Class 10

MP Board 10th English book Solutions Guide Pdf Free Download of Text Books The Rainbow, The Spring Blossom Questions and Answers, Notes, Summary are part of MP Board Class 10th Solutions. Here we have given Madhya Pradesh Syllabus MP Board Class 10 English Book Solutions Pdf of The Rainbow, The Spring Blossom Textbook Workbook Special, and General English Class 10th Solutions.

Students can also download MP Board 10th Model Papers and MP Board Class 10th English Important Questions to help you to revise the complete Syllabus and score more marks in your examinations.

MP Board Class 10th English Book Solutions

FREE downloadable MP Board Class 10th Special English The Rainbow Text Book, Workbook Solutions and MP Board Class 10th General English The Spring Blossom Text Book, Workbook Solutions and Answers Guide Pdf download.

The Rainbow Textbook Special English Class 10th Solutions

You can download MP Board The Rainbow Textbook (Special English) Class 10 Solutions, Questions and Answers, Notes, Summary, Lessons: Pronunciation, Translation, Word Meanings, Textual Exercises.

• Chapter 1 Goodwill (Yajurveda (V-S))
• Chapter 2 The Happy Prince (Oscar Wilde)
• Chapter 3 Of Expense (Francis Bacon)
• Chapter 4 To the Cuckoo (William Wordsworth)
• Chapter 5 Refund (Fritz Karinthy)
• Chapter 6 Two Gentlemen of Verona (A.J.Cronin)
• Chapter 7 Report of An Adjudged Case (William Cowper)
• Chapter 8 What is Culture? (Pt. Jawaharlal Nehru)
• Chapter 9 If (Rudyard Kipling)
• Chapter 10 The Last Leaf (O’ Henry)
• Chapter 11 Mending Wall (Robert Frost)
• Chapter 12 The Pie and the Tart (Hugh Chesterman)
• Chapter 13 Three Days to See (Hellen Keller)
• Chapter 14 Lake Isle of Innisfree (W.B. Yeats)
• Chapter 15 Journey of the Atlantic on a Papyrus Boat
• Chapter 16 All the World’s a Stage (William Shakespeare)
• Chapter 17 The Bet (Anton Chekhov Pavlovich)
• Chapter 18 The Bridge Builder (Will Allen Dromgoole)
• Chapter 19 If I Were You
• Chapter 20 Teaching in the Tolstoy Farm (M.K. Gandhi)
• Chapter 21 Gitanjali (Rabindranath Tagore)

The Rainbow Workbook Special English Class 10th Solutions

You can download MP Board The Rainbow Workbook(Special English) Class 10 Solutions and Answers.

• Chapter 1 Goodwill
• Chapter 2 The Happy Prince
• Chapter 3 Of Expense
• Chapter 4 To the Cuckoo
• Chapter 5 Refund
• Chapter 6 Two Gentlemen of Verona
• Chapter 7 Report of An Adjudged Case
• Chapter 8 What is Culture?
• Chapter 9 If
• Chapter 10 The Last Leaf
• Chapter 11 Mending Wall
• Chapter 12 The Pie and the Tart
• Chapter 13 Three Days to See
• Chapter 14 Lake Isle of Innisfree
• Chapter 15 Journey of the Atlantic on a Papyrus Boat
• Chapter 16 All the World’s a Stage
• Chapter 17 The Bet
• Chapter 18 The Bridge Builder
• Chapter 19 If I Were You
• Chapter 20 Teaching in the Tolstoy Farm
• Chapter 21 Gitanjali

MP Board Class 10th Special English Reading Skills

MP Board Class 10th Special English Reading Unseen Passages

• Unseen Passages Factual
• Unseen Passages Discursive
• Unseen Passages Literary

MP Board Class 10th Special English Writing Skills

Short Composition

• Short Composition with Guidance
• Notice Writing
• Message Writing
• Telegram Writing
• Postcard Writing

Long Composition

• Composition: Verbal Stimulus
• Composition: Visual Stimulus

MP Board Class 10th Special English Grammar

• Special English Grammar Type I
• Special English Grammar Type II
• Special English Grammar Type III
• Special English Grammar Type IV
• Special English Grammar Type V
• Special English Grammar Type VI
• Special English Grammar Type VII

The Spring Blossom Textbook General English Class 10th Solutions

You can download MP Board The Spring Blossom Textbook General English Class 10th Solutions, Questions and Answers, Notes, Summary, Lessons: Pronunciation, Translation, Word Meanings, Textual Exercises.

• Chapter 1 Light the Lamp of Thy Love
• Chapter 2 The Power of Determination
• Chapter 3 Salutation to the Nation
• Chapter 4 Children
• Chapter 5 The Sneeze
• Chapter 6 The New Player in the Team
• Chapter 7 Anger
• Chapter 8 One Step Ahead
• Chapter 9 The Tribute-I
• Chapter 10 The Tribute-II
• Chapter 11 Wind
• Chapter 12 Maharana Pratap
• Chapter 13 Opportunity for Youth
• Chapter 14 The Pot of Gold
• Chapter 15 In Memoriam
• Chapter 16 The Red Rice Granary
• Chapter 17 Torch Bearers
• Chapter 18 The Tables Turned

MP Board Class 10th General English Reading Skills

MP Board Class 10th General English Reading Unseen Passages

• Important Extracts from Prose
• Important Extracts from Poems
• Unseen Passages

MP Board Class 10th General English Writing Skills

• Letter Writing: Informal and Formal Letters
• Note Making and Summarising
• Writing Task (Based on Verbal / Visual Stimulus)
• Essay Writing

MP Board Class 10th General English Grammar

• Tenses
• Modals
• Sentence Connectors
• Non-finite Forms of Verbs
• Clauses
• Translation
• Supplementary Material (Based on ‘Reading Time’ Section of Work-Book)

MP Board Class 10th General English Syllabus & Marking Scheme

Time: 3.00 Hours
Maximum Marks: 100

Unitwise Weightage

Section A
Reading (15 Marks, 27 Periods)

A1, A2 & A3 Three unseen passages of total 450 words with a variety of questions including 3 marks for vocabulary.
The prose passage will be
Factual passage, e.g. instruction, description, report etc.
Discursive passage e.g. argumentative, interpretative, persuasive etc and
Literary passage, e.g. fiction, poetry, interview, biography etc in nature.
About 150 words in length (5 Marks)
About 150 words in length (5 Marks)
About 150 words in length (5 Marks)
There will be questions for local comprehension besides questions on vocabulary and comprehension of higher-level skills such as drawing inferences and conclusions.

Section B
Writing (20 Marks, 36 Periods)

B1 – Letter writing: One letter based on provided verbal stimulus and context. (6 Marks)
Types of Letter: informal: personal – such as to family and friends
Formal: letters of complaints, enquiries, requests, applications.

B2 – Note making and summarising: 6 Marks
(a) Students will be asked to make notes on the passage given (100 words). (3 Marks)
(b) The students will be asked to prepare a summary looking at the given notes. (3 Marks)

B3 – Composition: A short writing task based on a verbal and/or visual stimulus (diagram, picture, graph, map, chart, table, flow chart etc) (80 words) (8 Marks)
OR
an essay in about 200 words on topics of day to day life.
After given an example practice to a student to write an original composition for two or three years, the option of ‘Essay’ may be eliminated.

Section C
Grammar and Translation (20 Marks, 36 Periods)

A variety of short questions involving the use of particular structures within a context. Test Types used will include cloze, gap-filling, sentence completion, sentence- re-ordering, editing, dialogue-completion and sentence transformation. The grammar syllabus for this class will include the following areas for teaching:

1. Use of non-finite
2. Sentence connectives: as, since, while, then, just because, just until
3. Clauses with what, where and how
4. Past Tense
5. Modals: can, could, may, must, might
6. Translation (from Hindi to Eng) (05 marks)

Note: All other areas covered in Class IX will be tested in Class X as this is an integrated course for this area of learning.

