Class 9

Balbharti Maharashtra State Board Class 9 Sanskrit Solutions Anand Chapter 3 किं मिथ्या ? किं वास्तवम् ? Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Sanskrit Anand Solutions Chapter 3 किं मिथ्या ? किं वास्तवम् ?

Sanskrit Anand Std 9 Digest Chapter 3 किं मिथ्या ? किं वास्तवम् ? Textbook Questions and Answers

भाषाभ्यासः

1. एकवाक्येन उत्तरत।

प्रश्न अ.
चलभाषे काः क्रीडा: सन्ति ?
उत्तरम् :
चलभाषे यष्टिकन्दुकक्रीडा, पादकन्दुकक्रीडा वाहनक्रीडा, पत्रक्रीडा, चतुरङ्गक्रीडा इत्येता: क्रीडाः सन्ति।

प्रश्न आ.
चलभाषे कीदृशं विश्वम् ?
उत्तरम् :
चलभाषणस्य योग्य: उपयोग: करणीयः।

प्रश्न इ.
किं किं दृष्ट्रा तनयायाः मुखं लालायितम् ?
उत्तरम् :
पिझ्झा, पावभाजी, सिजलर्स इत्यादीनि खाद्यानि दृष्ट्वा तनायाया मुखं लालायितम्।

प्रश्न ई.
स्वास्थ्यरक्षणाय किं किम् आवश्यकम् ?
उत्तरम् :
स्वास्थ्यरक्षणाय सम्यक् क्रीडा, पौष्टिकम् अन्नं, रात्रौ निद्रा आवश्यकम्।

2. माध्यमभाषया उत्तरत।

प्रश्न अ.
माता तनयायाः विचारपरिवर्तनं कथं करोति ?
उत्तरम् :
“किं मिथ्या? किं वास्तव्यम्।” ह्या पाठात तनया व तिची आई यांच्या संवादातून तंत्रज्ञानाधारित काल्पनिक जग आणि वास्तव यांच्यातील फरक दाखवला आहे.

तनया नावाची मुलगी मोबाईलमध्ये आणि कुरकुरे खाण्यात मग्न असताना तिच्या मैत्रिणी खेळण्यासाठी तिला बोलवायला येतात. तनयाला मात्र घरात बसून मोबाईलवर गेम्स खेळणेच जास्त आवडते. ती तसे तिच्या मैत्रिणींना सांगते आणि घरातच बसून राहते. तिच्यामते मोबाईलवरच क्रिकेट, फुटबॉल ह्यासारखे खेळ खेळणे शक्य असताना बाहेर जायची काय गरज? असे म्हणून ती फोनमध्येच खेळण्यात मग्न होते.

थोड्यावेळाने जेव्हा तनयाला भूक लागते तेव्हाती आईकडे खायला मागते. आई तिला मोबाईल मधील पिझ्झा, पावभाजी इ. स्वादिष्ट खाद्यपदार्थाचे फोटो दाखवते. ते पाहून तनयाच्या तोंडाला पाणी सुटते. ती आईला काहीतरी खायला दे म्हणून हट्ट करते.

तेव्हा आई म्हणते की फोटो पाहिलेसना आत्ताच, मग आता खायची काय गरज? पण तनया हट्टाने म्हणते, “मोबाईवर फोटो बघून भूक कशी भागेल? तेव्हा तिची आई उत्तर देते, हे कळतंय तर मग मोबाइल वर खेळून व्यायाम होत नाही.

हे नाही कळलं तुला? पण तनयाचे असे म्हणणे होते की तिला मोबाईलवर खेळायला आवडते, आई म्हणते जे जे आवडते जे आरोग्यासाठी चांगले असतेच असे नाही. योग्य खेळ, पौष्टिक अन्न आणि रात्रीची झोप आरोग्यासाठी आवश्यक असते आणि ह्या सगळ्या गोष्टी प्रत्यक्षात करणे गरजेचे आहे, मोबाईलच्या आभासात्मक जगात नाही.

तंत्रज्ञानाचा वापर करावा, पण त्याची गरज ओळखून. ‘अति तेथे मती’ ह्या म्हणीनुसार तंत्रज्ञाचाही अतिवापर वाईटच. आईने तिच्याच शब्दांत समजवल्यामुळे तनयाला तिचे म्हणणे पटले. अशा प्रकारे आईने तनयाला तिचेच उदाहरण देऊन तिचे मतपरिवर्तन केले.

In this mother – daughter conversation “किं मिथ्या? कि वास्तवम्।” written by Smt. Mugdha Risbud, we see how mother cleverly teaches her daughter a valuable lesson of what is real and what is virtual in a very subtle manner.

The girl Tanaya refused to go out and play with her friends and instead preferred to play games on the cellphone. After playing mobile games when Tanaya entered the kitchen saying that she was hungry, her mother asked her to look at some pictures of food items on her mobile phone.

Looking at the pictures of pizza, pav bhaji and sizzlers Tanaya’s mouth watered and she asked for food. Her mother told her that after looking at the pictures there was no need to eat food. Yet Tanaya insistead to give her food.

Then her mother made her realise that if her hunger is not pacified with the pictures of food then how exercise can be done by playing games on the phone? So, for good health, playing games, nutritious food and regular sleep is necessary. The use of gadgets must be according to the requirement.

In this way mother changed Tanaya’s approach about being blindly reliable on gadgets.

प्रश्न आ.
किं मिथ्या ? किं वास्तवम् ? इति पाठस्य तात्पर्य लिखत ।
उत्तरम् :
“किं मिथ्या? किं वास्तवम्।” ह्या पाठात तनया आणि आई यांच्या संवादातून तंत्रज्ञानावर मयदिपेक्षा जास्त विसंबून राहणे कसे हानिकारक आहे हे सांगितले आहे.

तनया ही शाळेत जाणारी मुलगी आहे. ज्या वयात मैदानार जाऊन खेळणे, मित्रमैत्रिणींबरोबर गप्पा मारणे अशा गोष्टी करायच्या त्या वयात ती दिवसभर मोबाइलवरच ह्या सगळ्या गोष्टी करते.

घरी आरामात बसून हातात मोबाइल घेऊन जर खेळता येत असेल, चॅटींग करता येत असेल तर बाहेर जायची काय गरज असे तिला वाटते. मोबाइलवरचे आभासात्मक जग आणि वास्तवातील जग यांच्यातील भेद तिला कळत नाही.

