Class 12

Std 12 Geography Chapter 3 Question Answer Human Settlements and Land Use Maharashtra Board

Balbharti Maharashtra State Board Class 12 Geography Solutions Chapter 3 Human Settlements and Land Use Textbook Exercise Questions and Answers.

Class 12 Geography Chapter 3 Human Settlements and Land Use Question Answer Maharashtra Board

Geography Class 12 Chapter 3 Question Answer Maharashtra Board

1. Identify the correct correlation.

A : Assertion R : Reasoning
Question 1.
A – Settlements can be of various types.
R – Various physical factors affect the growth of settlements.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(d) Both A and R are correct but R is not the correct explanation of A.

Question 2.
A – When cities grow, their functions also grow.
R – Cities can have only one function.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is the correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(a) Only A is correct.

2. Give geographical reasons.

Question 1.
Not all rural settlements change into urban settlements.
Answer:

• The area between rural and urban is called rural-urban fringe.
• Villages are beyond the rural-urban fringe and cities have different land use pattern.
• In villages, land is mainly used for agriculture and related activities such as permanent pasture, grazing land, miscellaneous tree crops and groves, fallow land etc.
• In city areas land is mainly used for industries, residential purpose, recreation, transportation etc.
• Since the villages are far away from the city, they maintain their distinct identity and do not change into urban settlement.

Question 2.
In rural settlements, land use is related to agriculture.
Answer:

• Generally, the land in rural areas is used for agriculture and related activities.
• The classification of the land use in rural areas is done according to the Land Records Department.
• As per Land Records Department, the land in rural areas is mainly used for activities related to agriculture, such as some land is under permanent pastures and grazing lands or some under tree crops or culturable waste-land or fallow land etc.
• Thus, all the above types of land use around the rural settlements are related to mainly agriculture.

Question 3.
Rural-urban fringe have the characteristics of both urban and rural settlements.
Answer:

• The area between urban and rural areas is called rural-urban fringe.
• It has the characteristics of both urban as well as rural areas, since it is a transition zone between the two.
• Thus, in rural-urban fringe there is a mixture of urban-rural land use.
• In some rural areas apart from the land use for agriculture, some agricultural land has been converted into residential and industrial uses.
• The villages in rural fringe are partly affected by urbanization.
• Thus, rural-urban fringe has the characteristics of both urban and rural settlements.

Question 4.
Growth of urban areas is linked to land use.
Answer:

• Land use in urban areas is different from land use in rural areas.
• In rural areas the land use is closely related to agricultural activities like cultivation of different crops, plantation of trees, permanent pastures, grazing land, cultivation of tree crops, fallow land etc.
• On the other hand, land use in urban areas is varied and closely related to housing and economic activities.
• As the population of the urban area increases, more and more non-agricultural activities develop in urban areas.
• Growth of urban areas depends upon area under construction, industries, different types of institutions such as school, college, insurance companies, bank etc.
• Recreational activities, transportation are the other urban land uses.
• Thus, growth of urban areas is linked with land use.

3. Write short notes on.

Question 1.
Interrelationship between urban and rural settlement.
Answer:

• Settlements can be divided into urban and rural on the basis of their functions.
• Rural settlements are smaller than urban settlements.
• In rural areas agriculture and allied agricultural activities like livestock rearing, fishing, lumbering etc., are developed.
• In urban areas, industries, construction and economic activities like trade, transport and communication, banking and insurance are the important activities.
• The development industries need various raw material and services which are supplied by rural areas.
• For example, cotton grown in rural areas is supplied to cotton textile industries in urban areas. Everyday many rural people commute to urban areas to work in different activities.
• Thus, there is good interrelationship between urban and rural areas since they depend on each other.

Question 2.
Problems of urban settlements.
Answer:

• Most of the urban areas have very large size of population and density of population is also very high.
• When cities increase in size, many changes occur. These changes are related to land use and structure of the city.
• The large size of population and high density create number of socio-economic, cultural, infrastructural, administrative and environmental problems.
• The air, water and noise pollution, development of slums, traffic jam, overcrowding in trains and buses, waste disposal etc., are some of the serious problems in most of the urban settlements.

Question 3.
Suburbs
Answer:

• In the outer part of the urban areas there are small towns or small cities, they are known as suburbs.
• When big cities become overcrowded and overpopulated, the further development starts outside city area and thus suburbs develop.
• For example, Dombivali, Kalyan, Ambarnath etc., are the suburbs of Mumbai.
• Suburbs generally consists of residential housing and shops of low order, which act as central place for the local community.
• Often, suburbs are the most recent growth of an urban area and their end marks the urban fringe.
• With increase in population there is growth of suburb, the growth of suburbs may result in urban sprawl.

Question 4.
Mixed land use.
Answer:

• Mixed land use is observed in some of the urban areas.
• It is an area where different types of land use exist together.
• In some cities residential, industrial, commercial, administrative functions are found in an integrated manner.
• In many cities in developing countries one can find schools, clinics, houses, business shops at one place itself.
• Generally mixed land use is found in cities which are growing very fast, because land in the city is not sufficient to reserve certain areas for certain land use, therefore there is mixed land use.

4. Answer the following questions in detail.

Question 1.
Explain the characteristics of rural settlement.
Answer:

• On the basis of functions, settlements can be divided into two types – rural and urban.
• Agriculture and allied agricultural activities like fishing, livestock rearing, lumbering etc., are most economic activities in rural areas.
• The classification of land use in rural areas is done according to Land Records Department.
• As per Land Records Department the land in rural areas is mainly used for activities related to agriculture, such as some land is under agriculture, some under permanent pastures and grazing lands or some under tree crops or culturable waste-land or fallow land etc.
• Most of rural settlements are semi-clustered or fragmented and small in size.
• Primary activities like agriculture, lumbering, fishing, livestock rearing is more developed in rural settlements.
• In rural areas, sometimes agricultural areas are converted into residential or industrial areas.
• Many people in rural areas daily commute to city areas for work, thus rural and urban areas are connected with each other.
• There is an area between rural and urban area which is called rural-urban fringe, rural settlements are beyond the rural-urban fringe.

Question 2.
What factors are responsible for development of various patterns in settlement? Give examples.
Answer:
1. Patterns of settlements are affected by various physical factors like relief, soils, climate, availability of water supply etc.

2. Physical factors influence the type and spacing of settlements, which results into various patterns of settlements.

3. Type of soil and quality of soil are two important factors which affect rural settlements.

4. Fertile plains and valleys have thick, rich and fertile alluvial soil, which supports agriculture, so nucleated settlements develop in these areas. For example, most of the villages in Ganga plains have nucleated settlements.

5. Settlements develop as per the relief of that area. For example, Foothill settlements develop at the foot of mountain, hilltop settlements develop at the top of the hill. For example, hilltop settlement at Shimla or Manali.

6. In the areas of mountainous or hilly relief, due to inaccessibility, there are dispersed or isolated settlements. For example, dispersed or isolated settlements in Himalaya mountains.

7. Sometimes settlements develop along the coastline, canal, river, road, or railway line. They are in straight line; they are called linear settlements. For example, settlements along Konkan coastline and settlements along Mumbai-Pune road.

8. Climate also affects development of settlements. Areas of extreme climate are avoided by people. Therefore, such areas have dispersed settlements.

9. For example, due to extreme hot climate there are dispersed settlements in Rajasthan and due to extreme cold climate, there are isolated settlements in the polar regions.

10. Water is essential for human development. Therefore, many settlements develop around lakes or natural tanks, they are circular settlements.

11. Sometimes settlements develop at the confluence of two rivers, the settlements grow in all three sides, they are triangular settlements. For example, Karad town is on the confluence of Krishna and Koyna river.

12. Sometimes settlements grow around the mines, or any central object. This centre point provides source of livelihood to the people. Thus, radial pattern of settlement develops. For example, radial settlements around coal mines in the State of Bihar.

5. Differentiate between.

Question 1.
Land Use and Land Cover
Answer:

Land Use	Land Cover
(i) Land use describes the use of land by people for different activities, such as recreation, housing, agriculture, educational institutes etc.	(i) Land cover describes the physical surface covering the land such as forest rock, ice, sand, water etc.
(ii) For example, in any city, people may use some land for housing, some for recreation, some for sports ground etc.	(ii) For example, in any area land may be covered by natural factors such as vegetation, river, sand dune, snow covered mountain etc.
(iii) Land use may change from place to place depending upon type of land and need for the people.	(iii) Land cover is natural factor it hardly changes unless man purposefully makes changes in it.
(iv) Land use cannot be studied by the satellite imagery alone.	(iv) Land cover can be studied by satellite imagery.

Question 2.
Barren and Non-agricultural Land
Answer:

Barren Land	Non-agricultural Land
(i) The land which is not used for any human activity is called barren land. In other words, it is wasteland.	(i) The land which is used by man for various human activities except agricultural activity, is called as non-agricultural land.
(ii) For example, hilly or mountainous land, desert land, ravines, swampy and marshy land etc.	(ii) For example, land used for housing, industries, construction of roads, railways etc.
(iii) Generally barren land cannot be used for agriculture or other activities with available technology.	(iii) Due to the development of secondary or tertiary activities there is increase in the used of non-agricultural land.

Question 3.
Radial pattern and Circular Pattern
Answer:

Radial Pattern	Circular Pattern
(i) In radial pattern settlements grow around certain object.	(i) In circular pattern settlements grow around water body.
(ii) They develop near temple or a centre of commercial activity.	(ii) They develop near lake or natural tank.
(iii) For example, settlements near Vindhyachal in Uttar Pradesh.	(iii) For example, settlements around Nainital lake in Uttarakhand.

Question 4.
Nucleated and Dispersed Settlement
Answer:

Nucleated Settlement	Dispersed Settlement
(i) Settlement where buildings or houses are grouped or clustered around a central point or nucleus is known as nucleated settlements.	(i) Settlement where buildings or houses are scattered or dispersed, is known as dispersed settlement.
(ii) Houses or buildings are very close to each other. There may be common wall between two houses.	(ii) Houses or buildings are far from each other. One house may be at a distance of half a kilometre from the other.
(iii) Geographical factors are favourable such as spring or fertile river valley.	(iii) Geographical factors are not favourable such as extreme climate, barren land etc.
(iv) For example, settlements in river valleys.	(vi) For example, settlements in the desert of Rajasthan.

6. Draw a neat and labelled diagram

(1) Linear settlement
(2) Radial settlement
(3) Compact settlement
(4) Dispersed settlement
Answer:

7. Write a note in your own words about how land used in Lonar city has evolved. Refer the map on textbook page no. 30.

Answer:

• Two maps of Lonar city are given. One map is of the year 2005-06 and another is of 2015-16.
• These two maps show the changes in land use that have taken place in the span of 10 years.
• The following changes have been registered.
• There is no change in the size of Lonar lake.
• The area occupied by Lonar city has increased substantially.
• Forest area around the lake has increased.
• An area under waste land/ scrubs have increased.
• Around the temple in the southeast, on the waste scrub land a new rural hospital, government hostel, government ITI, Tahsil office etc., has been developed. This newly developed area has been named as Krishna Nagar, which is not in 2005-06 map.
• To the north-east of the Lonar city two more building have been constructed. One is central public school and another is civil and criminal court.
• To the south of the temple in the heart of Lonar city built up residential area is spreading over built up residential sparse.
• Overall built up residential sparse is increasing in all directions around Lonar lake and south-eat of the Lonar city and thus there is encroachment over agricultural land.

8. Read the given passage and answer the following questions.

Different types of human settlements include hamlets, villages, small towns, large towns, isolated places, cities and conurbations. In some systems, types of human settlements are broken up into urban, suburban and rural; for example, the U.S. Census Bureau divides settlements into urban or rural categories based on precise definitions. Small settlements, such as hamlets and villages, have low populations and restricted access to services, larger types of settlements, such as cities, have higher populations, higher densities and greater access to services.

For example, a village may have only one or two general stores, while a large metropolis may have many specialized stores and chain stores. These differences are known as low-order service settlements and high-order service settlements. Larger settlements also have a sphere of influence affecting surrounding settlements. Settlements may also be divided by the site chosen, such as sites selected based on resources, trading points, defensive sites, shelter and relationship to water resources. The functions of human settlements also differ, as settlements may be established as ports, market towns and resorts. Types of rural settlements may also be classified by function, such as proximity to farming, fishing and mining. Settlements that focus on one economic activity are called single functional settlements. Human settlements may be permanent or temporary. For example, a refugee camp is a temporary settlement, while a city is a permanent settlement.

Question 1.
Which human settlements are mentioned in the passage above?
Answer:
Hamlets, villages, small towns, large towns, isolated places, cities and conurbations are the types of settlements mentioned in the passage.

Question 2.
On what basis are urban and rural areas classified?
Answer:
Settlement are classified on the basis of the size of population, density, access to higher order and lower order services, site chosen, functions, permanent or temporary etc.

Question 3.
What are the functions carried out in rural settlement?
Answer:
Functions carried out in rural settlements are farming, fishing, mining, one or two general stores, etc.

Question 4.
Explain the difference between low-order services and higher order service settlements.
Answer:

• The hamlets and villages have low population and restricted access to lower order services.
  For example, a village may have only one or two general stores. They are called low order settlements.
• Large metropolitan cities have higher population, higher density and greater services of higher order.
• For example, large metropolitans may have chain stores, malls, departmental stores, super markets etc. They are called higher order settlements.

Class 12 Geography Chapter 3 Human Settlements and Land Use Intext Questions and Answers

Try These

Question 1.
Observe Fig 3.2 A to F (Textbook Page No. 22-23). They show various patterns of settlements. Try to understand the difference between them. Carefully read their characteristics in the second column. According to the applicable characteristics, write alphabet of the image settlement in the place provided below characteristics.
Answer:

Satellite image of the settlements	Characteristics of settlements
Photo A	C
Photo B	A
Photo C	F
Photo D	E
Photo E	D
Photo F	B

Question 2.
Can you identify problems faced by your city/town/village in terms of any of the following? (Textbook Page No. 27)

Types of problems	Problems / Issues
1. Economic
2. Social
3. Cultural
4. Environmental
5. Infrastructural
6. Governance and Administrative
7. Others

Answer:

Types of problems	Problems / Issues
1. Economic	Unemployment and poverty
2. Social	Religious conflicts and tension in society
3. Cultural	Commercialisation of festivals
4. Environmental	Air, water and noise pollution
5. Infrastructural	Inadequate roads, bridges and public transport
6. Governance and Administrative	Increase in crime rate and bribery
7. Others	Overcrowding, traffic jam, housing problem

Make friends with maps!

Question 1.
See map of Ichalkaranji city (Textbook Page No. 28) and observe how changes have occurred in the city over the years. Answer the questions that follow.
(i) Enlist the colours used for showing land uses in the index.
(ii) What do the blue and black lines show?
(iii) What is the name of the river in the map?
(iv) Name any two villages shown on the map.
(v) Which city is shown on the map?
(vi) Which periods do the map belongs to?
(vii) Which land covers have reduced? What are their colours?
(viii)Which landcovers seen to have increased? what are their colours?
(ix) Which land cover has been replaced by increased landcovers?
(x) Write a conclusive note comparing both the maps.
Answer:
(i)

• Yellow for agriculture
• Green for forest
• Pink for residential land use
• Dark blue for industrial land use
• Dark green for recreation
• Brown for mining/ quarry
• Violet for public/semi public

(ii) Blue lines show rivers and black lines show roads.

(iii) Panchaganaga river

(iv) Jambhali and Haroli villages

(v) Ichalkaranji

(vi) 2007 and 2017

(vii) Open space-dark green colour, forest – light green colour, wasteland – light violet colour residential sparse – light orange

(viii) Following landcovers have increased landcover for residential area – colour pink, Landcover for industrial area colour dark blue

(ix) Following land cover have been replaced by increased landcovers. Residential sparse areas are replaced by residential built up land use, open space is replaced by residential built up area and wastelands are replaced by industries.

(x) Map A shows land use and land cover map of Ichalkaranji city in the year 2007.

