Bhagya

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.10 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.10

Evaluate the following:

Question 1.
(i) \(\Upsilon\) (4)
Solution:
Γ(4) = Γ(3 + 1) = 3! = 6

(ii) \(\Upsilon\) (\(\frac { 9 }{2}\))
Solution:

(iii) \(\int_{0}^{∞}\) e^-mx x⁶ dx
Solution:
W.K.T \(\int_{0}^{∞}\) xⁿ e^-ax dx = \(\frac { n! }{a^{n+1}}\)
∴ \(\int_{0}^{∞}\) e^-mx x⁶ dx = \(\frac { 6! }{3^{6+1}}\) = \(\frac { 6! }{m^7}\)

(iv) \(\int_{0}^{∞}\) e^-4x x⁴ dx
Solution:

(v) \(\int_{0}^{∞}\) e^-x/2 x⁵ dx
Solution:

Question 2.
If f(x) = \(\left\{\begin{array}{l}
x^{2} e^{-2 x}, x \geq 0 \\
0, \text { otherwise }
\end{array}\right.\), then evaluate \(\int_{0}^{∞}\) f(x) dx
Solution:
Given

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 2 Integral Calculus I Ex 2.11 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.11

Evaluate the following integrals as the limit of the sum:

Question 1.
\(\int_{0}^{1}\) (x + 4) dx
Solution:

Question 2.
\(\int_{1}^{3}\) x dx
Solution:

Question 3.
\(\int_{1}^{3}\) (2x + 3) dx
Solution:

Question 4.
\(\int_{0}^{1}\) x² dx
Solution:

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Miscellaneous Problems

Question 1.
A manufacture’s marginal revenue functional is given by MR = 275 – x – 0.3x². Find the increase in the manufactures total revenue if the production increased from 10 to 20 units.
Solution:
MR = 275 – x – 0.3x²
The increase in the manufactures total revenue 20
T.R = ∫MR dx = (275 – x – 0.3x²) dx

= [5500 – 200 – 0.1 (8000)] – [2750 – 50 – 0.1(1000)]
= [5500 – 200 – 800] – [2750 – 50 – 100]
= 4500 – 2600
= Rs 1900

Question 2.
A company has determined that marginal cost function for product of a particular commodity is given by MC = 125 + 10x – \(\frac { 8 }{3}\). Where C is the cost of producing x units of the commodity. If the fixed cost Rs 250 what is cost of producing 15 units.
Solution:
MC = 125 + 10x – \(\frac { x^2 }{9}\)
Fixed cos t K = Rs 250
C = ∫MC dx – ∫(125 + 10x – \(\frac { x^2 }{9}\)) dx
C = 125x + \(\frac { 10x^2 }{9}\) – \(\frac { x^3 }{9×3}\) + k
C = 125x + 5x² – \(\frac { x^3 }{27}\) + 250
when x = 15
C = 125(15) + 5(15)² – \(\frac { (15)^3 }{27}\) + 250
= 1875 + 1125 – 125 + 250
C = Rs 3,125

Question 3.
The marginal revenue function for a firm given by MR = \(\frac { 2 }{x+3}\) – \(\frac { 2x }{(x+3)^2}\) + 5. Show that the demand function is P = \(\frac { 2x }{(x+3)^2}\) + 5
Solution:

Question 4.
For the marginal revenue function MR = 6 – 3x² – x³, Find the revenue function and demand function
Solution:
MR = 6 – 3x² – x³
Revenue function R = ∫MR dx
R = ∫(6 – 3x² – x³)dx

Question 5.
The marginal cost of production of a firm is given by C'(x) = 20 + \(\frac { x }{20}\) the marginal revenue is given by R’(x) = 30 and the fixed cost is Rs 100. Find the profit function.
Solution:
C'(x) = 20 + \(\frac { x }{20}\)
Fixed cost k₁ = Rs 100
C(x) = ∫C₁(x) dx + k₁

Question 6.
The demand equation for a product is P_d = 20 – 5x and the supply equation is P_s = 4x + 8. Determine the consumers surplus and producer’s surplus under market equilibrium.
Solution:
P_d = 20 – 5x and P_s = 4x + 8
At market equilibrium
P_d = P_d
20 – 5x = 4x + 8 ⇒ 20 – 8 = 4x + 5x
9x = 12 ⇒ x = \(\frac { 12 }{9}\)
∴ x = \(\frac { 4 }{3}\)

Question 7.
A company requires f(x) number of hours to produce 500 units. lt is represented by f(x) = 1800x^-0.4. Find out the number of hours required to produce additional 400 units. [(900)^0.6 = 59.22, (500)^0.6 = 41.63]
Solution:
f(x) number of hours to produce 500
f(x) = 1800 x^-0.4
The number of hours required to produce additions

Question 8.
The price elasticity of demand for a commodity is \(\frac { p }{x^3}\). Find the demand function if the quantity of demand is 3, When the price is Rs 2.
Solution:
Price elasticity of demand

Question 9.
Find the area of the region bounded by the curve between the parabola y = 8x² – 4x + 6 the y-axis and the ordinate at x = 2.
Solution:
Equation of the parabola
y = 8x² – 4x + 6

The required region is bounded by the y-axis and the ordinate at x = 2.
∴ Required Area A = \(\int_{0}^{2}\)ydx

Question 10.
Find the area of the region bounded by the curve y² = 27x³ and the line x = 0, y = 1 and y = 2
Solution:
Equation of the curve is y² = 27x³
⇒ x³ = \(\frac { y^2 }{27}\) = \(\frac { y^3 }{3^3}\)
∴ x = \(\frac { (y)^{2/3} }{3}\)
Since the Area of the region bounded by the given curve and the lines x = 0, y = 1 and y = 2

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Students can download 10th Social Science Geography Chapter 5 India: Population, Transport, Communication, and Trade Questions and Answers, Notes, Samacheer Kalvi 10th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Social Science Solutions Geography Chapter 5 India: Population, Transport, Communication, and Trade

Samacheer Kalvi 10th Social Science India: Population, Transport, Communication, and Trade Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
The scientific study of different aspects of population is called:
(a) Photography
(b) Demography
(c) Choreography
(d) Population density
Answer:
(b) Demography

Question 2.
The state with highest literacy rate as per 2011 census is ……..
(a) Tamil nadu
(b) Karnataka
(c) Kerala
(d) Uttarpradesh
Answer:
(c) Kerala

Question 3.
Human Development is measured in terms of:
(a) Human Resource Index
(b) Per capita index
(c) Human Development Index
(d) UNDP
Answer:
(c) Human Development Index

Question 4.
……… Transport provides door to door services.
(a) Railways
(b) Roadways
(c) Airways
(d) Waterways
Answer:
(b) Roadways

Question 5.
The length of Golden Quadrilateral superhighways in India is:
(a) 5846 km
(b) 5847 km
(c) 5849 km
(d) 5800 km
Answer:
(a) 5846 km

Question 6.
The length of navigable Inland waterways in India is ……..
(a) 17,500 km
(b) 5000 km
(c) 14,500 km
(d) 1000 km
Answer:
(c) 14,500 km

Question 7.
The National Remote sensing Centre(NRSC) is located at:
(a) Bengaluru
(b) Chennai
(c) Delhi
(d) Hyderabad
Answer:
(d) Hyderabad

Question 8.
The transport useful in the inaccessible areas is ……..
(a) Roadways
(b) Railways
(c) Airways
(d) Waterways
Answer:
(c) Airways

Question 9.
Which of the following is associated with helicopter service?
(a) Air India
(b) Indian Airlines
(c) Vayudoot
(d) Pavan Hans
Answer:
(d) Pavan Hans

Question 10.
The major import item of India is
(a) Cement
(b) Jewells
(c) Tea
(d) Petroleum
Answer:
(d) Petroleum

II. Match the following

Question 1.
Match the Column I with Column II.


Answer:
A. (v)
B. (i)
C. (iv)
D. (ii)
E. (iii)

III. Answer the following questions briefly

Question 1.
What is Human Development?
Answer:
It is a process of enlarging the range of people’s choice, increasing their opportunities for education, health care, income and empowerment. It covers a full range of human choices from a sound physical environment to economic, social and political freedom.

Question 2.
What is migration? State its types.
Answer:
Migration is the movement of people across regions and territories. It can be internal (within a country) or international (between the countries). Migration depends on

Push factor: Unemployment and underemployment in rural areas.

Pull factor: higher wages, employment opportunity, industrial development.

Question 3.
Write any four advantages of railways.
Answer:

1. Railways are the price mode of transport for goods and passengers in India.
2. They make it possible to conduct varied activities like business, sightseeing and pilgrimage along with transportation of goods over longer distances.
3. They are suitable for long-distance travel and play an important role in national integration.
4. They bind the economic life of the country as well as the accelerate the development of the industry and agriculture.

Question 4.
Write a note on Pipeline network transport in India.
Answer:
Pipelines are the very convenient transport to connect oil and natural gas fields, refineries and to the markets.
It can be laid through difficult Terrain as well as under water.
It ensures steady supply of goods and reduces the transhipment loss and delays.

Three important network of pipeline:

1. Oil fields in upper Assam and Kanpur.
2. From Salaya in Gujarat to Jalandhar in Punjab.
3. From Hazira in Gujarat to Jagadispur in Uttar Pradesh

Question 5.
State the major Inland waterways of India.
Answer:
The major inland waters ways of India are :

1. National Waterway I: It extends between Haldia and Allahabad, measures 1620 km and includes the stretches of the Ganga – Bhagirathi – Hooghly river system.
2. National Waterway 2: This waterway includes the stretch of the Brahmaputra river between Dhubri and Sadiya a distance of 891 km.
3. National Waterway 3: This waterway extends between Kollam and Kottappuram in the state of Kerala. It is the first national waterway in the country with 24 hour navigation facilities along its entire stretch of 205 km.

Question 6.
What is communication? What are its types?
Answer:
Communication is a process that involves exchange of information, thoughts , and ideas. lt is categorized into two types.

Personal Communication: Post and telegraph, telephone, mobile phone, short message services, fax, internet, e-mail etc.
Mass Communication: Electronic media-Radio, T.V, internet and print media-News papers.

Question 7.
Define “International trade”.
Answer:
When trade takes place between two countries it is known as international trade.