हा केवळ तनयाचा नाही तिच्या वयाच्या व त्याहून मोठ्या सगळ्या मुलांचासुद्धा प्रश्न आहे. तंत्रज्ञानामुळे घरी राहून सगळ्या गोष्टी शक्य असताना बाहेर जाण्याची गरज कोणालाच भासत नाही. त्यामुळेच व्हर्चुअल / आभासात्मक जग आणि वास्तवातील जग यांच्यातील भेद पुसट होत चालला आहे.

पण चांगले आरोग्यदायी आयुष्य जगण्यासाठी शरीराच्या वास्तवादी गरजा ओळखून त्या प्रकारची जीवनशैली आचारणात आणणे गरजेचे आहे. हेच तनयाची आई तनयाला समजावून सांगत आहे.

या पाठामध्ये तंत्रज्ञानावर आधारित आभासात्मक जग आणि वास्तवातील जग यांच्यातील भेद प्रत्यक्ष उदाहरणातून दाखवला
आहे.

Through the conversation “कि मिथ्या? किं वास्तवम्।” between a mother and her thirteen – year old daughter, Smt. Mugdha Risbud, tells us how excessive dependence on technology can be harmful.

Playing on the ground is the best exercise and it is necessary for good health while on the other hand games played on the cellphone are virtual. They don’t provide any benefit to the body.

So by showing the pictures of food and not actully giving food to Tanaya, her mother taught her how the vitural world and real world are different. Playing on the phone virtually, is not beneficial as playing on the ground.

Tanaya believes that when we can do everything just with a cellphone then why do we need to go out. This is not just Tanaya’s problem but that of the whole generation. Children like Tanaya have the same attitude.

Hence, the difference between real and virtual world has become difficult to identify In this lesson author tries to show this difference through the conversation between a mother and daughter and conveys the message that use of anything must be according to the requirement.

3. अ. सन्धिविग्रहं कुरुत।

प्रश्न 1.
क. तदेव = + …………. एव।
ख. नास्ति = न + …………. ।
उत्तरम् :
क. तदेव – तत् + एव।
ख. नास्ति – न + अस्ति ।

आ. कालवचनपरिवर्तनं कुरुत।

प्रश्न 1.

1. अहं न आगच्छामि । (लङ्लकारे परिवर्तयत।)
2. बुभुक्षिता अस्मि। (बहुवचने परिवर्तयत ।)
3. सख्यौ क्रीडार्थं प्रतिगच्छतः। (एकवचने लिखत ।)
4. एतानि छायाचित्राणि पश्य । (उत्तमपुरुषे परिवर्तयत ।)

उत्तरम् :

1. अहं न आगच्छम्।
2. बुभुक्षिताः स्मः।
3. सखी क्रीडार्थ प्रतिगच्छति।
4. एतानि छायाचित्राणि पश्यानि।

इ. शब्दस्य वर्णविग्रहं कुरुत।

प्रश्न 1.

1. द्वारम्
2. स्वास्थ्यम्
3. बुभुक्षा
4. पौष्टिकम्

उत्तरम् :

1. द्वारम् – द् + व् + आ + र् + अ + म्।
2. स्वास्थ्य म् – स् + व् + आ +स् + थ् + य् + अ + म् ।
3. बुभुक्षा – ब् + उ + भ + उ + क् + ष् + आ।
4. पौष्टिकम् – प् + औ + ष् + ट् + इ + क् + अ + म्।

ई. मेलनं करुत।

प्रश्न 1.

उत्तरम् :

विशेषणम्	विशेष्यम्
बुभुक्षा	शान्ता
सैनिकाः	मृताः
खेल:	समाप्तः
अन्नम्	पौष्टिकम्
विश्वम	आभासात्मकम्

उ. पाठात् त्वान्त-ल्यबन्त-अव्ययानि चिनुत लिखत च।

प्रश्न 1.
पाठात् त्वान्त-ल्यबन्त-अव्ययानि चिनुत लिखत च।
उत्तरम् :

त्वान्त अव्यय धातु + त्वा / ध्वा / ट्वा / ढ्वा / इत्वा अयित्वा	ल्यबन्त अव्यय उपसर्ग + धातु + य / त्य	तुमन्त अव्यय   थातु + तुम् / धुम् / टुम् / ढुम् / इतुम् / अयितुम्
स्थित्वा, दर्शयित्वा	आकर्ण्य, उपविश्य	–
दृष्ट्वा, क्रीडित्वा	–	–

4. अ. समानार्थकशब्दं चिनुत ।
भटः, क्रीडा, जननी, क्षुधा, स्वापः।

प्रश्न 1.
समानार्थकशब्दं चिनुत ।
भटः, क्रीडा, जननी, क्षुधा, स्वापः।
उत्तरम् :

• सैनिक – योद्धा, भटः।
• क्रीडा – खेलः।
• माता – जननी, अम्बा, जन्मदात्री।
• बुभुक्षा – अशना, क्षुध् ।
• स्वापः – निद्रा, शयनम्, स्वापः, स्वप्नः, संवेशः।

आ. विरुद्धार्थकशब्दं लिखत ।
समाप्तः, स्वास्थ्यम्, आभासात्मकम्, मृतः ।

प्रश्न 1.
विरुद्धार्थकशब्दं लिखत ।
समाप्तः, स्वास्थ्यम्, आभासात्मकम्, मृतः ।
उत्तरम् :

• प्रारम्भः × समाप्तः।
• अस्वास्थ्यम् × स्वास्थ्यम्।
• आभासात्मकम् × वास्तवम्।
• मृतः × जीवितः।

Sanskrit Anand Class 9 Textbook Solutions Chapter 3 किं मिथ्या ? किं वास्तवम् ? Additional Important Questions and Answers

उचितं पर्यायं चिनुत।

प्रश्न 1.
तनया कस्मिन् मग्ना?
(अ) चलभाषे
(आ) पुस्तके
(इ) चित्रपटे
(ई) अध्ययने
उत्तरम् :
(अ) चलभाषे

प्रश्न 2.
द्वारघण्टिकाम् आकर्ण्य का द्वारम् उद्घाटयति?
(अ) माता
(आ) तनया
(इ) सखी
(ई) पिता
उत्तरम् :
(अ) माता

एकवाक्येन उत्तरत।

प्रश्न 1.
श्रुतिः किमर्थं तनायाम् आह्वयति?
उत्तरम् :
श्रुतिः बहिः क्रीडनाय तनायाम् आह्वयति।

प्रश्न 2.
तनया किं पश्यति ?
उत्तरम् :
तनया खाद्यपदार्थानां छायाचित्राणि पश्यति ।

सत्यं वा असत्यं लिखत।

प्रश्न 1.

