Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 1.
What are biomolecules? Give examples of biomolecules.
Answer:
Biomolecules: The lifeless, complex organic molecules which combine in a specific manner to produce life or control biological reactions are called biomolecules.

Examples: Carbohydrates, lipids (fats and oils), nucleic acids, enzymes.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 2.
What is the importance of biomolecules?
Answer:
Biomolecules are organic molecules which combine in a particular fashion to give complex substances which help to sustain life and produce identical daughter cells and play an important role in the actions of an organism.

  • Carbohydrates are the major constituents of food and source of energy.
  • Proteins help in proper functioning of living beings. They are important constituents of skin, hair, muscles. Enzymes which catalyse chemical reactions that take place in cells are proteins.
  • Lipids (fats and oils) function as the storehouses of energy.
  • Nucleic acids, the ribonucleic acid (RNA), and deoxyribonucleic acid (DNA) are responsible for genetic characteristics and synthesis of proteins.

Question 3.
What are carbohydrates?
OR
Define the term : Carbohydrates.
Answer:
Carbohydrates : Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones, or the compounds which on hydrolysis produce polyhydroxy aldehydes or polyhydroxy ketones.

Examples : Glucose, sucrose, fructose.

Question 4.
What is monosaccharide?
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide. The monosaccharide is crystalline and soluble in water. E.g. Glucose, fructose, ribose.

Question 5.
Mention the names of monosaccharides or simple carbohydrates.
Answer:
Monosaccharides are (1) glucose (2) fructose (3) ribose.

Question 6.
State the basic unit of all carbohydrates.
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide.

Question 7.
How are carbohydrates classified?
OR
Classification of carbohydrates with examples.
Answer:
Carbohydrates are classified as monosaccharides oligosaccharides and polysaccharides.
(1) Monosaccharides : These carbohydrates cannot be further hydrolysed into smaller units. They are basic units of all carbohydrates, and are called monosaccharides.

Examples : Glucose, fructose, ribose

(2) Oligosaccharides : An oligosaccharide is a carbohydrate (sugar) which on hydrolysis gives two to ten monosaccharide units.
Depending on the number of monosaccharides produced on hydrolysis, oligosaccharides are further classified as :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 1

Oligosaccharide is homogeneous. In this, each molecule of oligosaccharide contains the same number of monosaccharide units joined together in the same order as every other molecule of the same oligosaccharide.

(3) Polysaccharides : These are carbohydrates which on hydrolysis give a large number of monosaccharides.

Polysaccharides are tasteless, amorphous, insoluble in water. They are long chain, naturally αcurring polymers of carbohydrates.

Example : Cellulose, starch, glycogen.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 8.
Classify the following carbohydrates into Monosaccharide, Disaccharide, Oligosaccharide, Polysaccharide.
(1) Glucose
(2) Starch
(3) Sucrose
(4) Maltose
(5) Galactose
(6) Lactose
(7) Ribose.
Answer:

Carbohydrates Class
(1) Glucose Monosaccharide
(2) Starch Polysaccharide
(3) Sucrose Disaccharide
(4) Maltose Disaccharide
(5) Galactose Monosaccharide
(6) Lactose Disaccharide
(7) Ribose Monosaccharide

Question 9.
Classify the following carbohydrates.
(1) Cellulose,
(2) Maltose,
(3) Raffinose,
(4) Fructose.
Answer:

Carbohydrates Class
(1)     Cellulose

(2)     Maltose

(3)     Raffinose

(4)     Fructose

Polysaccharide

Disaccharide

Trisaccharide

Monosaccharide

Question 10.
Classify the following into monosaccharides, oligosaccharides and polysaccharides.
(1) Starch
(2) Glucose
(3) Stachyose
(4) Maltose
(5) Raffinose
(6) Cellulose
(7) Sucrose
(8) Lactose.
Answer:

Monosaccharides Glucose
Oligosaccharides Stachyose, maltose, raffinose, sucrose, lactose
Polysaccharides Starch, cellulose

Question 11.
Classify the following into monosaccharides and disaccharides.
Ribose, maltose, galactose, fructose and lactose (~2 mark each)
Answer:

Monosaccharides Ribose, galactose, fructose
Disaccharides Maltose, lactose

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 12.
Give the preparation of glucose from sucrose or cane sugar.
OR
Describe the laboratory method of preparation of glucose.
Answer:
Preparation of glucose from sucrose (cane sugar) : Laboratory method.

Glucose is prepared in the laboratory by hydrolysis of sucrose by boiling it with dilute hydrαhloric acid or dilute sulphuric acid for about two hours. On hydrolysis, sucrose gives one molecule of glucose and one molecule of fructose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 5
Alcohol is added during cooling to separate glucose and fructose since, glucose is almost insoluble in alcohol, hence it crystallizes out first. Fructose remains in the solution as it is more soluble than glucose.

Crystals of glucose are separated out by filtration and purified by recrystallization.

Question 13.
Give the preparation of glucose from starch.
OR
How is glucose prepared on commercial scale?
Answer:
Commercially, on a large scale, glucose is prepared by hydrolysis of starch with dilute sulphuric acid. Starchy material is mixed with water and dilute sulphuric acid and heated at 393 K under 2 to 3-atm pressure. Starch is hydrolysed to give glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 6

Question 14.
Explain the structure of glucose.
Answer:
Molecular formula of glucose is C6H12O6.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 7
Glucose has an aldohexose structure. In other words, glucose molecule contains one aldehydic, that is, formyl group and the remaining five carbons carry one hydroxyl group (-OH) each. The six carbons in glucose form one straight chain.

Question 15.
Describe the action of following reagents on glucose :
(1) HI
(2) Hydroxyl amine (NH2OH)
(3) Hydrogen cyanide
(4) Bromine water
(5) dil. Nitric acid
(6) Acetic anhydride.
Answer:
(1) Action of HI : Glucose on prolonged heating with HI gives n-hexane, indicates that all the six carbon atoms are linked in straight chain.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 8

(2) Action of hydroxyl amine : Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of CHO group in glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 9

(3) Action of hydrogen cyanide : Glucose reacts with hydrogen cyanide to form glucose cyanohydrin.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 10

(4) Action of bromine water : Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid, which shows that the carbonyl group in glucose is aldehyde group.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 11

(5) Action of dll. nitric acid : Glucose on oxidation with dilute nitric acid forms dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic group (-CH2OH) in glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 12

(6) Action of acetic anhydride : When glucose is heated with acetic anhydride in the presence of catalyst pyridine, glucose penta acetate is formed. It indicates that glucose is a stable compound and contains five hydroxyl groups.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 13

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 16.
Write Fischer projection formulae for
(1) Glucose
(2) Gluconic acid
(3) Saccharic acid.
Answer:
Fischer projection formulae of glucose, gluconic acid and saccharic acid :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 17

Question 17.
Explain D and L configuration in sugars.
Answer:
The simplest carbohydrates glyceraldehyde is chosen as the standard, to assign D and L configuration to monosaccharides. Glyceraldehyde contains one asymmetric carbon atom and exist in two enantiomeric forms
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 18

The dextro entantiomer is represented as (+) glyceraldehyde and it is referred as D-configuration i.e., D-glyceraldehyde. The laevo enantiomer of glyceraldehyde is represented as ( -) glyceraldehyde and it corelated as L-configuration i.e., L-glyceraldehyde.

In Fischer projection formula, a monosaccharide is assigned D-configuration if the (- OH) hydroxyl group at the last chiral carbon and lies towards right hand side. On the other hand it is assigned L-configuration if the – OH group on the last chiral carbon atom and lies on the left hand side. In monosaccharides, the most oxidised carbon (i.e., -CHO) is at the top.

Examples :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 19

Question 18.
Write Fischer projection formulae for (a) L-( + )-erythrose (b) L-( +) ribulose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 23

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 19.
Is the following sugar, D-sugar or L-sugar?
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 24
Answer:
The compound is L-sugar.
The compound is L-sugar.

Question 20.
Assign D/L configuration to the following monosaccharides.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 25
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 26
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 27

Question 21.
Explain ring structure of glucose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 28
Glucose has two cyclic structures (II and III) which are in equilibrium with each other through the open chain structure (I) in aqueous solution.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

The ring structure of glucose is formed by reaction between the formyl ( – CHO) group and the alcoholic (- OH) group at C – 5. Thus, the ring structure is called a hemiacetal. The two hemiacetal structures (II and III) differ only in the configuration of C – I (Fig.), the additional chiral centre resulting from ring closure. The two ring structures are called α- and β- anomers of glucose and C-l is called the anomeric carbon. The ring of the cyclic structure of glucose contains five carbons and one oxygen. Thus, it is a six membered ring. It is called pyranose structure, in analogy with the six membered heterαyclic compound pyran (IV). Hence glucose is also called glucopyranose.

Question 22.
Write the structures of α-D-( + )-glucopyranose and β-D-( +) glucopyranose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 29

Question 23.
Explain Haworth formula of glycopyranose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 30
In the Haworth formula the pyranose ring is considered to be in a perpendicular plane with respect to the plane of paper. The carbons and oxygen in the ring are in the places as they appear in figure. The lower side of the ring is called α-side and the upper side is the β-side. The α-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the α-side, whereas the β-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the β-side. The groups which appear on right side in the Fischer projection formula appear on α-side in the Haworth formula, and the groups which appear on left side in the fischer projection formula appear on a β-side in the Haworth formula.

Question 24.
Explain the structure of fructose.
Answer:
Fructose has molecular formula C6H12O6. It contains ketonic functional group at carbon number 2 and six carbon atoms in straight chain. It belongs to D-series and is a laevo rotatory compound. It is written as D-( – )-fructose. Being an α-hydroxy keto compound fructose is a reducing sugar.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 31

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 25.
Draw mirror images of glucose and fructose.
Answer:
(1) Glucose
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 32
(2) Fructose
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 33

Question 26.
Write the two cyclic structures of α-D-( – )-fructofuranose and β-D-( – )-fructofuranose exist in equilibrium with open chain structure.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 34

Question 27.
Write the Haworth projection formulae for α -D-( -) – Fructofuranose and β – D – ( -) – Fructo- furanose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 35

Question 28.
Explain the structure of sucrose.
Answer:
Sucrose is a hexasaccharide and has molecular formula C12H22O11. The structure of shcrose contains glycosidic linkage between C – 1 of α-glucose and C – 2 of β-fructose. Since aldehyde and ketone groups of both monosaccharide units are involved in the formation of glycosidic bond, sucrose is a nonreducing sugar.

Sucrose is dextrorotatory, on hydrolysis with dilute acid or an enzyme invertase gives equimolar mixture of dextrorotatory glucose and laevorotatory fructose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 36
The solution is laevorotatory because laevo rotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.50), hence the sign of rotation is changed from (+) to (-) after hydrolysis, the product is called invert sugar.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 37

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 29.
Explain the structure of maltose.
Answer:
Maltose is another disaccharide obtained by partial hydrolysis of starch or made of two units of D-glucose. In maltose, C-l of one α-D-glucose is linked to C-4 of another α-D-glucose molecule by glycosidic linkage. The glucose ring which uses its hydroxyl group at C-1 is α – 1 → 4 glycosidic linkage. It is a reducing sugar because a free aldehyde group can be produced at C1 of second glucose molecule. Maltose on hydrolysis with dilute acids gives glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 38

Question 30.
Draw a neat diagram for Haworth formula of maltose.

Question 31.
Explain the structure of lactose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 39
Lactose (C12H22O11) is a disaccharide. It is found in milk, therefore, it is also known as milk sugar. It is formed from two monosaccharide units, namely D – galactose and D – glucose. The glycosidic linkage is formed between C-l of β-D-galactose and C -4 of glucose. Therefore the linkage in lactose is called β – 1,4 – glycosidic linkage. The hemiacetal group at C-l of the glucose unit is not involved in glycosidic linkage but is free. Hence lactose is a reducing sugar. The above figure shows Haworth formula of lactose.

Question 32.
What are the hydrolysis products of (1) lactose (2) sucrose?
Answer:
(1) Lactose on hydrolysis in presence of an acid or enzyme lactase gives one molecule each of glucose and galactose
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 40
(2) Sucrose on hydrolysis in the presence of dii. acid or the enzyme invertase gives one molecule each of glucose and fructose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 41

Question 33.
Explain the structure of starch.
Answer:
Starch is found in cereal grains, roots, tubers, potatoes, etc. It is a polymer of α-D-glucose and consists of two components, amylose and amylopectin.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Amylose is water soluble component forms blue coloured complex with iodine. It constitutes about 20 % of starch. Amylose contains 200 to 1000 α-D-glucose units linked together by glycosidic linkage between C-l of one unit and C-4 of another unit. i.e. α-1, 4 glycosidic linkages.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 43
Amylopectin is insoluble in water and constitutes about 80 % starch which forms blue-violet coloured complex with iodine. It is a branched chain polymer. In amylopectin, α-D-glucose molecules are linked together by glycosidic linkage between C1 – of one unit and C-4 of another unit to form long chain and branching αcurs by glycosidic linkage between C-l and C6 glycosidic linkage.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 44

Question 34.
What are polysaccharides?
Answer:
A large number of same or different monosaccharides are joined together by glycosidic linkages are called polysaccharides. They have general formula (C6H10O5)n.

Question 35.
Explain the structure of cellulose.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 45
Cellulose mainly αcurs in plants. Cell wall of plant cells is made up of cellulose. It is a long chain polymer. In cellulose, β-D-glucose units are linked by glycosidic linkage between C1-of one unit of glucose and C4 of another glucose unit. Thus cellulose contains 1 → 4β glycosidic linkages like those in cellobiose.

Question 36.
Explain the structure of glycogen.
Answer:
The glucose is stored in animal body in the form of glycogen. It is also known as animal starch because its structure is similar to amylopectin. Glycogen is highly branched. Whenever the body is required glucose, enzymes breaks the glycogen to glucose.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 37.
How is glycogen different from starch?
Answer:
Starch is the main storage molecules of plants whereas glycogen is the main storage molecule of animals. Starch is found in cereals, roots, tubers, etc. Glycogen is present in liver, muscles and brain.

Question 38.
What do you understand by the term glycosidic linkage?
Answer:
The linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 46

Question 39.
What is the basic structural difference between starch and cellulose?
Answer:
Starch is a polymer of a-glucose and consists of two components-amylose and amylopectin. In amylose α-D-D-( + )-glucose units held by C,-C4 glycosidic linkage and in amylopectin, α-D-glucose units held by C1-C4 glycosidic linkage whereas branching αcurs by C1-C6 glycosidic linkage. [Refer Question 35 (i) (ii) Fig.] Cellulose is a straight chain polysaccharide composed only of β-D-glucose units held by C1-C4 glycosidic linkage. (Refer Question 37 Fig.)

Question 40.
Define the term : Protein OR What are proteins?
Answer:
Chemically proteins are polyamides which are high molecular weight polymers of the monomer units i.e. α-amino acids. OR It can also be defined as Proteins are the biopolymers of a large number of a-amino acids and they are naturally occurring polymeric nitrogenous organic compounds containing 16% nitrogen and peptide linkages (-CO-NH-).

