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TN State Board 11th Economics Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3:00 Hours
Maximum Marks: 90

PART – I

Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
The cost incurred by producing one more unit of output is ……………………. cost.
(a) Variable
(b) Fixed
(c) Marginal
(d) Total
Answer:
(c) Marginal

Question 2.
Ordinal utility can be measured by ……………………
(a) Ranking
(b) Numbering
(c) Wording
(d) None of these
Answer:
(a) Ranking

Question 3.
A production function measures the relation between ……………………..
(a) Input prices and output prices
(b) Input prices and the quantity of output
(c) The quantity of inputs and the quantity of output
(d) The quantity of inputs and input prices
Answer:
(c) The quantity of inputs and the quantity of output

Question 4.
If average product is decreasing, then marginal product ……………………
(a) Must be greater than average product
(b) Must be less than average product
(c) Must be increasing
(d) Both (a) & (c)
Answer:
(b) Must be less than average product

Question 5.
Price discrimination will always lead to ………………………
(a) Increase in output
(b) Increase in profit
(c) Different prices
(d) (b) and (c)
Answer:
(d) (b) and (c)

Question 6.
Which of the following is a micro – economics statement?
(a) The real domestic output increased by 2.5 percent last year.
(b) Unemployment was 9.8 percent of the labour force last year.
(c) The price of wheat determines its demand
(d) The general price level increased by 4 percent last year.
Answer:
(c) The price of wheat determines its demand

Question 7.
The marginal productivity theory was developed by ……………………
(a) Clark
(b) Wickseed
(c) Walras
(d) All
Answer:
(c) Walras

Question 8.
The chief exponent of the cardinal utility approach was …………………..
(a) J.R. Hicks
(b) R.G. D. Allen
(c) Marshall
(d) Stigler
Answer:
(c) Marshall

Question 9.
Keynesian Theory of interest is popularly known as ……………………
(a) Abstinence theory
(b) Liquidity preference theory
(c) Loanable funds theory
(d) Agio theory
Answer:
(b) Liquidity preference theory

Question 10.
Total revenue is equal to total output sold multiplied by ………………….
(a) Price
(b) Total cost
(c) Marginal revenue
(d) Marginal cost
Answer:
(a) Price

Question 11.
In which of the following is not a type of market structure price will be very high?
(a) Perfect competition
(b) Monopoly
(c) Duopoly
(d) Oligopoly
Answer:
(b) Monopoly

Question 12.
The Father of Green Revolution in India was ……………………..
(a) M.S. Swaminathan
(b) Gandhi
(c) Visweswaraiah
(d) N.R. Viswanathan
Answer:
(a) M.S. Swaminathan

Question 13.
A Statement of equality between two quantities is called ……………………
(a) Inequality
(b) Equality
(c) Equations
(d) Functions
Answer:
(c) Equations

Question 14.
Product obtained from additional factors of production is terms as ………………….
(a) Marginal product
(b) Total product
(c) Average product
(d) Annual product
Answer:
(a) Marginal product

Question 15.
Rent is the reward for the use of ……………………..
(a) Capital
(b) Labour
(c) Land
(d) Organization
Answer:
(c) Land

Question 16.
Revenue received from the sale of additional unit is termed as …………………….. revenue.
(a) Profit
(b) Average
(c) Marginal
(d) Total
Answer:
(c) Marginal

Question 17.
Tamil Nadu is rich in ……………………
(a) Forest resource
(b) Human resource
(c) Mineral resource
(d) All the above
Answer:
(d) All the above

Question 18.
The New foreign trade policy was announced in the year ……………………….
(a) 2000
(b) 2002
(c) 2010
(d) 2015
Answer:
(a) 2000

Question 19.
The PQLI was developed by ……………………..
(a) Planning Commission
(b) Nehru
(c) Morris
(d) Morrisd.Biswajeet
Answer:
(c) Morris

Question 20.
In India’s total cement production, Tamil Nadu ranks ……………………….
(a) Third
(b) Fourth
(c) First
(d) Second
Answer:
(a) Third

PART – II

Answer any seven question in which Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
State the meaning of privatization?
Answer:
Privatization means transfer of ownership and management of enterprises from public sector to private sector.

Question 22.
What are the reasons for upward sloping supply curve?
Answer:

1. A supply curve represents the data given in the supply schedule. As the price of the commodity increases, the quantum supplied of the commodity also increases. Thus the supply curve has a positive slope from left to right.
2. The quantum supplied of commodity x is represented on X axis. And the price of the commodity is represented on the Y axis. The points such as e, d, c, b and a on the supply curve SS’, represent various quantities at different prices.

Question 23.
Give a short note on Sen’s “choice of Technique”?
Answer:
Sen’s “Choice of Technique” was a research work where he argued that in a labour surplus economy; generation of employment cannot be increased at the initial stage by the adaptation of capital – intensive technique.

Question 24.
What are major ports in Tamil Nadu?
Answer:

1. Tamil Nadu has three major ports; one each at Chennai, Ennore, and Thoothukudi as well as one intermediate port in Nagpattinam, and 23 minor ports.
2. All the minor ports are managed by the Tamil Nadu Maritime Board, Chennai Port.
3. Ennore port was recently converted from an intermediate port to a major port and handles all the coal and ore traffic in Tamil Nadu.

Question 25.
Define cost function?
Answer:
The function relationship between cost and output is expressed as ‘Cost Function’.
A Cost Function may be written as
C = f(Q)
Eg. TC = Q³ – 18Q² + 91Q + 12
Where, C = Cost and Q = Quantity of output. Cost functions are derived functions because they are derived from Production Functions.

Question 26.
What are the main menus of MS word?
Answer:

Where, C = Cost and Q = Quantity of output. Cost functions are derived functions because they are derived from Production Functions.

Question 27.
If 62 = 34 + 4 x what is x?
Answer:
62 – 34 = 4x ⇒ 28 = 4x
28 ÷ 5 – 4 = 1x
∴ x = 7

Question 28.
Define‘Rent’?
Answer:
Rent is the price or reward given for the use of land or house or a machine to the owner. But, in Economics, “Rent” or “Economic Rent” refers to that part of payment made by a tenant to his land lords for the use of land only.

Question 29.
What are conditions for producer’s equilibrium?
Answer:

1. Producer equilibrium implies the situation where producer maximizes his output.
2. It is also known as optimum combination of the factors of production.
3. In short, the producer manufactures a given amount of output with ‘least cost combination of factors’, with his given budget.

Question 30.
Write three policy initiative introduced in 1991 – 92 to correct the fiscal imbalance?
Answer:
The important policy initiatives introduced in the budget for the year 1991 – 92 for correcting the fiscal imbalance were:

1. Reduction in fertilizer subsidy.
2. Abolition of subsidy on sugar.
3. Disinvestment of a part of the government’s equity holdings in select public sector undertakings.

PART – III

Answer any seven question in which Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
Describe the performance of Tamil Nadu economy in health?
Answer:

1. Tamil Nadu has a three – tier health infrastructure comprising hospitals, primary health centres, health units, community health centres and sub – centres.
2. As of March 2015, the State had 34 district hospitals, 229 sub – divisional hospitals, 1,254 primary health centres, 7,555 sub – centres and 313 community health centres.

Question 32.
Explain the difference between internal and external economies?
Answer:

Question 33.
State and explain the elasticity of supply?
Answer:

1. Elasticity of supply may be defined as the degree of responsiveness of change in supply to change in price on the part of sellers.
2. It is Mathematically expressed as,

Elasticity of supply = Proportionate change in supply / Proportionate change in price
e_s = \(\frac { \Delta Q_{ S } }{ Qs } \)/\(\frac { \Delta P }{ P } \)
e_s = \(\frac { \Delta Q_{ S } }{ \Delta P } \) × \(\frac { P }{ Q_{ S } } \)
Where Q_s represents the supply, P represents price, ∆ denotes a change.

Question 34.
Write a note on mineral resources in TamilNadu?
Answer:
Mineral Resources in Tamil Nadu:

1. Tamil Nadu has a few mining projects based on Titanium, Lignite, Magnesite, Graphite, Limestone, Granite and Bauxite.
2. The first one is the Neyveli Lignite Corporation that has led development of large industrial complex around Neyveli in Cuddalore district with Thermal Power Plants, Fertilizer and Carbonisation plants.
3. Magnesite mining is at Salem from which mining of Bauxite ores are carried out at Yercaud, and this region is also rich in iron ore at Kanjamalai.
4. Molybdenum is found in Dharmapuri, and is the only source in the country.

Question 35.
What are the motives of demand for money?
Answer:
Motives of Demand for Money:
According to Keynes, there are three motives for liquidity preference. They are:

1. The Transaction Motive :

• The transaction motive relates to the desire of the people to hold cash for the current transactions.
• The amount saved under this motive depends on the level of income. M_t = f(y)

2. The Precautionary Motive:

• The precautionary motive relates to the desire of the people to hold cash to meet unexpected or unforeseen expenditures such as sickness, accidents, fire and theft.
• The amount saved for this motive also depends on the level of income. M_p – f(r)

3. The Speculative Motive:
The speculative motive relates to the desire of the people to hold cash in order to take advantage of market movements regarding the future changes in the price of bonds and securities in the capital market. The amount saved for this motive depends on the rate of interest.

M_S = f(i). There is inverse relation between liquidity preference and rate of interest.

Question 36.
State Ambedkar’s Economic Ideas on agricultural economics?
Answer:
In the 1918, Ambedkar published a paper “Small Holding in India and their Remedies”. Citing Adam Smith’s “Wealth of Nations”, he made a fine distinction between “Consolidation of Holdings” and Enlargement of Holdings”.

Question 37.
What are the functions of Entrepreneur?
Answer:
Functions of an Entrepreneur:

1. Initiation: An organizer is the initiator of the business, by considering the situation and availability of resources and planning the entire process of business of production.
2. Innovation: A successful entrepreneur is always an innovator. He introduces new methods in the production process.
3. Co – ordination: An organizer applies a particular combination of the factor of production to start and run the business or production.
4. Control, Direction and Supervision: An organiser controls so that nothing prevents the organisation from achieving its goal. He directs the factors to get better results and supervises for the efficient functioning of all the factors involved in the process of production.
5. Risk – taking and uncertainty – bearing: There are risk – taking and uncertainty-bearing obstacles. Risks may be insured but uncertainties cannot be insured. They reduce the profit.

Question 38.
Explain GSDP in Tamil Nadu?
Answer:

1. The Gross State Domestic Product (GSDP) refers to the total money value of all the goods and services produced annually in the state.
2. Tamil nadu is the second largest economy in India with a GSDP of $ 207.8 billion in 2016 – 2017 according to the Directorate of Economics and Statistics, Tamil Nadu.
3. The GSDP of Tamil Nadu is equal to the GDP of Kuwait on nominal term and GDP of UAE on PPP terms. Per capita GSDP would be better for intercountry or interstate comparisons.
4. Tamil Nadu GSDP = $207.8 billion in 2016 – 17.

Question 39.
Briefly explain the concept of consumer’s equilibrium?
Answer:
The consumer reaches equilibrium at the point where the budget line is tangent on the indifference curve. T is the point of equilibrium as budget line AB is tangent on indifference curve IC3 the upper IC which implies maximum possible level of satisfaction.

At equilibrium point, the slope of IC refers to MRS_XY and the slope of BL (Budget Line) refers to ratio of price of X to price of Y i.e. P_x / P_y . Therefore MRS_x,y = P_x / P_y.

Question 40.
List out the kinds of wages:
Answer:
Wages are divided into four types.

1. Nominal Wages or Money Wages: Nominal wages are referred to the wages paid in terms of money.
2. Real Wages: Real wages are the wages paid in terms of goods and services. Hence, real wages are the purchasing power of money wages.
3. Piece Wages: Wages that are paid on the basis of quantum of work done.
4. Time Wages: Wages that are paid on the basis of the amount of time that the worker works.

PART – IV

Answer all the questions. [7 × 5 = 35]

Question 41 (a).
Write a brief note on the Gandhian economic ideas?
Answer:
Gandhian Economics is based on ethical foundations. Gandhi wrote “ Economics that hurts the moral well-being of an individual or a nation is immoral, and therefore , Sinful”. Gandhi repeated the same belief “ that economy is untrue which ignores or disregards moral values”.

Salient features of Gandhian Economic Thought:

(I) Village Republics:

• India lives in villages.
• He was interested in developing the villages as self – sufficient units.
• He opposed extensive use of machinery, urbanization and industrialization.

(II) On Machinery:

• Gandhi described machinery as ‘Great sin’. He said that “Books could be written to demonstrate its evils”.
• It is necessary to realize that machinery is bad.
• Instead of welcoming machinery as a boon, we should look upon it as an evil.
• It would ultimately cease.

(III) Industrialism:

• Gandhi considered industrialism as a curse on mankind.
• He thought industrialism depended entirely on a country’s capacity to exploit.