Question 8.
State the merits of Roadways.
Answer:

1. Roadways can provide door to door services.
2. Easy and cheap to construct and maintain. Indian roads are cost-efficient.
3. The most universal mode of transport. It is used by all sections of people in society.
4. Can establish easy contact between farms, fields, factories and markets.

IV. Distinguish between

Question 1.
Density of population and Growth of population.
Answer:
Density of Population:

1. Density of population is the number of persons living per square kilometre.
2. It is calculated per 1000.
3. As per 2011 census the average density of population in India is 382 persons living in per sq.km.

Growth of Population:

1. Growth of population refers to the change in the number of in habitants of a country/territory in a specific period of time.
2. It is expressed in percentage.
3. The year 1921 is called the year of Great Demographic Divide.

Question 2.
Persona! communication and Mass communication.
Answer:
Personal communication:

1. Exchange of communication between the individuals.
2. Enables the user to establish direct contact.
3. It includes post and telegraph, telephones, mobile phones, short message service (SMS) fax, Internet, e-mail etc.

Mass communication:

1. Millions of people get the information at the same time.
2. Provide information through print media and electronic media.
3. It includes Radio, Television and Internet. Electronic media, News paper, Magazines, books, journals etc.

Question 3.
Print Media and Electronic Media.
Answer:
Print Media:

1. Most powerful means of communication.
2. Many news papers carry information on local national and international events to the people.
3. Knowledge of reading is essential.

Electronic Media:

1. Communication to millions of people through electronic gadgets at the same time.
2. Radio, Television, Internet provide and create awareness on various national policies and programme.
3. A powerful audio-visual medium.

Question 4.
Roadways and Railways.
Answer:
Roadways:

1. Highly suitable for short distance services.
2. Most common mode of transport, easy, cheap to construct and maintain.
3. Based on the construction and maintenance divided into National Highways, State Highways, District Roads, Village roads, Border roads, International highways, Express ways.
4. Second longest network in the world.
5. Door to door services possible

Railways:

1. Ideal for long distance services.
2. Construction depend upon the climatic and physical factors like terrain. Costly when compared with roadways.
3. On the basis of width of the track falls under 4 types, Broad gauge, meter gauge, narrow gauge and light gauge.
4. Largest in Asia and second largest in the World.
5. Cannot provide door to door services.

Question 5.
Waterways and Airways.
Answer:
Waterways:

1. Oldest and Cheapest means of transport.
2. Most suitable for carrying heavy and bulky materials.
3. Two types, inland waterways-river and backwaters and canals and oceanic routes connect coastal areas.

Airways:

1. Modem and costliest means of transport.
2. Only limited weight can be carried. Cannot carry heavy and bulky materials.
3. Domestic air sendee within the country and International Airways connect major cities of the World.

Question 6.
internal trade and International trade.
Answer:
Internal trade:

1. Trade is carried on within the domestic territory.
2. Land transport plays a major role (Roadways and Railways).
3. Local currency is used.
4. Helps to promote a balanced regional growth.

International trade:

1. Trade is carried on between two or more countries.
2. Waterways and Airways play a vital role.
3. Foreign currency is involved.
4. Helps to promote country’s economy and raise the standard of living

V. Answer the following in a paragraph

Question 1.
What is urbanization? Explain its impacts.
Answer:
The process of society’s transformation from rural to urban is known as urbanization. The level of urbanization of a place is assessed based on the size of population of the towns and cities and the proportion of population engaged in non agricultural sectors. These two are closely linked to the process of industrialization and expansion of the secondary and tertiary sectors of economy.
Impacts of urbanization:

• Urbanization and population concentration go hand – in – hand and are closely related to each other. A rapid rate of urbanization in a society is taken as an indicator of its economic development.
• Urbanization is increasing rapidly in the developing countries including India.
• Rural to urban migration leads to population explosion in urban areas. Metropolitan cities like Mumbai, Kolkatta and Delhi have more population than that can accommodate.
  The following are the major problem of urbanization in India.
• It creates urban sprawl.
• It makes overcrowding in urban centres.
• It leads to shortage of houses in urban areas.
• It leads to the formation of slums.
• It increases traffic congestion in cities.
• It creates water scarcity in cities.
• It creates drainage problems.
• It poses the problem of solid waste management.
• It increases the rate of crime.

Question 2.
Explain the importances of satellite communication in India.
Answer:

1. The communication through satellites emerged as a new era in the system of communication in our country.
2. The use of Satellite in getting a continuous and synoptic view of larger area has made this communication system very vital for the country.
3. Satellite images are used for weather forecasting.
4. Monitoring of natural calamities.
5. Surveillance of border areas.
6. The INSAT (series) is a multipurpose system for telecommunication, meterological observation and for various other programs.
7. They are used for relaying signals to television, telephone, radio and mobile phone.
8. And also useful in weather detection, internet and military applications.

Question 3.
Bring out the distribution and density of population in India.
Answer:
The term ‘Population Distribution’ refers to the way the people are spaced over the earth’s surface. The distribution of population in India is quite uneven because of the vast variation ki the availability of resources. The population is mostly concentrated in the regions of industrial centres and the good agricultural lands. On the other hand, the areas such as high mountains and lands thickly forested areas and some remote comers are very thinly populated and some areas are even uninhabited.

Terrain, climate, soil, water bodies, mineral resources, industries, transport and urbanization are the major factors which affect the distribution of population in our country. Population density is a better measure of understanding the variation in distribution of population. It is expressed as number of person per unit area usually per sq.km. According to 2011, the average density of population of India is 382 person per sq.km. India is one of the most thickly populated ten countries of the world. The most densely populated state of India is Bihar and the state with lest population density is Arunachal Pradesh. Among union territories, Delhi is the densely populated one with 11,297 per sq.km, while Andaman and Nicobar Islands have the lowest density of population.

Question 4.
Explain the process of measuring Human Development.
Answer:
Dr. Mahabub-ul-haq defined Human Development as “It is a process of enlarging the range of people’s choice, increasing their opportunities for education, health care, income and empowerment. It covers the full range of human choices from a. sound physical environment to economic, social and political freedom”.

Measuring of Human Development (HDI): Human Development Index is a composite index focusing on three basic dimensions of human development.

1. Health: Life expectancy at birth.
2. Education: Expected years of schooling for school age children and average years of schooling for the adult population.
3. Income: Measured by gross national income and per capita income.

Question 5.
Classify and explain the roadways in India.
Answer:

1. The Indian roads are cost-efficient and the most popular dominant mode of transport
   linking different parts of our country.
2. Roads stretch across the length of people in society.
3. It is used by all sections of people in society.
4. Road network in India is the second-longest in the world accounting for 3,314 millions of km.

Types of Roadways:

1. Village Roads:

• Village roads link different villages with towns.
• They are maintained by Village Panchayath.
• In India village roads run to a length of 26,50,000 kms.

2. District Roads:

• District roads link the towns with the district headquarters.
• They are maintained by the Corporations and Municipalities.
• It is used by all sections of people in the society.
• In India district roads run to a total length of 4,67,763 kms.

3. State Highways:

• State Highways links the State capitals with the different district headquarters.
• These roads are maintained by the State Public Works Department.
• The State Highways run to a length of 1,31,899 kms. (Ex: Cudalore – Chittor Road).

4. National Highways:

• National Highways link the State Capitals with the National Capital.
• They are the primary road system of our country and are maintained by the Central Public Works Department.
• It runs to a length of 70,548 kms. (Ex: NH47 is a National Highway which connects Tamil Nadu and Kerala)

5. Golden Quadrilateral Super Highways:

• It is a major road development project launched by the Government of India.
• It runs to a length of 14,846 kms connecting the major cities of India.
• The major objective of these roads is providing ‘connectivity’, ‘speed’ and ‘safety’.

6. Expressways:

• Expressways are the technologically improved high class roads in the Indian Road network.
• They are six lanes roads. They run to a length of more than 200 kms.
• New Mumbai – Pune Road is an example for Expressway.

7. Border Road:

• Border Roads are the roads constructed along the northern and north eastern borders of our country.
• These roads are constructed and maintained by Border Roads Organization.
• The Organization has constructed 46,780 km of roads in difficult terrain.

8. International Highways:
International Highways are the roads that link India with neighbouring countries for promoting harmonious relationship with them.

VI. On the outline map of India mark the following

Question 1.
National Highway NH-7.
Answer:

Question 2.
Major seaports in India.
Answer:

Question 3.
Major International Airports in India.
Answer:

Question 4.
Densely populated state of India – Uttarpradesh
Answer:

Question 5.
State of highest literacy in India – Kerala
Answer:

Question 6.
Railways zones of India.
Answer:

TB Page No. 151
Hots:

Question 1.
What could be the reasons for uneven distribution of population in India?
Answer:

• Topography, favourable climate, fertility of soils, availability of fresh water, minerals are major geographical factors affecting population density of a region.
• People prefer to live on plains more than mountains or plateaus and they live more in moderate climates than extreme hot or cold. From the agriculture point of view, fertile lands are preferred.
• Areas with mineral deposits are more populated.
• Some social factors that boost the density of population in a region are better housing, education and health facilities.
• Places with cultural or historical significances are usually populated.
• Employment opportunities is another attraction for large chunks of the population.

Question 2.
What are the reasons for the rapid growth of population in India?
Answer:

1. The reasons for the rapid growth of population in India is mainly the , high birth rate and low death rate.
2. Birth rate is high, death rate declined due to advanced medical facilities and immunization to dreadful diseases.
3. And also migration of population due to employment, education and industrial development.

TB Page No. 154

Question 3.
The sex ratio in our country is always unfavourable to females. Give reasons.
Answer:
The number of females per thousand men is called the sex ratio.
Reasons:

• Lesser care of female children.
• Greater risk to womens’ life especially at the time of child brith.
• Women are also killed or forced to die by the dowry seekers.
• Due to illiteracy.
• Lack of medical facilities for women, etc.

TB Page No. 158

Question 4.
Find out what are the functions of NHAI?
Answer:

1. National Highway Authority of India was established in 1995.
2. It is an autonomous body under the ministry of Surface Transport.
3. Management of all major National Highways and the highways entrusted to it.
4. Maintenance, development and operation of the National Highways is undertaken by NHAI.