• Map B shows land use and land cover map of Ichalkaranji city in the year 2017.
• Both the maps show that there is land use for agriculture, industries, transportation, residential purpose, recreational purpose, mining around the city, village settlements etc.
• In both the maps there is land is covered by Forest, waterbodies, wasteland/ scrubs, Panchaganaga river etc.
• The landcover in 2017 as compared to landcover in 2007 have been reduced for forest, residential sparse, open space and wasteland.
• The landcover in 2017 as compared to landcover in 2007 have been increased for industries and built up residential areas.
• Residential sparse areas are replaced by residential built up land use, open space is replaced by residential built up area and wasteland are replaced by industries.

Can you tell? (Textbook Page No. 27)

You know what is urban and what is rural. What will you call the area that lies between them?
Answer:

• The area between urban and rural areas is called rural-urban fringe.
• It is characterised by the urban as well as rural characteristics, since it is transition zone between the two.
• Thus, in rural-urban fringe there is a mixture of urban-rural land use.
• In some rural areas apart from the land use for agriculture, some agricultural land has been converted into residential and industrial uses.
• Thus, villages in rural fringe are partly affected by urbanization.
• Thus rural-urban and fringe have the characteristics of both urban and rural settlements.

Find out (Textbook Page No. 27)

Compare the cover page of Std. XII text book with Std. XI geography text book. Discuss and write a short paragraph about changes in land use / land cover in your own words.
Answer:
Geography Cover Page (Textbook of standard XI)

• Depicts the natural landscape.
• There are two snow covered mountains peaks and rivers having their source in these mountains.
• At the foot of the mountain there is fan shape deposit of silt.
• The river has developed number of meanders and an ox-bow lake.
• The slope of the mountains is covered with coniferous forest and on the lower ground at the foot of the mountain there is mixed forest.
• There is a sandy beach. Along the beach there are coconut trees.

Geography Cover Page (Textbook of standard XII)

• Depicts cultural/man made development super imposed on natural landscape.
• A quarry is developed at the foot of mountain.
• There is deforestation and development of two villages and a town on the right bank of the river.
• On this bank of river there is development of industry as well.
• Number of multi-story building have come up on the left bank of the river including a mall and hospital.
• Power line, concrete road and railway have developed in the last 10 years.
• On the beach hotels, rest houses, sport activities have been developed for tourists and therefore number of tourists are seen on the beach.
• The natural landscape on cover of the textbook of Std. XI changes into cultural landscape on the cover page of geography textbook of Std. XII.

Let’s recall (Textbook Page No. 24)

Can you differentiate between urban and rural settlements?
Answer:

• On the basis of functions, settlements are divided into two types – rural settlement and urban settlement.
• There is difference in land use in rural and urban settlements.
• In rural areas the land use is closely related to agricultural activities like cultivation of different crops, plantation of trees, permanent pastures, grazing land, cultivation of tree crops, fallow land etc.
• Where as in urban areas land use is for industries, construction and economic activities like trade, transport and communication, banking and insurance etc.
• Urban settlements are large and compact, since population is more compared to available land.
• Rural settlements are small and dispersed, since population is less compared to available land.

Think about it (Textbook Page No. 24)

Can a town have only one function? Why do the cities become multi-functional?
Answer:
1. Towns do not have only one function.

2. Some towns have one important and major function. They are known by that function. But they have many other functions also. For example, Shirdi in Maharashtra is known for religious function but it has other functions like tourism, education, commercial etc.

3. Cities become multifunctional as they grow. With increase in population demand for various functions increases. As cities grow in size many changes occur and therefore land use also changes.

4. For example, when any city develops as industrial centre, its main function is industries. But as people start coming to that city for employment opportunities, the city grows. Then other functions like educational institutes, business centres, recreational centres, etc., functions develop to fulfil the needs of increasing population. Thus, cities become multi-functional.

Question 1.
Observe Fig. (Textbook Page No. 21) and answer the following questions.

(i) Where are humans’ settlements likely to develop: A, B, C, D or E? Why?
(ii) In the above figure in which place human settlement is not likely to develop? Why?
(iii) Looking at the figure above, what factor do you think could contribute to the development of human settlements?
(iv) Can economic factors be important along with physical factors for the development of human settlements?
(v) Do physical factors affect the economic activity of human settlements?
(vi) Make a list of factors which affect development of settlements in an area.
Answer:
(i) Human settlement is likely to develop at C and D.
The most important factor responsible for the development of settlement is river. Hence, agriculture seems to be the most important activity and development of agriculture needs fertile soil and water supply which is readily available here.

(ii) Human settlement is not likely to develop at A and B. This is because of steep slope and rugged terrain.

(iii) The most important factors that could contribute to the development of human settlements are availability of water from the river and fertile soil on the bank of river.

(iv) Yes, economic factors are equally important along with physical factors for the development of human settlements. For example, industries need development of roads to carry raw material or agriculture also needs development of roads. Many settlements are developed along roads or railway lines.

(v) Yes, physical factors affect the economic activity in the development of human settlement. For example, agriculture is’ an important activity which is affected by relief and water supply. It can be developed where fertile soil and sufficient water supply is available.

(vi) The factors which affect development of settlements are relief, terrain, climate, soil, water supply, altitude, drainage, minerals etc.

Question 2.
Visit http://censusindia.gov.in/2011-prov-results/paper2/data_files/India2/1.%20Data%20 Highlight.pdf to know how cities are divided into various types in India on the basis of their populations. Also look for examples from Maharashtra. Refer to the website and complete the table as given below : (Textbook Page No. 24)
Answer:

Classification	Population
Class I	100,00 and above
Class II	50,000 to 99,999
Class III	20,000 to 49,999
Class IV	10,000 to 19,999
Class V	5,000 to 9,999
Class VI	Less than 5000

Question 3.
On the basis of dominant or specialised functions, Indian cities and towns can be broadly classified as follows. Complete the following table with examples from Maharashtra and India. (Textbook Page No. 24)
Answer:

Function	Name of city in Maharashtra	Name of cities outside Maharashtra
Administrative	Nashik	Gandhinagar
Industrial	Chinchwad	Jamshedpur
Transport	Nagpur	Bengaluru
Commercial	Mumbai	Surat
Mining	Chandrapur	Balaghat
Cantonment	Khadki	Agra
Educational	Pune	Kota
Religious	Pandarpur	Varanasi
Tourism	Matheran	Ooty

Class 12 Geography Solutions Digest

• Population Part 1 Geography Class 12 Question Answer
• Population Part 2 Geography Class 12 Question Answer
• Human Settlements and Land Use Geography Class 12 Question Answer
• Primary Economic Activities Geography Class 12 Question Answer
• Secondary Economic Activities Geography Class 12 Question Answer
• Tertiary Economic Activities Geography Class 12 Question Answer
• Region and Regional Development Geography Class 12 Question Answer 
• Geography: Nature and Scope Geography Class 12 Question Answer

Std 12 Geography Chapter 5 Question Answer Secondary Economic Activities Maharashtra Board

Balbharti Maharashtra State Board Class 12 Geography Solutions Chapter 5 Secondary Economic Activities Textbook Exercise Questions and Answers.

Class 12 Geography Chapter 5 Secondary Economic Activities Question Answer Maharashtra Board

Geography Class 12 Chapter 5 Question Answer Maharashtra Board

1. Complete the Chain

Question 1.

A	B	C
(1) Small scale industries	(1) Manual manufacturing	(1) Ceramics
(2) Cottage industries	(2) Skilled crafts person	(2) Tata Iron and Steel company
(3) Consumer goods	(3) Individual	(3) Potters
(4) Private	(4) Ready for direct consumption	(4) Pharmaceutical

Answer:

A	B	C
(1) Small scale industries	(1) Ready for direct consumption	(1) Potters
(2) Cottage industries	(2) Manual manufacturing	(2) Ceramics
(3) Consumer goods	(3) Skilled crafts person	(3) Pharmaceutical
(4) Private	(4) Individual	(4) Tata Iron and Steel company

2. Identify the correct correlation.

A : Assertion, R : Reasoning
Question 1.
A – The humid climate of Mumbai offered great scope for the development of cotton textile industries.
R – Industries require ample amount of water.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(d) Both A and R are correct but R is not the correct explanation of A.

Question 2.
A – In India, industries are found concentrated in few areas are available.
R – India is predominantly agrarian country.
(a) Only A is correct.
(b) Only R is correct.
(c) Both A and R are correct and R is correct explanation of A.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(c) Both A and R are correct and R is correct explanation of A.

3. Give geographical reasons.

Question 1.
Distribution of industries is uneven.
Answer:

• Distribution of industries depend upon various physical factors like climate, raw material, water and power supply, labour, transportation, land, etc., and economic factors like capital, market and government policies.
• Physical and economic factors vary from region to region and political factors vary from country to country.
• Industries are developed where physical and economic factors are favourable for their development.
• Thus, distribution of industries is uneven.
• For example, in India industries are concentrated in Mumbai, Chennai, Kolkata, Delhi, Chota Nagpur region and in the rest of India, agriculture is the only major economic activity.

Question 2.
Iron and steel industries are found in mineral rich area of Dhanbad.
Answer:

• Important raw materials used in iron and steel industry are iron-ore, limestone, dolomite, manganese and coal.
• All these raw materials are heavy, bulky and weight-loosing.
• The cost of transportation of these materials are very high.
• Dhanbad and its surrounding areas are the major iron-ore and coal producing areas.
• • Since this industry is using weight-losing raw materials which are available around Dhanbad, iron and steel industry is found at Dhanbad.

Question 3.
Fruit-processing industries are found in Ratnagiri and Sindhudurg districts of Konkan region.
Answer:

• Fruit processing industries use fruits as raw material.
• Ratnagiri and Sindhudurg districts are known for horticulture. They are major mango, jackfruit, cashew and kokum producing districts of Maharashtra.
• All these fruits are raw materials in fruit processing industry.
• These raw materials are bulky, weight-losing and perishable and hence the industry is located in Ratnagiri and Sindhudurg districts where fruits are grown in plenty.

Question 4.
What are the major factors which have hindered the growth of industries in South America?
Answer:

• Industries in the continent of South America are developed only in coastal areas of Brazil, Argentina, Chile and Peru.
• The hindrances in the growth of industries in South America are due to unfavourable physical factors such as dense forests in Brazil, deserts in interior parts of Argentina, Andes mountain range running along the west coast.
• The economic factors like limited capital, lack of modern technology and lack of transportation facilities etc., create hindrance in development of industries.
• Comparatively low density of population and lack of markets are the other factors responsible to have hindered the growth of industries in South America.

4. Write short notes on.

Question 1.
Footloose industries.
Answer:

• Footloose industry is a general term for an industry that can be placed and located at any location without effect of factors of location such as land, labour, climate and capital.
• The raw material used as well as finished products of these industries are very light so their location near the source of raw material or transport is not important.
• These industries often have spatially fixed cost, which means that costs of the product do not change despite where the product is assembled.
• Diamond cutting, computer chips and mobiles manufacturing are some of the examples of footloose industries.
• Advance technology like internet, email are used for raw material as well as marketing.
• These are generally non-polluted industries.

Question 2.
Public sector industries.
Answer:

• Public sector industries are owned by the state.
• The investments in public sector industries is done by the government. Also, marketing of goods is done through government agencies.
• Public sectors include public goods and governmental services such as military, law enforcement infrastructure (public roads, bridges, tunnels, water supply electrical grids, health, etc.).
• Bharat Heavy Electrical Limited (BHEL) is one of the largest engineering and manufacturing companies of its kind in India engaged in designing engineering construction.

Question 3.
Economics of scale.
Answer:
1. Economics of scale are the cost advantages that enterprises obtain because of various facilities
established due to basic industries in certain region.

2. Sometimes due to advantage of many favourable factors for industrial development in certain areas, there is concentration of industries in that area, which is called agglomeration of industries.

3. In these regions industries develop not due to any locational factors but due to economies of scale enjoyed because agglomeration of industries.

4. Due to the development of basic industries other ancillary industries which are complementary to each other also develop. For example, once cotton textile industry develops in any region, readymaide garment making industries, industries supplying dyes and chemicals, industries producing materials like thread, buttons, laces, etc., also develop.

5. Due to such agglomeration, the industries in that region get more profit compared to their investment due to economies of scale such as cheap transport, labour, financial facilities etc. For example, transport companies give concession, hence, the cost of transportation decreases.

6. Since industries in this region are complementary, it is easier to collect or supply goods from other industries of nearby areas. For example, dye making industries supply dyes to cotton textile industry and cotton textile industry supplies cloth to ready-made garment industries.

Question 4.
Role of transportation in industries.
Answer:

• Transportation facilities are important for the collection of raw materials and distribution of finished products.
• Low cost of transportation is the key factor in the location of industries.
• Transport network is crucial for sustained economic growth and development of a nation.
• Transport system helps to send raw material, fuel and machinery to different industries at the right time and runs the industry.
• Thus, transport plays a crucial role in the development of industries.

5. Differentiate between.

Question 1.
Weight-losing and Weight-gaining Industries.
Answer:

Weight-loosing	Weight-gaining
(i) Weight-losing industries are those industries in which raw material are relatively bulky but finished products are relatively light.	(i) Weight-gaining industries are those industries in which raw materials are relatively lighter than finished products.
(ii) Weight-losing industries are located near raw materials.	(ii) Weight-gaining industries are located near markets.
(iii) For example, sugar industries locate near sugarcane producing areas. 10 tons of sugarcane is required to produce 1 ton of sugar.	(iii) For example, cotton textile industries locate near the market because finished product cotton cloth is heavier than the raw material cotton.

Question 2.
Primary and Secondary Activities.
Answer:

Primary Activities	Secondary Activities
(i) Primary activities include activities, such as hunting, fishing, mining, agriculture.	(i) Secondary activities include manufacturing and constructions.
(ii) These activities are concerned with obtaining materials directly from nature. For example, fish from water or wood from trees.	(ii) These activities add value to the already existing products by changing their form, making them more suitable to satisfy our needs and wants.
(iii) Primary activities produce raw materials.	(iii) Secondary activities produce finished products.
(iv) Production of goods in primary activities depends upon natural factors.	(iv) Production of goods in secondary activities depends on technology, skill of labours and capital.

Question 3.
Basic Industries and Consumer Industries.
Answer:

Basic Industries	Consumer Industries
(i) The finished products of these industries are used by other industries.	(i) The finished products of these are used as raw material which are directly consumed by consumers.
(ii) For example, iron and steel industry provides steel as raw material to machine tool making or agricultural implements making industry.	(ii) For example, the medicines made in pharmaceutical industries are directly consumed by consumers.
(iii) These industries supply their products to others. Hence, they are called linkage products.	(iii) These industries supply their products to consumers. Hence, they are called as consumer goods industries.

6. Answer the following questions in detail.

Question 1.
Explain the physical factors affecting location of industries.
Answer:
Climate, raw material, water, power, labour, land and transport are important physical factors deciding the location of industry.

Extreme climate like too hot, too cold, dry and very humid climate are not suitable for industries. Generally, industries develop in areas of moderate climate.

Availability of raw material is essential for the development of industries. Industries using perishable, heavy, bulky and weight-losing raw materials are located in the areas of source of raw materials. For example, sugar mills near sugarcane growing areas.

Most of the industries require lot of water for processing, like cooling, washing etc. Such industries are located near lakes, rivers or other water bodies. For example, many industries are located around Great Lakes in USA.

Labour is another important factor deciding location of industries. Availability of skilled, semi-skilled and cheap labour is very important to work in industries. Hence, we generally find labour colonies are located where skilled and cheap labour is available.

Transport facilities are essential for transport of raw materials and finished products. Low cost transport is the key factor in the development of industries. Industries develop in the areas, where transportation network is well developed. For example , many industries are developed along Trans-Siberian railway line in Russia.

Industrial development need huge land for the construction of industrial units. Industries develop where cheap, plain and extensive land is available.

Industries need power supply, main sources of power supply are coal, oil, electricity and now a days wind energy, solar energy etc. Industries locate near coal mines, or availability of electricity. For example, industries concentrated around coal mining area of Ruhr region of Germany or industries concentrated near Khopoli hydro power station.

Question 2.
Explain the factors affecting location of sugar industries.
Answer:
The following are the factors affecting the location of sugar industries.