1. माता सुखासन्दे उपविश्य हस्तेन कुकुंरिकां खादति।
2. तनया चलभाषे युद्धक्रीडायां मग्ना अस्ति।
3. तनया द्वारम् उद्घाटयति।

उत्तरम् :

1. असत्यम्
2. सत्यम्
3. असत्यम्

प्रश्न 2.
1. चलभाषे भोजनं दृष्ट्वा बुभुक्षा शान्ता भवेत् ।
2. सिजलर्स दृष्ट्वा मुखं लालयितम् ।
उत्तरम् :
1. असत्यम्
2. सत्यम्

प्रश्न 3.
1. चलभाषे आभासात्मकं विश्वं न वास्तवम्।
2. सम्यक् क्रीडा, पौष्टिकम् अन्न, रात्रौ निद्रा अनारोग्याय भवति ।
उत्तरम् :
1. सत्यम्
2. असत्यम्

कः कं वदति।

प्रश्न 1.
1. सत्वरं कीडनाय आगच्छ, बहिः खेलामः।
2. किं खेलसि?
उत्तरम् :
1. श्रुति : तनयां वदति।
2. राजसी तनयां वदति।

प्रश्न 2.

1. अम्ब बुभुक्षिता अस्मि ।
2. एतानि छायाचित्राणि पश्य ।
3. चलभाषे दृष्टवती खलु ।

उत्तरम् :

1. तनया मातरं वदति ।
2. माता तनयां वदति ।
3. माता तनयां वदति ।

प्रश्न 3.
आवश्यकतानुसारं योग्य: उपयोग: करणीयः एव।
उत्तरम् :
माता तनयां वदति।

उचितं कारणं चित्वा वाक्यं पुनर्लिखत।

प्रश्न 1.
द्वारघण्टिकाम् आकर्ण्य माता द्वारम् उद्घाटयति यतः ……..।
a. तनया चलभाषे मग्ना अस्ति।
b. तनया गृहकार्ये मग्ना अस्ति।
उत्तरम् :
द्वारघण्टिकाम् आकर्ण्य माता द्वारम् उद्घाटयति यतः तनया चलभाषे मग्ना अस्ति।

प्रश्न 2.
क्रीडाङ्गणं गत्वा खेलनस्य आवश्यकता नास्ति यतः ……..।
a. चलभाषे क्रीडाः सन्ति।
b. व्यायमस्य आवश्यकता नास्ति।
उत्तरम् :
क्रीडाङ्गणं गत्वा खेलनस्य आवश्यकता नास्ति यतः चलभाषे क्रीडा: सन्ति।

विशेषण-विशेष्य-सम्बन्धः। (स्तम्भमेलनं कुरुत)।

प्रश्न 1.

विशेषणम्	विशेष्यम्
1. अपरेण	अ. तनया
2. सर्वाः	ब. हस्तेन
3. मग्ना	क. क्रीडाः
4. योग्यः	इ. उपयोग:

उत्तरम् :

विशेषणम्	विशेष्यम्
1. अपरेण	ब. हस्तेन
2. सर्वाः	क. क्रीडाः
3. मग्ना	अ. तनया
4. योग्यः	इ. उपयोग:

शब्दस्य वर्णविग्रहं कुरुत।

• कुकुरिकाम् – क् + उ + र् + क् + उ + र् + इ + क + आ + म्।
• द्वारघण्टिकाम् – द् + व् + आ + र् + अ + घ् + अ + ण् + ट् + इ + क् + आ + म्।
• आह्वयतः – आ + ह् + व् + अ + य् + अ + त् + अः।
• क्रीडाङ्गणम् – क् + र् + ई + ड + आ + इ + ग् + अ + ण् + अ + म्।
• समाप्तः – स् + अ + म् + आ + + त् + अः।
• पिडा – प् + इ + अ + अ + आ ।
• सिजलर्स – स् + इ + ज् + अ + ल् + अ + र + स् + ।
• नास्ति – न् + आ + स् + त् + इ।
• स्वास्थ्य करम् – स् + व् + आ + स् + थ् + य् + अ + क् + अ + र + अ + म् ।
• सम्यक् – स् + अ + म् + य् + अ + क्।
• निद्रा – न् + इ + द् + र् + आ।
• योग्यः – य + आ + ग् + य् + अः।

प्रश्ननिर्माणं कुरुत।

प्रश्न 1.

1. पिझ्झा, पावभाजी, सिजलर्स इत्यादीनि खाद्यानि दृष्ट्वा लालायितं मुखम्।
2. तनया खाद्यपदार्थानां छायाचित्राणि पश्यति।
3. तनया सुखासन्दे उपविशति ।
4. खेलनस्य आवश्यकता एव नास्ति ।

उत्तरम् :

1. किं किं दृष्ट्वा लालायितं मुखम्?
2. तनया केषां छायाचित्राणि पश्यति?
3. तनया कुत्र / कस्मिन् उपविशति?
4. कस्य आवश्यकता एव नास्ति?

विभक्त्यन्तरूपाणि।

• प्रथमा – अहम्, माता, क्रीडाः, तनया, यष्टिकन्दुकक्रीडा, पादुकन्दुकक्रीडा, वाहनक्रीडा, पत्रक्रीडा, चतुरङ्गक्रीडा, खेल:,
  सर्वे, सैनिकाः, सर्वाः, एषः, सैनिकः, स्वास्थरक्षणम्, सर्वे, सैनिकाः, खेल: अहम्, व्यायामः, उपयोगः, क्रीडनम्, हितकरम्, आरोग्यपूर्णम् , पौष्टिकम्, अन्नम्, आभासात्मकम, विश्वम्।
• द्वितीया – द्वारम्, चलभाषम्, पाकगृहम्, द्वारघण्टिकाम, ताम, क्रीडाङ्गणम्, स्वास्थ्यरक्षणम्, भोजनम्, पाकगृहम्, एतानि, छायाचित्राणि, उपयोगम्।
• तृतीया – हस्तेन, अपरेण, चलभाषेण।
• चतुर्थी – कीडनाय, मह्यम्, आरोग्याय।
• षष्ठी – खेलनस्य, तनयायाः, खाद्यपदार्थानाम्, चलभाषस्य।
• सप्तमी – सुखासन्दे, द्वारे, गृहे, एतस्मिन्, चलभाषे, युद्धक्रीडायाम्, एतस्याम, क्रीडायाम, चलभाषे, रात्रौ।
• सम्बोधन – अम्ब, तनये, मातः, तनये, मातः।

