MP Board Class 12th Biology Solutions जीव विज्ञान

MP Board Class 12th Biology Solutions Guide Pdf Free Download जीव विज्ञान in both Hindi Medium and English Medium are part of MP Board Class 12th Solutions. Here we have given NCERT Madhya Pradesh Syllabus MP Board Class 12 Biology Book Solutions Jiv Vigyan Pdf.

Students can also download MP Board 12th Model Papers to help you to revise the complete Syllabus and score more marks in your examinations.

MP Board Class 12th Biology Book Solutions in Hindi Medium

MP Board Class 12th Biology Book Solutions in English Medium

MP Board Class 12th Biology Syllabus and Marking Scheme

Latest Syllabus and Marks Distribution Biology Class XII for the academic year 2019 – 2020 Year Examination.

Biology
Class XII

Time : 3 Hours.
Maximum Marks: 70

Unit Wise Division of Marks

Unit Title Period Marks
6. Reproduction 30 14
7. Genetics and Evolution 40 18
8. Biology and Human Welfare 30 14
9. Biotechnology and Its Applications 30 10
10. Ecology and Environment 30 14
Total 160 70

We hope the given MP Board Class 12th Biology Solutions Guide Pdf Free Download जीव विज्ञान in both Hindi Medium and English Medium will help you. If you have any query regarding NCERT Madhya Pradesh Syllabus MP Board Class 12 Biology Book Solutions Jiv Vigyan Pdf, drop a comment below and we will get back to you at the earliest.

MP Board Class 12th Chemistry Solutions रसायन विज्ञान

MP Board Class 12th Chemistry Solutions Guide Pdf Free Download रसायन विज्ञान in both Hindi Medium and English Medium are part of MP Board Class 12th Solutions. Here we have given Madhya Pradesh Syllabus NCERT Solutions for Class 12 Chemistry Book Solutions Rasayan Vigyan Pdf.

Students can also download MP Board 12th Model Papers to help you to revise the complete Syllabus and score more marks in your examinations.

MP Board Class 12th Chemistry Book Solutions in Hindi Medium

MP Board Class 12th Chemistry Book Solutions in English Medium

MP Board Class 12th Chemistry Syllabus and Marking Scheme

Latest Syllabus and Marks Distribution Chemistry Class XII for the academic year 2019 – 2020 Year Examination.

Chemistry
Class XII

Time : 3 Hours.
Maximum Marks: 70

Unit Wise Division of Marks

MP Board Class 12th Chemistry Syllabus and Marking Scheme

MP Board Class 12th Chemistry Syllabus and Marking Scheme

1. Solid State
Classification of solids based on different binding forces : Molecular, ionic, covalent and metallic solids, amorphous and crystalline solids (elementary idea). Unit cell in two-dimensional and three-dimensional lattices, calculation of density of unit cell, packing in solids, packing efficiency, voids, number of atoms per unit cell in a cubic unit cell, point defects, electrical and magnetic properties. Band theory of metals, conductors, semiconductors and insulators and n and p type semiconductors.

2. Solutions
Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions, colligative properties : Relative lowering of vapour pressure, Raoult’s law, elevation in boiling point, depression in freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass, van’t Hoff factor.

3. Electrochemistry
Redox reactions, conductance in electrolytic solutions, specific and molar con-ductivity, variations of conductivity with concentration, Kohlrausch’s law, elec-trolysis and law of electrolysis (elementary idea), dry cell- electrolytic cells and Galvanic cells; lead accumulator, EMF of a cell, standard electrode potential, Nemst equation and its application to chemical cells, Relation between Gibbs energy change and EMF of a cell, fuel cells, corrosion.

4. Chemical Kinetics
Rate of a reaction (Average and instantaneous), factors affecting rate of reaction : concentration, temperature, catalyst; order and molecularity of a reaction, rate law and specific rate constant, integrated rate equations and half-life (only for zero and first order reactions); concept of collision theory (elementary idea, no mathematical treatment). Activation energy, Arrhenius equation. Surface Chemistry

5. Adsorption : Physisorption and chemisorption, factors affecting adsorption of gases on solids.
Catalysis: Homogeneous and heterogeneous, activity and selectivity; enzyme catalysis.
Colloidal state : Distinction between true solutions, colloids and suspension; lyophilic, lyophobic, multimolecular and macromolecular colloids; properties of colloids; Tyndall effect, Brownian movement, electrophoresis, coagulation. Emulsion : Types of emulsions.

6. General Principles and Processes of Isolation of Elements
Principle and methods of extraction; concentration, oxidation, reduction, elec-trolytic method and refining, occurrence and principles of extraction of aluminium, copper, zinc and iron.

7. p-block Elements
Group-15 elements : General indroduction, electronic configuration, occurrence, oxidation states, trends in physical and chemical properties; Nitrogen : Preparation, properties and uses; compounds of nitrogen, preparation and properties of ammonia and nitric acid, oxides of nitrogen (structure only); Phosphorus : Allotropic forms; compounds of phosphorus preparation and properties of phosphine, halides (PC13, PC15) and oxo-acids (elementary idea only).
Group-16 elements : General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; Dioxygen; preparation, properties and uses; classification of oxides; ozone, Sulphur: allotropic forms; compounds of sulphur; preparation, properties and uses of sulphur dioxide; sulphuric acid, industrial process of manufacture, properties and uses, oxo-acids of sulphur (structures only).
Group-17 elements : General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; compounds of halogens; preparation, properties and uses of chlorine, hydrochloric acid and interhalogen compounds, oxo-acids of halogens (structures only). Group-18 elements : General introduction, electronic configuration, occurrence, trends in physical ahd chemical properties, uses.

8. d- and f-block Elements
General introduction, electronic configuration, occurrence and characteristics of transition metals, general trends in properties of the first row transition metals : metallic character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation, preparation and properties of K2Cr207 and KMn04.
Lanthanoids : Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction and its consequences.
Actinoids : Electronic configuration, oxidation states and comparison with lanthanoids.

9. Co-ordination Compounds
Co-ordination compounds : Introduction, ligands, co-ordination number, colour, magnetic properties and shapes. IUPAC nomenclature of mononuclear co-ordination compounds. Bonding; Werner’s theory, VBT and CFT structure and stereo-isomerism, importance of co-ordination compounds (in qualitative inclusion, extraction of metals and biological system).

10. Haloalkanes and Haloarenes
Haloalkanes : Nomenclature, nature of C—X bond, physical and chemical properties, mechanism of substitution reactions, optical rotation.
Haloarenes : Nature of C—X bond, substitution reactions (Directive influence of halogen in monosubstituted compounds only). Uses and environmental effects of dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT.

11. Alcohols, Phenols and Ethers
Alcohols: Nomenclature, methods of preparation, physical and chemical prop-erties of (primary alcohols only); identification of primary, secondary and tertiary alcohols; mechanism of dehydration, uses with special reference to methanol and ethanol.
Phenols : Nomenclature; methods of preparation, physical and chemical prop-erties, acidic nature of phenol, electrophilic substitution reactions, uses of phenols.
Ethers : Nomenclature; methods of preparation, physical and chemical properties, uses.

12. Aldehydes, Ketones and Carboxylic Acids
Aldehydes and Ketones : Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic ad-dition, reactivity of alpha hydrogen in aldehydes; uses.
Carboxylic acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses.

13. Organic Compounds Containing Nitrogen.
Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, identification of primary, secondary and tertiary amines.
Cyanides and isocyanides : will be mentioned at relevant places in context. Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry.

14. Bio-molecules
Carbohydrates : Classification (aldoses and ketoses), monosaccharides (glucose and fructose), D-L configuration, oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen); importance.
Proteins : Elementary idea of -amino acids, peptide bond, polypeptides, proteins, structure of proteins : primary, secondary, tertiary and quaternary structures (qualitative idea only), denaturation of proteins, enzymes. Hormones : elementary idea excluding structure.
Vitamins : Classification and functions.
Nucleic acids : DNA and RNA.

15. Polymers
Classification : Natural and synthetic, methods of polymerization (addition and condensation), copolymerization, some important polymers : Natural and synthetic like polythene, nylon, polyesters, bakelite, rubber. Biodegradable and non-biodegradable polymers.

16. Chemistry in Everyday Life
Chemicals in medjcines : Analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamines. Chemicals in food : Preservatives, artificial sweetening agents, elementaiy idea of antioxidants.
Cleansing agents : Soaps and detergents, cleansing action.

We hope the given MP Board Class 12th Chemistry Solutions Guide Pdf Free Download रसायन विज्ञान in both Hindi Medium and English Medium will help you. If you have any query regarding Madhya Pradesh Syllabus MP Board Class 12 Chemistry Book Solutions Rasayan Vigyan Pdf, drop a comment below and we will get back to you at the earliest.

MP Board Class 12th Hindi Solutions मकरंद, स्वाति

MP Board Class 12th Hindi Solutions Guide Pdf Free Download of हिंदी मकरंद, स्वाति are part of MP Board Class 12th Solutions. Here we have given Madhya Pradesh Syllabus MP Board Class 12 Hindi Book Solutions Makrand, Swati Pdf Special and General.

Students can also download MP Board 12th Model Papers to help you to revise the complete Syllabus and score more marks in your examinations.

MP Board Class 12th Hindi Book Solutions Makrand

Here we have given MP Board Class 12 General Hindi Makrand Solutions Hindi Samanya Kaksha 12 (मकरंद हिंदी सामान्य कक्षा 12).

Makrand Hindi Book Class 12 Solutions

MP Board Class 12th General Hindi व्याकरण

MP Board Class 12th General Hindi Model Question Paper

MP Board Class 12th Hindi Book Solutions Swati

Here we have given MP Board Class 12 Special Hindi Swati Solutions Hindi Vishisht Kaksha 12 (स्वाति हिंदी विशिष्ट कक्षा 12).

Swati Hindi Book Class 12 Solutions

पद्य खण्ड : भाव सारांश,सम्पूर्ण पद्यांशों की सप्रसंग व्याख्या

गद्य खण्ड : सारांश, शब्दार्थ, व्याख्या एवं अभ्यास

MP Board Class 12th Hindi सहायक वाचन Solutions

MP Board Class 12th Special Hindi व्याकरण

भाषा-बोध

काव्य-बोध

अपठित-बोध

MP Board Class 12th Special Hindi Model Question Paper

MP Board Class 12 Special Hindi Syllabus & Marking Scheme

समय : 3 घण्टा
पूर्णांक : 100

क्रम विषय सामग्री अंक कालखण्ड
1. पद्य खण्ड-
पद्य साहित्य का विकास-आधुनिक काव्य प्रवृत्तियाँ, कवि परिचय, व्याख्या, सौन्दर्य बोध, विषय-बोध एवं मूल्य-बोध पर प्रश्न
(4 + 23 = 27) 40
2. गद्य खण्ड-
गद्य साहित्य : विविध विधाएँ-निबन्ध, नाटक, कहानी, लेखक परिचय, व्याख्या, विचार-बोध, विषय-बोध पर प्रश्न
(4 + 19 = 23) 35
3. सहायक वाचन-
विविध पाठों की विषय-वस्तु तथा आँचलिक भाषा के पाठों पर प्रश्न
10 15
4. भाषा बोध-
वाक्य बोध, शुद्ध वाक्य रचना, वाक्य परिवर्तन, भाव-विस्तार, मुहावरे/लोकोक्तियाँ, बोली, विभाषा. मातृभाषा, राजभाषा, राष्ट्रभाषा
10 15
5. काव्य बोध-
रस परिचय, अंग, रस भेद उदाहरण सहित, अलंकार, छन्द, काव्य की परिभाषा एवं काव्य के भेद, काव्य गुण
10 15
6. अपठित बोध 05 10
7. पत्र-लेखन 05 10
8. निबंध-लेखन 10 20
पुनरावृत्ति 20
सम्पूर्ण योग (पूर्णांक) 100 180

