and refrigerator for ₹ 18,000 and ₹ 15,000 respectively. She sold the refrigerator at a loss of 20% and the mobile at a profit of 20%. What is her overall profit or loss?
Solution:
C.P. of mobile phone = ₹ 18,000
Profit percentage on mobile phone = 20%
Selling price (S.P.) of mobile phone = 18,000 (1 + \(\frac{20}{100}\))
= 18,000 (1 + \(\frac{1}{5}\))
= 18,000 × \(\frac{6}{5}\)
= ₹ 21,600
C.P. of refrigerator = 15,000
Loss percentage on refrigerator = 20%
∴ Selling price (S.P.) = 15,000(1 – \(\frac{20}{100}\))
= 15,000(1 – \(\frac{1}{5}\))
= 15,000 × \(\frac{4}{5}\)
= ₹ 12,000
∴ Total gelling price for the transaction = 21,600 + 12,000 = ₹ 33,600
Total cost price (purchase price) for the transaction = 18,000 + 15,000 = ₹ 33,000
∴ Overall profit made by Shraddha = Total S.P. – Total C.P.
= 33,600 – 33,000
= ₹ 600
Thus, Shraddha made on overall profit of ₹ 600.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Question 3.
A vendor bought toffees at 6 for ₹ 10. How many for ₹ 10 must he sell to gain 20%?
Solution:
Vendor bought toffees at the rate of 6 for ₹ 10
∴ Cost price of one toffee = \(\frac{10}{6}\)
i.e. C.P. = \(\frac{10}{6}\) …….(i)
Let x be the number of toffees he must sell in ₹ 10 to gain 20%
i.e. S.P. = \(\frac{10}{x}\) …….(ii)
Profit percentage = \(\frac{\text { S.P. }-\text { C.P. }}{\text { C.P. }}\)
Using (i) and (ii) we have
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q3
∴ 30(6 – x) = 6x
∴ 180 – 30x = 6x
∴ 36x = 180
∴ x = 5
The vendor must sell 5 toffees for ₹ 10 in order to gain 20%.

Question 4.
The percentage profit earned by selling an article for ₹ 2,880 is equal to the percentage loss incurred by selling the same article for ₹ 1,920. At what price the article should be sold to earn a 25% profit?
Solution:
Let x be C.P. of the article
Let y % be both, the gain and loss made when article is sold at ₹ 2,880 and ₹ 1,920 respectively. Then
x + \(\frac{y}{100}\) x = 2880 ……(i)
x – \(\frac{y}{100}\) x = 1920 …..(ii)
Adding (i) and (ii), we get
2x = 4800
∴ x = 2400
i.e. C.P. of the article = ₹ 2400
Required profit percentage = 25%
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q4
∴ The article should be sold at ₹ 3000 to earn 25% profit.

Question 5.
A cloth merchant advertises for selling cloth at a 4% loss. By using a faulty meter scale, he is earning a profit of 20%. What is the actual length of the scale?
Solution:
Let the cost price of the cloth be ₹ ‘x’ per meter
He claims a loss of 4%
∴ Selling price of the cloth
S.P. = C.P.(1 – \(\frac{loss%}{100}\))
= x(1 – \(\frac{4}{100}\))
= 0.96x …..(i)
The actual cost price of the cloth is lower as the cloth is measured by a faulty meter scale.
Given that shopkeeper’s profit = 20%
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q5
∴ The actual cost price is 0.8 times the cost price as advertised.
In other words, the meter scale used for the fraud is 0.8 times the meter scale that should have been used.
∴ The length of the faulty meter scale used = 0.8 × 1 = 0.8 meter
∴ The actual length of the scale is 0.8 meters.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Question 6.
Sunil sells his bike worth ₹ 25,000 to Rohit at a profit of 20%. After 6 months Rohit sells the bike back to Sunil at a loss of 20%. Find the total profit percent of Sunil considering both the transactions.
Solution:
Sunil sells his bike to Rohit at 20% profit.
So S.P. of the bike for Sunil
= 25000 + \(\frac{20}{100}\) × 25000
= 25000 + 5000
= 30000
∴ Cost price of bike to Rohit = ₹ 30000
Rohit sells the bike back to Sunil at 20% loss
∴ S.P. of the bike for Rohit = 30000 – \(\frac{20}{100}\) × 30000
= 30000 – 6000
= 24000
∴ In second transaction Sunil pays 24000 to Rohit
In the first transaction, he had received 30000 from Rohit
∴ Sunil made a profit of ₹ (30000 – 24000) = ₹ 6000
Sunil earned this profit on the bike which costed him ₹ 25000
∴ Total profit % that Sunil makes = \(\frac{6000}{25000}\) × 100
= \(\frac{600}{25}\)
= 24
∴ Sunil makes 24% profit considering both the transactions.

