Category: Class 9

Balbharti Maharashtra State Board Class 9 Sanskrit Solutions Aamod Chapter 13 सरमाया: शीलम् Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Sanskrit Aamod Solutions Chapter 13 सरमाया: शीलम्

Sanskrit Aamod Std 9 Digest Chapter 13 सरमाया: शीलम् Textbook Questions and Answers

भाषाभ्यास:

माध्यमभाषया उत्तरत।

1. सरमायाः कर्तव्यपालने के विघ्नाः अभवन् ?

प्रश्न 1.
सरमायाः कर्तव्यपालने के विघ्नाः अभवन् ?
उत्तरम् :
देवांची कुत्री सरमा तिच्या स्वामिनिष्ठेमुळे वेदवाङ्मयात प्रसिद्ध आहे. पूर्वी पणी नामक असुरांनी देवांच्या गायी पळवल्या, त्या शोधण्यासाठी देवांचा गरुड सुपर्ण पणींच्या देशात गेला. मात्र पणींनी त्याला भेटवस्तू देऊन आपल्याकडे वळवले आणि गोधनाविषयी सुपर्ण याने देवांना काहीच माहिती दिली नाही. गायी शोधण्यासाठी देवांनी त्यांची कुत्री सरमा हिला पाठवले.

पणींच्या देशात जाण्यासाठी सरमेला रसा नदी पार करावी लागणार होती, मात्र ती रसा नदी अतिशय वेगवान होती. ती पार करणे सरमेला शक्य नव्हते; मात्र सोपवलेले काम पूर्ण करायच्या निश्चयाने सरमेने नदीला विनंती केली की तिने तिचा वेग कमी करावा तसेच पाणी उथळ करावे. ज्यामुळे नदी पोहून दुसऱ्या काठावर जाणे शक्य होईल.

मात्र रसा स्वत:ला सर्वश्रेष्ठ समजत असल्याने तिने सरमेची विनंती धुडकावली. रसासारखी वेगवान नदी असूनही सरमेने जराही न डगमगता तिच्यात उडी मारली आणि धैर्याने दुसऱ्या तीरावर पोहोचली. तिथे पोहचल्या नंतर सुद्धा गायी शोधणे सोपे नव्हते. तरीसुद्धा सरमेने चिकाटीने गोधनाचा शोध लावला. इंद्राकडे परतताना पणींनी तिला लालूच दाखवली, तिची निंदा केली. पण सरमेने त्यांना दाद दिली नाही. मार्गात आलेल्या सर्व अडचणींना नेटाने तोंड देऊन सरमेने आपले कर्तव्य बजावले.

In Vedic literature, thestory of सरमा, Indra’s dog is very famous. She is famous for her loyalty towards the Gods. Formerly, the demons named for stole the cattle-wealth of Gods. In order to search the cattle-wealth, the eagle सुपर्ण was sent but the पणिs bribed him. सुपर्ण did not inform Gods about the cattle though he had found it hidden in the cave. Hence, was appointed to search the cattle wealth.

सरमा had to cross the river रसा to reach the country of पणि, रसा was flowing swiftly. It was very difficult for HRAT to cross over; but she did not give up. She requested रसा modestly to slow down her speed so that she would reach the other bank; but the who considered herself superior declined her request.

Yet, without caring for life descended into the current and reached the other bank courageously. She also faced great difficulty in searching the cattle-welath. ufus tried to test her loyalty by trying to bribe her.

Though there were so many difficulties in fulfilling her duty, सरमा overcame those difficulties with determination and accomplished her task.

2. सरमा कर्तव्यपालने विघ्नान् कथं तरति ?

प्रश्न 2.
सरमा कर्तव्यपालने विघ्नान् कथं तरति ?
उत्तरम् :
देवांची कुत्री सरमा तिच्या स्वामिनिष्ठेमुळे वेदवाङ्मयात प्रसिद्ध आहे. पूर्वी पणी नामक असुरांनी देवांच्या गायी पळवल्या. त्या शोधण्यासाठी देवांचा गरुड सुपर्ण पणींच्या देशात गेला. मात्र पणींनी त्याला भेटवस्तू देऊन आपल्याकडे वळवले आणि गोधनाविषयी सुपर्ण याने देवांना काहीच माहिती दिली नाही. गायी शोधण्यासाठी देवांनी त्यांची कुत्री सरमा हिला पाठवले.

पणींच्या देशात जाण्यासाठी सरमेला रसा नदी पार करावी लागणार होती, मात्र ती रसा नदी अतिशय वेगवान होती. ती पार करणे सरमेला शक्य नव्हते; मात्र सोपवलेले काम पूर्ण करायच्या निश्ययाने सरमेने नदीला विनंती केली की तिने तिचा वेग कमी करावा तसेच पाणी उथळ करावे. ज्यामुळे नदी पोहून दुसऱ्या काठावर जाणे शक्य होईल.

मात्र रसा स्वत:ला सर्वश्रेष्ठ समजत असल्याने तिने सरमेची विनंती धुडकावली. रसासारखी वेगवान नदी असूनही सरमेने जराही न डगमगता तिच्यात उडी मारली आणि धैर्याने दुसऱ्या तीरावर पोहोचली. गोधनाच्या शोधासाठी सुद्धा सरमा पणींच्या देशात खूप भटकली. अनेकांना विचारले मात्र कुणीच काही बोलले नाही. शेवटी जंगलातल्या गुहेत शिरुन पणींनी लपवलेल्या गायींचा शोध लावण्यात ती यशस्वी झाली. इंद्राकडे परतताना पणीनी तिला लालूच दाखवली, तिची निंदा केली. पण सरमेने त्यांना दाद दिली नाही.

मार्गात आलेल्या सर्व अडचणींना नेटाने तोंड देऊन सरमेने आपले कर्तव्य बजावले.

In Vedic literature, the story of सरमा, Indra’s dog is very famous. She is famous for her loyalty towards the Gods. Formerly, the demons named for stole the cattle-wealth of Gods. In order to search the cattle-wealth, the eagle सुपर्ण was sent but the पणिs bribed him. सुपर्ण did not inform Gods about the cattle though he had found it hidden in the cave. Hence, सरमा was appointed to search the cattle-wealth.

सरमा had to cross the river रसा to reach the country of पणि. रसा was flowing swiftly. It was very difficult for 1 to cross over; but she did not give up. She requested the modestly to slow down her speed so that she would reach the other bank, but tai who considered herself superior declined her request.

Yet, it without caring for life descended the current and reached the other bank courageously. सरमा wandered here and there to find out the cattlewelath in the country of us. She asked many people there but no one was ready to help her. At the end, she was successful in searching the cattle wealth. wus tried to stop her in every possible manner but सरमा did not step back even a little.

सरमा overcame obstaclescourageously. She was extremly loyal and determined to accomplish her task by defeating all obstacles.

3. पणयः सरमाया: निन्दा कदा अकुर्वन् ?

प्रश्न 3.
पणयः सरमाया: निन्दा कदा अकुर्वन् ?
उत्तरम् :
देवांची कुवी सरमा तिच्या स्वामिनिष्ठेमुळे वेदवाङ्मयात प्रसिद्ध आहे. पूर्वी पणी नामक असुरांनी देवांच्या गायी पळवल्या. त्या शोधण्यासाठी देवांचा गरुड सुपर्ण पणींच्या देशात गेला. मात्र पणींनी त्याला भेटवस्तू देऊन आपल्याकडे वळवले आणि गोधनाविषयी सुपर्ण याने देवांना काहीच माहिती दिली नाही. गायी शोधण्यासाठी देवांनी त्यांची कुत्री सरमा हिला पाठवले.