Question 5.
What are the highlights and benefits of the Golden Quadrilateral Highways? Highlights:
Answer:

• It is the largest Highway project completed in India.
• It is the fifth largest highway project in the world.
• The overall length of the Golden Quadrilateral is 5,846 km.
• The Golden Quadrilateral passes through 13 states of India.
• It connects four major metropolitan cities of the country in four directions.

Benefits:

• Provides faster transport networks between major cities and ports.
• Provides connectivity to major agricultural industrial and cultural centres of India.
• Provides smoother movement of goods and people within the country.
• Enables industrial development and job creation in smaller town through access to varied markets.
• Farmers are able to transport their produce to major cities and towns for sale and export and there is less wastage and spoils.
• More economic growth through construction and indirect demand for steel, cement, and other construction materials.
• Giving an impetus to truck transports.

TB Page No. 162

Question 6.
Why is air travel preferred in the North Eastern State?
Answer:
The North Eastern States are mountainous and forested. So construction of roadways or railways is very difficult due to terrains. Air transport has made accessibility easier. So air travel is preferred in the North eastern states.

TB Page No. 165

Question 7.
Find out the major trade blocs which are useful for multilateral trade.
Answer:
Trading blocs are usually groups of countries in specific regions that manage and promote trade activities. Trading blocs lead to trade liberalisation and trade creation between members, since they are treated favourably in comparison to non-members. The World Trade Organisation (WTO) permits the existence of trading bloc provided that they result in lower protection against outside countries than existed before the creation trading bloc.
The most significant trading blocs currently are:

1. European Union (EU): A customs union, a single market and now with a single currency.
2. European Free Trade Area (EFTA)
3. North American Free Trade Agreement (NFATA) between the USA, Canada and Mexico.
4. Mercosur: A customs union between Argentina, Brazil, Paraguay, Uruguay, and Venezuela.
5. Association of Southeast Asian Nations (ASEAN)
6. Association of Free Trade Area (AFTA)
7. Common market of Eastern and Southern Africa (COMESA)
8. South Asian Free Trade Area (SAFTA) created in 2006 with countries such as India and Pakistan.
9. Pacific Alliance: 2013 – a regional trade agreement between Chile, Colombia, Mexico and Peru.

TB Page No. 159
Activity:

Question 1.
Prepare a seminar topic about “Role of Railways in Indian Economy”Key Points:
Answer:

1. Large scale movement
2. National integration promotion
3. Commercialisation of agriculture
4. Movement of perishable goods
5. Avoids traffic congestion how?
6. Engineering marvel
7. Quiz regarding railways

TB Page No. 165

Question 2.
Collect the countries names and make it as a table of Bilateral trade and multilateral trade countries.
Answer:
India has made bilateral trade agreement with these countries.

Bilateral Trade Countries: U.S.A . China, Hongkong, Singapore, United Kingdom, Germany, Bangladesh.

Multilateral Trade Countries: The members of the SAARC (South Asian Association for Regional Cooperation) Korea, Japan, Bangladesh, Bhutan, Maldives, Nepal, Pakistan, Afghanistan, Srilanka, U.S.A and United Kingdom.

Samacheer Kalvi 10th Social Science India: Population, Transport, Communication, and Trade Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
…………………. is the second-most populous country next to China.
(a) India
(b) Pakistan
(c) Bangladesh
(d) Nepal
Answer:
(a) India

Question 2.
The fast movement of traffic are established by ……
(a) national highways
(b) Express highways
(c) International highways
Answer:
(a) national highways

Question 3.
As per 2011 census the average density of population of India is …………………. persons per Sq.Km.
(a) 302
(b) 382
(c) 100
(d) 365
Answer:
(b) 382

Question 4.
The ………. have more railways than the Himalayan Mountains.
(a) Northern Plains
(b) Coastal Plains
(c) Deccan Plateau
Answer:
(a) Northern Plains

Question 5.
The Grand Trunk Road extends from …………………. to ………………….
(a) Delhi to Mumbai
(b) Amritsar to Kolkatta
(c) Mumbai to Thane
(d) Srinagar to Amritsar.
Answer:
(b) Amritsar to Kolkatta

Question 6.
The cheapest mode of transport is …….
(a) roadways
(b) railways
(c) waterways
Answer:
(c) waterways

Question 7.
…………………. are multi-lane good quality highways for high speed traffic.
(a) National Highways
(b) State highways
(c) Border roads
(d) Express highways.
Answer:
(d) Express highways.

Question 8.
Trade carried on within the domestic territory of a country is known as …… trade.
(a) External
(b) Foreign
(c) Internal
Answer:
(c) Internal

Question 9.
The …………………. railway accounts for the longest route length.
(a) Northern Railways
(b) Central Railway
(c) Eastern railway
(d) Southern Railway
Answer:
(a) Northern Railways

Question 10.
A cost-efficient and most popular mode of transport in our country is ……..
(a) Airways
(b) Roadways
(c) Waterways
Answer:
(b) Roadways

Question 11.
India is the …………………. largest ship owning country in Asia.
(a) First
(b) Second
(c) Third
(d) Fourth
Answer:
(b) Second

Question 12.
The costliest and most modem means of transport is ……..
(a) Air transport
(b) Road transport
(c) Rail transport
Answer:
(a) Air transport

Question 13.
The is the first communication Satellite in INSAT Series.
(a) G.SAT
(b) EDUSAT
(c) INSAT-IB
(d) Kalpana-1
Answer:
(c) INSAT-IB

Question 14.
The major ports are managed and controlled by …….
(a) National Ports Corporation
(b) Port Trust of India
(c) Indian Airlines
Answer:
(b) Port Trust of India

Question 15.
Indian space Research organisation was established in the year:
(a) 1959
(b) 1969
(c) 1979
(d) 1996
Answer:
(b) 1969

II. Match the following

Question 1.
Match the Column I with Column II.


Answer:
A. (v)
B. (iv)
C. (i)
D. (iii)
E. (ii)

Question 2.
Match the Column I with Column II.

Answer:
A. (iii)
B. (iv)
C. (v)
D. (i)
E. (ii)

III. Answer the following questions briefly

Question 1.
Describe the three population density zones of India.
Answer:
The three population density zones of India are :

1. High-density zone: The Northern plains above 500 people per sq.km. Northern plains and Kerala in the. South.
2. Moderate or Medium density zone: Mountain region 250-500 people per sq.km. Ex. Assam and Peninsular states.
3. Low-density zone: Plateau region below 250 people per sq.km. Ex. Jammu and Kashmir, Sikkim, Arunachal Pradesh.

Question 2.
What are the reasons for the uneven distribution of population in India?
Answer:
The uneven distribution of population in the country is the result of several factors such as physical, socio-economic and historical factors.

Physical factors: Relief, climate, water, natural vegetation, minerals and energy resources.

Socio-economic factors: Religion, culture, political issues, economy, human settlements, transport network, industrialization, urbanization, employment opportunity etc.

Question 3.
What is the major objective to develop the Super Highways?
Answer:
The major objective to develop the Super Highways is to reduce the time and distance between
the megacities of India facilitating the fast movement of traffic.

Question 4.
Which phase period in population growth of India is often referred as period of population explosion?
Answer:
During the third phase (1951-1981) the population of India grew from 361 million in 1951 to 683 million in 1981. Growth rate in this period is almost doubled. So this period (1951-1981) is often referred as the period of “population explosion”.

Question 5.
State some problems of road transport in our country.
Answer:

1. Distribution of road is not uniform in the country.
2. Keeping in view the volume of traffic and passengers, the road network is inadequate.
3. About half of the roads are unmetalled and this restricts their usage during the rainy season.
4. The roadways are highly congested in cities.
5. Poor maintenance is also a big problem.

Question 6.
What are the four major shipyards in India?
Answer:
India as four major shipyards:

1. Hindustan shipyard – Vishakapatnam
2. Garden Reach workshop – Kolkata
3. Mazagaon Dock – Mumbai
4. Kochi shipyard – Kochi

Question 7.
State the highlights of India’s foreign trade policy since 2004.
Answer:

1. Merchandise trade has been doubled.
2. Thrust is given for employment generation; especially in semi-urban and rural areas.
3. Trade procedure is simplified and transaction cost is reduced.
4. Special focus is given to make India a global hub.
5. A new scheme called Vishesh Krishi Upay Yojna has been introduced to boost exports of fruits, vegetables, flowers and minor forest products.

Question 8.
Mention the services provided by the Indian postal system in India.
Answer:
Pcstai system is one among the personal communication system. The postal service was open to the public in the country in 1837.

1. Collecting and delivering mail is the primary function of the postal department.
2. It introduced Quick Mail Service in 1975 on the basis ofPIN code (1972)
3. The premium products include the Money order. E-money order, Speed post. Express parcel post, Business post, Media post, Retail post, Data post, Satellite post, Greeting post, Speed net and Speed passport services,

Question 9.
What are the advantages of communication network?
Answer:

1. It has enhanced the efficiency of communication. Because it enables quick exchange of information with people any where in the world.
2. It leads to enormous growth of trade.
3. It helps the government to tackle various socio – economic problems in the society.
4. It improves the quality of human life.
5. It opens the door to the information age.
6. It promotes Edusat programs.
7. It plays a significant role in the economic and social growth of our country.

Question 10.
Write the full form of STD, ISD, PCO, Internet.
Answer:

• STD – Subscriber Trunk Dialing
• ISD – International Subscriber Dialing
• PCO – Public Call Office
• Internet – Inter connected network.

Question 11.
What are major ports along the West Coast and East Coast?
Answer:
West Coast Ports :

1. Kandla,
2. Mumbai,
3. Jawahar Lal Nehru,
4. Marmagoa,
5. New Mangalore and
6. Cochin East Cost Ports:
   • Tuticorin,
   • Chennai,
   • Ennore,
   • Vishakhapatnam,
   • Paradip,
   • Haldia and
   • Kolkota.

Question 12.
Name the major ports on the East coast and on the West Coast of India.
Answer:
There are 13 major ports in India. They are administered by the Central Government.

1. The major ports on the East coast are Kolkata, Haldia, Paradip,Visakhapatnam, Chennai, Ennore and Tuticorin.
2. The major ports on the West Coast are Kandla, Mumbai, Nhava Seva (Jawaharial Nehru Port), New Mangalore, Marmagao and Kochi.