Section D
Text Books (45 Marks, 81 Periods)

Prose (20 Marks)
D1 and D2 – Two extracts from different prose lessons included in the textbook (approximately 100 words each) (5 × 2 = 10 marks)
These extracts chosen from different lessons will be literary and discursive in nature. Each extract will be of 5 marks. One mark in each extract will be for vocabulary. 4 marks in each passage will be used for testing local and global comprehension besides a question on interpretation.

D3 – One out of two questions: extrapolative in nature based on any one of the prose lessons from the textbook to be answered in about 50 to 80 words. (6 Marks)

D4 – One out of two questions on drama text (local and global comprehension questions) (25-30 words) (4 Marks)

Poetry (10 Marks)
D5 – One out of two extracts from different poems from the prescribed reader, each followed by two or three questions to test the local and global comprehension of the set text. (3 Marks)

D6 – One out of two short answer type questions on the interpretation of themes and ideas contained in the poems to be answered in about 20-25 words each. (3 Marks)

D7 and D8 – Two out of three short answer type questions on the appreciation of the poems. (4 Marks) (15 Marks)

D9 – One out of two questions from supplementary Materials to interpret, evaluate and analyse character, the plot or situations occurring in the lessons to be answered in about 100 words. (7 Marks)

D10 & D11 – Two out of three short answer type questions of interpretative and evaluative nature based on lessons (2 × 2 = 4 Marks)

D12 & D13 – Two out of three short answer type questions based on factual aspects of the lessons. (2 × 2 = 4 Marks)

Book Prescribed:

1. Text Book – The Spring Blossom
2. WorkBook – The Spring Blossom

Compiled by M.P. Rajya Shiksha Kendra and Published by M.P. Text Book Corporation.

MP Board Class 10th Special English Syllabus & Marking Scheme

Time: 3.00 Hours
Maximum Marks: 100

Unitwise Weightage

Section A: Reading (30 Marks)

Three unseen passages with a variety of comprehension questions, including 5 marks for word attack skills such as word formation and inferring meaning: (54 Periods)

1. About 150 words in length (8 marks)
2. About 200 words in length (8 marks)
3. About 300 words in length (14 marks)

The total of the passages will be about 650 words. The passage will include one each of the following types:
Factual passage, e.g., instruction, description, report.
Discursive passage involving opinion, e.g., argumentative, persuasive or interpretative text.
Literary passage, e.g., extract from fiction, drama, poetry essay or biography.
In the case of a poetry extract, the text may be shorter than 150 words.

Section B: Writing (30 Marks)

Four writing tasks as indicated below: (54 Periods)

1. A linguistically controlled task, where a student builds up a composition with guidance (5 marks)
2. Short composition of not more than 50 words, e.g., a note or notice, message, telegram, short postcard.
3. Composition based on a verbal and/or visual stimuli such as an advertisement; notice, newspaper cutting, table, diary extracts, notes, letter or other forms of correspondence (10 marks)
4. Composition based on a verbal and/or visual stimuli such as a diagram, picture, graph, map, cartoon, or flow chart. (10 marks)

One of the longer (10 marks) compositions will draw on the thematic content of the Main Course Book…
At least one task will involve the production of a form of correspondence, e.g., a letter, postcard, note or notice.
One task will involve the production of discursive text in which the student is required to express his/her point of view on the topic given.

Section C: Grammar (15 Marks)

A variety of short questions involving the use of particular structures within a context (i.e., not in isolated sentences). Test types will include, for example., cloze (gap filling exercise with blanks at regular interval), sentences-completion, sentence-recording, editing, dialogue. Completion and sentence transformation. (27 Periods)
The grammar syllabus will be sampled each year, with marks allotted for different areas of the content of the syllabus prescribed for class IX as well as X as detailed below:
By the end of the course, students should be able to use the following accurately and appropriately in context.

1. Verbs forms
Tenses: Present/Past forms, Simple/Continuous forms, Perfect forms, Future time reference, Modals, Active and Passive Voice, Subject-Verb agreement, Non-finite verb forms (infinitive and participles)

2. Sentence Structure
Connectors, Types of Sentences: affirmative/interrogative sentences, negation exclamations, Types of phrases and clauses, Finite and non-finite clauses, noun clauses and phrases, adjective clauses and phrases, adverb and phrases, Indirect Speech, Comparison, Nominalisation

3. Other Areas
Determiners, Pronouns, Prepositions.

1. Two extracts from different poems from the prescribed readers, each followed by two or three questions to test local and global comprehension of the set text. Each extract will carry four marks. (8 marks)
2. One or two questions based on one of the drama texts from the prescribed reader to test local and global comprehension of the set text. An extract may or may not be used. (5 marks)
3. One questions based on one of the prose text from the prescribed reader to test global comprehension and extrapolation beyond the set text. (4 marks)
4. One extended questions based on one of the prose texts from the prescribed reader to test global comprehension and extrapolation beyond the set text. (8 marks)

Questions will test comprehension at different levels; literal inferential and evaluative.
At the end of class X, the Board’s three-hour examination will test reading and writing skills specified in the teaching/testing objectives together with representative samples of the literature and grammar objectives. The structure of the class X examination paper will be in accordance with the sample paper given in Grammar WorkBook.

Book Prescribed:

1. Textbook – The Rainbow
2. Workbook – The Rainbow

Compiled by M.P. Rajya Shiksha Kendra and Published by M.P. Text Book Corporation. Bhopal.

MP Board Class 10 Special English Blue Print of Question Paper

You can download MP Board Class 10th English Blueprint and Marking Scheme 2019-2020 in Hindi and English medium.

MP Board Class 10 General English Blue Print of Question Paper

MP Board Class 10 Special English Format of Question Paper Design

MP Board Class 10 General English Format of Question Paper Design

We hope the given MP Board Class 10th English Solutions Guide Pdf Free Download of Books The Rainbow, The Spring Blossom Solutions, Questions and Answers, Notes, Summary will help you. If you have any queries regarding Madhya Pradesh Syllabus MP Board Class 10 English Book Solutions Pdf of The Rainbow, The Spring Blossom Textbook Workbook Special and General English Class 10th Solutions, drop a comment below and we will get back to you at the earliest.

MP Board Class 10th Sanskrit Solutions Guide Pdf Free Download संस्कृत दूर्वा are part of MP Board Class 10th Solutions. Here we have given Madhya Pradesh Syllabus MP Board Durva Sanskrit Book Class 10 Solutions Pdf.

Students can also download MP Board 10th Model Papers to help you to revise the complete Syllabus and score more marks in your examinations.

MP Board Class 10 Science Solutions

MP Board Class 10th Sanskrit Book Solutions Durva

Durva Sanskrit Book Class 10 Solutions

• Chapter 1 शिवसङ्कल्पमस्तु (पद्यम्) (यजुर्वेदात्)
• Chapter 2 न्यग्रोधवृक्षः (कथा) (कथावल्लरीतः)
• Chapter 3 महर्षिः दयानन्दः (गद्यम्) (सङ्कलितम्)
• Chapter 4 सुभाषितानि (पद्यम्) (सङ्कलितम्)
• Chapter 5 कुटुम्बानुरक्तिः (नाट्यांशः) (मध्यमव्यायोगात्)
• Chapter 6 यशः शरीरम् (कथा) (सङ्कलिता)
• Chapter 7 विश्वभारतीयम् (संवादः) (सङ्कलितः)
• Chapter 8 सद्वृत्तम् (गद्यम्) (चरकसंहितातः)
• Chapter 9 मदपरिणामः (कथा) (पञ्चतन्त्रात्)
• Chapter 10 आह्वानम् (गीतम्) (सङ्कलितम्)
• Chapter 11 महाभिनिष्क्रमणम् (नाट्यांशः) (वाल्मीकिरामायणतः)
• Chapter 12 वर्षावर्णनम् (पद्यम्) (वाल्मीकिरामयनतः)
• Chapter 13 महाभारते विज्ञानम् (गद्यम्) (सङ्कलितम्)
• Chapter 14 समयस्य सदुपयोगः (संवादः) (सङ्कलितः)
• Chapter 15 भगीरथः (कथा) (सङ्कलिता)
• Chapter 16 ब्रह्मवादिनी मैत्रेयी (गद्यम्) (सङ्कलितम्)
• Chapter 17 चाणक्यनीतिः (पद्यम्) (चाणक्यनीतितः)
• Chapter 18 महाराजः छत्रसालः (गद्यम्) (सङ्कलितम्)
• Chapter 19 गुरुदक्षिणा (पद्यम्) (रघुवंशात्)
• Chapter 20 विदिशा (पत्रम्) (विदिशादिशातः)
• Chapter 21 सूक्तयः (स्फुट) (सङ्कलिताः)