• Sugar industry is and agro-based industry.
• Sugarcane is the raw material in sugar industry.
• Sugarcane is perishable and weight-losing raw material.
• It is also heavy and bulky raw material.
• Taking into consideration all above characteristics of raw material used in sugar industry, it is always located in the areas of sugarcane production.
• In Maharashtra, Satara, Kolhapur, Ahmednagar, Sangli etc., are the sugarcane producing districts and the same districts have concentration of sugar industry.

Question 3.
Explain the economic and political factors affecting location of industries.
Answer:
The following are the economic factors affecting the location of industries.
1. Neatness to market : Transport cost is one of the important costs included in production cost of goods. If industries locate near the market there is saving on transport cost and goods reach to the consumers as early as possible. Some industries whose finished products are perishable or bulky or heavy, locate near the market. For example, ice cream making, furniture making or air craft making industries.

2. Availability of capital : Capital is required for land, construction, equipment, labour, transport etc. Therefore, industries are located in areas where banking and financial facilities are available. Generally, these facilities are available in big cities, therefore in India many industries are located in big cities like Mumbai, Chennai, Bengaluru, Ahmedabad, Kolkata etc.

The following are the political factors affecting the location of industries.
Government policies : The government policy of encouragement or discouragement directly affects location of industry.

• Sometimes government gives encouragement for development of industries in economically backward areas or to reduce the overcrowding in nearby cities by providing land, water, power at cheaper rate. For example, to reduce the overcrowding in Mumbai city, the government provided land, water, electricity at a cheaper rate in Navi Mumbai.
• The government policy regarding import, export, taxes, subsidies, etc., also affect location of industries. For example, liberal policy for industries in the State of Gujarat attracted many industrialists to set up industries.
• Development of SEZ : Due to the development of Special Economic Zone (SEZ) many industries are attracted in SEZ area. SEZ are developed to set up public sector or private sector industries, specially to increase export quality production in the country.

7. Show the following on a map of the world with suitable index

(1) Ruhr industrial region
(2) An industrial region in Japan
(3) An industrial region in South Africa
(4) An industrial region in Australia
(5) Industrial region near Great Lakes
Answer:

Class 12 Geography Chapter 5 Secondary Economic Activities Intext Questions and Answers

Observe the pictures given in the textbook figure. Identify the activities with which these figures are associated and complete the table.

Answer:

Try These

Question 1.
Look at the map of Solapur district. It shows the location of some sugar industries. Shamrao is a farmer at location A. He has just harvested his sugarcane crop. Study the map and tell where should he send his crop? What factors will Shamrao consider for this? (Textbook Page No. 43)
Answer:
(i) Shamrao has two options, he can sell his sugarcane to Pimpalner Sugar Factory or Virag Sugar Factory, since both the factories almost at equal distance from A.
(ii) The most important factor of consideration is the distance and perishability of sugarcane as a raw material.
(iii) Third factor he should take into consideration is from A, the road condition. The factory to which he should send sugarcane must be in good condition; so that sugarcane can be transported to industry as early as possible.

Question 2.
Firoz’s son has done an advanced course in Bakery Management from the nearby city. He wants his son to start a bakery in their village but his son says it should be in the city, which is around 20 kms from their village. Who is correct? Firoz or his son? Why? (Textbook Page No. 44)
Answer:
The decision of Firoz’s son is correct. He should start a bakery in the city. First thing is that bakery products are perishable and secondly there is loss of weight of raw materials in the production of bakery products.

Question 3.
Shantaram is a young boy from a tribal area in Maharashtra. He wants to start a honey processing unit as he has access to good honeycombs in the forest. The city is around 35 kms away. He has his own land in the village and is also getting land in the cit3\ Where should he establish the honey factory? In the village or in the city? Why? (Textbook Page No. 44)
Answer:
(i) Shantaram should start his honey factory in the city. Raw material in his factory will be honeycombs.
(ii) From honeycombs he will get two finished products – honey and wax. The weight of honeycombs after extracting honey and wax will be the same and therefore ideal location is city.

Can you tell?

Question 1.
Find example of market-oriented industries. (Textbook Page No. 45)
Answer:
(i) All textile industries are market-oriented industries Cotton Textile, Silk Textile, Woollen Textile.
(ii) All assembly industries are market-oriented industries Automobile, Air-craft, shipbuilding industries.

Use your brain power!

Question 1.
Following is a list of few industries. Think about the factors of location of these industries and tell whether they are footloose industries or not. Complete the table accordingly. (Textbook Page No. 47)
(i) Cotton textile industries
(ii) Cement industries
(iii) Diamond industries
(iv) Mobile manufacturing units
(v) Paper industry
(vi) Sugar industry
(vii) Food processing industry
(viii) IT (Hardware) industry
(ix) Oil refinery
(x) Hairpins industry
Answer:

Question 2.
(i) Refer to the lumbering map of the world in fig. 4.1 and 5.1, tell which industries will be located in the northernmost island of Japan?
(ii) How have the Great Lakes been instrumented in development of industrial regions in the USA?
(iii) How has the Trans-Siberian Railway contributed to the development of industries in Russia? (Textbook Page No. 51)
Answer:
(i) Paper and pulp industry.
(ii) The Great Lakes have been instrumented in the development of industrial regions in the USA through concentration of minerals and coal producing areas, local market, capital and technological development.

(iii) The Trans-Siberian Railway contributed to the development of industries in Russia is

• Trans-Siberian Railway is the longest railway system connecting Petersburg in Western Russia to Vladivostok on the pacific coast.
• This railway connects all major cities in Russia.
• It passes through major iron-ore and coal producing areas, which has helped to the development of iron and steel industry in Russia.
• This railway line helped for the movement of raw material and finished goods.
• Therefore, many industries have been developed along this railway route.
• It has also helped the movement of passenger and goods traffic in East-West direction.

Make friends with maps!

Question 1.
Study the given map in fig 5.3 and answer the following questions. (Textbook Pages 47 & 48)

(i) In which hemisphere do you find more concentration of industries?
(ii) In which part of North America has the industrial region been mainly developed?
(iii) In which part of Europe is concentration of industries mainly found?
(iv) Why is less industrial development found in other parts of African continent except coastal areas?
(v) In which part of India do you find concentration of industries?
(vi) Why do you think coastal areas have higher concentration industries?
(vii) Write a concluding paragraph on latitudinal distribution of industries in the world.
Answer:
(i) Northern Hemisphere.
(ii) North eastern Region.
(iii) North western, South and Central part
(iv) Dense forest, deserts, mountains, grasslands in other parts of African continent are responsible for less development of industries.
(v) Portal cities like Mumbai, Chennai, Kolkata and other areas like Delhi, Bangalore, Chota Nagpur etc.

(vi) Coastal areas have better ports and inland water transport facilities. Even road and rail transport are well developed.

• Extensive level land.
• Industries which depend on imported raw material like oil prefer to develop on coastal areas.
• Industries which need humid climate develop in coastal areas for example textile industry.

(vii) Major industrial regions are concentrated in Northern Hemisphere in the latitudinal belt of 30° North to 50° North latitudes. However, there is an exception of Siberian industrial region in Russia.

• North America, major European countries, Japan, China and parts of Russia falls in this latitudinal belt.
• Major cause of the development of industries in this latitudinal belt is the Industrial Revolution in Europe.
• In this region both mineral based and agro-based industries have developed.
• Development of science and technology, skilled labour, well developed transportation facilities, high standard of living and extensive local market in this latitudinal belt are the other factors responsible for the development of industries.
• However, outside this major industrial belt in Northern Hemisphere* only exception is India, industries have developed in latitudinal belt 8° North to 20° North.
• Even in Southern Hemisphere industries have developed in the same latitudinal belt i.e. 30° South to 50° South latitudinal belt in South Africa, South America and South Australia.

Give it a try

Question 1.
Look at the given logo. Find out more information about it and write a short note on it. (Textbook Page No. 53)
Answer:
Digital India.
(i) Digital India is a campaign launched by the Government of India in order to ensure the government’s services are made available to citizens electronically by improved online infrastructure and by increasing internet connectivity or making the country digitally empowered in the field of technology.
(ii) The initiative includes plans to connect rural India with high speed internet networks.
(iii) Digital India consists of secure and stable digital infrastructure, delivering government services digitally and universal digital literacy.

Make in India:
On 25 September 2014, the Government of India launched a movement ‘Make in India’, like Swadeshi movement, for 25 sectors of Indian Economy. The main purpose to start this movement was to provide encouragement to businessmen in India to set up companies to manufacture their products in India and to investment more for the same.

In line with make in India, individual states too launched their own local initiative, such as ‘Magnetic Maharashtra’, ‘Make in Odisha’, Vibrant Gujarat, etc.

Startup India
(i) Startup India is an initiative of the government of India was first announce by Indian Prime minister, Narendra Modi during his 15 August 2015 address from the Red Fort in New Delhi.

(ii) The action plan of this initiative is focussing on three areas:

• Simplification and Handholding.
• Funding support and incentives.
• Industry-Academia partnership incubation.

(iii) An additional area relating to this initiative is to discard restrictive state Government Policies within this domain, such an ‘License Raj’, Land permissions, Foreign Investment Proposals, and Environmental clearance. It was organised by Department for promotion of industry and internal trade.

Question 2.
Find at least 2 examples of each of the types of industries from India.
Answer:
(i) Industries in Digital India are:
DigiLocker-
DigiLocker is an online service provided by the government wherein any Aadhar card holder can access a cloud with their authentic documents/certificates. For example, driving license, vehicle registration, academic mark sheet etc.

MyGov-
MyGov is an initiative started by the Indian government to engage citizens in governance. The portal can be accessed via a website or a dedicated mobile app.

(ii) Industries in Startup India are:
Zomato:
Zomato is an Indian restaurant aggregator and food delivery startup that was launched in 2008. It is one of the most successful food tech startups.

Paytm:
It is an Indian e-commerce payment system based in Noida. It was founded in August 2010.

Ola Cabs:
It is founded in 2010 and giving tough competition to Uber.

(iii) Industries in Make in India:
Make in India focuses on the following industries viz. Automobiles, Aviation, Biotechnology, Chemicals, Defence manufacturing, etc.

Class 12 Geography Solutions Digest

• Population Part 1 Geography Class 12 Question Answer
• Population Part 2 Geography Class 12 Question Answer
• Human Settlements and Land Use Geography Class 12 Question Answer
• Primary Economic Activities Geography Class 12 Question Answer
• Secondary Economic Activities Geography Class 12 Question Answer
• Tertiary Economic Activities Geography Class 12 Question Answer
• Region and Regional Development Geography Class 12 Question Answer 
• Geography: Nature and Scope Geography Class 12 Question Answer

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 8 Binomial Distribution Miscellaneous Exercise 8 Questions and Answers.

12th Maths Part 2 Binomial Distribution Miscellaneous Exercise 8 Questions And Answers Maharashtra Board

(I) Choose the correct option from the given alternatives:

Question 1.
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
(a) √50
(b) 5
(c) 25
(d) 10
Answer:
(b) 5

Question 2.
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probablity of 2 successes is
(a) \(\frac{128}{256}\)
(b) \(\frac{219}{256}\)
(c) \(\frac{37}{256}\)
(d) \(\frac{28}{256}\)
Answer:
(d) \(\frac{28}{256}\)

Question 3.
For a binomial distribution, n = 5. If P(X = 4) = P(X = 3) then p = ___________
(a) \(\frac{1}{3}\)
(b) \(\frac{3}{4}\)
(c) 1
(d) \(\frac{2}{3}\)
Answer:
(d) \(\frac{2}{3}\)

Question 4.
In a binomial distribution, n = 4. If 2 P(X = 3) = 3 P(X = 2) then p = ___________
(a) \(\frac{4}{13}\)
(b) \(\frac{5}{13}\)
(c) \(\frac{9}{13}\)
(d) \(\frac{6}{13}\)
Answer:
(c) \(\frac{9}{13}\)

Question 5.
If X ~ B (4, p) and P (X = 0) = \(\frac{16}{81}\), then P (X = 4) = ___________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{81}\)
(c) \(\frac{1}{27}\)
(d) \(\frac{1}{8}\)
Answer:
(b) \(\frac{1}{81}\)

Question 6.
The probability of a shooter hitting a target is \(\frac{3}{4}\). How many minimum numbers of times must he fie so that the probability of hitting the target at least once is more than 0·99?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 7.
If the mean and variance of a binomial distribution are 18 and 12 respectively, then n = ___________
(a) 36
(b) 54
(c) 18
(d) 27
Answer:
(b) 54

(II) Solve the following:

Question 1.
Let X ~ B(10, 0.2). Find
(i) P(X = 1)
(ii) P(X ≥ 1)
(iii) P(X ≤ 8).
Solution:
X ~ B(10, 0.2)
∴ n = 10, p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
The p,m.f. of X is given by

Question 2.
Let X ~ B(n, p).
(i) If n = 10, E(X) = 5, find p and Var(X).
(ii) If E(X) = 5 and Var(X) = 2.5, find n and p.
Solution:
X ~ B(n, p)
(i) Given: n = 10 and E(X) = 5
But E(X) = np
∴ np = 5.
∴ 10p = 5
∴ p = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Var(X) = npq = 10(\(\frac{1}{2}\))(\(\frac{1}{2}\)) = 2.5.
Hence, p = \(\frac{1}{2}\) and Var(X) = 2.5

(ii) Given: E(X) = 5 and Var(X) = 2.5
∴ np = 5 and npq = 2.5
∴ \(\frac{n p q}{n p}=\frac{2.5}{5}\)
∴ q = 0.5 = \(\frac{5}{10}=\frac{1}{2}\)
∴ p = 1 – q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Substituting p = \(\frac{1}{2}\) in np = 5, we get
n(\(\frac{1}{2}\)) = 5
∴ n = 10
Hence, n = 10 and p = \(\frac{1}{2}\)

Question 3.
If a fair coin is tossed 10 times and the probability that it shows heads (i) 5 times (ii) in the first four tosses and tail in the last six tosses.
Solution:
Let X = number of heads.
p = probability that coin tossed shows a head
∴ p = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Given: n = 10
∴ X ~ B(10, \(\frac{1}{2}\))
The p.m.f. of X is given by
P(X = x) = \({ }^{n} C_{x} P^{x} q^{n-x}\)

(i) P(coin shows heads 5 times) = P[X = 5]

Hence, the probability that can shows heads exactly 5 times = \(\frac{63}{256}\)

(ii) P(getting heads in first four tosses and tails in last six tosses) = P(X = 4)

Hence, the probability that getting heads in first four tosses and tails in last six tosses = \(\frac{105}{512}\).

Question 4.
The probability that a bomb will hit a target is 0.8. Find the probability that out of 10 bombs dropped, exactly 2 will miss the target.
Solution:
Let X = the number of bombs hitting the target.
p = probability that bomb will hit the target
∴ p = 0.8 = \(\frac{8}{10}=\frac{4}{5}\)
∴ q = 1 – p = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Given: n = 10
∴ X ~ B(10, \(\frac{4}{5}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e.p(x) = \({ }^{10} \mathrm{C}_{x}\left(\frac{4}{5}\right)^{x}\left(\frac{1}{5}\right)^{10-x}\)
P(exactly 2 bombs will miss the target) = P(exactly 8 bombs will hit the target)
= P[X = 8]
= p(8)

Hence, the probability that exactly 2 bombs will miss the target = 45\(\left(\frac{2^{16}}{5^{10}}\right)\)

Question 5.
The probability that a mountain bike travelling along a certain track will have a tire burst is 0.05. Find the probability that among 17 riders:
(i) exactly one has a burst tyre
(ii) at most three have a burst tyre
(iii) two or more have burst tyres.
Solution:
Let X = number of burst tyres.
p = probability that a mountain bike travelling along a certain track will have a tyre burst.
∴ p = 0.05
∴ q = 1 – p = 1 – 0.05 = 0.95
Given: n = 17
∴ X ~ B(17, 0.05)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} P^{x} q^{n-x}\)
i.e.(x) = \({ }^{17} \mathrm{C}_{x}(0.05)^{x}(0.95)^{17-x}\), x = 0, 1, 2, ……, 17
(i) P(exactly one has a burst tyre)
P(X = 1) = p(1) = \({ }^{17} \mathrm{C}_{1}\) (0.05)¹ (0.95)^17-1
= 17(0.05) (0.95)¹⁶
= 0.85(0.95)¹⁶
Hence, the probability that riders has exactly one burst tyre = (0.85)(0.95)¹⁶

(ii) P(at most three have a burst tyre) = P(X ≤ 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= p(0) + p(1) + p(2) + p(3)


Hence, the probability that at most three riders have burst tyre = (2.0325)(0.95)¹⁴.