लकारं लिखत।

• खादति – खाद् धातुः प्रथमगण: परस्मैपदं लट्लकार: प्रथमपुरुष: एकवचनम्।
• आह्वयतः – आ + ह्वे धातुः प्रथमगण: परस्मैपदं लट्लकार: प्रथमपुरुष: द्विवचनम्।
• खेलामः – खेल् धातुः प्रथमगण: परस्मैपदं लट्लकार: उत्तमपुरुष: बहुवचनम्।
• आगच्छामि – आ + गम् – गच्छ धातुः प्रथमगणः परस्मैपदं लट्लकार : उत्तमपुरुष: एकवचनम्।
• पश्य : दृश् – पश्य् धातुः प्रथमगणः परस्मैपदं लोट्लकार: मध्यमपुरुषः एकवचनम्।
• अस्मि – अस् धातुः द्वितीयगणः परस्मैपदं लट्लकार: उत्तमपुरुष: एकवचनम्।
• यच्छामि – दा-यच्छ् धातुः प्रथमगणः परस्मैपदं लट्लकारः उत्तमपुरुष: एकवचनम्।
• भवेत् – भू-भव् धातुः प्रथमगण: परस्मैपदं विधिलिङ्लकार: उत्तमपुरुषः एकवचनम्।
• रोचते – रुच् – रोचधातुः प्रथमगण: आत्मनेपदं लट्लकार: प्रथमपुरुष: एकवचनम्।
• वर्जयेत् – वृज् – वर्जु धातुः दशमगण: उभयपदं अत्र परस्मैपदं विधिलिङ्लकारः प्रथमपुरुषः एकवचनम्।
• करोमि – कृ धातुः अष्टमगण: उभयपदं अत्र परस्मैपदं लट्लकार : उत्तमपुरुष: एकवचनम्।

समानार्थकशब्दं योजयित्वा वाक्यं पुनर्लिखत।

प्रश्न 1.
सैनिक; – क्रीडायां सर्वे सैनिका: मृताः।
उत्तरम् :
क्रीडायां सर्वे योद्धार:/भटाः मृताः ।

व्याकरणम् :

शब्दानां पृथक्करणम्

समासाः।

समासनाम	अर्थ:	समासविग्रहः	समासनाम्
सुखासन्दे	seat for pleasure	सुखाय आसन्दः, तस्मिन् ।	चतुर्थी तत्पुरुष समास
द्वारघण्टिकाम्	bell of door	द्वारस्य घण्टिका, ताम् ।	षष्ठी तत्पुरुष समास
पाकगृहम्	place for cooking	पाका य गृहम् ।	चतुर्थी तत्पुरुष समास

किं मिथ्या ? किं वास्तवम् ? Summary in Marathi and English

प्रस्तावना :

आधुनिक तंत्रज्ञानाने मानवाच्या भौतिकविकासासाठी विस्तीर्ण क्षेत्र खुले केले आहे यात शंकाच नाही. परंतु, तंत्रज्ञानयुक्त साधनांच्या अत्याधुनिक वापरामुळे नवीन प्रश्न, विघ्ने आणि सामाजिक समस्या निर्माण होतात. आभासात्मक आणि वास्तव यामधील फरक आपण जाणून घेतला पाहिजे. आधुनिक काळाची हीच आवश्यकता आहे.

भ्रमणध्वनीचा आवश्यक तेवढाच वापर करावा हा महत्त्वपूर्ण संदेश लेखिका श्रीमती मुग्धा रिसबूड यांनी यांनी तेरा वर्षांची मुलगी आणि तिची आई या दोघींच्या सहज संवादातून पटवून दिला आहे. ‘अति तेथे माती’ या म्हणीनुसार तंत्रज्ञानाचा अति वापर वाईटच हा महत्वाचा विचार संवादरूपाने मांडला आहे.

It is indeed true that modern technology has opened the wide field of materialistic development for humans. Nevertheless, due to the excess usage of technological gadgets new questions and several new social problems have been created.

Yet technology is the necessity of the modern age. This beautiful conversation between a thirteen year old girl and her mother is presented to us by the author Shrimati Mugdha Risbud, where the difference between the real and virtual world is communicated. It tells us that though technology is an important part of our life today, excess should always be avoided.

परिच्छेद : 1

त्रयोदशवर्षीया ……………… युद्धक्रीडायां मग्ना।) (त्रयोदशवर्षीया तनया सुखासन्दे उपविश्य एकेन हस्तेन कुर्कुरिका खादति । अपरेण हस्तेन चलभाषं चालयति। तदानीं द्वारघण्टिकाम् आकर्य
माता द्वारम् उद्घाटयति। तनयाया: सख्यौ द्वारे एव स्थित्वा ताम् आह्वयतः ।)
श्रुतिः – तनये, सत्वरं क्रीडनाय आगच्छ, बहिः खेलामः।
तनया – अहं गृहे एव खेलामि।
राजसी – किं खेलसि?
तनया – (चलभाषं दर्शयित्वा) एतस्मिन् चलभाषे यष्टिकन्दुकक्रीडा, पादकन्दुकक्रीडा, वाहनक्रीडा, पत्रक्रीडा, चतुरङ्गक्रीडा, इत्यादयः सर्वाः क्रीडाः सन्ति। क्रीडाङ्गणं गत्वा खेलनस्य आवश्यकता एव नास्ति। अहं न आगच्छामि। (सख्यौ क्रीडा निर्गच्छतः। माता पाकगृहं गच्छति । तनया चलभाषे युद्धक्रीडायां मग्ना।

अनुवादः

(तेरा वर्षांची मुलगी तनया सोफ्यावर बसून एका हाताने कुरकुरे खात आहे. आणि दुसऱ्या हाताने भ्रमणध्वनी चालवत आहे. तेव्हाच दाराची घंटी (बेल) ऐकून आई दरवाजा उघडते. तनयाच्या दोन मैत्रिणी दारात उभ्या राहून तिला बोलावतात.)
श्रुती – तनया, लवकर खेळायला ये, बाहेर खेळूया.
तनया – मी घरीच खेळते.
राजसी – काय खेळतेस?
तनया – (भ्रमणध्वनी दाखवून) या भ्रमणध्वनीवर हॉकी, फुटबॉल, गाड्यांचा खेळ, पत्त्यांचा खेळ, बुद्धिबळ इ. खेळ आहेत. मैदानावर जाऊन खेळण्याची आवश्यकताच नाही.
मी येणार नाही.
(मैत्रिणी खेळण्यासाठी निघून जातात.
आई स्वयंपाकघरात जाते.
तनया भ्रमणध्वनीवर युद्धाचा खेळ खेळण्यात मग्न होते.)