MP Board Class 10th Sanskrit व्याकरणखण्डः

• सन्धि-प्रकरण
• समास-प्रकरण
• कारक एवं उपपद विभक्ति प्रकरण
• प्रत्यय-प्रकरण
• अव्यय-प्रकरण
• संख्या बोध प्रकरण
• समय ज्ञान-प्रकरण
• अशुद्धिसंशोधन-प्रकरण
• उपसर्ग-प्रकरण
• शब्द रूप-प्रकरण
• धातु रूप-प्रकरण
• पत्र-लेखन प्रकरण
• संवाद-लेखनम्
• चित्र-आधारितम् वर्णनम्
• अपठित-अवबोधन प्रकरण
• कथाक्रमसंयोजनम्
• निबन्ध-लेखन प्रकरण

MP Board Class 10th Sanskrit Syllabus

‘क’ खण्डः (अपठितअवबोधनम्)
(सरल-गद्यांश-आधारितं कार्य-गद्यांशद्वयम्)

1. 40-50 शब्दपरिमितः गद्यांशः (एकसरलगद्यांशः)

• एकपदेन पूर्णवाक्येन च प्रश्नोत्तारिण।
• भाषिक-कार्यम्।

2. 80-130 शब्दपरिमितः गद्यांशः (एकसरलगद्यांशः) (सरलकथा-घटनावर्णनं वा)

• एकपदेन पूर्णवाक्येन च प्रश्नोत्तराणि।
• समुचितशीर्षकप्रदानम्।
• व्युत्क्रमेण कथालेखनमपि।

भाषिक-कार्यम्

• वाक्ये क्रियापदचयनम्।
• विशेषण-विशेष्य-अन्टितिः।
• अनुच्छेदे प्रदत्तानां पर्याय-विलोमपदचयनम्।
• विशेषाः विशेष्य, कर्ता-क्रिया, अन्विति, पर्यायवाचिनः विलोमशब्दाः संज्ञास्थाने, सर्वनाम-प्रयोगः।
• कर्तृक्रिया अन्वितिः।
• संज्ञास्थाने सर्वनामप्रयोग अथवा सर्वनामस्थाने संज्ञाप्रयोगः।

‘ख’ खण्डः (अनुप्रयुक्तव्याकरणम्)
(प्रस्तावितपाठ्यपुस्तकाधारितम्)

1. सङ्केताधारितम् आनौपचारिकपत्रम्।
2. सङ्केताधारितं संवादलेखनम्।
3. चित्राधारितं वर्णनम्।

‘ग’ खण्डः (अनुप्रयुक्तव्याकरणम्)
(प्रस्तावितपाठ्यपुस्तकाधारितम्)

1. संधिकार्यम

• स्वरसन्धिः – दीर्घः, गुणः, वृद्धिः, यण, अयादि, पूर्वरूपम्।
• व्यञ्जनसन्धिः – परसवर्णः, छत्वः, तुक्, आगमः, अनुस्वरः, वर्गीयप्रथमाक्षराणां तृतीयवणे परिवर्तनम्, प्रथमवर्णस्य पञ्चं वर्णे परिवर्तनम्।
• विसर्गसन्धिः – विसर्गस्य उत्वं. रत्वं, लोपः विसर्गस्थाने स, श।

2. समास: (वाक्येषु समस्तपदानां विग्रहः विग्रहपदानां च समासः)

• तत्पुरुषः (विभक्तिः नन्, उपपदः)।
• कर्मधारयः (विशेषणम्-विशेष्यम्, उपमान-उपमेयम्)।
• द्विग:
• द्वन्द्वः (इतरेतर, समाहारः एकशेषः)।
• बहुव्रीहिः (समानाधिकरणम्)
• अव्ययीभावः (अनु, उप, स, निर, प्रति, यथा।)

3. कारकाणां प्रयोगः (सोदाहरणम्)।

4. प्रत्ययाः अधोलिखितप्रत्यययोगैः वाक्यसंयोजनं रिक्तस्थानपूर्तिः।

• कृदन्ताः – तव्यत्, अनीयर्, शतृ, शानच्, क्त, क्तवतु, क्त्वा, ल्यप्, तुमुन्, क्तिन्।
• तद्धिता – मतुप, इन्, ठक्, त्व, त्रल्।
• स्त्रीप्रत्ययाः – टाप, डीए, ङीप्, ङीन्।

5. अन्यपदानि (कथयाम् अनुच्छेदे संवादे वा अव्यानां प्रयोगः)।
अपि, इस, मा, इतस्ततः, यत्-तत्, अत्र-तत्र-तत्र, यदा-कदा, यथा-तथा, यावत्-तावत्, शनै।

6. घाटिका-चित्र- अङ्गानां स्थाने समयनेखम् (सामान्य, संपाद, मार्थ, पादोन)

7. सङ्ख्या एकतः पञ्चपर्यन्तं वाक्यप्रयोगः, एकतः शत्पर्यन्तम् सङ्ख्याज्ञानम्।

8. वचन-लिङ्ग-पुरुष-लकार-दृष्टया संशोधनम्।

9. उपपदविभक्तिनां प्रयोगः।

• द्वितीय: अभितः, परितः, उभयतः, समया, निकषा, प्रति, धिक्, विना।
• तृतीया: विना, अलम्, सह, हीनः।
• चतुर्थी: नमः, स्वस्ति, स्वाहा, स्वधा, अलम्, रुच, दा, क्रुध, स्पृह, असूय।

10. प्रादयः द्वाविंशतिः उपसर्गाः।

11. शब्दरूपाणि

• अजन्त: राम, रमा, कवि, साधु, पितृ।
• हलन्त: नामन्, भवत्, राजन्।
• सर्वनाम: अस्मद्, युष्मद्, तत्, एतत्, किम्, लिङ्ग त्रये।

12. धातुरूपाणि

• पञ्चलकाराः लट्, लोट्, लङ्, विधिलिङ्, लुट च।
• भ्वादिगणीयः (प्रथमगणः) धातवः।
• परस्मैपदी-भू (भव), गम् (गच्छ्), दृश् (पश्य), पच्, पा (पिब्)।
• आत्मनेपदी-लभ, सेव्, वृध, वृत्।
• उभयपदी-नी, ह, याच्।

‘घ’ खण्डः (पठितअवबोधनम्)

1. पठित-सामग्रीम् आधृत्य अवबोधनकार्यम्

• गद्यांशः
• पद्यांशः
• नाट्यांशः
• प्रति-अंशम् आधारितम् अवबोधनकार्यम् (एकपदेन पूर्णवाक्येन च प्रश्नोत्तराणि, रिवतस्थानपूर्तिः)।
• भाषिक-कार्यम्।
  (अ) वाक्ये क्रियापदचयनम्।
  (ब) कर्तृक्रिया-अन्वितिः
  (स) विशेषण-विशेष्य-अन्वितिः।
  (द) संज्ञास्थाने सर्वनाम-प्रयोगः अथवा सर्वनाम स्थाने संज्ञाप्रयोगः।
  (ङ) अनुच्छेदे प्रदत्तानां पर्याय-विलोम-पदचयनम्।

2. भावावबोधनम् (अंशद्वयम्)
(रिक्तस्थानपूर्ति द्वारा, विकल्पचयनेन, शुद्ध-अशुद्धमाध्येन, सूक्तिमाध्यमे च)।
3. अन्वये रिक्तस्थानपूर्तिः।
4. प्रश्ननिर्माणम्
5. क्रमरहित कथाक्रमसंयोजनं कथापूर्तिः वा।
6. सन्दर्भशब्दानां प्रयोगः शब्दार्थ-मेलनम् च (उत्तराणि केवलं संस्कतेन लेखितव्यानि)।

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MP Board Class 10th Books Solutions Guide Pdf download all subjects in both Hindi Medium and English Medium are part of MP Board Solutions. Here we have given NCERT Madhya Pradesh Syllabus MP Board Class 10 Text Book Solutions Pdf.