(iii) P(two or more have tyre burst) = P(X ≥ 2)
= 1 – P(X < 2)
= 1 – [P(X = 0) + P(X = 1)]
= 1 – [p(0) + p(1)]
= 1 – [\({ }^{17} \mathrm{C}_{0}\) (0.05)⁰ (0.95)¹⁷ + \({ }^{17} \mathrm{C}_{1}\) (0.05)(0.95)¹⁶]
= 1 – [1(1)(0.95)¹⁷ + 17(0.05)(0.95)¹⁶]
= 1 – (0.95)¹⁶[0.95 + 0.85]
= 1 – (1.80)(0.95)¹⁶
= 1 – (1.8)(0.95)¹⁶
Hence, the probability that two or more riders have tyre burst = 1 – (1.8)(0.95)¹⁶.

Question 6.
The probability that a lamp in a classroom will be burnt out is 0.3. Six such lamps are fitted in the classroom. If it is known that the classroom is unusable if the number of lamps burning in it is less than four, find the probability that the classroom cannot be used on a random occasion.
Solution:
Let X = number of lamps burnt out in the classroom.
p = probability of a lamp in a classroom will be burnt
∴ p = 0.3 = \(\frac{3}{10}\)
∴ q = 1 – p = 1 – \(\frac{3}{10}\) = \(\frac{7}{10}\)
Given: n = 6
∴ X ~ B(6, \(\frac{3}{10}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{6} \mathrm{C}_{x}\left(\frac{3}{10}\right)^{x}\left(\frac{7}{10}\right)^{6-x}\)
Since the classroom is unusable if the number of lamps burning in it is less than four, therefore
P(classroom cannot be used) = P[X < 4]
= P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3]
= p(0) + p(1) + p(2) + p(3)

Hence, the probability that the classroom cannot be used on a random occasion is 0.92953.

Question 7.
A lot of 100 items contain 10 defective items. Five items are selected at random from the lot and sent to the retail store. What is the probability that the store will receive at most one defective item?
Solution:
Let X = number of defective items.
p = probability that item is defective
∴ p = \(\frac{10}{100}=\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{10}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{5} C_{x}\left(\frac{1}{10}\right)^{x}\left(\frac{9}{10}\right)^{5-x}\)
P (store will receive at most one defective item) = P[X ≤ 1]
=P[X = 0] + P[X = 1]
= p(0) + p(1)

Hence, the probability that the store will receive at most one defective item is (1.4)(0.9)⁴.

Question 8.
A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. The inspector of the retailer picks 20 items from a shipment. What is the probability that the store will receive at most one defective item?
Solution:
Let X = number of defective electronic devices.
p = probability that device is defective
∴ p = 3% = \(\frac{3}{100}\)
∴ q = 1 – p = 1 – \(\frac{3}{100}\) = \(\frac{97}{100}\)
Given: n = 20
∴ X ~ B(20, \(\frac{3}{100}\))
The p.m.f. of X is given as:

Hence, the probability that the store will receive at most one defective item = (1.57)(0.97)¹⁹.

Question 9.
The probability that a certain kind of component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 tested components tested survive.
Solution:
Let X = number of tested components survive.
p = probability that the component survives the check test

Hence, the probability that exactly 2 of the 4 tested components survive is 0.3456.

Question 10.
An examination consists of 10 multiple choice questions, in each of which a candidate has to deduce which one of five suggested answers is correct. A completely unprepared student guesses each answer completely randomly. What is the probability that this student gets 8 or more questions correct? Draw the appropriate moral.
Solution:
Let X = number of correct answers.
p = probability that student gets correct answer
∴ p = \(\frac{1}{5}\)
∴ q = 1 – p = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Given: n = 10 (number of total questions)
∴ X ~ B(10, \(\frac{1}{5}\))
The p.m.f. of X is given by

Hence, the probability that student gets 8 or more questions correct = \(\frac{30.44}{5^{8}}\)

Question 11.
The probability that a machine will produce all bolts in a production run within specification is 0.998. A sample of 8 machines is taken at random. Calculate the probability that (i) all 8 machines (ii) 7 or 8 machines (iii) at most 6 machines will produce all bolts within specification.
Solution:
Let X = number of machines which produce the bolts within specification.
p = probability that a machine produce bolts within specification
p = 0.998 and q = 1 – p = 1 – 0.998 = 0.002
Given: n = 8
∴ X ~ B(8, 0.998)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{8} \mathrm{C}_{x}(0.998)^{x}(0.002)^{8-x}\), x = 0, 1, 2, …, 8
(i) P(all 8 machines will produce all bolts within specification) = P[X = 8]
= p(8)
= \({ }^{8} \mathrm{C}_{8}\) (0.998)⁸ (0.002)^8-8
= 1(0.998)⁸ . (1)
= (0.998)⁸
Hence, the probability that all 8 machines produce all bolts with specification = (0.998)⁸.

(ii) P(7 or 8 machines will produce all bolts within i specification) = P (X = 7) + P (X = 8)

Hence, the probability that 7 or 8 machines produce all bolts within specification = (1.014)(0.998)⁷.

(iii) P(at most 6 machines will produce all bolts with specification) = P[X ≤ 6]
= 1 – P[x > 6]
= 1 – [P(X = 7) + P(X = 8)]
= 1 – [P(7) + P(8)]
= 1 – (1.014)(0.998)⁷
Hence, the probability that at most 6 machines will produce all bolts with specification = 1 – (1.014)(0.998)⁷.

Question 12.
The probability that a machine develops a fault within the first 3 years of use is 0.003. If 40 machines are selected at random, calculate the probability that 38 or more will develop any faults within the first 3 years of use.
Solution:
Let X = the number of machines who develop a fault.
p = probability that a machine develops a fait within the first 3 years of use
∴ p = 0.003 and q = 1 – p = 1 – 0.003 = 0.997
Given: n = 40
∴ X ~ B(40, 0.003)
The p.m.f. of X is given by

Hence, the probability that 38 or more machines will develop the fault within 3 years of use = (775.44)(0.003)³⁸.

Question 13.
A computer installation has 10 terminals. Independently, the probability that anyone terminal will require attention during a week is 0.1. Find the probabilities that (i) 0 (ii) 1 (iii) 2 (iv) 3 or more, terminals will require attention during the next week.
Solution:
Let X = number of terminals which required attention during a week.
p = probability that any terminal will require attention during a week
∴ p = 0.1 and q = 1 – p = 1 – 0.1 = 0.9
Given: n = 10
∴ X ~ B(10, 0.1)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{10} C_{x}(0.1)^{x}(0.9)^{10-x}\), x = 0, 1, 2, …, 10
(i) P(no terminal will require attention) = P(X = 0)

Hence, the probability that no terminal requires attention = (0.9)¹⁰

(ii) P(1 terminal will require attention)

Hence, the probability that 1 terminal requires attention = (0.9)⁹.

(iii) P(2 terminals will require attention)

Hence, the probability that 2 terminals require attention = (0.45)(0.9)⁸.

(iv) P(3 or more terminals will require attention)

Hence, the probability that 3 or more terminals require attention = 1 – (2.16) × (0.9)⁸.

Question 14.
In a large school, 80% of the pupil like Mathematics. A visitor to the school asks each of 4 pupils, chosen at random, whether they like Mathematics.
(i) Calculate the probabilities of obtaining an answer yes from 0, 1, 2, 3, 4 of the pupils.
(ii) Find the probability that the visitor obtains answer yes from at least 2 pupils:
(a) when the number of pupils questioned remains at 4.
(b) when the number of pupils questioned is increased to 8.
Solution:
Let X = number of pupils like Mathematics.
p = probability that pupils like Mathematics

(i) The probabilities of obtaining an answer yes from 0, 1, 2, 3, 4 of pupils are P(X = 0), P(X = 1), P(X = 2), P(X = 3) and P(X = 4) respectively

(ii) (a) P(visitor obtains the answer yes from at least 2 pupils when the number of pupils questioned remains at 4) = P(X ≥ 2)
= P(X = 2) + P(X = 3) + P(X = 4)

(b) P(the visitor obtains the answer yes from at least 2 pupils when number of pupils questioned is increased to 8)

Question 15.
It is observed that it rains 12 days out of 30 days. Find the probability that
(i) it rains exactly 3 days of the week.
(ii) it will rain at least 2 days of a given week.
Solution:
Let X = the number of days it rains in a week.
p = probability that it rains

(i) P(it rains exactly 3 days of week) = P(X = 3)

Hence, the probability that it rains exactly 3 days of week = 0.2903.

(ii) P(it will rain at least 2 days of the given week)

Hence, the probability that it rains at least 2 days of a given week = 0.8414

Question 16.
If the probability of success in a single trial is 0.01. How many trials are required in order to have a probability greater than 0.5 of getting at least one success?
Solution:
Let X = number of successes.
p = probability of success in a single trial
∴ p = 0.01
and q = 1 – p = 1 – 0.01 = 0.99
∴ X ~ B(n, 0.01)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)

Hence, the number of trials required in order to have a probability greater than 0.5 of getting at least one success is \(\frac{\log 0.5}{\log 0.99}\) or 68.

Question 17.
In binomial distribution with five Bernoulli’s trials, the probability of one and two success are 0.4096 and 0.2048 respectively. Find the probability of success.
Solution:
Given: X ~ B(n = 5, p)
The probability of X success is


Hence, the probability of success is \(\frac{1}{5}\).

Class 12 Maharashtra State Board Maths Solution 

• Exercise 7.1 Class 12 Maths 2
• Exercise 7.2 Class 12 Maths 2
• Miscellaneous Exercise 7 Class 12 Maths 2
• Exercise 8.1 Class 12 Maths 2
• Miscellaneous Exercise 8 Class 12 Maths 2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 8 Binomial Distribution Ex 8.1 Questions and Answers.

12th Maths Part 2 Binomial Distribution Exercise 8.1 Questions And Answers Maharashtra Board

Question 1.
A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of
(i) 5 successes
(ii) at least 5 successes
(iii) at most 5 successes.
Solution:
Let X = number of successes, i.e. number of odd numbers.
p = probability of getting an odd number in a single throw of a die
∴ p = \(\frac{3}{6}=\frac{1}{2}\) and q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Given: n = 6
∴ X ~ B(6, \(\frac{1}{2}\))
The p.m.f. of X is given by

Hence, the probability of 5 successes is \(\frac{3}{32}\).

(ii) P(at least 5 successes) = P[X ≥ 5]
= p(5) + p(6)

Hence, the probability of at least 5 successes is \(\frac{7}{64}\).

(iii) P(at most 5 successes) = P[X ≤ 5]
= 1 – P[X > 5]

Hence, the probability of at most 5 successes is \(\frac{63}{64}\).

Question 2.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
Let X = number of doublets.
p = probability of getting a doublet when a pair of dice is thrown
∴ p = \(\frac{6}{36}=\frac{1}{6}\) and
q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Given: n = 4
∴ X ~ B(4, \(\frac{1}{6}\))
The p.m.f. of X is given by

Hence, the probability of two successes is \(\frac{25}{216}\).

Question 3.
There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?
Solution:
Let X = number of defective items.
p = probability of defective item
∴ p = 5% = \(\frac{5}{100}=\frac{1}{20}\)
and q = 1 – p = 1 – \(\frac{1}{20}\) = \(\frac{19}{20}\)
∴ X ~ B(10, \(\frac{1}{20}\))
The p.m.f. of X is given by

P(sample of 10 items will include not more than one defective item) = P[X ≤ 1]

Hence, the probability that a sample of 10 items will include not more than one defective item = 29\(\left(\frac{19^{9}}{20^{10}}\right)\).

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards, find the probability that
(i) all the five cards are spades
(ii) only 3 cards are spades
(iii) none is a spade.
Solution:
Let X = number of spade cards.
p = probability of drawing a spade card from a pack of 52 cards.
Since there are 13 spade cards in the pack of 52 cards.
∴ p = \(\frac{13}{52}=\frac{1}{4}\) and
q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{4}\))
The p.m.f. of X is given by

(i) P(all five cards are spade)

Hence, the probability of all the five cards are spades = \(\frac{1}{1024}\)

(ii) P(only 3 cards are spade) = P[X = 3]

Hence, the probability of only 3 cards are spades = \(\frac{45}{512}\)

(iii) P(none of cards is spade) = P[X = 0]

Hence, the probability of none of the cards is a spade = \(\frac{243}{1024}\)

Question 5.
The probability of a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one, will fuse after 150 days of use.
Solution:
Let X = number of fuse bulbs.
p = probability of a bulb produced by a factory will fuse after 150 days of use.
∴ p = 0.05
and q = 1 – p = 1 – 0.05 = 0.95
Given: n = 5
∴ X ~ B(5, 0.05)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{5} C_{x}(0.05)^{x}(0.95)^{5-x}\), x = 0, 1, 2, 3, 4, 5
(i) P(none of a bulb produced by a factory will fuse after 150 days of use) = P[X = 0]
= p(0)
= \({ }^{5} \mathrm{C}_{0}(0.05)^{0}(0.95)^{5-0}\)
= 1 × 1 × (0.95)⁵
= (0.95)⁵
Hence, the probability that none of the bulbs will fuse after 150 days = (0.95)⁵.

(ii) P(not more than one bulb will fuse after 150 days of j use) = P[X ≤ 1]
= p(0) + p(1)
= \({ }^{5} \mathrm{C}_{0} \cdot(0.05)^{0}(0.95)^{5-0}+{ }^{5} \mathrm{C}_{1}(0.05)^{1}(0.95)^{4}\)
= 1 × 1 × (0.95)⁵ + 5 × (0.05) × (0.95)⁴
= (0.95)⁴ [0.95 + 5(0.05)]
= (0.95)⁴ (0.95 + 0.25)
= (0.95)⁴ (1.20)
= (1.2) (0.95)⁴
Hence, the probability that not more than one bulb will fuse after 150 days = (1.2)(0.95)⁴.

(iii) P(more than one bulb fuse after 150 days)
= P[X > 1]
= 1 – P[X ≤ 1]
= 1 – (1.2)(0.95)⁴
Hence, the probability that more than one bulb fuse after 150 days = 1 – (1.2)(0.95)⁴.

(iv) P(at least one bulb fuse after 150 days)
= P[X ≥ 1]
= 1 – P[X = 0]
= 1 – p(0)
= 1 – \({ }^{5} C_{0}(0.05)^{0}(0.95)^{5-0}\)
= 1 – 1 × 1 × (0.95)⁵
= 1 – (0.95)⁵
Hence, the probability that at least one bulb fuses after 150 days = 1 – (0.95)⁵.

Question 6.
A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Solution:
Let X = number of balls marked with digit 0.
p = probability of drawing a ball from 10 balls marked with the digit 0.
∴ p = \(\frac{1}{10}\)
and q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
The p.m.f. of X is given by

P(none of the ball marked with digit 0) = P(X = 0)

Hence, the probability that none of the bulb marked with digit 0 is \(\left(\frac{9}{10}\right)^{4}\)

Question 7.
On a multiple-choice examination with three possible answers for each of the five questions. What is the probability that a candidate would get four or more correct answers just by guessing?
Solution:
Let X = number of correct answers.
p = probability that a candidate gets a correct answer from three possible answers.
∴ p = \(\frac{1}{3}\) and q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{3}\))
The p.m.f. of X is given by

P(four or more correct answers) = P[X ≥ 4]
= p(4) + p(5)

Hence, the probability of getting four or more correct answers = \(\frac{11}{243}\).