(IV) Decentralization:
Gandhi advocated a decentralized economy i. e., production at a large number of places on a small scale or production in the people’s homes.

(V) Village Sarvodaya:

• According to Gandhi, “ Real India was to be found in villages and not in towns or cities”.
• He suggested, self – dependent villages.

(VI) Bread Labour:

• Gandhi realized the dignity of human labour.
• He believed that God created man to eat his bread by the sweat of his brow.
• Bread labour or body labour was the expression that Gandhi used to mean manual labour.

(VII) The Doctrine of Trusteeship:
Trusteeship provides a means of transforming the present capitalist order of society into an egalitarian one.

(VIII) On the Food Problem:

• Gandhi was against any sort of food controls.
• Once India was begging for food grain, but now India tops the world with very large production of food grains, fruits, vegetables, milk, egg, meat etc.

(IX) On Population:

• Gandhi opposed the method of population control through contraceptives.
• He was, however, in favour of birth control through Brahmacharya or self – control.
• He considered self – control as a sovereign remedy to the problem of Over population.

(X) On Prohibition:

• Gandhi regarded the use of liquor as a disease rather than a vice.
• He felt that it was better for India to be poor than to have thousands of drunkards.
• Many states depend on revenue from liquor sales.

[OR]

(b) Analyse the causes for Rural Indebtedness?
Answer:
The Causes for Rural Indebtedness:

(I) Poverty of Farmers:

• The vicious circle of poverty forces the farmers to borrow for consumption and cultivation.
• Thus poverty, debt and high rates of interest hold the farmer in the grip of money lenders.

(II) Failure of Monsoon:

• Frequent failure of monsoon is a curse to the farmers and they have to suffer due to the failure of nature.
• Farmers find it difficult to identify good years to repay their debts.

(III) Litigation:

• Due to land disputes litigation in the court compels them to borrow heavily.
• Being uneducated and ignorant they are caught in the litigation process and dry away their savings and resources.

(IV) Money Lenders and High Rate of Interest:
The rate of interest charged by the local money lenders is very high and the compounding of interest leads to perpetuate indebtedness of the farmer.

Question 42 (a).
Explain various divisions of Economics?
Answer:
Economics sub divisions are:

(I) Consumption:

• Human wants coming under consumption is the starting point of economic activity.
• In this section the characteristics of human wants based on the behaviour of the consumer, the diminishing marginal utility and consumer’s surplus are dealt with.

(II) Production:

• Production is the process of transformation of inputs into output.
• This division covers the characteristics and role of the factors of production namely land, labour, capital and organization.

(III) Exchange:

• Exchange is concerned with price determination in different market forms.
• This division covers trade and commerce.
• Consumption is possible only if the produced commodity is placed in the hands of the consumer.

(IV) Distribution:

• Production is the result of the coordination of factors of production.
• Since a commodity is produced with the efforts of land, labour, capital and organization, the produced wealth has to be distributed among the cooperating factors.
• The reward for factors of production is studied in this division under rent, wages, interest and profit.
• Distribution studies about the pricing of factors of production.

[OR]

(b) Explain the law of demand and its exceptions?
Answer:
Definitions:
The Law of Demand says as “the quantity demanded increases with a fall in price and diminishes with a rise in price”. – Marshall
“The Law of Demand states that people will buy more at lower price and buy less at higher prices, other things remaining the same”. – Samuelson Assumptions of Law of Demand:

1. The income of the consumer remains constant.
2. The taste, habit and preference of the consumer remain the same.
3. The prices of other related goods should not change.
4. There should be no substitutes for the commodity in study.
5. The demand for the commodity must be continuous.
6. There should not be any change in the quality of the commodity.

Given these assumptions, the law of demand operates. If there is change even in one of these assumptions, the law will not operate.

Explanation:
The law of demand explains the relationship between the price of a commodity and the quantity demanded of it. This law states that quantity demanded of a commodity expands with a fall in price and contracts with a rise in price.

In other words, a rise in price of a commodity is followed by a contraction demand and a fall in price is followed by extension in demand. Therefore, the law of demand states that there is an inverse relationship between the price and the quantity demanded of a commodity.

Exceptions to the law of demand:
Normally, the demand curve slopes downwards from left to right. But there are some unusual demand curves which do not obey the law and the reverse occurs.

A fall in price brings about a contraction of demand and a rise in price results in an extension of demand. Therefore the demand curve slopes upwards from left to right. It is known as exceptional demand curve.

In the above diagram, DD is the demand curve which slopes upwards from left to right. It shows that when price is OP₁,, OQ₁, is the demand and when the price rises to OP₂, demand also extends to OQ₂.

Question 43 (a).
Describe the degree of price discrimination?
Answer:
Degrees of Price Discrimination:
Price discrimination has become widespread in almost all monopoly markets. According to A.C.Pigou, there are three degrees of price discrimination.

(I) First degree price discrimination:
A monopolist charges the maximum price that a buyer is willing to pay. This is called as perfect price discrimination. This price wipes out the entire consumer’s surplus. This is maximum exploitation of consumers. Joan Robinson named it as “Perfect Discriminating Monopoly”.

(II) Second degree price discrimination:
Under this degree, buyers are charged prices in such a way that a part of their consumer’s surplus is taken away by the sellers. This is called as imperfect price discrimination. Joan Robinson named it as “Imperfect Discriminating Monopoly”.

Under this degree, buyers are divided into different groups and a different price is charged for each group. For example, in cinema theatres, prices are charged for same film show from viewers of different classes. In a theatre the difference between the first row of first class and the last row in the second class is smaller as compared to the differences in charges.

(III) Third degree price discrimination:
The monopolist splits the entire market into a few sub – market and charges different price in each sub – market. The groups are divided on the basis of age, sex and location. For example, railways charge lower fares from senior citizens. Students get discounts in museums, and exhibitions.

[OR]

(b) Write a short note on Total Revenue?
Answer:
Total revenue is the amount of income received by the firm from the sale of its products. It is obtained by multiplying the price of the commodity by the number of units sold.

where, TR denotes Total Revenue, P denotes Price and Q denotes Quantity sold.
For example, a cell – phone company sold 100 cell – phones at the price of ₹500 each.
TR is ₹50,000. (TR = 500 × 100 = 50,000).

When price is constant, the behaviour of TR is shown in the above table and diagram, assuming P = 5. When P = 5; TR = PQ

When price is declining with increase in quantity sold. (Eg. Imperfect Competition on the goods market) the behaviour of TR is shown in the following table and diagram. TR can be obtained from Demand fuction: If Q = 11 – P,
When P = 1, Q = 10

TR = PQ = 1 × 10 = 10
When P = 3, Q = 8, TR = 24
When P = 0, Q = 1, TR = 10

Question 44 (a).
Discuss the problems of Rural Economy?
Answer:
Rural areas are facing number of problems relating to,

1. People
2. Agriculture
3. Infrastructure
4. Economy
5. Society and Culture
6. Leadership and
7. Administration.

(I) People Related Problems:
The problems related to individuals and their standard of living consist of illiteracy, lack of technical know how, low level of confidence, dependence on sentiments and beliefs etc.

(II) Agriculture Related Problems:
The problems related to agriculture include as follows,

• Lack of expected awareness, knowledge, skill and attitude.
• Unavailability of inputs.
• Poor marketing facility.

(III) Infrastructural Related problems:
Poor infrastructure facilities like, water, electricity, transport, educational institutions, communication, health, employment, storage facility, banking and insurance are found in rural areas.

(IV) Economics related problems:
The economic problems related to rural areas are,

• Inability to adopt high cost technology
• High cost of inputs
• Under privileged rural industries.
• Low income
• Indebtedness
• Existence of inequality in land holdings and assets

(V) Leadership Related Problems:
The Specific leadership related problems found in rural areas are:

• Leadership among the hands of inactive and incompetent people
• Self – interest of leaders
• Biased political will
• Less bargaining power
• Negation skills and dominance of political leaders

(VI) Administrative Problems:
The rural administrative problems are,

• Political interference
• Lack of motivation and interest
• Low wages in villages
• Improper utilization of budget
• Absence of monitoring
• Implementation of rural development programme.

[OR]

(b) Mention the causes for Rural poverty?
Answer:
Causes for Rural Poverty:
1. Unequal distribution of Land:
The distribution of land is highly skewed in rural areas. Therefore, majority of rural people work as hired labour to support their families.

2. Lack of Non – Farm Employment:
Non – farm employment opportunities do not match the increasing labour force. The excess supply of labour in rural areas reduces the wages and increases the incidence of poverty.

3. Lack of public sector Investment:
The root cause of rural poverty in our country is lack of public sector investment on human resource development.

4. Inflation:
Steady increase in prices affects the purchasing power of the rural poor leading to rural poverty.

5. Low productivity:
Low productivity of rural labour and farm activities is a cause as well as the effect of poverty.

6. Unequal Benefit of Growth:
Major gains of economic development are enjoyed by the urban rich people leading to concentration of wealth. Due to defective economic structure and policies, gains of growth are not reaching the poor and the contributions of poor people are not accounted properly.

7. Low Rate of Economic Growth:
The rate of growth of India is always below the target and it has benefited the rich. The poor are always denied of the benefits of the achieved growth and development of the country.

8. More Emphasis on Large Industries:
Huge investment in large industries catering to the needs of middle and upper classes in urban areas are made in India. Such industries are capital-intensive and do not generate more employment opportunities. Therefore, poor are not in a position to get employed and to come out from the poverty in villages.

9. Social evils:
Social evils prevalent in the society like custom, beliefs, etc., increase unproductive expenditure.

Question 45 (a).
Bring out the relationship between TR, AR, MR and Elasticity of demand?
Answer:
Relationship among TR, AR and MR Curves:
When marginal revenue is positive, total revenue rises, when MR is zero the total revenue becomes maximum. When marginal revenue becomes negative total revenue starts falling. When AR and MR both are falling, then MR falls at a faster rate than AR.

TR, AR, MR and Elasticity of demand:

1. The relationship among AR, MR and elasticity of demand (e) is stated as follows.
   MR = AR (e – 1/e)
2. The relationship between The AR curve and MR curve depends upon the elasticity of AR curve [AR = DD = Price]

• When price elasticity of demand is greater than one, MR is positive and TR is increasing.
• When price elasticity of demand is less than one, MR is negative and TR is decreasing.
• When price elasticity of demand is equal to one, MR is equal to Zero and TR is maximum and constant.

It is to be noted that, the output range of 1 to 5 units, the price elasticity of demand is greater than one according to total outlay method. Hence, TR is increasing and MR is positive.
TR, AR, MR & Elasticity

At the output range of 5 to 6 units, the price elasticity of demand is equal to one. Hence, TR is maximum and MR is equals to zero. At the output range of 6 units to 10 units, the price elasticity of demand is less than unity. Hence, TR is decreasing and MR is negative.

[OR]

(b) Explain the monetary and financial sector reforms?
Answer:
Monetary reforms aimed at doing away with interest rate distortions and rationalizing the structure of lending rates. The new policy tried in many ways to make the banking system more efficient. Some of the measures undertaken were:

(I) Reserve Requirements:

• Reduction in Statutory liquidity ratio [SLR] and the cash reserve ratio [CRR] were the recommendations by the Narasimham Committee Report, 1991.
• It was proposed to cut down the SLR from 38.5 percent to 25 percent within a time span of three years.

(II) Interest Rate Liberalisation:
RBI controlled:

• The interest rates payable on deposits. The interest rate which could be charged for bank loans.
• Greater competition among public sector, private sector and foreign banks and elimination of administrative constraints.
• Liberalisation of bank branch licensing policy in order to rationalize the existing branch network.
• Banks were given freedom to relocate branches and open specialized branches.
• Guidelines for opening new private sector banks.
• New accounting norms regarding classification of. assets and provisions of bad debt were introduced in tune with the Narasimham Committee Report.

Question 46 (a).
Examine the law of variable proportions with the help of a diagram?
Answer:
The law states that if all other factors are fixed and one input is varied in the short run, the total output will increase at an increasing rate at first instance^ be constant at a point and then eventually decrease. Marginal product will become negative at last.

According to G.Stigler, “As equal increments of one input are added, the inputs of other productive services being held constant, beyond a certain point, the resulting increments of product will decrease, i.e., the marginal product will diminish”.

Assumptions:
The Law of Variable Proportions is based on the following assumptions.

• Only one factor is variable while others are held constant.
• All units of the variable factor are homogeneous.
• The product is measured in physical units.
• There is no change in the state of technology.
• There is no change in the price of the product.

Total Product (TP)
Total product refers to the total amount of commodity produced by the combination of all inputs in a given period of time.