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In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 समान्तर श्रेढ़ियाँ Ex 5.2 Pdf, 5.2 Class 10 Hindi Medium, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 समान्तर श्रेढ़ियाँ Ex 5.2

Class 10 Math Chapter 5.2 In Hindi प्रश्न 1.
निम्नलिखित सारणी में रिक्त स्थानों को भरिए, जहाँ A.P. का प्रथम पद a, सार्वान्तर d और n वाँ पद a_n है :
हल:

हल:
(i) ∵ a_n = a + (n – 1) × d
⇒ a_n = 7 + (8 – 1) × 3
⇒ a_n = 7 + 7 × 3
⇒ a_n = 7 + 21 = 28

(ii) ∵ a_n = a + (n – 1) × d
⇒ 0 = – 18 + (10 – 1) × d
⇒ 0 = – 18 + 9d
⇒ 9d = 18
⇒ d = \(\frac { 18 }{ 9 } \) = 2.

(iii) ∵ a_n = a + (n – 1) × d
⇒ -5 = a + (18 – 1) (-3)
⇒ -5 = a+ 17(-3)
⇒ -5 = a – 51
⇒ a = 51 – 5 = 46.

(iv) ∵ a_n = a + (n – 1) × d
⇒ 3.6 = 18.9 + (n – 1) × 25
⇒ 3.6 = – 18.9 + 2.5n – 2.5
⇒ 2.5n = 18.9 + 3.6 + 2.5
⇒ 2.5 n = 25.0
⇒ n = \(\frac { 25 }{ 2.5 } \) = 10

(v) ∵ a_n = a + (n – 1) × d
⇒ a_n = 35 + (105 – 1) × 0
⇒ a_n = 35 + 104 × 0
⇒ a_n = 3.5
अतः अत: a_n = 3.5

Exercise 5.2class 10 In Hindi प्रश्न 2.
निम्नलिखित में सही उत्तर चुनिए और उसका औचित्य दीजिए:
(i) AP: 19,7, 4 ….. 30 वाँ पद है :
(a) 97
(b) 77
(c) -77
(d) -87

(ii) AP -3, –\(\frac { 1 }{ 2 } \), 2, …….. का 11 वाँ पद है:
(a) 28
(b) 22
(c) -38
(d) -48 \(\frac { 1 }{ 2 } \)
हल:
(i) सही उत्तर (C) -77 है, क्योंकि a = 10, d = -3, n = 30
एवं a_n = a + (n – 1)d ⇒ a₃₀ = 10 + (30 – 1) (-3)
⇒ a₃₀ = 10 – 29 × 3 ⇒ 10 – 87 = -77

(ii) सही उत्तर (B) 22 है, क्योंकि a = -3, d = 2\(\frac { 1 }{ 2 } \), n = 11
एवं a_n = a + (n – 1)d ⇒ a_n = -3 + (11 – 1) (2.5)
⇒ a₁₁ = -3 + 10 × 2.5 = -3 + 25 = 22

Class 10 Math 5.2 In Hindi प्रश्न 3.
निम्नलिखित सामान्तर श्रेढ़ियों में रिक्त स्थानों (boxes) के पदों को ज्ञात कीजिए।
(i) 2, [], 26
(ii) [], 13, [], 3
(iii) 5, [], [], 9\(\frac { 1 }{ 2 } \)
(iv) -4, [], [], [], [], 6
(v) [], 38, [], [], [],-22
हल:
(i) प्रश्नानुसार, a = 2, a₃ = 26 एवं n = 3.
चूंकि an = a + (n – 1) (d)
⇒ 26 = 2 + (3 – 1)d
⇒ 26 = 2 + 2d ⇒ 2d = 26 – 2 = 24 ⇒ d = \(\frac { 24 }{ 2 } \) = 12
अतः रिक्त स्थान a₂ = a + d = 2 + 12 = 14
अत: अभीष्ट रिक्त स्थान (box) में पद 14 होगा।

(ii) प्रश्नानुसार, a₂ = 13 एवं a₄ = 3
⇒ 13 = a + (2 – 1)d ⇒ a + d = 13 ….(1)
एवं 3 = a+ (4 – 1)d ⇒ a + 3d = 3 …..(2)
⇒ 2d = 3 – 13 = – 10 ⇒ d = –\(\frac { 10 }{ 2 } \) = -5
[समी. (2) – समी. (1) से]
d = -5 का मान समीकरण (1) में रखने पर,
a – 5 = 13 ⇒ a = 13 + 5 = 18
एवं a₃ = a + (3 – 1) (-5) = 18 + 2(-5) = 18 – 10 = 8
अतः अभीष्ट रिक्त स्थान में पद क्रमशः 18 एवं 8 होंगे।

(iii) प्रश्नानुसार, a = 5 एवं a₄ = 9 \(\frac { 1 }{ 2 } \)
⇒ 9\(\frac { 1 }{ 2 } \) = 5 + (4 – 1) (d) ⇒ 9\(\frac { 1 }{ 2 } \) = 5 + 3d
⇒ 3d = 9 \(\frac { 1 }{ 2 } \) – 5 = 4 \(\frac { 1 }{ 2 } \) ⇒ d = \(\frac { 1 }{ 3 } \) × 4\(\frac { 1 }{ 2 } \) = 1\(\frac { 1 }{ 2 } \)
⇒ a2 = 5 + 1 \(\frac { 1 }{ 2 } \) = 6\(\frac { 1 }{ 2 } \)
एवं a3 = 5 + 2 × 1 \(\frac { 1 }{ 2 } \) = 5 + 3 = 8
अत: अभीष्ट रिक्त स्थानों के अभीष्ट पद क्रमशः 6\(\frac { 1 }{ 2 } \) एवं 8 हैं।

(iv) प्रश्नानुसार, a = – 4 एवं a₆ = 6
⇒ 6 = -4 + (6 – 1)d = 6 = -4 + 5d
⇒ 5d = 6 + 4 = 10 ⇒ d = \(\frac { 10 }{ 5 } \) = 2
अब a₂ = -4 + 2 = -2
a₃ = -4 + 2 × 2 = -4 + 4 = 0
a₄ = -4 + 3 × 2 = -4 + 6 = 2
a₅ = -4 + 4 × 2 = -4 + 8 = 4
अतः अभीष्ट पद क्रमशः – 2,0,2 एवं 4 हैं।

(v) प्रश्नानुसार, a₂ = 38 एवं a₆ = -22
⇒ 38 = a + d ⇒ a + d = 38 ….(i)
एवं -22 = a + 5d ⇒ a + 5d = – 22 ….(ii)
⇒ 4d = -60 [समीकरण (2)- समीकरण (1) से]
⇒ d = – \(\frac { 60 }{ 4 } \) = -15
d =-15 का मान समीकरण (1) में रखने पर,
a + (-15) = 38 ⇒ a = 38 + 15 = 53
अब a₃ = 53 + 2 × (-15) = 53 – 30 = 23
a₄ = 53 + 3 × (-15) = 53 – 45 = 8
a₅ = 53 + 4 × (-15) = 53 – 60 = -7
a₆ = 53 + 5 × (- 15) = 53 – 75 = – 22
अत: अभीष्ट रिक्त स्थान में पद क्रमशः 53, 23, 8, -7 होंगे।

5.2 Class 10 In Hindi प्रश्न 4.
AP : 3, 8, 13, 18, ……….. का कौन-सा पद 78 है?
हल:
प्रश्नानुसार, a = 3,d = 8 – 3 = 5, a_n = 78
एवं a_n = a + (n – 1) × d
⇒ 78 = 3 + (n – 1) × 5 = 3 + 5n – 5
⇒ 5n = 78 + 5 – 3 = 80
⇒ n = \(\frac { 80 }{ 5 } \) = 16
अत: अभीष्ट 16वाँ पद 78 है।