Question 8.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \(\frac{1}{100}\), find the probability that he will win a prize
(i) at least once
(ii) exactly once
(iii) at least twice.
Solution:
Let X = number of winning prizes.
p = probability of winning a prize
∴ p = \(\frac{1}{100}\)
and q = 1 – p = 1 – \(\frac{1}{100}\) = \(\frac{99}{100}\)
Given: n = 50
∴ X ~ B(50, \(\frac{1}{100}\))
The p.m.f. of X is given by
\(P(X=x)={ }^{n} C_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{50} \mathrm{C}_{x}\left(\frac{1}{100}\right)^{x}\left(\frac{99}{100}\right)^{50-x}\), x = 0, 1, 2,… 50
(i) P(a person wins a prize at least once)

Hence, probability of winning a prize at least once = 1 – \(\left(\frac{99}{100}\right)^{50}\)

(ii) P(a person wins exactly one prize) = P[X = 1] = p(1)

Hence, probability of winning a prize exactly once = \(\frac{1}{2}\left(\frac{99}{100}\right)^{49}\)

(iii) P(a persons wins the prize at least twice) = P[X ≥ 2]
= 1 – P[X < 2]
= 1 – [p(0) + p(1)]

Hence, the probability of winning the prize at least twice = 1 – 149\(\left(\frac{99^{49}}{100^{50}}\right)\).

Question 9.
In a box of floppy discs, it is known that 95% will work. A sample of three of the discs is selected at random. Find the probability that (i) none (ii) 1 (iii) 2 (iv) all 3 of the sample will work.
Solution:
Let X = number of working discs.
p = probability that a floppy disc works
∴ p = 95% = \(\frac{95}{100}=\frac{19}{20}\)
and q = 1 – p = 1 – \(\frac{19}{20}\) = \(\frac{1}{20}\)
Given: n = 3
∴ X ~ B(3, \(\frac{19}{20}\))
The p.m.f. of X is given by

(i) P(none of the floppy discs work) = P(X = 0)

Hence, the probability that none of the floppy disc will work = \(\frac{1}{20^{3}}\).

(ii) P(exactly one floppy disc works) = P(X = 1)

Hence, the probability that exactly one floppy disc works = 3\(\left(\frac{19}{20^{3}}\right)\)

(iii) P(exactly two floppy discs work) = P(X = 2)

Hence, the probability that exactly 2 floppy discs work = 3\(\left(\frac{19^{2}}{20^{3}}\right)\)

(iv) P(all 3 floppy discs work) = P(X = 3)

Hence, the probability that all 3 floppy discs work = \(\left(\frac{19}{20}\right)^{3}\).

Question 10.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
Let X = number of sixes.
p = probability that a die shows six in a single throw
∴ p = \(\frac{1}{6}\)
and q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Given: n = 6
∴ X ~ B(6, \(\frac{1}{6}\))
The p.m.f. of X is given by


Hence, probability of throwing at most 2 sixes = \(\frac{7}{3}\left(\frac{5}{6}\right)^{5}\).

Question 11.
It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?
Solution:
Let X = number of defective articles.
p = probability of defective articles.
∴ p = 10% = \(\frac{10}{100}=\frac{1}{10}\)
and q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
Given: n = 12
∴ X ~ B(12, \(\frac{1}{10}\))
The p.m.f. of X is given by

Hence, the probability of getting 9 defective articles = \(22\left(\frac{9^{3}}{10^{11}}\right)\)

Question 12.
Given X ~ B(n, P)
(i) If n = 10 and p = 0.4, find E(x) and Var(X).
(ii) If p = 0.6 and E(X) = 6, find n and Var(X).
(iii) If n = 25, E(X) = 10, find p and SD(X).
(iv) If n = 10, E(X) = 8, find Var(X).
Solution:
(i) Given: n = 10 and p = 0.4
∴ q = 1 – p = 1 – 0.4 = 0.6
∴ E(X) = np = 10(0.4) = 4
Var(X) = npq = 10(0.4)(0.6) = 2.4
Hence, E(X) = 4, Var(X) = 2.4.

(ii) Given: p = 0.6, E (X) = 6
E(X) = np
6 = n(0.6)
n = \(\frac{6}{0.6}\) = 10
Now, q = 1 – p = 1 – 0.6 = 0.4
∴ Var(X) = npq = 10(0.6)(0.4) = 2.4
Hence, n = 10 and Var(X) = 2.4.

(iii) Given: n = 25, E(X) = 10
E(X) = np
10 = 25p
p = \(\frac{10}{25}=\frac{2}{5}\)
∴ q = 1 – p = 1 – \(\frac{2}{5}\) = \(\frac{3}{5}\)
Var(X) = npq = \(25 \times \frac{2}{5} \times \frac{3}{5}\) = 6
∴ SD(X) = √Var(X) = √6
Hence, p = \(\frac{2}{5}\) and S.D.(X) = √6.

(iv) Given: n = 10, E(X) = 8
E(X) = np
8 = 10p
p = \(\frac{8}{10}=\frac{4}{5}\)
q = 1 – p = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Var(X) = npq = \(10\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)=\frac{8}{5}\)
Hence, Var(X) = \(\frac{8}{5}\).

Class 12 Maharashtra State Board Maths Solution 

• Exercise 7.1 Class 12 Maths 2
• Exercise 7.2 Class 12 Maths 2
• Miscellaneous Exercise 7 Class 12 Maths 2
• Exercise 8.1 Class 12 Maths 2
• Miscellaneous Exercise 8 Class 12 Maths 2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Miscellaneous Exercise 7 Questions and Answers.

12th Maths Part 2 Probability Distributions Miscellaneous Exercise 7 Questions And Answers Maharashtra Board

(I) Choose the correct option from the given alternatives:

Question 1.
P.d.f. of a c.r.v. X is f(x) = 6x(1 – x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere) If P(X < a) = P(X > a), then a =
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 2.
If the p.d.f. of a c.r.v. X is f(x) = 3(1 – 2x²), for 0 < x < 1 and = 0, otherwise (elsewhere), then the c.d.f. of X is F(x) =
(a) 2x – 3x²
(b) 3x – 4x³
(c) 3x – 2x³
(d) 2x³ – 3x
Answer:
(c) 3x – 2x³

Question 3.
If the p.d.f. of a c.r.v. X is f(x) = \(\frac{x^{2}}{18}\), for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) =
(a) \(\frac{1}{27}\)
(b) \(\frac{1}{28}\)
(c) \(\frac{1}{29}\)
(d) \(\frac{1}{26}\)
Answer:
(a) \(\frac{1}{27}\)

Question 4.
If p.m.f. of a d.r.v. X takes values 0, 1, 2, 3, … which probability P(X = x) = k(x +1) . 5-x, where k is a constant, then P(X = 0) =
(a) \(\frac{7}{25}\)
(b) \(\frac{16}{25}\)
(c) \(\frac{18}{25}\)
(d) \(\frac{19}{25}\)
Answer:
(b) \(\frac{16}{25}\)

Question 5.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{\left({ }^{5} \mathrm{C}_{x}\right)}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. If a = P(X ≤ 2) and b = P(X ≥ 3), then
(a) a < b
(b) a > b
(c) a = b
(d) a + b
Answer:
(c) a = b

Question 6.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{x}{n(n+1)}\), for x = 1, 2, 3, ……, n and = 0, otherwise, then E(X) =
(a) \(\frac{n}{1}+\frac{1}{2}\)
(b) \(\frac{n}{3}+\frac{1}{6}\)
(c) \(\frac{n}{2}+\frac{1}{5}\)
(d) \(\frac{n}{1}+\frac{1}{3}\)
Answer:
(b) \(\frac{n}{3}+\frac{1}{6}\)

Question 7.
If p.m.f. of a d.r.v. X is P(x) = \(\frac{c}{x^{3}}\), for x = 1, 2, 3 and = 0, otherwise (elsewhere), then E(X) =
(a) \(\frac{343}{297}\)
(b) \(\frac{294}{251}\)
(c) \(\frac{297}{294}\)
(d) \(\frac{294}{297}\)
Answer:
(b) \(\frac{294}{251}\)

Question 8.
If the d.r.v. X has the following probability distribution:

then P(X = -1) =
(a) \(\frac{1}{10}\)
(b) \(\frac{2}{10}\)
(c) \(\frac{3}{10}\)
(d) \(\frac{4}{10}\)
Answer:
(a) \(\frac{1}{10}\)

Question 9.
If the d.r.v. X has the following probability distribution:

then k =
(a) \(\frac{1}{7}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{10}\)
Answer:
(d) \(\frac{1}{10}\)

Question 10.
Find the expected value of X for the following p.m.f.

(a) 0.85
(b) -0.35
(c) 0.15
(d) -0.15
Answer:
(b) -0.35

(II) Solve the following:

Question 1.
Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values:
(i) An economist is interested in the number of unemployed graduates in the town of population 1 lakh.
(ii) Amount of syrup prescribed by a physician.
(iii) The person on a high protein diet is interesting to gain weight in a week.
(iv) 20 white rats are available for an experiment. Twelve rats are males. A scientist randomly selects 5 rats, the number of female rats selected on a specific day.
(v) A highway-safety group is interested in studying the speed (in km/hr) of a car at a checkpoint.
Solution:
(i) Let X = number of unemployed graduates in a town.
Since the population of the town is 1 lakh, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, …, 99999, 100000}.

(ii) Let X = amount of syrup prescribed by a physician.
Then X takes uncountable infinite values.
∴ random variable X is continuous.

(iii) Let X = gain of weight in a week
Then X takes uncountable infinite values
∴ random variable X is continuous.

(iv) Let X = number of female rats selected on a specific day.
Since the total number of rats is 20 which includes 12 males and 8 females, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, 3, 4, 5}

(v) Let X = speed of .the car in km/hr.
Then X takes uncountable infinite values
∴ random variable X is continuous.

Question 2.
The probability distribution of discrete r.v. X is as follows:

(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:

Question 3.
The following is the probability distribution of X:

Find the probability that
(i) X is positive
(ii) X is non-negative
(iii) X is odd
(iv) X is even.
Solution:
(i) P(X is positive) = P(X = 1) + P(X = 2) + P(X = 3)
= 0.25 + 0.15 + 0.1
= 0.50

(ii) P(X is non-negative)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.20 + 0.25 + 0.15 + 0.1
= 0.70

(iii) P(X is odd)
= P(X = -3) + P(X = -1) + P(X = 1) + P(X = 3)
= 0.05 + 0.15 + 0.25 + 0.1
= 0.55

(iv) P(X is even)
= P(X = -2) + P(X = 0) + P(X = 2)
= 0.10 + 0.20 + 0.15
= 0.45.

Question 4.
The p.m.f. of a r.v. X is given by P(X = x) = x = \(\frac{{ }^{5} \mathrm{C}_{\mathrm{x}}}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. Then show that P(X ≤ 2) = P(X ≥ 3).
Solution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{2}}{2^{5}}\)
= \(\frac{{ }^{5} \mathrm{C}_{5}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{4}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}\) ………[latex]{ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}[/latex]
= P(X = 5) + P(X = 4) + P(X = 3)
= P(X ≥ 3)
∴ P(X ≤ 2) = P(X ≥ 3).

Question 5.
In the p.m.f. of r.v. X

Find a and obtain c.d.f. of X.
Solution:
For p.m.f. of a r.v. X
\(\sum_{i=1}^{5} P(X=x)=1\)
∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

Let F(x) be the c.d.f. of X.
Then F(x) = P(X ≤ x)
∴ F(1) = P(X ≤ 1) = P(X = 1) = \(\frac{1}{20}\)
F(2) = P(X ≤ 2) = P(X = 1) + P (X = 2)
\(=\frac{1}{20}+\frac{3}{20}=\frac{4}{20}=\frac{1}{5}\)
P(3) = P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}=\frac{9}{20}\)
F(4) = P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}=\frac{19}{20}\)
F(5) = P(X ≤ 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}+\frac{1}{20}=\frac{20}{20}=1\)
Hence, the c.d.f. of the random variable X is as follows:

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Write down the probability distribution of X. Also, find the formula for p.m.f. of X.
Solution:
When a fair coin is tossed 4 times then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
X denotes the number of heads.
∴ X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n (X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{0}}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{1}}{16}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{{ }^{4} \mathrm{C}_{2}}{16}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{3}}{16}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{4}}{16}\)
∴ the probability distribution of X is as follows:

Also, the formula for p.m.f. of X is
P(x) = \(\frac{{ }^{4} \mathrm{C}_{x}}{16}\), x = 0, 1, 2, 3, 4 and = 0, otherwise.

Question 7.
Find the probability distribution of the number of successes in two tosses of a die, where success is defined as
(i) number greater than 4
(ii) six appear on at least one die.
Solution:
When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.
Let X be the random variable, which represents the number of successes.
Here, success refers to the number greater than 4.
P(X = 0) = P(number less than or equal to 4 on both the tosses)
= \(\frac{4}{6} \times \frac{4}{6}=\frac{16}{36}=\frac{4}{9}\)
P(X = 1) = P(number less than or equal to 4 on first toss and greater than 4 on second toss) + P(number greater than 4 on first toss and less than or equal to 4 on second toss)
= \(\frac{4}{6} \times \frac{2}{6}+\frac{4}{6} \times \frac{2}{6}\)
= \(\frac{8}{36}+\frac{8}{36}\)
= \(\frac{16}{36}\)
= \(\frac{4}{9}\)
P(X = 2) = P(number greater than 4 on both the tosses)
= \(\frac{2}{6} \times \frac{2}{6}=\frac{4}{36}=\frac{1}{9}\)
Thus, the probability distribution is as follows:

(ii) Here, success means six appears on at least one die.
P(Y = 0) = P(six appears on none of the dice) = \(\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}\)
P(Y = 1) = P(six appears on none of the dice x six appears on at least one of the dice ) + P(six appears on none of the dice x six appears on at least one of the dice)
= \(\frac{1}{6} \times \frac{5}{6}+\frac{1}{6} \times \frac{5}{6}=\frac{5}{36}+\frac{5}{36}=\frac{10}{36}\)
P(Y = 2) = P(six appears on at least one of the dice) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)
Thus, the required probability distribution is as follows:

Question 8.
A random variable X has the following probability distribution:

Determine:
(i) k
(ii) P(X > 6)
(iii) P(0 < X < 3).

Question 9.
The following is the c.d.f. of a r.v. X:

Find
(i) p.m.f. of X
(ii) P( -1 ≤ X ≤ 2)
(iii) P(X ≤ X > 0).
Solution:
(i) From the given table
F(-3) = 0.1, F(-2) = 0.3, F(-1) = 0.5
F(0) = 0.65, f(1) = 0.75, F(2) = 0.85
F(3) = 0.9, F(4) = 1
P(X = -3) = F(-3) = 0.1
P(X = -2) = F(-2) – F(-3) = 0.3 – 0.1 = 0.2
P(X = -1) = F(-1) – F(-2) = 0.5 – 0.3 = 0.2
P(X = 0) = F(0) – F(-1) = 0.65 – 0.5 = 0.15
P(X = 1) = F(1) – F(0) = 0.75 – 0.65 = 0.1
P(X = 2) = F(2) – F(1) = 0.85 – 0.75 = 0.1
P(X = 3) = F(3) – F(2) = 0.9 – 0.85 = 0.1
P(X = 4) = F(4) – F(3) = 1 – 0.9 = 0.1
∴ the p.m.f of X is as follows:

(ii) P(-1 ≤ X ≤ 2) = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2)
= 0.2 + 0.15 + 0.1 + 0.1
= 0.55

(iii) (X ≤ 3) ∩ (X > 0)
= { -3, -2, -1, 0, 1, 2, 3} n {1, 2, 3, 4}
= {1, 2, 3}

Question 10.
Find the expected value, variance, and standard deviation of the random variable whose p.m.f’s are given below:

Solution:
(i) We construct the following table to find the expected value, variance, and standard deviation:

(ii) We construct the following table to find the expected value, variance, and standard deviation:



(iii) We construct the following table to find the expected value, variance, and standard deviation:



(iv) We construct the following table to find the expected value, variance, and standard deviation:

Question 11.
A player tosses two wins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of the winning amount.
Solution:
When a coin is tossed twice, the sample space is
S = {HH, HT, TH, HH}
Let X denote the amount he wins.
Then X takes values 10, 5, 2.
P(X = 10) = P(2 heads appear) = \(\frac{1}{4}\)
P(X = 5) = P(1 head appears) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(X = 2) = P(no head appears) = \(\frac{1}{4}\)
We construct the following table to calculate the mean and the variance of X:

From the table Σx_i . P(x_i) = 5.5, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 38.5
E(X) = Σx_i . P(x_i) = 5.5
Var(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) – [E(X)]²
= 38.5 – (5.5)²
= 38.5 – 30.25
= 8.25
∴ Hence, expected winning amount = ₹ 5.5 and variance of winning amount = ₹ 8.25.