(Thirteen-year-old Tanaya sitting on the sofa is eating Kurkure with one hand. She is operating the mobile with the other hand.
Then hearing the doorbell ring, her mother opens the door. Standing at the door itself, Tanaya’s friends call out to her.)
Shruti – Tanaya, come to play quickly. We shall play outside.
Tanaya – I shall play at home itself.
Rajasi – What are you playing?
Tanaya – (Showing the mobile) there are many games on this mobile-hockey, football, car games, card games, chess etc.
There is no need to go to the playground to play.
I am not coming
(The friends go to play
The mother goes to the kitchen.
Tanaya is engrossed playing a war game.)

परिच्छेद: 2

तनया – (स्वगतम्) …………………. स्वास्थ्यरक्षणं न भवति?

• तनया – (स्वगतम्) एतस्यां क्रीडायाम् एषः सैनिक: मृत: अपर: च मारणीयः। …… सर्वे सैनिकाः मता:। हेड ……. जितं मया। खेलः
• समाप्तः । (तनया पाकगृहं गच्छति ।) अम्ब, बुभुक्षिता अस्मि।
• माता – तनये, मम चलभाषे एतानि छायाचित्राणि पश्य। (तनया खाद्यपदार्थानां छायाचित्राणि पश्यति।)
• तनया – मातः, पिझ्झा, पावभाजी, सिजलर्स इत्यादीनि खाद्यानि दृष्ट्वा लालायितं मे मुखम्। कृपया किमपि यच्छतु।
• माता – चलभाष दृष्टवती खलु? अधुना खाद्यपदार्थानाम् आवश्यकता एव नास्ति। अहं न बच्छामि।
• तनया – चलभाषे भोजनं दृष्ट्वा बुभुक्षा शान्ता कथं भवेत्? .
• माता – यदि एतत् सत्यं, तर्हि कथं न जानासि यत् चलभाषेण क्रीडित्वा व्यायामः न भवति? स्वास्थ्यरक्षणं न भवति?

अनुवादः

तनया – (स्वत:शीच) या खेळात हा सैनिक मेला आता दुसऱ्याला मारूया…. सगळे सैनिक मेले. हे ………… मी जिंकले.
खेळ संपला.
(तनया स्वयंपाकघरात जाते.)
आई, मला भूक लागली आहे.
माता – तनया, माझ्या भ्रमणध्वनीवर ही छायाचित्रे (फोटो) बघ.
(तनया खाद्यपदार्थांची छायाचित्रे पाहते.)
तनया – आई, पिझ्झा, पावभाजी, सिजलर्स इत्यादी खाद्यपदार्थ पाहून माझ्या तोंडाला पाणी सुटले आहे.
कृपया (प्लीज) काहीतरी दे.
माता – भ्रमणध्वनीवर पाहिलेस ना?
आता खाद्यपदार्थांची आवश्यकताच नाही.
मी देणार नाही:
तनया – भ्रमणध्वनीवर जेवणाचे पदार्थ पाहून भूक कशी भागेल?
माता – जे हे खरे असेल, तर मग तुला हे माहीत नाही का की, भ्रमणध्वनीवर खेळून व्यायाम होत नाही? आरोग्याची काळजी घेतली जात नाही?

Tanaya – (to herself) In this game, this soldier is dead and the other should be killed……all soldiers are dead… hey…I have won. . The game is over.
(Tanaya goes to the kitchen).
Mother, I am hungry.
Mother – Tanaya, see these images on my mobile.
(Tanaya sees the images of the foodstuff).
Tanaya – Mother, my mouth is watering seeing the pizza, pavbhaji, Sizzlers etc.
Please give me something
Mother – You saw them on the mobile right?
Now there is no need of the foodstuff.
I shan’t give.
Tanaya – How will hunger be pacified by seeing the food on the mobile?
Mother – If this is true, then how do you not know that by playing on the mobile you don’t exercise?
Health is not taken care off?

परिच्छेद : 3

तनया – परं मात :………………. करोमि?

• तनया – परं मातः मह्यं रोचते तत् चलभाषण कीडनम्।
• माता – तनये, यद् यद् रोचते तत् तत् सर्व स्वास्थ्यकरं, हितकरम्, आरोग्यपूर्ण न भवति । सम्यक् क्रीडा, पौष्टिकम् अन्न, रात्री
• निद्रा आरोग्याय भवति। चलभाषे आभासात्मकं विश्वं न वास्तवम्।
• तनया – तर्हि किं चलभाषस्य उपयोगः एव न करणीयः ।
• माता – तथा न। आवश्यकतानुसार योग्य: उपयोग: करणीयः एव। किन्तु अति सर्वत्र वर्जयेत्।
• तनया – आम् मातः । इतः परं चलभाषस्य योग्यम् उपयोगं करोमि।

अनुवादः

• तनया – परंतु माता, मला ते भ्रमणध्वनीवर खेळणे आवडते.
• माता – तनया, जे जे आवडते ते ते सर्व स्वास्थकारक, हितकारक, आरोग्यदायी नसते.
• योग्य प्रकारे खेळणे, पौष्टिक अन्न, रात्रीची झोप आरोग्यासाठी महत्त्वपूर्ण आहे. भ्रमणध्वनीवरील आभासात्मक विश्व वास्तव नाही.
• तनया – तर मग काय भ्रमणध्वनीचा उपयोगच करायचा नाही?
• माता – तसे नव्हे.
• आवश्यकतेनुसार योग्य उपयोग करावाच.
• परंतु अतिरेक सर्वथा टाव्यवा.
• तनया – ठीक आहे आई.
• यापुढे (मी) भ्रमणध्वनीचा योग्य उपयोग करेन.
• Tanaya – But mother, I like that-playing with the mobile.
• Mother – Tanaya, all that you like is not healthy, beneficial and healthy. Proper games, nutritious food,
• sleep at night will make you healthy. The virtual world on the mobile is not reality.
• Tanaya – Then should the mobile not be used at all?
• Mother – Not so.
• Certainly it should be properly used as required. But excess should be avoided everywhere.
• Tanaya – Yes mother.
• Henceforth, I shall use the mobile appropriately.