Question 7.
By selling a book at ₹ 405 bookseller incurs a loss of 25%. Find the cost price of the book.
Solution:
S.P. = ₹ 405
Loss% = 25
S.P. when there is a loss is given by
S.P. = C.P. × \(\frac{\text { Loss } \%}{100}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q7
∴ The cost price of the book is ₹ 540.

Question 8.
A cloth costs ₹ 675. If it is sold at a loss of 20%, what is its cost price as a percentage of its selling price?
Solution:
C.P. = ₹ 675
Loss% = 20%
∴ Loss made in selling = \(\frac{20}{100}\) × 675 = ₹ 135
S.P. = C.P. – Loss
= 675 – 135
= ₹ 540
Let C.P. be x % S.P.,
Then 675 = \(\frac{x}{100}\) × 540
∴ x = \(\frac{675 \times 100}{540}\) = 125
∴ Cost price is 125% of the selling price.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Question 9.
Ashwin buys an article for ₹ 500. He marks it for sale at 75% more than the cost price. He offers a 25% discount on the marked price to his customer. Calculate the actual percentage of profit made by Ashwin.
Solution:
C.P. = ₹ 500
Marked price = C.P. + \(\frac{75}{100}\) × C.P.
= \(\frac{75}{100}\) × 500
= 500 + 75 × 5
= 500 + 375
= 875
25% discount was given on marked price
∴ Discount = \(\frac{25}{100}\) × 875 = \(\frac{875}{4}\)
Selling price = marked price – discount
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q9
∴ Ashwin makes 31.25% profit.

Question 10.
The combined cost price of a refrigerator and a mixer is ₹ 12,400. If the refrigerator costs 600% more than the mixer, find the cost price of the mixer.
Solution:
Let ₹ x be the cost price of the mixer.
The cost price of the refrigerator = x + \(\frac{600}{100}\) x
= x + 6x
= 7x
Total cost price =12400 …..[Given]
i.e. x + 7x = 12400
i.e. 8x = 12400
∴ x = \(\frac{12400}{8}\) = 1550
∴ The cost price of mixer is ₹ 1550.

Question 11.
Find the single discount equivalent to the discount series of 5%, 7%, and 9%.
Solution:
Let the marked price be ₹ 100
After 1st discount the price = 100(1 – \(\frac{5}{100}\)) = 95
After 2nd discount the price = 95(1 – \(\frac{7}{100}\)) = \(\frac{95 \times 93}{100}\)
After 3rd discount the price = \(\frac{95 \times 93}{100}\left(1-\frac{9}{100}\right)\)
= \(\frac{95 \times 93 \times 91}{100 \times 100}\)
= \(\frac{803985}{10000}\)
= 80.3985 ~ 80.4
Selling price after 3 discounts is ₹ 80.4
Single equivalent discount = Marked price – Selling price
= 100 – 80.4
= ₹ 19.6
∴ Single equivalent discount is 19.6%.