Question 41.
Write the common sources of protein.
Answer:
Common sources of proteins are milk, pulses, peanuts, eggs, fishes, cheese, cereals, etc. They are also the principal materials of muscle, nerves, tendons, skin, blood, enzymes, many hormones and antibiotics.

Question 42.
What are the products of hydrolysis of proteins?
Answer:
On hydrolysis, proteins give a mixture of α-anlino acids.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 47
The α-carbon in α-amino acids ohtained by hydrolysis of proteins has ‘L’ configuration.

Question 43.
What are the a-amino acids?
Answer:
α-Amino acids are carboxylic acids having an amino (- NH2) group bonded to the α-carbon, i.e. the carbon next to the carboxyl (- COOH) group.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 48
α-amino acids are derivatives of carboxylic acids, obtained by replacing – H atom by amino group. They are bifunctional compounds containing acidic Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 49 and basic – NH2 groups.
Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 50 (where R is an alkyl group or aryl group).

The amino acids are colourless, crystalline, water soluble, high melting solids. These acids in their aqueous solutions behave like salts due to presence of both acidic, Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 51 and basic. (- NH2) groups in the same molecule.

Such a doubly charged ion is known as zwitter ion. Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 52
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 53

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 44.
What are the final products of hydrolysis of proteins?
Answer:
Proteins on hydrolysis with dilute solution of acids, alkalies or enzymes give a mixture of large number of a-amino acids as final products.

For example :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 54

Question 45.
Write the classification of amino acids, giving examples.
Answer:
The amino acids are of three types : acidic, basic and neutral. The symbol ‘R’ in the structure of a-amino acids represents side chain and may contain additional functional groups.

(1) Acidic amino acids : If ‘R’ contains a carboxyl (- COOH) group the amino acid is acidic amino acid, i.e. If carboxyl groups are more in number than amino groups, then amino acids are acidic in nature.

Examples : Glutamic acid HOOC-CH2-CH2-; Aspartic acid HOO-CH2

(2) Basic amino acids : If ‘R’ contains an amino (1°, 2°, or 3°) group, it is called basic amino acid i.e. If amino groups are more in number than carboxyl groups then amino acids are basic in nature.

Examples : Arginine Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 55

(3) Neutral amino acids : The other amino acids having neutral or no functional group in ‘R’ are called neutral amino acids, i.e. The amino acids having equal number of amino and carboxyl groups are neutral amino acids.

Examples : Alanine CH3-; Valine (CH3)2-CH

Question 46.
What are essential and non-essential amino acids? Give two examples of each.
Answer:
The amino acids, which cannot be synthesised in the body and are supplied through diet are called essential amino acids. Examples : Lysine H2N-(CH2)4-; Valine (CH3)2CH- The amino acids which are synthesized in the body are called non-essential amino acids.

Examples : Glutamic acid HOO-CH2-CH2-; Serine HO-CH2

Question 47.
What is meant by Zwitter ion?
Answer:
An a-amino acid molecule contains both acidic carboxyl ( – COOH) group as well as basic amino (- NH2) group. Proton transfer from acidic group to basic group of amino acid forms a salt, which is a dipolar ion called a zwitterion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 61

Question 48.
Draw zwitter ion of alanine and other two forms.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 62

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 49.
What is a peptide bond (peptide linkage)?
OR
Define peptide bond.
Answer:
Proteins are the polymers of a-amino acids and they are connected to each other. The bond that connects a-amino acids to each other is called peptide bond (peptide linkage, – CONH -).

Question 50.
How is peptide linkage (dipeptide linkage) formed in proteins? How is tripeptide formed?
Answer:
Peptide linkage is formed by condensation of acidic Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 63 group of one molecule of a-amino acid and basic -NH2 group of other molecule of α-amino acid with elimination of water.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 64

When one more molecule of amino acid combines with dipeptide, it forms tripeptide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 65
Thus, it forms tetra, penta and finally a polypeptide chain i.e. proteins. Hence, proteins are basically polypeptides.

Question 51.
Write the structures of all possible dipeptides which can be obtained from glycine and alanine.
Answer:
(1) Dipeptide from glycine :
Carboxylic group of glycine reacting with amino group another molecule of glycine to form dipeptide
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 73
(2) Dipeptide from alanine :
Carboxylic goup of alanine reaction with amino goup of another molecule of alamine to form dipeptide
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 74
(3) Dipeptide from glycine and alanine :
Carboxylic group of glycine reacting with amino group another molecule of alanine to form dipeptide
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 75

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 52.
How are proteins classified on the basis of molecular shapes?
Answer:
On the basis of their molecular shapes proteins are classified as :
(1) Fibrous proteins : The proteins in which the polypeptide chains lie parallel (side by side) to form fibre-like structure, are called fibrous proteins. The polypeptide chains held together by hydrogen bonds. These proteins are insoluble in water.

The fibrous proteins are tough and insoluble in water, and dilute acids or bases.

Example : myαin (in muscles), keratin (in hair, nails, skin), fibroin (in silk), collagen (in tendons), etc.

(2) Globular proteins : The proteins have spherical shape. This shape results from coiling around of the polypeptide chain of protein, and have intramolecular hydrogen bonding are called globular proteins.

They are soluble in water and dilute acids or bases.

Example : Haemoglobin (in blood), albumin (in eggs), insulin (in pancreas), etc.

Question 53.
Distinguish between globular and fibrous proteins.
Answer:

Globular proteins Fibrous proteins
(1) The chains of polypeptides of protein coil around to give a spherical shape.
(2) Globular proteins are soluble in water.
(3) They are sensitive to small changes of temperature and pH.
(4) They possess biological activity.
(1) The proteins in which the polypeptide chains lie parallel to form fibre like structure.
(2) Fibrous proteins are insoluble in water.
(3) They are stable to moderate changes of temperature and pH.
(4) They do not possess biological activity.

Question 54.
Draw a neat labelled diagram for the secondary structure of protein.
Answer:
Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 78
(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 79

  • The C = O and N – H bonds lie in the planes of the sheet.
  • Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
  • The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

Question 55.
What is denaturation of proteins? How is denaturation brought about?
OR
What is the effect of denaturation on the structure of proteins?
Answer:
The prαess by which the molecular shape of protein changes without breaking the amide / peptide bonds that form the primary structure is called denaturation. OR Proteins gets easily precipitated. It is an irreversible change and the prαess is called denaturation of proteins.

Denaturation uncoils the protein and destroys the shape and thus loses their characteristic biological activity. Denaturation is brought about by heating the protein with alcohol, concentrated inorganic acids or by salts of heavy metals. During denaturation secondary and tertiary and quternary structures are destroyed but primary structure remains intact.

Example : Boiling of egg to coagulate egg white, conversion of milk to curd.

Question 56.
Define : Enzymes
Answer:
All biological reactions are catalysed by bio-catalyst in living organisms called enzymes.

Question 57.
What are enzymes? Explain with suitable example.
Answer:
All biological or bio-catalysts which catalyse the reactions in living organisms are called enzymes. Chemically all enzymes are proteins. They are required in very small quantities as they are catalyst also they reduce the activation energy for a particular reaction.

Example : Enzyme maltase converts maltose to glucose.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 84

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 58.
Explain the catalytic action of enzymes.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 85
Answer:
Mechanism of enzyme catalysis : Action of an enzyme on a substrate is known as lock-and-key mechanism.

Accordingly, the enzyme has active site on its surface. A substrate molecule can attach to this active site only if it has the right size and shape. Once in the active site, the substrate is held in the correct orientation, enzymes provide functional group which will attack the substrate and forms the products of reaction. The products leave the active site and the enzyme is ready to act as catalyst again.

Question 59.
Give examples of industrial application of enzyme catalysis.
Answer:

  • Glucose Isomerase (enzyme) is used in conversion of glucose to sweet-tasting fructose.
  • New antibiotics are manufactured using penicillin acylase (enzyme).
  • Laundry detergentts are manufactured using proteases (enzyme).
  • Esters used in cosmetics are manufactured using genetically engineered enzyme.

Question 60.
Draw a neat diagram for enzyme catalysis.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 85

Question 61.
State the main functions of enzymes.
Answer:
Enzymes are biological catalyst and they are highly specific in nature. The two main functions are as follows :
(1) They lower the requirement of activation energy.
(2) They speed up the rate of reaction.
E.g. Enzyme maltase catalyses maltose to glucose.
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Maltase }}{\longrightarrow} 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 62.
What are nucleic acids?
Answer:
Nucleic acids are unbranched polymers of repeating monomers i.e. nucleotides. In other words, nucleic acids have a polynucleotide structure which in turn consists of a base, a pentose sugar and phosphate moiety.

Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins or chromosomes.

(Nucleoproteins = Proteins + Nucleic acid)
(prosthetic group)

Question 63.
State the types of nucleic acids.
Answer:
The types of nucleic acids are : Ribonucleic acid (RNA) and deoxy ribonucleic acid (DNA). DNA molecules contain several million nucleotides while RNA molecules contain a few thousand nucleotides.

Question 64.
Explain chemical composition of nucleic acids.
Answer:
Nucleic acids have a polynucleotide structure. Nucleic acids (RNA and DNA) consists of three components :
(1) monosaccharide (sugar)
(2) nitrogen containing base and
(3) phosphate group.

(1) Monosaccharides : Nucleotides of both RNA consist of five membered monosaccharide ring (furanose), called as simply sugar component.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 86
In RNA, the sugar component of nucleotide units is D-ribose and in DNA 2-deoxy-D-ribose.
2 – deoxy means no – OH group at C2 position.

(2) Nitrogen containing base : Total five nitrogen – containing bases are present in nucleic acids. Three bases with one ring (cytosine, uracil and thymine) are derived from the parent compound pyrimidine. Two bases with two rings (adenine and guanine) are derived from the parent compound purine. Each base in designated by a one-letter symbol. Uracil (U) αcurs only in RNA while thymine (T) ocurs only in DNA.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 87

(3) Phosphate group : The sugar units are joined to phosphate through C3 and C5 hydroxyl groups.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 65.
What is meant by nucleosides?
OR
Write the structure of nucleoside. Give examples.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 89
A nucleoside contains two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base.

A nucleoside is formed when 1 -position of a pyrimidine (cytosine, thymine or uracil) or 9-position of guanine or adenine base is attached to C- l of sugar by β-linkage.

Examples: Formation of nucleoside:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 90

Question 66.
What is meant by nucleotide?
OR
Write the structure of nucleotide. Give example.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 91
A nucleotide contains all three basic components of nucleic acids i.e., a pentose sugar, a phosphoric acid and a nitrogenous base. These are obtained by esterification of \(\mathrm{C}_{5}^{1}-\mathrm{OH}\) group of the pentose sugar by phosphoric acid. Nucleotides are joined together through phosphate ester linkage. Thus, nucleotides are monophosphates of nucleosides. Abridged names of some nucleotides are AMP, dAMP, UMP, dTMP and so on. Here, the first capital letter is derived from the corresponding base. MP stands for monophosphate. Small letter ‘d’ in the beginning indicates deoxyribose in the nucleotide.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Example :
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 92

 

Question 67.
Write the structure of nucleic acids.
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 95
Answer:
Nucleic acids, both DNA and RNA, are polymers of nucleotides, formed by joining the 3′ – OH group of one nucleotide with 5′ – phosphate of another nucleotide. Two ends of polynucleotide chain are distinct from each other. One end having free phosphate group of 5′ position is called 5′ end. The other end is 3′ end and has free OH – group at 3′ position.

Question 68.
Draw a schematic representation of polynucleotide structure of nucleic acids.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 96

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

Question 69.
Explain double helix.
OR
State the salient features of the Watson and Crick mode of DNA.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 98
The Salient features are :

  1. DNA is made of two polynucleotide strands that wind into a right-handed double helix.
  2. The two strands run in opposite directions: one from the Y end to the 3’ end, while the other from the 3’ end to the Y end.
  3. Pcrpcndicular to the axis of the helix, the sugar – phosphate backbone lies on the outside of the helix and the bases lic on the inside.
  4. The hydrogen bonding between the hases of the two DNA strands stabilizes the double helix. This gives rise to a ladder-like structure of DNA double helix.
  5. Adenine always forms two hydrogen bonds with thymine and guanine forms three hydrogen bonds with cytosinc. Thus A – T arid C – G arc complementary hase pairs and the Two strands of the double helix arc complementary to each other.

Question 70.
Give scientific reasons :
1. In the preparation of glucose from sucrose, ethyl alcohol is added at the time of cooling.
Answer:
Hydrolysis of sucrose with dilute hydrαhloric acid gives glucose along with fructose.

Ethyl alcohol is added at the time of cooling in the preparation of glucose, to separate glucose from fructose. Glucose being insoluble in alcohol, crystallizes out first, while fructose being more soluble in alcohol, remains in the solution.

Question 71.
Answer in one sentence :

(1) How is glucose stored in the animal body?
Answer:
Glucose is stored in the form of glycogen in the animal body.

(2) Write other term used for carbohydrates.
Answer:
Carbohydrates are often termed as saccharides or sugars.

(3) How many moles of acetic anhydride will be required to form glucose penta acetate from 1 mole of glucose?
Answer:
10 moles of acetic anhydride.

(4) What are reducing sugars?
Answer:
Reducing sugars : Carbohydrates which reduce Fehling solution to red ppt of Cu20 or Tollen’s reagent to shining metallic silver are called reducing sugars. All monosaccharides and oligosaccharides except sucrose are reducing sugars.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

(5) What are non-reducing sugars?
Answer:
Non-reducing sugars : Carbohydrates which do not reduce Fehling solution and Tollen’s reagent are called non-reducing sugars. E.g. sucrose.

(6) Give an example each of reducing and non-reducing sugars.
Answer:
Reducing sugars : Maltose or lactose
Non-reducing sugars : Sucrose.

(7) Name the linkage which joins two monosaccharide units through oxygen atom.
Answer:
The linkage which joins two monosaccharide units through oxygen atom is called glycosidic linkage.

(8) Name the sugar present in DNA.
Answer:
The sugar present in DNA is deoxyribose.

(9) A nucleotide from DNA containing thymine is hydrolysed. What are the products formed?
Answer:
When nucleotide from DNA containing thymine is hydrolysed, 2-deoxy-D-ribose, thymine and phosphoric acid is obtained.

(10) How is zwitterion formed?
Answer:
In aqueous solution, the carboxyl group loses a proton while the amino group accepts it, as a result, a dipolar or zwitter ion is formed.

(11) Name the amino acids which are synthesized in the body.
Answer:
The amino acids which are synthesized in the body are called non-essential amino acids. Examples : Glutamic acid, serine.

(12) Name the four bases present in DNA which of these is not present in RNA.
Answer:
Purines-adenine (A) and guanine (G); Pyrimidines-thymine (T) and cytosine (C), these four bases are present in DNA. Out of these, thymine (T) is not present in RNA.

(13) What are different types of RNA which are found in the cell?
Answer:
There are three different types of RNA found in the cell. (1) The messenger RNA which carries the message to the ribosome (2) Ribosomal RNA where synthesis of protein takes place (3) The transport RNA.