पद्य खण्ड-
पद्य साहित्य का विकास : आधुनिक काल की प्रमुख प्रवृत्तियाँ- छायावाद, रहस्यवाद, प्रगतिवाद, प्रयोगवाद, नई कविता पर आधारित दो प्रश्न, पद्य पाठों पर आधारित कवि का साहित्यिक परिचय (2 + 2 = 4 Marks)
(रचनाएँ, काव्यगत विशेषताएँ-भावपक्ष, कलापक्ष) पर एक प्रश्न (5 Marks)
दो पद्यांश में से एक पद्यांश की प्रसंग सहित व्याख्या (5 Marks)
सौन्दर्य-बोध पर आधारित दो प्रश्न (4 + 3 = 7 Marks)
विषय-वस्तु एवं मूल्य-बोध पर दो प्रश्न (3 + 3 = 6 Marks)

गद्य खण्ड-
गद्य साहित्य : विविध विधाओं (निबन्ध, कहानी, नाटक) पर एक प्रश्न (4 Marks)
लेखक परिचय (रचनाएँ, भाषा-शैली) पर एक प्रश्न (5 Marks)
दो में से एक गद्यांश की सप्रसंग व्याख्या पर एक प्रश्न (5 Marks)
विचारबोध पर आधारित दो प्रश्न ( 3 + 2 = 5 Marks)
विषय-वस्तु पर आधारित एक प्रश्न (4 Marks)

सहायकवाचन-
पाठों की विषय-वस्तु पर दो प्रश्न (3 + 3 = 6 Marks)
आँचलिक भाषा के पाठों पर दो प्रश्न (2 + 2 = 4 Marks)

भाषा बोध-
वाक्य के प्रकार-रचना के आधार पर, अर्थ के आधार पर, वाच्य के आधार पर, वाक्य परिवर्तन पर एक प्रश्न (2 Marks)
बोली, विभाषा, मातृभाषा, राजभाषा, राष्ट्रभाषा पर एक प्रश्न (2 Marks)
मुहावरे/लोकोक्तियों का वाक्य में प्रयोग पर एक प्रश्न (2 Marks)
भाव विस्तार पर एक प्रश्न (4 Marks)

काव्य बोध-
काव्य की परिभाषा, काव्य भेद, प्रबन्ध काव्य, मुक्तक काव्य पर एक प्रश्न
काव्य गुण पर एक प्रश्न
रस-परिचय, अंग, भेद एवं उदाहरण पर एक प्रश्न
अलंकार-यमक,श्लेष, व्याजस्तुति, व्याज निन्दा, विभावना, विशेषोक्ति पर एक प्रश्न
छन्द-छप्पय, कवित्त, सवैया (मत्तगयन्द, दुर्मिल) पर एक प्रश्न

अपठित बोध-
गद्यांश/पद्यांश (शीर्षक, सारांश एवं प्रश्न) पर आधारित एक प्रश्न (5 Marks)

पत्र-लेखन-
आवेदन-पत्र, स्थानीय निकाय से सम्बन्धित पत्र, सम्पादक के नाम पत्र-अपनी रचना प्रकाशन हेतु, समसामयिक विषयों पर परिचर्चा, सम-सामयिक समस्याओं के समाधान हेतु पत्र पर आधारित एक प्रश्न (5 Marks)

निबन्ध-लेखन-
विचारात्मक, भावात्मक, ललित निबन्ध एवं सम-सामयिक विषयों पर निबन्ध-लेखन पर एक प्रश्न (10 Marks)

प्रायोजना कार्य
1. क्षेत्रीय बोली-पहेलियाँ, चुटकुले, लोकगीत, लोक कथाओं का परिचय तथा खड़ी बोली में उनका अनुवाद
2. दूरदर्शन/आकाशवाणी के कार्यक्रम पर प्रतिक्रियाएँ विश्लेषण।
3. हिन्दी साहित्य का स्वतन्त्र पठन/टिप्पणी एवं प्रेरणाएँ।
4. हस्तलिखित पत्रिका तैयार करना।
5. म. प्र. से प्रकाशित होने वाली हिन्दी भाषा की पत्र-पत्रिकाओं की जानकारी।

टिप्पणी- प्रायोजना कार्य से सम्बन्धित विषय-वस्तु पर (अंक आबंटित न होने के कारण) परीक्षा में प्रश्न पूछे जाना अपेक्षित नहीं है।

MP Board Class 12 Special Hindi Blue Print of Question Paper

You can download MP Board Class 12th Hindi Blueprint and Marking Scheme 2019-2020 in Hindi and English medium.

MP Board Class 12 Hindi Blue Print of Question Paper 1

MP Board Class 12 General Hindi Blue Print of Question Paper

MP Board Class 12 Hindi Blue Print of Question Paper 2

We hope the given MP Board Class 12th Hindi Solutions Guide Pdf Free Download of हिंदी मकरंद, स्वाति will help you. If you have any query regarding Madhya Pradesh Syllabus MP Board Class 12 Hindi Book Solutions Makrand, Swati Pdf Special and General, drop a comment below and we will get back to you at the earliest.

MP Board Class 6th Special English Revision Exercises 2

MP Board Class 6th Special English Solutions Revision Exercises 2

Word Power

(a) Match the Following

MP Board Class 6th Special English Revision Exercises 2 img-1
Answer:
1. → 3
2. → 1
3. → 2
4. → 6
5. → 4
6. → 5

MP Board Solutions

Fill in the blanks with help of given words:

(friendship, favour, treasure, wasteful, righteousness, justice, praises, virtues, leader, outlaws)

  1. Saladin, the great, found one day that he had spent all his …………… in wars and ……………….. living.
  2. To the end of his days, the Jew enjoyed Saladin’s ………….. and ……………
  3. The king ruled the Kingdom with …………… and ……………
  4. He only heard …………….. of his ……………. from everyone.
  5. Robin Hood was the ………………. of a company of ………………. who lived in Sherwood Forest.

Answer:

  1. treasure, wasteful
  2. friendship, favour
  3. justice, righteousness
  4. praises, virtues
  5. leader, outlaws.

(c) Make sentences with the given words and phrases.
(inherit, willingly, countryside, virtue, passing through, ignorant, put on)

Inherit                 : He has inherited huge property from his ancestors.
Willingly              : I do my work willingly.
Countryside        : The natural scene of the countryside is very beautiful.
Virtue                  : Everyone praises him for his virtues.
Passing through : I got afraid while I was passing through a dense forest.
Ignorant              : A child is ignorant of the ways of the worlds.
Put on                 : Small children put on new clothes on festive occasions.

Comprehension

Answer these questions.

Class 6 English Revision Exercise 2 Question 1.
Why did Saladin think of a trick to get some money from the Jew?
Answer:
Saladin thought of a trick to get some money from the Jew because he was so miserly that he would never lend money willingly.

Class 6th English Revision Exercise 2 Question 2.
Why was it not possible to settle the questions of inheritance?
Answer:
It was not possible to settle the question of inheritance because all the three rings were so similar that no one could tell which was the true one.

Class 5 English Revision Exercise 2 Question 3.
What was Brahmadatta’s object in leaving the city in disguise?
Answer:
Brahmadatta left the city in disguise to search a person who could find any fault in him.

Revision Exercise 2 Class 6 Question 4.
Why were the two chariots not able to move on?
Answer:
The two chariots were not able to move on because the track was narrow and neither of the charioteers was willing to let the other pass.

Class 5 English Revision Exercise 2 MP Board Question 5.
Has anyone seen the wind?
Answer:
No, no one has seen the wind.

Class 5 English Revision Exercise 2 MP Board Solutions Question 6.
What do the leaves do when the wind is passing through?
Answer:
The leaves hang trembling when the wind passing through.

Class 6 Revision Exercise 2 Question 7.
Why was Robin Hood a terror to the rich people?
Answer:
Robin Hood used to rob the rich people to help the poor with the booty. Hence they were afraid of him (Robin Hood).

Head Make Sentence For Class 1 Question 8.
When did Robin Hood let the Sheriff go?
Answer:
Robin Hood allowed the Sheriff to leave only after snatching all the money he had.

MP Board Solutions

Grammar

(a) Combine the following pairs of sentences using “So ……. that.”
1. This book is not very difficult.
We can read it easily.

2. This bag is not very heavy.
I can lift it.

3. These pens are not very expensive.
We can buy them.

4. That bus is not very crowded.
We can travel in it comfortably.
Answer:
1. This book is not so difficult that we cannot read it easily.
2. This bag is not so heavy that I cannot lift it.
3. These pens are not so expensive that we cannot buy them
4. That bus is not so crowded that we cannot travel in it comfortably.

(b) Change the voice.

  1. They are writing the letters.
  2. He is playing cricket.
  3. He has posted the letters.
  4. She has answered the questions satisfactorily.

Answer:

  1. The letters are being written by them.
  2. Cricket is being played by him.
  3. The letters has been posted by him.
  4. The questions have been answered satisfactorily by her.

(c) Combine the following pairs of sentences using neither ………. nor.

1. Bholu didn’t play cricket yesterday.
Bholu didn’t play hockey yesterday.

2. Ritu didn’t take coffee.
Ritu didn’t take tea.
Answer:
1. Bholu played neither cricket nor hockey yesterday.
2. Ritu took neither coffee nor tea.

Let’s Write

Move Sentence For Class 2 Question 1.
Write a paragraph on “The Story of the True Ring.” What lesson does it teach?
Answer:
A rich man had a very beautiful ring. It was declared that before he died, he would pass the ring on to the worthiest of his sons. The son who received the ring, would inherit his property and be the head of the family after his death.

In this way the ring passed from father to son through several hundred years. At last it came into the hands of a man who had three sons.

The man loved his sons equally. He promised the ring to each of them. So he got two rings made. The three rings were alike. When the man was dying, he secretly gave each of the sons a ring. So after his death each claimed the inheritance. In order to prove their claim, each showed a ring. Everyone found the rings so similar that no one could tell which was the true ring. So the question of inheritance could not be solved. It is clear from the story that all the religions are true.

MP Board Solutions

Revision Test 2 Class 6 Question 2.
Describe in a paragraph how Brahmadatta ruled his kingdom.
Answer:
Brahmadatta was a just, impartial and righteous ruler. He gave fearless judgements. The people were satisfied under his rule. They settled their disputes without violence. People remained peaceful. He asked his ministers and the courtiers about his faults, but no one pointed his faults, as he had none. The people praised him highly for his virtues.

Revision Exercises Question 3.
Write a letter to your friend congratulating him/her on his/her birthday.
Answer:
F-20, Patel Nagar
Indore …….
May 15, 2005

Dear friend,
Accept my heartiest congratulation on your birthday. I had wished to join you on this occasion. But I could not came. Anyway hope you will manage everything in a grand manner. Your parents are very enthusiastic to arrange for such occasion, your brothers and sisters are there to enjoy and give you company. Once again I congratulate you and wish all my best for the occasion.

Yours
Ashish

MP Board Class 6th English Solutions

MP Board Class 11th Biology Solutions जीव विज्ञान

MP Board Class 11th Biology Solutions Guide Pdf Free Download जीव विज्ञान in both Hindi Medium and English Medium are part of MP Board Class 11th Solutions. Here we have given NCERT Madhya Pradesh Syllabus MP Board Class 11 Biology Book Solutions Jiv Vigyan Pdf.

MP Board Class 11th Biology Book Solutions in Hindi Medium

MP Board Class 11th Biology Book Solutions in English Medium

  • Chapter 1 The Living World
  • Chapter 2 Biological Classification
  • Chapter 3 Plant Kingdom
  • Chapter 4 Animal Kingdom
  • Chapter 5 Morphology of Flowering Plants
  • Chapter 6 Anatomy of Flowering Plants
  • Chapter 7 Structural Organisation in Animals
  • Chapter 8 Cell: The Unit of Life
  • Chapter 9 Biomolecules
  • Chapter 10 Cell Cycle and Cell Division
  • Chapter 11 Transport in Plants
  • Chapter 12 Mineral Nutrition
  • Chapter 13 Photosynthesis in Higher Plants
  • Chapter 14 Respiration in Plants
  • Chapter 15 Plant Growth and Development
  • Chapter 16 Digestion and Absorption
  • Chapter 17 Breathing and Exchange of Gases
  • Chapter 18 Body Fluids and Circulation
  • Chapter 19 Excretory Products and their Elimination
  • Chapter 20 Locomotion and Movement
  • Chapter 21 Neural Control and Coordination
  • Chapter 22 Chemical Coordination and Integration

MP Board Class 11th Biology Important Questions in English Medium

We hope the given MP Board Class 11th Biology Solutions Guide Pdf Free Download जीव विज्ञान in both Hindi Medium and English Medium will help you. If you have any query regarding NCERT Madhya Pradesh Syllabus MP Board Class 11 Biology Book Solutions Jiv Vigyan Pdf, drop a comment below and we will get back to you at the earliest.

MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1

MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1

MP Board Class 8 Maths Chapter 8 प्रश्न 1.
निम्नलिखित का अनुपात ज्ञात कीजिए –

(a) एक साइकिल की 15 km प्रति घण्टे की गति का एक स्कूटर की 30 km प्रति घण्टे की गति से।
(b) 5 m का 10 km से
(c) 50 पैसे का ₹ 5 से।

हल:
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-1

Rashiyon Ki Tulna Class 8 प्रश्न 2.
निम्नलिखित अनुपातों को प्रतिशत में परिवर्तित कीजिए –

(a) 3 : 4
(b) 2 : 3

हल:
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-2

Class 8 Maths Chapter 8 Exercise 8.1 In Hindi प्रश्न 3.
25 विद्यार्थियों में से 72% विद्यार्थी गणित में अच्छे हैं। कितने प्रतिशत विद्यार्थी गणित में अच्छे नहीं हैं?
हल:
25 में से 72% विद्यार्थी गणित में अच्छे हैं। ऐसे विद्यार्थी जो गणित में अच्छे नहीं हैं –
= (100 – 72)% = 28%
अत: 28% विद्यार्थी गणित में अच्छे नहीं हैं।

MP Board Solutions

MP Board Class 8 Maths Chapter 8 Exercise 8.1 प्रश्न 4.
एक फुटबॉल टीम ने जितने मैच खेले उनमें से 10 में जीत हासिल की। यदि उनकी जीत का प्रतिशत 40 था तो उस टीम ने कुल कितने मैच खेले?
हल:
माना कि फुटबॉल टीम ने कुल x मैच खेले।
अब, क्योंकि कुल मैचों का 40% = 10 है।
∴ x का 40% = 10
या x × \(\frac{40}{100}\) = 10
या x = \(\frac{10×100}{40}\) = 25
अतः कुल 25 मैच खेले।

MP Board Class 8 Maths Solutions English Medium प्रश्न 5.
यदि चमेली के पास अपने धन का 75% खर्च करने के बाद ₹ 600 बचे तो ज्ञात कीजिए कि उसके पास शुरू में कितने ₹ थे?
हल:
माना कि चमेली के पास शुरू में x रुपये थे।
चमेली द्वारा खर्च किए ₹ = x का 75%
खर्च करने के बाद बचे ₹
= x का (100 – 75)%
= x का 25%
∴ प्रश्नानुसार, x का 25% = ₹ 600
या x × \(\frac{25}{100}\) = 600
या x = \(\frac{600×100}{25}\) = ₹ 2400
25 अत: चमेली के पास शुरू में ₹ 2400 थे।

Rashiyo Ki Tulna Class 8 प्रश्न 6.
यदि किसी शहर में 60% व्यक्ति क्रिकेट पसन्द करते हैं, 30% फुटबॉल पसन्द करते हैं और शेष अन्य खेल पसन्द करते हैं। यदि कुल व्यक्ति 50 लाख हैं, तो प्रत्येक प्रकार के खेल को पसन्द करने वाले व्यक्तियों की यथार्थ संख्या ज्ञात कीजिए।
हल:
कुल व्यक्तियों की संख्या = 5000000
∴ क्रिकेट पसन्द करने वाले व्यक्तियों की संख्या = \(\frac{60}{100}\) x 5000000
= 3000000 = 30 लाख
फुटबॉल पसन्द करने वाले व्यक्तियों की संख्या \(\frac{30}{100}\) x 5000000
= 1500000 = 15 लाख
अन्य खेल पसन्द करने वाले व्यक्ति = 100 – (60 + 30)%
= (100 – 90)% = 10%
∴ अन्य खेल पसन्द करने वाले व्यक्तियों की संख्या = \(\frac{10}{100}\) x 5000000
= 500000 = 5 लाख

पाठ्य-पुस्तक पृष्ठ संख्या # 129 – 130

प्रयास कीजिए (क्रमांक 8.2)

MP Board Solutions

Class 8 Maths 8.1 Hindi Medium प्रश्न 1.
एक दुकान 20% बट्टा देती है। निम्नलिखित में से प्रत्येक का विक्रय मूल्य क्या होगा?

  1. ₹ 120 अंकित वाली एक पोशाक।
  2. ₹ 750 अंकित वाले एक जोड़ी जूते।
  3. ₹ 250 अंकित मूल्य वाला एक थैला।

हल:
1. अंकित मूल्य = ₹ 120, बट्टा = 20%
∴ बट्टा = ₹ 120 का 20%
= ₹ 120 x \(\frac{20}{100}\) = ₹ 24
∴ विक्रय मूल्य = अंकित मूल्य – बट्टा
= ₹ 120 – ₹ 24
= ₹ 96

2. अंकित मूल्य = ₹ 750, बट्टा = 20%
∴ बट्टा = ₹ 750 का 20%
= ₹ 750 x \(\frac{20}{100}\) = ₹ 150
∴ विक्रय मूल्य = अंकित मूल्य – बट्टा
= ₹750 – ₹ 150
= ₹ 600

3. अंकित मूल्य = ₹ 250, बट्टा = 20%
∴ बट्टा= ₹ 250 का 20%
= ₹ 250 x 20 = ₹ 50
∴ विक्रय मूल्य = अंकित मूल्य – बट्टा
= ₹ 250 – ₹ 50
= ₹ 200

MP Board Class 8 Maths Solutions प्रश्न 2.
₹15,000 अंकित मूल्य वाली एक मेज ₹ 14,400. में उपलब्ध है। बट्टा और बट्टा प्रतिशत ज्ञात कीजिए।
हल:
अंकित मूल्य = ₹ 15,000, विक्रय मूल्य = ₹ 14,400
बट्टा = अंकित मूल्य – विक्रय मूल्य
= ₹ 15000 – ₹ 14,400 = ₹600
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-3

Class 8 Maths 8.1 In Hindi प्रश्न 3.
एक अलमारी 5% बट्टे पर ₹ 5,225 में बेची जाती है। अलमारी का अंकित मूल्य ज्ञात कीजिए।
हल:
बट्टा प्रतिशत = 5%,
विक्रय मूल्य = ₹ 5,225
माना कि अलमारी का अंकित मूल्य = ₹ x है।
∴ बट्टा = ₹ x का 5%
= ₹ x × \(\frac{5}{100}\) = ₹ \(\frac{5x}{2}\)
∴ बट्टा = अंकित मूल्य – विक्रय मूल्य
₹ \(\frac{5x}{100}\) = ₹ x – ₹ 5225
5x = 100x – 5225 x 100
या 100x – 5x = 5225 x 100
या 95x = 5225 x 100
या x = \(\frac{5225×100}{95}\) = ₹ 5,500
अतः अलमारी का अंकित मूल्य ₹ 5,500 है।

MP Board Solutions

पाठ्य-पुस्तक पृष्ठ संख्या # 130

प्रतिशत में आकलन

Class 8 Rashiyon Ki Tulna प्रश्न 1.
यदि दुकान पर आपका बिल ₹ 577.80 है।

  1. इस बिल राशि का 20% बट्टे से आकलन करने का प्रयास कीजिए।
  2. ₹ 375 का 15% ज्ञात करने का प्रयास कीजिए।

हल:
1. बिल को ₹ 577.80 की निकटतम दहाई में पूर्णांकित करने पर
∴ ₹ 577.80 = ₹ 580
अतः बट्टा = ₹ 580 का 20%
= ₹ 580 x \(\frac{20}{100}\) = ₹ 116
∴ बिल की राशि का सन्निकट मान = ₹ 580 – ₹ 116
= ₹ 464

2. ₹ 375 का 15% = ₹ 375 x = \(\frac{15}{100}\)
= ₹ 56.25

पाठ्य-पुस्तक पृष्ठ संख्या # 131

प्रयास कीजिए (क्रमांक 8.3)

MP Board Class 8 Maths प्रश्न 1.
यदि लाभ की दर 5% है, तो निम्नलिखित का विक्रय मूल्य ज्ञात कीजिए –

  1. ₹ 700 की एक साइकिल जिस पर ऊपरी खर्च ₹ 50 है।
  2. ₹ 1,150 में खरीदा गया एक घास काटने का यन्त्र जिस पर ₹ 50 परिवहन व्यय के रूप में खर्च किए गए हैं।
  3. ₹ 560 में खरीदा गया एक पंखा जिस पर ₹ 40 मरम्मत के लिए खर्च किए गए हैं।

हल:
1. साइकिल का क्रय मूल्य = ₹ 700
ऊपरी खर्च = ₹50
∴ साइकिल का कुल क्रय मूल्य = ₹ 700 + ₹ 500
= ₹ 750 लाभ = 5%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-4
= ₹ 787.50

2. घास काटने के यन्त्र का क्रय मूल्य = ₹ 1,150
परिवहन व्यय = ₹ 50
∴ अब, यन्त्र का क्रय मूल्य = ₹ 1,150 + ₹ 50
=₹ 1,200
लाभ = 5%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-5
= ₹ 1,260

3. पंखे का क्रय मूल्य = ₹ 560
ऊपरी खर्चा = ₹40
∴ अब, यन्त्र का क्रय मूल्य = ₹ 560 + ₹ 40 = ₹ 600
लाभ = 5%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-6
= ₹ 630

पाठ्य-पुस्तक पृष्ठ संख्या # 132

प्रयास कीजिए (क्रमांक 8.4)

Class 8 Maths Chapter 8.1 In Hindi प्रश्न 1.
एक दुकानदार ने दो टेलीविजन सेट ₹ 10,000 प्रति सेट की दर से खरीदे। उसने एक को 10% हानि से और दूसरे को 10% लाभ से बेच दिया। ज्ञात कीजिए कि कुल मिलाकर, उसे इस सौदे में लाभ हुआ अथवा हानि?
हल:
पहले टेलीविजन के लिए,
क्रय मूल्य = ₹ 10,000 तथा लाभ = 10%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-7
= ₹ 110 x 100
=₹ 11,000
दूसरे टेलीविजन के लिए,
क्रय मूल्य = ₹ 10,000, हानि = 10%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-8
= ₹ 90 x 100 = ₹ 9000
दोनों टेलीविजन का कुल क्रय मूल्य
= ₹ 10000 + ₹ 10000 = ₹ 20000
कुल विक्रय मूल्य = ₹ 11000 + ₹ 9000 = ₹ 20000
∴ विक्रय मूल्य = क्रय मूल्य है
अतः दुकानदार को न तो लाभ हुआ और न ही हानि।

पाठ्य-पुस्तक पृष्ठ संख्या # 133

प्रयास कीजिए (क्रमांक 8.5)

Maths Class 8 MP Board प्रश्न 1.
निम्नलिखित वस्तुओं को खरीदने पर यदि 5% बिक्री कर जुड़ता है तो प्रत्येक का खरीद (विक्रय) मूल्य ज्ञात कीजिए –

  1. ₹ 50 वाला एक तौलिया।
  2. साबुन की दो टिकिया जिनमें से प्रत्येक का मूल्य ₹ 35 है।
  3. ₹ 15 प्रति किलोग्राम की दर से 5 kg आटा।

हल:
1. तौलिया का मूल्य = ₹ 50
बिक्री कर दी दर = 5%
बिक्री कर = ₹ 50 x \(\frac{5}{100}\) = ₹ 2.50
∴ तौलिया का खरीद मूल्य = ₹ 50 + ₹ 2.50
= ₹ 52.50

2. साबुन की दो टिकियाओं का मूल्य = ₹ 35 x 2 = ₹ 70
बिक्री कर = ₹ 70 का 5%
= ₹ 70 x \(\frac{5}{100}\) = ₹ 3.50
∴ साबुन की दो टिकियाओं का खरीद मूल्य
= ₹ 70 + ₹ 3.50
= ₹ 73.50