Summation of marginal products, i.e.
TP = ΣMP
where, TP = Total Product,
MP = Marginal Product

Average Product (AP)
Average Product is the result of the total product divided by the total units of the input employed. In other words, it refers to the output per unit of the input.
Mathematically, AP = TP/N
Where, AP = Average Product
TP = Total Product
N = Total units of inputs employed

Marginal Product (MP)
Marginal Product is the addition or the increment made to the total product when one more unit of the variable input is employed. In other words, it is the ratio of the change in the total product to the change in the units of the input. It is expressed as
MP = ∆TP/∆N

where,
MP = Marginal Product
∆TP = Change in total product
∆N = Change in units of input It is also expressed as
MP = TP(n) – TP (n – 1)
Where, MP = Marginal Product
TP(n) = Total product of employing n^th unit of a factor.
TP(n – 1) = Total product of employing the previous unit of a factor, that is, (n – 1)^th unit of a factor.
The Law of Variable Proportions is explained with the help of the following schedule and diagram:

In above table, units of variable factor (labour) are employed along with other fixed factors of production. The table illustrates that there are three stages of production. Though total product increases steadily at first instant, constant at the maximum point and then diminishes, it is always positive forever.

While total product increases, marginal product increases up to a point and then decreases. Total product increases up to the point where the marginal product is zero. When total product tends to diminish marginal product becomes negative.

In diagram, the number of workers is measured on X axis while TP_L, AP_L and MP_L are denoted on Y axis. The diagram explains the three stages of production as given in the above table.

Stage I:
In the first stage MP_L increases up to third labourer and it is higher than the average product, so that total product is increasing at an increasing rate. The tendency of total product to increase at an increasing rate stops at the point A and it begins to increase at a decreasing rate. This point is known as ‘Point of Inflexion’.

Stage II:
In the second stage, MP_L decreases up to sixth unit of labour where MP_L curve intersects the X – axis. At fourth unit of labor MP_L = AP_L. After this, MP_L curve is lower than the AP_L. TP_L increases at a decreasing rate.

Stage III:
Third stage of production shows that the sixth unit of labour is marked by negative MP_L, the AP_L continues to fall but remains positive. After the sixth unit, TP_L declines with the employment of more units of variable factor, labour.

[OR]

(b) What are the types of elasticity of supply?
Answer:
There are five types of elasticity of supply

1. Relatively Elastic Supply:
The co – efficient of elastic supply is greater than 1 [E_S > 1], One percent change in the price of a commodity causes more than one percent change in the quantity supplied of the commodity.

2. Unitary Elastic Supply:
The co – efficient of elastic supply is equal to 1 [E_g = 1]. One percent change in the price of a commodity causes an equal [one percent ] change in the quantity supplied of the commodity.

3. Relatively Inelastic Supply:
The co-efficient of elasticity is less than one [ES < 1]. One percent change in the price of a commodity causes and less than one percent change in the quantity supply supplied of the commodity.

4. Perfectly Inelastic Supply:
The co – efficient of elasticity is equal to zero [E_S = 0], One percent change in the price of a commodity causes no change in the quantity supplied of the commodity.

5. Perfectly Elastic Supply:
The co – efficient of elasticity of supply is infinity [E_S = 3]. One percent change in the price of a commodity causes an infinite change in the quantity supplied of the commodity.

Question 47 (a).
Analyse the causes for Rural Indebtedness?
Answer:
The Causes for Rural Indebtedness:
(I) Poverty of Farmers:

• The vicious circle of poverty forces the farmers to borrow for consumption and cultivation.
• Thus poverty, debt and high rates of interest hold the farmer in the grip of money lenders.

(II) Failure of Monsoon:

• Frequent failure of monsoon is a curse to the farmers and they have to suffer due to the failure of nature.
• Farmers find it difficult to identify good years to repay their debts.

(III) Litigation:

• Due to land disputes litigation in the court compels them to borrow heavily.
• Being uneducated and ignorant they are caught in the litigation process and dry away their savings and resources.

(IV) Money Lenders and High Rate of Interest:
The rate of interest charged by the local money lenders is very high and the compounding of interest leads to perpetuate indebtedness of the farmer.

[OR]

(b) Explain the public transport system in Tamil Nadu?
Answer:
Transport:
Tamil Nadu has a well established transportation system that connects all parts of the State. This is partly responsible for the investment in the State.

Tamil Nadu is served by an extensive road network in terms of its spread and quality, providing links between urban centres, agricultural market-places and rural habitations in the countryside. However, there is scope for improvement.

Road Transport:

1. There are 28 national highways in the State covering a total distance of 5,036 km.
2. The State has a total road length of 167,000 km, of which 60,628 km are maintained by Highways Department.

Rail Transport:

1. Tamil Nadu has a well-developed rail network as part of Southern Railway, Head quartered at Chennai.
2. Tamil Nadu has a total railway track length of 6,693 km and there are 690 railway stations in the State.
3. Main rail junctions in the State include Chennai, Coimbatore, Erode, Madurai, Salem, Tiruchirapalli and Tirunelveli.
4. Chennai has a well – established suburban Railway network, a Mass Rapid Transport system and is currently developing a Metro System, with its first underground stretch operational since May 2017.

Air Transport:

1. Tamil Nadu has four major international airports.
2. Chennai International Airport is currently the third largest airport in India.
3. Other international Airports in Tamil Nadu include Coimbatore International Airports, Madurai International Airport and Tiruchirapalli International Airport.
4. It also has domestic airports at Tuticorin, Salem, and Madurai.
5. Increased industrial activity has given rise to an increase in passenger traffic as well as freight movement which has been growing at over 18% per year.

Ports:

1. Tamil Nadu has three major ports; one each at Chennai, Ennore and Tuticorin, as well as one intermediate port in Nagapattinam, and 23 minor ports.
2. The ports are currently capable of handling over 73 million metric tonnes of cargo annually (24 % share of India).
3. All the minor ports are managed by the Tamil Nadu Maritime Board, Chennai Port.
4. This is an artificial harbour and the second principal port in the country for handling containers.
5. It is currently being upgraded to have a dedicated terminal for cars capable of handling 4,00,000 vehicles.
6. Ennore port was recently converted from an intermediate port to a major port and handles all the coal and ore traffic in Tamil Nadu.

Students can download Maths Chapter 8 Statistics and Probability Ex 8.4 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 8 Statistics and Probability Ex 8.4

Question 1.
If P (A) = \(\frac{2}{3}\), P(B) = \(\frac{2}{5}\), P(A ∪ B) = \(\frac{1}{3}\), then find P(A ∩ B).
Answer:
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)
\(\frac{1}{3}=\frac{2}{3}+\frac{2}{5}\) – P (A ∩ B)

Question 2.
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A∩B)=016. Find (i) P(not A)
(ii) P(not B)
(iii) P(A or B)
Solution:
(a) P(A) = 0.42 ;
P(B) = 0.48
P(A∩B) = 0.16
(i) P(not A) = P(\(\overline{\mathbf{A}}\)) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = P(\(\overline{\mathbf{B}}\)) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.42 + 0.48 – 0.16
= 0.74

Question 3.
If A and B are two mutually exclusive events of a random experiment and P (not A) = 0.45, P (A ∪ B) = 0.65, then find P(B).
Answer:
P(not A) = 0.45
1 – P (A) = 0.45
P (A) = 1 – 0.45 = 0.55
P(A ∪ B) = P (A) + P (B)
0. 65 = 0.55 + P(B)
0. 65 – 0.55 = P(B)
0.10 = P (B)
P(B) = 0.1

Question 4.
The probability that atleast one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P(\(\overline{\mathbf{A}}\)) + P(\(\overline{\mathbf{B}}\)).
Solution:
P(A∪B) = 0.6
P(A∩B) = 0.2
P(A) + P(B) = [1 – P(A∪B)] + [1 – P(A∩B)] = [1 – 0.6] + [1 – 0.2]
= 0.4 + 0.8 = 1.2

Question 5.
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B happen.
Answer:
Here P(A) = 0.5, P (B) = 0.3
P(A ∪ B) = P (A) + P(B) [A and B are mutually exclusive]
= 0.5 + 0.3
= 0.8
Probability of neither A nor [P(A ∪ B)’] = 1 – P(A ∪ B) = 1 – 0.8 = 0.2

Question 6.
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Answer:
Sample space = {(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 A be the event of getting an even number on the first time
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (A) = 18
\(P(A)=\frac{n(A)}{n(S)}=\frac{18}{36}\)
(ii) Let B be the event of getting a total of face sum 8.
B = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{36}\)
A ∩ B = {(2, 6) (4, 4) (6, 2)}
n(A ∩ B) = 3
P(A ∩ B) = \(\frac{3}{36}\)
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)

The required probability = \(\frac{5}{9}\)

Question 7.
From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of its being either a red king or a black queen.
Answer:
n(S) = 52
Let A be the event of getting a red king
n(A) = 2
\(P(A)=\frac{n(A)}{n(S)}=\frac{2}{52}\)
Let B be the event of getting a black Queen king
n(B) = 2
\(P^{\prime}(B)=\frac{n(B)}{n(S)}=\frac{2}{52}\)
It A and B are mutually exclusive
P(A ∪ B) = P(A) + P(B)
\(=\frac{2}{52}+\frac{2}{52}=\frac{4}{52}=\frac{1}{13}\)
The required probability is \(\frac{1}{13}\)

Question 8.
A box contains cards numbered 3, 5, 7, 9,… 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.
Answer:
Sample space = {3, 5, 7, 9,…….,35, 37}
n(S) = 18
Let A be the event of getting a multiple of 7
A = {7, 21, 35}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{18}\)
Let B be the event of getting a prime number
B = {3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37}
n(B) = 11

Probability of getting a multiple of 7 or a prime number = \(\frac{13}{18}\)

Question 9.
Three unbiased coins are tossed once. Find the probability of getting atmost 2 tails or atleast 2 heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting atmost 2 tails.
A = {HTT, THT, TTH, HHT, HTH, THH, HHH}
n(A) = 7

Probability of getting atmost two tails or atleast 2 heads = \(\frac{7}{8}\)

Question 10.
The probability that a person will get an electrification contract is \(\frac{3}{5}\) and the probability that he will not get plumbing contract is \(\frac{5}{8}\). The probability of getting atleast one contract is \(\frac{5}{7}\). What is the probability that he will get both?
Answer:
Let A and B represent the event of getting electrification control and plumbing contract.


Probability of getting both the job is \(\frac{73}{280}\)

Question 11.
In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?
Answer:
Total number of people in a town is 8000.
n(S) = 8000
Total number of females = 3000
Let A be the event of getting number of females
n(A) = 3000
\(P(A)=\frac{n(A)}{n(S)}=\frac{3000}{8000}\)
Number of people over 50 years = 1300
Let B be the event of getting number of people over 50 years.
n(B) = 1300
\(P(B)=\frac{n(B)}{n(S)}=\frac{1300}{8000}\)
Given 30% of the females are over 50 years.


Proability of getting either a female or over 50 years = \(\frac{17}{40}\)

Question 12.
A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting exactly two heads.
A = {HHT, HTH, THH}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}\)
Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}\)
Let C be the event of getting consecutively
C = {HHH, HHT, THH}
n(C) = 3



The probability is 1.

Question 13.
If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P (A ∩ B) = \(\frac{1}{6}\), P(B ∩ C) = \(\frac{1}{4}\), P(A ∩ C) = \(\frac{1}{8}\), P(A ∪ B ∪ C) = \(\frac{9}{10}\) and P (A ∩ B ∩ C) = \(\frac{1}{15}\), then find P(A), P(B) and P(C)?
Answer:
By the given condition,
P(B) = 2 P(A), P(C) = 3 P(A)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

Question 14.
In a class of 35, students are numbered from 1 to 35. The ratio of boys and girls is 4 : 3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.
Answer:
Sample space (S) = {1, 2, 3,… ,35}
n(S) = 35
Total number of students = 35
Number of boys = \(\frac{4}{7}\) × 35 = 20 [Boys Numbers = {1, 2, 3,…, 20}]
Number of girls = \(\frac{3}{7}\) × 35 = 15 [Girls Numbers = { 21, 22,…, 35}]
Let A be the event of getting a boy role number with prime number
A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\frac{8}{35}\)
Let B be the event of getting girls roll number with composite number.
B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}
n(B) = 12
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}\) = \(\frac{12}{35}\)
Let C be the event of getting an even roll number.
C = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}
n(C) = 17


Probability of getting roll number is \(\frac{29}{35}\)

Students can download Maths Chapter 7 Mensuration Unit Exercise 7 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 7 Mensuration Unit Exercise 7

Question 1.
The barrel of a fountain-pen cylindrical in shape is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one-fifth of a litre?
Answer:

Question 2.
A hemispherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litres per second. How much time will it take to empty the tank completely?
Answer:
Radius of the hemispherical tank = 1.75 m
Volume of the tank = \(\frac{2}{3} \pi r^{3}\) cu.units

Time taken = \(\frac{11229.17}{7}\) = 1604.17 seconds = 26.74 minutes = 27 minutes (approximately)

Question 3.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Answer:
Radius of a cone = Radius of a hemisphere = r unit

Height of a cone = r units
(height of the cone = radius of a hemisphere)
Maximum volume of the cone