10 Class Math 5.2 In Hindi प्रश्न 5.
निम्नलिखित समान्तर श्रेढ़ियों में से प्रत्येक श्रेणी में कितने पद हैं?
(i) 7, 13, 19, ………. 205
(ii) 18, 15\(\frac { 1 }{ 2 } \),13, …..(-47)
हल:
(i) चूँकि A.P : 7, 13, 19, ……. 205 (दी गयी है।)
प्रश्नानुसार, a = 7,d = 13 – 7 = 6 एवं a_n = 205
चँकि a_n = a + (n – 1) (d)
⇒ 205 = 7 + (n – 1) (6)
⇒ 205 = 7 + 6n – 6
⇒ 6n = 205 + 6 – 7 = 204
⇒ n = \(\frac { 204 }{ 6 } \) = 34
अतः श्रेणी में अभीष्ट 34 पद हैं।

(ii) चूँकि AP : 18, 15\(\frac { 1 }{ 2 } \), 13, ….., (-47) (दी गयी है)
प्रश्नानुसार, a = 18, d = 15, – 18 = -2\(\frac { 1 }{ 2 } \), एवं a_n = -47
चूँकि a_n = a + (n – 1) (d)
⇒ -47 = 18 + (n – 1) (-2\(\frac { 1 }{ 2 } \))
⇒ – 47 = 18 – 2\(\frac { 1 }{ 2 } \)n + 2\(\frac { 1 }{ 2 } \)
⇒ \(\frac { 5 }{ 2 } \)n = 47 + 18 + 2\(\frac { 1 }{ 2 } \) = 67\(\frac { 1 }{ 2 } \) = \(\frac { 135 }{ 2 } \)
⇒ n = \(\frac { 135 }{ 2 } \) × \(\frac { 2 }{ 5 } \) = 27
अतः श्रेणी में अभीष्ट 27 पद हैं।

5.2 Maths Class 10 Hindi Medium प्रश्न 6.
क्या AP. 11,8, 5, 2 ……… का एक पद — 150 है? क्यों?
हल:
प्रश्नानुसार, a = 11, d = 8 – 11 = -3, a_n = – 150.
चूँकि a_n = a + (n – 1) (d)
⇒ – 150 = 11 + (n – 1) (- 3) = 11 – 3n + 3
⇒ 3n = 150 + 11 + 3 = 164
⇒ n = \(\frac { 164 }{ 3 } \) = 54 \(\frac { 2 }{ 3 } \) जो एक पूर्णांक नहीं है।
अत: दत्त AP का कोई भी पद -150 नहीं होगा।

Class 10 Maths Chapter 5 Exercise 5.2 Hindi Medium प्रश्न 7.
उस AP का 31वाँ पद ज्ञात कीजिए जिसका 11वाँ पद 38 है और 16वाँ पद 73 है। (2019)
हल:
प्रश्नानुसार, n₁₁ = 38 एवं n₁₆ = 73.
⇒ 38 = a + 10d ⇒ a + 10d = 38 …..(1)
एवं 73 = a + 15d ⇒ a + 15d = 73 …..(2)
⇒ 5d = 35 [समीकरण (2) – समीकरण (1) से]
⇒ d = \(\frac { 35 }{ 5 } \) = 7
अब d = 7 का मान समीकरण (1) में रखने पर,
a + 10 × 7 = 38 ⇒ a = 38 – 70 = – 32
अब a₃₁ = a + 30d = – 32 + 30 × 7
⇒ a₃₁ = -32 + 210 = 178
अंतः अभीष्ट 31वाँ पद = 178 है।

Class 10 Maths 5.2 Solutions In Hindi प्रश्न 8.
एक AP में 50 पद है, जिसका तीसरा पद 12 है और अन्तिम पद 106 है। इसका 29वाँ पद ज्ञात कीजिए।
हल:
प्रश्नानुसार, n = 50, a₃ = 12 एवं a₅₀ = 106.
चूँकि a_n = a + (n – 1)d
⇒ 106 = a + 49d ⇒ a + 49d = 106 …..(1)
एवं 12 = a + 2d ⇒ a + 2d = 12 …..(2)
⇒ 47d = 94 [समीकरण (2) – समीकरण (1) से
⇒ d = \(\frac { 94 }{ 47 } \) = 2
d = 2 का मान समीकरण (2) में रखने पर,
a + 2 × 2 = 12 ⇒ a = 12 – 4 = 8
अब n₂₉ = 8 + 28 × 2 = 8 + 56 = 64
अतः अभीष्ट 29वाँ पद = 64 है।

Prashnavali 5.2 Class 10 प्रश्न 9.
यदि किसी AP के तीसरे और नौवें पद क्रमशः 4 और – 8 हैं, तो इसका कौन-सा पद शून्य होगा?
हल:
प्रश्नानुसार, a₃ = 4 एवं a₉ = – 8 है।
⇒ a₃ = a + 2d = 4 …..(1)
एवं a₉ = a + 8d = – 8 …..(2)
⇒ 6d = -12 [समीकरण (2)- समीकरण (1) से]
⇒ d = –\(\frac { 12 }{ 6 } \) = -2
d = – 2 का मान समीकरण (1) में रखने पर
चूँकि a + 2(-2) = 4 ⇒ a = 4 + 4 = 8.
अब a_n = a + (n – 1)d
⇒ 0 = 8 + (n – 1)(-2) ⇒ 0 = 8 – 2n + 2
⇒ 2n = 8 + 2 = 10 ⇒ n = \(\frac { 10 }{ 2 } \) = 5
अतः अभीष्ट पाँचवाँ पद शून्य होगा।

10th Class Math 5.2 In Hindi प्रश्न 10.
किसी AP का 17वाँ पद उसके 10वें पद से 7 अधिक है। इसका सार्वान्तर ज्ञात कीजिए।
हल:
प्रश्नानुसार, (a + 16d) – (a + 9d) = 7
⇒ 16d – 9d = 7 ⇒ 7d = 7 ⇒ d = \(\frac { 7 }{ 7 } \) = 1
अतः d का अभीष्ट मान = 1 है।

MP Board Class 10 Maths Solutions प्रश्न 11.
AP3 , 15, 27, 39, …………. का कौन-सा पद उसके 54वें पद से 132 अधिक होगा?
हल:
प्रश्नानुसार, a = 3, d = 15 – 3 = 12 एवं a_n – a₅₄ = 132
⇒ [3 + (n – 1) (12)] – [3 + (54 – 1) (12)] = 132
⇒ (3 + 12n – 12) – (3 + 53 × 12) = 132
⇒ 12n – 12 – 636 = 132
⇒ 12n = 132 + 12 + 636 = 780
⇒ n = \(\frac { 780 }{ 12 } \) = 65
अतः अभीष्ट 65वाँ पद होगा।

Class 10 Maths Ex 5.2 Solutions In Hindi प्रश्न 12.
दो समान्तर श्रेढ़ियों का सार्वान्तर समान है। यदि इनके 100वें पदों का अन्तर 100 है, तो इनके 1000 वें पदों का अन्तर क्या होगा?
हल:
प्रश्नानुसार, दो समान्तर श्रेढ़ियाँ क्रमशः a, a + d, a + 2d, …………., a + (n – 1)d
एवं b, b + d, b + 2d, …………, b + (n – 1)d
एवं [a+ (100 – 1)d] – [b + (100 – 1)d] = 100
⇒ (a + 99d) – (b + 99a) = 100
⇒ a – b = 100 ….(1)
अब [a + (1000 – 1)d] – [b+ (1000 – 1)d]
= (a + 999d) – (b + 999d)
= a – b = 100 [समीकरण (1) से]
अतः हजारवें पदों का अभीष्ट अन्तर = 100 होगा।