Question 13.
Write a short note on Internet.
Answer:

1. Internet is a vast network of computers, Internet means interconnected network of net , works.
2. It connects many of the world’s business institutions and individuals.
3. It enables computer users throughout the world to send and receive messages and information in a variety of forms.
4. It is fully a multimedia based system with capacity to deliver pictures, images, video and audio.
5. The basic services of internet are e-mail. The World Wide Web and Internet Phone.

IV. Distinguish between

Question 1.
Exports and Imports.
Exports:

1. Goods and services sold for foreign currency.
2. Major exports of India are tea, ores and minerals, marine products, textiles etc.
3. Value of exports are more than the value of imports favourable balance of trade.
4. India exports goods to nearly 190 countries of the world.

Imports:

1. Goods and services bought from overseas producers.
2. Major imports are petroleum products, gold, telecom instruments.
3. Value of imports exceeds value of exports unfavourable balance of trade.
4. Imports we get from nearly 140 countries.

Question 2.
Low Density and High Density of population.
Answer:
Low Density of Population:

1. Areas that have 150 to 300 persons living per sq.km.
2. Extreme climate, high mountains, remote areas, forested regions have low density of population.
3. In India Arunachal Pradesh, Sikkim, Mizoram, Andaman and Nicobar islands are some states with low population density.

High Density of Population:

1. Areas that have 500 to 1000 persons living per sq.km.
2. Favourable climate, plains, employment opportunities, industrial centres area have high density of population.
3. Punjab, Tamil Nadu, Uttar Pradesh and Kerala are the states with high population density.

Question 3.
National and State Highways.
Answer:
National Highways:

1. Connects Capitals of States, major ports, rail junctions, industrial and tourist centres.
2. Ministry of Road transport and Highways of India is responsible development and maintenance.
3. Runs to a length of 1,01,011 kms as of 2016.

State Highways:

1. Link important cities, towns and district head quarters within the State and connect them with national highways or highways of neighbouring states.
2. Administered and financed by state Governments.
3. Runs to the length of 1,76,166 kms as of 2016.

Question 4.
Domestic and International Airways.
Answer:
Domestic Airways:

1. Fly within the boundaries of the country.
2. Indian Airlines provides the domestic air services. NACIL (I)
3. There are about 80 domestic airports and about 25 civil enclaves at defence air fields.

International Airways:

1. Fly across the world and connect major cities of the world.
2. Air India provides international air services. NACIL (A)
3. There are 19 designated international airports in the country.

Question 5.
Harbour and Port.
Answer:
Harbour:

1. Extensive stretch of deep water near the seashore.
2. Vessels can anchor securely.

Port:

1. Commercial part of a harbour.
2. Facility of loading and unloading of goods and space for the storage of cargo
3. Cuddalore, Ennore.

Question 6.
Birth and Death.
Answer:
Birth Rate:

1. No. of live Births per 1000 people in a year.
2. Also known as Nationality rate.

Death Rate:

1. No. of Deaths per 1000 people in a year.
2. Also known as Mortality rate.

V. Answer the following in a paragraph

Question 1.
What is the importance of railway transport?
Answer:

1. The Indian Railways have been a great integrating force for more than 150 years.
2. They have a network of about 7,112 stations spread over a route length of 66,687 km with a fleet of 11,122 locomotives, 70,241 passenger coaches and 2,54,000 wagons.
3. Railways are the prime mode of transport for goods and passengers in India.
4. They make it possible to conduct varied activities like business, sightseeing and pilgrimage along with transportation of goods over longer distances.
5. They are suitable for long distance travel and play an important role in national integration.
6. They bind the economic life of the country as well as accelerate the development of the industry and agriculture.
7. The Indian Railways is the largest public sector undertaking in the country. The first train streamed of from Mumbai to Thane in 1853, covering a distance of 34 km.

Question 2.
What do you mean by Census? How it is useful?
Answer:

1. Population census is the total process of collecting, compiling, analyzing or otherwise disseminating demographic, economic and social data pertaining at a specific time, of all persons in a country or a well defined part of a country.
2. Census is being taken in an interval of ten years.
3. The data collected through the census are used for administration, planning, policy making, management and evaluation of various programmes by the government.
4. In India 1st census was carried out in the year 1872.
5. But the first complete and synchronous census was conducted in 1881 The 2011 census represents the fifteenth census of India.

Question 3.
Describe the importance of pipelines in India. Name three important networks of pipelines transportation in the country.
Answer:
Pipeline transport network is a new arrival on the transportation of India.

1. Transport of crude oil, petroleum products and natural gas from oil and natural gas fields to refineries, fertiliser factories and big thermal power plants. ‘
2. Even solids can be transported through pipelines when converted into slurry.
3. The far inland locations of refineries and gas based fertiliser plants could be transported.
4. Initial cost of lying pipelines is high but subsequent running costs are tninimal.
5. It rules out trans-shipment losses or delays.

Three important networks of pipeline transportation in the country:

• From oil field in the upper Assam to Kanpur (UP), via Guwahati, Baraumi and Allahabad.
• From Salaya in Gujarat to Jalandhar in Punjab. Via Viramgam, Mathura, Delhi and Sonipat.
• Gas pipeline from Hazira in Gujarat connects Jagdishpur in UP, via Vijaipur in MP.

Question 4.
What are the problems created by over population in India?
Answer:
In India, growing pressure of Population on resource base, created many socio economic, cultural, political, ecological and environmental problems. The population problem varies in space and time and differ from region to region.

Major issues created by the over population are – overcrowding, unemployment and under employment, low standard of living mal nutrition, mismanagement of natural and agricultural resources, unhealthy environment etc.

Question 5.
Write about the significance of Indian Railways.
Answer:

1. Indian Railway system is the main artery of the country’s inland transport.
2. Indian Railway network is the largest in Asia and second largest in the world.
3. Railways are considered as the back bone of the surface transport system of India.
4. Significances: It caters to the needs of large scale movement of traffic both for freight and passengers, there by contributing to economic growth.
5. It promotes trade, tourism, education etc.
6. It promotes national integration by bringing people together.
7. Railways help in the commercialization of the agricultural sector by facilitating the quick movements of perishable goods.
8. It provides invaluable service by transporting raw materials to industries and finished goods to market.

Question 6.
What is trade? What are the two types of trade? State its components:
Answer:

1. Trade is an act (or) process of buying selling or exchanging of goods and services. Trade in general is of two types. They are Internal trade and International trade.
2. Internal Trade: Trade carried on within the domestic territory of a country.
3. International trade: Trade carried on between two or more countries.
4. Exports and imports are the two components of international trade.
5. The difference between value of exports and imports is called balance of trade.
6. If the value of exports exceeds the value of imports trade is said to be favourable balance of trade.
7. If the value of Imports exceeds value of exports it is said to be unfavourable balance of trade.
8. The value of currency of a country depends upon the balance of trade of that country.

VI. On the outline map of India mark the following

Question 1.
North-South corridor
Answer:

Question 2.
East-West corridor
Answer:

Question 3.
Head quarters of Indian Railway – Delhi.
Answer:

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.3

Question 1.
Calculate consumer’s surplus if the demand function p = 50 – 2x and x = 20
Solution:
Demand function p = 50 – 2x and x = 20
when x = 20, p = 50 – 2(20)
p = 50 – 40 = 10
∴ p₀ = 10
CS = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)
= \(\int _{0}^{20}\) (50 – 2x)dx – (10 × 20)
= [50x – 2(\(\frac { x^2 }{2}\))]\( _{0}^{20}\) – 200
= [50x – x²]\( _{0}^{20}\) – 200
= {50(20) – (20)² – [0]} – 200
= (1000 – 400) – 200
= 600 – 200
∴ C.S = 400 units

Question 2.
Calculate consumer’s surplus if the demand function p = 122 – 5x – 2x² and x = 6.
Solution:
Demand function p = 122 – 5x – 2x² and x = 6
when x = 6; p = 122 – 5(6) – 2(6)²
= 122 – 30 – 2 (36)
= 122 – 102 = 20
∴ p₀ = 20
C.S = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)
= \(\int _{0}^{6}\)(122 – 5x – 2x²) dx – (20 × 6)

[732 – 5(18) – 2(72)] – 120
= 732 – 90 – 144 – 120
= 732 – 354 = 378
∴ CS = 378 units

Question 3.
The demand function p = 85 – 5x and supply function p = 3x – 3. Calculate the equilibrium price and quantity demanded. Also calculate consumer’s surplus.
Solution:
Demand function p = 85 – 5x
Supply function p = 3x – 35
W.K.T. at equilibrium prices p_d = p_s
85 – 5x = 3x – 35
85 + 35 = 3x + 5x
120 = 8x ⇒ x = \(\frac { 120 }{8}\)
∴ x = 15
when x = 15 p₀ = 85 – 5(15) = 85 – 75 = 10
C.S = \(\int _{0}^{x}\) f(x) dx – x₀p₀
= \(\int _{0}^{x}\) (85 – 5x) dx – (15)(10)

= 1275 – \(\frac { 1125 }{2}\) – 150
= 1275 – 562.50 – 150
= 1275 – 712.50
∴ CS = 562.50 units

Question 4.
The demand function for a commodity is p = e^-x. Find the consumer’s surplus when p = 0.5
Solution:
The demand function p = e^-x
when p = 0.5 ⇒ 0.5 = e^-x
\(\frac { 1 }{2}\) = \(\frac { 1 }{e^x}\) ⇒ e^x = 2
∴ x = log 2
∴ Consumer’s surplus
C.S = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)

Question 5.
Calculate the producer’s surplus at x = 5 for the supply function p = 7 + x
Solution:
The supply function p = 7 + x
when x = 5 ⇒ p = 7 + 5 = 12
∴ x₀ = 5 and p₀ = 12
Producer’s surplus

Question 6.
If the supply function for a product is p = 3x + 5x². Find the producer’s surplus when x = 4
Solution:
The supply function p = 3x + 5x²
when x = 4 ⇒ p = 3(4) + 5(4)²
p = 12 + 5(16)= 12 + 80
p = 92
∴ x₀ = 4 and p₀ = 92
Producer’s Surplus