Question 4.
An oil funnel of the tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion by 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Answer:

Total height of oil funnel = 22 cm
Height of the cylindrical portion = 10 cm
Height of the frustum (h) = 22 – 10 = 12 cm
Radius of the cylindrical portion = 4 cm
Radius of the bottom of the frustum = 4 cm
Top radius of the funnel (frustum) = \(\frac{18}{2}\) = 9 cm
Area of the tin sheet required = C.S.A of the frustum + C.S.A of the cylinder
= π (R + r) l + 2πrh sq. units.
= [π(9 + 4) \(\sqrt{12^{2}+(9-4)^{2}}\) + 2π × 4 × 10] cm²
= π [13 × \(\sqrt{144+25}\) + 25 + 80] cm²
= \(\frac{22}{7}\) [13 × 13 + 80] cm²
= \(\frac{22}{7}\) [169 + 80] cm²
= \(\frac{22}{7}\) × 249 cm²
= 782.57 cm²
Area of sheet required to make the funnel = 782.57 cm²

Question 5.
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer:
Radius of the cylinder = \(\frac{4.5}{2}\) cm
Height of the cylinder = 10 cm
Volume of the cylinder = πr²h cu. units


Number of coins = 450

Question 6.
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and the whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of a solid cylinder.
Answer:
External radius of the hollow cylinder R = 4.3 cm
Internal radius of the hollow cylinder r = 1.1 cm
Length of the cylinder (h) = 4 cm
Length of the solid cylinder (H) = 12 cm
Let the radius of the solid cylinder be “x”
Volume of the solid cylinder = Volume of the hollow cylinder

Diameter of the solid cylinder = 2 × 2.4 = 4.8 cm

Question 7.
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at ₹ 100 per sq. m.
Answer:
Slant height of a frustum (l) = 4 m
Perimeter of the top part = 18 m
2πR = 18


Cost of painting = ₹ 100 × 68 = ₹ 6800

Question 8.
A hemispherical hollow bowl has material of volume cubic \(\frac{436 \pi}{3}\) cubic cm. Its external diameter is 14 cm. Find its thickness.
Answer:
External radius of a hemisphere (R) = 7 cm
Volume of a hemi-spherical bowl = \(\frac{436 \pi}{3}\) cm³

Internal radius = 5 cm
Thickness of the hemisphere = (7 – 5) cm = 2 cm

Question 9.
The volume of a cone is 1005\(\frac{5}{7}\) cu. cm. The area of its base is 201\(\frac{1}{7}\) sq. cm. Find the slant height of the cone.
Answer:
Area of the base of a cone = 201\(\frac{1}{7}\) sq. cm

Question 10.
A metallic sheet in the form of a sector of a circle of radius 21 cm has a central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Answer:

Radius of a cone (r) = 21 cm
Central angle (θ) = 216°
Let “R” be the radius of a cone
Circumference of the base of a cone = arc length of the sector
2πR = \(\frac{\theta}{360} \times 2 \pi r\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.3 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 8 Statistics and Probability Ex 8.3

Question 1.
Write the sample space for tossing three coins using tree diagram.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Question 2.
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Answer:
Sample space = {(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)}

Question 3.
If A is an event of a random experiment such that P(A) : P(\(\bar{A}\)) = 17 : 15 and n(s) = 640 then find (i) P(\(\bar{A}\))
(ii) n(A)
Answer:

Question 4.
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting consecutive tails
A = {HTT, TTH, TTT}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{8}\)

Question 5.
At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize
(ii) the second player wins a prize if the first has won?
Solution:
22² = 484
31² = 961
23² = 529
32² = 1024
23, 24, 25, 26, 27, 28, 29, 30, 31 has squares below 500 × 1000.
(i) P(first player wins a prize) = \(\frac{9}{1000}\)
(ii) P(second player ins if first has won) = \(\frac{8}{999}\)

Question 6.
A bag contains 12 blue balls and x red balls. If one ball is drawn at random (i) what is the probability that it will be a red ball? (ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Answer:
Sample space = 12 + x
n(S) = x + 12
(i) Let A be the event of getting a red ball
n(A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\left(\frac{x}{x+12}\right)\)
(ii) 8 more red balls are added
Sample space = x + 12 + 8 = x + 20
Number of red balls = x + 8
Probability of drawing red ball = \(\frac{x+8}{x+20}\)
By the given condition
\(\frac{x+8}{x+20}=2\left(\frac{x}{x+12}\right)\)
(x + 8)(x + 12) = 2x(x + 20)
x² + 20x + 96 = 2x² + 40x
x² + 20x – 96 = 0
(x + 24)(x – 4) = 0
x = -24 (or) x = 4
The value of x = 4 (Number of balls will not be negative)
The probability of getting red balls = \(\left(\frac{x}{x+12}\right)=\frac{4}{16}=\frac{1}{4}\)

Question 7.
Two unbiased dice are rolled once. Find the probability of getting
(i) a doublet (equal numbers on both dice)
(ii) the product as a prime number
(iii) the sum as a prime number
(iv) the sum as 1
Answer:
(i) Sample space = {(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 A be the event of getting doublet
A = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of getting a product is a prime number.
B = {(1, 2) (1, 3) (1, 5) (2, 1) (3, 1) (5, 1)}
n(B) = 6
\(P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting a sum is a prime number
C = {(1, 1) (1, 2) (1, 4) (1, 6) (2, 1) (2, 3) (2, 5) (3, 2), (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)}
n(C) = 15
\(P(C)=\frac{n(C)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
(iv) Let D be the event of getting a sum is 1
n(D) = 0
\(P(D)=\frac{n(D)}{n(S)}=\frac{0}{36}=0\)
Probability of getting a sum is 1 is 0

Question 8.
Three fair coins are tossed together. Find the probability of getting
(i) all heads
(ii) atleast one tail
(iii) atmost one head
(iv) atmost two tails
Solution:
Possible outcomes = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
No. of possible outcomes = 2 × 2 × 2 = 8
(i) Prob(all heads) = \(\frac{1}{8}\)
(ii) Atleast one tail = {(THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
Prob(atleast one tail) = \(\frac{7}{8}\)
(iii) Atmost one head = {(HTT), (THT), (TTH), (TTT)}
∴ Prob(atmost one head) = \(\frac{4}{8}=\frac{1}{2}\)
(iv) Atmost two tail = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT)}
∴ Prob(atmost two tail) = \(\frac{7}{8}\)

Question 9.
Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Answer:
1st dice A = {1, 2, 3, 4, 5, 6}
2nd dice B = {1, 1, 2, 2, 3, 3}
Sample Space (S) = {(1, 1), (1, 1), (1, 2), (1, 2), (1, 3), (1, 3), (2, 1), (2, 1), (2, 2), (2, 2), (2, 3), (2, 3), (3, 1), (3, 1), (3, 2), (3, 2), (3, 3), (3, 3), (4, 1), (4, 1), (4, 2), (4, 2), (4, 3), (4, 3), (5, 1), (5, 1), (5, 2), (5, 2), (5, 3), (5, 3),(6, 1), (6, 1), (6, 2), (6, 2), (6, 3), (6, 3)}
n(S) = 36
(i) Let A₁ be the event of getting sum is 2
A1 = {(1, 1) (1, 1)}
n(A₁) = 2
\(P\left(A_{1}\right)=\frac{n\left(A_{1}\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}\)
(ii) Let A₂ be the event of getting a sum is 3.
A₂ = {(1, 2) (1, 2) (2, 1) (2, 1)}
n(A₂) = 4
\(P\left(A_{2}\right)=\frac{4}{36}=\frac{1}{9}\)
(iii) Let A₃ be the event of getting a sum is 4.
A₃ = {(1, 3) (1, 3) (2, 2) (2, 2) (3, 1) (3, 1)}
n(A₃) = 6
\(P\left(A_{3}\right)=\frac{6}{36}=\frac{1}{6}\)
(iv) Let A₄ be the event of getting a sum is 5.
A₄ = {(2, 3) (2, 3) (3, 2) (3, 2) (4, 1) (4, 1)}
n(A₄) = 6
\(P\left(A_{4}\right)=\frac{6}{36}=\frac{1}{6}\)
(v) Let A₅ be the event of getting a sum is 6.
A₅ = {(3, 3) (3, 3) (4, 2) (4, 2) (5, 1) (5, 1)}
n(A₅) = 6
\(P\left(A_{5}\right)=\frac{6}{36}=\frac{1}{6}\)
(vi) Let A₆ be the event of getting a sum is 7.
A₆ = {(4, 3) (4, 3) (5, 2) (5, 2) (6, 1) (6, 1)}
n(A₆) = 6
\(P\left(A_{6}\right)=\frac{6}{36}=\frac{1}{6}\)
(vii) Let A₇ be the event of getting a sum is 8.
A₇ = {(5, 3) (5, 3) (6, 2) (6, 2)}
n(A₇) = 4
\(P\left(A_{7}\right)=\frac{4}{36}=\frac{1}{9}\)
(viii) Let A₈ be the event of getting a sum is 9.
A₈ = {(6, 3) (6, 3)}
n(A₈) = 2
\(P\left(A_{8}\right)=\frac{2}{36}=\frac{1}{18}\)

Question 10.
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Answer:
Sample space (S) = 5 + 6 + 7 + 8
n(S) = 26
(i) Let A be the event of getting a white ball
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(P(A)=\frac{6}{26}=\frac{3}{13}\)
(ii) Let A be the event of getting a black ball
n(A) = 8
\(P(A)=\frac{n(A)}{n(S)}=\frac{8}{26}\)
Let B be the event of getting a red ball
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{26}\)
Probability of getting black or red ball
P(A ∪ B) = P (A) + P (B)
= \(\frac{8}{26}+\frac{5}{26}=\frac{13}{26}=\frac{1}{2}\)
(iii) Not white probability of getting white ball
P(A) = \(\frac{3}{13}\) from (i)
Probability of not getting white ball P(\(\bar{A}\)) = 1 – P(A)
\(1-\frac{3}{13}=\frac{13-3}{13}=\frac{10}{13}\)
(iv) Probability of getting a white ball.
P(A) = \(\frac{6}{26}\) (from 1)
Let B be the event of getting a black ball
n(B) = 8
\(P(B)=\frac{n(B)}{n(S)}=\frac{8}{26}\)
P(A ∪ B) = P(A) + P(B) = \(\frac{6}{26}+\frac{8}{26}=\frac{14}{26}\)
Probability of neither white nor black P(A ∪ B)’ = 1 – P(A ∪ B)
= \(1-\frac{14}{26}\)
= \(\frac{26-14}{26}=\frac{12}{26}=\frac{6}{13}\)

Question 11.
In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is \(\frac{3}{8}\) then, find the number of defective bulbs.
Answer:
Let the number of defective bulbs be “x”
Sample space (S) = 20 + x
n(S) = 20 + x
Let A be the event of getting to be defective
n(A) = x
\(P(A)=\frac{n(A)}{n(S)}\)
⇒ \(\frac{3}{8}=\frac{x}{20+x}\)
⇒ 8x = 3(20 + x) = (60 + 3x)
⇒ 8x – 3x = 60
⇒ 5x = 60
⇒ x = \(\frac{60}{5}\)
⇒ x = 12
Number of defective bulbs = 12

Question 12.
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card
Answer:
King diamond + Queen diamonds = 1 + 1 = 2 …….(1)
Queen hearts + Jack of hearts = 1 + 1 = 2 …….(2)
Jack spade + King of spades =1 + 1 = 2 …….(3)
Remaining number of cards = 52 – (6)
n(S) = 46
(i) Let A be the event of getting a clavor
n (A) = 13
\(P(A)=\frac{n(A)}{n(S)}=\frac{13}{46}\)
(ii) Let B be the event of getting a queen of red card
n(B) = 2
But the above two cards are removed from (1) and (2)
n(B) = 0
\(P(B)=\frac{n(B)}{n(S)}=\frac{0}{46}=0\)
(iii) Let B be the event of getting a king of black card
n(B) = (2 – 1) [from (3) one black card is removed]
n (B) = 1
\(P(B)=\frac{n(B)}{n(S)}=\frac{1}{46}\)

Question 13.
Some boys are playing a game, in which the stone thrown by them landing in a circular region given in the figure is considered as win and landing other than the circular region is considered as a loss. What is the probability to win the game?
Answer:

Area of a rectangle = l × b sq. feet = 3 × 4 sq. feet = 12 sq. feet
sample space (S) = 12
n(S) = 12
Let A be the event of getting the stone landing in a circular region
n(A) = Area of a circle
= πr²
= π × 1 × 1 (radius of a circle = 1 feet)
= π

Probability to win the game = \(\frac{11}{42}\) (or) \(\frac{157}{600}\)