Question 12.
Let the p.m.f. of r.v. X be P(x) = \(\frac{3-x}{10}\), for x = -1, 0, 1, 2 and = 0, otherwise.
Calculate E(X) and Var(X).
Solution:
P(X) = \(\frac{3-x}{10}\)
X takes values -1, 0, 1, 2
P(X = -1) = P(-1) = \(\frac{3+1}{10}=\frac{4}{10}\)
P(X = 0) = P(0) = \(\frac{3-0}{10}=\frac{3}{10}\)
P(X = 1) = P(1) = \(\frac{3-1}{10}=\frac{2}{10}\)
P(X = 2) = P(2) = \(\frac{3-2}{10}=\frac{1}{10}\)
We construct the following table to calculate the mean and variance of X:

From the table
Σx_iP(x_i) = 0 and \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) = 1
E(X) = Σx_iP(x_i) = 0
Var(X) = \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) – [E(X)]²
= 1 – 0
= 1
Hence, E(X) = 0, Var (X) = 1.

Question 13.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x²), -2 ≤ x ≤ 2 and = 0 otherwise.
Compute
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(X < -0.5 or X > 0.5).
Solution:
(i) P(X > 0)


(ii) P(-1 < X < 1)


(iii) P(X < -0.5 or X > 0.5)

Question 14.
The p.d.f. of a continuous r.v. X is given by f(x) = \(\frac{1}{2 a}\), for 0 < x < 2a and = 0, otherwise. Show that P( X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:

Question 15.
The p.d.f. of r.v. X is given by f(x) = \(\frac{k}{\sqrt{x}}\), for 0 < x < 4 and = 0, otherwise. Determine k. Determine c.d.f. of X and hence find P(X ≤ 2) and P(X ≤ 1).
Solution:
Since f is p.d.f. of the r.v. X,

Class 12 Maharashtra State Board Maths Solution 

• Exercise 7.1 Class 12 Maths 2
• Exercise 7.2 Class 12 Maths 2
• Miscellaneous Exercise 7 Class 12 Maths 2
• Exercise 8.1 Class 12 Maths 2
• Miscellaneous Exercise 8 Class 12 Maths 2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Ex 7.2 Questions and Answers.

12th Maths Part 2 Probability Distributions Exercise 7.2 Questions And Answers Maharashtra Board

Question 1.
Verify which of the following is p.d.f. of r.v. X:
(i) f(x) = sin x, for 0 ≤ x ≤ \(\frac{\pi}{2}\)
(ii) f(x) = x, for 0 ≤ x ≤ 1 and -2 – x for 1 < x < 2
(iii) fix) = 2, for 0 ≤ x ≤ 1.
Solution:
f(x) is the p.d.f. of r.v. X if
(a) f(x) ≥ 0 for all x ∈ R and

Hence, f(x) is the p.d.f. of X.

(ii) f(x) = x ≥ 0 if 0 ≤ x ≤ 1
For 1 < x < 2, -2 < -x < -1
-2 – 2 < -2 – x < -2 – 1
i.e. -4 < f(x) < -3 if 1 < x < 2
Hence, f(x) is not p.d.f. of X.

(iii) (a) f(x) = 2 ≥ 0 for 0 ≤ x ≤ 1

Hence, f(x) is not p.d.f. of X.

Question 2.
The following is the p.d.f. of r.v. X:
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
Find
(a) P(x < 1.5)
(b) P(1 < x < 2) (c) P(x > 2).
Solution:

Question 3.
It is known that error in measurement of reaction temperature (in 0°C) in a certain experiment is continuous r.v. given by
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0. otherwise.
(i) Verify whether f(x) is p.d.f. of r.v. X
(ii) Find P(0 < x ≤ 1)
(iii) Find the probability that X is negative.
Solution:

Question 4.
Find k if the following function represents p.d.f. of r.v. X
(i) f(x) = kx. for 0 < x < 2 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{3}{2}\)).
(ii) f(x) = kx(1 – x), for 0 < x < 1 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{1}{2}\)), P(x < \(\frac{1}{2}\)).
Solution:
(i) Since, the function f is p.d.f. of X

(ii) Since, the function f is the p.d.f. of X,

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected students and suppose X has p.d.f.
f(x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.
Calculate:
(i) P(x ≤ 1)
(ii) P(0.5 ≤ x ≤ 1.5)
(iii) P(x ≥ 1.5).
Solution:
(i) P(x ≤ 1)

(ii) P(0.5 ≤ x ≤ 1.5)

(iii) P(x ≥ 1.5)

Question 6.
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = \(\frac{1}{5}\), for 0 ≤ x ≤ 5 and = 0 otherwise. Find the probability that
(i) waiting time is between 1 and 3
(ii) waiting time is more than 4 minutes.
Solution:
(i) Required probability = P(1 < X < 3)

(ii) Required probability = P(X > 4)

Question 7.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x²), -2 ≤ x ≤ 2 and 0 otherwise.
Compute:
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(-0.5 < X or X > 0.5).
Solution:
Since, f is the p.d.f. of X,

Question 8.
The following is the p.d.f. of continuous r.v. X
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
(i) Find expression for c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7 and 5.
Solution:
(i) Let F(x) be the c.d.f. of X

Question 9.
Given the p.d.f. of a continuous random r.v. X, f(x) = \(\frac{x^{2}}{3}\), for -1 < x < 2 and = 0 otherwise. Determine c.d.f. of X and hence find P(X < 1); P(X < -2), P(X > 0), P(1 < X < 2).
Solution:

Question 10.
If a r.v. X has p.d.f.
f(x) = \(\frac{c}{x}\) for 1 < x < 3, c > 0. Find c, E(X), Var (X).
Solution:
Since f(x) is p.d.f of r.v. X

Class 12 Maharashtra State Board Maths Solution 

• Exercise 7.1 Class 12 Maths 2
• Exercise 7.2 Class 12 Maths 2
• Miscellaneous Exercise 7 Class 12 Maths 2
• Exercise 8.1 Class 12 Maths 2
• Miscellaneous Exercise 8 Class 12 Maths 2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Ex 7.1 Questions and Answers.

12th Maths Part 2 Probability Distributions Exercise 7.1 Questions And Answers Maharashtra Board

Question 1.
Let X represent the difference between a number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?
Solution:
When a coin is tossed 6 times, the number of heads can be 0, 1, 2, 3, 4, 5, 6.
The corresponding number of tails will be 6, 5, 4, 3, 2, 1, 0.
∴ X can take values 0 – 6, 1 – 5, 2 – 4, 3 – 3, 4 – 2, 5 – 1, 6 – 0
i.e. -6, -4, -2, 0, 2, 4, 6.
∴ X = {-6, -4, -2, 0, 2, 4, 6}.

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
The urn contains 5 red and 2 black balls.
If two balls are drawn from the urn, it contains either 0 or 1 or 2 black balls.
X can take values 0, 1, 2.
∴ X = {0, 1, 2}.

Question 3.
State which of the following are not the probability mass function of a random variable. Give reasons for your answer.

Solution:
P.m.f. of random variable should satisfy the following conditions:
(a) 0 ≤ p_i ≤ 1
(b) Σp_i = 1.

(i)

(a) Here 0 ≤ p_i ≤ 1
(b) Σp_i = 0.4 + 0.4 + 0.2 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(ii)

P(X = 3) = -0.1, i.e. P_i < 0 which does not satisfy 0 ≤ P_i ≤ 1
Hence, P(X) cannot be regarded as p.m.f. of the random variable X.

(iii)

(a) Here 0 ≤ p_i ≤ 1
(b) ∑p_i = 0.1 + 0.6 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(iv)

Here ∑p_i = 0.3 + 0.2 + 0.4 + 0 + 0.05 = 0.95 ≠ 1
Hence, P(Z) cannot be regarded as p.m.f. of the random variable Z.

(v)

Here ∑p_i = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence, P(Y) cannot be regarded as p.m.f. of the random variable Y.

(vi)

(a) Here 0 ≤ p_i ≤ 1
(b) ∑p_i = 0.3 + 0.4 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Solution:
(i) For two tosses of a coin the sample space is {HH, HT, TH, TT}
Let X denote the number of heads in two tosses of a coin.
Then X can take values 0, 1, 2.
∴ P[X = 0] = P(0) = \(\frac{1}{4}\)
P[X = 1] = P(1) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P[X = 2] = P(2) = \(\frac{1}{4}\)
∴ the required probability distribution is

(ii) When three coins are tossed simultaneously, then the sample space is
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let X denotes the number of tails.
Then X can take the value 0, 1, 2, 3.
∴ P[X = 0] = P(0) = \(\frac{1}{8}\)
P[X = 1] = P(1) = \(\frac{3}{8}\)
P[X = 2] = P(2) = \(\frac{3}{8}\)
P[X = 3] = P(3) = \(\frac{1}{8}\)
∴ the required probability distribution is

(iii) When a fair coin is tossed 4 times, then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
Let X denotes the number of heads.
Then X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n(X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}\)
∴ the probability distribution of X is as follows:

Question 5.
Find the probability distribution of a number of successes in two tosses of a die, where success is defined as a number greater than 4 appearing on at least one die.
Solution:
When a die is tossed twice, the sample space s has 6 × 6 = 36 sample points.
∴ n(S) = 36
The trial will be a success if the number on at least one die is 5 or 6.
Let X denote the number of dice on which 5 or 6 appears.
Then X can take values 0, 1, 2.
When X = 0 i.e., 5 or 6 do not appear on any of the dice, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ n(X) = 16.
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{16}{36}=\frac{4}{9}\)
When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then
X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}
∴ n(X) = 16
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{16}{36}=\frac{4}{9}\)
When X = 2, i.e. 5 or 6 appear on both of the dice, then
X = {(5, 5), (5, 6), (6, 5), (6, 6)}
∴ n(X) = 4
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)
∴ the required probability distribution is

Question 6.
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Solution:
Here, the number of defective bulbs is the random variable.
Let the number of defective bulbs be denoted by X.
∴ X can take the value 0, 1, 2, 3, 4.
Since the draws are done with replacement, therefore the four draws are independent experiments.
Total number of bulbs is 30 which include 6 defectives.
∴ P(X = 0) = P(0) = P(all 4 non-defective bulbs)
= \(\frac{24}{30} \times \frac{24}{30} \times \frac{24}{30} \times \frac{24}{30}\)
= \(\frac{256}{625}\)
P(X = 1) = P (1) = P (1 defective and 3 non-defective bulbs)

P(X = 2) = P(2) = P(2 defective and 2 non-defective)

P(X = 3) = P(3) = P(3 defectives and 1 non-defective)

P(X = 4) = P(4) = P(all 4 defectives)
= \(\frac{6}{30} \times \frac{6}{30} \times \frac{6}{30} \times \frac{6}{30}\)
= \(\frac{1}{625}\)
∴ the required probability distribution is

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. If the coin is tossed twice. Find the probability distribution of a number of tails.
Solution:
Given a biased coin such that heads is 3 times as likely as tails.
∴ P(H) = \(\frac{3}{4}\) and P(T) = \(\frac{1}{4}\)
The coin is tossed twice.
Let X can be the random variable for the number of tails.
Then X can take the value 0, 1, 2.
∴ P(X = 0) = P(HH) = \(\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}\)
P(X = 1) = P(HT, TH) = \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}=\frac{6}{16}=\frac{3}{8}\)
P(X = 2) = P(TT) = \(\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)
∴ the required probability distribution is

Question 8.
A random variable X has the following probability distribution:

Determine:
(i) k
(ii) P(X < 3) (iii) P(X > 4)
Solution:
(i) Since P (x) is a probability distribution of x,
\(\sum_{x=0}^{7} P(x)=1\)
⇒ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1
⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
⇒ 10k² + 9k – 1 = 0
⇒ 10k² + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(10k – 1) = 0
⇒ 10k – 1 = 0 ……..[∵ k ≠ -1]
⇒ k = \(\frac{1}{10}\)

(ii) P(X< 3) = P(0) + P(1) + P(2)
= 0 + k + 2k
= 3k
= 3(\(\frac{1}{10}\))
= \(\frac{3}{10}\)

(iii) P(0 < X < 3) = P (1) + P (2)
= k + 2k
= 3k
= 3(\(\frac{1}{10}\))
= \(\frac{3}{10}\)

Question 9.
Find expected value and variance of X for the following p.m.f.:

Solution:
We construct the following table to calculate E(X) and V(X):

From the table,
Σx_ip_i = -0.05 and \(\Sigma x_{i}^{2} \cdot p_{i}\) = 2.25
∴ E(X) = Σx_ip_i = -0.05
and V(X) = \(\Sigma x_{i}^{2}+p_{i}-\left(\sum x_{i}+p_{i}\right)^{2}\)
= 2.25 – (-0.05)²
= 2.25 – 0.0025
= 2.2475
Hence, E(X) = -0.05 and V(X) = 2.2475.

Question 10.
Find expected value and variance of X, where X is the number obtained on the uppermost face when a fair die is thrown.
Solution:
If a die is tossed, then the sample space for the random variable X is
S = {1, 2, 3, 4, 5, 6}
∴ P(X) = \(\frac{1}{6}\); X = 1, 2, 3, 4, 5, 6.

Hence, E(X) = 3.5 and V(X) = 2.9167.

Question 11.
Find the mean number of heads in three tosses of a fair coin.
Solution:
When three coins are tossed the sample space is {HHH, HHT, THH, HTH, HTT, THT, TTH, TTT}
∴ n(S) = 8
Let X denote the number of heads when three coins are tossed.
Then X can take values 0, 1, 2, 3
P(X = 0) = P(0) = \(\frac{1}{8}\)
P(X = 1) = P(1) = \(\frac{3}{8}\)
P(X = 2) = P(2) = \(\frac{3}{8}\)
P(X = 3) = P(3) = \(\frac{1}{8}\)
∴ mean = E(X) = Σx_iP(x_i)
= \(0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}\)
= \(0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}\)
= \(\frac{12}{8}\)
= 1.5

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
When two dice are thrown, the sample space S has 6 × 6 = 36 sample points.
∴ n(S) = 36
Let X denote the number of sixes when two dice are thrown.
Then X can take values 0, 1, 2
When X = 0, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}
∴ n(X) = 25
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{25}{36}\)
When X = 1, then
X = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
∴ n(X) = 10
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{10}{36}\)
When X = 2, then X = {(6, 6)}
∴ n(X) = 1
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{1}{36}\)
∴ E(X) = ΣxiP(xi)
= \(0 \times \frac{25}{36}+1 \times \frac{10}{36}+2 \times \frac{1}{36}\)
= \(0+\frac{10}{36}+\frac{2}{36}\)
= \(\frac{1}{3}\)

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:
Two numbers are chosen from the first 6 positive integers.
∴ n(S) = \({ }^{6} C_{2}=\frac{6 \times 5}{1 \times 2}\) = 15
Let X denote the larger of the two numbers.
Then X can take values 2, 3, 4, 5, 6.
When X = 2, the other positive number which is less than 2 is 1.
∴ n(X) = 1
∴ P(X = 2) = P(2) = \(\frac{n(X)}{n(S)}=\frac{1}{15}\)
When X = 3, the other positive number less than 3 can be 1 or 2 and hence can be chosen in 2 ways.
∴ n(X) = 2
P(X = 3) = P(3) = \(\frac{n(X)}{n(S)}=\frac{2}{15}\)
Similarly, P(X = 4) = P(4) = \(\frac{3}{15}\)
P(X = 5) = P(5) = \(\frac{4}{15}\)
P(X = 6) = P(6) = \(\frac{5}{15}\)
∴ E(X) = Σx_iP(x_i)
= \(2 \times \frac{1}{15}+3 \times \frac{2}{15}+4 \times \frac{3}{15}+5 \times \frac{4}{15}+6 \times \frac{5}{15}\)
= \(\frac{2+6+12+20+30}{15}\)
= \(\frac{70}{15}\)
= \(\frac{14}{3}\)

Question 14.
Let X denote the sum of numbers obtained when two fair dice are rolled. Find the standard deviation of X.
Solution:
If two fair dice are rolled then the sample space S of this experiment is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let X denote the sum of the numbers on uppermost faces.
Then X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

∴ the probability distribution of X is given by

Question 15.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the student is recorded. What is the probability distribution of the random variable X? Find mean, variance, and standard deviation of X.
Solution:
Let X denote the age of the chosen student. Then X can take values 14, 15, 16, 17, 18, 19, 20, 21.
We make a frequency table to find the number of students with age X:

The chances of any student selected are equally likely.
If there are m students with age X, then P(X) = \(\frac{m}{15}\)
Using this, the following is the probability distribution of X:


Variance = V(X) = \(\Sigma x_{i}^{2}\) . P(x_i) – [E(X)]²
= 312.2 – (17.53)²
= 312.2 – 307.3
= 4.9
Standard deviation = √V(X) = √4.9 = 2.21
Hence, mean = 17.53, variance = 4.9 and standard deviation = 2.21.