पणींच्या देशात पोहोचल्यावर सरमैने अतिशय कष्टाने गायींना शोधले. गायींना पाहिल्यावर आनंद झालेली सरमा इंद्राला सांगण्यासाठी निघाली. हे पाहिल्यावर पणींनी तिला अनेक प्रकारे फितवण्याचा, आपल्याकडे वळवण्याचा प्रयत्न केला. तिला दूध, तूप, दही खाण्याचा आग्रह केला. इतकेच नाही तर गोधनातील काही भागही देऊ केला, मात्र सरमा बधली नाही. तेव्हा पणींनी तिची निंदा करण्यास सुरुवात केली. जेणेकरून सरमा इंद्राकडे जाण्यापासून परावृत्त होईल.

In Vedic literature, the story of सरमा, Indra’s dog is very famous. She is famous for her loyalty towards the Gods. Formerly, the demons named for stole the cattle-wealth of Gods. In order to search the cattle-wealth, the eagle you was sent but the पणी bribed him. सुपर्ण did not inform Gods about the cattle though he had found it hidden in the cave. Hence, सरमा was appointed to search the cattle-wealth.

सरमा searched for the cattle-wealth rigorously. There was no help available. She succeeded in finding out the cattle-wealth hidden in one of the caves in the forest. Immediately she started to return to inform the gods. quit tried to bribe her. They offered her milk, butter, ghee. They were ready to share some of the cattle-welath with her; but she refused everything and remained loyal to her duty. Then the Panis started to criticise her to stop from going back to Indra.

4. पणयः सरमां किमर्थं निन्दन्ति ?

प्रश्न 4.
पणयः सरमां किमर्थं निन्दन्ति ?
उत्तरम् :
देवांची कुत्री सरमा तिच्या स्वामिनिष्ठेमुळे वेदवाङ्मयात प्रसिद्ध आहे. पूर्वी पणी नामक असुरांनी देवांच्या गायी पळवल्या. त्या शोधण्यासाठी देवांचा गरुड सुपर्ण पणींच्या देशात गेला. मात्र पणींनी त्याला भेटवस्तू देऊन आपल्याकडे वळवले आणि गोधनाविषयी सुपर्ण याने देवांना काहीच माहिती दिली नाही. गायी शोधण्यासाठी देवांनी त्यांची कुत्री सरमा हिला पाठवले.

देवांच्या गायी पळवून आणल्यानंतर पणींनी त्या गुहेत दडवून ठेवल्या. देवांची कुत्री सरमा हिने अतिशय कष्टाने त्या शोधल्या आणि त्यांची माहिती देण्यासाठी की इंद्राकडे निघाली. सुपर्ण गरुडाप्रमाणे सरमेला आपल्याकडे वळवून घेण्याचा प्रयत्न पणींनी केला मात्र सरमा त्याला बदली नाही. तिने मी लोभाला बळी पडणार नाही. मी इंद्राची स्वामिनिष्ठ सेविका आहे असे ठणकावले.

हे ऐकताच पणी खवळले आणि सरमेचा अपमान केला. सरमा ही मांस खाते असा अपप्रचार केला. यामुळे सरमेचे मनोबल बळेल, असे पणींना वाटले, मात्र अविचल कार्यनिष्ठा असलेली सरमा पणींकडे दुर्लक्ष करून निघाली.

In Vedic literature, the story of सरमा, Indra’s dog is very famous. She is famous for her loyalty towards the Gods. Formerly, the demons named for stole the cattle-wealth of Gods. The was appointed to search the cattle wealth.

सरमा searched for the cattle-wealth rigorously. There was no help available. She succeeded in finding out the cattle-wealth hidden in one of the caves in the forest. Immediately she started to return to inform ths gods. for tried to bribe her.

They offered her milk, butter, ghee. They were ready to share some of the cattle-welath with her; but सरमा with undisturbed mind igonored पणि, After that पणि criticized सरमा that she consumes meat. for thought that he would get discouraged due to insult and would not inform gods.

Sanskrit Aamod Class 9 Textbook Solutions Chapter 13 सरमाया: शीलम् Additional Important Questions and Answers

धातुसाधितविशेषणानि।

धातुसाधित – विशेषणम्	विशेष्यम्
दृष्टः	किङ्करः
नियुक्ता, आज्ञप्ता, प्रस्थिता, प्राप्ता, प्रमुदिता, अवरोधिता, निन्दिता	सरमा
दृष्टा, पराभूता	रसा
रुद्धम्	गोधनसर्वस्वम्
समाप्तम्	अन्वेषणकार्यम्
अन्विष्टाः	धेनवः
अपिहितम्	कर्णद्वयम्

सरमाया: शीलम् Summary in Marathi and English

प्रस्तावना :

प्राण्यांच्या चातुर्यकथा, नीतीकथा जगभरातील साहित्यात खूप प्रसिद्ध आहेत. रामायणातसुद्धा सागरी सेतू बांधण्यासाठी श्रीरामाला मदत करणाऱ्या खारीची गोष्ट येते. सरमा नामक कुत्रीची, तिच्या अविचल स्वामिनिष्ठेची कथा प्रसिद्ध आहे. पूर्वी यणी नामक असुरांनी देवांचे गोधन चोरले. ते शोधण्यासाठी सुपर्ण गरुडास देवांनी पणींच्या देशात पाठवले मात्र पणींनी सुपर्णाला लाच देऊन आपल्याकडे वळवले त्यामुळ देवांना गोधनाविषयी माहिती मिळाली नाही.

देवांनी या कामी सरमा नावाच्या कुत्रीला पाठवले. मार्गात येणाऱ्या वेगवान रसा नदीला ओलांडून, पणींच्या देशात निर्भयतेने शोध घेऊन सरमेने गोधन शोधले. पणींच्या लोभाला बळी न पडता सरमेने इंद्राला सर्व वृत्तांत सांगितला. कामात यशस्वी झालेल्या सरमेला इंद्राने वर दिला तर द्रोही गरुडाला शाप. सरमा ही कुत्री असूनही कार्यनिष्ठा, स्वामिनिष्ठा, धैर्य, स्वाभिमान या गुणांच्या जोरावर आदर्श ठरली तिच्या या कर्तृत्वाचे वर्णन पं. प्रभाकर भातखंडे यांनी संस्कृत पोवाड्यातून केले आहे. पोवाडा हा महाराष्ट्रातील लोककलाप्रकार असून त्यात वीररसाचे महत्त्वाचे स्थान आहे. शूरांच्या गाथा सांगण्यासाठी पोवाड्याची रचना केली जाते.

Stories advising morals through animals is very famous idea in literature all over the world. Infact several values are taught through the medium of animals and birds. This is dates back to the vedic period.

In the Ramayana too a story of squirrel helping the monkeys while building the bridge is known. The story of सरमा a dog from the ऋग्वेद is famous for her firm loyalty. It is said that some demons called ufuis once stole the cattle wealth of gods. Indra sent their eagle yup to look for the cattle. But the ufus bribed it and turned it on their side.

So Indra did not get any information about their cattle. Then Indra sent सरमा for the same task. सरमा accomplished her task fighting against all odds. Indra blessed ERCAT with a boon and cursed the eagle सुपर्ण..

सरमा set an example of courage, loyalty and commitment even if she was a bitch (female dog). This पोवाडा composed by पं. प्रभाकर भातखंडे celebrates T’s courageous behaviour. पोवाडाis a type of Marathi folk songs. The essence of bravery (a) is highlighted in these kind of songs. This present day creation is a beautiful example of Vedic story in a form that can reach the modern generation is very attractive manner.