3. 5 kg आटे का मूल्य = ₹ 15 x 5 = ₹ 75
बिक्री कर = ₹ 75 का 5%
= ₹ 75 x \(\frac{5}{100}\) = ₹ 3.75
∴ 5 kg आटा का खरीद मूल्य = ₹ 75 + ₹ 3.75
= ₹ 78.75

MP Board Solutions

Class 8 Maths Chapter 8 Exercise 8.1 Solutions In Hindi प्रश्न 2.
निम्नलिखित वस्तुओं के मूल्य में यदि 8% वैट सम्मिलित है तो वास्तविक मूल्य ज्ञात कीजिए –

  1. ₹ 14,500 में खरीदा गया एक टेलीविजन
  2. ₹ 180 में खरीदी गई शैम्पू की एक शीशी।

हल:
1. माना कि टेलीविजन का वास्तविक मूल्य = ₹ x है।
8% की दर से ₹ x पर वैट = ₹ \(\frac{8}{100}\) × x
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-9
= ₹ 13,425.93
अतः टेलीविजन का वास्तविक मूल्य = ₹ 13,425.93

2. माना कि शैम्पू की शीशी का वास्तविक मूल्य = ₹x है।
8% की दर से ₹ x पर वैट = ₹x × \(\frac{8}{100}\) = ₹ \(\frac{8x}{100}\)
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-10
= ₹ 166.67
अतः शैम्पू की शीशी का वास्तविक मूल्य = ₹ 166.67

सोचिए, चर्चा कीजिए और लिखिए (क्रमांक 8.1)

Math Class 8 MP Board प्रश्न 1.
किसी संख्या को दुगुना करने पर उस संख्या में 100% वृद्धि होनी है। यदि हम उस संख्या को आधा कर दें तो कितना प्रतिशत ह्रास होगा?
हल:
माना कि कोई संख्या x है, तब इसका आधा = \(\frac{1}{2}\)x
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-11
= 50%

MP Board Class 8 Maths प्रश्न 2.
₹ 2,400 की तुलना में ₹ 2,000 कितना प्रतिशत कम है ? क्या यह प्रतिशत उतना ही है, जितना ₹ 2,000 की तुलना में ₹ 2,400 अधिक है?
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-12
= 20%
नहीं, दोनों प्रतिशत समान नहीं हैं।

MP Board Class 8th Maths Solutions

MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms

MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms

Respiration in Organisms Intex Questions

Question 1.
Bhoojho wants to know if cockroaches, snails, fish, earthworms, ants and mosquitoes also have lungs?
Answer:
Yes.

Table 10.1 Changes In Breathing Rate Under Different Conditions Question 2.
Boojho has seen in television programmes that whales and dolphins often come up to the water surface? They even release a fountain of water sometimes while moving upwards? Why do they do so?
Answer:
Whales and dolphins take in air during inhalation. They exhale out the air on the surface. The water vapour condenses and we see the condensed water vapour as the fountain.

MP Board Solutions

Question 3.
Paheli wants to know whether roots, which are under ground also take in oxygen? If so, how?
Answer:
Yes. Roots take up air from the air spaces preseht between the soil particles.

Activities

Activity – 1
If you try you can count your rate of breathing. Breathe in and out normally. Find out how many times you breathe in and breathe out in a minute? Did you inhale the same number of times as you exhaled? Now count your breathing rate (number of breaths/minute) after brisk walk and after running. Record your breathing rate as soon as you finish and also after complete rest. Tabulate your findings and compare your breathing rates under different conditions with those of your classmates.

Table:
Changes in breathing rate under different conditions:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 1

Activity – 2
Figure shows the various activities carried out by a person during a normal day. Can you say in which activity, the rate of breathing will be the slowest and in which it will be the fastest? Assign numbers to the pictures in the order of increasing rate of breathing according to your experience.
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 2

Activity – 3
Take a deep breath. Measure the size of the chest with a measuring tape and record your observations in Table. Measure the size of the chest again when expanded and indicate which classmate shows the maximum expansion of the chest.
Answer:
Effect of breathing on the chest size of some classmates:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 3 - Copy

Respiration in Organisms Text Book Exercise

Class 7 Science Chapter 10 Table 10.1 Question 1.
Why does an athlete breathe faster and deeper than usual after finishing the race?
Answer:
During the race, the athlete has to run very fast. The demand for energy at that time increases, which increase the demand for more supply of oxygen. Thus, athlete has to breathe faster and deep to inhale more oxygen.

Question 2.
List the similarities and differences between aerobic and anaerobic respiration?
Answer:
Similarities:
Both aerobic and anaerobic respiration produce energy and give out carbon dioxide.

Differences:
Aerobic respiration require oxygen while anaerobic respiration does not require oxygen. In aerobic respiration large amount of energy is released while in anaerobic respiration small amount of energy is released.

Question 3.
Why do we often sneeze when we inhale a lot of dust laden air?
Answer:
We sneeze to get rid of the unwanted particles like dust from air body. It allows only clean and dust free air to enter our body.

MP Board Solutions

MP Board Class 7 Science Chapter 10 Question 4.
Take three test – tubes. Fill \(\frac{1}{2}\)th of each with water. Label them A, B and C. Keep a snail in test – tube A, a water plant in test – tube B and in C, keep snail and plant both. Which test – tube would have the highest concentration of CO2?
Answer:
Test – tube A.

Question 5.
Tick the correct answer:

Question (a)
In cockroaches, air enters the body through?
(a) lungs
(b) gills
(c) spiracles
(d) skin.
Answer:
(c) spiracles

Class 7 Science Chapter 10 Question (b)
During heavy exercise, we get cramps in the legs due to the accumulation of?
(a) carbon dioxide
(b) lactic acid
(c) alcohol
(d) water.
Answer:
(b) lactic acid

Question (c)
Normal range of breathing rate per minute in an average adult person at rest is?
(a) 9 – 12
(b) 15 – 18
(c) 21 – 24
(d) 30 – 33
Answer:
(b) 15 – 18

Question (d)
During exhalation, the ribs?
(a) move outwards
(b) move downwards
(c) move upwards
(d) do not move
Answer:
(b) move downwards

MP Board Class 7th Science Chapter 10 Question 6.
Match the items in Column I with those in Column II:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 4
Answer:

(a) – (iii)
(b) – (iv)
(c) – (i)
(d) – (v)
(e) – (ii)
(f) – (vi)

Table 10.1 Class 7 Science Question 7.
Mark if the statement is true and if it is false:

  1. During heavy exercise the breathing rate of a person slows down.
  2. Plants carry out photosynthesis only during the day and respiration only at night.
  3. Frogs breathe through their skins as well as their lungs.
  4. The fishes have lungs for respiration.
  5. The size of the chest cavity increases during inhalation.

Answer:

  1. False
  2. False
  3. True
  4. False
  5. True

MP Board Solutions

Question 8.
Given below is a square of letters in which are hidden different words related to respiration in organisms. These words may be present in any direction – upwards, downwards, or along the diagonals. Find the words for your respiratory system. Clues about those words are given below the square?

  1. The air tubes of insects
  2. Skeletal structures surrounding chest cavity
  3. Muscular floor of chest cavity
  4. Tiny pores on the surface of leaf
  5. Small openings on the sides of the body of an insect
  6. The respiratory organs of human beings
  7. The openings through which we inhale
  8. An anaerobic organism
  9. An organism with tracheal system

Answer:

  1. Trachea
  2. Rib
  3. Diaphragm
  4. Stomats
  5. Spiracles
  6. Lung
  7. Strils
  8. Yeast
  9. Ant.

MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 5

Class 7th Science Chapter 10 Question 10.
The mountaineers carry oxygen with them because:
(a) At an altitude of more than 5 km there is no air.
(b) The amount of air available to a person is less than that available on the ground.
(c) The temperature of air is higher than that on the ground.
(d) The pressure of air is higher than that on the ground.
Answer:
(b) The amount of air available to a person is less than that available on the ground.

Extended Learning – Activities and Projects

Question 1.
Observe fish in an aquarium. You will find flap like structures on both sides of their heads. These are flaps which cover the gills. These flaps open and close alternately. On the basis of these observations, explain the process of respiration in the fish?
Answer:
Do with the help of your subject teacher.

Question 2.
Visit a local doctor. Learn about the harmful effects of smoking. You can also collect material on this topic from other sources. You can seek help of your teacher or parents. Find out the percentage of people of your area who smoke. If you have a smoker in your family, confront him with the material that you have collected?
Answer:
Do yourself.

Class 7 Science Table 10.1 Question 3.
Visit a doctor. Find out about artificial respiration? Ask the doctor:
(a) When does a person need artificial respiration?
(b) Does the person need to be kept on artificial respiration temporarily or permanently?
(c) From where can the person get supply of oxygen for artificial respiration?
Answer:
Do yourself.

MP Board Solutions

Question 4.
Measure the breathing rate of the members of your family and some of your friends? Investigate:
(a) If the breathing rate of children is different from that of adults?
(b) If the breathing rate of males is different from that of females?
If there is a difference in any of these cases, try to find the reason?
Answer:
Do with the help of your parents.

Respiration in Organisms Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Class 7th Science Chapter 10 Question Answer Question (a)
The life processes that provide energy are?
(a) respiration
(b) nutrition
(c) both respiration and nutrition
(d) none of these.
Answer:
(c) both respiration and nutrition

Question (b)
In …………………… respiration, there is an exchange of gases between the cells and the blood?
(a) aerobic
(b) anaerobic
(c) external
(d) internal.
Answer:
(d) internal.

Class 7 Science Ch 10 Question (c)
In the cell, the food (glucose) is broken down into carbon dioxide and water using?
(a) hydrogen
(b) nitrogen
(c) oxygen
(d) none of these.
Answer:
(c) oxygen

MP Board Solutions

Question (d)
Which of the following is not a feature of respiration?
(a) involvement of enzymes
(b) occur outside the cells
(c) release of energy
(d) is a chemical process.
Answer:
(d) is a chemical process.

Std 7 Science Lesson No 10 Question Answer Question (e)
During heavy exercise, the breathing rate in an adult can increase upto?
(a) 25 times per minute
(b) 30 times per minute
(c) 35 times per minute
(d) none of these.
Answer:
(a) 25 times per minute

Question (f)
The percentage of oxygen and carbon dioxide in inhaled air is –
(a) 21%, 0.04%
(b) 20%, 0.10%
(c) 21%, 1.0%
(d) 20%, 1.0%.
Answer:
(a) 21%, 0.04%

Science Class 7 Chapter 10 Question Answer Question (g)
The percentage of oxygen and carbondioxide in exhalted air is –
(a) 16.4%, 4.4%
(b) 21%, 1.0%
(c) 16.4%, 3.4%
(d) 16.4%, 0.04%.
Answer:
(a) 16.4%, 4.4%

Question 1.
Fill in the blanks:

  1. All living organisms require ………………………… to perform various life process.
  2. The liver and ……………………… are found near the stomach.
  3. The exhaled air has a higher percentage of carbon dioxide as compared to the ……………………………… air.
  4. Breathing is a process at organ levels, whereas respiration is a ……………………….. process.
  5. When breakdown of glucose occurs with the use of oxygen, it is called …………………….. respiration.
  6. The taking in of air rich in oxygen into the body is called …………………………
  7. The number of times a person breathes in a minute is termed as the ………………………….
  8. Lungs are present in the ……………………….. cavity.
  9. During inhalation, ribs move up and outwards and diaphragm moves ………………………..
  10. Insects have a network of air tubes called ……………………….. for gas exchange.
  11. Like all other living cells of the plants, the root cells also need oxygen to ………………………. energy.

Answer:

  1. energy
  2. pancreas
  3. inhaled
  4. cellular
  5. aerobic
  6. inhalation
  7. breathing rate
  8. chest
  9. down
  10. tracheae
  11. generate.