Class 10 Maths Chapter 5.2 In Hindi प्रश्न 13.
तीन अंकों वाली कितनी संख्याएँ 7 से विभाज्य हैं।
हल:
7 से विभाज्य तीन अंकों वाली संख्याओं की सूची है।
105, 112, 119, ……………., 994
जहाँ, a = 105, d = 112 – 105 = 7 एवं a_n = 994
चूँकि an = a + (n – 1)d
⇒ 994 = 105+ (n – 1) × 7
⇒ 994 = 105 + 7n – 7
⇒ 7n = 994 + 7 – 105
⇒ 7n = 1001 – 105 = 896
⇒ n = \(\frac { 896 }{ 7 } \) = 128
अतः 7 से विभाज्य तीन अंकों वाली कुल अभीष्ट संख्याएँ 128 हैं।

Class 10 Maths Chapter 5.2 Hindi Medium प्रश्न 14.
10 और 250 के बीच में 4 के कितने गुणज हैं?
हल:
10 और 250 के बीच 4 के गुणजों की सूची है :
12, 16, 20, 24, …………, 248
जहाँ, a = 12,d = 16 – 12 = 4 एवं a_n = 248
चूँकि a_n = a + (n – 1)d
⇒ 248 = 12 + (n – 1) (4)
⇒ 248 = 12 + 4n – 4 = 4n + 8
⇒ 4n = 248 – 8 = 240
⇒ n = \(\frac { 240 }{ 4 } \) = 60
अत: 10 और 250 के बीच 4 के गुणजों की अभीष्ट संख्या 60 है।

Class 10 Math Chapter 5.2 Solutions In Hindi प्रश्न 15.
n के किस मान के लिए दोनों समान्तर श्रेढ़ियों 63, 65,67,………….. और 3, 10, 17,……… के nवें पद बराबर होंगे?
हल:
चूँकि प्रथम AP का a = 63 एवं d = 65 – 63 = 2, एवं द्वितीय A.P. का a’ = 3 एवं d’ = 10 – 3 = 7 है, तो प्रश्नानुसार,
63 + (n – 1) (2) = 3 + (n – 1) (7)
63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n -4
⇒ 7n – 2n = 61 + 4
⇒ 5n = 65 ⇒ n= \(\frac { 65 }{ 5 } \) = 13
अतः n के अभीष्ट मान 13 के लिए दोनों श्रेढ़ियों के nवें पद बराबर होंगे।

Class 10 Maths Chapter 5 Exercise 5.2 In Hindi Medium प्रश्न 16.
वह AP ज्ञात कीजिए जिसकी तीसरा पद 16 है और 7वाँ पद 5वें पद से 12 अधिक है।
हल:
मान लीजिए कि AP का प्रथम पद a तथा सार्वान्तर d है, तो
प्रश्नानुसार, a₃ = 16 ⇒ a + 2d = 16 ….(i)
एवं a₇ – a₅ = 12 ⇒ (a + 6d) – (a + 4d) = 12
⇒ 2d = 12 ⇒ d = \(\frac { 12 }{ 2 } \) = 6 …..(2)
d का मान समीकरण (2) से समीकरण (1) में रखने पर,
a + 2 × 6 = 16 ⇒ a + 12 = 16 ⇒ a = 16 – 12 = 4
अतः अभीष्ट AP = 4, 10, 16, 22, ……… है।

MP Board Class 10th Maths Solutions प्रश्न 17.
AP 3,8, 13, ……………, 253 में अन्तिम पद से 20वाँ पद ज्ञात कीजिए।
हल:
AP को घटते क्रम में लिखने पर,
253, 248, 243, ………… 13, 8, 3.
जहाँ a = 253 एवं d = (248 – 253) = -5
⇒ a₂₀ = 253 + (20 – 1) (-5)
= 253 + 19 (-5) = 253 – 95
= 158
अतः दत्त AP के अन्तिम पद से अभीष्ट 20वाँ पद = 158 है।

10 Class Ka Math 5.2 In Hindi प्रश्न 18.
किसी AP के चौथे और 8वें पदों का योग 24 है तथा 6वें और 10वें पदों का योग 44 है। इस AP के प्रथम तीन पद ज्ञात कीजिए।
हल:
मान लीजिए a, a + d, a + 2d, a + 3d, ……., समान्तर श्रेढ़ी में हैं, तब प्रश्नानुसार,
∵ a₄ + a₈ = 24
⇒ (a + 3d) + (a + 7a) = 24
⇒ 2a + 10d = 24 ⇒ a + 5d = 12 ……(1)
एवं a₆ + a₁₀ = 44
⇒ (a + 5a) + (a + 9d) = 44
⇒ 2a + 14d = 44 ⇒ a + 7d = 22 …..(2)
⇒ 2d = 10 [समीकरण (2) – (1) से]
⇒ d = \(\frac { 10 }{ 2 } \) = 5
d का मान समीकरण (1) में रखने पर,
a + 5 × 5 = 12 ⇒ a + 25 = 12 ⇒ a = 12 – 25 = – 13
⇒ a2 = a + d = -13 + 5 = -8
एवं a = a + 2d = – 13 + 5 × 2 = – 13 + 10 = -3
अतः दी हुई समान्तर श्रेढ़ी के अभीष्ट प्रथम तीन पद क्रमश: -13, -8 एवं -3 हैं।

Math 5.2 Class 10 In Hindi प्रश्न 19.
सुब्बाराव ने 1995 में ₹ 5,000 के मासिक वेतन पर कार्य प्रारम्भ किया ओर प्रत्येक वर्ष ₹200 की वेतन वृद्धि प्राप्त की। किस वर्ष में उसका वेतन ₹ 7000 हो गया?
हल:
सुब्बाराव के प्रतिवर्ष के वेतन की सूची एक AP का निर्माण करेगी, जिसमें a = ₹ 5,000, d = ₹ 200 एवं an = ₹ 7,000 होगा।
इसलिए प्रश्नानुसार,
a_n = a + (n – 1)d
⇒ 7000 = 5000 + (n – 1) × 200
⇒ 7000 = 5000 + 200n – 200
⇒ 7000 = 4800 + 200n
⇒ 200n = 7000 – 4800 = 2200
⇒ n = \(\frac { 2200 }{ 200 } \) = 11
अत: सुब्बाराव का अभीष्ट वेतन 11वें वर्ष में होगा।

10th Math 5.2 In Hindi प्रश्न 20.
रामकली ने किसी वर्ष के प्रथम सप्ताह में ₹5 की बचत की और फिर अपनी साप्ताहिक बचत में ₹ 1.75 बढ़ाती गयी। यदि वें सप्ताह में उसकी बचत ₹ 20.75 हो जाती है, तो n ज्ञात
कीजिए।
हल:
रामकली के साप्ताहिक बचत की सूची एक AP का निर्माण करती है जिसमें a = ₹5 एवं d = ₹ 1.75 तथा a_n = ₹ 20.75, तो प्रश्नानुसार,
a_n = a + (n – 1)d
⇒ 20.75 = 5 + (n – 1) (1.75)
⇒ 20.75 = 5 + 1.75n – 1.75
⇒ 1.75n = 20.75 + 1.75 – 5
⇒ 1.75n = 22.50 – 5 = 17.50
⇒ n = \(\frac { 17.50 }{ 1.75 } \) = 10
अतः n का अभीष्ट मान = 10 है।

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.2 Pdf, Class 10 Maths Chapter 3 Exercise 3.2 Solutions, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

3.2 Maths Class 10 MP Board Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Let the number of boys = x and Number of girls = y
∴ x + y = 10 …. (1)
Also, Number of girls = [Number of boys] + 4
∴ y = x + 4 ⇒ x – y = – 4 … (2)
Now, from the equation (1), we have :
l₁ : x + y = 10 ⇒ y = 10 – x

And from the equation (2), we have
l₂ : x – y = -4 ⇒ y = x + 4

The lines l₁ and l₂ intersects at the point (3, 7)
∴ The solution of the pair of linear equations is : x = 3, y = 7
⇒ Number of boys = 3
and number of girls = 7
Let the cost of a pencil = ₹ x
and cost of a pen = ₹ y
Since, cost of 5 pencils + Cost of 7 pens = ₹ 50
5 x + 7y = 50 … (1)
Also, cost of 7 pencils + cost of 5 pens = ₹ 46
7x + 5y = 46 …. (2)
Now, from equation (1), we have

And from equation (2), we have

Plotting the points (10, 0), (3, 5) and (-4, 10), we get a straight line l₁ and plotting the points (8, -2), (3, 5) and (0, 9.2) we get a straight line l₂.