शाहिरः – सभायां ………………… हे जी जी जी जी जी ।।

शाहिरः – सभायां देवराज इन्द्रः
चिन्तयति कः प्रेषितव्यः।
किकरो नैकोऽपि दृष्टः
यो भवेत् सदा स्वामिनिष्ठः।।

गायकवृन्दः – तदा?
शाहिरः – सेवायां तत्परा, देवगणप्रिया,
नियुक्ता गोधनप्राप्त्यर्थम्।
मघवता आज्ञप्ता सरमा,
प्रस्थिता देवशुनी सरमा। हे जी जी जी जी।।

गायकवृन्दः – प्रस्थिता देवशुनी सरमा। हे जी जी जी जी जी।
शाहिरः – तदा मार्गे दृष्टा, तटरुहतरुत्पाटिनी रसा
जलौघसमृद्ध्या पृथुलरभसा याति सरसा।।

रसां दृष्टा सरमया चिन्तितम् –
“अवश्यं गच्छेयं नदी यदि भवेत् अद्य सुसहा।
अतस्तां याचेऽहं विनयवचसा नम्रशिरसा।।
रसे, कुरु कृपाम्।
जलं तवेदम् । महागभीरम्।
कुरुष्व गाधम् । येन तरेयम्।
परं च तीरम् । अहं व्रजेयम् ।
स्वामिनः शोभनकार्यार्थम्,
आर्यगणमङ्गलकार्यार्थम्।

गायकवृन्दः – आर्यगणमङ्गलकार्यार्थम्। हे जी जी जी जी।।
रसा उवाच – “अहं च श्रेष्ठा, त्वं तु क्षुद्रा
न हि त्वदर्थ, जलप्रवाहम्।
करोमि अल्पम्, अपसर अपसर।।
श्रुत्वा तद् वाक्यं सरमया गर्जितम्।
“पश्य मे सामर्थ्य, कार्यनिष्ठासामर्थ्यम्
इति उक्त्वा सा जले अपतत्

शाहिरः – सा गता परं तीरम्।
रसा सा अधोमुखी भूता।
लज्जया म्लानमुखी जाता।
क्षुद्रया कथं पराभूता।
पश्यति क्षुद्रशुनीं सरमाम्।

गायकवृन्दः – पश्यति क्षुद्रशुनीं सरमाम्।
हे जी जी जी जी जी।।

गद्यम्: – एवं सरमा परतीरं प्राप्ता।
इतः पश्यति, ततः पश्यति
वलं पृच्छति, पणीन् पृच्छति।।
वनं गच्छति, गुहां प्रविशति।
गुहायां पणिभिर्यद्रुद्धं प्रभोः गोधनसर्वस्वम्।
समाप्तमन्वेषणकार्य, समाप्तमन्वेषणकार्यम्।।

गायकवृन्दः – प्रमुदिता इन्द्रशुनी सरमा हे जी जी जी जी जी।।
गद्यम्: – सरमया धेनवः अन्विष्टाः।
एतां वार्ताम् इन्द्रं वक्तुं प्रस्थिता सा।
परं वलैः सा अवरोधिता।

गायकवन्दः – हे सरमे, हे सरमे, दुग्धं पिबतु, दधि खादतु।
घृतं पिबतु, अत्र तिष्ठतु
गोधनस्य भागं तुभ्यं दद्यः।।

सरमा – सेविका नाहं लोभस्य,सेविका नाहं मोहस्य।
गायकवृन्दः – सेविका स्वामिनः इन्द्रस्य।
हे जी जी जी जी जी।।
सेविका स्वामिनः इन्द्रस्य।
हे जी जी जी जी जी।।

गद्यम्: – अन्ते पणिभि: सरमा निन्दिता।
“एषा सरमा जारु खादति।
“एषा सरमा जारु खादति।
सरमया किश्चत् तन्त्र श्रुतम्।
अपिहितं दीर्घ कर्णद्वयम्।
“वृत्तान्तं वक्तुं गमनमारभे।
प्रस्थिता सत्यप्रिया सरमा।

गायकवृन्दः – प्रस्थिता देवशुनी सरमा।
हे जी जी जी जी जी।।

अनुवादः

शाहीर – सभेमध्ये देवराज इंद्र (बसला आहे) कोणाला पाठवावे याचा विचार करत आहे. त्याला एकही सेवक दिसला
नाही – जो सदैव स्वामिनिष्ठ असेल.
गायकगण – तेव्हा मग पुढे?
शाहीर – सेवेमध्ये तत्पर असलेली, देवांची लाडकी, गोधन मिळवण्यासाठी नियुक्त झालेली, इंद्राने आशा दिलेली, सरमा देवांची कुत्री सरमा निघाली.
गायकगण – देवांची कुत्री सरमा निघाली.
शाहीर – तेव्हा तिने वाटेत काठावर वाढलेल्या झाडांना उन्मळून (पाडून) टाकणारी रसा (नदी) पाहिली. पाण्याच्या समृद्ध प्रवाहने (ती) रसा वेगाने वाहत होती. रसेला पाहून सरमेने विचार केला, “आज नदी सुसहा (तरुन जाण्यायोग्य झाली) झाली तर मी अवश्य जाईन. म्हणून मी तिला नम्र बोलण्याने, डोके टेकवून विनंती करते. रसे, कृपा कर, तुझे पाणी खूप खोल आहे. ते उथळ कर. जेणेकरुन मी पोहून जाईन. दुसऱ्या तीरावर मी जाईन. स्वामींच्या (इंद्राच्या) कार्यासाठी आर्यांच्या (देवांच्या) मंगल कामासाठी (मी जात आहे.)
गायकगण – आर्यांच्या (देवांच्या) मंगल कामासाठी (मी जात आहे.)
रसा म्हणाली – मी श्रेष्ठ, तू क्षुद्र जलप्रवाह तुझ्यासाठी नाही करणार उथळ! (कमी) मागे हो, मागे हो. ते वाक्य ऐकून सरमा गर्जली. माझे सामर्थ्य बघ, कार्य निष्ठेची शक्ती बघ असे म्हणून ती पाण्यात उतरली.
शाहीर – ती दुसऱ्या तीरावर गेली. त्या रसेने तोंड झुकवले. लाजेने तिचे तोंड उतरले. मी क्षुद्र कुत्रीकडून कशी पराभूत झाले? (असे म्हणत ती) क्षुद्र सरमेला पाहू लागली.
गायकगण – क्षुद्र सरमेला पाहू लागली.
गद्य – अशा रीतीने सरमा दुसऱ्या तीरावर गेली. इथे पाहिले, तिथे पाहिले. वलांना विचारले, पणींना विचारले, वनात गेली, गुहेत शिरली, गुहेमध्ये देवांचे गोधन पणींनी दडवून ठेवले होते. शोधकार्य संपले, शोधकार्य संपले.
गायकगण – इंद्राची कुत्री सरमा आनंदली.
गद्य – सरमेने गायी शोधल्या ही बातमी इंद्राला सांगण्यासाठी ती निघाली मात्र वलांनी तिला अडवले.
गायकगण – अगं सरमे, दूध पी, दही खा. तूप पी, इथेच थांब गोधनाचा भाग तुलाही देतो आम्ही.
सरमा – मी लोभाची सेविका नाही. मी मोहाची सेविका नाही. (मी लोभामोहाने बधणार नाही.)
गायकगण – इंद्राची सेविका.
इंद्राची सेविका.
गद्य – शेवटी पणींनी सरमेची निंदा केली ही सरमा मांस (क्षुद्रान) खाते. सरमेने ते काहीएक ऐकून घेतले नाही. तिने आपले दोन मोठे कान बंद केले. वृत्तांत सांगण्यासाठी मी जात आहे. सत्यप्रिय सरमा निघाली.
गायकगण – देवांची कुत्री सरमा निघाली.