MP Board Solutions

Std 7 Science Chapter 10 Question Answer Question 3.
Which of the following statements are true (T) or false (F):

  1. Respiration is a type of combustion at ordinary temperature.
  2. Breathing is a process that takes place at the cellular level.
  3. Oxygen is released during the process of respiration.
  4. During respiration the plants – take CO2 and release Or
  5. Respiration involves on exchange of gases.
  6. Cellular respiration takes place in the cells of all organisms.
  7. Anaerobic respiration do not takes places in the muscle cells to fulfill the demand of energy.
  8. The giving out of air rich in carbon dioxide is known as exhalation.
  9. A breathe means one inhalation plus one exhalation.
  10. On an average, an adult human being at rest breathes in and out 15 to 18 times in a minute.
  11. A cockroach has small openings on the sides of its body.
  12. Gills are not supplied with blood vessels for exchange of gases.
  13. The end product of anaerobic respiration are carbon dioxide and water.
  14. In earthworm, the exchange of gases occurs through the moist skin.

Answer:

  1. True
  2. False
  3. False
  4. True
  5. True
  6. True
  7. False
  8. True
  9. True
  10. True
  11. True
  12. False
  13. True
  14. True

Respiration in Organisms Very Short Answer Type Questions

Question 1.
Where does cellular respiration takes place?
Answer:
Cells of organisms.

Question 2.
Write the equation for breakdown of food in anaerobic respiration?
Answer:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 6 - Copy

Up Board Class 7 Science Chapter 10 Question 3.
Write the equation for breakdown of food in aerobic respiration?
Answer:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms img t

Question 4.
What are anaerobes?
Answer:
There are some organisms such as yeast that can service in the absence of air. They are called anaerobes.

Question 5.
What is yeast?
Answer:
Yeast is single – celled organisms.

Respiration In Organisms Question 6.
Write the uses of yeast?
Answer:
Yeast respire anaerobically and during this process yield alcohol. So, they are used to make beer and wine.

Question 7.
Which chemical reaction takes place in internal respiration?
Answer:
Glucose + Oxygen → Carbon dioxide + Water + Energy.

Question 8.
Give an example of an oxygen respiration?
Answer:
In human beings.

Class 7 Science Chapter 10 Short Answer Question 9.
Name the parts of digestive system of humans?
Answer:
The parts of digestive systems are mouth, oesophagus, stomach, intestine and anus.

MP Board Solutions

Question 10.
Name two processes of respiration?
Answer:
Inhalation and exhalation are the two processes of respiration.

Question 11.
Name the parts of respiratory system of human?
Answer:

  1. Nostrils
  2. Trachea
  3. Lungs with alveali, and
  4. Diaphragm.

Class 7 Science Chapter 10 Question Answer Question 12.
What waste materials are produced during respiration?
Answer:
Carbon dioxide is produced as the waste material during respiration.

Question 13.
Define respiration?
Answer:
The process of breaking down of food by using oxygen, to form carbon dioxide and release energy required for various life activities, is called respiration.

Question 14.
Name the fuels used for the production of energy during respiration?
Answer:
Glucose is oxidized to give out energy.

Class 7 Chapter 10 Science Question 15.
Which organs of plants participate in respiration?
Answer:
There is no special organ in plants for breathing.

Question 16.
Write the names of organs in human respiratory system in sequence?
Answer:
Nostrils → Nasal cavity → Pharynx → Tracheae → Lungs.

Question 17.
What is importance of hairs present in the noise?
Answer:
These small hairs present in nose act as filters. These prevent dust particles and harmful germs to enter into respiratory track.

Chapter 10 Science Class 7th Question 18.
Name the organs of the body from which blood freshly enriched with oxygen goes into the heart?
Answer:
The lungs helps the blood to get freshly enrichment of oxygen.

Question 19.
What happens during breathing?
Answer:
During breathing, oxygen enriched air is inhaled which reaches lungs. Here, oxygen centers blood and unwanted water vapour and carbon dioxide are released out during breathing.

MP Board Solutions

Question 20.
What is breathing?
Answer:
The process of taking oxygen and leaving of carbon dioxide during respiration is called breathing.

Question 21.
Write the name of gases which are involved in breathing?
Answer:
Carbon dioxide and oxygen.

Class 7 Science Lesson 10 Question Answer Question 22.
How will you prove that we exhale CO2 gas during respiration?
Answer:
Pass the exhaled air given out by us into lime water. It will turn milky in colour. We know that CO2 gas turn lime water milky This confirms that we exhale CO2 gas in respiration.

Question 23.
What do we exhale?
Answer:
We exhale air rich in carbon dioxide.

Question 24.
Do we exhale only carbon dioxide or a mixture of gases along with it?
Answer:
We exhale a mixture of gases along with carbon dioxide.

Question 25.
How do ribs and diaphragm move during inhalation?
Answer:
During inhalation, ribs move up and outwards and diaphragm moves down.

MP Board Solutions

Question 26.
Which is the respiratory organ for earthworm?
Answer:
Skin.

Question 27.
Can we survive in water?
Answer:
No.

Question 28.
How do fish breathe under water?
Answer:
Gills in fish help them to use oxygen dissolved in water. Gills are projections of the skin. Gills are well supplied with blood vessels for exchange of gases.

Question 29.
What is the function of gills?
Answer:
The fishes and other aquatic animals respire through gills or similar structure.

Question 30.
Can you guess what would happen if a potted plant is over watered?
Answer:
The roots will not get air to respire, so the roots will die and hence the whole plant will also die.

Respiration in Organisms Short Answer Type Questions

Question 1.
What is respiration?
Answer:
All the living organisms perform a number of vital activities. The energy is obtained by the oxidation of food or respiration. When the living organisms are completely at the stage of rest, even then they require some minimum amount of energy for the maintainance of cells and tissues. Thus, respiration is one of the most important process for living organisms.

Question 2.
Define respiration with the help of chemical equation?
Answer:
The process in which the oxidation of absorbed food is takes place by the CO2 which is inhaled by breathing and energy is released out is called respiration. Chemical equation of nutrition is as follows:
C6H12O2 + 6O2 → 6CO2 + 6H2O + 673 kcal (Energy)

Question 3.
Why does our body need a transporting system?
Answer:
Our body needs a transporting system to:

  1. Transport oxygen to body cells from lungs.
  2. Transport food to body cells from liver.
  3. Transport waste material from body cells to excretory organs.
  4. Maintains body temperature constant.

MP Board Solutions

Question 4.
Write difference between oxy – respiration and anoxy – respiration?
Answer:
The differences between the oxy – respiration and anoxy – respiration:
oxy – respiration:

  1. It takes place in presence of O2
  2. The end products are CO2 and H2O.
  3. The energy released is more.

Anoxy – respiration:

  1. It takes place without oxygen.
  2. The end products are ethyl alcohol and CO2.
  3. The energy released

Question 5.
Describe the various types of respiratory organs found in animals?
Answer:
In animals, there are definite respiratory organs for exchange of gases.

  1. In earthworm and leech exchange of gases takes place through moist, thin and vascular skin.
  2. In insects the trachea are the repiratory organs.
  3. In fishes the gills are the respiratory organs.
  4. Higher animals like mammal and birds including man have lungs for respiration.

Question 6.
Describe the importance of respiration in plants?
Answer:

  1. It takes place in the presence of oxygen.
  2. It is completed in cytoplasm and mitochondria of cell.
  3. It involves the complete oxidation of glucose into CO2, and
  4. H2O + C2H2O2 + 6O2 → 6CO2 + energy + 6H2O.
  5. It occurs in all the living cell of the organisms.
  6. It is day night process.
  7. Energy is released in this process.

Question 7.
Name the organs associated with the following functions:

  1. digestion
  2. absorption of minerals
  3. respiration, and
  4. excretion of carbon – dioxide in man.

Answer:

1. Digestion. Mouth, stomach, oesophagus, pharynx, small intestine, large intestine.

2. Absorption of minerals.
In Animals : Small intestine
In plants : Root hairs.

3. Respiration.
In Animals : Nose, trachaea, larynx, lung.
In plants. Stomata and lenticells.

4. Excretion of carbon – dioxide.
In Animals : Kidneys, ureters, urinary bladder, urethra.

MP Board Solutions

Question 8.
State the difference between respiration and breathing?
Answer:
Differences between Respiration and Breathing:
Respiration:

  1. It takes place inside the cells.
  2. The exchange of gases takes place between blood and the tissues of the body.
  3. In this process nutrients are oxidised to liberate energy.

Breathing:

  1. It takes place at the surface of the respiratory organs.
  2. The exchange of gases takes place between the blood and the external environment.
  3. The nutrients are not oxidised to liberate energy.

Question 9.
Name the major components of urine?
Answer:
The kidney, ureter, bladder, and urethra are the organs used in the removal of urine from the body. Renal artery carry urea and uric acid along with the large amount of water with the blood into the kindneys, when this blood enters into glomerulus, the solid wastes filter here while the water is diffused out from the network of blood capillaries into the uriniferous tubules. The ques mixture is called as urine.

Question 10.
If a person drinks very little water per day? The volume of urine decreases? In what ways does it affect the body?
Answer:
If a person drinks very little water per day. The volume of urine decreases. Large amount of water will dissolve large quantity of urea in it and large amount of urine will pass out from the body. If someone drinks lesser amounts of water, the concentration of urea in the cells and its larger quantity is very harmful for the body.

Question 11.
What is saliva? What are the functions of saliva?
Answer:
Saliva is a digestive secretion produced by three layered salivary glands, the paroted submaxillary and sunblingual present in our mouth cavity. This soften and lubricants food for easier swallowing and converts starch into reducing sugars. The salivary amylase. enzyme of saliva acts of starch in a neutral medium.

Question 12.
How does the food digested in the stomach?
Answer:
After some time from mouth the food reach inside the stomach. The gastric glands of stomach secrete the gastric juices. The gastric juice containes three enzymes. These are pepsin, renin and HCI. The HCI makes the medium acidic, and inhibites the bacterial growth and prevents the food. The renin curdiles the milk protein to be hydrololysed by pepsin. The pepsin reacts with proties and changes into peptides.

Respiration in Organisms Long Answer Type Questions

Question 1.
Define the respiratory organs in animals?
Answer:
The process of respiratory system in animals possess following organs:

  1. Nasal cavity
  2. Larynx
  3. Trachea
  4. Bronchi
  5. Lungs.

In animals some organs like gills and lungs are developed for the purpose of exchange of gases. The amount of CO2 produced after respiration cannot be utilised by animals as in plants. This is also true for the production of oxygen which is required for respiration. Mitochondria are the site of respiration in both plants and animals.

The process of break – down of glucose in the presence of oxygen and some enzymes into CO2. And water, which is accompanied with release of energy is very complex. Materials like proteins and fats are also consumed during respiration to produce energy.

MP Board Solutions

Question 2.
Write difference between photosynthesis and respiration?
Answer:
The difference between respiration and photosynthesis are:
Photosynthesis:

  1. It takes place only in green plants.
  2. It requires energy.
  3. It requires CO2 and H2O.
  4. It releases oxygen and make food.
  5. It is a building up process.
  6. It takes place in the chloroplast of the plant cell.

Respiration:

  1. It takes place in all plants and animals.
  2. It releases energy.
  3. It releases CO2 and H2O.
  4. It requires oxygen and oxidise the food.
  5. It is a breaking down process.
  6. It takes place in mitochondira of a cell.

Question 3.
Draw the labelled diagram to show respiratory system in man?
Answer:
The process of respiration is aimed to release energy
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 8

Question 4.
What is the difference in the amount of carbon choxide in the inhaled and the exhaled air? How will you test the presence of CO2 in the exhaled air?
Answer:
Excess carbon dioxide is present in exhaled air as compared inhaled air.

Test to indicate the presence of CO2 in the exhaled air:
Take two test tubes. Fill each of them half with freshly prepared lime water. Fix stoppers with two holes in both the test tubes. Insert glass tubes in both the stoppers. The lime water through which exhaled air is passed turns milky. This shows that more CO2 is present in exhaled air.
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 9

Question 5.
Draw diagrams to show movements of rib and diaphragm during breathing?
Answer:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 10

Question 6.
How do the following organism breathe? Amoeba, fish, frog, grasshopper, earthworm?
Answer:
Amoeba:
Amoeba breathes by diffusion of gases in between body surface and the water.