Question 14.
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) the same day
(ii) different days
(iii) consecutive days?
Answer:
Sample space (S) = 6 × 6 = 36
n(S) = 36
[priya and Amuthan are visiting a particular shop in any one of 6 days is 6 × 6 = 36]
(i) Let A be the event of getting both are shopping on the same day
A = {(Mon, Mon) (Tue, Tue) (Wed, Wed) (Thu, Thu) (Fri, Fri) (Sat, Sat)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of shopping in different days.
n(B) = 36 – 6 = 30
\(P(B)=\frac{n(B)}{n(S)}\)
\(=\frac{30}{36}=\frac{5}{6}\)
(iii) Let C be the event of shopping consecutive days
C = {(Mon, Tue) (Tue, Wed) (Wed, Thu) (Thu, Fri) (Fri, Sat)}
n(C) = 5
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}\) = \(\frac{5}{36}\)

Question 15.
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Answer:
Sample space (S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting double entry fee (only getting 3 heads)
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting her entry fee (one or two heads to show)
n(B) = Probability of one head + Probability of 2 head
= \(\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\)
(iii) To loss the entry means not getting the head (only tail)
n(C) = 1
\(P(C)=\frac{n(C)}{n(S)}=\frac{1}{8}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.2 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 8 Statistics and Probability Ex 8.2

Question 1.
The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer:
Standard deviation of a data (σ) = 6.5
Mean of the data (\(\bar{x}\)) = 12.5
Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100 \%\)
= \(\frac{6.5}{12.5} \times 100 \%=52 \%\)
Coefficient of variation = 52%

Question 2.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Answer:
Standard deviation (σ) = 1.2
Coefficient of variation = 25.6

Question 3.
If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Answer:
Mean (\(\bar{x}\)) = 15
Co efficient of variation = 48

Question 4.
If n = 5, \(\bar{x}\) = 6, Σx² = 765, then calculate the coefficient of variation.
Answer:

Question 5.
Find the coefficient of variation of 24, 26, 33, 37, 29, 31.
Answer:
Arrange in ascending order we get 24, 26, 29, 31, 33, 37
Assumed mean = 29

Question 6.
The time taken (in minutes) to complete homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Answer:
Arrange in ascending order we get, 38, 40, 43, 44, 46, 47, 49, 53.
Assumed mean = 46


= \(\frac{453}{45}\)%
= 10.066
Coefficient of variation = 10.07%

Question 7.
The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation of 4.6 and 2.4 respectively. Who is more consistent in performance?
Answer:
Total marks scored by sathya = 460
Total marks scored by vidhya = 480
Number of subjects = 5
Mean marks of sathya = \(\frac{460}{5}\)
\(\bar{x}\) = 92%
Given standard deviation, (σ) = 4.6

Vidhya coefficient of variation is less than Sathya.
Vidhya is more consistent.

Question 8.
The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below.

Which of the three subjects shows the highest variation and which shows the lowest variation in marks?
Answer:
(i) Mathematics:
Mean (\(\bar{x}\)) = 56
Standard deviation (σ) = 12
Coefficient variation (CV₁) = \(\frac{\sigma}{\bar{x}} \times 100=\frac{12}{56} \times 100\)

Science shows the highest variation
Social science shows the lowest variation

Question 9.
The temperature of two cities A and B in the winter season are given below.

Find which city is more consistent in temperature changes?
Answer:
(i) city A:
Assumed mean = 22




C.V of city A < C.V of city B
City A is more consistent in temperature change.

Students can download Maths Chapter 8 Statistics and Probability Ex 8.1 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 8 Statistics and Probability Ex 8.1

Question 1.
Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Answer:
(i) Here the largest value (L) = 125
The smallest value (S) = 63
Range = L – S = 125 – 63 = 62

Question 2.
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Solution:
If the range = 36.8 and
the smallest value = 13.4 then
the largest value = L = R + S
= 36.8 + 13.4 = 50.2

Question 3.
Calculate the range of the following data.

Answer:
Smallest value (S) = 400
Largest value (L) = 650
Range = L – S = 650 – 400 = 250

Question 4.
A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
Answer:
The remaining number of pages to be completed is 60 – 32; 60 – 35; 60 – 37; 60 – 30; 60 – 33; 60 – 36; 60 – 35 and 60 – 37
The pages to be completed are, 28, 25, 23, 30, 27, 24, 25, and 23
Arrange in ascending order we get, 23, 23, 24, 25, 25, 27, 28 and 30

Question 5.
Find the variance and standard deviation of the wages of 9 workers given below:
₹ 310, ₹ 290, ₹ 320, ₹ 280, ₹ 300, ₹ 290, ₹ 320, ₹ 310, ₹ 280.
Answer:
Arrange in ascending order we get,
280, 280, 290, 290, 300, 310, 310, 320 and 320
Assumed mean = 300


Variance = 222.222
Standard deviation = √Variance = √222.222 = 14.907 = 14.91
Variance = 222.22
Standard deviation = 14.91

Question 6.
A wall clock strikes the bell once at 1 o’clock, 2 times at 2 o’clock, 3 times at 3 o’clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell make a day.
Answer:
Wall clock strikes the bell in 12 hours
1, 2, 3, 4, 5,… ,12
Wall clock strikes in a day (24 hours)
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Assumed mean = 14



The standard deviation of bell strike in a day is 6.9

Question 7.
Find the standard deviation of the first 21 natural numbers.
Answer:
Here n = 21
The standard deviation of the first ‘n’ natural numbers,

The standard deviation of the first 21 natural numbers = 6.06

Question 8.
If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Solution:
If the standard deviation of a data is 4.5 and each value of the data decreased by 5, the new standard deviation does not change and it is also 4.5.

Question 9.
If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Answer:
The standard deviation of the data = 3.6
Each value of the data is divided by 3
New standard deviation = \(\frac{3.6}{3}\) = 1.2
New Variance = (1.2)² = 1.44 [∴ Variance = (S.D)²]
New standard Deviation = 1.2
New variance = 1.44

Question 10.
The rainfall recorded in various places of five districts in a week are given below.

Find its standard deviation.
Answer:
Assumed mean = 60

Question 11.
In a study about viral fever, the number of people affected in a town were noted as

Find its standard deviation.
Answer:
Assumed mean = 35

Question 12.
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation.

Answer:
Assumed mean = 34.5

Question 13.
The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Answer:
Assumed mean = 11


Standard deviation (σ) = 1.24

Question 14.
For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Answer:
Number of candidates = 100
n = 100
Mean (\(\bar{x}\)) = 60
standard deviation (σ) = 15
Mean (\(\bar{x}\)) = \(\frac{\Sigma x}{n} \Rightarrow 60=\frac{\Sigma x}{100}\)
Σx = 6000
Correct total = 6000 + (45 – 40) + ( 72 – 27) = 6000 + 5 + 45 = 6050
Correct mean (\(\bar{x}\)) = \(\frac{6050}{100}\) = 60.5
Given standard deviation = 15


Correct mean = 60.5
Correct standard deviation (σ) = 14. 61

Question 15.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Answer:
Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8

Students can download Maths Chapter 7 Mensuration Additional 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 7 Mensuration Additional Questions

I. Multiple Choice Questions:

Question 1.
The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to ______
(1) π cm²
(2) 2π cm²
(3) 3π cm²
(4) 2 cm²
Answer:
(2) 2π cm²
Hint:
C.S.A of a cylinder = 2πrh sq. units = 2 × π × 1 × 1 cm² = 2π cm²

Question 2.
The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to ______ sq. units.
(1) \(\frac{3}{2} \pi h\)
(2) \(\frac{2}{3} \pi h^{2}\)
(3) \(\frac{3}{2} \pi h^{2}\)
(4) \(\frac{2}{3} \pi h\)
Answer:
(3) \(\frac{3}{2} \pi h^{2}\)
Hint:
T.S.A = 2πr(h + r)
[radius is half of the height]
= \(2 \pi \times \frac{h}{2}\left(\frac{h}{2}+h\right)\)
= \(=\pi h\left(\frac{3 h}{2}\right)=\frac{3}{2} \pi h^{2}\) sq. units

Question 3.
Base area of a right circular cylinder is 80 cm². If its height is 5 cm, then the volume is equal to _______
(1) 400 cm³
(2) 16 cm³
(3) 200 cm³
(4) \(\frac{400}{3}\) cm³
Answer:
(1) 400 cm³
Hint:
Volume of a cylinder = πr²h cu. units
[Base area (πr²) = 80 cm² = 80 × 5 cm³ = 400 cm³

Question 4.
If the total surface area of a solid right circular cylinder is 200π cm² and its radius is 5 cm, then the sum of its height and radius is ______
(1) 20 cm
(2) 25 cm
(3) 30 cm
(4) 15 cm
Answer:
(1) 20 cm
Hint:
T.S.A of a cylinder = 200π cm²
2πr (h + r) = 200π
2 × 5 (h + r) = 200
(h + r) = 20 cm

Question 5.
The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to ______
(1) πa²b sq.cm
(2) 2πab sq.cm
(3) 2π sq.cm
(4) 2 sq.cm
Answer:
(2) 2πab sq.cm .
Hint:
C.S.A. of a cylinder = 2πrh sq. units = 2 × π × a × b sq. cm = 2πab sq. cm

Question 6.
Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm³, then the volume of the cone is equal to _______
(1) 1200 cm³
(2) 360 cm³
(3) 40 cm³
(4) 90 cm³
Answer:
(3) 40 cm³
Hint:
Volume of the cone = \(\frac{1}{3}\) × volume of the cylinder
= \(\frac{1}{3}\) × 120 cm³
= 40 cm³

Question 7.
If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is
(1) 10 cm
(2) 20 cm
(3) 30 cm
(4) 96 cm
Answer:
(1) 10 cm
Hint:
Slant height of a cone

Question 8.
If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to ______
(1) 1200π cm²
(2) 600π cm²
(3) 300π cm²
(4) 600 cm²
Answer:
(2) 600π cm²
Hint:
Circumference (2πr) = 120π cm
Slant height (l) = 10 cm;
Curved surface area of a cone = πrl sq. units
= \(\frac{120 \pi}{2}\) × 10 cm² = 600π cm²

Question 9.
If the volume and the base area of a right circular cone are 48π cm and 12π cm respectively, then the height of the cone is equal to ______
(1) 6 cm
(2) 8 cm
(3) 10 cm
(4) 12 cm
Answer:
(4) 12 cm
Hint:
Volume of a cone = 48π cm³
[Base area (πr²) = 12π]
\(\frac{1}{3}\) πr²h = 48π
\(\frac{1}{3}\) × 12π × h = 48π
[Substitute πr² = 12π]
h = \(\frac{48}{4}\) = 12 cm

Question 10.
If the height and the base area of a right circular cone are 5 cm and 48 sq.cm respectively, then the volume of the cone is equal to _______
(1) 240 cm³
(2) 120 cm³
(3) 80 cm³
(4) 480 cm³
Answer:
(3) 80 cm³
Hint:
Volume of a cone (V) = \(\frac{1}{3}\) πr²h sq. units
Base area (πr²) = 48 sq. cm
V = \(\frac{1}{3}\) × 48 × 5 = 80 cm³

Question 11.
The ratios of the respective heights and the respective radii of two cylinders are 1 : 2 and 2 : 1 respectively. Then their respective volumes are in the ratio _______
(1) 4 : 1
(2) 1 : 4
(3) 2 : 1
(4) 1 : 2
Answer:
(3) 2 : 1
Hint:
h₁ : h₂ = 1 : 2
r₁ : r₂ = 2 : 1
Ratio of their volumes
= \(\frac{1}{3} \pi r_{1}^{2} h_{1}: \frac{1}{3} \pi r_{2}^{2} h_{2}\)
= 22 × 1 : 12 × 2 = 4 : 2 = 2 : 1

Question 12.
If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to ________
(1) 8π cm²
(2) 16 cm²
(3) 12π cm²
(4) 16π cm²
Answer:
(4) 16π cm²
Hint:
C.S.A of a sphere = 4πr² sq. units
[radius = 2 cm]
= 4 × π × 22 cm²
= 16π cm²

Question 13.
The total surface area of a solid hemisphere of diameter 2 cm is equal to _______
(1) 12 cm²
(2) 12π cm²
(3) 4π cm²
(4) 3π cm²
Answer:
(4) 3π cm²
Hint:
Radius of a hemisphere = \(\frac{2}{2}\) = 1 cm
Total surface area of a hemisphere = 3πr² sq. units = 3 × π × 12 cm² = 3π cm²

Question 14.
If the volume of a sphere is \(\frac{9}{16} \pi\) cu.cm, then its radius is ________
(1) \(\frac{4}{3}\) cm
(2) \(\frac{3}{4}\) cm
(3) \(\frac{3}{2}\) cm
(4) \(\frac{2}{3}\) cm
Answer:
(2) \(\frac{3}{4}\) cm
Hint:
Volume of the sphere = \(\frac{9}{16} \pi\)

Question 15.
The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio _______
(1) 81 : 625
(2) 729 : 15625
(3) 27 : 75
(4) 27 : 125
Answer:
(4) 27 : 125
Hint:
Ratio of their surface area = 9 : 25