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Question 12.
The printed price of a shirt is ₹ 390. Lokesh pays ₹ 175.50 for it after getting two successive discounts. If the first discount is 10%, find the second discount.
Solution:
Marked price = ₹ 390
After the first discount of 10%, the price of the shirt
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q12
∴ x = 50
∴ Second discount is 50%

Question 13.
Amar, a manufacturer, gives a discount of 25% on the list price to his distributor Akbar, Akbar sells at a 10% discount on the list price to his customer Anthony. Anthony paid ₹ 540 for the article. What is the profit percentage of Akbar on his cost price?
Solution:
Let ₹ ‘x’ be the list price of the article.
Amar gives a discount of 25% on the list price.
∴ Selling price for Amar = \(x\left(1-\frac{25}{100}\right)\)
= \(x\left(1-\frac{1}{4}\right)\)
= ₹ \(\frac{3 x}{4}\)
Amar sells the article to Akbar
Cost price of article for Akbar = ₹ \(\frac{3 x}{4}\) ……(i)
Akbar sells the article to Anthony at 10% discount on list price
∴ Selling price for Akbar = \(x\left(1-\frac{10}{100}\right)\)
= \(x\left(1-\frac{1}{10}\right)\)
= ₹ \(\frac{9 x}{10}\) …..(ii)
Profit percentage = \(\frac{\text { S.P. }-\text { C.P. }}{\text { C.P. }} \times 100\)
Using (i) and (ii), we have the profit percentage for Akbar as,
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q13
∴ Akbar gets a profit of 20% on his cost price.

Question 14.
A man sells an article at a profit of 25%. If he had bought it at a 10% loss and sold it for ₹ 7 less, he would have gained 35%. Find the cost price of the article.
Solution:
Let ₹ ‘x’ be the C.P. of the article
∴ Article was sold at 25% profit
∴ S.P. of the article = \(x\left(1+\frac{25}{100}\right)\)
= \(x\left(1+\frac{1}{4}\right)\)
= 1.25x
If the article was bought at 10% loss
i.e., the new C.P. = \(x\left(1-\frac{10}{100}\right)\)
= \(x\left(\frac{9}{10}\right)\)
= 0.9x
and sold at ₹ 7 less
∴ New S.P. = 1.25x – 7
Then, the profit would have been 35%
Using profit percentage = \(\frac{\text { S.P.-C.P. }}{\text { C.P. }} \times 100\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2 Q14
∴ Cost price of the article is ₹ 200

Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Question 15.
Mr. Mehta sold his two luxury cars at ₹ 39,10,000 each. On one he gains 15% but on the other, he loses 15%. How much does he gain or lose in the whole transaction?
Solution:
Let x, y be the C.P. of two cars.
S.P. of both the cars = 39,10,000 …..[Given]
∴ One car is sold at 15% loss
∴ S.P. of the first car = x – \(\frac{15}{100}\)x
∴ \(\frac{85}{100}\)x = 39,10,000
∴ x = \(\frac{39,10,000 \times 100}{85}\)
∴ x = 46,000 × 100
∴ x = 46,00,000
Other car is sold at 15% gain
∴ S.P. of second car = y + \(\frac{15}{100}\) y
∴ y + \(\frac{15}{100}\) y = 39,10,000
∴ \(\frac{115}{100}\)y = 39,10,000
∴ y = \(\frac{39,10,000 \times 100}{115}\)
∴ y = 34,000 × 100
∴ y = 34,00,000
x + y = Total C.P. of two cars
= 46,00,000 + 34,00,000
= 80,00,000
Total S.P. = 39,10,000 + 39,10,000 = 78,20,000
∴ S.P. < C.P.
∴ There is a loss of ₹ (80,00,000 – 78,20,000) = ₹ 1,80,000
∴ Loss % = \(\frac{1,80,000}{80,00,000} \times 100\)
= \(\frac{18}{8}\)
= 2.25
∴ Mr. Mehta bears a 2.25% loss in the whole transaction.