Question 7.
The demand function for a commodity is p = \(\frac { 36 }{x+4}\) Find the producer’s surplus when the prevailing market price is Rs 6.
Solution:
The demand function for a commodity
p = \(\frac { 36 }{x+4}\)
when p = 6 ⇒ 6 = \(\frac { 36 }{x+4}\)
x + 4 = \(\frac { 36 }{6}\) ⇒ x + 4 = 6
x = 2
∴ p₀ = 6 and x₀ = 2
The consumer’s surplus
C.S = \(\int _{0}^{x}\) f(x) dx – x₀p₀
= \(\int _{0}^{2}\) (\(\frac { 36 }{x+4}\)) dx – 2(6)
= 36 [log (x + 4)]\(_{0}^{2}\) – 12
= 36 [log (2 + 4) – log (0 + 4)] – 12
= 36 [log6 – log4] – 12
= 36[log(\(\frac { 6 }{4}\))] – 12
∴ CS = 36 log(\(\frac { 6 }{4}\)) – 12 units

Question 8.
The demand and supply functions under perfect competition are pd = 1600 – x² and p_s = 2x² + 400 respectively, find the producer’s surplus.
Solution:
pd = 1600 – x² and p_s = 2x² + 400
Under the perfect competition p_d = p_s
1600 – x² = 2x² + 400
1600 – 400 = 2x² + x² ⇒ 1200 = 3x²
⇒ x² – 400 ⇒ x = 20 or -20
The value of x cannot be negative, x = 20 when x₀ = 20;
p₀ = 1600 – (20)² = 1600 – 400
P₀ = 1200
PS = x₀p₀ – \(\int _{0}^{x_0}\) g(x) dx

Question 9.
Under perfect competition for a commodity the demand and supply laws are P_d = \(\frac { 8 }{x+1}\) – 2 and Ps = \(\frac { x+3 }{2}\) respectively. Find the consumer’s and producer’s surplus.
Solution:

16 – (x² + 3x + x + 3) = 2 [2(x + 1)]
16 – (x² + 4x + 3) = 4(x + 1)
16 – x² – 4x – 3 = 4x + 4
x² + 4x + 4x + 4 + 3 – 16 = 0
x² + 8x – 9 = 0
(x + 9) (x – 1) = 0 ⇒ x = -9 (or) x = 1
The value of x cannot be negative x = 1 when x₀ = 1
p₀ = \(\frac { 8 }{1+1}\) – 2 ⇒ p₀ = \(\frac { 8 }{2}\) – 2
p₀ = 4 – 2 ⇒ p₀ = 2
CS = \(\int _{0}^{x}\) f(x) dx – x₀p₀
= \(\int _{0}^{1}\) (\(\frac { 8 }{x+1}\) – 2) dx – (1) (2)
= {8[log(x + 1)] – 2x}\(\int _{0}^{1}\) – 2
= 8 {[log (1 + 1) – 2(1)] – 8 [log (0 + 1) – 2(0)]} – 2
= [8 log (2) – 2 – 8 log1] – 2
= 8 log(\(\frac { 8 }{2}\)) – 2 – 2
C.S = (8 log 2 – 4) units
P.S = x₀p₀ – \(\int _{0}^{x_0}\) g(x) dx

Question 10.
The demand equation for a products is x = \(\sqrt {100-p}\) and the supply equation is x = \(\frac {p}{2}\) – 10. Determine the consumer’s surplus and producer’s, under market equilibrium.
Solution:
p_d = \(\sqrt {100-p}\) and p_s = \(\sqrt {100-p}\)
Under market equilibrium, p_d = p_s
\(\sqrt {100-p}\) = \(\frac {p}{2}\) – 10
Squaring on both sides

36p = p² ⇒ p² – 36 p = 0
p (p – 36) = 0 ⇒ p = 0 or p = 36
The value of p cannot be zero, ∴p₀ = 36 when p₀ = 36
x₀ = \(\sqrt {100-36}\) = \(\sqrt {64}\)
∴ x₀ = 8

= 288 – [x² + 20x]\( _{0}^{8}\)
= 288 – { [(8)² + 20(8)] – [0]}
= 288 – [64 + 160]
= 288 – 224 = 64
PS = 64 Units

Question 11.
Find the consumer’s surplus and producer’s surplus for the demand function pd = 25 – 3x and supply function ps = 5 + 2x
Solution:
p_d = 25 – 3x and p_s = 5 + 2x
Under market equilibrium, p_d = p_s
25 – 3x = 5 + 2x
25 – 5 = 2x + 3x ⇒ 5x = 20
∴ x = 4
when x = 4
P₀ = 25 – 3(4)
= 25 – 12 = 13
p₀ = 13

= 52 – [5x + x²]\( _{0}^{4}\)
= 52 – (5(4) + (4)²) – (0)}
= 52 – [20 + 16]
= 52 – 36
∴ PS = 16 units

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.4 Text Book Back Questions and Answers, Notes.

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Choose the most suitable answer from the given four alternatives:

Question 1.
Area bounded by the curve y = x (4 – x) between the limits 0 and 4 with x-axis is
(a) \(\frac { 30 }{3}\) sq.unit
(b) \(\frac { 31 }{2}\) sq.unit
(c) \(\frac { 32 }{3}\) sq.unit
(d) \(\frac { 15 }{3}\) sq.unit
Solution:
(c) \(\frac { 32 }{3}\) sq.unit
Hint:

Question 2.
Area bounded by the curve y = e^-2x between the limits 0 < x < ∞ is
(a) 1 sq.units
(b) \(\frac { 1 }{2}\) sq.units
(c) 5 sq.units
(d) 2 sq.units
Solution:
(b) \(\frac { 1 }{2}\) sq.units
Hint:

Question 3.
Area bounded by the curve y = \(\frac { 1 }{x}\) between the limits 1 and 2 is
(a) log 2 sq.units
(b) log 5 sq.units
(c) log 3 sq.units
(d) log 4 sq.units
Solution:
(a) log 2 sq.units
Hint:
Area = \(\int_{1}^{2} \frac{1}{x} d x\)
= \((\log x)_{1}^{2}\)
= log 2 – log 1
= log 2 (Since log 1 = 0)

Question 4.
If the marginal revenue function of a firm is MR = e^{\(\frac { -x }{10}\)} then revenue is
(a) 1 – e^-x/10
(b) e^-x/10 + 10
(c) 10(1 – e^-x/10)
(d) -10e^-x/10
Solution:
(c) 10(1 – e^-x/10)
Hint:
MR = e^{\(\frac { -x }{10}\)} then R = ∫MR dx
R = ∫e^-x/10 dx = \(\frac { e^{-x/10} }{(-1/10)}\) + k
R = -10e^-x/10 + k when x = 0, R = 0
⇒ 0 = -10e⁰ + k
0 = -10(1) + k
∴ k = 10
R = -10e^-x/10 + 10 = 10(1 – e^-x/10)

Question 5.
If MR and MC denotes the marginal revenue and marginal cost functions, then the profit functions is
(a) P = ∫(MR – MC) dx + k
(b) P = ∫(R – C) dx + k
(c) P = ∫(MR + MC)dx + k
(d) P = ∫(MR) (MC) dx + k
Solution:
(a) P = ∫(MR – MC) dx + k
Hint:
Profit = Revenue – Cost

Question 6.
The demand and supply functions are given by D(x) = 16 – x² and S(x) = 2x² + 4 are under perfect competition, then the equilibrium price x is
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(a) 2
Hint:
D(x) =16 – x² and S(x) = 2x² + 4
Under perfect competition D(x) = S(x)
16 – x² = 2x² + 4; 16 – 4 = 2x² + x²
3x² = 12 ⇒ x² = \(\frac { 12 }{3}\) = 4
∴ x = ± 2, x cannot be in negative
∴ x = 2

Question 7.
The marginal revenue and marginal coast functions of a company are MR = 30 – 6x and MC = -24 + 3x where x is the product, profit function is
(a) 9x² + 54x
(b) 9x² – 54x
(c) 54x – \(\frac { 9x^2 }{2}\)
(d) 54x – \(\frac { 9x^2 }{2}\) + k
Solution:
(d) 54x – \(\frac { 9x^2 }{2}\) + k
Hint:
Profit = ∫(MR – MC) dx + k
= ∫(30 – 60) – (-24 + 3x) dx + k
= ∫(54 – 9x) dx + k
= 54x – \(\frac{9 x^{2}}{2}\) + k

Question 8.
The given demand and supply function are given by D(x) = 20 – 5x and S(x) = 4x + 8 if they are under perfect competition then the equilibrium demand is
(a) 40
(b) \(\frac { 41 }{2}\)
(c) \(\frac { 40 }{3}\)
(d) \(\frac { 41 }{5}\)
Solution:
(c) \(\frac { 40 }{3}\)
Hint:
Under perfect competition D(x) = S(x)
20 – 5x = 4x + 8
20 – 8 = 4x + 5x ⇒ 9x = 12
x = \(\frac { 4 }{3}\)
when x = \(\frac { 4 }{3}\); D(x) = 20 – 5(\(\frac { 4 }{3}\)) = 20 – \(\frac { 20 }{3}\)
= \(\frac { 40 }{3}\)

Question 9.
If the marginal revenue MR = 35 +7x – 3x², then the average revenue AR is.
(a) 35x + \(\frac { 7x^2 }{2}\) – x³
(b) 35x + \(\frac { 7x }{2}\) – x²
(c) 35x + \(\frac { 7x }{2}\) + x²
(d) 35x + 7x + x²
Solution:
(c) \(\frac { 40 }{3}\)
Hint:
R = ∫MR dx = ∫(35 + 7x – 3x²) dx

Question 10.
The profit of a function p(x) is maximum when
(a) MC – MR = 0
(b) MC = 0
(c) MR = 0
(d) MC + MR = 0
Solution:
(a) MC – MR = 0
Hint:
P = Revenue – Cost
P is maximum when \(\frac{d p}{d x}\) = 0
\(\frac{d p}{d x}\) = R'(x) – C'(x) = MR – MC = 0

Question 11.
For the demand function p(x), the elasticity of demand with respect to price is unity then.
(a) revenue is constant
(b) a cost function is constant
(c) profit is constant
(d) none of these
Solution:
(a) Revenue is constant