Shahir : In the assembly, Indra the king of Gods thought who should be sent. He didn’t see any servant who would be ever loyal.
Chorus : Then?
Shahir : Sarama, who was prompt in service, liked by the gods was appointed to bring the cows. Sarama the dog of the Gods who was ordered by Indra set off.
Chorus : The dog of the Gods Sarama set off.
Shahir : Then on the way she saw Rasa (river) who uprooted trees that grow on the banks, flowing with force with a strong current. Seeing Rasa Sarama thought, I shall surely go today if the river becomes tolerable (less forceful). So, I shall ask her modestly with modest words) and head bent in modestly. ‘Oh, Rasa, please do a favour. Your water is very deep. Please make it shallow. By which I shall cross to the other bank. I shall go for the good task of the master, for the auspicious task of the noble ones.
Chorus : For the auspicious task of the noble ones.
Rasa said : “I am superior and you are inferior. I shall not reduce the water current, move away, move away.” Listening to those words, Sarama roared, “See my capacity, the power of my dedication.” saying this she fell (jumped) into the water.
Shahir : She went to the other bank. Rasa was ashamed. She became sad due to shame. How was she defeated by some one inferior. Looks at that lowly dog Sarama.
Chorus : (She) looks at that lowly dog sarama.
Prose : Thus Sarama reached the other bank. She looks here, she looks there. She asks the Vala, she asks the demons. She goes to the forest, she enters the cave. In the cave is all the Lord’s cattle wealth trapped by the demons. The search was over, the search was over.
Chorus: Sarama the God’sdog was delighted.
Prose : Sarama had found the cattle. She set off to inform this news to Indra. But she was stopped by the Valas.
Chorus: O Sarama, O Sarama. Drink milk, eat curds, drink ghee, wait here. We shall give you a part of the cattle wealth.
Sarama : I am not servant of greed. I am not servant of illusion
Chorus : Servant of Lord Indra.
Servant of Lord Indra.
Prose : Finally Sarama was criticised by the demons. “This Sarama eats flesh” But. Sarama paid no heed to all that. She closed the two long ears and started to depart to narrate the account. The truth lover Sarama set off.
Chorus : Sarama the god’s dog set off.

शब्दार्थाः

1. उपहार: – gift – बक्षीस, नजराणा
2. प्रेषितव्यः – should be sent – पाठवण्याजोगा
3. किड्करः – servant – नोकर, सेवक
4. गोधनम् – cattle wealth – गोधन
5. मघवान् – Lord Indra – इंद्रदेव
6. ‘तटरुहतरुः – tree that grows on the banks – झाड
7. जलौघः – water-current – पाण्याचा प्रवाह
8. पृथुलरभसा – with tremendous speed – अत्यंत वेगाने
9. विनयवचसा – with modest speech – विनयपूर्ण वाणीने
10. गभीरम् – deep – खोल
11. म्लानमुखी – sad – खिन्न
12. परतीरम् – to other bank – पलीकडच्या काठावर
13. वलम् – community of demons Pani – पणि नावाचे असूर
14. अन्विष्टाः – found – सापडल्या
15. घृतम् – ghee – तूप
16. जारुः – flesh/bad food – मांस, निकृष्ट खाणे

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Practice Set 4.5 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1.
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Solution:
Let the number to be subtracted be x.
∴ (12 – x), (16 – x) and (21 – x) are in continued proportion.

∴ 84 – 4x = 80 – 5x
∴ 5x – 4x = 80 – 84
∴ x = -4
∴ -4 should be subtracted from 12,16 and 21 so that the resultant numbers in continued proportion.

Question 2.
If (28 – x) is the mean proportional of (23 – x) and (19 – x), then find the value ofx.
Solution:
(28 – x) is the mean proportional of (23 – x) and (19-x). …[Given]

∴ -5(19 – x) = 9(28 – x)
∴ -95 + 5x = 252 – 9x
∴ 5x + 9x = 252 + 95
∴ 14x = 347
∴ x = \(\frac { 347 }{ 14 }\)

Question 3.
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Solution:
Let the first number be x.
∴ Third number = 26 – x
12 is the mean proportional of x and (26 – x).
∴ \(\frac { x }{ 12 }\) = \(\frac { 12 }{ 26 – x }\)
∴ x(26 – x) = 12 x 12
∴ 26x – x² = 144
∴ x² – 26x + 144 = 0
∴ x² – 18x – 8x + 144 = 0
∴ x(x – 18) – 8(x – 18) = 0
∴ (x – 18) (x – 8) = 0
∴ x = 18 or x = 8
∴ Third number = 26 – x = 26 – 18 = 8 or 26 – x = 26 – 8 = 18
∴ The numbers are 18, 12, 8 or 8, 12, 18.

Question 4.
If (a + b + c)(a – b + c) = a² + b² + c², show that a, b, c are in continued proportion.
Solution:
(a + b + c)(a – b + c) = a² + b² + c² …[Given]
∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2
∴ a² – ab + ac + ab – b² + be + ac – be + c² = a² + b² + c²
∴ a² + 2ac – b² + c² = a² + b² + c²
∴ 2ac – b² = b²
∴ 2ac = 2b²
∴ ac = b²
∴ b² = ac
∴ a, b, c are in continued proportion.

Question 5.
If \(\frac { a }{ b }\) = \(\frac { b }{ c }\) and a, b, c > 0, then show that,
i. (a + b + c)(b – c) = ab – c²
ii. (a² + b²)(b² + c²) = (ab + be)²
iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)
Solution:
Let \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k
∴ b = ck
∴ a = bk =(ck)k
∴ a = ck² …(ii)

i. (a + b + c)(b – c) = ab – c²
L.H.S = (a + b + c) (b – c)
= [ck² + ck + c] [ck – c] … [From (i) and (ii)]
= c(k² + k + 1) c (k – 1)
= c² (k² + k + 1) (k – 1)
R.H.S = ab – c²
= (ck²) (ck) – c² … [From (i) and (ii)]
= c²k³ – c²
= c²(k³ – 1)
= c² (k – 1) (k² + k + 1) … [a³ – b³ = (a – b) (a² + ab + b²]
∴ L.H.S = R.H.S
∴ (a + b + c) (b – c) = ab – c²

ii. (a² + b²)(b² + c²) = (ab + bc)²
b = ck; a = ck²
L.H.S = (a² + b²) (b² + c²)
= [(ck²) + (ck)²] [(ck)² + c²] … [From (i) and (ii)]
= [c²k⁴ + c²k²] [c²k² + c²]
= c²k² (k² + 1) c² (k² + 1)
= c4k² (k² + 1)²
R.H.S = (ab + bc)²
= [(ck²) (ck) + (ck)c]² …[From (i) and (ii)]
= [c²k³ + c²k]²
= [c²k (k² + 1)]2 = c⁴(k² + 1)²
∴ L.H.S = R.H.S
∴ (a² + b²) (b² + c²) = (ab + bc)²

iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)