सन्धिविग्रहः

• इत्यादयः – इति + आदयः
• किमपि – किम् + अपि।
• इत्यादीनि – इति + आदीनि ।

समानार्थकशब्दाः

1. मग्ना – रता, निमग्ना, लीना।
2. हस्तः – करः, पाणिः।
3. गृहम् – गेहम, सदनम्, आलयः।
4. आकर्ण्य – श्रुत्वा, निशम्य।
5. सत्वरम् – शीघ्रम्, झटिति ।
6. अधुना – सम्प्रति, इदानीम्।
7. मुखम् – वदनम्, आननम्, आस्यम्।
8. चलभाष: – भ्रमणध्वनिः।
9. लोक: – जगत्, भुवनम्, विश्वम्।
10. निशाः – रजनी, रात्रिः, शर्वरी, यामिनी।

विरुद्धार्थकशब्दाः

• निर्गच्छति × प्रविशति, आगच्छति।
• सत्वरम् × शनैः शनैः।
• अनावश्यकता × आवश्यकता।
• हितकरम् × अहितकरम्।
• उपयोगः × निरुपयोगः।
• आरोग्यम् × अनारोग्यम्।

शब्दार्थाः

1. उपविश्य – after stitting – बसून
2. चलभाष: – mobile – भ्रमणध्वनी
3. आकर्य – after listening – ऐकून
4. उद्घाटयति – opens – उघडते
5. यष्टिकन्दुकक्रीडा – hockey – हॉकी
6. सुखासन्दः – sofa – कोच
7. पादकन्दुकक्रीडा – football – फुटबॉल
8. वाहनक्रीडा – car games – वाहनक्रीडा
9. पत्रक्रीडा – cards – पत्ते
10. चतुरङ्गक्रीडा – chess – बुद्धिबळ
11. आह्वयत: – invite – बोलावतात
12. द्वारघण्टिका – door bell – दरवाजाची घंटी
13. बुभुक्षिता – hungry – भुकेलेली
14. लालयितम् – watering – पाणी सुटले
15. दृष्टवती – saw – पाहिले
16. स्वास्थ्यरक्षणम् – care of health – आरोग्याची काळजी
17. मृतः – dead – मृत झालेलेc
18. अपर: – another – दुसरा
19. मारणीयः – worth killing – मारण्याजोगा
20. बुभुक्षा – hunger – भूक
21. हितकरम् – beneficial – हितावह
22. आरोग्यपूर्णम् – healthy – आरोग्यपूर्ण
23. आभासात्मकम् – virtual – काल्पनिक
24. वर्जयेत् – should avoid – वयं करावे
25. पौष्टिकम् – healthy – पौष्टिक

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 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 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

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Practice Set 7.1 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
The following table shows the number of Buses and Trucks in nearest lakh units. Draw percentage bar diagram. (Approximate the percentages to the nearest integer)

Solution:


Question 2.
In the table given below, the information is given about roads. Using this draw sub-divided and percentage bar diagram (Approximate the percentages to the nearest integer)

Solution:
i. Sub-divided bar diagram:

ii. Percentage bar diagram:

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.1 Intext Questions and Activities

Question 1.
A farmer has produced Wheat and Jowar in his field. The following joint bar diagram shows the production of Wheat and Jowar. From the gken diagram answer the following questions: (Textbook pg. no. 108)
i. Which crop production has increased consistently in 3 years?
ii. By how many quintals the production ofjowar has reduced in 2012 as compared to 2011?
iii. What is the difference between the production of wheat in 2010 and 2012 ?
iv. Complete the following table using this diagram.


Solution:
i. The crop production of wheat has increased consistently in 3 years.
ii. The production of jowar has reduced by 3 quintals in 2012 as compared to 2011.
iii. The difference between the production of wheat in 2010 and 2012 = 48 – 30 = 18 quintals
iv.

Question 2.
In the following table, the information of number of girls per 1000 boys is given for different states. Fill In the blanks and complete the table. (Textbook pg. no. 111)

Solution:
Draw percentage bar diagram from this information and discuss the findings from the diagram.

Question 3.
For the above given activity, the information of number of girls per 1000 boys is given for five states. The literacy percentage of these five states is given below. Assam (73%), Bihar (64%), Punjab (77%), Kerala (94%), Maharashtra (83%). Think of the number of girls and the literacy percentages in the respective states. Can you draw any conclusions from it? (Textbook pg. no. 112)
Solution:
By observing the number of girls per 1000 boys and literacy percentages in the given respective states, we can conclude that the literacy rate of girls is least in Bihar and is highest in Kerala.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Practice Set 3.4 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
For x = 0, find the value of the polynomial x² – 5x + 5.
Solution:
p(x) = x² – 5x + 5
Put x = 0 in the given polynomial.
∴ P(0) = (0)² – 5(0) + 5
= 0 – 0 + 5
∴ p(0) = 5

Question 2.
If p(y) = y² – 3√2 + 1, then find p( 3√2 ).
Solution:
p(y) = y² – 3√2 y + 1
Putp= 3√2 in the given polynomial.
∴ p( 3√2 ) = (3√2 )² – 3√2 (3√2 ) + 1
= 9 x 2 – 9 x 2 + 1
= 18 – 18 + 1
∴ p( 3√2 ) = 1

Question 3.
If p(m) = m³ + 2m² – m + 10, then P(a) + p(-a) = ?
Solution:
p(m) = m³ + 2m² – m + 10
Put m = a in the given polynomial.
∴ p(a) = a³ + 2a² – a + 10 …(i)
Put m = -a in the given polynomial.
p(-a) = (-a)³ + 2(-a)² – (-a) +10
∴ p (-a) = -a³ + 2a² + a + 10 …(ii)
Adding (i) and (ii),
p(a) + p(-a) = (a³ + 2a² – a + 10) + (-a³ + 2a² + a + 10)
= a³ – a³ + 2a² + 2a² – a + a + 10 + 10
∴ p(a) + p(-a) = 4a² + 20

Question 4.
If p(y) = 2y³ – 6y² – 5y + 7, then find p(2).
Solution:
p(y) = 2y³ – 6y² – 5y + 7
Put y = 2 in the given polynomial.
∴ p(2) = 2(2)³ – 6(2)² – 5(2) + 7
= 2 x 8 – 6 x 4 – 10 + 7
= 16 – 24 – 10 + 7
∴ P(2) = -11

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Practice Set 3.3 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder.
i. (2m² – 3m + 10) ÷ (m – 5)
ii. (x⁴ + 2x³ + 3x² + 4x + 5) ÷ (x + 2)
iii. (y³ – 216) ÷ (y – 6)
iv. (2x⁴ + 3x³ + 4x – 2x²) ÷ (x + 3)
v. (x⁴ – 3x² – 8) ÷ (x + 4)
vi. (y³ – 3y² + 5y – 1) ÷ (y – 1)
Solution:
i. Synthetic division:
(2m² – 3m + 10) ÷ (m – 5)
Dividend = 2m² – 3m + 10
∴ Coefficient form of dividend = (2, -3, 10)
Divisor = m – 5
∴ Opposite of -5 is 5.