(14) State the functions of RNA and DNA.
Answer:
RNA and DNA are responsible for generic characteristics : DNA preserves the information and uses it by producing duplicate identical DNA molecules. RNA carries messages and transports them.

Multiple Choice Questions

Question 72.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which of the following is not sugar?
(a) Sucrose
(b) Starch
(c) Fructose
(d) Glucose
Answer:
(b) Starch

2. Which of the following is the example of disaccharide?
(a) Glucose
(b) Raffinose
(c) Cellulose
(d) Sucrose
Answer:
(d) Sucrose

3. Fructose is
(a) aldopentose
(b) aldohexose
(c) ketopentose
(d) ketohexose
Answer:
(d) ketohexose

4. Oxidation product of glucose with bromine water is
(a) sorbitol
(b) gluconic acid
(c) glutamic acid
(d) saccharic acid
Answer:
(b) gluconic acid

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

5. The general formula of carbohydrates is
(a) C(H2O)
(b) Cx(H2O)y
(c) Cx(H2O)
(d) Cx(H2O)x
Answer:
(b) Cx(H2O)y

6. Monosaccharides containing aldehyde group are called
(a) aldoses
(b) ketoses
(c) polysaccharides
(d) disaccharides
Answer:
(a) aldoses

7. Which of the following sugars can be used to prepare glucose on a large scale?
(a) Cellulose
(b) Cane sugar
(c) Galactose
(d) Starch
Answer:
(d) Starch

8. Which of the following carbohydrates cannot undergo hydrolysis?
(a) Glucose
(b) Sucrose
(c) Cellulose
(d) Maltose
Answer:
(a) Glucose

9. Glucose differs from fructose in
(a) the functional group
(b) the number of chiral carbon atoms
(c) the number of carbon atoms
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

10. The example of aldopentose is
(a) arabinose
(b) glucose
(c) fructose
(d) sucrose
Answer:
(a) arabinose

11. Dextrose, grape sugar and blood sugar αcurs in
(a) fructose
(b) glucose
(c) sucrose
(d) starch
Answer:
(b) glucose

12. The example of ketopentose is
(a) galactose
(b) ribose
(c) raffinose
(d) maltose
Answer:
(b) ribose

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

13. Cane sugar on hydrolysis gives
(a) glucose and maltose
(b) glucose and lactose
(c) glucose and fructose
(d) only glucose
Answer:
(c) glucose and fructose

14. On commerical scale, glucose is prepared from
(a) starch
(b) potato pulp
(c) sucrose
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

15. The number of monosaccharide units formed on hydrolysis of glucose are
(a) zero
(b) one
(c) two
(d) three
Answer:
(a) zero

16. Which of the following is NOT TRUE about glucose?
(a) It is monosaccharide
(b) It is a polyhydroxy aldehyde
(c) It is polyhydroxy ketone
(d) It contains six carbon atoms
Answer:
(c) It is polyhydroxy ketone

17. Final hydrolysis product of simple protein is
(a) carboxylic acid
(b) α-amino acid
(c) mineral acid
(d) acetic acid
Answer:
(b) α-amino acid

18. Haemoglobin is the example of-
(a) simple protein
(b) derived protein
(c) fibrous protein
(d) conjugated protein
Answer:
(d) conjugated protein

19. Protein are also called
(a) polysaccharides
(b) polypeptides
(c) polyglycerides
(d) polyster
Answer:
(b) polypeptides

20. The simplest amino acid is
(a) glycine
(b) oxalic acid
(c) adipic acid
(d) caprolactam
Answer:
(a) glycine

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

21. Amino acids usually exist in the form of zwitter ion which consist of
(a) the basic group-NH2 and the acidic group -COOH
(b) the acidic group -N+H3 and the basic group COO
(c) the acidic group -COO+ and the acidic group NH3-
(d) acidic or basic group
Answer:
(b) the acidic group -N+H3 and the basic group COO-

22. The water insoluble protein is
(a) casein of milk
(b) albumin
(c) serum albumin
(d) keratin of hair
Answer:
(d) keratin of hair

23. The main structural feature of a protein molecule is the presence of
(a) an ester linkage
(b) an ether linkage
(c) a peptide linkage
(d) all of these
Answer:
(c) a peptide linkage

24. Milk sugar is
(a) sucrose
(b) lactose
(c) maltose
(d) glucose
Answer:
(b) lactose

25. The carbohydrates used for silvering of mirror is
(a) fructose
(b) starch
(c) glucose
(d) cellulose
Answer:
(c) glucose

26. Which one of the following is NOT produced by human body?
(a) DNA
(b) Hormones
(c) Enzymes
(d) Vitamins
Answer:
(c) Enzymes

27. A biological catalyst is essentially
(a) an amino acid
(b) an enzyme
(c) a nitrogen molecule
(d) a carbohydrate
Answer:
(d) a carbohydrate

28. Which one of the following is not a constituent of RNA?
(a) Ribose
(b) Uracil
(c) Thymine
(d) Phosphate
Answer:
(b) Uracil

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

29. DNA is a polymer of units of
(a) sugars
(b) ribose
(c) amino acids
(d) nucleotides
Answer:
(c) amino acids

30. Which one of the following molecules will form zwitter ion?
(a) CH3COOH
(b) CH3CH2NH2
(c) CCl3NO2
(d) NH2CH2COOH
Answer:
(d) NH2CH2COOH

31. In metabolic prαess the maximum energy is given by
(a) carbohydrates
(b) proteins
(c) vitamins
(d) fats
Answer:
(d) fats

32. DNA has a structure of helix was reported by
(a) Herman Fischer
(b) Fedrick Sauger
(c) Andreas Marggraf
(d) Watson and Crick
Answer:
(d) Watson and Crick

33. The secondary structure of a protein is determined by
(a) co-ordinate bond
(b) covalent bond
(c) ionic bond
(d) hydrogen bond
Answer:
(d) hydrogen bond

34. In maltose, glycosidic linkage is present between the two glucose units at positions
(a) 1, 2
(b) 1, 1
(c) 1, 3
(d) 1, 4
Answer:
(d) 1, 4

35. Which of the following amino acids is basic in nature?
(a) Valine
(b) Tyrosine
(c) Arginine
(d) Luecine
Answer:
(c) Arginine

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

36. Sucrose molecules consists of
(a) a glucofuranose and a fructopyranose
(b) a glucofuranose and a fructofuranose
(c) a glucopyranose and a fructopyranose
(d) a glucopyranose and a’ fructofuranose
Answer:
(d) a glucopyranose and a’ fructofuranose

37. Which one of the following statements is not correct about DNA molecule?
(a) It has double helix structure
(b) It serves as hereditary material
(c) The two DNA strands are exactly similar
(d) Its replication is called semi-conservative mode of replication
Answer:
(c) The two DNA strands are exactly similar

38. Glycine on heating forms
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 107
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 108
Answer:
(a)

39. Acidic amino acid is
(a) Glutamine
(b) Glutamic acid
(c) Tyrosine
(d) Lysine
Answer:
(b) Glutamic acid

40. Basic amino acid is
(a) Lysine
(b) Glycine
(c) Cystine
(d) Alanine
Answer:
(a) Lysine

41. Precipitation of protein is referred to as
(a) destruction of proteins
(b) separation of proteins
(c) denaturation of proteins
(d) fragmentation of proteins
Answer:
(c) denaturation of proteins

42. An amino acid containing sulphur is
(a) serine
(b) cysteine
(c) valine
(d) asparagine
Answer:
(b) cysteine

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

43. Rhamnose is a
(a) carbohydrate
(b) protein
(c) lipid
(d) vitamin
Answer:
(a) carbohydrate

44. Lactose on hydrolysis gives
(a) glucose + glucose
(b) glucose + fructose
(c) glucose + galactose
(d) fructose + galactose
Answer:
(c) glucose + galactose

45. Raffinose on hydrolysis gives
(a) glucose + glucose + galactose
(b) glucose + fructose + galactose
(c) glucose + galactose + galactose
(d) fructose + galactose + galactose
Answer:
(b) glucose + fructose + galactose

46. Naturally αcurring glucose is
(a) dextro rotatory
(b) laevo rotatory
(c) racemic mixture
(d) all of these
Answer:
(a) dextro rotatory

47. Amylopectin is
(a) soluble in water and constitutes about 80% of starch
(b) insoluble in water and constitutes about 80% of starch
(c) Soluble in alcohol and constitutes about 60% of starch
(d) in soluble in alcohol and constitutes about 60% of starch
Answer:
(b) insoluble in water and constitutes about 80% of starch

48. Insulin contains
(a) 51 amino acids
(b) 151 amino acids
(c) 15 amino acids
(d) 115 amino acids
Answer:
(a) 51 amino acids

49. Pyranose structure of glucose is
(a) an open chain structure of glucose
(b) a structure of reduction product of glucose
(c) a cyclic six-membered structure of glucose
(d) a four-membered cyclic form of glucose
Answer:
(c) a cyclic six-membered structure of glucose

50. The number of – OH groups present in ribulose is
(a) 3
(b) 4
(c) 6
(d) 5
Answer:
(b) 4

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

51. Peptide linkage is
Maharashtra Board Class 12 Chemistry Solutions Chapter 14 Biomolecules 109
Answer:
(d)

52. Stachyose is an example of
(a) monosaccharides
(b) disaccharides
(c) trisaccharides
(d) tetrasaccharides
Answer:
(d) tetrasaccharides

53. How many moles of (CH3CO)2O will be required to form glucose pentaacetate form 2 moles of glucose?
(a) 2
(b) 5
(c) 10
(d) 2.5
Answer:
(c) 10

54. Which of the following NOT present in DNA?
(a) Adenine
(b) Guanine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

Maharashtra Board Class 12 Chemistry Important Questions Chapter 14 Biomolecules

55. Maltose is a
(a) polysaccharide
(b) disaccharide
(c) trisaccharide
(d) monosaccharide
Answer:
(b) disaccharide

12th Chemistry Chapter 4 Exercise Chemical Thermodynamics Solutions Maharashtra Board

Class 12 Chemistry Chapter 4

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 4 Chemical Thermodynamics Textbook Exercise Questions and Answers.

Chemical Thermodynamics Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 4 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 4 Exercise Solutions

1. Select the most apropriate option.

Question 1.
The correct thermodynamic conditions for the spontaneous reaction at all temperatures are
(a) ΔH < 0 and ΔS > 0
(b) ΔH > 0 and ΔS < 0
(c) ΔH < 0 and ΔS < 0
(d) ΔH < 0 and ΔS = 0
Answer:
(a) ΔH < 0 and ΔS > 0

Question ii.
A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, ΔU of the gas will be
(a) -500 J
(b) +500J
(c) -1013 J
(d) +1013 J
Answer:
(a) -500 J

Question iii.
In which of the following, entropy of the system decreases ?
(a) Crystallisation of liquid into solid
(b) Temperature of crystalline solid is increased from 0 K to 115 K
(c) H2(g) → 2H(g)
(d) 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
Answer:
(a) Crystallisation of liquid into solid

Question iv.
The enthalpy of formation for all elements in their standard states is
(a) unity
(b) zero
(c) less than zero
(d) different elements
Answer:
(b) zero

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question v.
Which of the following reactions is exothermic ?
(a) H2(g) → 2H(g)
(b) C(s) → C(g)
(c) 2Cl(g) → Cl2(g)
(d) H2O(s) → H2O(l)
Answer:
(c) 2Cl(g) → Cl2(g)

Question vi.
6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat. Enthalpy of vaporization of ethanol will be
(a) 43.4 kJ mol-1
(b) 60.2 kJ mol-1
(c) 38.9 kJ mol-1
(d) 20.4 kJ mol-1
Answer:
(a) 43.4 kJ mol-1

Question vii.
If the standard enthalpy of formation of methanol is -238.9 kJ mol-1 then entropy change of the surroundings will be
(a) -801.7 JK-1
(b) 801.7 JK-1
(c) 0.8017 JK-1
(d) -0.8017 JK-1
Answer:
(b) 801.7 JK-1

Question viii.
Which of the following are not state functions ?
1. Q + W 2. Q 3. W 4. H-TS
(a) 1, 2 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2, 3 and 4
Answer:
(b) 2 and 3

Question ix.
For vaporization of water at 1 bar, ΔH = 40.63 kJ mol-1 and ΔS =108.8 JK-1 mol-1. At what temperature, ΔG = 0?
(a) 273.4 K
(b) 393.4 K
(c) 373.4 K
(d) 293.4 K
Answer:
(c) 373.4 K

Question x.
Bond enthalpies of H – H, Cl – Cl and H – Cl bonds are 434 kJ mol-1, 242 kJ mol-2 and 431 kJ mol-1, respectively. Enthalpy of formation of HCl is
(a) 245 kJ mol-1
(b) -93 kJ mol-1
(c) -245 kJ mol-1
(d) 93 kJ mol-1
Answer:
(b) -93 kJ mol-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

2. Answer the following in one or two sentences.

Question i.
Comment on the statement: No work is involved in an expansion of a gas in vacuum.
Answer:
(1) When a gas expands against an external pressure Pex, changing the volume from V1 to V2, the work obtained is given by
W = -Pex (V2 – V1).
(2) Hence the work is performed by the system when it experiences the opposing force or pressure.
(3) Greater the opposing force, more is the work.
(4) In free expansion, the gas expands in vaccum where it does not experience opposing force, (P = 0). Since external pressure is zero, no work is obtained.
∴ W = -Pex (V2 – V1)
= -0 × (V2 – V1)
= 0
(5) Since during expansion in vacuum no energy is expended, it is called free expansion.

Question ii.
State the first law of thermodynamics.
Answer:
The first law of thermodynamics is based on the principle of conservation of energy and can be stated in different ways as follows :

  1. Energy can neither be created nor destroyed, however, it may be converted from one form into another.
  2. Whenever, a quantity of one kind of energy is consumed or disappears, an equivalent amount of another kind of energy appears.
  3. The total mass and energy of an isolated system remain constant, although there may be interconservation of energy from one form to another.
  4. The total energy of the universe remains constant.

Question iii.
What is enthalpy of fusion?
Answer:
Enthalpy of fusion (ΔfusH) : The enthalpy change that accompanies the fusion of one mole of a solid into a liquid at constant temperature and pressure is called enthalpy of fusion.
For example,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 1
This equation describes that when one mole of ice melts (fuses) at 0 °C (273 K) and 1 atmosphere, 6.1 kJ of heat will be absorbed.

Question iv.
What is standard state of a substance?
Answer:
The thermodynamic standard state of a substance (compound) is the most stable physical state of it at 298 K and 1 atmosphere (or 1 bar). The enthalpy of the substance in the standard state is represented as ΔfH0.

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question v.
State whether ∆S is positive, negative or zero for the reaction 2H(g) → H2(g). Explain.
Answer:
(i) The given reaction, 2H(g) → H2(g) is the formation of H2(g) from free atoms.
(ii) Since two H atoms form one H2 molecule, ∆n = 1 – 2= -1 and disorder is decreased. Hence entropy change ∆S < 0 (or negative).