9th Standard Algebra Practice Set 4.5 Question 6. Find mean proportional of \(\frac{x+y}{x-y}, \frac{x^{2}-y^{2}}{x^{2} y^{2}}\).
Solution:
Let a be the mean proportional of \(\frac{x+y}{x-y}\) and \(\frac{x^{2}-y^{2}}{x^{2} y^{2}}\)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Practice Set 4.1 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1.
From the following pairs of numbers, find the reduced form of ratio of first number to second number.
i. 72,60
ii. 38,57
iii. 52,78
Solution:
i. 72, 60
\(\text { Ratio }=\frac{72}{60}=\frac{12 \times 6}{12 \times 5}=\frac{6}{5}=6 : 5\)

ii. 38, 57
\(\text { Ratio }=\frac{38}{57}=\frac{19 \times 2}{19 \times 3}=\frac{2}{3}=2 : 3\)

iii. 52, 78
\(\text { Ratio }=\frac{52}{78}=\frac{26 \times 2}{26 \times 3}=\frac{2}{3}=2 : 3\)

Question 2.
Find the reduced form of the ratio of the first quantity to second quantity.
i. ₹ 700, ₹ 308
ii. ₹ 14, ₹ 12 and 40 paise
iii. 5 litres, 2500 ml
iv. 3 years 4 months, 5 years 8 months
v. 3.8 kg, 1900 gm
vi. 7 minutes 20 seconds, 5 minutes 6 seconds
Solution:
i. ₹ 700, ₹ 308
\( \text { Ratio }=\frac{700}{308}=\frac{28 \times 25}{28 \times 11}=\frac{25}{11}=25 : 11\)

ii. ₹ 14, ₹12 and 40 paise
₹ 14 = 14 x 100 paise = 1400 paise
₹ 12 and 40 paise = 12 x 100 paise + 40 paise
= (1200 + 40) paise
= 1240 paise

iii. 5 litres, 2500 ml
5 litres = 5 x 1000 ml = 5000ml

iv. 3 years 4 months, 5 years 8 months
3 years 4 months = 3×12 months + 4 months
= (36 + 4) months
= 40 months
5 years 8 months = 5 x 12 months + 8 months
= (60 + 8) months
= 68 months

v. 3.8 kg, 1900 gm
3.8 kg = 3.8 x 1000 gm = 3800 gm

vi. 7 minutes 20 seconds, 5 minutes 6 seconds
7 minutes 20 seconds = 7 x 60 seconds + 20 seconds
= (420 + 20) seconds
= 440 seconds
5 minutes 6 seconds = 5 x 60 seconds + 6 seconds
= (300 + 6) seconds
= 306 seconds

Question 3.
Express the following percentages as ratios
i. 75 : 100
ii. 44 : 100
iii. 6.25%
iv. 52: 100
v. 0.64%
Solution:

Question 4.
Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
Solution:
Let the persons required to build a house in 6 days be x.
Days required to build a house and number of persons are in inverse proportion.
∴ 6 × x = 8 × 3
∴ 6 x = 24
∴ x = 4
∴ 4 persons are required to build the house in 6 days.

Question 5.
Convert the following ratios into percentages.
i. 15 : 25
ii. 47 : 50
iii. \(\frac { 7 }{ 10 }\)
iv. \(\frac { 546 }{ 600 }\)
v. \(\frac { 7 }{ 16 }\)
Solution:
Let 15 : 25 = x %

∴ 15 : 25 = 60 %

ii. Let 47 : 50 = x%

∴ 47 : 50 = 94 %

iii. Let \(\frac { 7 }{ 10 }\) = x %

iv. Let \(\frac { 546 }{ 600 }\) = x %

v. Let \(\frac { 7 }{ 16 }\) = x %

Question 6.
The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha’s birth her mothers age was 27 years. Find the present ages of Abha and her mother.
Solution:
The ratio of ages of Abha and her mother is 2 : 5.
Let the common multiple be x.
∴ Present age of Abha = 2x years and
Present age of Abha’s mother = 5x years
According to the given condition, the age of Abha’s mother at the time of Abha’s birth = 27 years
∴ 5x – 2x = 27
∴ 3x = 27
∴ x = 9
∴ Present age of Abha = 2x = 2 x 9 = 18 years
∴ Present age of Abha’s mother = 5x =5 x 9 = 45 years
The present ages of Abha and her mother are 18 years and 45 years respectively.

Question 7.
Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?
Solution:
Present age of Vatsala = 14 years
Present age of Sara = 10 years
After x years,
Vatsala’s age = (14 + x) years
Sara’s age = (10 + x) years
According to the given condition,
After x years the ratio of their ages will become 5 : 4
∴ \(\frac { 14 + x }{ 10 + x }\) = \(\frac { 5 }{ 4 }\)
∴ 4(14 + x) = 5(10 + x)
∴ 56 + 4x = 50 + 5x
∴ 56 – 50 = 5x – 4x
∴ 6 = x
∴ x = 6
∴ After 6 years, the ratio of their ages will become 5 : 4.

Question 8.
The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana’s present age ?
Solution:
The ratio of present ages of Rehana and her mother is 2 : 7
Let the common multiple be x.
∴ Present age of Rehana = 2x years and Present age of Rehana’s mother = 7x years
After 2 years,
Rehana’s age = (2x + 2) years
Age of Rehana’s mother = (7x + 2) years
According to the given condition,
After 2 years, the ratio of their ages will be 1 : 3
∴ \(\frac { 2x + 2 }{ 7x + 2 }\) = \(\frac { 1 }{ 3 }\)
∴ 3(2x + 2) = 1(7x + 2)
∴ 6x + 6 = 7x + 2
∴ 6 – 2 = 7x – 6x
∴ 4 = x
∴ x = 4
∴ Rehana’s present age = 2x = 2 x 4 = 8 years
∴ Rehana’s present age is 8 years.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Problem Set 7 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
Write the correct alternative answer for each of the following questions.

i. Which of the following data is not primary ?
(A) By visiting a certain class, gathering information about attendance of students.
(B) By actual visit to homes, to find number of family members.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
(D) Review the cleanliness status of canals by actually visiting them.
Answer:
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

ii. What is the upper class limit for the class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer:
(B) 35

iii. What is the class-mark of class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer:
(D) 30

iv. If the classes are 0 – 10, 10 – 20, 20 – 30, …, then in which class should the observation 10 be included?
(A) 0 – 10
(B) 10 – 20
(C) 0 – 10 and 10-20 in these 2 classes
(D) 20 – 30
Answer:
(B) 10 – 20

v. If \(\overline { x }\) is the mean of x₁, x₂, ……. , x_n and \(\overline { y }\) is the mean of y₁, y₂, ….. y_n and \(\overline { z }\) is the mean of x₁,x₂, …… , x_n , y₁, y₂, …. y_n , then z = ?

Answer:
x₁, x₂, x₃, ……. , x_n
∴ \(\overline{x}=\frac{\sum x}{\mathrm{n}}\)
∴ n\(\overline{x}\) = ∑x
Similarly, n\(\overline{y}\) = ∑y
Now,

\(\text { (A) } \frac{\overline{x}+\overline{y}}{2}\)

vi. The mean of five numbers is 50, out of which mean of 4 numbers is 46, find the 5th number.
(A) 4
(B) 20
(C) 434
(D) 66
Answer:
5^th number = Sum of five numbers – Sum of four numbers
= (5 x 50) – (4 x 46)
= 250 – 184
= 66
(D) 66

vii. Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.
(A) 40.6
(B) 40.4
(C) 40.3
(D) 40.7
Answer:
New mean = \(\frac { 4000-30+70 }{ 100 }\)
= 40.4
(B) 40.4

viii. What is the mode of 19, 19, 15, 20, 25, 15, 20, 15?
(A) 15
(B) 20
(C) 19
(D) 25
Answer:
(A) 15

ix. What is the median of 7, 10, 7, 5, 9, 10 ?
(A) 7
(B) 9
(C) 8
(D) 10
Answer:
(C) 8

x. From following table, what is the cumulative frequency of less than type for the class 30 – 40?