Question 16.
The total surface area of a solid hemisphere whose radius is a units, is equal to ________
(1) 2πa² sq. units
(2) 3πa² sq. units
(3) 3πa sq. units
(4) 3a² sq. units
Answer:
(2) 3πa² sq. units
Hint:
T.S.A. of a solid hemisphere = 3πr² sq. units
= 3 × π × a × a sq.units
= 3πa² sq. units

Question 17.
If the surface area of a sphere is 100π cm², then its radius is equal to ______
(1) 25 cm
(2) 100 cm
(3) 5 cm
(4) 10 cm
Answer:
(3) 5 cm
Hint:
Surface area of a sphere = 100π cm²
4πr² = 100π
r² = 25
r = √25 = 5 cm

Question 18.
If the surface area of a sphere is 36π cm², then the volume of the sphere is equal to _______
(1) 12π cm³
(2) 36π cm³
(3) 72π cm³
(4) 108π cm³
Answer:
(2) 36π cm³
Hint:
Surface area of a sphere = 36π cm²
4πr² = 36π
r² = 9
r = 3 cm
Volume of a sphere = \(\frac{4}{3} \pi r^{3}\) cu. units
= \(\frac{4}{3} \pi\) × 3 × 3 × 3 cm³ = 36π cm³

Question 19.
If the total surface area of a solid hemisphere is 12π cm² then its curved surface area is equal to ______
(1) 6π cm²
(2) 24π cm²
(3) 36π cm²
(4) 8π cm²
Answer:
(4) 8π cm²
Hint:
T.S.A of a hemisphere = 12π cm²
3πr² = 12π
r² = 4
r = 2
Curved surface area of a hemisphere = 2πr² = 2 × π × 4 = 8π cm²

Question 20.
If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio _____
(1) 1 : 8
(2) 2 : 1
(3) 1 : 2
(4) 8 : 1
Answer:
(1) 1 : 8
Hint:
\(r_{1}=\frac{r_{2}}{2} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{1}{2} \Rightarrow r_{1}: r_{2}=1: 2\)

II. Answer the following questions:

Question 1.
Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.
Answer:
Given, Circumference of the base of a cylinder = 110 cm
2πr = 110 ……. (1)
Curved surface area = 4400 cm²
2πrh = 4400 cm² ……. (2)
From (1) & (2), \(\frac{(2)}{(1)} \Rightarrow \frac{2 \pi r h}{2 \pi r}=\frac{4400}{110}=40 \mathrm{cm}\)
Height of the cylinder (h) = 40 cm
From (1), 2πr = 110
2 × \(\frac{22}{7}\) × r = 110
r = \(\frac{35}{2}\)
We know that, diameter (d) = 2 × radius
d = 2 × \(\frac{35}{2}\) = 35 cm
Diameter of the Circular cylinder = 35 cm

Question 2.
A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost of painting the lateral surface of the pillars at ₹ 20 per square metre.
Answer:
Given, Radius of a cylinder (r) = 50 cm = 0.5 m
Height of a cylinder (h) = 3.5 m
Curved surface area of a pillar = 2πrh sq. units
Curved surface area of 12 pillars = 12 × 2πrh
= 12 × 2 × \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 132 sq. m.
Cost for painting the lateral surface of pillars per metre = ₹ 20
Cost of painting = 132 × ₹ 20 = ₹ 2640

Question 3.
The total surface area of a solid right circular cylinder is 231 cm². Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.
Answer:
Given, Total surface area of a cylinder (T.S. A) = 231 sq.cm
Curved surface area = \(\frac{2}{3}\) × T.S.A = \(\frac{2}{3}\) × 231 = 154 cm²
2πrh = 154 cm² …… (1)
Total surface area = 231 cm²
2πr (h + r) = 231
2πrh + 2πr²= 231
154 + 2πr² = 231 [from (1)]
2πr² = 231 – 154 = 77

Radius of the cylinder = 3.5 cm
Height of the cylinder = 7 cm

Question 4.
The total surface area of a solid right circular cylinder is 1540 cm². If the height is four times the radius of the base, then find the height of the cylinder.
Answer:
Given, Let the radius of the cylinder be ‘r’
Height of a cylinder (h) = 4r (by given condition)
Total surface area = 1540 cm²
2πr(h + r) = 1540 cm²

Height of the cylinder = 4r = 4 × 7 = 28 cm

Question 5.
If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.
Answer:
Given, In the figure, OAB is a cone and OC ⊥ AB
∠AOC = \(\frac{60^{\circ}}{2}\) = 30°
In the right ∆OAC, tan 30° = \(\frac{\mathrm{AC}}{\mathrm{OC}}\)

Slant Height of the cone (l) = 15 × 2 = 30 cm

Question 6.
The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.
Answer:

Given, Radius of a sector (r) = 21 cm
The angle of the sector (θ) = 180°
Let “R” be the radius of the cone.
Circumference of the base of a cone = Arc length of the sector

Radius of the cone (R) = 10.5 cm

Question 7.
If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.
Answer:
Given, the Curved surface area of a solid hemisphere = 2772 cm²
2πr² = 2772

Total surface area = 3πr² sq. units
= 3 × \(\frac{22}{7}\) × 21 × 21
= 4158 sq.cm
Aliter:
C.S.A of a hemisphere = 2772 cm²
2πr² = 2772 cm²
πr² = \(\frac{2772}{2}\) = 1386 cm
T.S.A of a hemisphere = 3πr² sq.units = 3 × 1386 cm² = 4158 cm²

Question 8.
An inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of ₹ 5 per sq. m.
Answer:
Given, Circumference of the dome = 17.6 m
2πr = 17.6
\(r=\frac{17.6 \times 7}{2 \times 22}=\frac{8.8 \times 7}{22}=2.8 \mathrm{m}\)
Curved surface area of the dome = 2πr² sq. units
= 2 × \(\frac{22}{7}\) × 2.8 × 2.8 m²
= 49.28 m²
Cost of painting for one sq.metre = ₹ 5
Cost of painting the curved surface = 49.28 × ₹ 5 = ₹ 246.40

Question 9.
Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.
Answer:
Given, Height of a cylinder (h) = 4.5 cm
Volume of a solid cylinder = 62.37 cu. cm

Radius of a cylinder (r) = 2.1 cm

Question 10.
A rectangular sheet of metal foil with dimension 66 cm × 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.
Answer:
After rolling the rectangular sheet into a cylinder

Volume of the cylinder = 4158 cm³

Question 11.
The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.
Answer:
Given, Height of the wooden solid cone (h) = 12 m
Circumference of the base = 44 m
2πr = 44
r = \(\frac{44 \times 7}{2 \times 22}\) = 7 m
Volume of the wooden solid = \(\frac{1}{3} \pi r^{2} h\) cu. units
= \(\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12 \mathrm{m}^{3}\)
= 88 × 7
= 616 m³
Volume of the solid = 616 m³

Question 12.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.
Answer:
Given, Edge of the cube = 14 cm
The largest circular cone is cut out from the cube.
Radius of the cone (r) = \(\frac{14}{2}\) = 7 cm
Height of the cone (h) = 14 cm
Volume of a cone

Volume of a cone = 718.67 cm³

Question 13.
The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. (Take π = \(\frac{22}{7}\))
Answer:
Let r, R and w be the inner and outer radii and thickness of the hemispherical bowl respectively.

Given that r = 5 cm, w = 0.25 cm
R = r + w = 5 + 0.25 = 5.25 cm
Now, outer surface area of the bowl = 2πR²
= 2 × \(\frac{22}{7}\) × 5.25 × 5.25
= 173.25 sq. cm
Thus, the outer surface area of the bowl = 173.25 sq. cm

Question 14.
Volume of a hollow sphere is \(\frac{11352}{7}\) cm3. If the outer radius is 8 cm, find the inner radius of the sphere. (Take π = \(\frac{22}{7}\))
Answer:
Let R and r be the outer and inner radii of the hollow sphere respectively.
Let V be the volume of the hollow sphere.

Hence, the inner radius r = 5 cm

Question 15.
How many litres of water will a hemispherical tank whose diameter is 4.2 m?
Answer:
Radius of the tank = \(\frac{4.2}{2}\) = 2.1 m
Volume of the hemisphere
= \(\frac{2}{3} \pi r^{3}\) cu.units
= \(\frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{m}^{3}\)
= 19.404 m³
= 19.404 x 1000 lit
= 19,404 litres

III. Answer the following questions.

Question 1.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Answer:

For cylindrical part:
Radius (r) = 7 cm
Height (h) = 6 cm
Curved surface area = 2πrh = 2 × \(\frac{22}{7}\) × 7 × 6 cm² = 264 cm²
For hemispherical part:
Radius (r) = 7 cm
Surface area (h) = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7 cm²
= 308 cm²
Total surface area = (264 + 308) = 572 cm²

Question 2.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:

Question 3.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Answer:

For cylinderical part:
Height (h) = 2.4 cm
Diameter (d) = 1.4 cm
Radius (r) = 0.7 cm
Total surface area of the cylindrical part

For conical part:
Base area (r) = 0.7 cm
Height (h) = 2.4 cm

Question 4.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Answer:
Diameter of the cylindrical well = 7 m
Radius of the cylinder (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
Volume = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \mathrm{m}^{3}\)
= 22 × 7 × 5 m³
Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having
Length (l) = 22 m
Breadth (b) = 14 m
Let ‘h’ be the height of the platform.
Volume of the platform = 22 × 14 × h m³
Volume of the platform = Volume of the earth taken out
22 × 14 × h = 22 × 7 × 5
\(h=\frac{22 \times 7 \times 5}{22 \times 14}=\frac{5}{2} \mathrm{m}=2.5 \mathrm{m}\)
Thus, the required height of the platform is 2.5 m.

Question 5.
The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum is 8 cm, find the whole surface area of the frustum. [Use π = 3.14]
Answer:

Let the radii of circular ends are R and r [R > r]
Perimeter of circular ends are 207.24 cm and 169.56 cm
2πR = 207.24 cm

The whole surface area of the frustum = π [(R² + r²) + (R + r) l]
Required whole surface area of the frustum
= 3.14 [33² + 27² + (33 + 27) × 10] cm²
= 3.14 [1089 + 729 + 600] cm²
= 3.14 [2418] cm²
= 7592.52 cm²

Question 6.
A cuboid-shaped slab of iron whose dimensions are 55 cm × 40 cm × 15 cm is melted and recast into a pipe. The outer diameter and thickness of the pipe are 8 cm and 1 cm respectively. Find the length of the pipe. (Take π = \(\frac {22}{7}\))
Answer:

Let h₁ be the length of the pipe
Let R and r be the outer and inner radii of the pipe respectively.
Iron slab:
Volume = lbh = 55 × 40 × 15 cm³
Iron pipe:
Outer diameter, 2R = 8 cm
Outer radius, R = 4 cm
Thickness, w = 1 cm
Inner radius, r = R – w = 4 – 1 = 3 cm
Now, the volume of the iron pipe = Volume of the iron slab

Time is taken by the pipe to empty half of the tank = 3 hours 12 minutes.

Question 7.
The perimeter of the ends of a frustum of a cone are 44 cm and 8.4π cm. If the depth is 14 cm., then find its volume.
Answer:
Given let the radius of the top of the frustum be “R” and the radius of the bottom of the frustum be “r”

Question 8.
A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.
Answer:

Given, Total height of solid = 13.5 cm
Diameter of the cylinder (d) = 28 m
Height of a cylinder (h) = 3 m
Height of a conical portion = 13.5 – 3 = 10.5 m
From the diagram, Radius of a cone = Radius of a cylinder
Radius (r) = 14 m

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
40 ml of methane is completely burnt using 80ml of oxygen at room temperature. The volume of gas left after cooling to room temperature. The volume of gas left after cooling to room temperature is …………………….
(a) 40 ml of CO₂ gas
(b) 40 ml of CO₂ gas and 80 ml of H₂O gas
(c) 60 ml of CO₂ gas and 60 ml H₂O gas
(d) 120 ml of CO₂ gas
Solution:
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l)

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.
Answer:
(a) 40 ml of CO₂ gas

Question 2.
What are the values of n, l, m and s for 3p_x electron?
(a) 3, 2, 1,0
(b) 3, 1,-1, +\(\frac{1}{2}\)
(c) 3, 2, +1, –\(\frac{1}{2}\)
(d) 3, 0, 0, +\(\frac{1}{2}\)
Solution:
3p_x electron; n = 3 (main shell)
for p_x orbital, l = 1, m = -1, s = \(\frac{1}{2}\)
Answer:
(b) 3, 1,-1, +\(\frac{1}{2}\)

Question 3.
Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.
Answer:
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.