Question vi.
State second law of thermodynamics in terms of entropy.
Answer:
The second law of thermodynamics states that the total entropy of the system and its surroundings (universe) increases in a spontaneous process.
OR
Since all the natural processes are spontaneous, the entropy of the universe increases.
It is expressed mathematically as
∆ STotal = ∆ Ssystem + ∆Ssurr > 0
∆ SUniverse = ∆ Ssystem + ∆ Ssurr > 0

Question vii.
If the enthalpy change of a reaction is ∆H how will you calculate entropy of surroundings?
Answer:
(i) For endothermic reaction, ∆H > 0. This shows the system absorbs heat from surroundings.
∴ ∆surr H < 0.
∵ Entropy change = ∆surr S = \(\frac{-\Delta_{\text {surr }} H}{T}\)
There is decrease in entropy of surroundings.
(ii) For exothermic reaction, ∆H < 0, hence for surroundings, ∆surr H > 0

∴ ∆surr > 0.

Question viii.
Comment on spontaneity of reactions for which ∆H is positive and ∆S is negative.
Answer:
Since ∆H is +ve and ∆S is -ve, ∆G will be +ve at all temperatures. Hence reactions will be non-spontaneous at all temperatures.

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

3. Answer in brief.

Question i.
Obtain the relationship between ∆G° of a reaction and the equilibrium constant.
Answer:
Consider following reversible reaction, aA + bB ⇌ cC + dD
The reaction quotient Q is,
Q = \(\frac{[\mathrm{C}]^{c} \times[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a} \times[\mathrm{B}]^{b}}\)
The free energy change ∆G for the reaction is ∆G = ∆G° + RT in Q
Where ∆G° is the standard free energy change.
At equilibrium
Q = \(\frac{[\mathrm{C}]_{e}^{c} \times[\mathrm{D}]_{e}^{d}}{[\mathrm{~A}]_{e}^{a} \times[\mathrm{B}]_{e}^{b}}=\mathrm{K}\)
∴ ∆G = ∆G° + RT In K
∵ at equilibrium ∆G = 0
∴ 0 = AG° + RT In K
∴ ∆G° = -RT In K
∴ ∆G°= -2.303 RT log10K.

Question ii.
What is entropy? Give its units.
Answer:
(i) Entropy : Being a state function and thermodynamic function it is defined as entropy change (∆S) of a system in a process which is equal to the amount of heat transferred in a reversible manner (Qrev) divided by the absolute temperature (T), at which the heat is absorbed. Thus,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 2
(ii) Units of entropy are JK-1 in SI unit and cal K-1 in c.g.s. units. It is also expressed in terms of entropy unit (e.u.). Hence 1 e.u. = 1 JK-1.
(iii) Entropy is a measure of disorder in the system. Higher the disorder, more is entropy of the system.

Question iii.
How will you calculate reaction enthalpy from data on bond enthalpies?
Answer:
(i) In chemical reactions, bonds are broken in the reactant molecules and bonds are formed in the product molecules.
(ii) Energy is always required to break a chemical bond while energy is always released in the formation of the bond.
(iii) The enthalpy change of a gaseous reactions (ΔfH0) involving substances with covalent bonds can be calculated with the help of bond enthalpies of reactants and products. (In case of solids we need lattice energy or heat of sublimation while in case of liquids we need heat of evaporation.)
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 3
If the energy required to break the bonds of reacting molecules is more than the energy released in the bond formation of the products, then the reaction will be endothermic and ∆H0 reaction will be positive. On the other hand if the energy released in the bond formation of the products is more than the energy required to break the bonds of reacting molecules then the reaction will be exothermic and ∆H0 reaction will be negative.

Question iv.
What is the standard enthalpy of combustion ? Give an example.
Answer:
Standard enthalpy of combustion or standard heat of combustion : it is defined as the enthalpy change when one mole of a substance in the standard state undergoes complete combustion in a sufficient amount of oxygen at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by ∆cH0.
E.g. CH3OH(l) + \(\frac {3}{2}\) O2(g) = CO2(g) + 2H2O
cH0= -726 kJ mol-1
(∆cH0 is always negative.)
[Note : Calorific value : It is the enthalpy change or amount of heat liberated when one gram of a substance undergoes combustion.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 4

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question v.
What is the enthalpy of atomization? Give an example.
Answer:
Enthalpy of atomisation (∆atomH) : it is the enthalpy change accompanying the dissociation of one mole of gaseous substance into its atoms at constant temperature and pressure.
For example : CH4(g) → C(g) + 4H(g)atomH = 1660 kJ mol-1

Question vi.
Obtain the expression for work done in chemical reaction.
Answer:
Consider n1 moles of gaseous reactants A of volume V1 change to n2 moles of gaseous products B of volume V2 at temperature T and pressure P.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 5
In the initial state, PV1 = n1RT
In the final state, PV2 = n2RT
PV2 – PV1 = n2RT – n1RT = (n2 – n1)RT = ∆nRT
where ∆n is the change in number of moles of gaseous products and gaseous reactants.
Due to net changes in gaseous moles, there arises change in volume against constant pressure resulting in mechanical work, -P∆V.
∴ W = -P∆V = -P(V2 – V1) = – ∆nRT
(i) If n1 – n2, ∆n = 0, W = 0. No work is performed.
(ii) If n2 > n1, ∆n > 0, there is a work of expansion by the system and W is negative.
(iii) If n2 < n1, ∆n < 0, there is a work of compression, hence work is done on the system and W is positive.

Question vii.
Derive the expression for PV work.
Answer:
Consider a certain amount of an ideal gas enclosed in an ideal cylinder fitted with massless, frictionless rigid movable piston at pressure P, occupying volume V1 at temperature T.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 6
Fig. 4.8 : Work of expansion
As the gas expands, it pushes the piston upward through a distance d against external force F, pushing the surroundings.
The work done by the gas is,
W = opposing force × distance = -F × d
-ve sign indicates the lowering of energy of the system during expansion.
If a is the cross section area of the cylinder or piston, then,
W = \(-\frac{F}{a}\) × d × a
Now the pressure is Pex = \(\frac{F}{a}\)
while volume change is, ΔV = d × a
∴ W = -Pex × ΔV
If during the expansion, the volume changes from V1 and V2 then, ΔV = V2 – V1
∴ W= -Pex(V2 – V1)
During compression, the work W is +ve, since the energy of the system is increased,
W = +Pex(V2 – V1)

Question viii.
What are intensive properties? Explain why density is intensive property.
Answer:
(A) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.
Explanation :

  1. Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
  2. The intensive properties are not additive.

(B) Density is a ratio of two extensive properties namely, mass and volume. Since the ratio of two extensive properties represents an intensive property, density is an intensive property. It does not depend on the amount of a substance.

Question ix.
How much heat is evolved when 12 g of CO reacts with NO2 ? The reaction is :
4 CO(g) + 2 NO2(g) → 4CO2(g) + N2(g), ∆H0 = -1200 kJ

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

4. Answer the following questions.

Question i.
Derive the expression for the maximum work.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 7
Consider n moles of an ideal gas enclosed in an ideal cylinder fitted with a massless and frictionless movable rigid piston. Let V be the volume of the gas at a pressure P and a temperature T.
If in an infinitesimal change pressure changes from P to P – dP and volume increases from V to V + dV. Then the work obtained is, dW = -(P-dP) dV
= -PdV + dPdV
Since dP.dV is negligibly small relative to PdV
dW= -PdV
Let the state of the system change from A(P1, L1) to B (P2, V2) isothermally and reversibly, at temperature T involving number of infinitesimal steps.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 8
Then the total work or maximum work in the process is obtained by integrating above equation.
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 9
At constant temperature,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 10
where n, P, V and T represent number of moles, pressure, volume and temperature respectively. For the process,
ΔU = 0, ΔH = 0.
The heat absorbed in reversible manner
Qrev, is completely converted into work.
Qrev = -Wmax.
Hence work obtained is maximum.

Question ii.
Obtain the relatioship between ∆H and ∆U for gas phase reactions.
Answer:
Consider a reaction in which n1 moles of gaseous reactant in initial state change to n2 moles of gaseous product in the final state.
Let H1, U1, P1, V1 and H2, U2, P2, V2 represent enthalpies, internal energies, pressures and volumes in the initial and final states respectively then,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 11
The heat of reaction is given by enthalpy change ΔH as,
ΔH = H2 – H1
By definition, H = U + PV
∴ H1 = U1 + P1V1 and H2 = U2+ P2V2
∴ ΔH = (U2 + P2V2) – (U1 + P1V1)
= (U2 – U1) + (P2V2 – P1V1)
Now, ΔU = U2 – U1
Since PV = nRT,
For initial state, P1V1= n1RT
For final state, P2V2 = n2RT
∴ P2V2 – P1V1 = n2RT – n1RT
= (n2 – n1) RT
= ΔnRT
where Δn
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 12
∴ ΔH = ΔU + ΔnRT
If QP and QV are the heats involved in the reaction at constant pressure and constant volume respectively, then since QP = ΔH and QV = ΔU.
∴ QP = QV = ΔnRT

Question iii.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
Answer:
Statement of law of constant heat summation : It states that, the heat of a reaction or the enthalpy change in a chemical reaction depends upon initial state of reactants and final state of products and independent of the path by which the reaction is brought about (i.e. in single step or in series of steps).
OR
Heat of reaction is same whether it is carried out in one step or in several steps.
Explanation :
Consider the formation of CO2(g).
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 13
Hess’s law treats thermochemical equations mathematically i.e., they can be added, subtracted or multiplied by numerical factors like algebraic equations.

Applications : Hess’s law is used for :

  1. To calculate heat of formation, combustion, neutralisation, ionization, etc.
  2. To calculate the heat of reactions which may not take place normally or directly.
  3. To calculate heats of extremely slow or fast reactions.
  4. To calculate enthalpies of reactants and products.

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question iv.
Although ∆S for the formation of two moles of water from H2 and O2 is -327 JK-1, it is spontaneous. Explain. (Given ∆H for the reaction is -572 kJ).
Answer:
Given : ΔS= -327 JK-1; ΔH = -572 kJ
ΔG = ΔH – TΔS, and ΔH << ΔS
∴ ΔG < 0, and hence the formation of H2O(l) is spontaneous.

Question v.
Obtain the relation between ∆G and ∆STotal. Comment on spontaneity of the reaction.
Answer:
(i) Gibbs free energy, G is defined as,
G = H – TS
where H is the enthalpy, S is the entropy of the system at absolute temperature T.
Since H, T and S are state functions, G is a state function and a thermodynamic function.

(ii) At constant temperature and pressure, change in free energy ΔG for the system is represented as, ∆G = ∆H – T∆S
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 14
This is called Gibbs free energy equation for ∆G. In this ∆S is total entropy change, i.e., ∆STotal.

(iii) The SI units of ∆G are J or kJ (or Jmol-1 or kJmol-1).
The c.g.s. units of ∆G are cal or kcal (or cal mol-1 or kcal mol-1.)

The second law explains the conditions of spontaneity as below :
(i) ∆Stotal > 0 and ∆G < 0, the process is spontaneous.
(ii) ∆Stotal < 0 and ∆G > 0, the process is nonspontaneous.
(iii) ∆Stotal = 0 and ∆G = 0, the process is at equilibrium.

Question vi.
One mole of an ideal gas is compressed from 500 cm3 against a constant external pressure of 1.2 × 105 Pa. The work involved in the process is 36.0 J. Calculate the final volume.
Answer:
Given : V1 = 500 cm3 = 0.5 dm3;
Pex = 1.2 × 105 Pa = 1.2 bar; W= 36 J;
1 dm3 bar = 100 J; V2 = ?
W = -Pex (V2 – V1)
36 J = – 1.2 (V2 – 0.5) dm3 bar
= -1.2 × 100 (V2 – 0.5) J
∴ V2 – 0.5 = \(\frac{-36}{1.2 \times 100}=-0.3\)
∴ V2 = 0.5 -0.3 = 0.2 dm3 = 200 cm3
Ans. Final volume = 200 cm3.

Question vii.
Calculate the maximum work when 24g of O2 are expanded isothermally and reversibly from the pressure of 1.6 bar to 1 bar at 298 K.
Answer:
Given : W02 = 24 g, P1 = 1.6 bar, P2 = 1 bar
T = 298 K, Wmax = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 15

Question viii.
Calculate the work done in the decomposition of 132 g of NH4NO3 at 100 °C.
NH4NO3(s) → N2O(g) + 2 H2O(g)
State whether work is done on the system or by the system.
Answer:
NH44NO3(s) → N2O(g) + 2 H2O(g)
mNH4NO3 = 132 g; MNH4NO3 = 80 g mol-1
T = 273 + 100 = 373 K; Δn = ?
For the reaction,
Δn = Σn2 gaseous products – Σn1 gaseous reactants
= 3 – 0 = 3 mol
Since there is an increase in number of gaseous moles, the work is done by the system.
nNH4NO3 = \(\frac{m_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}{M_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}\)
= \(\frac{132}{80}\)
= 1.65 mol
For 1 mol NH4NO3(s) Δn = 3 mol
∴ For 1.65 mol NH4NO3(s) Δn = 3 × 1.65 = 4.95 mol
W = -ΔnRT = -4.95 × 8.314 × 373
= – 15350 J
= – 15.35 kJ
Ans. Work is done by the system.
Work done = – 15.35 kJ

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question ix.
Calculate standard enthalpy of reaction,
Fe2O3(s) + 3CO(g) → 2 Fe(s) + 3CO2(g),
from the following data.
fH0(Fe2O3) = -824 kJ/mol,
fH0(CO) = -110 kJ/mol,
fH0(CO2) = -393 kJ/mol
Answer:
Given : ∆fH0Fe2O3 = -824 kJ/mol-1;
fH0(CO) = – 110 kJ mol-1
fH0(CO2) = – 393 kJ/mol-1; ∆fH0 = ?
Required equation,
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
∆H1 = ? – (I)
Given equations :
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 16
= -(-824) -3 (-110) + 3(-393)
= 824 + 330 – 1179
fH0 = -25 kJ
Ans. ∆fH0 = -25 kJ

Question x.
For a certain reaction ∆H0 =219 kJ and ∆S0 = -21 J/K. Determine whether the reaction is spontaneous or nonspontaneous.
Answer:
Given : ∆H0 = 219 kJ; ∆S0 = -21 J/K, ∆G0 = ?
For standard state, T = 298 K
∆G0 = ∆H0 – T∆S0
= 219 – 298 × (-21) × 10-3
= 219 + 6.258
= 225.3 kJ
Since ∆S < 0 and ∆G0 > 0, the reaction is non-spontaneous.

Question xi.
Determine whether the following reaction is spontaneous under standard state conditions.
2 H2O(l) + O2(g) → 2H2O2(l)
if ∆H0 = 196 kJ, ∆S0 = -126 J/K
Does it have a cross-over temperature?
Answer:
Given : 2H2O(l) + O2(g) → 2H2O2(l)
∆H0 = +196 kJ
∆S0 = -126 JK-1 =0.126 kj K-1
T= 298 K
∆G0 = ?
Cross over temperature = T = ?
∆G0 = ∆H0 – T∆S0
= 196 – 298 (-0.126)
= 196 + 37.55
= + 233.55 kJ
∵ ∆G0 > 0, the reaction is non-spontaneous.
∆H0 > 0, ∆S0 < 0,
Since at all temperatures, ∆G0 > 0, there is no cross over temperature.
Ans. The reaction is non-spontaneous.
There is no cross-over temperature for the reaction.