Coefficient form of quotient = (2, 7)
∴ Quotient = 2m + 7,
Remainder = 45
Linear division method:
2m² – 3m + 10
To get the term 2m², multiply (m – 5) by 2m and add 10m,
= 2m(m – 5) + 10m- 3m + 10
= 2m(m – 5) + 7m + 10
To get the term 7m, multiply (m – 5) by 7 and add 35
= 2m(m – 5) + 7(m- 5) + 35+ 10
= (m – 5) (2m + 7) + 45
∴ Quotient = 2m + 7,
Remainder = 45

ii. Synthetic division:
(x⁴ + 2x³ + 3x² + 4x + 5) ÷ (x + 2)
Dividend = x⁴ + 2x³ + 3x² + 4x + 5
∴ Coefficient form of dividend = (1, 2, 3, 4, 5)
Divisor = x + 2
∴ Opposite of + 2 is -2.

Coefficient form of quotient = (1, 0, 3, -2)
∴ Quotient = x³ + 3x – 2,
Remainder = 9

Linear division method:
x⁴ + 2x³ + 3x² + 4x + 5
To get the term x⁴, multiply (x + 2) by x³ and subtract 2x³,
= x³(x + 2) – 2x³ + 2x³ + 3x² + 4x + 5
= x³(x + 2) + 3x² + 4x + 5
To get the term 3x², multiply (x + 2) by 3x and subtract 6x,
= x³(x + 2) + 3x(x + 2) – 6x + 4x + 5
= x³(x + 2) + 3x(x + 2) – 2x + 5
To get the term -2x, multiply (x + 2) by -2 and add 4,
= x³(x + 2) + 3x(x + 2) – 2(x + 2) + 4 + 5
= (x + 2) (x3 + 3x – 2) + 9
∴ Quotient = x³ + 3x – 2,
Remainder – 9

iii. Synthetic division:
(y³ – 216) ÷ (y – 6)
Dividend = y³ – 216
∴ Index form = y³ + 0y³ + 0y – 216
∴ Coefficient form of dividend = (1, 0, 0, -216)
Divisor = y – 6
∴ Opposite of – 6 is 6.

Coefficient form of quotient = (1, 6, 36)
∴ Quotient = y² + 6y + 36,
Remainder = 0

Linear division method:
y³ – 216
To get the term y³, multiply (y – 6) by y² and add 6y²,
= y²(y – 6) + 6y² – 216
= y²(y – 6) + 6ysup>2 – 216
To get the, term 6 y² multiply (y – 6) by 6y and add 36y,
= y²(y – 6) + 6y(y – 6) + 36y – 216
= y²(y – 6) + 6y(y – 6) + 36y – 216
To get the term 36y, multiply (y- 6) by 36 and add 216,
= y² (y – 6) + 6y(y – 6) + 36(y – 6) + 216 – 216
= (y – 6) (y² + 6y + 36) + 0
Quotient = y² + 6y + 36
Remainder = 0

iv. Synthetic division:
(2x⁴ + 3x³ + 4x – 2x²) ÷ (x + 3)
Dividend = 2x⁴ + 3x³ + 4x – 2x²
∴ Index form = 2x⁴ + 3x³ – 2x² + 4x + 0
∴ Coefficient form of the dividend = (2,3, -2,4,0)
Divisor = x + 3
∴ Opposite of + 3 is -3

Coefficient form of quotient = (2, -3, 7, -17)
∴ Quotient = 2x³ – 3x² + 7x – 17,
Remainder = 51

Linear division method:
2x⁴ + 3x³ + 4x – 2x² = 2x² + 3x³ – 2x² + 4x
To get the term 2x⁴, multiply (x + 3) by 2x³ and subtract 6x³,
= 2x³(x + 31 – 6x³ + 3x³ – 2x² + 4x
= 2x³(x + 3) – 3x³ – 2x² + 4x

To get the term – 3x³, multiply (x + 3) by -3x² and add 9x²,
= 2x³(x + 3) – 3x²(x + 3) + 9x² – 2x² + 4x
= 2x³(x + 3) – 3x²(x + 3) + 7x² + 4x

To get the term 7x², multiply (x + 3) by 7x and subtract 21x,
= 2x³(x + 3) – 3x²(x + 3) + 7x(x + 3) – 21x + 4x
= 2x³(x + 3) – 3x²(x + 3) + 7x(x + 3) – 17x

To get the term -17x, multiply (x + 3) by -17 and add 51,
= 2x³(x + 3) – 3x²(x + 3) + 7x(x+3) – 17(x + 3) + 51
= (x + 3) (2x³ – 3x² + 7x- 17) + 51
∴ Quotient = 2x³ – 3x² + 7x – 17,
Remainder = 51

v. Synthetic division:
(x⁴ – 3x² – 8) + (x + 4)
Dividend = x⁴ – 3x² – 8
∴ Index form = x⁴ + 0x³ – 3x² + 0x – 8
∴ Coefficient form of the dividend = (1,0, -3,0, -8)
Divisor = x + 4
∴ Opposite of + 4 is -4

∴ Coefficient form of quotient = (1, -4, 13, -52)
∴ Quotient = x³ – 4x² + 13x – 52,
Remainder = 200

Linear division method:
x⁴ – 3x² – 8
To get the term x⁴, multiply (x + 4) by x³ and subtract 4x³,
= x³(x + 4) – 4x³ – 3x² – 8
= x³(x + 4) – 4x³ – 3x² – 8
To get the term – 4x³, multiply (x + 4) by -4x² and add 16x²,
= x³(x + 4) – 4x² (x + 4) + 16x² – 3x² – 8
= x³(x + 4) – 4x² (x + 4) + 13x² – 8
To get the term 13x², multiply (x + 4) by 13x and subtract 52x,
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52x – 8
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52x – 8
To get the term -52x, multiply (x + 4) by – 52 and add 208,
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52(x + 4) + 208 – 8
= (x + 4) (x³ – 4x² + 13x – 52) + 200
∴ Quotient = x³ – 4x² + 13x – 52,
Remainder 200

vi. Synthetic division:
(y³ – 3y² + 5y – 1) ÷ (y – 1)
Dividend = y³ – 3y² + 5y – 1
Coefficient form of the dividend = (1, -3, 5, -1)
Divisor = y – 1
∴Opposite of -1 is 1.