Question 4.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 5.
Lithium shows diagonal relationship with ………………………
(a) Sodium
(b) Magnesium
(c) Calcium
(d) Ahuninium
Answer:
(b) Magnesium

Question 6.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\) \((\frac { \partial V }{ \partial T } )\)
(a) T
(b) 1/T
(c) P
(d) None of these
Solution:

Answer:
(b) 1/T

Question 7.
Heat of combustion is always …………………….
(a) Positive
(b) Negative
(c) Zero
(d) Either positive or negative
Answer:
(b) Negative

Question 8.
If in a mixture where Q = K, then what happens?
(a) The reaction shift towards products
(b) The reaction shift towards reactants
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate
(d) Nothing happens
Answer:
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate

Question 9.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N₂
(b) He
(c) CO₂
(d) H₂
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.
Answer:
(c) CO₂

Question 10.
In the molecule O_A = C = O_B, the formal charge on O_A, C and O_B are respectively.
(a) – 1, 0, + 1
(b) + 1, 0, -1
(c) – 2, 0, + 2
(d) 0, 0, 0
Solution:

Formal charge of O_A/O_B = N_V – (N_e + \(\frac { N_{ b } }{ 2 } \)) = 6 – (4 + \(\frac{4}{2}\)) = 6 – 6 = 0
Formal charge of C = 4 – (0 + \(\frac{8}{2}\)) = 4 – 4 = 0
Answer:
(d) 0, 0, 0

Question 11.
In the hydrocarbon CH₃ – CH₃ – CH = CH – CH₂ – C = CH the state of hybridisation of carbon 1, 2, 3, 4 and 7 are in the following sequence.
(a) sp, sp, sp³, sp², sp³
(b) sp², sp, sp³, sp², sp³
(c) sp, sp, sp², sp, sp³
(d) None of these
Answer:
(a) sp, sp, sp³, sp², sp³

Question 12.
Enzyme present in apple is ………………………..
(a) Polyphenol oxidase
(b) Polyphenol reductase
(c) Polyphenol
(d) Polyphenol hydrolase
Answer:
(a) Polyphenol oxidase

Question 13.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is ………………………….
(a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 14.
In Finkelstein reaction, the mechanism followed is ……………………….
(a) S_N1
(b) E₁
(c) E₂
(d) S_N2
Answer:
(d) S_N2

Question 15.
Which sequence for greenhouse gases is based on GWP?
(a) CFC > N₂O > CO₂ > CH₄
(b) CFC > CO₂ > N₂O > CH₄
(c) CFC > N₂O > CH₄ > CO₂
(d) CFC > CH₄ > N₂O > CO₂
Answer:
(c) CFC > N₂O > CH₄ > CO₂

PART – II

Answer any six questions in which question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
Define Avogadro Number?
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 × 10²³.

Question 17.
Explain the meaning of the symbol 4P. Write all the four quantum numbers for these electrons?
Answer:
4f²: It means that the element has 2 electrons in outermost 4f shell.
Quantum number values are,
n = principal quantum number = 4
l = azimuthal quantum number = 3
m = magnetic quantum number = – 3, -2
s = spin quantum number = +\(\frac{1}{2}\), –\(\frac{1}{2}\)

Question 18.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”?
Answer:
No, It is not correct. Thq accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 19.
What are the uses of calcium hydroxide?
Calcium hydroxide is used
Answer:

1. In the preparation of mortar, a building material.
2. In white wash due to its disinfectant nature.
3. In glass making and tanning industry.
4. For the preparation of bleaching powder and for the purification of sugar.

Question 20.
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1. Calculate the melting point of sodium chloride?
Answer:
Given:
∆H_f = 30.4 kJ = 30400 J mol^-1
∆S_f (NaCl) = 28.4 kJ^-1 mol^-1
T_f = ?
∆S_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}}\); T_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{S}_{\mathrm{f}}}\)
T_f = \(\frac{30400 \mathrm{J} \mathrm{mol}^{-1}}{28.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}}\) = 1070.4 K

Question 21.
How is a gas-solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecules in the gaseous state and those dissolved in the liquid.
Example: In carbonate beverages the following equilibrium exists.
CO₂(g) ⇄ CO₂ (solution).

Question 22.
What type of hybridisations are possible in the following geometeries?

1. Octahedral
2. Tetrahedral
3. Square planar

Answer:

1. Octahedral geometry is possible by sp³d² (or) d²sp³ hybridisation.
2. Tetrahedral geometry is possible by sp³ hybridisation.
3. Square planar geometry is possible by dsp² hybridisation.

Question 23.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

1. A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
2. The tube is heated till reaction ceases and become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
3. The contents of the dist is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 24.
Complete the reactions and identify the products?

Answer:

PART – III

Answer any six questions in which question No. 32 is compulsory. [6 × 3 = 18]

Question 25.
The balanced equation for a reaction is given below 2x + 3y → 41 + m
When 8 moles of x react with 15 moles of y, then

1. Which is the limiting reagent?
2. Calculate the amount of products formed.
3. Calculate the amount of excess reactant left at the end of the reaction.

Answer:
2x + 3y → 41 + m
1. 2x reacts with 3y to give products.
8x reacts with 15y means, y is the excess because 8 moles of x should react with 4 × 3y = 12y moles of y to give products.
In this reaction 15y moles are used.
Therefore, 3 moles of y is excess and x is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 × 41 i.e. 161 and 4m as product.
8x+ 12y → 161 + 4m

3. At the end of the reaction, the excess reactant left is 3 moles of y.

Question 26.
Which would you expect to have a higher melting point, magnesium oxide or magnesium fluoride? Explain your reasoning.
Answer:

1. Magnesium oxide has very strong ionic bonds as compared to magnesium fluoride.
2. Mg²⁺ and O^2- have charges of +2 and -2, respectively.
3. Oxygen ion is smaller than fluoride ion.
4. The smaller the ionic radii, the smaller the bond length in MgO and the bond is stronger than MgF₂.
5. Due to more strong bond nature in MgO, it has high melting point than MgF₂.

Question 27.
A sample of solid KClO₃ (potassium chlorate) was heated in a test tube to obtain O₂ according to the reaction
2KClO_3(s) → 2KCl + 3O₂
The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:

Question 28.
List the characteristics of entropy?
Characteristics of entropy:
Answer:

• Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
• In general, the entropy of gaseous system is greater than liquids and greater than solids. The symbol of entropy is S.
• Entropy is defined as “for a reversible change taking place at a constant temperature (T), the change in entropy (∆S) of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).
  \(\Delta S_{\mathrm{sys}}=\frac{q_{\mathrm{rev}}}{T}\)
• If heat is absorbed, then ∆S is positive and there will be increase in entropy. If heat is evolved, ∆S is negative and there is a decrease in entropy,
• The change in entropy of a process represented by ∆S and is given by the equation,
  ∆S_sys = S_f – S_i
• If S_f > S_i, ∆S is positive, the reaction is spontaneous and reversible.
  If S_f < S_i, ∆S is negative, the reaction is non-spontaneous and irreversible.
• Unit of entropy: SI unit of entropy is J K^-1.

Question 29.
Derive the values of K_C and K_P for the synthesis of HI.
Answer:
H₂(g) + I₂(g) ⇄ 2HI(g)
Let us consider the formation of HI in which V moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’.
Let ‘x’ moles of each of H₂ and I₂ react together to form 2x moles of HI.

Applying mass of action

Caluculation of K_P: K_P = K_C.RT^∆ng
Here ∆n_g = n_p – n_r = 2 -2 = 0
Hence, K_P = K_C
K_P = \(\frac{4 x^{2}}{(a-x)(b-x)}\)

Question 30.
Describe the classification of organic compounds based on their structure?
Answer:
Classification of organic compounds based on the structure

Question 31.
For the following bond cleavages use curved-arrows to show the electron flow and classify each as homolytic or heterolytic fission. Identify reactive intermediate produced as free radical, carbocation and carbanion?

Answer:

Question 32.
An alkyl halide with molecular formula C₆H₁₂Br on dehydrohalogenation gave two isomeric alkenes X and,Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH₂CHO and (CH₃)₂ CHCHO. Find the alkyl halide?
Answer:

1. C₆ H₁₃ Br is 3 – Bromo – 4 – methylpentane.
2. 3 – Bromo – 4 – methylpentane on dehydration give two isomers X and Y as follows:

Therefore C₆H₁₃ Br is 3 – Bromo – 4 – methylpentane

Question 33.
What do you mean by ozone hole? What are its consequences?
Answer:
Depletion of ozone layer creates some sort of holes in the blanket of ozone which surrounds us in the atmosphere and this is known as ozone hole.

1. With the depletion of the ozone layer, UV radiations filters into the troposphere which leads to ageing of skin, cataract, sunburn, skin cancer etc.
2. By killing many of the phytoplanktons, it can damage the fish productivity.
3. Evaporation rate increases through the surface and stomata of leaves which can decrease the moisture content of the soil.

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) Balance the following equations by ion electron method?
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
(II) Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?

[OR]

(b)
(I) How many unpaired electrons are present in the ground state of Fe³⁺ (z = 26), Mn²⁺ (z = 25) and argon (z = 18)?
(II) Explain about the significance of de Broglie equation?
Answer:
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
Oxidation half reaction: (loss of electron)

Reduction of halfa reaction: (gain of electron)

Add H₂O to balance oxygen atoms

Add Hcl to balance hydrogen atoms
KMnO₄ + 5e^– + 8 HCl → MnCl₂ + 4H₂O …………………. (4)

To equalize the number of electrons equation (1) × 5 and equation (2) × 2

(II) Molecular mass of H₃BO₃ = (1 × 3) + (11 × 1) + (16 × 3) = 62
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass x mole
= 62 × 0.543
= 33.66 g

[OR]

(b) (I) Fe → Fe³⁺ + 3e^–
Fe (Z = 26)
Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25) Electronic configuration is
1s² 2s² 2p⁵ 3s² 3p⁶ 4s² 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

(II) Significance of de Broglie equation:

1. λ = \(\frac{h}{mv}\) This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
2. For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
3. For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.
4. For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 35 (a).
(I) Mention any two anomalous properties of second period elements?
(II) Arrange Na⁺, Mg²⁺ and Al³⁺ in the increasing order of ionic radii. Give reason?

[OR]

(b)
(I) How do you expect the metallic hydrides to be useful for hydrogen storage?
(II) Write a note about ortho water and para water?
Answer:
(a) (I) Anomalous properties of second period elements:

1. In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
2. In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

(II) Na⁺, Mg²⁺ and Al³⁺ are isoelectronic cations.

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is, \(\mathbf{r}_{\mathrm{Na}^{+}}\), > \(\mathbf{r}_{\mathrm{Mg}^{2+}}\), > \(\mathbf{r}_{\mathrm{Al}^{3+}}\).

[OR]

(b)
(I) In metallic hydrides, hydrogen is adsorbed as H-atoms. Due to the adsorption of H atoms the metal lattice expands and become unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.

(II)
1. Water exists in space in the interstellar clouds, in proto-planetary disks, in the comets and icy satellites on the solar system, and on the Earth.
2. In particular, the ortho-to-para ratio (OPR) of water in space has recently received attention. Like hydrogen, water can also be classified into ortho – H₂O and para – H₂O, in which the directions are antiparallel.

3. At the temperature conditions of the Earth (300 K), the OPR of H₂O is 3.
4. At low temperatures below (< 50 K) the amount of para – H₂O increases. It is known that the OPR of water in interstellar clouds and comets has more para – H₂O (OPR = 2.5) than on Earth.

Question 36 (a).
Derive the values of critical constants from the Van der Waals constants?

[OR]

(b) Derive the values of K_p and K_C for dissociation of PCl₅?
Answer:
(a) Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT for 1 mole.
From this equation, the values of critical constant P_C, V_C and T_C are derived in terms of a and b the Vander Waals constants.

The above equation (4) is an cubic equation of V, which can have three roots. At the critical point, all the three values of V are equal to the critical volume Vc. i.e. V = V_C„
i.e; V = V_C
V – V_C = 0 ………………… (5)
(V – V_C)³ ………………….. (6)
(V³ – 3V_CV²) + 3V_C²V – V_C³ …………………. (7)
As the equation (4) is identical with equation (7), comparing the ‘V’ terms in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

Substituting the values of V_C and P_C in equation (9)

Critical constants a and b can be calculated rising Vander Waals constants as follows:

(b) Consider that V moles of PCl₅is taken in container of volume‘V’
Let x moles of PCl₅ be dissociated into x moles of PCl₃ and x moles of Cl₂.

Applying law of mass action

K_p caluculation: K_p = K_C. \(\mathrm{RT}^{\mathrm{An}}\); ∆n_g = 2 – 1 = 1
We know that PV = nRT
RT = \(\frac{PV}{n}\)
Where ‘n’ is the total number of moles at equilibrium
n = a – x + x + x = a + x

Question 37 (a).
(I) Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement?
(II) Explain how non-ideal solutions shows positive deviation from Raoult’s law?