Question xii.
Calculate ∆U at 298 K for the reaction,
C2H4(g) + HCl(g) → C2H5Cl(g), ∆H = -72.3 kJ
How much PV work is done?
Answer:
Given : C2H4(g) + HCl(g) → C2H5Cl(g)
T = 298 K; ∆H = -72.3 kJ; PV = ?;
∆U = ?
∆n = Σn2gaseous products – Σn1gaseous reactants
= 1 – (1 + 1)= -1 mol
For PV work :
W = -∆nRT
= – (- 1) × 8.314 × 298
= 2478 J = 2.478 kJ
∆H = ∆U + ∆nRT
∴ ∆U = ∆H – ∆nRT
= – 72.3 – (-2.478)
= – 69.82 kJ
Ans. PV work = 2.478 kJ
∆U = -69.82 kJ

Question xiii.
Calculate the work done during synthesis of NH3 in which volume changes from 8.0 dm3 to 4.0 dm3 at a constant external pressure of 43 bar. In what direction the work energy flows?
Answer:
Given : V1 = 8.0 dm3; V2 = 4.0 dm3; Pex = 43 bar
W = ? What direction work energy flows ?
W = -Pex(V2 – V1)
= -43 (4 – 8)
= 172 dm3 bar
= 172 × 100 J
= 17200 J
= 17.2 kJ
In this compression process, the work is done on the system and work energy flows into the system.

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question xiv.
Calculate the amount of work done in the
(a) oxidation of 1 mole HCl(g) at 200 °C according to reaction.
4HCl(g) + O2(g) → 2 Cl2(g) + 2 H2O(g)
(b) decomposition of one mole of NO at 300 °C for the reaction
2 NO(g) → N2(g) + O2
Answer:
Given :
(a) 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)
nHCl = 1 mol; T = 273 + 200 = 473 K, W = ?
For 4 mol HCl ∆n = (2 + 2) – (4 + 1) = – 1 mol
∴ For 1 mol HCl ∆n = –\(\frac {1}{4}\) = -0.25 mol
W = -∆nRT = – (-0.25) × 8.314 × 473 = 983.11
(b) ∆n = (1 + 1) – 2 = 0 mol
W = -∆nRT = -(0) × 8.314 × 473 = 0
Ans. (a) W = 983.1 J
(b) W = 0.0 J

Question xv.
When 6.0 g of O2 reacts with CIF as per
2CIF(g) + O2(g) → Cl2O(g) + OF2(g)
The enthalpy change is 38.55 kJ. What is standard enthalpy of the reaction ?
Answer:
Given : The given reaction is for 1 mol O2 or 32 g O2.
∵ For 6.0 g O2
∆ H0 = 38.55 kJ
∴ For 32 g O2
∆ H0 = \(\frac{32 \times 38.55}{6}\)
= 205.6 kJ
Ans. ∆H0 = 205.6 kJ

Question xvi.
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
i. CH3OH(l) + \(\frac {3}{2}\) O2(g) → CO2(g) + 2H2O(l), ∆H0 = -726 kJ mol-1
ii. C (Graphite) + O2(g) → CO2(g), ∆cH0 = -393 kJ mol-1
iii. H2(g) + \(\frac {1}{2}\) O2(g) → H2O(l), ∆fH0 = -286 kJ mol-1
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 17
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 18
∴ ∆H0
= –\(\Delta H_{2}^{0}\) + \(\Delta H_{3}^{0}\) + 2∆\(\Delta H_{4}^{0}\)
= – (- 726) + (- 393) + 2(- 286)
= 726 – 393 – 572
= – 239 kJ mol-1
Ans. Standard enthalpy of formation = ∆fH0= -239 kJ mol-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Question xvii.
Calculate ∆H0 for the following reaction at 298 K
H2B4O7(s) + H2O(l) → 4HBO2(aq)
i. 2H3BO3(aq) → B2O3(s) + 3H2O(l), ∆H0 = 14.4 kJ mol-1
ii. H3BO3(aq) → HBO2(aq) + H2O(l), ∆H0 = -0.02 kJ mol-1
iii. H2B4O7(s) → 2B2O3(s) + H2O(l), ∆H0 = 17.3 kJ mol-1
Answer:
Given equations :
i. 2H3BO3(aq) → B2O3(s) + 3H2O(l), ……….(i)
∆H0 = 14.4 kJ mol-1
ii. H3BO3(aq) → HBO2(aq) + H2O(l) ……….(ii)
∆H0 = -0.02 kJ mol-1
iii. H2B4O7(s) → 2B2O3(s) + H2O(l), ……….(iii)
∆H0 = 17.3 kJ mol-1
Required equation :
(iv) H2B4O7(s) + H2O(l) → 4HBO2(aq) ……. (iv)
\(\Delta H_{4}^{0}=?\)
To obtain eq. (iv) add 4 times equation (ii) and eq.
(iii) and subtract 2 times equation (i).
∴ eq. (iv) = 4 eq. (ii) + eq. (iii) – 2eq. (i)
∴ \(\Delta H_{4}^{0}=4 \Delta H_{2}^{0}+\Delta H_{3}^{0}-2 \Delta H_{1}^{0}\)
= 4(-0.02) + 17.3 – 2(14.4)
= -0.08 + 17.3 – 28.8
= -11.58 kJ
∴ Enthalpy change for the reaction
= ∆rH0 = -11.58 kJ
Ans. ∆rH0 for the given reaction = -11.58 kJ

Question xviii.
Calculate the total heat required (a) to melt 180 g of ice at 0 °C, (b) heat it to 100 °C and then (c) vapourise it at that temperature. Given ∆fusH(ice) = 6.01 kJ mol-1 at 0 °C, ∆vapH(H2O) = 40.7 kJ mol-1 at 100 °C specific heat of water is 4.18 J g-1 K-1.
Answer:
Given : Mass of ice = m = 180 g
T1 = 273 + 0 °C = 273 K
T2 = 273 + 100 °C = 373 K
fusH(ice) = ∆fusH(H2O)(s) = 6.01 kJ mol-1
vapHH2O(l) = 40.7 kJ mol-1
Specific heat of water = C = 4.18 J g-1 K-1
For converting 180 g ice into vapour, ∆ HTotal = ?
Number of moles of H2O = \(\frac {180}{18}\) = 10 mol
The total process can be represented as,
Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics 19
(i) ∆H1 = ∆fusH = 10 mol × 6.01 kJ mol-1
= 60.1 kJ
(ii) When the temperature of water is raised from 0 °C to 100 °C (i.e., 273 K to 373 K), then
∆ H2 = m × C × ∆T
= m × C × (T2 – T1)
= 180 g × 4.18 Jg-1K-1 × (373 – 273) × 10-3 kJ = 75.24 kJ
∆ H3 = ∆vapH = 10 mol × 40.7 kJ mol-1 = 407 kJ
Hence total enthalpy change,
∆ HTotal = ∆H1 + ∆H2 + ∆H3
= 60.1 + 75.24 + 407
= 542.34 kJ
Ans. Total heat required = 542.34 kJ

Question xix.
The enthalpy change for the reaction,
C2H4(g) + H2(g) → C2H6(g)
is -620 J when 100 ml of ethylene and 100 mL of H2 react at 1 bar pressure. Calculate the pressure volume type of work and ∆U for the reaction.
Answer:
Given :
\(\begin{aligned}
&\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})} \\
&100 \mathrm{~mL} \quad 100 \mathrm{ml} \quad 100 \mathrm{ml}
\end{aligned}\)
∆H = – 620 J; VC2H4 = 100 mL; VH2 = 100 mL
Pex= 1 bar; W=?; ∆U = ?
∆V = 100 – (100 + 100) = -100 mL = -0.1 dm3
W = -Pex(V2 – V1)
= -Pex × ∆V
= -1 × (-0.1)
= 0.1 dm3 bar
= 0.1 × 100 J
= +10 J
∆H = ∆U + P∆V
∴ ∆U = ∆H – P∆V = -620 – (+10) = -610 J
Ans. W = +10 J; ∆U = -610 J

Question xx.
Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of NH4NO3 at 100 °C
NH4NO3(s) → N2O(g) + 2H2O(g)
Answer:
Given : NH4NO3(s) → N2O(g) + 2H2O(g)
nNH4NO3 = 2 mol; T = 273 + 100 = 373 K
W = ? Comment on work = ?
∆nreaction = (1 + 2) – 0 = 3 mol
∵ For 1 mol of NH4NO3 ∆nreaction = 3 mol
∴ For 2 mol of NH4NO3 ∆nreaction = 6 mol
Due to 6 moles of gaseous products from 2 mol NH4NO3, there is work of expansion, hence work is done by the system.
W = -∆nRT
= – 6 × 8.314 × 373 = -18606 J
= -18.606 kJ
Ans. Work is done by the system.
W= -18.606 kJ

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

12th Chemistry Digest Chapter 4 Chemical Thermodynamics Intext Questions and Answers

(Textbook page No. 73)

Question 1.
Under what conditions ∆H = ∆U ?
Answer:
(a) ∆H = ∆U + P∆V
when ∆V = 0, ∆H = ∆U
(b) ∆H = ∆U + ∆nRT
when ∆n = 0, ∆H = ∆U

Try this… (Textbook page No. 71)

Question 1.
25 kJ of work is done on the system and it releases 10 kJ of heat. What is ∆U?
Answer:
W = 25 kJ; Q= -10 kJ
∆U = Q + W = -10 + 25
∆U = + 15 kJ

Try this… (Textbook page No. 75)

Question 1.
For KCl, ∆LH = 699 kJ/mol-1 and ∆hydH = -681.8 kJ/mol-1. What will be its enthalpy of solution?
Answer:
Enthalpy of solution :
solnH = ∆LH + ∆hydH
= 699 + (-681.8)
solnH = +17.2 kJ mol-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Try this… (Textbook page No. 76)

Question 1.
Given the thermochemical equation,
C2H2(g) + \(\frac {5}{2}\) O2(g) → 2CO2(g)+ H2O(l), ∆rH0 = -1300 kJ
Write thermochemical equations when
i. Coefficients of substances are multiplied by 2.
ii. equation is reversed.
Answer:
(i) 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)
rH0 = -2 × 1300 kJ
= – 2600 kJ
(ii) 2CO2(g) + H2O(l) → C2H2(g) + \(\frac {5}{2}\)O2(g)
rH0 = +1300 KJ

Try this… (Textbook page No. 78)

Question 1.
(i) Write thermochemical equation for complete oxidation of one mole of H2(g). Standard enthalpy change of the reaction is -286 kJ.
(ii) Is the value -286 kJ, enthalpy of formation or enthalpy of combustion or both? Explain.
Answer:
(i) H2(g) + \(\frac {1}{2}\)O2(g) → H2O(l) ∆cH0 = -286 KJ mol-1
(ii) The value -286 kJ is the standard enthalpy of formation of H2O(l) or standard enthalpy of combustion of H2(g).

Question 2.
Write equation for bond enthalpy of Cl-Cl bond in Cl2 molecule ∆rH0 for dissociation of Cl2 molecule is 242.7 kJ.
Answer:
Equation for bond enthalpy :
Cl2(g) → 2Cl(g)rH0 = 242.7 kJ mol-1
∴ Bond enthalpy of Cl2 = 242.7 kJ mol-1

Maharashtra Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

Try this… (Textbook page No. 82)

Question 1.
State whether ∆S is positive, negative or zero for the following reactions.
i. 2H2(g) + O2(g) → 2H2O(l)
ii. CaCO3(s) → CaO(s) + CO2(g)
Answer:
(i) 2H2(g) + O2(g) → 2H2O(l)
Since the system is converted from gaseous state to a liquid state, the disorder is decreased, hence ∆S < O (negative).

(ii) CaCO3(s) → CaO(s) + CO2(g)
Since molecules of solid CaCO3 break giving gaseous CO2, disorder is increased hence ∆S > O (positive).

12th Std Chemistry Questions And Answers:

Renaissance in Europe and Development of Science Question Answer Class 12 History Chapter 1 Maharashtra Board

Balbharti Maharashtra State Board Class 12 History Solutions Chapter 12 India Transformed Part 2 Textbook Exercise Questions and Answers.

Std 12 History Chapter 1 Question Answer Renaissance in Europe and Development of Science Maharashtra Board

Class 12 History Chapter 1 Renaissance in Europe and Development of Science Question Answer Maharashtra Board

History Class 12 Chapter 1 Question Answer Maharashtra Board

1A. Choose the correct alternative and rewrite the statement.

Question 1.
In 1995, the health department of the Indian government launched the campaign, dubbed as __________
(a) Measles-Rubella
(b) Pulse Polio
(c) B.C.G.
(d) Triple vaccine
Answer:
(b) Pulse Polio

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

Question 2.
The first district in __________ to become completely literate was Ernakulum.
(a) Gujrat
(b) Kerala
(c) Karnataka
(d) Tamil Nadu
Answer:
(b) Kerala

1B. Find the incorrect pair from group ‘B’ and write the corrected one.

Question 1.

Group ‘A’ Group ‘B’
(a) National Human Rights Commission Protection of Human Rights
(b) Centre for Science and Environment Study of Pollution in Delhi
(c) SEESCAP Institute for conservation of turtles
(d) INTACH Organisation creating awareness for conservation of heritage

Answer:
SEESCAP – Institute for conservation of long-billed vultures

2. Write the names of historical places/persons/events.

Question 1.
Commission established vide Human Right Protection Act –
Answer:
National Human Rights Commission

Question 2.
The player who was awarded the highest title, ‘Bharat Ratna’ –
Answer:
Mr. Sachin Tendulkar

3. Complete the following concept map.

Question 1.
Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2 Q3
Answer:
Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2 Q3.1

4. Write short notes.

Question 1.
Speed Post
Answer:

  • The Indian postal department started the service of speed post in 1986 which changed the nature of post service.
  • This service was used by the majority of Indians, more than three crore letters and parcels were delivered from this service.
  • The customer will get the message of successful delivery of their items.
  • The postal department now offers courier services like passport delivery, business parcels, cash-on-delivery, logistics posts, and air freights.
  • The post office even offers a packaging service. Over one lakh and fifty thousand post offices are offering the services like paying bills, sending festive cards and other objects.
  • Since 2016, the postal department has started service of delivering Ganges water from Rishikesh and Gangotri at personal addresses.
  • Buying personalised postage stamps with personal photographs and special schemes for philatelists are also available at ‘Post Shops’ opened at 80 post offices.

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

Question 2.
Air Pollution
Answer:

  • The entire world, nowadays, is facing the problem of air pollution which is mainly done by vehicles.
  • All most every state in India is facing the problem of air pollution.
  • The ‘Centre for Science and Environment’, studied and proved that the increasing number of automobiles and vehicles of poor maintenance are the causes of air pollution in Delhi.
  • The institution recommended the following suggestions to curb air pollution:
    • Use of CNG (Compressed Natural Gas) as fuel.
    • It was decided that vehicles without a ‘PUC’ certificate will not qualify for insurance and renewal of insurance.
  • In this way, we can curb air pollution and make our environment clean and pure.