Question 16.
In a meeting, 70% of the member’s favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed and X = 1 if he is in favour. Find E(X) and Var(X).
Solution:
X takes values 0 and 1.
It is given that
P(X = 0) = P(0) = 30% = \(\frac{30}{100}\) = 0.3
P(X = 1) = P(1) = 70% = \(\frac{70}{100}\) = 0.7
∴ E(X) = Σx_i . P(x_i) = 0 × 0.3 + 1 × 0.7 = 0.7
Also, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 0 × 0.3 + 1 × 0.7 = 0.7
∴ Variance = V(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)-[E(X)]^{2}\)
= 0.7 – (0.7)²
= 0.7 – 0.49
= 0.21
Hence, E(X) = 0.7 and Var(X) = 0.21.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 7.1 Class 12 Maths 2
• Exercise 7.2 Class 12 Maths 2
• Miscellaneous Exercise 7 Class 12 Maths 2
• Exercise 8.1 Class 12 Maths 2
• Miscellaneous Exercise 8 Class 12 Maths 2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Miscellaneous Exercise 6 Questions and Answers.

12th Maths Part 2 Differential Equations Miscellaneous Exercise 6 Questions And Answers Maharashtra Board

(I) Choose the correct option from the given alternatives:

Question 1.
The order and degree of the differential equation \(\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}\) are respectively……..
(a) 2, 1
(b) 1, 2
(c) 3, 2
(d) 2, 3
Answer:
(d) 2, 3

Question 2.
The differential equation of y = c² + \(\frac{c}{x}\) is…….
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)
(b) \(\frac{d y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
(c) \(x^{3}\left(\frac{d y}{d x}\right)^{2}+x \frac{d y}{d x}=y\)
(d) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
Answer:
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)

Question 3.
x² + y² = a² is a solution of ………
(a) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
(b) \(y=x \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}+a^{2} y\)
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)
(d) \(\frac{d^{2} y}{d x^{2}}=(x+1) \frac{d y}{d x}\)
Answer:
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)

Question 4.
The differential equation of all circles having their centres on the line y = 5 and touching the X-axis is
(a) \(y^{2}\left(1+\frac{d y}{d x}\right)=25\)
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(c) \((y-5)^{2}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(d) \((y-5)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=25\)
Answer:
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)

Question 5.
The differential equation y \(\frac{d y}{d x}\) + x = 0 represents family of ………
(a) circles
(b) parabolas
(c) ellipses
(d) hyperbolas
Answer:
(a) circles

Hint:
y \(\frac{d y}{d x}\) + x = 0
∴ ∫y dy + ∫x dx = c
∴ \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=c\)
∴ x² + y² = 2c which is a circle.

Question 6.
The solution of \(\frac{1}{x} \cdot \frac{d y}{d x}=\tan ^{-1} x\) is……
(a) \(\frac{x^{2} \tan ^{-1} x}{2}+c=0\)
(b) x tan^-1x + c = 0
(c) x – tan^-1x = c
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)
Answer:
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)

Question 7.
The solution of (x + y)² \(\frac{d y}{d x}\) = 1 is…….
(a) x = tan^-1(x + y) + c
(b) y tan^-1(\(\frac{x}{y}\)) = c
(c) y = tan^-1(x + y) + c
(d) y + tan^-1(x + y) = c
Answer:
(c) y = tan^-1(x + y) + c

Question 8.
The Solution of \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}-y^{2}}}{2}\) is……
(a) sin^-1(\(\frac{y}{x}\)) = 2 log |x| + c
(b) sin^-1(\(\frac{y}{x}\)) = log |x| + c
(c) sin(\(\frac{x}{y}\)) = log |x| + c
(d) sin(\(\frac{y}{x}\)) = log |y| + c
Answer:
(b) sin^-1(\(\frac{y}{x}\)) = log |x| + c

Question 9.
The solution of \(\frac{d y}{d x}\) + y = cos x – sin x is……
(a) y e^x = cos x + c
(b) y e^x + e^x cos x = c
(c) y e^x = e^x cos x + c
(d) y² e^x = e^x cos x + c
Answer:
(c) y e^x = e^x cos x + c
Hint:
\(\frac{d y}{d x}\) + y = cos x – sin x
I.F. = \(e^{\int 1 d x}=e^{x}\)
∴ the solution is y . e^x = ∫(cos x – sin x) e^x + c
∴ y . e^x = e^x cos x + c

Question 10.
The integrating factor of linear differential equation x \(\frac{d y}{d x}\) + 2y = x² log x is……..
(a) \(\frac{1}{x}\)
(b) k
(c) \(\frac{1}{n^{2}}\)
(d) x²
Answer:
(d) x²
Hint:
I.F. = \(e^{\int \frac{2}{x} d x}\)
= e^{2 log x}
= x²

Question 11.
The solution of the differential equation \(\frac{d y}{d x}\) = sec x – y tan x is…….
(a) y sec x + tan x = c
(b) y sec x = tan x + c
(c) sec x + y tan x = c
(d) sec x = y tan x + c
Answer:
(b) y sec x = tan x + c

Hint:
\(\frac{d y}{d x}\) = sec x – y tan x
∴ \(\frac{d y}{d x}\) + y tan x = sec x
I.F. = \(e^{\int \tan x d x}=e^{\log \sec x}\) = sec x
∴ the solution is
y . sec x = ∫sec x . sec x dx + c
∴ y sec x = tan x + c

Question 12.
The particular solution of \(\frac{d y}{d x}=x e^{y-x}\), when x = y = 0 is……
(a) e^x-y = x + 1
(b) e^x+y = x + 1
(c) e^x + e^y = x + 1
(d) e^y-x = x – 1
Answer:
(a) e^x-y = x + 1

Question 13.
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is a solution of……..
(a) \(\frac{d^{2} y}{d x^{2}}+y x+\left(\frac{d y}{d x}\right)^{2}=0\)
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)
(c) \(y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+y=0\)
(d) \(x y \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}=0\)
Answer:
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)

Question 14.
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially, there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is……
(a) 5\(\frac{2}{3}\) grams
(b) 5\(\frac{1}{3}\) grams
(c) 5.1 grams
(d) 5 grams
Answer:
(b) 5\(\frac{1}{3}\) grams

Question 15.
If the surrounding air is kept at 20°C and the body cools from 80°C to 70°C in 5 minutes, the temperature of the body after 15 minutes will be…..
(a) 51.7°C
(b) 54.7°C
(c) 52.7°C
(d) 50.7°C
Answer:
(b) 54.7°C

(II) Solve the following:

Question 1.
Determine the order and degree of the following differential equations:
(i) \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(ii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2 \times 5}=1+\frac{d y}{d x}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{10}=1+\frac{d y}{d x}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 10.
∴ the given D.E. is of order 3 and degree 10.

(iii) \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
On cubing both sides, we get
\(1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 3.
∴ the given D.E. is of order 2 and degree 3.

(iv) \(\frac{d y}{d x}=3 y+\sqrt[4]{1+5\left(\frac{d y}{d x}\right)^{2}}\)
Solution:
The given D.E. is

This D.E. has the highest order derivative \(\frac{d y}{d x}\) with power 4.
∴ the given D.E. is of order 1 and degree 4.

(v) \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\).
∴ order = 4
Since this D.E. cannot be expressed as a polynomial in differential coefficient, the degree is not defined.

Question 2.
In each of the following examples verify that the given function is a solution of the differential equation.
(i) \(x^{2}+y^{2}=r^{2} ; x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)
Solution:
x² + y² = r² ……. (1)
Differentiating both sides w.r.t. x, we get

Hence, x² + y² = r² is a solution of the D.E.
\(x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)

(ii) y = e^ax sin bx; \(\frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+\left(a^{2}+b^{2}\right) y=0\)
Solution:

(iii) y = 3 cos(log x) + 4 sin(log x); \(x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
Solution:
y = 3 cos(log x) + 4 sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get

(iv) xy = ae^x + be^-x + x²; \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+x^{2}=x y+2\)
Solution:

(v) x² = 2y² log y, x² + y² = xy \(\frac{d x}{d y}\)
Solution:
x² = 2y² log y ……(1)
Differentiating both sides w.r.t. y, we get

∴ x² + y² = xy \(\frac{d x}{d y}\)
Hence, x² = 2y² log y is a solution of the D.E.
x² + y² = xy \(\frac{d x}{d y}\)

Question 3.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) y² = a(b – x)(b + x)
Solution:
y² = a(b – x)(b + x) = a(b² – x²)
Differentiating both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = a(0 – 2x) = -2ax
∴ y \(\frac{d y}{d x}\) = -ax …….(1)
Differentiating again w.r.t. x, we get

This is the required D.E.

(ii) y = a sin(x + b)
Solution:
y = a sin(x + b)

This is the required D.E.

(iii) (y – a)² = b(x + 4)
Solution:
(y – a)² = b(x + 4) …….(1)
Differentiating both sides w.r.t. x, we get
\(2(y-a) \cdot \frac{d}{d x}(y-a)=b \frac{d}{d x}(x+4)\)

(iv) y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
Solution:
y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
∴ y² = a cos (log x) + b sin (log x) …….(1)
Differentiating both sides w.r.t. x, we get

(v) y = Ae^3x+1 + Be^-3x+1
Solution:
y = Ae^3x+1 + Be^-3x+1 …… (1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

Question 4.
Form the differential equation of:
(i) all circles which pass through the origin and whose centres lie on X-axis.
Solution:

Let C (h, 0) be the centre of the circle which pass through the origin. Then radius of the circle is h.
∴ equation of the circle is (x – h)² + (y – 0)² = h²
∴ x² – 2hx + h² + y² = h²
∴ x² + y² = 2hx ……..(1)
Differentiating both sides w.r.t. x, we get
2x + 2y \(\frac{d y}{d x}\) = 2h
Substituting the value of 2h in equation (1), we get
x² + y² = (2x + 2y \(\frac{d y}{d x}\)) x
∴ x² + y² = 2x² + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) + x² – y² = 0
This is the required D.E.

(ii) all parabolas which have 4b as latus rectum and whose axis is parallel to Y-axis.
Solution:
Let A(h, k) be the vertex of the parabola which has 4b as latus rectum and whose axis is parallel to the Y-axis.
Then equation of the parabola is
(x – h)² = 4b(y – k) ……. (1)
where h and k are arbitrary constants.

Differentiating both sides of (1) w.r.t. x, we get
2(x – h). \(\frac{d}{d x}\)(x – h) = 4b . \(\frac{d}{d x}\)(y – k)
∴ 2(x – h) x (1 – 0) = 4b(\(\frac{d y}{d x}\) – 0)
∴ (x – h) = 2b \(\frac{d y}{d x}\)
Differentiating again w.r.t. x, we get
1 – 0 = 2b \(\frac{d^{2} y}{d x^{2}}\)
∴ 2b \(\frac{d^{2} y}{d x^{2}}\) – 1 = 0
This is the required D.E.

(iii) an ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of the major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ x² + 4y² = 4b²
Differentiating w.r.t. x, we get
2x + 4 × 2y \(\frac{d y}{d x}\) = 0
∴ x + 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

(iv) all the lines which are normal to the line 3x + 2y + 7 = 0.
Solution:
Slope of the line 3x – 2y + 7 = 0 is \(\frac{-3}{-2}=\frac{3}{2}\).
∴ slope of normal to this line is \(-\frac{2}{3}\)
Then the equation of the normal is
y = \(-\frac{2}{3}\)x + k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{2}{3} \times 1+0\)
∴ 3\(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

(v) the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\).
Solution:
The equation of the hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\)
i.e., \(\frac{x^{2}}{16 k}-\frac{y^{2}}{36 k}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 16k, b² = 36k
∴ a = 4√k, b = 6√k
∴ l(transverse axis) = 2a = 8√k
and l(conjugate axis) = 2b = 12√k
Let 2A and 2B be the lengths of the transverse and conjugate axes of the required hyperbola.
Then according to the given condition
2A = a = 4√k and 2B = b = 6√k
∴ A = 2√k and B = 3√k
∴ equation of the required hyperbola is
\(\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1\)
i.e., \(\frac{x^{2}}{4 k}-\frac{y^{2}}{9 k}=1\)
∴ 9x² – 4y² = 36k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
9 × 2x – 4 × 2y \(\frac{d y}{d x}\) = 0
∴ 9x – 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

Question 5.
Solve the following differential equations:
(i) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:

(ii) \(\frac{d y}{d x}\) = x²y + y
Solution:

(iii) \(\frac{d y}{d x}=\frac{2 y-x}{2 y+x}\)
Solution:

(iv) x dy = (x + y + 1) dx
Solution:

(v) \(\frac{d y}{d x}\) + y cot x = x² cot x + 2x
Solution:
\(\frac{d y}{d x}\) + y cot x = x cot x + 2x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = cot x and Q = x² cot x + 2x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \cot x d x}\)
= \(e^{\log (\sin x)}\)
= sin x
∴ the solution of (1) is given by
y(I.F.) = ∫Q . (I.F.) dx + c
∴ y sin x = ∫(x² cot x + 2x) sin x dx + c
∴ y sinx = ∫(x² cot x . sin x + 2x sin x) dx + c
∴ y sinx = ∫x² cos x dx + 2∫x sin x dx + c
∴ y sinx = x² ∫cos x dx – ∫[\(\frac{d}{d x}\left(x^{2}\right)\) ∫cos x dx] dx + 2∫x sin x dx + c
∴ y sin x = x² (sin x) – ∫2x(sin x) dx + 2∫x sin x dx + c
∴ y sin x = x² sin x – 2∫x sin x dx + 2∫x sin x dx + c
∴ y sin x = x² sin x + c
∴ y = x² + c cosec x
This is the general solution.

(vi) y log y = (log y² – x) \(\frac{d y}{d x}\)
Solution:

(vii) 4 \(\frac{d x}{d y}\) + 8x = 5e^-3y
Solution:

Question 6.
Find the particular solution of the following differential equations:
(i) y(1 + log x) = (log xx) \(\frac{d y}{d x}\), when y(e) = e²
Solution:

(ii) (x + 2y²) \(\frac{d y}{d x}\) = y, when x = 2, y = 1
Solution:


This is the general solution.
When x = 2, y = 1, we have
2 = 2(1)² + c(1)
∴ c = 0
∴ the particular solution is x = 2y².

(iii) \(\frac{d y}{d x}\) – 3y cot x = sin 2x, when y(\(\frac{\pi}{2}\)) = 2
Solution:
\(\frac{d y}{d x}\) – 3y cot x = sin 2x
\(\frac{d y}{d x}\) = (3 cot x) y = sin 2x ……..(1)
This is the linear differential equation of the form

(iv) (x + y) dy + (x – y) dx = 0; when x = 1 = y
Solution:

(v) \(2 e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0\), when y(0) = 1
Solution:

Question 7.
Show that the general solution of defferential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\) is given by (x + y + 1) = c(1 – x – y – 2xy).
Solution:

Question 8.
The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.
Solution:
Let P(x, y) be a point on the curve y = f(x).
Then slope of the normal to the curve is \(-\frac{1}{\left(\frac{d y}{d x}\right)}\)
∴ equation of the normal is

This is the general equation of the curve.
Since, the required curve passed through the point (2, 3), we get
2² + 3² = 4(2) + c
∴ c = 5
∴ equation of the required curve is x² + y² = 4x + 5.