These two straight lines intersects at (3, 5).
∴ Cost of a pencil = ₹ 3 and cost of a pen = ₹ 5.

Class 10 Math Chapter 3.2 Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\) find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
Comparing the given equations with
a₁x + b₁y + C₁ = 0, a₂x + b₂y + c₂ = 0, we have

(i) For, 5x – 4y + 8 = 0, 7x + 6y – 9 = 0,
a₁ = 5, b₁ = -4, C₁ = 8; a₂ = 7, b₂ = 6, c₂ = -9

So, the lines are intersecting, i.e., they intersect at a point.

(ii) For, 9x + 3y + 12 = 0,18x + 6y + 24 = 0,
a₁ = 9, b₁ = 3, C₁ = 12; a₂ = 18, b₂ = 6, c₂ = 24

So, the lines are coincident.

(iii) For, 6x – 3y +10 = 0, 2x – y + 9 = 0

So, the lines are parallel.

Exercise 3.2 Class 10 Maths Question 3.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y = 8; 4x – 6y = 9
(iii) \(\frac{3}{2} x+\frac{5}{3} y\) = 7; 9x-10y = 14
(iv) 5x – 3y = 11; -10x + 6y = -22
Solution:
Comparing the given equations with

So, lines are intersecting i.e., they intersect at a point.
∴ It is consistent pair of equations.

(ii) For 2x – 3y = 8, 4x – 6y = 9, 2x – 3y – 8 = 0 and 4x – 6y – 9 = 0

So, lines are parallel i.e., the given pair of linear equations has no solution.
∴ It is inconsistent pair of equations.

(iii)

⇒ The given pair of linear equations has exactly one solution.
∴ It is a consistent pair of equations.

(iv)

So, lines are coincident.
⇒ The given pair of linear equations has infinitely many solutions.
Thus, it is consistent pair of equations

(v)

So, the lines are coincident.
⇒ The given pair of linear equations have infinitely many solutions.
Thus it is consistent pair of equations.

3.2 Maths Class 10 Question 4.
Which of the following pairs of linear equations are consistent / inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5; 2x + 2y = 10
(ii) x – y = 8; 3x – 3y = 16
(iii) 2x + y – 6 = 0; 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
Solution:
(i) For, x + y = 5, 2x + 2y = 10 ⇒ x + y – 5 = 0 and 2x + 2y – 10 = 0


So, lines l₁ and l₂ are coinciding. i.e., They have infinitely many solutions.

(ii) For, x – y = 8,
3x – 3y = 16
⇒ x – y – 8 = 0
and 3x – 3y = 16

so, line are parallel
∴ The pair of linear equations are inconsistent

(iii) For 2x + y – 6 = 0, 4x – 2y – 4 = 0


The pair of linear equations are inconsistent

Consistent And Inconsistent Equations Class 10 Question 5.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let the width of the garden = x m
and the length of the garden = y m
According to question,


The lines l₁ and l₂ intersect at (16, 20).
∴ x = 16 and y = 20
So, Length = 20m, and Width = 16m

Class 10 Maths Chapter 3 Exercise 3.2 Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Let the pair of linear equations be 2x + 3y – 8 = 0
⇒ a₁ = 2, b₁ = 3 and C₁ = -8 and a₂x + b₂y + c₂ = 0.

(i) For intersecting lines, we have:

∴ We can have a₂ = 3, b₂ = 2 and c₂ = – 7
∴ The required equation can be 3x + 2y – 7 = 0

(ii) For parallel lines, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ Any line parallel to 2x + 3y – 8 = 0, can be taken as 2x + 3y -12 = 0

(iii) For coincident lines, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
∴ Any line parallel to 2x + 3y – 8 = 0 can be 2(2x + 3y – 8 = 0)
⇒ 4x + 6y – 16 = 0

Class 10 Maths Exercise 3.2 Solutions Question 7.
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
Since, we have

The lines l₁ and l₂ intersect at (2, 3). Thus, co-ordinates of the vertices of the shaded triangular region are (4, 0), (-1, 0) and (2, 3).

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Class 10 Maths 8.1 Question 1.
In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
In right angle ∆ABC, we have
AB = 24 cm, BC = 7 cm

∴ Using Pythagoras theorem,
AC² = AB² + BC²
⇒ AC² = 24² + 7² = 576 + 49 = 625 = 25²
⇒ AC = 25 cm

Class 10 Math 8.1 Question 2.
In the figure, find tan P – cotR?

Solution:
In right angle ∆PQR
Using the Pythagoras theorem, we get
QR² = PR² – PQ²
⇒ QR² = 13² – 12² = (13 – 12)(13 + 12) = 1 × 25 = 25
∴ QR = \(\sqrt{25}\) = 5 cm
Now, tanP = \(\frac{Q R}{P Q}=\frac{5}{12}\) , cotR = \(\frac{Q R}{P Q}=\frac{5}{12}\)
∴ tanP – cotR = \(\frac{5}{12}-\frac{5}{12}\) = 0.

Ex 8.1 Class 10 Question 3.
If sinA = \(\frac{3}{4}\), calculate cosA and tanA.
Solution:
Let us consider, the right angle ∆ABC, we have
Perpendicular = BC and Hypotenuse = AC

Class 10 Maths Chapter 8 Exercise 8.1 Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
In the right angle triangle ABC, we have 15 cot A = 8

Introduction To Trigonometry Class 10 Exercise 8.1 Question 5.
Given sec θ = \(\frac { 13 }{ 12 } \), calculate all other trigonometric ratios.
Solution:

Class 10th Maths Chapter 8 Exercise 8.1 Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:

From equations (i) and (ii) we get:
\(\frac{\mathrm{CD}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{BF}}\)
⇒ ∆CDA ~ ∆EFB [By SSS similarity]
⇒ ∠A = ∠B Hence Proved

Class 10th Exercise 8.1 Question 7.
If cot θ = \(\frac { 7 }{ 8 }\), evaluate:
(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right)}\)
(ii) cot²θ
Solution:

Class 10th Math Exercise 8.1 Question 8.
If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.
Solution:
In right angled ∆ABC, ∠B = 90°
For ∠A, we have:
Base = AB and perpendicular = BC. Also, Hypotenuse = AC

Maths Class 10 Chapter 8 Exercise 8.1 Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\) find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
In right ∆ABC, ∠B = 90°
For ∠A, we have
Base = AB, Perpendicular = BC,
Hypotenuse = AC

Class 10 Chapter 8.1 Question 10.
In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.
Solution:
In right ∆PQR, Q = 90°

PR + QR = 25 cm and PQ = 5 cm
Let QR = x cm ⇒ PR = (25 – x) cm
∴ By Pythagoras theorem, we have
PR² = QR² + PQ²
⇒ (25 – x)² = x² + 5²
⇒ 625 – 50x + x² = x² + 25
⇒ – 50x = – 600
⇒ x = \(\frac{-600}{-50}\) = 12 i.e., QR = 12 cm
⇒ RP = 25 – 12 = 13 cm

Class 10 Chapter 8.1 Maths Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) secA = \(\frac{12}{5}\) for some value of angle A.
(iii) cosA is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sinθ = \(\frac{4}{3}\) for some angle θ.
Solution:
(i) False
∵ A tangent of an angle is the ratio of sides other than hypotenuse, which may be equal or unequal to each other.
(ii) True
∵ cos A is always less than 1.
∴ \(\frac{1}{\cos A}\) i.e., sec A will always be greater than 1.
(iii) False
∵ ‘cosine A’ is abbreviated as ‘cosA’
(iv) False
∵ ‘cot A’ is a single and meaningful term whereas ‘cot’ alone has no meaning.
(v) False
∵ \(\frac{4}{3}\) is greater than 1 and sinθ cannot be greater than 1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Class 10 Maths Chapter 5 Exercise 5.1 Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic
progression, and why’
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and 8 for each
additional km.
(ii) The amount of air present In a cylinder when a vacuum pump removes 1/4 of the
air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs 150 for
the first metre and rises by 50 for each subsequent metre.
(iv) The amount of money in the account every year, when 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Let us consider, the first term
(T₁) Fare for the first 1 km = ₹ 15
Since, the taxi fare beyond the first 1 km is ₹ 8 for each additional km.
∴ Fare for 2 km = ₹ 15 + 1 × ₹ 8
T₂ = a + 8 [where a = 15]
Fare for 3 km = ₹ 15 + 2 × ₹ 8
⇒ T₃ = a + 16
Fare for 4 km= ₹ 15 + 3 × ₹ 8
⇒ T₄ = a + 24
Fare for 5 km = ₹ 15 + 4 × ₹ 8
⇒ T₅ = a + 32
Fare for n km = ₹ 15 + (n – 1)8
⇒ T_n = a + (n – 1)8
We see that above terms form an AP with common difference 8.