[OR]

(b)
(I) How will you distinguish between electrophiles and nucleophiles?
(II) Complete the following reactions and identify the products?
(a)
(b)
Answer:
(a) (I) When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.
(II)

1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water
2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alco¬hol and water-water interaction).
3. This results in the increased evaporation of both components from the aqueous solution of ethanol.
4. Consequently, the vapour pressure of the solution is greater than the vapour pressure
   predicted by Raoult’s law.
5. Here, the mixing process is endothermic i.e., DHmixing > 0 and there will be a slight increase in volume (DVmixing > 0)
6. 

(b)
(I)
Electrophiles:

1. They are electron deficient.
2. They are cations.
3. They are lewis acids.
4. Accept an electron pair.
5. Attack on electron rich sites.

Nucleophiles:

1. They are electron rich.
2. They are anions.
3. They are lewis bases.
4. Donate an electron pair.
5. Attack on electron deficient sites.

(II)
(a)
(b)

Question 38 (a).
(I) Is it possible to prepare methane by Kolbe’s electrolytic method?
(II) Explain how 2-butyne reacts with
(a) Lindlar’s catalyst
(b) Sodium in liquid ammonia.

[OR]

(b) (I) Discuss the aromatic nucleophilic substitution reactions of chlorobenzene?
(II) CCl₄ > CHCl₃ > CH₂C₁₂ > CH₃Cl is the decreasing order of boiling point in haloalkanes. Give reason?
Answer:
(a) (I) Kolbe’s electrolytic method is suitable for the preparation of symmetrical alkanes, that is alkanes containing even number of carbon atoms. Methane has only one carbon, hence it cannot be prepared by Kolbe’s electrolytic method.
(II) (a) 2-butyne reacts with Lindlar’s catalyst: 2-butyne can be reduced to cis – 2 butene using CaCO₃ supported in Pd -metal partially deactivated with sulphur.
This reaction is stereo specific giving only the cis – 2 – butene.

(b) 2-butyne reacts with sodium in liquid ammonia:
2-butyne can also be reduced to trans – 2 – butene using sodium in liquid ammonia. This reaction is stereospecific giving only the trans – 2 – butene.

[OR]

(b) (I) Aromatic nucleophilic substitution reactions:
Dow’s process:

1.

2.

3.

(II) The boiling point of chloro, bromo and iodoalkanes increases with increase in the number of halogen atoms. So the correct decreasing order of boiling point of haloalkanes is:
CCl₄ > CHCl₃ > CH₂Cl₂ > CH₃Cl.

Students can download Maths Chapter 8 Statistics and Probability Unit Exercise 8 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 8 Statistics and Probability Unit Exercise 8

Question 1.
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f₁ and f₂.

Answer:
Arithmetic Mean (\(\bar{x}\)) = 62.8
Sum of all the frequencies (Σf_i) = 50
Let the missing frequencies be f₁ and f₂


Substitute of the value of f₂ in (1)
f₁ + 12 = 20
⇒ f₁ = 20 – 12 = 8
The Missing frequency is 8 and 12.

Question 2.
The diameter of circles (in mm) drawn in the design are given below.

Calculate the standard deviation.
Answer:
Assumed mean = 42. 5

Question 3.
The frequency distribution is given below

In table k is a positive integer, has a variance of 160. Determine the value of k.
Answer:
Assumed mean = 3k



The value of k = 7

Question 4.
The standard deviation of some temperature data in degree Celsius (°C) is 5. If the data were converted into degree Fahrenheit (°F) then what is the variance?
Solution:
F° = (C° × 1.8) + 32
\(\begin{array}{l}{\sigma_{c}=5^{\circ} \mathrm{C}} \\ {\sigma_{\mathrm{F}}=\left(1.8 \times 5^{\circ} \mathrm{C}\right) \cdot 9^{\circ} \mathrm{F}}\end{array}\)
Adding value to data doesn’t affect standard deviation.
New variance = \(\sigma_{\mathrm{F}}^{2}=81^{\circ} \mathrm{F}\)

Question 5.
If for a distribution, Σ (x – 5) = 3, Σ (x – 5)² = 43, and total number of observations is 18, find the mean and standard deviation.
Answer:


(i) Arithmetic mean (\(\bar{x}\)) = 5.17
(ii) Standard deviation (σ) = 1.53

Question 6.
Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Answer:
Coefficients of variation of prices in city A.
Arrange in ascending order we get, 16, 19, 20, 22, 23
Assumed mean = 20

Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100\)
= \(\frac{2.45}{20} \times 100\)
= 12. 25%
Coefficient of variation = 12.25%
Coefficients of variation of prices in city B.
Arrange in ascending order we get, 10, 12, 15, 18, 20
Assumed mean = 15


Prices in city A is more stable (since 12.25 < 24.6 %)

Question 7.
If the range and coefficient of range of the data are 20 and 0.2 respectively, then find the largest and smallest values of the data.
Answer:
Range of the data (R) = 20
L – S = 20 ……(1)
Coefficient of range = 0.2
Coefficient of range = \(\frac{L-S}{L+S}\)
0.2 = \(\frac{20}{L+S}\)

substituting the value of L = 60 in (2)
60 + S = 100
S = 100 – 60 = 40
Largest value = 60
Smallest value = 40

Question 8.
If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.
Answer:
Sample space = {(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
(i) Let A be the event of getting product of face value 6.
A = {(1, 6), (2, 3), (3, 2) (6, 1)}
n(A) = 4
P (A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{4}{36}\)
(ii) Let B be the event of getting difference of face value is 5.
B = {(6, 1)}
n(B) = 1

The probability is \(\frac{1}{9}\)

Question 9.
In a two children family, find the probability that there is at least one girl in a family.
Answer:
Sample space (S) = {(Boy, Boy) (Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(S) = 4
Let A be the event of getting atleast one Girl
A = {(Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}\)
Probability of atleast one girl in a family is \(\frac{3}{4}\)

Question 10.
A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.
Answer:
Let the number of black balls be “x”
Sample space (S) = x + 5
n(S) = x + 5
Let A be the event of drawing a black ball
n (A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{x}{x+5}\)
Let B be the event of getting a white ball
n(B) = 5
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{5}{x+5}\)
By the given condition,
\(\frac{x}{x+5}=2 \times\left(\frac{5}{x+5}\right)\)
⇒ \(\frac{x}{x+5}=\left(\frac{10}{x+5}\right)\)
⇒ 10x + 50 = x² + 5x
⇒ x² + 5x – 10x – 50 = 0
⇒ x² – 5x – 50 = 0
⇒ (x – 10)(x + 5) = 0
⇒ x = 10 or x = -5
Number of balls will not be negative.
Number of black balls = 10

Question 11.
The probability that a student will pass the final examination in both English and Tamil is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Tamil examination?
Answer:
Let A be the event of getting student pass in English
Let B be the event of getting student pass in Tamil


Probability of passing the tamil examination is \(\frac{13}{20}\)

Question 12.
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5.
Answer:
Total number of cards = 52
3 cards are removed
Remaining number of cards = 52 – 3 = 49
n(S) = 49
(i) Let A be the event of getting a diamond card.
n(A) = 13
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{13}{49}\)
(ii) Let B be the event of getting a queen card.
n(B) = (4 – 1) = 3
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{3}{49}\)
(iii) Let C be the event of getting a spade card.
n(C) = (13 – 3) = 10
P(C) = \(\frac{n(C)}{n(S)}=\frac{10}{49}\)
(iv) Let D be the event of getting a 5 of heart card.
n(D) = 1
P(D) = \(\frac{n(\mathrm{D})}{n(\mathrm{S})}=\frac{1}{49}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.5 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 8 Statistics and Probability Ex 8.5

Multiple Choice Questions.

Question 1.
Which of the following is not a measure of dispersion?
(1) Range
(2) Standard deviation
(3) Arithmetic mean
(4) Variance
Answer:
(3) Arithmetic mean
Hint:
Measures of dispersion are,
(i) Range
(ii) Mean deviation
(iii) Quartile deviation
(iv) Standard deviation
(v) Variance
(vi) coefficient of variation

Question 2.
The range of the data 8, 8, 8, 8, 8.. . 8 is
(1) 0
(2) 1
(3) 8
(4) 3
Solution:
(1) 0
Hint
R = L – S = 8 – 8 = 0

Question 3.
The sum of all deviations of the data from its mean is _______
(1) always positive
(2) always negative
(3) zero
(4) non-zero integer
Answer:
(3) zero

Question 4.
The mean of 100 observations is 40 and their standard deviation is 3. The sum of squares of all deviations is
(1) 40000
(2) 160900
(3) 160000
(4) 30000
Solution:
(2) 160900
Hint:
σ = 3

Question 5.
Variance of first 20 natural numbers is ______
(1) 32.25
(2) 44.25
(3) 33.25
(4) 30
Answer:
(3) 33.25
Hint:
Variance of 20 natural numbers is

Question 6.
The standard deviation of a data Is 3. If each value is multiplied by 5 then the new variance is
(1) 3
(2) 15
(3) 5
(4) 225
Solution:
σ = 3. 1f each is multiplied by 5. The new standard variation is also multiplied by 3.
∴ The new S.D = 5 × 3 = 15
Variance = 15² = 225

Question 7.
If the standard deviation of x, y, z is p then the standard deviation of 3x + 5, 3y + 5, 3z + 5 is ________
(1) 3p + 5
(2) 3p
(3) p + 5
(4) 9p + 15
Answer:
(2) 3p
Hint:
(i) Each value is added by any constant there is no change in standard deviation.
(ii) Each value is multiplied by 3 standard deviations also multiplied by 3.
The standard deviation is 3p.

Question 8.
If the mean and coefficient of variation of a data are 4 and 87.5% then the standard
deviation is
(1) 3.5
(2) 3
(3) 4.5
(4) 2.5
Solution:
(1) 3.5
Hint:
\(\bar{x}\) = 4, coefficient of variation is = 87. 5%

Question 9.
Which of the following is incorrect?
(1) P(A) > 1
(2) 0 ≤ P(A) ≤ 1
(3) P(Φ) = 0
(4) P(A) + P(\(\bar{A}\)) = 1
Answer:
(1) P(A) > 1
Hint:
Probability is always less than one or equal to one.

Question 10.
The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is _________
(1) \(\frac{q}{p+q+r}\)
(2) \(\frac{p}{p+q+r}\)
(3) \(\frac{p+q}{p+q+r}\)
(4) \(\frac{p+r}{p+q+r}\)
Answer:
(1) \(\frac{q}{p+q+r}\)
Hint:
Sample spaces = p + q + r
Let A be the event of getting red
n(A) = p
P(A) = \(\frac{q}{p+q+r}\)

Question 11.
A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is _______
(1) \(\frac{3}{10}\)
(2) \(\frac{7}{10}\)
(3) \(\frac{3}{9}\)
(4) \(\frac{7}{9}\)
Answer:
(2) \(\frac{7}{10}\)
Hint:
Here n(S)= 10 (given digit at imit place. It has two digit)
n(A) = 7
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{7}{10}\)

Question 12.
The probability of getting a job for a person is \(\frac{x}{3}\). If the probability of not getting the job is \(\frac{2}{3}\) then the value of x is.
(1) 2
(2) 1
(3) 3
(4) 1.5
Solution:
(2) 1
Hint:
Probability of getting a job = \(\frac{x}{3}\)
Probability of not getting a job = 1 – \(\frac{x}{3}\)

Question 13.
Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \(\frac{1}{9}\), then the number of tickets bought by Kamalam is _______
(1) 5
(2) 10
(3) 15
(4) 20
Answer:
(3) 15
Hint:
n(S) = 135
P(A) = \(\frac{1}{9}\)

Question 14.
If a letter is chosen at random from the English alphabets {a, b, …, z}, then the probability that the letter chosen precedes x.
(1) \(\frac{12}{13}\)
(2) \(\frac{1}{13}\)
(3) \(\frac{23}{26}\)
(4) \(\frac{3}{26}\)
Solution:
(3) \(\frac{23}{26}\)
Hint:
n(S) = 26
Let A denote the letter chosen precedes x
A= {a, b, c, d, …, x}
n(A) = 23
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{23}{26}\)

Question 15.
A purse contains 10 notes of ₹ 2000, 15 notes of ₹ 500, and 25 notes of ₹ 200. One note is drawn at random. What is the probability that the note is either a ₹ 500 note or ₹ 200 note?
(1) \(\frac{1}{5}\)
(2) \(\frac{3}{10}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{4}{5}\)
Answer:
(4) \(\frac{4}{5}\)
Hint:
Sample space (S) = 10 + 15 + 25 = 50
n(S) = 50
Let A be the event of getting ₹ 500 note
n (A) = 15
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{15}{50}\)
Let B be the event of getting ₹ 200 note
n (B) = 25
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{25}{50}\)
Probability of the note is either a ₹ 500 note or ₹ 200 note
P(A) + P(B) = \(\frac{15}{50}+\frac{25}{50}\) = \(\frac{40}{50}\) = \(\frac{4}{5}\)