Question 3.
Eradication of Polio.
Answer:

  • The government of India launched a campaign for the eradication of polio from India.
  • The health department of the Indian government launched the campaign known as ‘Pulse Polio’ which was started with the joint sponsorship of‘World Health Organisation – WHO, Rotary International, UNICEF, and Indian Government.’
  • The objective of the campaign was not to leave a single child under the age of five years without administering the polio vaccine.
  • Awareness camps, home visits, and extensive advertising made this campaign successful.

5. Answer the following questions in detail.

Question 1.
Explain the sports policy of the Government of India.
Answer:
India is home to a diverse population playing and showing their talent in a variety of different sports. Every nation needs a well-knit sports policy.

  • In 2001, the Government of India announced its sports policy. The main objective of this policy is:
    • To take sports to all parts of India.
    • To help the players to develop special skills.
    • To build supportive and fundamental sports facilities.
    • To help the National Sports Federation of India and associated institutions to search for sports talent.
    • To initiate co-operation from industries, corporate and private institutions for the cause of sports.
    • To create awareness of the importance of sports and interest in sports among people.
  • In 2011, the Indian government announced a novel scheme for sports named ‘Come and Play’.
  • Sports Authority of India gave permission to use five sports complexes in Delhi to local youths.
  • They were also provided an opportunity to train under Sports Authority of India (SAI) coaches.
  • The National Sports University was founded in Manipur in 2018.
  • This university offers different courses from Bachelor and Masters to M.Phil. and Ph.D. Apart from this, sports universities also offer courses in sports, education, sports management, sports psychology, coaching, etc. Research in sports is also encouraged in the sports university.
  • Khelo India.

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

Question 2.
Which programmes and facilities are created by the Ministry of Tourism of the Indian government to attract tourists?
Answer:
India has a rich cultural heritage. Several tourists visit India each year to see its cultural heritage, monuments, etc. The tourism industry is a continuously growing industry which also gives foreign exchange to the country. The Indian Government has adopted different policies to attract tourists. They are as follows:

  • The government provides different facilities to the tourists who visit India. The three ministries of home affairs, tourism, and foreign affairs together created the facility of e-visa which included e-business visas, e-medical visas, and e-visa.
  • A facility of giving information to tourists on mobile in Hindi and 10 other foreign languages, for 24 hours throughout the week was started.
  • To avail of this facility, tourists have to dial 1363. The information regarding cruise tours, health, and sports tourism, eco-friendly tourism, adventurous sports tourism, film festivals are given through this service.
  • Hotels with amenities and premium quality accommodation including luxurious services are available to the tourists.
  • The Institutes which offer training courses in ‘Hospitality and Hotel Management’ have been established in major cities of India.
  • An advertisement campaign called ‘Atulya Bharat’ was designed to attract tourists.
  • A travel show entitled ‘GONORTHEAST’ was released on the Discovery channel to boost the tourism of beautiful places in the northeastern states of India.
  • The government took the help of electronic and digital media channels like Discovery, BBC, History is showcasing various programs introducing India’s historical and cultural heritage.
  • Swadesh’ and ‘Prasad’ schemes were launched by the Ministry of Tourism of Government of India to encourage visits to 95 pilgrimage and spiritual centers in India.
  • The Ministry of Tourism, Government of India, and Federation of Associations in Indian Tourism and Hospitality (FAITH) organized Indian Tourism Mart – 2018. This was the first event based on the model of International Tourism Marts in other countries.

6. State your opinion.

Question 1.
Joint military practice sessions are beneficial for both participant countries.
Answer:

  • Such sessions are extremely helpful for both countries because of the technological exchange that happens on these occasions.
  • It also helps the armies of both the countries to know, learn and practice new methods of resolving problems.
  • The process of modernization of arms gives impetus to further research.
  • Due to the development of science and technology, there is increasing scope for the exchange of the latest technology to fight terrorism, to augment our own competencies, and optimum use of modern technology for the end of terrorism.
  • The Indian army carried out exercises with different countries e.g., exercise with Oman army at Bakloh, there were combined exercises of Mangolian army and Jammu and Kashmir Rifles.

Question 2.
All of us have the responsibility of taking care of our heritage places.
Answer:

  • India has an extraordinary and vast cultural heritage. It is in the form of ancient monuments, buildings, and other archaeological sites and remains.
  • These monuments are the living witnesses of our golden historic era.
  • It is our duty to preserve the monuments for the next generations. A little initiate from our side can save our heritage.
  • The tourism industry generated foreign exchange on a large scale. Tourists come to India every year to see its cultural heritage. Therefore, it is our duty to preserve and protect our heritage.
  • As a citizen of India, we should spread awareness among the people about the importance of the preservation of monuments. A little effort on our side can create desirable changes which will make past, present, and future generations of the country and the entire world proud of us.

Class 12 History Chapter 12 India Transformed Part 2 Intext Questions and Answers

Try to do this: (Textbook Page No. 100)

NRHM – Make a list of the benefits of the National Rural Health Mission to people.
Answer:

  • The Indian government launched National Rural Health Mission (NRHM) in April 2005 with an aim to strengthen the health systems in rural and urban areas. The list of the benefits of the NRHM are as follows:
  • It aims to provide equitable, affordable, and quality healthcare services.
  • It has strengthened the healthcare infrastructure.
  • It has brought down the maternal mortality rate among poor pregnant women.
  • The prevalence of tobacco use and the number of tobacco users have been reduced.
  • The Janani Shishu Suraksha Karyakram entitles pregnant women to give birth in public health institutions at no expense.
  • The government launched different schemes for community participation under NRHM. Rogi Kalyan Samiti is responsible for maintaining the facilities and ensuring the provision of better facilities for the patients in the hospital.
  • Established the Global Knowledge Hub for smokeless tobacco. It also issued an advisory to ban Electronic Nicotine Delivery Systems.
  • After the implementation of various initiates under NHRM many states have shown improved progress in healthcare facilities.

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

Try to do this: (Textbook Page No. 100)

Make a list of solutions to reduce the levels of air pollution.
Answer:
Air pollution is the biggest threat to the environment and to all living species. Following are some of the solutions to reduce the level of air pollution:

  • Public Transport: Encourage greater use of public transport i.e. the use of railways, bus services or metros, etc. Instead of using private vehicles, people should encourage to use public transport.
  • Use of CNG Vehicles: Citizens should also be encouraged to use CNG vehicles as it is a much cleaner fuel than petrol or diesel. New registration should be discouraged by increasing registration charges of vehicles.
  • Use bicycles: Using bicycles is the best way to reduce air pollution. The government should mark out bicycle lanes in residential colonies as well as on roads.
  • Plant and care for trees: Trees filter pollutants and absorb carbon dioxide. Trees release oxygen into the atmosphere. The practice of planting trees provides more benefits to the environment.
  • Use less energy: Use energy-efficient appliances. Turn off electrical appliances when not in use. Get an energy audit done and follow the advice.
  • Avoid the use of crackers: Avoid the use of crackers during festivals and weddings as it creates a lot of air pollution and is harmful to birds and animals including human beings.
  • Reuse, Reduce and Recycle: The three ‘Rs’ are the best way to reduce air pollution.
  • Avoid using chemical products: Avoid using chemical products like paint, perfumes, sprays, etc as they contain harmful products. Try to use products with less chemical content.
  • Prevention of forest fires and burning of garbage: Don’t burn garbage or leaves because it releases harmful smoke in the air which decreases the quality of air.

Try to do this: (Textbook Page No. 102)

1990 was the ‘International Year of Literacy’. Make a list of similarly declared international years for special causes and campaigns launched on the occasion.
Answer:
The following are the international years currently observed by the United Nations.
2024:

  • International Year of Camelids

2022:

  • International Year of Artisanal Fisheries and Aquaculture

2021:

  • International Year of Peace and Trust
  • International Year of Creative Economy for Sustainable Development
  • International Year of Fruits and Vegetables
  • International Year of Eliminations of Child Labour

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

2020:

  • International Year of Planet Health
  • International Year of the Nurse and Midwife

2019:

  • International Year of Indigenous Languages
  • International Year of Moderation
  • International Year of Periodic Table of Chemical Elements

2017:

  • International Year of Sustainable Tourism for Development.

2016:

  • International Year of Pulses

2015:

  • International Year of Light and Light-based Technologies
  • International Year of Soils

2014:

  • International Year of Solidarity with the Palestinian People
  • International Year of Small Island Developing States
  • International Year of Crystallography
  • International Year of Family Farming

2013:

  • International Year of Water cooperation
  • International Year of Quinoa

2012:

  • International Year of Cooperatives
  • International Year of Sustainable Energy for All

2011:

  • International Year of Forest
  • International Year of Chemistry
  • The International Year of African Descent

2010:

  • The International Year of Biodiversity
  • The International Year for the Rapprochement of cultures
  • The International Year of Youth

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

2009:

  • The International Year of Astronomy
  • The International Year of Human Rights Learning
  • The International Year of Natural Fibres
  • The International Year of Reconciliation

2008:

  • The International Year of Languages
  • The International Year of Planet Earth
  • The International Year of the Potato
  • The International Year of Sanitation

2007-08:

  • International Polar Year

2006:

  • International Year of Deserts and Desertification

2005:

  • International Year of Physics
  • International Year of Sport and Physical Education
  • International Year of Microcredit

2004:

  • International Year of Rice
  • International Year to Commemorate the Struggle against Slavery and its Abolition

2003:

  • International Year of Freshwater
  • Year of Kyrgyz Statehood

2002:

  • International Year of Mountains
  • International Year of Eco-tourism
  • United Nations Year for Cultural Heritage

2001:

  • International Year of Volunteers
  • International Year of Mobilisation against Racism, Racial Discrimination, Xenophobia, and Related Intolerance
  • United Nations Year of Dialogue among Civilisation

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

2000:

  • International Year of Thanksgiving
  • International Year for the Culture of Peace

Find out and tell us (Textbook Page No. 106)

Make a list of employment opportunities generated by the tourism industry.
Answer:
The travel and tourism industry in India is growing rapidly so many opportunities are available in the coming years in this field. Some of the opportunities that are available in the tourism industry are as follows:
(i) Hotels: Many job perspectives are available in the hotel industry. Some fields of hotel industries are

  • Manager
  • Operations
  • Housekeeping
  • Food and Beverage
  • Front office
  • Gardener
  • Security officer/personnel etc.

(ii) Airlines: One can take up the following job in airlines

  • Pilot
  • Ground staff (Traffic Assistant, Counter staff, Booking, and Reservation)
  • Flight Attendant

(iii) Tourism Department:

  • Tour guides
  • Tour planner
  • Information assistants
  • Reservation and counter staffs
  • Sales and Marketing
  • Interpreters
  • Translators

(iv) Transportation Industry:
This is an ever-growing industry where one can have many job opportunities. Job opportunities are available in all types of transportation i.e.

  • Railway service
  • Bus service
  • Cruise service or ferry service
  • Private transportation – Cars, Rickshaws, Horse riding, etc.

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

(v) The employment opportunities are also available in the management of adventure sports, theme parks, amusement parks, water sports, mountaineering, children’s fantasy land, etc.

(vi) The other job opportunities in this field include Destination Manager, Itinerary Planner, Travel Agent, Foreign Exchange, Counselor, etc.

Find out and tell us (Textbook Page No. 106)

Suggest ways and means to enhance the heritage and historical tourism in India.
Answer:
Heritage and historical tourism tend to attract many tourists. Different ways and means to enhance, this type of tourism are as follows:

  • Build better roads and access points to the heritage sites, packaged train travel, proper bus connections should be provided to the tourists.
  • Aggressive tourism and marketing strategies are also necessary whether it is broadcasting the ‘Incredible India’ campaign abroad, holding different seminars, or offering Indian locations with facilities to promote foreign film productions in the country are some of the strategies.
  • The Government of India has already started making a lot of improvements in this area. We need to capitalize on India as a destination. The government should make a documentary based on the heritage of India. One can even design a promotional campaign to enhance historical tourism.
  • The historical places always have a story to tell. They offer amazing aesthetics and attract people from all over the world. If tourism is accompanied by mind-blowing hotels, resorts, good public transportation facilities, and delectable cuisine then it becomes an added advantage to the tourist.

Project (Textbook Page No. 108)

Make a list of various business opportunities available at tourist destinations.
Answer:
Introduction: Tourism is a continuously growing industry. By the early 21st century, international tourism had become one of the world’s most important activities. Various business opportunities are available at the tourist destinations and they are as follows:
(i) Travel Agency: Everyone needs a platform where customers can come to and take advice about tour packages of different places which travel agency provides. The travel agency helps tourists in planning their travel for which they charge their commission.

(ii) Hotel: Hotel business is the best opportunity from which you can gain a high rate of return as well as expand your chain of hotels in the future.

(iii) Online Travel Business: In the modern era, the way of doing business is changing. Now everything is available online. With the change in technology, the way of doing business has also changed. One can start an online business portal where customers have easy access to everything online.

Maharashtra Board Class 12 History Solutions Chapter 12 India Transformed Part 2

(iv) Photography: Tourists always want to save the memories of the places they visit and a photographer helps them to restore their memories through their art of photography. There is a huge demand for photographers.

(v) Vehicle Renting: Some tourists like to plan their tours according to their own wishes. They don’t want any unknown person to be a part of their travel for safety reasons. Such tourists search for a vehicle that they can get on rent and go in the direction that they want. To invest in this business is profitable.

(vi) Tour Guide: A Tourist guide is a person who guides visitors in the language of their choice and interprets the cultural and natural history of the particular place. This is a challenging field with an increase in tours and travels.

(vii) Executive Chef: The executive chef is in charge of a restaurant’s kitchen and is responsible for managing the kitchen staff, planning the menu, and making sure that food hygiene is maintained. This is a well suitable job for people who are passionate about cooking.

12th Std History Questions And Answers:

The Cop and the Anthem 12th Question Answer English Chapter 1.3 Maharashtra Board

Class 12 English Chapter 1.3

Balbharti Yuvakbharati English 12th Digest Chapter 1.3 The Cop and the Anthem Notes, Textbook Exercise Important Questions and Answers.

Class 12 English Chapter 1.3 The Cop and the Anthem Question Answer Maharashtra Board

12th Std English Chapter 1.3 Brainstorming Question Answer

Yuvakbharati English Navneet 12th Digest PDF Free Download Maharashtra Board

Question 1.
Suppose you have gone to a place where the winter season is very severe. Discuss with your partner the ways in which you would protect yourself in the cold climate. (The answer is given and underlined.)
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 1.3 The Cop and the Anthem 1
Answer:
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 1.3 The Cop and the Anthem 2

Maharashtra Board Solutions

Question 2.
When you see a cop approaching, you feel either ‘relieved’ or ‘scared’. Discuss with your partner and write down the situations when you feel ‘relieved’ or ‘scared’.
Answer:
Relieved:
(a) You are walking alone in a dark street.
(b) There is a fight taking place near you.
(c) A group of rough-looking people are coming towards you on a lonely road.
Scared:
(a) You are riding a bike without a valid driving licence.
(b) You have broken a traffic signal.
(c) You have been involved in a fight.