(ii) Let the amount of air present in the cylinder be x

The above terms are not in A.P.

(iii) Here, the cost of digging for first 1 metre = ₹ 150
The cost of digging for first 2 metres = ₹ 150 + ₹ 50 = ₹ 200
The cost of digging for first 3 metres = ₹ 150 + (₹ 50) × 2 = ₹ 250
The cost of digging for first 4 metres = ₹ 150 + (₹ 50) × 3 = ₹ 300
∴ The terms are: 150, 200, 250, 300,…..
Since, 200 – 150 = 50 and 250 – 200 = 50
(200 – 150) (250 – 200) = 50
∴ The above terms form an AP with common difference 50.

(iv)

Exercise 5.1 Class 10 Solutions Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given
as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = 1/2
(y) a = -1.25, d = -0.25
Solution:
(i) ∵ T_n = a + (n – 1)d
∴ For a = 10 and d = 10, we have:
T₁ =10 + (1 – 1) × 10 = 10 + 0 = 10
T₂ = 10 + (2 – 1) × 10 = 10 + 10 – 20
T₃ = 10 + ( 3 – 1) × 10 = 10 + 20 = 30
T₄ = 10 + (4 – 1) × 10 = 10 + 30 = 40
Thus, the first four terms are:
10, 20, 30, 40

(ii) ∵ T_n = a + (n – 1)d
∴ For a = -2 and d = 0,we have:
T₁ = -2 + (1 – 1) × 0 = -2 + 0 = -2
T₂ = -2 + (2 – 1) × 0 = -2 + 0 = -2
T₃ = -2 + (3 – 1) × 0= -2 + 0 = -2
T₄ = -2 + (4 – 1) × 0 = -2 + 0 = -2
∴ Thus, the first four terms are: -2, -2, -2, -2

(iii) ∵ T_n = a + (n – 1)d
For a = 4 and d = -3, we have:
T₁ = 4 + (1 – 1) × (-3) = 4 + 0 = 4
T₂ = 4 + (2 – 1) × (-3) = 4 + (-3) = 1
T₃ = 4 + (3 – 1) × (-3) = 4 + (-6) = -2
T₄ = 4 + (4 – 1) × (-3) = 4 + (-9) = -5
Thus, the first four terms are:
4, 1, -2, -5

(iv)

(v) ∵ T_n = a + (n – 1)d
∴ For a = -1.25 and d = -0.25, we have
T₁ = -1.25 + (1 – 1) × (-0.25) = -1.25 + 0
= -1.25
T₂ = -1.25 + (2 – 1) × (-0.25) = -1.25 + (-0.25) = -1.50
T₃ = -1.25 + (3 – 1) × (-0.25) = -1.25 + (-0.50) = -1.75
T₄ = -1.25 + (4 – 1) × (-0.25) = -1.25 + (-0.75) = -2.0

Class 10th Maths Ex 5.1 Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3,…..
(ii) -5, -1, 3, 7,….
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots .\)
(iv) 0.6, 1.7, 2.8, 3.9, …..
Solution:
(i) We have : 3, 1, -1, -3, …
⇒ T₁ ⇒ 3 = a = 3
T₂ = 1, T₃= -1, T₄= -3
∴ T₂ – T₁ = 1 – 3 = -2
T₄ = T₃ = -3 – (-1) = -3 + 1 = -2 ⇒ d = -2

(ii) We have: -5, -1, 3, 7,….
⇒ T₁ = -5 ⇒ a = -5, T₂ = -1, T₃ = 3, T₄ = 7
∴ T₂ – T₁ = -1 – (-5) = -1 + 5 = 4
and T₄ – T₃ = 7 – 3 = 4 ⇒ d = 4
Thus a = -5, d = 4

(iii)

Class 10 Maths Chapter 5 Question 4.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16,
(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ….
(iii) -1.2, -3.2, -5.2, -7.2, ……
(iv) -10, -6, -2, 2, ….
(v) \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots\)
(vi) 0.2, 0.22, 0.222, 0.2222, ….
(vii) 0, -4, -8, -12,
(ix) 1, 3, 9, 27, …..
(x) o, 2a, 3a, 4a,…
(xi) a, a₂, a₃, a₄,….
(xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\)
(xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots .\)
(xiv) 1², 3², 5², 7², ….
(xv) 1², 5², 7², 73, …..
Solution:
(i)

∴ The given numbers do not form an AP

(ii)

(iii) We have: -1.2, -3.2, -5.2, -7.2,
∴ T₁ = -1.2, T₂ = -3.2, T₃ = -5.2, T₄ = -7.2
T₂ – T₁ = -3.2 + 1.2 = -2
T₃ – T₂ = -5.2 + 3.2 = -2
T₄ – T₃ = -7.2 + 5.2 = -2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = -2 d = -2
∴ The given numbers form an AP such that d = -2.
Now, T₅ = T₄ + (-2) = -7.2 + (-2) = -9.2,
T₆ = T₅ + (-2) = -9.2 + (-2) = -11.2 and
T₇ = T₆ + (-2) = -11.2 + (-2) = -13.2
Thus, d = -2 and T₅ = -9.2, T₅ = -11.2 and T₆ = -13.2

(iv) We have: -10, -6, -2, 2,
T₁ = -10, T₂ = -6, T₃ = -2, T₄ = 2
T₂ – T₁ = -6 + 10 = 4
T₃ – T₂ = -2 + 6 = 4
T₄ – T₃ = 2 + 2 = 4
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 4 ⇒ d = 4
∴ The given numbers form an AP Now, T₅ = T₄ + 4 = 2 + 4 = 6,
T₆ = T₅ + 4 = 6 + 4 = 10,
T₇ = T₆ + 4 = 10 + 4 = 14
Thus, d = 4 and T₅ = 6, T₆ = 10, T₇ = 14

(v) We have :
3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),….

(vi) We have 0.2, 0.22, 0.222, 0.2222,….

∴ The given numbers do not form an AP

(vii) We have : 0, – 4, – 8, – 12,
∴ T₁ = 0, T₂ = -4, T₃ = -8, T₄ = -12 T₂ – T₁ = -4 – 0 = -4
T₃ – T₂ = – 8 + 4 = – 4
T₄ – T₃ = -12 + 8 = -4
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = -4 ⇒ d = -4
∴ The given numbers form an AP
Now, T₅ = T₄ + (- 4) = -12 + (- 4) = -16
T₆ = T₅ + (-4) = -16 + (-4)= -20
T₇ = T₆ + (- 4) = -20 + (- 4) = -24
Thus, d = – 4 and T₅ = -16, T₆ = -20,
T₇ = -24.

(viii)

(ix) We have, 1, 3, 9, 27,….

(x) We have : a, 2a, 3a, 4a,

(xi)

(xii)

(xiii)

(xiv)
We have 1², 3₂, 5₂, 7₂,….

(xv) We have 1₂ , 5₂, 7₂, 7₂,….