Question 12.
The demand function for the marginal function MR = 100 – 9x² is
(a) 100 – 3x²
(b) 100x – 3x²
(c) 100x – 9x²
(d) 100 + 9x²
Solution:
(a) 100 – 3x²
Hint:
R = ∫(MR) dx + c₁
R = ∫(100 – 9x²) dx + c₁
R = 100x – 3x³ + c₁
When R = 0, x = 0, c₁ = 0
R = 100x – 3x³
Demand function is \(\frac{R}{x}\) = 100 – 3x²

Question 13.
When x₀ = 5 and p₀ = 3 the consumer’s surplus for the demand function p_d = 28 – x²
(a) 250 units
(b) \(\frac { 250 }{3}\) units
(c) \(\frac { 251 }{2}\) units
(d) \(\frac { 251 }{3}\) units
Solution:
(b) \(\frac { 250 }{3}\) units
Hint:

Question 14.
When x₀ = 2 and P₀ = 12 the producer’s surplus for the supply function P₀ = 2x² + 4 is
(a) \(\frac { 31 }{5}\) units
(b) \(\frac { 31 }{2}\) units
(c) \(\frac { 32 }{2}\) units
(d) \(\frac { 30 }{7}\) units
Solution:
(c) \(\frac { 32 }{2}\) units
Hint:
Producer’s Surplus

Question 15.
Area bounded by y = x between the lines y = 1, y = 2 with y = axis is
(a) \(\frac { 1 }{2}\) sq units
(b) \(\frac { 5 }{2}\) sq units
(c) \(\frac { 3 }{2}\) sq units
(d) 1 sq units
Solution:
(c) \(\frac { 3 }{2}\) sq units
Hint:

Question 16.
The producer’s surplus when supply the function for a commodity is p = 3 + x and x₀ = 3 is
(a) \(\frac { 1 }{2}\)
(b) \(\frac { 9 }{2}\)
(c) \(\frac { 3 }{2}\)
(d) \(\frac { 7 }{2}\)
Solution:
(b) \(\frac { 9 }{2}\)
Hint:
p = 3 + x and x₀ = 3
then p₀ = 3 + 3 = 6

Question 17.
The marginal cost function is MC = 100√x find AC given that TC = 0 when the out put is zero is
(a) \(\frac { 200 }{3}\) x^1/2
(b) \(\frac { 200 }{3}\) x^3/2
(c) \(\frac { 200 }{3x^{3/2}}\)
(d) \(\frac { 200 }{3x^{1/2}}\)
Solution:
(a) \(\frac { 200 }{3}\) x^1/2
Hint:
TC = ∫MC dx = ∫100√x dx = 100 ∫(x)^1/2 dx

Question 18.
The demand and supply function of a commodity are P(x) = (x – 5)² and S(x) = x² + x + 3 then the equilibrium quantity x₀ is
(a) 5
(b) 2
(c) 3
(d) 10
Solution:
(b) 2
Hint:
At equilibrium, P(x) = S(x)
⇒ (x – 5)² = x² + x + 3
⇒ x² – 10x + 25 = x² + x + 3
⇒ 11x = 22
⇒ x = 2

Question 19.
The demand and supply function of a commodity are D(x) = 25 – 2x and S(x) = \(\frac { 10+x }{2}\) then the equilibrium price P₀ is
(a) 2
(b) 2
(c) 3
(d) 10
Solution:
(a) 2
Hint:
At equilibrium, D(x) = S(x)
25 – 2x = \(\frac{10+x}{4}\)
⇒ 100 – 8x = 10 + x
⇒ x = 10
That is x₀ = 10
P₀ = 25 – 2(x₀) = 25 – 20 = 5

Question 20.
If MR and MC denote the marginal revenue and marginal cost and MR – MC = 36x – 3x² – 81, then maximum profit at x equal to
(a) 3
(b) 6
(c) 9
(d) 10
Solution
(c) 9
Hint:
Profit P = ∫(MR – MC) dx = ∫(36x – 3x² – 81) dx
P = [\(\frac { 36x^2 }{2}\) – \(\frac { 3x^3 }{3}\) – 81x] = 18x² – x³ – 81x
when p = 0; 18x² – x³ – 81x = 0 ⇒ x² – 18x + 81 = 0
(x – 9)² = 0 ⇒ x – 9 = 0
∴ x = 9

Question 21.
If the marginal revenue of a firm is constant, then the demand function is
(a) MR
(b) MC
(c) C(x)
(d) AC
Solution:
(a) MR
Hint:
MR = k (constant)
Revenue function R = ∫(MR) dx + c₁
= ∫kdx + c₁
= kx + c₁
When R = 0, x = 0, ⇒ c₁ = 0
R = kx
Demand function p = \(\frac{R}{x}=\frac{k x}{x}\) = k constant
⇒ p = MR

Question 22.
For a demand function p, if ∫\(\frac { dp }{p}\) = k ∫\(\frac { dx }{x}\) then k is equal to
(a) n_d
(b) -n_d
(c) \(\frac { -1 }{n_d}\)
(d) \(\frac { 1 }{n_d}\)
Solution:
(c) \(\frac { -1 }{n_d}\)

Question 23.
The area bounded by y = e^x between the limits 0 to 1 is
(a) (e – 1) sq.units
(b) (e + 1) sq.units
(c) (1 – \(\frac { 1 }{e}\)) sq.units
(d) (1 + \(\frac { 1 }{e}\)) sq.units
Solution:
(a) (e – 1) sq.units
Hint:
Area A = \(\int_{a}^{b}\)ydx = \(\int_{0}^{1}\)e^xdx = [e^x]\(_{0}^{1}\)
= [e^x – e°] = [e – 1]

Question 24.
The area bounded by the parabola y² = 4x bounded by its latus rectum is
(a) \(\frac { 16 }{3}\) sq units
(b) \(\frac { 8 }{3}\) sq units
(c) \(\frac { 72 }{3}\) sq units
(d) \(\frac { 1 }{3}\) sq units
Solution:
(b) \(\frac { 8 }{3}\) sq units
Hint:

y² = 4x ⇒ y = \(\sqrt { 4x}\) 2√x = 2(x)^1/2
In this parabola 4a = 4 ⇒ a = 1 and vertex V(0, 0)

Question 25.
The area bounded by y = |x| between the limits 0 and 2 is
(a) 1 sq.units
(b) 3 sq.units
(c) 2 sq.units
(d) 4 sq.units
Solution:
(c) 2 sq.units
Hint:
Area A = \(\int_{a}^{b}\)ydx = \(\int_{0}^{2}\)x dx = [ \(\frac { x^2 }{2}\) ]\(_{0}^{2}\)
= \(\frac { (2)^2 }{2}\) – (0) = \(\frac { 4 }{2}\) = 2

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Students can download Maths Chapter 8 Statistics Ex 8.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.1

Question 1.
In a week, temperature of a certain place is measured during winter are as follows 26°C, 24°C, 28°C, 31°C, 30°C, 26°C, 24°C. Find the mean temperature of the week.
Solution:
Mean temperature of the week

= 27°C
Mean temperature of the week 27° C

Question 2.
The mean weight of 4 members of a family is 60 kg. Three of them have the weight 56 kg, 68 kg and 72 kg respectively. Find the weight of the fourth member.
Solution:
Weight of 4 members = 4 × 60 kg
= 240 kg
Weight of three members = 56 kg + 68 kg + 72 kg
= 196 kg
Weight of the fourth member = 240 kg – 196 kg
= 44 kg

Question 3.
In a class test in mathematics, 10 students scored 75 marks, 12 students scored 60 marks, 8 students scored 40 marks and 3 students scored 30 marks. Find the mean of their score.
Solution:
Total marks of 10 students = 10 × 75 = 750
Total marks of 12 students = 12 × 60 = 720
Total marks of 8 students = 8 × 40 = 320
Total marks of 3 students = 3 × 30 = 90
Total marks of (10 + 12 + 8 + 3) 33 students
= 750 + 720 + 320 + 90
= 1880
Mean of marks = \(\frac{1880}{33}\)
= 56.97 (or) 57 approximately
Aliter:
Total number of students = 10+12 + 8 + 3
= 33
Mean of their marks

= 56.97 (or) 57 approximately

Question 4.
In a research laboratory scientists treated 6 mice with lung cancer using natural medicine. Ten days later, they measured the volume of the tumor in each mouse and given the results in the table.

find the mean.
Solution:

\(\bar { x }\) = \(\frac{Σfx}{Σf}\)
= \(\frac{2966}{21}\)
= 141.238
= 141.24
The Arithmetic mean = 141.24 mm³

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

Solution:

\(\bar { x }\) = \(\frac{Σfx}{Σf}\)
20.2 = \(\frac{610+20p}{30+p}\)
610 + 20 p = 20.2 (30 + p)
610 + 20 p = 606 + 20.2 p
610 – 606 = 20.2 p – 20 p
4 = 0.2 p
p = \(\frac{4}{0.2}\)
= \(\frac{4×10}{2}\)
= 20
The value of p = 20

Question 6.
In the class, weight of students is measured for the class records. Calculate mean weight of the class students using direct method.

Solution:

Arithmetic mean \(\bar { x }\) = \(\frac{Σfx}{Σf}\)
= \(\frac{2010}{50}\)
= 40.2
Arithmetic mean = 40.2

Question 7.
Calculate the mean of the following distribution using Assumed Mean Method.

Solution:
Assumed Mean (A) = 25

Arithmetic mean (\(\bar { x }\)) = A+\(\frac{Σfd}{Σf}\)
= 25 + \(\frac{270}{63}\)
= 25 + 4. 29
Assumed Mean = 29.29

Question 8.
Find the Arithmetic Mean of the following data using Step Deviation Method:

Solution:
Assumed Mean (A) = 32

Arithmetic mean = A+\(\frac{Σfd}{Σf}\) × c
= 32 + (\(\frac{-77.5}{105}\)×4)
= 32 – 2.95
= 29.05
Arithmetic mean = 29.05

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.
Check whether the which triangles are similar and find the value of x.


Solution:
(i) In ∆ABC and ∆AED
\(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)
\(\frac { 8 }{ 3 } \) = \(\frac{11}{\frac{2}{2}}\)
\(\frac { 8 }{ 3 } \) = \(\frac { 11 }{ 4 } \) ⇒ 32 ≠ 33
∴ The two triangles are not similar.