Question 3.
Discuss some of the motivating things that can change a person’s life :
Answer:
(a) Listening to an inspiring speech
(b) Reading motivating books and biographies of great people
(c) Reading epics and religious books
(d) Observing successful/happy people or watching biopics of their lives.

(A1)

Question (i)
Discuss with your partner and find out the different ways in which Soapy tried ; to get arrested. The first one is given.
Answer:
(a) Tried to enter a luxurious cafe.
(b) Threw a stone and broke a shop window.
(c) Ate heartily at a restaurant and then said he had no money.
(d) Shouted and howled and raved and danced on the sidewalk.
(e) Stole an umbrella.

Question (ii)
Describe the atmosphere when Soapy reached near the Church.
Answer:
(a) A soft light glowed through the violet-stained window.
(b) Sweet music drifted out of the quaint, old church.
(c) There was a full, radiant moon, and few vehicles and pedestrians.
(d) Sparrows twittered sleepily in the eaves.

Maharashtra Board Solutions

(A2)

Question (i)
Read the story and match the incidents given in Column A with the consequences given in Column B.
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 1.3 The Cop and the Anthem 3
Answer:

  1. Soapy tried to enter a cafe – Strong and ready hands of the head waiter turned him around.
  2. Soapy broke a glass window – The cop ran after another man.
  3. Two waiters pitched Soapy on the callous pavement – He stood up slowly beating the i dust from his clothes.
  4. Soapy heard the anthem being played in the Church – Suddenly a wonderful change came in his heart.
  5. Cop arrests Soapy for hanging around. – Dream of turning around in life was shattered.

Question (ii)
Give reasons and complete the following: (The answers are given directly and underlined.)
Answer:
(a) Soapy had confidence in himself because he was shaven, his coat was trim and he had a neat, black bow. The portion of him that showed above the table looked respectable and would raise no doubt in the waiter’s mind.
(b) The head waiter of the luxurious cafe did not allow Soapy to enter because he saw Soapy’s tattered trousers and old, worn out shoes, and knew that Soapy would not have money to pay for a meal.
(c) The cop did not arrest Soapy for breaking the glass window because Soapy was standing calmly and talking to him. The policeman felt that men who smash glass windows do not remain to chat with the police.
(d) The cop did not arrest Soapy for shouting and dancing because it was the time of celebrations for the local college boys. They were generally noisy but harmless, and he had been told by his superiors to let them be.

(iii) Pick out the lines from the text which show that:

Question (a)
Soapy wants to enter the cafe for two reasons.
Answer:
1. A roasted mallard duck, thought Soapy, would be about the thing with a bottle of wine and then some cheese, a cup of coffee and a cigar.
2. The meat would leave him filled and happy for the journey to his winter island.

Maharashtra Board Solutions

Question (b)
Soapy was afraid that he won’t be able to enter the prison.
Answer:
It seemed that his route to the coveted island was not to be an easy one. Some other way of entering the limbo must be devised.

Question (c)
Soapy was not caught by the cop for throwing stones at the glass.
Answer:
1. The policeman refused to accept Soapy even as a clue.
2. The policeman saw a man half-way down the block running to catch a car. With drawn club he joined in the pursuit.

Question (d)
Soapy actually did not want the umbrella.
Answer:
He hurled the umbrella angrily into the excavation.

Question (e)
Listening to the anthem, Soapy remembered his good old days.
Answer:
He had known it well in the days when his life contained such things as mothers and roses and ambitions and friends and immaculate thoughts and collars.

Question (iv)
‘He would make a man of himself again’ – The word ‘man’ in the sentence means ……………….. .
Answer:
‘He would make a man of himself again’ – The word ‘man’ in the sentence means a responsible and worthy human being.

Maharashtra Board Solutions

Question (v)
Soapy’s earlier life was much different from his present life. Complete the table to show this contrast. One is done for you.
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 1.3 The Cop and the Anthem 4
Answer:

Earlier life Present life
(a) contained friends and roses (a) unworthy desires
(b) eager ambitions (b) dead hopes, degraded days
(c) clean thoughts and clothes (c) wrecked faculties and base motives

Question (vi)
After listening to the sweet and solemn organ notes, Soapy decides to:
Answer:
1. pull himself out of the mire, conquer the evil that had enslaved him and make a man of himself again
2. resurrect his old eager ambitions and pursue them without faltering
3. go into the roaring downtown district and find work

Question (vii)
Write an incident in which you did something wrong and repented for it later. Give reasons.
Answer:
A lady who stayed in my building used to shout at me for playing noisily under her window. One day, she shouted at me as usual from her window and went inside. I suddenly got angry and threw a stone at her window. The stone hit the glass which broke. I heard a loud shout of pain and ran away. I later came to know that she had been badly injured by the shattered glass. I repented for what I had done. Though I did not tell her that I was the culprit, I was very good to her after that.

Maharashtra Board Solutions

(A3)

Question (i)
O’Henry has used different words to indicate prison where Soapy wants to reach. Make a list of those words from the extract.
Answer:
the island

Question (ii)
Find out the words used for the ‘degraded state of Soapy’.
Answer:

  1. the pit into which he had tumbled
  2. the degraded days
  3. unworthy desires
  4. dead hopes
  5. wrecked faculties
  6. base motives
  7. mire
  8. evil that had enslaved him.

Question (iii)
The specific meaning of word ‘anthem’ in the content of the story is:
Answer:
Anthem – a rousing or uplifting song.

(A4)

Question (i)
Convert the following sentences into the negative without changing their meanings:
(a) The policeman refused to accept Soapy even as a clue.
(b) Soapy drifted along, twice unsuccessful.
(c) Soapy stopped his unavailing racket.
(d) The island seemed very far away.
(e) The island seemed an unattainable Arcadia.
Answer:
(a) The policeman did not accept Soapy even as a clue.
(b) Soapy drifted along, twice not successful.
(c) Soapy stopped his racket which was not successful.
(d) The island seemed not at all near.
(e) The island seemed an Arcadia which was not attainable.

Maharashtra Board Solutions

Question (ii)
Convert the following sentences into the affirmative without changing their meanings :
(a) Men who smash windows do not remain to chat with the police.
(b) On the opposite side of the street was a restaurant of no great pretensions.
(c) Why don’t you call a cop?
(d) Noisy; but no harm.
(e) They seemed to regard him as a King who could do no wrong.
Answer:
(a) Men who smash windows refrain from remaining to chat with the police.
(b) On the opposite side of the street was a very ordinary restaurant.
(c) Please call a cop.
(d) Noisy; but harmless.
(e) They seemed to regard him as a King who was always right.

(A5)

Question (i)
‘Forgiveness is often better than punishment’. Write two paragraphs – one for and another against this notion.
Answer:
1. To err is human, to forgive is divine.

We all make mistakes. Nobody is perfect. That is why we are human. However, mistakes should be forgiven if there is sufficient repentance. Forgiveness will make the guilty person feel ashamed of his conduct and he will not repeat it. It is easy to punish but very difficult to forgive someone. It needs a big heart and a lot of kindness. If we punish the guilty person we will only make him more defiant. If punishing someone could have solved the problem, criminals who have been jailed would never have repeated the crime. But this is not found to be so. Just as God forgives us our mistakes, we should forgive others their mistakes too.

2. Punishment is the only answer

Forgiveness may work in certain cases, but there are hardened criminals who will not respond to forgiveness. They will only be stopped by punishment. If a person has murdered another in cold blood, will he improve by forgiveness? Never. He has to be punished severely so that he does not repeat it and society feels safe.

People only fear punishment. It can be easily seen at traffic signals – if there is no policeman to punish you, most people will break the signal. Then there will be chaos. If there is no punishment and no prisons, people will do whatever they want – rob, kill, etc. – and go off freely. No, in a society where it is not possible to expect everyone to have high values, punishment for misdeeds is the only solution.

Maharashtra Board Solutions

Question (ii)
You are the class representative and you have been asked by the Principal to conduct an interview of a cop. Frame 8-10 questions with the help of the following points, give introduction and conclusion.

  • reasons for joining the department
  • special trainings
  • developing the skill to identify and locate criminals
  • dealing with criminals
  • achievements and awards

Answer:
Good morning, Mr. Pawar. Congratulations on your excellent work in finding the bank robbers. May I ask you a few questions about your life? Thank you.

  1. When did you join the police department?
  2. Which examinations did you have to clear for the post?
  3. What were your reasons for joining the department?
  4. Did you have to go through any special training sessions?
  5. What type of criminals do you come across most in this area-thieves, killers, molesters,
    etc?
  6. How do you identify or locate criminals?
  7. Once you catch the culprit-say a thief-how do you deal with him?
  8. Can you tell me something about your achievements and awards?
  9. How can you motivate others to join the force?
  10. Any message to college students?

Thank you, Sir, for sparing the time for this interview. It will be published in our school magazine. Good day.

(A6)

Question (i)
Make a list of jobs which would give you an opportunity to help the society or serve the country. Also mention the different ways in which they can be beneficial to the people and also the country,

Maharashtra Board Solutions

Question (ii)
Go to your school/college library and read some other stories by O’Henry like, ‘The Gift of the Magi’, ‘The Last Leaf and ‘After Twenty years’. Write the stories in short in your notebook.

Yuvakbharati English 12th Digest Chapter 1.3 The Cop and the Anthem Additional Important Questions and Answers

Read the extract and complete the activities given below:

Global Understanding:

Question 1.
Name the following:
Answer:

  1. This is where Soapy stopped at a luxurious cafe
  2. This had been a gift to Soapy: a neat, black bow
  3. This is what Soapy wanted to eat: a roasted mallard duck, some cheese a bottle of wine, a cup of coffee
  4. This is what Soapy wanted to drink: Soapy’s tattered
  5. This is what the head waiter noticed:trousers and old shoes
  6. This is where Soapy was left by the head: on the sidewalk

Question 1.
Write if the following sentences are True or False. Correct the False sentences:
1. Soapy broke the glass of the shop window.
2. Nobody heard the breaking of the window.
3. The policeman chased Soapy.
4. Soapy did not run away from the place.
5. The restaurant Soapy entered was an ordinary one.
6. The policeman ate beefsteak, flapjacks, doughnuts and pie.
Answer:
True sentences:
1. Soapy broke the glass of the shop window.
4. Soapy did not run away from the place.
5. The restaurant Soapy entered was an ordinary one.

False sentences:
2. Nobody heard the breaking of the window.
3. The policeman chased Soapy.
6. The policeman ate beefsteak, flapjacks, doughnuts and pie.

Corrected sentences:
1. A policeman as well as some people heard the breaking of the window.
2. The policeman chased a man running to catch a car.
3. Soapy ate beefsteak, flapjacks, doughnuts and pie.

Maharashtra Board Solutions

Question 2.
Complete the table:
(The answers is given directly and underlined.)
Answer:

The Words mo said To whom
1. Noisy, but no harm A policeman A citizen
2. I took it. Soapy The umbrella man
3. You know how these mistakes occur. The umbrella man Soapy
4. Of course it’s mine! Soapy The umbrella man
5. We’ve instructions to let them be. A policeman A citizen
6. I hope you’ll excuse me. The umbrella man Soapy

Question 3.
Rearrange the following sentences according to their occurrence in the extract:

  1. Soapy decided to go into the downtown district and find work.
  2. A policeman caught Soapy’s arm.
  3. Soapy saw a quaint old church.
  4. Soapy’s ears caught sweet music.

Answer:

  1. Soapy saw a quaint old church.
  2. Soapy’s ears caught sweet music.
  3. Soapy decided to go into the downtown district and find work.
  4. A policeman caught Soapy’s arm.

Question 4.
Complete the following:
(The answers are given directly and underlined.)
Answer:

  1. A soft light glowed through one violet-stained window.
  2. Soapy came to a standstill on an unusually quiet corner.
  3. Soapy stood without moving near the iron fence listening to the anthem that the organist played.
  4. Soapy planned to resurrect his old eager ambitions.

Maharashtra Board Solutions

Answer the following in a few words each:

Question 1.
Who was lighting a cigar?
Answer:
A well-dressed man

Question 2.
Who twirled his club?
Answer:
A policeman

Question 3.
Who grabbed the umbrella?
Answer:
Soapy

Question 4.
Whom did the policeman help?
Answer:
A tall blonde

Complex Factual:

Question 1.
Give reasons and complete the following:
(The answers are given directly and underlined.)
Answer:
Soapy was disgusted with the policeman because he refused to accept that Soapy had broken the window, and he rushed off to chase another man.

Maharashtra Board Solutions

Question 2.
Complete the web:
(The answers are given directly and underlined.)
Answer:
Maharashtra Board Class 12 English Yuvakbharati Solutions Chapter 1.3 The Cop and the Anthem 5

Question 3.
Describe Soapy’s behaviour on the sidewalk.
Answer:
On the sidewalk Soapy began to yell drunken meaningless things at the top of his harsh voice. He danced, howled, spoke wildly and made a big disturbance.

Question 4.
Describe the wonderful change in Soapy’s soul.
Answer:
After listening to the anthem, Soapy remembers his earlier life, and is horrified to realize that he has indeed become a degraded person. He decides to pull himself out of the pit into which he has fallen and make a man of himself again. He determines to bring back to his life his old eager ambitions and pursue them. He makes up his mind to take up a job.

Question 5.
Describe the end of the story in your own words.
Answer:
Initially Soapy had felt that he would like to pass the winter months in prison, and he makes several efforts to get himself arrested. The notes of anthem transform him from within and he decides to give up his evil ways and become a man again. He resolves to work hard. At that very moment, ironically, a policeman arrests him for loitering and he is sent to prison for three months. Soapy faces the irony of fate as the moment he realizes that real freedom lies in a virtuous life, he is taken into confinement.

Maharashtra Board Solutions

Question 6.
Complete the following:
Soapy was angry because ………..
Answer:
Soapy was angry because when he wanted to fall into the clutches of the policemen. In order to be arrested, they seemed to regard him as a king who could do nothing wrong.

Inference/Interpretation/Analysis:

Question 1.
Complete the following:
(The answers are given directly and underlined.)
Answer:
Soapy took a stone because he wanted to break the glass of the shop window. This would result in a policeman arresting him for this act, and he would be imprisoned for the winter, which was exactly what he wanted.

Discuss the hidden meaning in the expressions/sentences.

Question 1.
It catered to large appetites and modest purses.
Answer:
The restaurant prepared food for ordinary workers who had large appetites but very little money.

Question 2.
He told the waiter the fact that the minutest coin and himself were total strangers.
Answer:
He told the waiter that he did not have any money.

Maharashtra Board Solutions

Question 3.
A voice like butter cakes and an eye like the cherry in the Manhattan cocktail.
Answer:
A very smooth voice and a hard, stony eye (a tough person).