(A) 13
(B) 15
(C) 35
(D) 22
Answer:
Cumulative frequency of less than type for the class 30 – 40 = 7 + 3 + 12 + 13 = 35
(C) 35

Question 2.
The mean salary of 20 workers is ₹10,250. If the salary of office superintendent is added, the mean will increase by ₹ 750. Find the salary of the office superintendent.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
The mean salary of 20 workers is ₹ 10,250.
∴ Sum of the salaries of 20 workers
= 20 x 10,250
= ₹ 2,05,000 …(i)
If the superintendent’s salary is added, then mean increases by 750
new mean = 10, 250 + 750 = 11,000
Total number of people after adding superintendent = 20 + 1 = 21
∴ Sum of the salaries including the superintendent’s salary = 21 x 11,000 = ₹ 2,31,000 …(ii)
∴ Superintendent salary = sum of the salaries including superintendent’s salary – sum of salaries of 20 workers
= 2, 31,00 – 2,05,000 …[From (i) and (ii)]
= 26,000
∴ The salary of the office superintendent is ₹ 26,000.

Question 3.
The mean of nine numbers is 77. If one more number is added to it, then the mean increases by 5. Find the number added in the data.
Solution:
∴ \( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations mean of nine numbers is 77
∴ sum of 9 numbers = 11 x 9 = 693 …(i)
If one more number is added, then mean increases by 5
mean of 10 numbers = 77 + 5 = 82
∴ sum of the 10 numbers = 82 x 10 = 820 …(ii)
∴ Number added = sum of the 10 numbers – sum of the 9 numbers = 820 – 693 … [From (i) and (ii)]
= 127
∴ The number added in the data is 127.

Question 4.
The monthly maximum temperature of a city is given in degree Celsius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5.29.0. 29.5.29.9.33.2.30.2
From the table, answer the following questions.
i. For how many days the maximum temperature was less than 34°C?
ii. For how many days the maximum temperature was 34°C or more than 34°C?
Solution:

i. Number of days for which the maximum temperature was less than 34°C
= 8 + 8 + 8 = 24
ii. Number of days for which the maximum temperature was 34°C or more than 34°C
= 5 + 1 = 6

Question 5.
If the mean of the following data is 20.2, then find the value of p.

Solution:

∴ 20.2 (30 + p) = 610 + 20p
∴ 606 + 20.2p = 610 + 20p
∴ 20.2p – 20p = 610 – 606
∴ 0.2p = 4
∴ p = \(\frac { 4 }{ 0.2 }\) = \(\frac { 40 }{ 2 }\) = 20
∴ p = 20

Question 6.
There are 68 students of 9th standard from Model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68,
80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43,
36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55,
56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78,
80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37,
45, 42, 70, 37,45, 66, 56, 47
By taking classes 30 – 40, 40 – 50, …. prepare the less than type cumulative frequency table. Using the table, answer the following questions:

i. How many students have scored marks less than 80?
ii. How many students have scored marks less than 40?
iii. How many students have scored marks less than 60?
Solution:
Class

i. 66 students have scored marks less than 80.
ii. 14 students have scored marks less than 40.
iii. 45 students have scored marks less than 60.

Question 7.
By using data in example (6), and taking classes 30 – 40, 40 – 50,… prepare equal to or more than type cumulative frequency table and answer the following questions based on it.
i. How many students have scored marks 70 or more than 70?
ii. How many students have scored marks 30 or more than 30?
Solution:

i. 11 students have scored marks 70 or more than 70.
ii. 68 students have scored marks 30 or more than 30.

Question 8.
There are 10 observations arranged in ascending order as given below.
45, 47, 50, 52, JC, JC + 2, 60, 62, 63, 74. The median of these observations is 53.
Find the value of JC. Also find the mean and the mode of the data.
Solution:
i. Given data in ascending order:
45,47, 50, 52, x, JC+2, 60, 62, 63, 74.
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
∴
∴ 106 = 2x + 2
∴ 106 – 2 = 2x
∴ 104 = 2x
∴ x = 52
∴ The given data becomes:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.

∴ The mean of the given data is 55.9.

iii. Given data in ascending order:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
∴ The observation repeated maximum number of times = 52
∴ The mode of the given data is 52.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Problem Set 7 Intext Questions and Activities

Question 1.
To show following information diagrammatically, which type of bar- diagram is suitable?
i. Literacy percentage of four villages.
ii. The expenses of a family on various items.
iii. The numbers of girls and boys in each of five divisions.
iv. The number of people visiting a science exhibition on each of three days.
v. The maximum and minimum temperature of your town during the months from January to June.
vi. While driving a two-wheeler, number of people wearing helmets and not wearing helmet in 100 families.
(Textbook pg. no. 112)
Solution:
i. Percentage bar diagram
ii. Sub-divided bar diagram
iii. Sub-divided bar diagram
iv. Sub-divided bar diagram
v. Sub-divided bar diagram
vi. Sub-divided bar diagram

Question 2.
You gather information for several reasons. Take a few examples and discuss whether the data is primary or secondary.
(Textbook pg. no, 113)
[Students should attempt the above activity on their own.]

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Practice Set 4.2 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1.
Using the property \(\frac { a }{ b }\) = \(\frac { ak }{ bk }\), fill in the blanks by substituting proper numbers in the following.

Solution:

Question 2.
Find the following ratios.
i. The ratio of radius to circumference of the circle.
ii. The ratio of circumference of circle with radius r to its area.
iii. The ratio of diagonal of a square to its side, if the length of side is 7 cm.
iv. The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of numbers denoting its perimeter to area.
Solution:
i. Let the radius of circle be r.
then, its circumference = 2πr
Ratio of radius to circumference of the circle

The ratio of radius to circumference of the circle is 1 : 2π.

ii. Let the radius of the circle is r.
∴ circumference = 2πr and area = πr²
Ratio of circumference to the area of circle

∴ The ratio of circumference of circle with radius r to its area is 2 : r.

iii. Length of side of square = 7 cm
∴ Diagonal of square = √2 x side
= √2 x 7
= 7 √2 cm
Ratio of diagonal of a square to its side

∴ The ratio of diagonal of a square to its side is √2 : 1.

iv. Length of rectangle = (l) = 5 cm,
Breadth of rectangle = (b) = 3.5 cm
Perimeter of the rectangle = 2(l + b)
= 2(5 + 3.5)
= 2 x 8.5
= 17 cm
Area of the rectangle = l x b
= 5 x 3.5
= 17.5 cm²
Ratio of numbers denoting perimeter to the area of rectangle

∴ Ratio of numbers denoting perimeter to the area of rectangle is 34 : 35.