(ii) In ∆ABC and ∆PQC
∠PQC = 70°
∠ABC = ∠PQC = 70°
∠ACB = ∠PCQ (common)
∆ABC ~ ∆PQC
\(\frac { 5 }{ X } \) = \(\frac { 6 }{ 3 } \)
6x = 15
x = \(\frac { 15 }{ 6 } \) = \(\frac { 5 }{ 2 } \)
∴ x = 2.5
∆ ABC and ∆PQC are similar. The value of x = 2.5

Question 2.
A girl looks the reflection of the top of the lamp post on the mirror which is 66 m away from the foot of the lamppost. The girl whose height is 12.5 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post.
Solution:
Let the height of the tower ED be “x” m. In ∆ABC and ∆EDC.
∠ABC = ∠CED = 90° (vertical Pole)
∠ACB = ∠ECD (Laws of reflection)
∆ ABC ~ ∆DEC
\(\frac { AB }{ DE } \) = \(\frac { BC }{ EC } \)
\(\frac { 1.5 }{ x } \) = \(\frac { 0.4 }{ 87.6 } \)

x = \(\frac{1.5 \times 87.6}{0.4}\) = \(\frac{1.5 \times 876}{4}\)
= 1.5 × 219 = 328.5
The height of the Lamp Post = 328.5 m

Question 3.
A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.
Solution:
In ∆ABC and ∆PQR,
∠ABC = ∠PQR = 90° (Vertical Stick)
∠ACB = ∠PRQ (Same time casts shadow)
∆BCA ~ ∆QRP

\(\frac { AB }{ PQ } \) = \(\frac { BC }{ QR } \)
\(\frac { 6 }{ x } \) = \(\frac { 4 }{ 28 } \)
4x = 6 × 28 ⇒ x = \(\frac{6 \times 28}{4}\) = 42
Length of the lamp post = 42m

Question 4.
Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.
Solution:
In ∆PQT and ∆STR we have
∠P = ∠S = 90° (Given)
∠PTQ = ∠STR (Vertically opposite angle)

By AA similarity
∆PTQ ~ ∆STR we get
\(\frac { PT }{ ST } \) = \(\frac { TQ }{ TR } \)
PT × TR = ST × TQ
Hence it is proved.

Question 5.
In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.

Solution:
In ∆ABC and ∆ADE
∠ACB = ∠AED = 90°
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE (By AA similarity)
\(\frac { BC }{ DE } \) = \(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)
\(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)
In ∆ABC, AB² = BC² + AC²
= 122 + 52 = 144 + 25 = 169
AB = \(\sqrt { 169 }\) = 13
Consider, \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)
∴ AE = \(\frac{5 \times 3}{13}\) = \(\frac { 15 }{ 13 } \)
AE = \(\frac { 15 }{ 13 } \) and DE = \(\frac { 36 }{ 13 } \)
Consider, \(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \)
DE = \(\frac{12 \times 3}{13}\) = \(\frac { 36 }{ 13 } \)

Question 6.
In the adjacent figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.

Solution:
Given ∆ACB ~ ∆APQ
\(\frac { AC }{ AP } \) = \(\frac { BC }{ PQ } \) = \(\frac { AB }{ AQ } \)
\(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)
Consider \(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \)
4 AC = 8 × 2.8
AC = \(\frac{8 \times 2.8}{4}\) = 5.6 cm
Consider \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)
8 AQ = 4 × 6.5
AQ = \(\frac{4 \times 6.5}{8}\) = 3.25 cm
Length of AC = 5.6 cm; Length of AQ = 3.25 cm

Question 7.
If figure OPRQ is a square and ∠MLN = 90°. Prove that

(i) ∆LOP ~ ∆QMO
(ii) ∆LOP ~ ∆RPN
(iii) ∆QMO ~ ∆RPN
(iv) QR2 = MQ × RN.
Solution:
(i) In ∆LOP and ∆QMO
∠OLP = ∠OQM = 90°
∠LOP = ∠OMQ (Since OQRP is a square OP || MN)
∴ ∆LOP~ ∆QMO (By AA similarity)

(ii) In ∆LOP and ∆RPN
∠OLP = ∠PRN = 90°
∠LPO = ∠PNR (OP || MN) .
∴ ∆LOP ~ ∆RPN (By AA similarity)

(iii) In ∆QMO and ∆RPN
∠MQO = ∠NRP = 90°
∠RPN = ∠QOM (OP || MN)
∴ ∆QMO ~ ∆RPN (By AA similarity)

(iv) We have ∆QMO ~ ∆RPN
\(\frac { MQ }{ PR } \) = \(\frac { QO }{ RN } \)
\(\frac { MQ }{ QR } \) = \(\frac { QR }{ RN } \)
QR² = MQ × RN
Hence it is proved.

Question 8.
If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm² and the area of ∆DEF is 16 cm² and BC = 2.1 cm. Find the length of EF.
Solution:
Given ∆ABC ~ ∆DEF

\(\frac { 9 }{ 16 } \) = \(\frac{(2.1)^{2}}{\mathrm{E} \mathrm{F}^{2}}\)
(\(\frac { 3 }{ 4 } \))² = (\(\frac { 2.1 }{ EF } \))²
\(\frac { 3 }{ 4 } \) = \(\frac { 2.1 }{ EF } \)
EF = \(\frac{4 \times 2.1}{3}\) = 2.8 cm
Legth of EF = 2.8 cm

Question 9.
Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y.

Solution:
In the ∆PAC and ∆BQC
∠PAC = ∠QBC = 90°
∠C is common
∆PAC ~ QBC
\(\frac { AP }{ BQ } \) = \(\frac { AC }{ BC } \)
\(\frac { 6 }{ y } \) = \(\frac { AC }{ BC } \)
∴ \(\frac { BC }{ AC } \) = \(\frac { y }{ 6 } \) …..(1)
In the ∆ACR and ∆QBC
∠ACR = ∠QBC = 90°
∠A is common
∆ACR ~ ABQ
\(\frac { RC }{ QB } \) = \(\frac { AC }{ AB } \)
\(\frac { 3 }{ y } \) = \(\frac { AC }{ AB } \)
\(\frac { AB }{ AC } \) = \(\frac { y }{ 3 } \) ……..(2)
By adding (1) and (2)
\(\frac { BC }{ AC } \) + \(\frac { AB }{ AC } \) = \(\frac { y }{ 6 } \) + \(\frac { y }{ 3 } \)
1 = \(\frac{3 y+6 y}{18}\)
9y = 18 ⇒ y = \(\frac { 18 }{ 9 } \) = 2
The Value of y = 2m

Question 10.
Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 2 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 2 }{ 3 } \) ).
Solution:
Given ∆PQR, we are required to construct another triangle whose sides are \(\frac { 2 }{ 3 } \) of the corresponding sides of the ∆PQR

Steps of construction:
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 3 points Q₁, Q₂ and Q₃ on QX.
So that QQ₁ = Q₁Q₂ = Q₂Q₃
(iv) Join Q₃ R and draw a line through Q₂ parallel to Q₃ R to intersect QR at R’.
(v) Draw a line through R’ parallel to the line RP to intersect QP at P’. Then ∆ P’QR’ is the required triangle.

Question 11.
Construct a triangle similar to a given triangle LMN with its sides equal to \(\frac { 4 }{ 5 } \) of the corresponding sides of the triangle LMN (scale factor \(\frac { 4 }{ 5 } \) ).
Solution:
Given a triangle LMN, we are required to construct another ∆ whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of the ∆LMN.

Steps of Construction:

1. Construct a ∆LMN with any measurement.
2. Draw a ray MX making an acute angle with MN on the side opposite to the vertex L.
3. Locate 5 Points Q₁, Q₂, Q₃, Q₄, Q₅ on MX.
   So that MQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₄Q₅
4. Join Q₅ N and draw a line through Q₄. Parallel to Q₅N to intersect MN at N’.
5. Draw a line through N’ parallel to the line LN to intersect ML at L’.
   ∴ ∆L’ MN’ is the required triangle.

Question 12.
Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac { 6 }{ 5 } \) of the corresponding sides of the triangle ABC (scale factor \(\frac { 6 }{ 4 } \)).
Solution:
Given triangle ∆ABC, we are required to construct another triangle whose sides are \(\frac { 6 }{ 5 } \) of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct an ∆ABC with any measurement.
(ii) Draw a ray BX making an acute angle with BC.
(iii) Locate 6 points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆ on BX such that
BQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₅Q₆

(iv) Join Q₅ to C and draw a line through Q₆ parallel to Q₅ C intersecting the extended line BC at C’.
(v) Draw a line through C’ parallel to AC intersecting the extended line segment AB at A’.
∆A’BC’ is the required triangle.

Question 13.
Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 7 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 7 }{ 3 } \)).
Solution:
Given triangle ABC, we are required to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 7 points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆, Q₇ on QX.
So that
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₅Q₆ = Q₆Q₇

(iv) Join Q₃ to R and draw a line through Q₇ parallel to Q₃R intersecting the extended line segment QR at R’.
(v) Draw a line through parallel to RP.
Intersecting the extended line segment QP at P’.
∴ ∆P’QR’ is the required triangle.

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
Solution:
Let the initial position of the man be “O” and his final
position be “B”.
By Pythagoras theorem
In the right ∆ OAB,

OB² = OA² + AB²
= 18² + 24²
= 324 + 576 = 900
OB = \(\sqrt { 900 }\) = 30
The distance of his current position is 30 m

Question 2.
There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).

Solution:
Distance between Sarah House and James House using “C street”.
AC² = AB² + BC²
= 2² + 1.5²
= 4 + 2.25 = 6.25
AC = \(\sqrt { 6.25 }\)
AC = 2.5 miles

Distance covered by using “A Street” and “B Street”
= (2 + 1.5) miles = 3.5 miles
Difference in distance = 3.5 miles – 2.5 miles = 1 mile

Question 3.
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
Solution:
In the right ∆ABC,
By Pythagoras theorem
AC²= AB² + BC² = 34² + 41²
= 1156 + 1681 = 2837
AC = \(\sqrt { 2837 }\)
= 53.26 m

Through A one must walk (34m + 41m) 75 m to reach C.
The difference in Distance = 75 – 53.26
= 21.74 m

Question 4.
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
Calculate the length and breadth of the rectangle?