Fish:
Fish breathe with gills. The gills are special organs. They help fish to extract dissolved oxygen from the water.

Frog:
Frog can breathe through skin and lungs. In water if breathes through skin whereas in air through lungs.

Grasshopper:
The grasshopper and other insects have holes and air tubes those help them to breathe.

Earthworm:
It breathes through its moist body surface.

MP Board Solutions

Question 7.
Describe the breathing in cockroach with diagram?
Answer:
A cockroach has small openings on the sides of its body. Other insects also have similar openings. These openings are called spiracles, bisects have a network of air tubes called tracheae for gas exchange. Oxygen rich air rushes through spiracles into the tracheal tubes, diffuses into the body tissue, and reaches every cell of the body. Similarly, carbon dioxide from the cells goes into the tracheal tubes and moves out through spiracles. These air tubes or tracheae are found only in insects and not in any other group of animals.
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms img u

MP Board Class 7th Science Solutions

MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3

MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3

Class 7 Maths Chapter 2 Exercise 2.3 Solutions Hindi Medium प्रश्न 1.
ज्ञात कीजिए :
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 1
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 1a
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 1b
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 1c

Class 7 Maths Chapter 2 Exercise 2.3 Solutions In Hindi प्रश्न 2.
गुणा कीजिए और न्यूनतम रूप में बदलिए (यदि सम्भव है) :
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 2
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 2a

MP Board Solutions

MP Board Class 7 Maths Solutions English Medium प्रश्न 3.
निम्नलिखित भिन्नों को गुणा कीजिए-
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 3
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 3a
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 3b

MP Board Class 7 Maths Solutions प्रश्न 4.
कौन बड़ा है:
(i) \(\frac { 3 }{ 4 }\) का \(\frac { 2 }{ 7 }\) अथवा \(\frac { 5 }{ 8 }\) का \(\frac { 3 }{ 5 }\)
(ii) \(\frac { 6 }{ 7 }\) का \(\frac { 1 }{ 2 }\) अथवा \(\frac { 3 }{ 7 }\) का \(\frac { 2 }{ 3 }\)
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 4

MP Board Solutions

Class 7 Maths Chapter 2.3 In Hindi प्रश्न 5.
सैली अपने बगीचे में चार छोटे पौधे एक पंक्ति में लगाती है। दो क्रमागत छोटे पौधों के बीच की दूरी – मीटर है। प्रथम और अन्तिम पौधे की बीच की दूरी ज्ञात कीजिए।
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 5
दो क्रमागत पौधों के बीच की दूरी = \(\frac { 3 }{ 4 }\) मीटर
∴ प्रथम और अन्तिम (चौथे) पौधे के बीच की दूरी = 3 x \(\frac { 3 }{ 4 }\) मीटर
= \(\frac { 3\times3 }{ 4 }\) मीटर = \(\frac { 9 }{ 4 }\) मीटर = 2\(\frac { 1 }{ 4 }\) मीटर

Class 7 Maths Exercise 2.3 Solutions In Hindi प्रश्न 6.
लिपिका एक पुस्तक को प्रतिदिन 1\(\frac { 3 }{ 4 }\) घण्टे पढ़ती है। वह सम्पूर्ण पुस्तक को 6 दिनों में पढ़ती है। उस पुस्तक को पढ़ने में उसने कुल कितने घण्टे लगाए ?
हल:
प्रतिदिन पढ़ने का समय = 1\(\frac { 3 }{ 4 }\) घण्टे = \(\frac { 7 }{ 4 }\) घण्टे
वह पुस्तक को पढ़ती है = 6 दिनों में
= \(\frac { 6\times7 }{ 4 }\) = \(\frac { 42 }{ 4 }\)
उसने पुस्तक पढ़ने में 10\(\frac { 1 }{ 2 }\) घण्टे लगाए।

Class 7 Maths 2.3 In Hindi प्रश्न 7.
एक कार 1 लीटर पैट्रोल में 16 किमी दौड़ती है। 2\(\frac { 3 }{ 4 }\) लीटर पैट्रोल में यह कार कुल कितनी दूरी तय करेगी?
हल:
∵ कार 1 लीटर में जाती है = 16 किमी
∴ कार द्वारा 23 लीटर में चली गई दूरी
= 16 x 2\(\frac { 3 }{ 4 }\) किमी
= 16 x \(\frac { 11 }{ 4 }\) = 4 x 11 = 44 किमी

MP Board Solutions

Class 7 Math 2.3 In Hindi प्रश्न 8.
(a) (i) बक्सा □ में संख्या लिखिए, ताकि \(\frac{2}{3}\) x □ = \(\frac{10}{30}\)
(ii) □ में प्राप्त संख्या का न्यूनतम रूप….. है।

(b) (i) बक्सा □ में संख्या लिखिए, ताकि \(\frac{3}{5}\) x □ = \(\frac{24}{75}\)
(ii) □ में प्राप्त संख्या का न्यूनतम रूप… है।
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 8
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 8a

पाठ्य-पुस्तक पृष्ठ संख्या # 44

ज्ञात कीजिए-
3 ÷ \(\frac { 1 }{ 2 }\) और 3 x \(\frac { 2 }{ 1 }\)
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 9

भिन्न का व्युत्क्रम

रिक्त स्थानों की पूर्ति कीजिए-
हल :
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 9a
ऐसी शून्येतर संख्याएँ जिनका परस्पर गुणनफल 1 है, एक-दूसरे के व्युत्क्रम कहलाते हैं।

ऐसे पाँच और युग्मों को गुणा कीजिए।
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 10
सोचिए, चर्चा कीजिए एवं लिखिए

MP Board Solutions

प्रश्न
(i) क्या एक उचित भिन्न का व्युत्क्रम भी उचित भिन्न होगी?
(ii) क्या एक विषम भिन्न का व्युत्क्रम भी एक विषम भिन्न होगा?
(ii) इसलिए हम कह सकते हैं कि
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 10a
हल:
(i) नहीं, एक उचित भिन्न का व्युत्क्रम उचित भिन्न नहीं होगा।
(ii) नहीं, एक विषम भिन्न का व्युत्क्रम भी विषम भिन्न नहीं होगा।
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 10b

MP Board Solutions

पाठ्य-पुस्तक पृष्ठ संख्या # 45
प्रयास कीजिए

Class 7th Bhinn Ka Guna Question प्रश्न 1.
ज्ञात कीजिए:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 11
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 11a

Kaksha 7 Prashnawali 2 Point 3 प्रश्न:
हल कीजिए:
(i) 4 ÷ 2\(\frac { 2 }{ 5 }\) = 4 ÷ \(\frac { 12 }{ 5 }\) = ?
(ii) 5 ÷ 3\(\frac { 1 }{ 3 }\) = 5 ÷ \(\frac { 10 }{ 3 }\) = ?
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 12

प्रयास कीजिए

Class 7 Maths MP Board प्रश्न 1.
ज्ञात कीजिए:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 13
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 13a

पूर्ण संख्या से भिन्न को भाग
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 14

MP Board Solutions

मिश्रित भिन्न को पूर्ण संख्या से भाग
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 15

एक भिन्न को दूसरी भिन्न से भाग
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 16

MP Board Solutions

पाठ्य-पुस्तक पृष्ठ संख्या # 46
प्रयास कीजिए

MP Board Class 7th Maths प्रश्न 1.
ज्ञात कीजिए :
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 17
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 17a
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 17b

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1

MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1

MP Board Class 7 Maths Solutions Chapter 1 प्रश्न 1.
किसी विशिष्ट दिन विभिन्न स्थानों के तापमानों को डिग्री सोल्सियस (°C) में निम्नलिखित संख्या रेखा पर दर्शाया गया है:
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
(a) इस संख्या रेखा को देखिए और इस पर अंकित स्थानों के तापमान लिखिए।
(b) उपर्युक्त स्थानों में से सबसे गर्म और सबसे ठण्डे स्थानों के तापमानों में क्या अन्तर है?
(c) लाहुलस्पीती एवं श्रीनगर के तापमानों में क्या अन्तर है?
(d) क्या हम कह सकते हैं कि शिमला और श्रीनगर के तापमानों का योग शिमला के तापमान से कम है? क्या इन दोनों स्थानों के तापमानों का योग श्रीनगर के तापमान से भी कम है?
हल:
(a)
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1

(b) यहाँ, सबसे गर्म स्थान बैंगलोर (22°C) और सबसे ठण्डा स्थान लाहुलस्पीती (-8°C) है।
∴ सबसे गर्म और सबसे ठण्डे स्थानों के तापमानों का अन्तर
= 22°C – (-8°C) = 22°C + 8°C = 30°C

(c) लाहुलस्पीती का तापमान = – 8°C, श्रीनगर का तापमान = -2°C
∴ अभीष्ट अन्तर = -2°C – (-8°C)
= -2°C + 8°C = 6°C

(d) शिमला और श्रीनगर के तापमानों का योग = 5°C + (-2°C) = 3°C
अतएव, हम कह सकते हैं कि
हाँ, शिमला और श्रीनगर के तापमानों का योग शिमला के तापमान से कम है।
श्रीनगर का तापमान = -2°C
∵ 3°C > -2°C नहीं, इन दोनों स्थानों के तापमानों का योग श्रीनगर के तापमान से कम नहीं है।

MP Board Solutions

MP Board Class 7th Maths Chapter 1 प्रश्न 2.
किसी प्रश्नोत्तरी में सही उत्तर के लिए धनात्मक अंक दिए जाते हैं और गलत उत्तर के लिए ऋणात्मक अंक दिए जाते हैं। यदि पाँच उत्तरोत्तर चक्करों (rounds) में जैक द्वारा प्राप्त किए गए अंक 25, -5, – 10, 15 और 10 थे, तो बताइए अन्त में उसके अंकों का योग कितना था ?
हल:
∵ पहले चक्कर में अंक = 25
दूसरे चक्कर में अंक = -5
तीसरे चक्कर में अंक = -10
चौथे चक्कर में अंक = 15
पाँचवें चक्कर में अंक = 10
∴ कुल अंक = 25 + (-5) + (-10) + 15 + 10
= 50 – 15 = 35
अतएव, जैक के अंकों का योग = 35

MP Board Class 7 Maths Solutions English Medium प्रश्न 3.
सोमवार को श्रीनगर का तापमान -5°C था और मंगलवार को तापमान 2°C कम हो गया। मंगलवार को श्रीनगर का तापमान क्या था? बुधवार को तापमान 4°C बढ़ गया। बुधवार को तापमान कितना था ?
हल:
सोमवार को श्रीनगर का तापमान = – 5°C
∵ 2°C तापमान कम हो गया।
∴ मंगलवार को तापमान = – 5°C + (-2°C)= – 7°C
बुधवार को तापमान 4°C बढ़ गय
∴ बुधवार को तापमान = – 7°C +4°C = -3°C

MP Board Class 7 Maths Solutions Hindi Medium प्रश्न 4.
एक हवाई जहाज समुद्र तल से 5000 मीटर की ऊँचाई पर उड़ रहा है। एक विशिष्ट बिन्दु पर यह हवाई जहाज समुद्र तल से 1200 मीटर नीचे तैरती हुई पनडुब्बी के ठीक ऊपर है। पनडुब्बी और हवाई जहाज के बीच की ऊर्ध्वाधर दूरी कितनी है ?
पाठ्य-पुस्तक में दिये गये चित्र के अनुसार, समुद्र तल 0 मीटर पर है और हवाई जहाज समुद्र तल से 5000 मीटर की ऊँचाई पर है।
हल:
समुद्र तल और हवाई जहाज के बीच की दूरी = 5000 मीटर
और पनडुब्बी समुद्र तल से 1200 मीटर नीचे है।
∴ समुद्र तल और पनडुब्बी के बीच की दूरी = 1200 मीटर
∴ हवाई जहाज और पनडुब्बी के बीच दूरी
= 5000 मीटर + 1200 मीटर
= 6200 मीटर