Question 3.
Compare the following

Solution:

Question 4.
Solve.
ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5 : 4. FInd the measure of ∠B. [2 Marksl
Solution:
Ratio of ∠A and ∠B for given parallelogram is 5 : 4
Let the common multiple be x.

m∠A = 5x°and m∠B=4x°
Now, m∠A + m∠B = 180° …[Adjacent angles of a parallelogram arc supplementary]
∴ 5x° + 4x°= 180°
∴ 9x° = 180°
∴ x° = 20°
∴ m∠B=4x°= 4 x 20° = 80°
∴ The measure of ∠B is 800.

ii. The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
Solution:
The ratio of present ages of Albert and Salim is 5 : 9
Let the common multiple be x.
∴ Present age of Albert = 5x years and
Present age of Salim = 9x years
After 5 years,
Albert’s age = (5x + 5) years and
Salim’s age = (9x + 5) years
According to the given condition,
Five years hence ratio of their ages will be 3 : 5
\(\frac{5 x+5}{9 x+5}=\frac{3}{5}\)
∴ 5(5x + 5) = 3(9x + 5)
∴ 25x + 25 = 27x + 15
∴ 25 – 15 = 27 x – 25 x
∴ 10 = 2x
∴ x = 5
∴ Present age of Albert = 5x = 5 x 5 = 25 years
Present age of Salim = 9x = 9 x 5 = 45 years
∴ The present ages of Albert and Salim are 25 years and 45 years respectively.

iii. The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
Solution:
The ratio of length and breadth of a rectangle is 3 : 1
Let the common multiple be x.
Length of the rectangle (l) = 3x cm
and Breadth of the rectangle (b) = x cm
Given, perimeter of the rectangle = 36 cm
Since, Perimeter of the rectangle = 2(l + b)
∴ 36 = 2(3x + x)
∴ 36 = 2(4x)
∴ 36 = 8x
∴ \(x=\frac{36}{8}=\frac{9}{2}=4.5\)
Length of the rectangle = 3x = 3 x 4.5 = 13.5 cm
∴ The length of the rectangle is 13.5 cm and its breadth is 4.5 cm.

iv. The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
Solution:
The ratio of two numbers is 31 : 23
Let the common multiple be x.
∴ First number = 31x and
Second number = 23x
According to the given condition,
Sum of the numbers is 216
∴ 31x + 23x = 216
∴ 54x = 216
∴ x = 4
∴ First number = 31x = 31 x 4 = 124
Second number = 23x = 23 x 4 = 92
∴ The two numbers are 124 and 92.

v. If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
Solution:
Ratio of two numbers is 10 : 9
Let the common multiple be x.
∴ First number = 10x and
Second number = 9x
According to the given condition,
Product of two numbers is 360
∴ (10x) (9x) = 360
∴ 90x² = 360
∴ x² = 4
∴ x = 2 …. [Taking positive square root on both sides]
∴ First number = 10x = 10x² = 20
Second number = 9x = 9x² = 18
∴ The two numbers are 20 and 18.

Question 5.
If a : b = 3 : 1 and b : c = 5 : 1, then find the value of [3 Marks each]

Solution:
Given, a : b = 3 : 1
∴ \(\frac { a }{ b }\) = \(\frac { 3 }{ 1 }\)
∴ a = 3b ….(i)
and b : c = 5 : 1
∴ \(\frac { b }{ c }\) = \(\frac { 5 }{ 1 }\)
b = 5c …..(ii)
Substituting (ii) in (i),
we get a = 3(5c)
∴ a = 15c …(iii)

Ratio and Proportion 9th Class Practice Set 4.1 Question 6. If \(\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}\) , then find the ratio \(\frac { a }{ b }\).
Solution:
\(\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}\) … [Given]
∴ 0.04 x 0.4 x a = (0.4)² x (0.04)² x b … [Squaring both sides]

9th Algebra Practice Set 4.2 Question 7. (x + 3) : (x + 11) = (x – 2) : (x + 1), then find the value of x.
Solution:
(x + 3) : (x + 11) = (x- 2) : (x+ 1)
\(\quad \frac{x+3}{x+11}=\frac{x-2}{x+1}\)
∴ (x + 3)(x +1) = (x – 2)(x + 11)
∴ x(x +1) + 3(x + 1) = x(x + 11) – 2(x + 11)
∴ x² + x + 3x + 3 = x² + 1 lx – 2x – 22
∴ x² + 4x + 3 = x² + 9x – 22
∴ 4x + 3 = 9x – 22
∴ 3 + 22 = 9x – 4x
∴ 25 = 5x
∴ x = 5

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 2.3 8th Std Maths Answers Solutions Chapter 2 Parallel Lines and Transversals.

Practice Set 2.3 8th Std Maths Answers Chapter 2 Parallel Lines and Transversals

Question 1.
Draw a line l. Take a point A outside the line. Through point A draw a line parallel to line l.
Solution:
Steps of construction:

1. Draw a line l and take any point A outside the line.
2. Place a set-square, such that one arm of the right angle passes through A and the other arm is on line l.
3. Place the second set-square as shown in the figure such that the vertex of the right angle is at point A.
4. Hold the two set-squares in place and draw a line parallel to line l through the edge of the second set-square. Name the line as m.

Line m is the required line parallel to line l and passing through point A.

Question 2.
Draw a line l. Take a point T outside the line. Through point T draw a line parallel to line l.
Solution:
Steps of construction:

1. Draw a line l and take any point T outside the line.
2. Place a set-square, such that one arm of the right angle passes through T and the other arm is on line l.
3. Place the second set-square as shown in the figure such that the vertex of the right angle is at point T.
4. Hold the two set-squares in place and draw a line parallel to line l through the edge of the second set-square. Name the line as m.

Line m is the required line parallel to line l and passing through point T.

Question 3.
Draw a line m. Draw a line n which is parallel to line m at a distance of 4 cm from it.
Solution:
Steps of construction:

1. Draw a line m and take any two points M and N on the line.
2. Draw perpendiculars to line m at points M and N.
3. On the perpendicular lines take points S and T at a distance 4 cm from points M and N respectively.
4. Draw a line through points S and T. Name the line as n.

Line n is parallel to line m at a distance of 4 cm from it.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.2 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Practice Set 1.2 8th Std Maths Answers Chapter 1 Rational and Irrational Numbers

Question 1.
Compare the following numbers.
i. 7, -2
ii. 0, \(\frac { -9 }{ 5 }\)
iii. \(\frac { 8 }{ 7 }\), 0
iv. \(-\frac{5}{4}, \frac{1}{4}\)
v. \(\frac{40}{29}, \frac{141}{29}\)
vi. \(-\frac{17}{20},-\frac{13}{20}\)
vii. \(\frac{15}{12}, \frac{7}{16}\)
viii. \(-\frac{25}{8},-\frac{9}{4}\)
ix. \(\frac{12}{15}, \frac{3}{5}\)
x. \(-\frac{7}{11},-\frac{3}{4}\)
Solution:
i. 7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7 ∴ -2 > -7

ii. 0, \(\frac { -9 }{ 5 }\)
On a number line, \(\frac { -9 }{ 5 }\) is to the left of zero.
∴ 0 > \(\frac { -9 }{ 5 }\)

iii. \(\frac { 8 }{ 7 }\), 0
On a number line, zero is to the left of \(\frac { 8 }{ 7 }\) .
∴ \(\frac { 8 }{ 7 }\) > 0

iv. \(-\frac{5}{4}, \frac{1}{4}\)
We know that, a negative number is always less than a positive number.
∴ \(-\frac{5}{4}<\frac{1}{4}\)

v. \(\frac{40}{29}, \frac{141}{29}\)
Here, the denominators of the given numbers are the same.
Since, 40 < 141
∴ \(\frac{40}{29}<\frac{141}{29}\)

vi. \(-\frac{17}{20},-\frac{13}{20}\)
Here, the denominators of the given numbers are the same.
Since, 17 < 13
∴ -17 < -13
∴ \(-\frac{17}{20}<-\frac{13}{20}\)

vii. \(\frac{15}{12}, \frac{7}{16}\)
Here, the denominators of the given numbers are not the same.
LCM of 12 and 16 = 48