Question 9.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Solution:
Let r be the radius and V be the volume of the spherical balloon at any time t.
Then the rate of change in volume of the spherical balloon is \(\frac{d V}{d t}\) which is a constant.


Hence, the radius of the spherical balloon after t seconds is \((63 t+27)^{\frac{1}{3}}\) units.

Question 10.
A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning are ₹ 10 lakhs and they dwindle down to ₹ 10,000 after 2 years, show that the person will be bankrupt in 2\(\frac{2}{9}\) years from the start.
Solution:
Let x be the assets of the presort at time t years.
Then the rate of reduction is \(\frac{d x}{d t}\) which is proportional to √x.
∴ \(\frac{d x}{d t}\) ∝ √x
∴ \(\frac{d x}{d t}\) = -k√x, where k > 0
∴ \(\frac{d x}{\sqrt{x}}\) = -k dt
Integrating both sides, we get
\(\int x^{-\frac{1}{2}} d x\) = -k∫dt
∴ \(\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\) = -kt + c
∴ 2√x = -kt + c
At the beginning, i.e. at t = 0, x = 10,00,000
2√10,00,000 = -k(0) + c
∴ c = 2000
∴ 2√x = -kt + 2000 ……..(1)
Also, when t = 2, x = 10,000
∴ 2√10000 = -k × 2 + 2000
∴ 2k = 1800
∴ k = 900
∴ (1) becomes,
∴ 2√x = -900t + 2000
When the person will be bankrupt, x = 0
∴ 0 = -900t + 2000
∴ 900t = 2000
∴ t = \(\frac{20}{9}=2 \frac{2}{9}\)
Hence, the person will be bankrupt in \(2 \frac{2}{9}\) years.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 12 Maths 2
• Exercise 6.2 Class 12 Maths 2
• Exercise 6.3 Class 12 Maths 2
• Exercise 6.4 Class 12 Maths 2
• Exercise 6.5 Class 12 Maths 2
• Exercise 6.6 Class 12 Maths 2
• Miscellaneous Exercise 6 Class 12 Maths 2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.6 Questions and Answers.

12th Maths Part 2 Differential Equations Exercise 6.6 Questions And Answers Maharashtra Board

Question 1.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let x be the number of bacteria in the culture at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
\(\int \frac{d x}{x}\) = k∫dt + c
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x₀
log x₀ = k × 0 + c
∴ c = log x₀
∴ log x = kt + log x₀
∴ log x – log x₀ = kt
∴ log(\(\frac{x}{x_{0}}\)) = kt ………(1)
Since the number doubles in 4 hours, i.e. when t = 4, x = 2x₀

∴ the number of bacteria will be 8 times the original number in 12 hours.

Question 2.
If the population of a country doubles in 60 years; in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?
[Given log 2 = 0.6912, log 3 = 1.0986]
Solution:
Let P be the population at time t years.
Then \(\frac{d P}{d t}\), the rate of increase of population is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
\(\int \frac{d P}{P}\) = k∫dt + c
∴ log P = kt + c
Initially i.e. when t = 0, let P = P₀
∴ log P₀ = k x 0 + c
∴ c = log P₀
∴ log P = kt + log P₀
∴ log P – log P₀ = kt
∴ log(\(\frac{P}{P_{0}}\)) = kt ……(1)
Since, the population doubles in 60 years, i.e. when t = 60, P = 2P₀

∴ the population becomes triple in 95.4 years (approximately).

Question 3.
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
Solution:
Let θ°C be the temperature of the body at time t minutes.
The room temperature is given to be 25°C.
Then by Newton’s law of cooling, \(\frac{d \theta}{d t}\), the rate of change of temperature, is proportional to (θ – 25).


∴ the temperature of the body will be 36.36°C after 1 hour.

Question 4.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after 2½ hours. [Take √2 = 1.414]
Solution:
Let x be the number of bacteria at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
\(\int \frac{d x}{x}\) = k∫dt + c
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c
∴ c = log 1000
∴ log x = kt + log 1000
∴ log x – log 1000 = kt
∴ log(\(\frac{x}{1000}\)) = kt ……(1)
Now, when t = 1, x = 2 × 1000 = 2000
∴ log(\(\frac{2000}{1000}\)) = k
∴ k = log 2

∴ the number of bacteria after 2½ hours = 5656.

Question 5.
The rate of disintegration of a radioactive element at any time t is proportional to its mass at that time. Find the time during which the original mass of 1.5 gm will disintegrate into its mass of 0.5 gm.
Solution:
Let m be the mass of the radioactive element at time t.
Then the rate of disintegration is \(\frac{d m}{d t}\) which is proportional to m.


∴ log(3)^-1 = -kt
∴ -log 3 = -kt
∴ t = \(\frac{1}{k}\) log 3
∴ the original mass will disintegrate to 0.5 gm when t = \(\frac{1}{k}\) log 3

Question 6.
The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.
Solution:
Let x gm be the amount of the substance left at time t.
Then the rate of decay is \(\frac{d x}{d t}\), which is proportional to x.


∴ \(\frac{x}{25}=\frac{27}{125}\)
∴ x = \(\frac{27}{5}\)
∴ the amount left after 3 hours \(\frac{27}{5}\) gm.

Question 7.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant.
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
\(\int \frac{1}{P} d P\) = k∫dt + c
∴ log P = kt + c
Initially, i.e. when t = 0, P = 30000
∴ log 30000 = k × 0 + c
∴ c = log 30000
∴ log P = kt + log 30000
∴ log P – log 30000 = kt
∴ log(\(\frac{P}{30000}\)) = kt …….(1)
Now, when t = 40, P = 40000

∴ the population of the city at time t = 30000\(\left(\frac{4}{3}\right)^{\frac{t}{40}}\).

Question 8.
A body cools according to Newton’s law from 100°C to 60°C in 20 minutes. The temperature of the surroundings is 20°C. How long will it take to cool down to 30°C?
Solution:
Let θ°C be the temperature of the body at time t.
The temperature of the surrounding is given to be 20°C.
According to Newton’s law of cooling


∴ the body will cool down to 30°C in 60 minutes, i.e. in 1 hour.

Question 9.
A right circular cone has a height of 9 cm and a radius of the base of 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec, where A is the area of the water surface
at that instant, show that the vessel will be full in 75 seconds.
Solution:

Let r be the radius of the water surface and h be the height of the water at time t.
∴ area of the water surface A = πr² sq cm.
Since height of the right circular cone is 9 cm and radius of the base is 5 cm.
\(\frac{r}{h}=\frac{5}{9}\)
∴ r = \(\frac{5}{9} h\)
∴ area of water surface, i.e. A = \(\pi\left(\frac{5}{9} h\right)^{2}\)
∴ A = \(\frac{25 \pi h^{2}}{81}\) ……..(1)
The water level, i.e. the rate of change of h is \(\frac{d h}{d t}\) rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec.

∴ t = \(\frac{81 \times 9 \times 25}{3 \times 81}\) = 75
Hence, the vessel will be full in 75 seconds.

Question 10.
Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.
Solution:
Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t.
Then V = \(\frac{4}{3}\)πr³ and S = 4πr²
The rate at which the raindrop evaporates is \(\frac{d V}{d t}\) which is proportional to the surface area.
∴ \(\frac{d V}{d t}\) ∝ S
∴ \(\frac{d V}{d t}\) = -kS, where k > 0 ………(1)

On integrating, we get
∫dr = -k∫dt + c
∴ r = -kt + c
Initially, i.e. when t = 0, r = 3
∴ 3 = -k × 0 + c
∴ c = 3
∴ r = -kt + 3
When t = 1, r = 2
∴ 2 = -k × 1 + 3
∴ k = 1
∴ r = -t + 3
∴ r = 3 – t, where 0 ≤ t ≤ 3.
This is the required expression for the radius of the raindrop at any time t.

Question 11.
The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city, it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years, if the initial population is 10,000. [Take e = 2.7182]
Solution:
Let P be the population of the city at time t.
Then the rate of growth of population is \(\frac{d P}{d t}\) which is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k = 0.04
∴ \(\frac{d P}{d t}\) = (0.04)P
∴ \(\frac{1}{P}\) dP = (0.04)dt
On integrating, we get
\(\int \frac{1}{P} d P\) = (0.04) ∫dt + c
∴ log P = (0.04)t + c
Initially, i.e., when t = 0, P = 10000
∴ log 10000 = (0.04) × 0 + c
∴ c = log 10000
∴ log P = (0.04)t + log 10000
∴ log P – log 10000 = (0.04)t
∴ log(\(\frac{P}{10000}\)) = (0.04)t
When t = 25, then
∴ log(\(\frac{P}{10000}\)) = 0.04 × 25 = 1
∴ log(\(\frac{P}{10000}\)) = log e ……[∵ log e = 1]
∴ \(\frac{P}{10000}\) = e = 2.7182
∴ P = 2.7182 × 10000 = 27182
∴ the population of the city after 25 years will be 27,182.

Question 12.
Radium decomposes at a rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?
Solution:
Let x be the amount of the radium at time t.
Then the rate of decomposition is \(\frac{d x}{d t}\) which is proportional to x.


Hence, \(\left(10-\frac{p}{10}\right)^{2} \%\) of the amount will be left after 2 years.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 12 Maths 2
• Exercise 6.2 Class 12 Maths 2
• Exercise 6.3 Class 12 Maths 2
• Exercise 6.4 Class 12 Maths 2
• Exercise 6.5 Class 12 Maths 2
• Exercise 6.6 Class 12 Maths 2
• Miscellaneous Exercise 6 Class 12 Maths 2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.5 Questions and Answers.

12th Maths Part 2 Differential Equations Exercise 6.5 Questions And Answers Maharashtra Board

Question 1.
Solve the following differential equations:
(i) \(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\)
Solution:
\(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\) …….(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = \(\frac{1}{x}\) and Q = x³ – 3

This is the general solution.

(ii) cos²x . \(\frac{d y}{d x}\) + y = tan x
Solution:

(iii) (x + 2y³) \(\frac{d y}{d x}\) = y
Solution:

(iv) \(\frac{d y}{d x}\) + y . sec x = tan x
Solution:
\(\frac{d y}{d x}\) + y sec x = tan x
∴ \(\frac{d y}{d x}\) + (sec x) . y = tan x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = sec x and Q = tan x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \sec x d x}\)
= \(e^{\log (\sec x+\tan x)}\)
= sec x + tan x
∴ the solution of (1) is given by
y (I.F.) = ∫Q . (I.F.) dx + c
∴ y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + tan²x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + sec²x – 1) dx + c
∴ (sec x + tan x) . y = sec x + tan x – x + c
∴ y(sec x + tan x) = sec x + tan x – x + c
This is the general solution.

(v) x \(\frac{d y}{d x}\) + 2y = x² . log x
Solution:

(vi) (x + y) \(\frac{d y}{d x}\) = 1
Solution:
(x + y) \(\frac{d y}{d x}\) = 1
∴ \(\frac{d x}{d y}\) = x + y
∴ \(\frac{d x}{d y}\) – x = y
∴ \(\frac{d x}{d y}\) + (-1) x = y ……….(1)
This is the linear differential equation of the form

This is the general solution.

(vii) (x + a) \(\frac{d y}{d x}\) – 3y = (x + a)⁵
Solution:

(viii) dr + (2r cot θ + sin 2θ) dθ = 0
Solution:
dr + (2r cot θ + sin 2θ) dθ = 0
∴ \(\frac{d r}{d \theta}\) + (2r cot θ + sin 2θ) = 0
∴ \(\frac{d r}{d \theta}\) + (2 cot θ)r = -sin 2θ ………(1)
This is the linear differential equation of the form dr
\(\frac{d r}{d \theta}\) + P . r = Q, where P = 2 cot θ and Q = -sin 2θ

This is the general solution.

(ix) y dx + (x – y²) dy = 0
Solution:
y dx + (x – y²) dy = 0
∴ y dx = -(x – y²) dy
∴ \(\frac{d x}{d y}=-\frac{\left(x-y^{2}\right)}{y}=-\frac{x}{y}+y\)
∴ \(\frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y\) ………(1)
This is the linear differential equation of the form

This is the general solution.

(x) \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}}\)
Solution:

(xi) \(\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}\)
Solution:

Question 2.
Find the equation of the curve which passes through the origin and has the slope x + 3y – 1 at any point (x, y) on it.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition,
\(\frac{d y}{d x}\) = x + 3y – 1
∴ \(\frac{d y}{d x}\) – 3y = x – 1 ………(1)
This is the linear differential equation of the form

This is the general equation of the curve.
But the required curve is passing through the origin (0, 0).
∴ by putting x = 0 and y = 0 in (2), we get
0 = 2 + c
∴ c = -2
∴ from (2), the equation of the required curve is 3(x + 3y) = 2 – 2e^3x i.e. 3(x + 3y) = 2 (1 – e^3x).

Question 3.
Find the equation of the curve passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\) having slope of the tangent to the curve at any point (x, y) is \(-\frac{4 x}{9 y}\).
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then the slope of the tangent to the curve at point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}=-\frac{4 x}{9 y}\)
∴ y dy = \(-\frac{4}{9}\) x dx
Integrating both sides, we get
∫y dy= \(-\frac{4}{9}\) ∫x dx
∴ \(\frac{y^{2}}{2}=-\frac{4}{9} \cdot \frac{x^{2}}{2}+c_{1}\)
∴ 9y² = -4x² + 18c₁
∴ 4x² + 9y² = c where c = 18c₁
This is the general equation of the curve.
But the required curve is passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\).
∴ by putting x = \(\frac{3}{\sqrt{2}}\) and y = √2 in (1), we get
\(4\left(\frac{3}{\sqrt{2}}\right)^{2}+9(\sqrt{2})^{2}=c\)
∴ 18 + 18 = c
∴ c = 36
∴ from (1), the equation of the required curve is 4x² + 9y² = 36.

Question 4.
The curve passes through the point (0, 2). The sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at any point by 5. Find the equation of the curve.
Solution:
Let A(x, y) be any point on the curve.
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
x + y = \(\frac{d y}{d x}\) + 5
∴ \(\frac{d y}{d x}\) – y = x – 5 ………(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = -1 and Q = x – 5
∴ I.F. = \(e^{\int P d x}=e^{\int-1 d x}=e^{-x}\)
∴ the solution of (1) is given by

This is the general equation of the curve.
But the required curve is passing through the point (0, 2).
∴ by putting x = 0, y = 2 in (2), we get
2 = 4 – 0 + c
∴ c = -2
∴ from (2), the equation of the required curve is y = 4 – x – 2e^x.

Question 5.
If the slope of the tangent to the curve at each of its point is equal to the sum of abscissa and the product of the abscissa and ordinate of the point. Also, the curve passes through the point (0, 1). Find the equation of the curve.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}\) = x + xy
∴ \(\frac{d y}{d x}\) – xy = x ……….. (1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = -x and Q = x
∴ I.F. = \(e^{\int P d x}=e^{\int-x d x}=e^{-\frac{x^{2}}{2}}\)
∴ the solution of (1) is given by
y . (I.F.) = ∫Q . (I.F.) dx + c

This is the general equation of the curve.
But the required curve is passing through the point (0, 1).
∴ by putting x = 0 and y = 1 in (2), we get
1 + 1 = c
∴ c = 2
∴ from (2), the equation of the required curve is 1 + y = \(2 e^{\frac{x^{2}}{2}}\).

Class 12 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 12 Maths 2
• Exercise 6.2 Class 12 Maths 2
• Exercise 6.3 Class 12 Maths 2
• Exercise 6.4 Class 12 Maths 2
• Exercise 6.5 Class 12 Maths 2
• Exercise 6.6 Class 12 Maths 2
• Miscellaneous Exercise 6 Class 12 Maths 2