Class 7th Maths MP Board प्रश्न 5.
मोहन अपने बैंक खाते में ₹ 2000 जमा करता है और अगले दिन इसमें से ₹1642 निकाल लेता है। यदिखाते में से निकाली गई राशि को ऋणात्मक संख्या से निरूपित किया जाता है, तो खाते में जमा की गई राशि को आप कैसे निरूपित करोगे? निकासी के पश्चात् मोहन के खाते में शेष राशि ज्ञात कीजिए।
हल:
चूँकि बैंक से निकासी की राशि बैंक में जमा की गई राशि के विपरीत है, अतएव जमा की गई राशि को धनात्मक संख्या से निरूपित करेंगे।
उत्तर जमा की गई राशि = ₹ 2000
निकाली गई राशि = – ₹ 1642
अतएव निकालने के बाद शेष राशि
= ₹ 2000 – ₹ 1642
= ₹ 358

MP Board Solutions

Class 7 Math MP Board प्रश्न 6.
रीता बिन्दु A से पूर्व की ओर बिन्दु B तक 20 किलोमीटर की दूरी तय करती है। उसी सड़क के अनुदिश बिन्दु B से वह 30 किलोमीटर की दूरी पश्चिम की ओर तय करती है। यदि पूर्व की ओर तय की गई दूरी को धनात्मक पूर्णांक से निरूपित किया जाता है, तो पश्चिम की ओर तय की गई दूरी को आप कैसे निरूपित करोगे? बिन्दु A से उसकी अन्तिम स्थिति को किस पूर्णांक से निरूपित करोगे?
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
हल:
पूर्व और पश्चिम की दिशाएँ एक-दूसरे की विपरीत है।
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
यदि पूर्व की ओर चली दूरी को धनात्मक संख्या मानें, तो पश्चिम की ओर चली गई दूरी को ऋणात्मक संख्या से निरूपित करेंगे।
अब, पूर्व की ओर तय की गई दूरी = + 20 किलोमीटर
और पश्चिम की ओर तय की गई दूरी = – 30 किलोमीटर
∴ बिन्दु A से उसकी अन्तिम स्थिति = – 30 + (+ 20) = -10 किलोमीटर

Class 7 Maths Solutions MP Board प्रश्न 7.
किसी मायावी वर्ग में प्रत्येक पंक्ति, प्रत्येक स्तम्भ एवं प्रत्येक विकर्ण की संख्याओंकायोगसमान होता है। बताइए निम्नलिखित में से कौन-सा वर्ग एक मायावी वर्ग है।
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
हल:
मायावी वर्ग (i)
अंकों का योग:
पहली पंक्ति = 5 + (-1) + (-4) = + 5 + (-5) = 0
दूसरी पंक्ति = (-5) + (-2) +7 = -7 + 7 = 0
तीसरी पंक्ति = 0 + 3 + (-3) = 3 + (-3) = 0
प्रथम स्तम्भ = 5 + (-5) + 0 = 5 + (-5) = 0
द्वितीय स्तम्भ = (-1) + (-2) + 3 = – 3 + 3 = 0
तृतीय स्तम्भ = (-4) + 7 + (-3) = – 7 +7 = 0
प्रथम विकर्ण = 5 + (-2) + (-3) = 5 + (-5) = 0
द्वितीय विकर्ण = 0 + (-2) + (-4) = – 2 – 4 = – 6
चूँकि वर्ग (i) में द्वितीय विकर्ण का योग अन्य पंक्ति, स्तम्भ एवं विकर्ण के बराबर (- 6 ≠ 0) नहीं है,
अतः वर्ग (i) मायावी वर्ग नहीं है।

मायावी वर्ग (ii)
अंकों का योग : पहली पंक्ति = 1 + (- 10) + 0 = 1 – 10 = –9
द्वितीय पंक्ति = (-4) + (-3) + (-2) = -9
तृतीय पंक्ति = – 6 + 4 + (-7) = – 13 + 4 = -9
प्रथम स्तम्भ = 1 + (-4) + (-6) = 1 + (- 10) = -9
द्वितीय स्तम्भ = (-10) + (-3) + 4 = – 13 + 4 = -9
तृतीय स्तम्भ = 0 + (-2) + (-7) = 0 – 9 = -9
प्रथम विकर्ण = 1 + (-3) + (-7)= 1 – 10 = -9
द्वितीय विकर्ण = (-6) + (-3) + 0 = – 6 – 3 = -9
मायावी वर्ग (ii) में प्रत्येक पंक्ति, स्तम्भ और विकर्ण का योग बराबर (-9) है।
अतः वर्ग (ii) एक मायावी वर्ग है।

MP Board Solutions

Class 7th Maths MP Board English Medium प्रश्न 8.
a और b के निम्नलिखित मानों के लिए a – (- b) = a + b का सत्यापन कीजिए :
(i) a = 21, b = 18;
(ii) a = 118, b = 125;
(iii) a = 75, b = 84;
(iv) a = 28, b = 11
हल:
(i) यहाँ, a = 21, b = 18
∴ L.H.S. = a – (-b)
= 21 – (-18)
= 21 + 18 = 39
R.H.S. = a + b
= 21 + 18 = 39
∵ L.H.S. = R.H.S
∴ a – (-b) = a + b सत्यापित है।

(ii) यहाँ, a = 118, b = 125
∴ L.H.S. = a – (-b)
= 118 – (- 125)
= 118 + 125 = 243.
और R.H.S. = a + b
= 118 + 125 = 243
∵ L.H.S. = R.H.S
∴ a – (-b) = a + b सत्यापित है।

(iii) यहाँ, a = 75, b = 84 .
∴ L.H.S. = a – (-b)
= 75 – (-84)
= 75 + 84 = 159
R.H.S. = a + b
= 75 + 84 = 159
∵ L.H.S. = R.H.S
∴ a – (-b) = a + b सत्यापित है।

(iv) यहाँ, a = 28, b = 11
∵ L.H.S. = a – (-b)
= 28 – (-11)
= 28 +11 = 391
और R.H.S: = a + b
= 28 + 11 = 39
∵ L.H.S. = R.H.S
∴ a – (-b) = a + b सत्यापित है।

MP Board Solutions

MP Board Class 7 Maths Solutions प्रश्न 9.
निम्नलिखित कथनों को सत्य बताने के लिए, बॉक्स में संकेत >, < अथवा = का उपयोग कीजिए-
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
हल:
(a) -8+ (-4) = -8 – 4 = – 12
तथा -8 – (-4) = -8 + 4 = -4
∴ (-12) < (-4)
अतः -8 + (-4) < -8 – (-4)

(b) -3 +7 – (19) = – 3 +7 – 19 = – 15
तथा 15 – 8 + (-9) = – 2
(-15) < (-2)
अतः -3 +7- (19) < 15 – 8 + (-9) (c) 23 – 41 + 11 = 34 – 41 = -7 तथा 23 – 41 – 11 = 23 – 52 = – 29 ∴ (-7) > (-29)
अतः 23 – 41 + 11 > 223 – 41 – 11

(d) 39+ (-24)- (15) = 39 – 24 – 15 = 0
तथा 36 + (-52) – (-36) = 36 – 52 + 36 = 20
∴ 0 < 20
अतः 39 + (-24) – (15) < 36 + (-52) – (-36) (e) -231 + 79 + 51 = – 231 + 130 = – 101 तथा -399 + 159+ 81 = -399 + 240 = -159 (-101) > (- 159)
अतः – 231 + 79 + 51 > -399 + 159 + 81

MP Board 7th Class Maths Book Solutions प्रश्न 10.
पानी के एक तालाब के अन्दर की ओर सीढ़ियाँ हैं। एक बन्दर सबसे ऊपर वाली सीढ़ी (यानी पहली सीढ़ी) पर बैठा हुआ है। पानी नौवीं सीढ़ी पर है।
(i) वह एक छलांग में तीन सीढ़ियाँ नीचे की ओर और अगली छलाँग में दो सीढ़ियाँ ऊपर की ओर जाता है। कितनी छलाँगों में वह पानी के स्तर तक पहुँच पाएगा?
(ii) पानी पीने के पश्चात् वह वापस जाना चाहता है। इस कार्य के लिए वह एक छलाँग में 4 सीढ़ियाँ ऊपर की ओर और अगली छलाँग में 2 सीढ़ियाँ नीचे की ओर जाता है। कितनी छलाँगों में वह वापस सबसे ऊपर वाली सीढ़ी पर पहुँच जाएगा?
(iii) यदि नीचे की ओर पार की गई संख्या को ऋणात्मक पूर्णांक से निरूपित किया जाता है और ऊपर की ओर पार की गई सीढ़ियों की संख्या को धनात्मक पूर्णांक से निरूपित किया जाता है तो निम्नलिखित को करते हुए भाग (i) और (ii) में उसकी गति को निरूपित कीजिए :
(a) – 3 + 2 – … = -8
(b) 4 – 2 +… = 8
(a) में योग (-8) आठ सीढ़ियाँ नीचे जाने को निरूपित करता है, तो (b) में योग 8 किसको निरूपित करेगा?
हल:
(i) बन्दर पहली सीढ़ी पर बैठा हुआ है।
∴ छलाँगों के बाद बन्दर की स्थिति निम्न प्रकार होगी : पहली छलाँग में वह चौथी सीढ़ी पर होगा।
दूसरी छलाँग में वह दूसरी सीढ़ी पर होगा। (∵4 – 2 = 2)
तीसरी छलाँग में वह पाँचवीं सीढ़ी पर होगा। (∵ 2 + 3 = 5)
चौथी छलाँग में वह तीसरी सीढ़ी पर होगा। (∵ 5 – 2 = 3)
पाँचवीं छलाँग में वह छठी सीढ़ी पर होगा। (∵ 3 + 3 = 6)
छठवी छलाँग में वह चौथी सीढ़ी पर होगा। (∵ 6 – 2 = 4)
सातवीं छलाँग में वह सातवीं सीढ़ी पर होगा। (∵ 4 + 3 = 7)
आठवीं छलाँग में वह पाँचवीं सीढ़ी पर होगा। (∵ 7 – 2 = 5)
नौवीं छलाँग में वह आठवीं सीढ़ी पर होगा। (∵ 5 + 3 = 8)
दसवीं छलाँग में वह छठवीं सीढ़ी पर होगा। (∵ 8 – 2 = 6)
ग्यारहवीं छलाँग में वह नौवीं सीढ़ी पर होगा। (∵ 6 + 3 = 9) जो कि पानी का स्तर है।
अत: पानी के स्तर तक वह 11 छलाँगों में पहुँच पाएगा।

(ii) यहाँ, बन्दर की स्थिति पानी के स्तर अर्थात् नौवीं सीढ़ी पर है।
∴ छलाँगों के बाद बन्दर की ऊपर वाली सीढ़ी से स्थिति निम्न प्रकार होगी :
पहली छलाँग में वह पाँचवीं सीढ़ी पर होगा। (∵ 9 – 4 = 5)
दूसरी छलाँग में वह सातवीं सीढ़ी पर होगा। (∵ 5 + 2 = 7)
तीसरी छलाँग में वह तीसरी सीढ़ी पर होगा। (7 – 4 = 3)
चौथी छलाँग में वह पाँचवीं सीढ़ी पर होगा। (3 + 2 = 5)
पाँचवीं छलाँग में वह पहली (ऊपर की) सीढ़ी पर होगा। (5 – 4 = 1)
∴ अभीष्ट छलाँगों की संख्या = 5

(iii) चूँकि बन्दर द्वारा ऊपर की ओर पार की गई सीढ़ियों की संख्या को धनात्मक पूर्णांक से निरूपित किया जाता है और नीचे की ओर पार की गई सीढ़ियों की संख्या को ऋणात्मक पूर्णांक से निरूपित किया जाता है। अतः भाग (i) में बन्दर की गति
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
(a) – 3 + 2 -3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 = -8

भाग (ii) में बन्दर की गति
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
(b) 4 – 2 + 4 – 2 + 4 = 8
(b) में योग 8 ऊपर की ओर 8 सीढ़ियाँ चढ़ने को निरूपित करता है।