Solution:
Let the length of the rectangle be “a” and the breadth of the rectangle be “b”.
XY + YZ = 17 cm
b + a = 17 …….. (1)
In the right ∆ WXZ,
XZ² = WX² + WZ²
(XZ)² = a² + b²

XZ = \(\sqrt{a^{2}+b^{2}}\)
Similarly WY = \(\sqrt{a^{2}+b^{2}}\) ⇒ XZ + WY = 26
2 \(\sqrt{a^{2}+b^{2}}\) = 26 ⇒ \(\sqrt{a^{2}+b^{2}}\) = 13
Squaring on both sides
a² + b² = 169
(a + b)² – 2ab = 169
17² – 2ab = 169 ⇒ 289 – 169 = 2 ab
120 = 2 ab ⇒ ∴ ab = 60
a = \(\frac { 60 }{ b } \) ….. (2)
Substituting the value of a = \(\frac { 60 }{ b } \) in (1)
\(\frac { 60 }{ b } \) + b = 17
b² – 17b + 60 = 0
(b – 2) (b – 5) = 0
b = 12 or b = 5

If b = 12 ⇒ a = 5
If b = 6 ⇒ a = 12
Lenght = 12 m and breadth = 5 m

Question 5.
The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
Solution:
Let the shortest side of the right ∆ be x.
∴ Hypotenuse = 6 + 2x
Third side = 2x + 6 – 2
= 2x + 4

In the right triangle ABC,
AC² = AB² + BC²
(2x + 6)² = x² + (2x + 4)²
4x² + 36 + 24x = x² + 4x² + 16 + 16x
0 = x² – 24x + 16x – 36 + 16
∴ x² – 8x – 20 = 0
(x – 10) (x + 2) = 0
x – 10 = 0 or x + 2 = 0

x = 10 or x = -2 (Negative value will be omitted)
The side AB = 10 m
The side BC = 2 (10) + 4 = 24 m
Hypotenuse AC = 2(10) + 6 = 26 m

Question 6.
5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
“C” is the position of the foot of the ladder “A” is the position of the top of the ladder.
In the right ∆ABC,
BC² = AC² – AB² = 5² – 4²
= 25 – 16 = 9
BC = \(\sqrt { 9 }\) = 3m.
When the foot of the ladder moved 1.6 m toward the wall.
The distance between the foot of the ladder to the ground is
BE = 3 – 1.6 m
= 1.4 m

Let the distance moved upward on the wall be “h” m
The ladder touch the wall at (4 + h) M
In the right triangle BED,
ED² = AB² + BE²
5² = (4 + h)² + (1.4)²
25 – 1.96= (4 + h)²
∴ 4 + h = \(\sqrt { 23.04 }\)
4 + h = 4. 8 m
h = 4.8 – 4
= 0.8 m
Distance moved upward on the wall = 0.8 m

Question 7.
The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ² = 2PR² + QR².
Solution:
Given QS = 3SR
QR = QS + SR
= 3SR + SR = 4SR
SR = \(\frac { 1 }{ 4 } \) QR …..(1)
QS = 3SR
SR = \(\frac { QS }{ 3 } \) ……..(2)

From (1) and (2) we get
\(\frac { 1 }{ 4 } \) QR = \(\frac { QS }{ 3 } \)
∴ QS = \(\frac { 3 }{ 4 } \) QR ………(3)
In the right ∆ PQS,
PQ² = PS² + QS² ……….(4)
Similarly in ∆ PSR
PR² = PS² + SR² ………..(5)
Subtract (4) and (5)
PQ² – PR² = PS² + QS² – PS² – SR²
= QS² – SR²

PQ² – PR² = \(\frac { 1 }{ 2 } \) QR²
2PQ² – 2PR² = QR²
2PQ² = 2PR² + QR²
Hence the proved.

Question 8.
In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE² = 3AC² + 5AD².
Solution:
Since the Points D, E trisect BC.
BD = DE = CE
Let BD = DE = CE = x
BE = 2x and BC = 3x

In the right ∆ABD,
AD² = AB² + BD²
AD² = AB² + x² ……….(1)
In the right ∆ABE,
AE² = AB² + 2BE²
AE² = AB² + 4X² ………..(2) (BE = 2x)
In the right ∆ABC
AC² = AB² + BC²
AC² = AB² + 9x² …………… (3) (BC = 3x)
R.H.S = 3AC² + 5AD²
= 3[AB² + 9x²] + 5 [AB² + x²] [From (1) and (3)]
= 3AB² + 27x² + 5AB² + 5x²
= 8AB² + 32x²
= 8 (AB² + 4 x²)
= 8AE² [From (2)]
= R.H.S.
∴ 8AE² = 3AC² + 5AD²

Students can download Maths Chapter 7 Mensuration Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Additional Questions

I. Choose the Correct Answer

Question 1.
If the sides of a triangles are 5 cm, 6 cm and 7 cm then the area is ……..
(a) 18 cm²
(b) 6 √2 cm²
(c) 6 √6 cm²
(d) 6 √3 cm²
Solution:
(c) 6 √6 cm²

Question 2.
The perimeter of an equilateral triangle is 60 cm then the area is ………
(a) 60 √3 cm²
(b) 20 √3 cm²
(c) 50 √3 cm²
(d) 100 √3 cm²
Solution:
(d) 100 √3 cm²

Question 3.
The total surface area of the cuboid with dimension 20 cm × 30 cm × 15 cm is ………
(a) 2700 cm²
(b) 1500 cm²
(c) 2500 cm²
(d) 3000 cm²
Solution:
(a) 2700 cm²

Question 4.
The number of bricks each measuring 70 cm × 80 cm × 40 cm that will be required to build a wall whose dimensions are 7 m × 8 m × 4 m is ……..
(a) 4000
(b) 3000
(c) 2000
(d) 1000
Solution:
(d) 1000

Question 5.
The volume of a cube is 4913 m² then the length of its side is ……..
(a) 13 m
(b) 17 m
(c) 34 m
(d) 27 m
Solution:
(b) 17 m

II. Answer the Following Questions

Question 6.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
The non parallel sides are 13 m and 14 m. Draw BE || AD. Such that BE = 13 m
∴ ABED is a parallelogram

To find Area of a ΔBCE
a = 13 m, b = 15 m and c = 14 m
s = \(\frac{a+b+c}{2}\)
= \(\frac{13+15+14}{2}\)
= \(\frac{42}{2}\)
= 21 m
s – a = 21 – 13 = 8 m
s – b = 21 – 15 = 6 m
s – c = 21 – 14 = 7 m
Area of a ΔBCE

= 2² × 3 × 7
= 84 m²
Let the height of the triangle BF be x
Area of the ΔBEC = 84 m²
= \(\frac{1}{2}\) × b × h = 84
= \(\frac{1}{2}\) × 15 × h = 84
x = \(\frac{84×2}{15}\)
= \(\frac{56}{5}\) m
= 11.2 m
Area of parallelogram ABED = base × height sq. units
= 10 × 11.2 m²
= 112 m²
∴ Area of the field = Area of ΔBCE + Area of parallelogram ABED
= 84 m² + 112 m²
= 196 m²
(OR)
Area of the field = Area of the trapezium ABCD
= \(\frac{1}{2}\) h (a + b)
= \(\frac{1}{2}\) × 11.2 (25 + 10)
= \(\frac{1}{2}\) × 11.2 (35)
= 196 m²

Question 7.
Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm and ⌊B = 90°.
Solution:

In ΔABC, ⌊B = 90°
∴ ABC is a right angle triangle
Area of the right ΔABC = \(\frac{1}{2}\) × AB × BC sq.units
= \(\frac{1}{2}\) × 8 × 6 cm²
= 24 cm²
In ΔACD a = 10 cm, b = 8 cm and c = 10 cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{10+8+10}{2}\)
= \(\frac{28}{2}\)
= 14 cm
s – a = 14 – 10 = 4 cm
s – b = 14 – 8 = 6 cm
s – c = 14 – 10 = 4 cm
Area of ΔACD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{14×4×6×4}\)
= \(\sqrt{2×7×4×2×3×4}\)
= 4 × 2 \(\sqrt{21}\) cm²
= 8\(\sqrt{21}\) cm²
= 8 × 4.58
= 36.64 cm²
Area of the quadrilateral ABCD
= Area of ΔABC + Area of ΔACD
= 24 cm² + 36.64 cm²
= 60.64 cm²
Area of the quadrilateral = 60.64 cm²

Question 8.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m².
Solution:
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
Area for white washing = Lateral surface area of four walls + Area of the ceiling
= 2(l + b) × h + (l × b)
= 2(5 + 4) × 3 + (5 × 4) m²
= (2 × 9 × 3 + 20) m²
= (54 + 20) m²
= 74 m²
Cost of white washing for one m² = Rs 7.50
Cost of white washing for 74 m² = Rs 74 × 7.50
= Rs 555
The required cost = Rs 555

Question 9.
How many hollow blocks of size 30 cm × 15 cm × 20 cm are needed to construct a wall 60 m in length 0.3 m in breadth and 2 m in height.
Solution:
Length of a wall = 60 m = 6000 cm
Breadth of a wall = 0.3 m = 30 cm
Height of a wall = 2 m = 200 cm
Volume of the wall = l × b × h sq. unit
= 6000 × 30 × 200 cm³
For hollow block
l = 30 cm, b = 15 cm, h = 20 cm
Volume of one hollow block = l × b × h
= 30 × 15 × 20 cm²
Number of hollow blocks required

= 4000
∴ Number of bricks = 4000

Question 10.
Find the number of cubes of side 3 cm that can be cut from a cuboid of dimensions 10 cm × 9 cm × 6 cm.
Solution:
Side of a cube = 3 cm
Volume of a cube = a³ cm
= 3 × 3 × 3 cm³
Length of the cuboid (l) = 10 cm
Breadth of the cuboid (b) = 9 cm
Height of the cuboid (h) = 6 cm
Volume of the cuboid = l × b × h cu. unit
= 10 × 9 × 6 cm
Number of cubes

∴ Number of cubes = 20