∴ Coefficient form of quotient = (1, -2, 3)
∴ Quotient = y² – 2y + 3,
Remainder = 2

Linear division method:
y³ -3y² + 5y – 1
To get the term y³ , multiply (y – 1) by y² and add y²
= y² (y – 1) + y² – 3y² + 5y – 1
= y² (y – 1) – 2y² + 5y – 1
To get the term -2y², multiply (y – 1) by -2y and subtract 2y,
= y² (y – 1) – 2y(y – 1) – 2y + 5y – 1
= y² (y – 1) – 2y(y – 1) + 3y – 1
To get the term 3y, multiply (y – 1) by 3 and add 3,
= y² (y – 1) – 2y(y – 1) + 3(y- 1) + 3 – 1
= (y – 1)(y² – 2y + 3) + 2
∴ Quotient = y² – 2y + 3,
Remainder = 2.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Practice Set 3.2 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.
Use the given letters to write the answer.
i. There are ‘a’ trees in the village Lat. If the number of trees increases every year by ’b‘. then how many trees will there be after ‘x’ years?
ii. For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?
iii. The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Solution:
i. Number of trees in the village Lat = a
Number of trees increasing each year = b
∴ Number of trees after x years = a + bx
∴ There will be a + bx trees in the village Lat after x years.

ii. Total rows = x
Number of students in each row = y
∴ Total students = Total rows × Number of students in each row
= x × y
= xy
∴ There are in all xy students for the parade.

iii. Digit in units place = n
Digit in tens place = m
∴ The two digit number = 10 x digit in tens place + digit in units place
= 10m + n
∴ The polynomial representing the two digit number is 10m + n.

Question 2.
Add the given polynomials.
i. x³ – 2x² – 9; 5x³ + 2x + 9
ii. -7m⁴+ 5m³ + √2 ; 5m⁴ – 3m³ + 2m² + 3m – 6
iii. 2y² + 7y + 5; 3y + 9; 3y² – 4y – 3
Solution:
i. (x³ – 2x² – 9) + (5x³ + 2x + 9)
= x³ – 2x² – 9 + 5x³ + 2x + 9
= x³ + 5x³ – 2x² + 2x – 9 + 9
= 6x³ – 2x² + 2x

ii. (-7m⁴ + 5m³ + √2 ) + (5m⁴ – 3m³ + 2m² + 3m – 6)
= -7m⁴ + 5m³ + √2 + 5m⁴ – 3m³ + 2m² + 3m – 6
= -7m⁴ + 5m⁴ + 5m³ – 3m³ + 2m² + 3m +√2 – 6
= -2m⁴ + 2m³ + 2m² + 3m + √2 – 6

iii. (2y² + 7y + 5) + (3y + 9) + (3y² – 4y – 3)
= 2y² + 7y + 5 + 3y + 9 + 3y² – 4y – 3
= 2y² + 3y² + 7y + 3y – 4y + 5 + 9 – 3
= 5y² + 6y + 11

Question 3.
Subtract the second polynomial from the first.
i. x² – 9x + √3 ; – 19x + √3 + 7x²
ii. 2ab² + 3a²b – 4ab; 3ab – 8ab² + 2a²b
Solution:
i. x² – 9x + √3 -(- 19x + √3 + 7x²)
= x² – 9x + √3 + 19x – √ 3 – 7x²
= x² – 7x² – 9x + 19x + √3 – √3
= – 6x² + 10x

ii. (2ab² + 3a²b – 4ab) – (3ab – 8ab² + 2a²b)
= 2ab² + 3a²b – 4ab – 3ab + 8ab² – 2a²b
= 2ab² + 8ab² + 3a²b – 2a²b – 4ab – 3ab
= 10ab² + a²b – 7ab

Question 4.
Multiply the given polynomials.
i. 2x; x² – 2x – 1
ii. x⁵ – 1; x³ + 2x² + 2
iii. 2y +1; y² – 2y + 3y
Solution:
i. (2x) x (x² – 2x – 1) = 2x³ – 4x² – 2x

ii. (x⁵ – 1) × (x³ + 2x² + 2)
= x⁵ (x³ + 2x² + 2) -1(x³ + 2x² + 2)
= x⁸ + 2x⁷ + 2x⁵ – x³ – 2x² – 2

iii. (2y + 1) × (y² – 2y³ + 3y)
= 2y(y² – 2y³ + 3y) + 1(y² – 2y³ + 3y)
= 2y³ – 4y⁴ + 6y² + y² – 2y³ + 3y
= -4y⁴ + 2y³ – 2y³ + 6y² + y² + 3y
= -4y⁴ + 7y² + 3y

Question 5.
Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor x Quotient + Remainder’.
i. x³ – 64; x – 4
ii. 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 ; x² – x
Solution:
i. x³ – 64 = x3 + 0x² + 0x – 64

∴ Quotient = x² + 4x + 16, Remainder = 0
Now, Dividend = Divisor x Quotient + Remainder
∴ x³ – 64 = (x – 4)(x² + 4x + 16) + 0

ii. 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 = 5x⁵ + 4x⁴ – 3x³ + 2x + 0x + 2

∴ Quotient = 5x³ + 9x² + 6x + 8,
Remainder = 8x + 2
Now, Dividend = Divisor x Quotient + Remainder
∴ 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 = (x² – x)(5x³ + 9x² + 6x + 8) + (8x + 2)

Question 6.
Write down the information in the form of algebraic expression and simplify.
There is a rectangular farm with length (2a² + 3b²) metre and breadth (a² + b²) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a2 – b2) metre. What is the area of the remaining part of the farm? [4 Marks]
Solution:
Length of the rectangular farm = (2a² + 3b²) m
Breadth of the rectangular farm = (a² + b²) m
Area of the farm = length x breadth = (2a² + 3b²) x (a² + b²)
= 2a²(a² + b²) + 3b²(a² + b²)
= 2a² + 2a²b² + 3a²b² + 3b⁴
= (2a⁴ + 5a²b² + 3b⁴) sq. m … (i)
The farmer used a square shaped plot of the farm to build a house.
Side of the square shaped plot = (a² – b²) m
∴ Area of the plot = (side)²
= (a² – b²)²
= (a⁴ – 2a²b² + b⁴) sq m… .(ii)

∴ Area of the remaining farm = Area of the farm – Area of the plot
= (2a⁴ + 5a²b² + 3b⁴) – (a⁴ – 2a²b² + b⁴) … [From (i) and (ii)]
= 2a⁴ + 5a²b² + 3b⁴ – a⁴ + 2a²b² – b⁴
= 2a⁴ – a⁴ + 5a²b² + 2a²b² + 3b⁴ – b⁴
= a⁴ + 7a²b² + 2b⁴
∴ The area of the remaining farm is (a⁴ + 7a²b² + 2b⁴) sq. m.