पाठ्य-पुस्तक पृष्ठ संख्या # 06

नीचे दी हुई सारणी को देखिए और इसे पूरा कीजिए :
आप क्या देखते हैं? क्या दो पूर्णांकों का योग हमेशा एक | पूर्णांक प्राप्त होता है? क्या आपको पूर्णांकों का ऐसा युग्म मिला जिसका योग पूर्णांक नहीं है। क्या पूर्णांक योग के अन्तर्गत संवृत होते हैं।
हल:
कथन                                            प्रेक्षण
(i) 17 + 23 = 40                   परिणाम एक पूर्णांक है।
(ii) (-10) +3 =-7                  परिणाम एक पूर्णांक है।
(iii) (-75) + 18 = – 57         परिणाम एक पूर्णांक है।
(iv) 19 + (-25) = – 6           परिणाम एक पूर्णांक है।
(v) 27 + (-27) = 0              परिणाम एक पूर्णांक है।
(vi) (-20) + 0 = – 20         परिणाम एक पूर्णांक है।
(vii) (-35) + (-10) = – 45  परिणाम एक पूर्णांक है।

(a) हम देखते हैं कि किन्हीं दो पूर्णांकों का योग हमेशा एक पूर्णांक होता है।
(b) हाँ, दो पूर्णांकों का योग सदैव पूर्णांक होता है।
(c) हमें यहाँ पूर्णांकों का ऐसा युग्म प्राप्त नहीं हुआ जिनका योग एक पूर्णांक न हो।
(d) अतः पूर्णांकों का योग पूर्णांक ही होता है। इसलिए हम कहते हैं कि पूर्णांक योग के अन्तर्गत संवृत (closed) होते हैं।

पाठ्य-पुस्तक पृष्ठ संख्या # 07

निम्नलिखित सारणी को देखिए और इसे पूरा कीजिए :
आप क्या देखते हैं? क्या पूर्णांकों का कोई ऐसा युग्म है जिसका अन्तर पूर्णांक नहीं है?
क्या हम कह सकते हैं कि पूर्णांक व्यवकलन के अन्तर्गत संवृत होते हैं?
हल:
कथन                                              प्रेक्षण
(i) 7 – 9 = – 2                         परिणाम एक पूर्णांक है।
(ii) 17 – (-21) = 38               परिणाम एक पूर्णांक है।
(iii) (-8) – (-14) = 6             परिणाम एक पूर्णांक है।
(iv) (-21) – (-10) = – 11       परिणाम एक पूर्णांक है।
(v) 32 – (-17) = 49              परिणाम एक पूर्णांक है।
(vi) (-18) – (-18)= 0           परिणाम एक पूर्णांक है।
(vii) (-29 )- 0 = – 29          परिणाम एक पूर्णांक है।

(a) हम देखते हैं कि दो पूर्णांकों का व्यवकलन भी एक पूर्णांक होता है।
(b) नहीं, पूर्णांकों का ऐसा कोई युग्म नहीं है जिसका अन्तर पूर्णांक नहीं है।
(c) हाँ, हम कह सकते हैं कि पूर्णांक व्यवकलन के अन्तर्गत संवृत होते हैं।

MP Board Solutions

MP Board Maths Class 7 प्रश्न 1.
क्या पूर्ण संख्याएँ भी इस गुण को सन्तुष्ट करती हैं ?
उत्तर:
नहीं, पूर्ण संख्याएँ इस गुण को सन्तुष्ट नहीं करतीं।

Class 7 MP Board Maths प्रश्न 2.
क्या निम्नलिखित समान हैं ?
(i) (-8) + (-9) और (-9) + (-8)
(ii) (-23) + 32 और 32 + (-23)
(iii) (-45) + 0 और 0 + (-45)
पाँच अन्य पूर्णांकों के युग्मों के लिए ऐसा प्रयास कीजिए। क्या आपको पूर्णांकों का कोई ऐसा युग्म मिलता है जिसके लिए पूर्णांकों का क्रम बदल देने में उनका योग भी बदल जाता है।
हल:
(i) (-8) + (-9)= – 8 – 9 = – 17
और (-9) + (-8) = -9 – 8 = – 17
हाँ, (-8) + (-9) और (-9) + (-8) समान हैं।

(ii) (-23) + 32 = 32 – 23 = 9
और 32 + (-23) = 32 – 23 = 9
हाँ, (-23) + 32 और 32 + (-23) समान हैं।

(iii) (-45) + 0 = -45 + 0 = -45
और 0 + (-45) = 0 – 45 = – 45
हाँ, (-45) + 0 और 0 + (-45) समान हैं।
उदाहरण:
(a) (-36) + (-15) = -51
और (-15) + (-36) = – 51
∴ (-36) + (-15) = (-15) + (-36)

(b) (-10) + 6 = – 10 + 6 = -4
और 6 + (-10) = 6 – 10 = -4
∴ (-10) + 6 = 6 + (-10)

(c) (-118) + 0 = – 118 + 0 = – 118
और 0 + (-118) = 0 – 118 = – 118
∴ (-118)+ 0 = 0+ (- 118)

(d) (-12) + 10 = – 12 + 10 = – 2
और 10 + (-12) = 10 – 12 = -2
∴ (-12) + 10 = 10 + (-12)

(e) 53 + (-26) = 53 – 26 = 27
और (-26)+ 53 = – 26 + 53 = 27
∴ 53 + (-26) = (-26) + 53

पाठ्य-पुस्तक पृष्ठ संख्या # 08

पूर्णांकों के कम-से-कम पाँच विभिन्न युग्म लीजिए और इस कथन की जाँच कीजिए कि व्यवकलन पूर्णांकों के लिए क्रम-विनिमेय नहीं हैं।
हल:
(i) 100 – 86 और 86 – 100
100 – 86 = 14 और 86 – 100 = – 14.
∴ 100 – 86 ≠ 86 – 100

(ii) (-19) और 5
(-19) – 5 = – 24 और 5-(-19) = 24
∴ (-19 – 5 ≠ 5- (-19)

(iii) (-17) और (-19)
(-17) – (- 19) = – 17 + 19 = 2
और (-19) – (-17) = – 19 + 17 = – 2
∴ (-17) – (-19) ≠ (-19) – (-17)

(iv) 69 और 0
69 – 0 = 69 और 0 – 69 = – 69
∴ 69 – 0 ≠ 0 – 69

(v) 118 और (-56)
118 – (-56) = 118 + 56 = 174
और (-56)- 118 = – 174
118 – (-56) ≠ (-56) – 118
उपर्युक्त उदाहरणों से स्पष्ट है कि व्यवकलन पूर्णांकों के लिए क्रम-विनिमेय नहीं है।
अर्थात् a – b # b – a.
इसी प्रकार – 3, 1 और – 7 को लीजिए।
-3 + [1 + (-7)] = -3 + (-6) = -9
[(-3) + 1] + (-7) = – 2 + (-7) = -9
इसी प्रकार के पाँच और उदाहरण लीजिए :
उदाहरण;
(i) -9, -4 और 6
(-9) + [(-4) + 6] = -9+ 2 = -7
और [(-9) + (-4)] + 6 = – 13 + 6 = -7
अतः (-9) + [(-4) + 6] = [(-9) + (-4)] + 6

(ii) -2, 10 और 5
= 8 + 5 = 13
और (-2) + [ 10 + 5] = -2 + 15 = 13
अतः [(-2) + 10] + 5 = – 2 + [10 + 5]

(iii) 13, – 12 और -7
[13 + (-12)] + (-7) = 1-7 = – 6
और 13 + [(-12) + (-7)] = 13 – 19 = -6
अतः [13 + (-12)] + (-7) = 13 + [(-12) + (-7)]

(iv) -4, 15 और -3
[(-4) + 15] + (-3) = 11 – 3 = 8 और -4 + [15 + (-3)] = – 4 + 12 = 8
अतः [(-4) + 15] + (-3) = -4+ [15 + (-3)]

(v) – 12, 19 और 15
[(-12)+ 19] + 15 = 7 + 15 = 22
और -12 + [19 + 15] = – 12 + 34 = 22
अतः [(-12) + 19] + 15 = – 12 + [19+ 15]
अतएव पूर्णांकों के लिए योग सहचारी (Associative) है।
∴ (a + b) + c = a + (b + c)

पाठ्य-पुस्तक पृष्ठ संख्या # 09

निम्नलिखित को देखिए और रिक्त स्थानों की पर्ति कीजिए-
हल:
(i) (-8) + 0 = – 8
(ii) 0 + (-8) = -8
(iii) (-23)+ 0 = – 23
(iv) 0 + (-37) = -37
(v) 0+ (-59) = -59
(vi) 0+ (-43) = – 43
(vii) -61 + 0 = – 61
(viii) -45 + 0 = – 45
उपर्युक्त उदाहरण दर्शाते हैं कि शून्य और ऋणात्मक पूर्णांकों का योग सदैव उसी पूर्णांक के बराबर होता है। अतएव किसी पूर्णांक a के लिए शून्य योज्य तत्समक है।
a + 0 = 0 + a = a

प्रयास कीजिए

Class 7 Maths Chapter 1 Exercise 1.1 Solutions Hindi Medium प्रश्न 1.
एक ऐसा पूर्णांक युग्म लिखिए जिसके योग से हमें निम्नलिखित प्राप्त होता है
(a) एक ऋणात्मक पूर्णांक
(b) शून्य
(c) दोनों पूर्णांकों से छोटा एक पूर्णांक
(d) दोनों पूर्णांकों में से केवल किसी एक से छोटा पूर्णांक
(e) दोनों पूर्णांकों से बड़ा एक पूर्णांक।
हल:
(a) -25 और 9
योग- (-25) + 9 = -16; -16 एक ऋणात्मक पूर्णांक है।
(b) – 27 और 27
योग -(-27) + 27 = 0

(c) – 16 और -4
योग – (-16) + (-4) = – 20; – 20 पूर्णांक – 16 और -4 से छोटा है।

(d) 4 और -6
योग-4+ (-6) = -2 ; – 2 केवल 4 से छोटा है।

(e) 29 और 11
योग-29 + 11 = 40; 40 पूर्णांक 29 और 11 से बड़ा है।

MP Board Solutions

MP Board Class 7 Maths Chapter 1 प्रश्न 2.
एक ऐसा पूर्णांक युग्म लिखिए जिसके अन्तर से हमें निम्नलिखित प्राप्त होता है
(a) एक ऋणात्मक पूर्णांक
(b) शून्य
(c) दोनों पूर्णांकों से छोटा एक पूर्णांक
(d) दोनों पूर्णांकों में से केवल किसी एक से बड़ा पूर्णांक
(e) दोनों पूर्णांकों से बड़ा एक पूर्णांक
हल:
(a) 13 और – 8
अन्तर – ( – 8) – 13 = – 21; – 21 एक ऋणात्मक पूर्णांक है।
(b) – 13 और – 13
अन्तर – (-13) – (- 13) = – 13 + 13 = 0
(c) 15 और 19
अन्तर -19 – 15 = 4; 4 पूर्णांक 19 और 15 से छोटा
(d) 16 और 7
अन्तर -16 – 7 = 9; 9 पूर्णांक 7 से बड़ा है।
(e) 18 और -6
अन्तर – 18 – (-6) = 24; 24 पूर्णांक 18 और -6 से बड़ा है।

MP Board Class 7th Maths Solutions

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

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 an है :
हल:
MP Board Class 10th Maths Solutions Chapter 5 समान्तर श्रेढ़ियाँ Ex 5.2 1
हल:
(i) ∵ an = a + (n – 1) × d
⇒ an = 7 + (8 – 1) × 3
⇒ an = 7 + 7 × 3
⇒ an = 7 + 21 = 28

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

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

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

MP Board Solutions

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 an = 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 an = a + (n – 1)d ⇒ an = -3 + (11 – 1) (2.5)
⇒ a11 = -3 + 10 × 2.5 = -3 + 25 = 22

MP Board Solutions

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, an = 78
एवं an = 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।

MP Board Solutions

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 an = 205
चँकि an = 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 an = -47
चूँकि an = 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, an = – 150.
चूँकि an = 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।

MP Board Solutions

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 an = 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 an = 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 an – 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।

MP Board Solutions

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 an = 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 an = 248
चूँकि an = 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।

MP Board Solutions

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,
an = 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 an = ₹ 20.75, तो प्रश्नानुसार,
an = 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।