Alternate method:
15 × 16 = 240
12 × 7 = 84
Since, 240 > 84
∴ 15 × 16 > 12 × 7

viii. \(-\frac{25}{8},-\frac{9}{4}\)
Here, the denominators of the given numbers are not the same.
LCM of 8 and 4 = 8

ix. \(\frac{12}{15}, \frac{3}{5}\)
Here, the denominators of the given numbers are not the same.
LCM of 15 and 5 = 15

x. \(-\frac{7}{11},-\frac{3}{4}\)
Here, the denominators of the given numbers are not the same.
LCM of 11 and 4 = 44

Maharashtra Board Class 8 Maths Solutions Chapter 1 Rational and Irrational Numbers Practice Set 1.2 Questions and Activities

Question 1.
Verify the following comparisons using a number line. (Textbook pg. no, .3)
i. 2 < 3 but – 2 > – 3
ii. \(\frac{5}{4}<\frac{7}{4}\) but \(\frac{-5}{4}<\frac{-7}{4}\)
Solution:

We know that, on a number line the number to the left is smaller than the other.
∴ 2 < 3 and -3 < -2
i.e. 2 < 3 and -2 > -3

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Practice Set 7.3 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
For class interval 20 – 25 write the lower class limit and the upper class limit.
Answer:
Lower class limit = 20
Upper class limit = 25

Question 2.
Find the class-mark of the class 35-40.
Solution:
Class-mark

∴ Class-mark of the class 35 – 40 is 37.5

Question 3.
If class-mark is 10 and class width is 6, then find the class.
Solution:
Let the upper class limit be x and the lower class limit be y.
Class mark = 10 …[Given]
Class-mark

∴ x + y = 20 …(i)
Class width = 6 … [Given]
Class width = Upper class limit – Lower class limit
∴ x – y = 6 …(ii)
Adding equations (i) and (ii),
x + y = 20
x – y = 6
2x = 26
∴ x = 13
Substituting x = 13 in equation (i),
13 + y = 20
∴ y = 20 – 13
∴ y = 7
∴ The required class is 7 – 13.

Question 4.
Complete the following table.

Solution:
Let frequency of the class 14 – 15 be x then, from table,
5 + 14 + x + 4 = 35
∴ 23 + x = 35
∴ x = 35 – 23
∴ x = 12

Question 5.
In a ‘tree plantation’ project of a certain school there are 45 students of ‘Harit Sena.’ The record of trees planted by each student is given below:
3, 5, 7, 6, 4, 3, 5, 4, 3, 5, 4, 7, 5, 3, 6, 6, 5, 3, 4, 5, 7, 3, 5, 6, 4, 4, 3, 5, 6, 6, 4, 3, 5 ,7, 3, 4, 5, 7, 6, 4, 3, 5, 4, 4, 7.
Prepare a frequency distribution table of the data.
Solution:

Question 6.
The value of n upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.
Solution:

Question 7.
In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.
i.

ii.

Solution:
i. Let the Lower class limit and upper class limit of the class mark 5 be x and y respectively.

∴x + y = 10
Here, class width = 15 – 5 = 10
But, Class width = Upper class limit – Lower class limit
∴ y – x = 10
∴ -x + y = 10 …(ii)
Adding equations (i) and (ii),
x+ y = 10
-x + y = 10
∴ 2y = 20
∴ y = 10
Substituting y = 10 in equation (i),
∴ x + 10 = 10
∴ x = 0
∴ class with class-mark 5 is 0 – 10
Similarly, we can find the remaining classes.
∴ frequency table taking inclusive and exclusive classes.

ii. Let the lower class limit and upper class limit of the class mark 22 be x andy respectively.

∴ x + y = 44 …(i)
Here, class width = 24 – 22 = 2
But, Class width = Upper class limit – Lower class limit
∴ y – x = 2
∴ -x + y = 2 …. (ii)
Adding equations (i) and (ii),
x + y = 44
– x + y= 2
2y = 46
∴ y = 23
Substituting y = 23 in equation (i),
∴ x + 23 = 44
∴ x = 21
∴ class with class-mark 22 is 21 – 23
Similarly, we can find the remaining classes
∴ frequency table taking inclusive and exclusive classes.

Question 8.
In a school, 46 students of 9th standard, were told to measure the lengths of the pencils in their compass-boxes in Centimetres. The data collected was as follows:
16, 15, 7, 4.5, 8.5, 5.5, 5, 6.5, 6, 10, 12, 13,
4.5, 4.9, 16, 11, 9.2, 7.3, 11.4, 12.7, 13.9, 16,
5.5, 9.9, 8.4, 11.4, 13.1, 15, 4.8, 10, 7.5, 8.5,
6.5, 7.2, 4.5, 5.7, 16, 5.7, 6.9, 8.9, 9.2, 10.2, 12.3, 13.7, 14.5, 10
By taking exclusive classes 0-5, 5-10, 10-15,…. prepare a grouped frequency distribution table.
Solution:

Question 9.
In a village, the milk was collected from 50 milkmen at a collection center in litres as given below:
27, 75, 5, 99, 70, 12, 15, 20, 30, 35, 45, 80, 77,
90, 92, 72, 4, 33, 22, 15, 20, 28, 29, 14, 16, 20,
72, 81, 85, 10, 16, 9, 25, 23, 26, 46, 55, 56, 66,
67, 51, 57, 44, 43, 6, 65, 42, 36, 7, 35
By taking suitable classes, prepare grouped frequency distribution table.
Solution:

Question 10.
38 people donated to an organisation working for differently abled persons. The amount in rupees were as follows:
101, 500, 401, 201, 301, 160, 210, 125, 175,
190, 450, 151, 101, 351, 251, 451, 151, 260,
360, 410, 150, 125, 161, 195, 351, 170, 225,
260, 290, 310, 360, 425, 420, 100, 105, 170, 250, 100
i. By taking classes 100 – 149, 150 – 199, 200 – 249… prepare grouped frequency distribution table.
ii. From the table, find the number of people who donated ₹350 or more.
Solution:
i.

ii. Number of people who donated ₹ 350 or more = 4 + 4 + 2 + 1 = 11

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.3 Intext Questions and Activities

Question 1.
The record of marks out of 20 in Mathematics in the first unit test is as follows:
20,6, 14, 10, 13, 15, 12, 14, 17. 17, 18, 1119,
9, 16. 18, 14, 7, 17, 20, 8, 15, 16, 10, 15, 12.
18, 17, 12, 11, 11, 10, 16, 14, 16, 18, 10, 7, 17,
14, 20, 17, 13, 15, 18, 20, 12, 12, 15, 10
Answer the following questions, from the above information.
a. How many students scored 15 marks?
b. How many students scored more than 15 marks?
c. How many students scored less than 15 marks?
d. What is the lowest score of the group?
e. What is the highest score of the group? (Textbook pg. no. 114)
Solution:
a. 5 students scored 15 marks.
b. 20 students scored more than 15 marks.
c. 25 students scored less than 15 marks.
d. 6 is the lowest score of the group.
e. 20 is the highest score of the group.

Question 2.
For the above Question prepare Frequency Distribution Table. (Textbook pg. no. 115)
Solution:

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Practice Set 7.2 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
Classify following information as primary or secondary data.
i. Information of attendance of every student collected by visiting every class in a school
ii. The information of heights of students was gathered from school records and sent to the head office, as it was to be sent urgently.
iii. In the village Nandpur, the information collected from every house regarding students not attending school.
iv. For science project, information of trees gathered by visiting a forest.
Answer:
i. Primary data
ii. Secondary data
iii. Primary data
iv. Primary data