Question 4.
Discuss the meaning in the context:
He caught at the immediate straw of ‘disorderly conduct’.
Answer:
Soapy wanted to be arrested by a policeman and imprisoned. However, his efforts towards this end had been unsuccessful, and he was worried that he would continue to be unsuccessful. When he suddenly came upon a policeman lounging in front of a theatre, an idea struck him. He felt that if he shouted and screamed and made a lot of noise, he would be arrested for behaving in a dangerous and disturbing way in public, and would be imprisoned, which was what he wanted.

Question 5.
Complete the following:
(The answer is given directly and underlined.)
Answer:
Soapy was angry because even after he had stolen a man’s umbrella, the man did not report him to the police, but instead apologized and said that perhaps he (the umbrella man) had made a mistake.

Question 6.
The umbrella man did not call a policeman. Give reasons for this.
Answer:
The umbrella man had himself probably stolen the umbrella from somewhere. When Soapy picked up the umbrella, the man first thought that he could get it back. But when Soapy spoke about calling a policeman, the man thought that the umbrella was actually Soapy’s, and Soapy would hand him over to the police. Hence, he apologized quickly and walked away without calling a policeman.

Maharashtra Board Solutions

Personal Response:

Question 1.
Give your opinion about Soapy’s desire to enter prison.
Answer:
I find it very strange that a person can actually want to enter prison, whatever be the reason. At least, from what I know of prisons, they are terrible places, and one has a very difficult time there. However, Soapy has obviously been to prison before, and probably enjoys the free food and protection from the winter that he gets there.

Question 2.
Have you ever bought/eaten something and then found that you did not have enough money to pay for it? Describe your feeling at that time.
Answer:
Yes, it happened to me once. I went to a mall and bought a jacket for myself. I had been looking at a lot of jackets and I got confused with the prices. Finally, when the cashier was making the bill, I found that the jacket I had chosen was very expensive and I did not have enough money to pay for it. I was very embarrassed to tell the cashier this, but I had to. He gave me an angry look.

Question 3.
Have you ever stolen/wanted to steal anything? Narrate in brief.
Answer:
Yes, when I was about 12 years old, I stole my friend’s remote-controlled toy car, which his uncle had sent him from abroad. It was a beautiful car. However, when I was playing with it at home my mother saw me, and she made me return the car. I later felt very ashamed of myself, but fortunately my friend forgave me.

Language Study.

Question 1.
If only he could reach a table in the restaurant unsuspected, success would be his.
(Rewrite using ‘unless’.)
Answer:
Unless he could reach a table in the restaurant unsuspected, success would not be his.

Maharashtra Board Solutions

Question 2.
One dollar for the cigar would be enough.
(Add a question tag.)
Answer:
One dollar for the cigar would be enough, wouldn’t it?

Question 3.
Some other way of entering the limbo must be devised.
(Use an infinitive in place of a gerund.)
Answer:
Some other way to enter the limbo must be devised.

Question 4.
He had set his silk umbrella by the door on entering. (Rewrite using the verb form of the underlined word.)
Answer:
He had set his silk umbrella by the door when he entered.

Question 5.
At length Soapy reached one of the avenues to the east. (Rewrite using another adverb phrase with the same meaning as the underlined phrase.)
Answer:
After a long time Soapy reached one of the avenues to the east.

Question 6.
On an unusually quiet corner, Soapy came to a standstill. (Rewrite using ‘that’.)
Answer:
Soapy came to a standstill on a corner that was unusually quiet.

Maharashtra Board Solutions

Vocabulary:

Question 1.
Guess the meaning of the following in the context:
1. winter island
2. eye fell upon
Answer:
1. winter island – prison.
2. eye fell upon – saw or noticed.

Question 2.
O’Henry has used different words to indicate prison, where Soapy wants to reach. Make a list of those words from the extract.
Answer:

  1. winter island
  2. coveted island
  3. limbo

Question 3.
Make sentences using the following words/expressions :
1. eye fell upon
2. strolled
Answer:
1. My eye fell upon the clock, and I sat up with shock.
2. Seema strolled along the beach, enjoying the breeze.

Question 4.
Guess the meaning of:

  1. napery
  2. betook
  3. brass buttons

Answer:

  1. napery – table linen.
  2. betook – to cause oneself to go.
  3. brass buttons – the police.

Maharashtra Board Solutions

Question 5.
O’Henry has used different words to indicate prison where Soapy wants to reach. Make a list of those words from the extract:
Answer:
the island

Question 6.
Fill in the blanks with the correct nouns from the extract:

  1. friendly
  2. electric
  3. large
  4. callous

Answer:

  1. friendly voice
  2. electric lights
  3. large appetites
  4. callous pavement

Question 7.
O’Henry has used different words to indicate prison where Soapy wants to reach. Make a list of those words from the extract.
Answer:
the island, Arcadia

Maharashtra Board Solutions

Question 8.
Pick out four verbs in the simple past tense from the extract.
Answer:
danced, howled, raved, disturbed

Question 9.
Match the words in Column A with the meanings in Column B :
Answer:

  1. disconsolate – very unhappy
  2. sauntered – walked in a relaxed manner
  3. raved – spoke wildly
  4. rendered – made

Non-Textual Grammar

Do as directed:

Question 1.
Shivani found a small box and dropped her bangles inside.
(Rewrite the sentence, beginning ‘Finding …)
Answer:
Finding a small box, Shivani dropped her bangles inside.

Maharashtra Board Solutions

Question 2.
Sunlight from the window made her black hair appear brown. (Rewrite using ‘that’.)
Answer:
Sunlight that came from the window made her black hair appear brown.

Question 3.
On the day the school closed for the summer, no student was more delighted than Rithik.
(Change the degree.)
Answer:
1. On the day the school closed for the summer, Rithik was the most delighted student. – Superlative degree
2. On the day the school closed for the summer, Rithik was more delighted than any other student. – Comparative degree

Spot the error in the following sentences and rewrite them correctly:

Question 1.
There is room for much boxes in this cupboard.
Answer:
There is room for many boxes in this cupboard.

Maharashtra Board Solutions

Question 2.
If I requires help for him in public places, I was not embarrassed to seek it from people around.
Answer:
If I required help for him in public places, I was not embarrassed to seek it from people around.

12th Std English Questions And Answers:

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

(I) Choose the correct alternative.

Question 1.
F(x) is c.d.f. of discreter r.v. X whose p.m.f. is given by P(x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\), for x = 0, 1, 2, 3, 4 & P(x) = 0 otherwise then F(5) = __________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{4}\)
(d) 1
Answer:
(d) 1

Question 2.
F(x) is c.d.f. of discrete r.v. X whose distribution is
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 I Q2
then F(-3) = __________
(a) 0
(b) 1
(c) 0.2
(d) 0.15
Answer:
(a) 0

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 3.
X : number obtained on uppermost face when a fair die is thrown then E(X) = __________
(a) 3.0
(b) 3.5
(c) 4.0
(d) 4.5
Answer:
(b) 3.5

Question 4.
If p.m.f. of r.v. X is given below.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 I Q4
then Var(X) = __________
(a) p2
(b) q2
(c) pq
(d) 2pq
Answer:
(d) 2pq

Question 5.
The expected value of the sum of two numbers obtained when two fair dice are rolled is __________
(a) 5
(b) 6
(c) 7
(d) 8
Answer:
(c) 7

Question 6.
Given p.d.f. of a continuous r.v. X as
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0 otherwise then F(1) =
(a) \(\frac{1}{9}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{3}{9}\)
(d) \(\frac{4}{9}\)
Answer:
(b) \(\frac{2}{9}\)

Question 7.
X is r.v. with p.d.f.
f(x) = \(\frac{k}{\sqrt{x}}\), 0 < x < 4
= 0 otherwise then E(X) = __________
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{2}{3}\)
(d) 1
Answer:
(b) \(\frac{4}{3}\)

Question 8.
If X follows B(20, \(\frac{1}{10}\)) then E(X) = __________
(a) 2
(b) 5
(c) 4
(d) 3
Answer:
(a) 2

Question 9.
If E(X) = m and Var(X) = m then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
Answer:
(b) Possion distribution

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 10.
If E(X) > Var(X) then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
Answer:
(a) Binomial distribution

(II) Fill in the blanks.

Question 1.
The values of discrete r.v. are generally obtained by __________
Answer:
counting

Question 2.
The values of continuous r.v. are generally obtained by __________
Answer:
measurement

Question 3.
If X is dicrete random variable takes the values x1, x2, x3, …… xn then \(\sum_{i=1}^{n} p\left(x_{i}\right)\) = __________
Answer:
1

Question 4.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = \(\frac{x-1}{3}\) for x = 1, 2, 3, and p(x) = 0 otherwise then F(4) = __________
Answer:
1

Question 5.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\) for x = 0, 1, 2, 3, 4, and p(x) = 0 otherwise then F(-1) = __________
Answer:
0

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 6.
E(X) is considered to be __________ of the probability distribution of X.
Answer:
centre of gravity

Question 7.
If X is continuous r.v. and f(xi) = P(X ≤ xi) = \(\int_{-\infty}^{x_{i}} f(x) d x\) then f(x) is called __________
Answer:
Cumulative Distribution Function

Question 8.
In Binomial distribution probability of success ________ from trial to trial.
Answer:
remains constant/independent

Question 9.
In Binomial distribution, if n is very large and probability success of p is very small such that np = m (constant) then ________ distribution is applied.
Answer:
Possion

(III) State whether each of the following is True or False.

Question 1.
If P(X = x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\) for x = 0, 1, 2, 3, 4, then F(5) = \(\frac{1}{4}\) when f(x) is c.d.f.
Answer:
False

Question 2.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 III Q2
If F(x) is c.d.f. of discrete r.v. X then F(-3) = 0.
Answer:
True

Question 3.
X is the number obtained on the uppermost face when a die is thrown the E(X) = 3.5.
Answer:
True

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 4.
If p.m.f. of discrete r.v.X is
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 III Q4
then E(X) = 2p.
Answer:
True

Question 5.
The p.m.f. of a r.v. X is p(x) = \(\frac{2 x}{n(n+1)}\), x = 1, 2,……n
= 0 otherwise,
Then E(X) = \(\frac{2 n+1}{3}\)
Answer:
True

Question 6.
If f(x) = kx (1 – x) for 0 < x < 1
= 0 otherwise then k = 12
Answer:
False

Question 7.
If X ~ B(n, p) and n = 6 and P(X = 4) = P(X = 2) then p = \(\frac{1}{2}\).
Answer:
True

Question 8.
If r.v. X assumes values 1, 2, 3,………, n with equal probabilities then E(X) = \(\frac{(n+1)}{2}\)
Answer:
True

Question 9.
If r.v. X assumes the values 1, 2, 3,………, 9 with equal probabilities, E(X) = 5.
Answer:
True

(IV) Solve the following problems.

Part – I

Question 1.
Identify the random variable as discrete or continuous in each of the following. Identify its range if it is discrete.
(i) An economist is interested in knowing the number of unemployed graduates in the town with a population of 1 lakh.
Solution:
X = No. of unemployed graduates in a town.
∵ The population of the town is 1 lakh
∴ X takes finite values
∴ X is a Discrete Random Variable
∴ Range of = {0, 1, 2, 4, …. 1,00,000}

(ii) Amount of syrup prescribed by a physician.
Solution:
X : Amount of syrup prescribed.
∴ X Takes infinite values
∴ X is a Continuous Random Variable.

(iii) A person on a high protein diet is interested in the weight gained in a week.
Solution:
X : Gain in weight in a week.
X takes infinite values
∴ X is a Continuous Random Variable.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

(iv) Twelve of 20 white rats available for an experiment are male. A scientist randomly selects 5 rats and counts the number of female rats among them.
Solution:
X : No. of female rats selected
X takes finite values.
∴ X is a Discrete Random Variable.
Range of X = {0, 1, 2, 3, 4, 5}

(v) A highway safety group is interested in the speed (km/hrs) of a car at a checkpoint.
Solution:
X : Speed of car in km/hr
X takes infinite values
∴ X is a Continuous Random Variable.

Question 2.
The probability distribution of a discrete r.v. X is as follows.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q2
(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:
(i) Assuming that the given distribution is a p.m.f. of X
∴ Each P(X = x) ≥ 0 for x = 1, 2, 3, 4, 5, 6
k ≥ 0
ΣP(X = x) = 1 and
k + 2k + 3k + 4k + 5k + 6k = 1
∴ 21k = 1 ∴ k = \(\frac{1}{21}\)

(ii) P(X ≤ 4) = 1 – P(X > 4)
= 1 – [P(X = 5) + P(X = 6)]
= 1 – [latex]\frac{5}{21}+\frac{6}{21}[/latex]
= 1 – \(\frac{11}{21}\)
= \(\frac{10}{21}\)
P(2 < X < 6) = p(3) + p(4) + p(5)
= 3k + 4k + 5k
= \(\frac{3}{21}+\frac{4}{21}+\frac{5}{21}\)
= \(\frac{12}{21}\)
= \(\frac{4}{7}\)

(iii) P(X ≥ 3) = p(3) + p(4) + p(5) + p(6)
= 3k + 4k + 5k + 6k
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q2.1

Question 3.
Following is the probability distribution of an r.v. X.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q3
Find the probability that
(i) X is positive.
(ii) X is non-negative.
(iii) X is odd.
(iv) X is even.
Solution:
(i) P(X is positive)
P(X = 0) = p(1) + p(2) + p(3)
= 0.25 + 0.15 + 0.10
= 0.50

(ii) P(X is non-negative)
P(X ≥ 0) = p(0) + p(1) + p(2) + p(3)
= 0.20 + 0.25 + 0.15 + 0.10
= 0.70

(iii) P(X is odd)
P(X = -3, -1, 1, 3)
= p(- 3) +p(-1) + p(1) + p(3)
= 0.05 + 0.15 + 0.25 + 0.10
= 0.55

(iv) P(X is even)
= 1 – P(X is odd)
= 1 – 0.55
= 0.45

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 4.
The p.m.f of a r.v. X is given by
\(P(X=x)= \begin{cases}\left(\begin{array}{l}
5 \\
x
\end{array}\right) \frac{1}{2^{5}}, & x=0,1,2,3,4,5 . \\
0 & \text { otherwise }\end{cases}\)
Show that P(X ≤ 2) = P(X ≥ 3).
Solution:
For x = 0, 1, 2, 3, 4, 5
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q4

Question 5.
In the following probability distribution of an r.v. X
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q5
Find a and obtain the c.d.f. of X.
Solution:
Given distribution is p.m.f. of r.v. X
ΣP(X = x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) = 1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q5.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q5.2

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p.m.f. of X.
Solution:
A fair coin is tossed 4 times
∴ Sample space contains 16 outcomes
Let X = Number of heads obtained
∴ X takes the values x = 0, 1, 2, 3, 4.
∴ The number of heads obtained in a toss is an even
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q6

Question 7.
Find the probability of the number of successes in two tosses of a die, where success is defined as (i) number greater than 4 (ii) six appearing in at least one toss.
Solution:
S : A die is tossed two times
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
(i) X : No. is greater than 4
Range of X = {0, 1, 2}
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q7

(ii) X : Six appears on aleast one die.
Range of X = {0, 1, 2}
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q7.1

Question 8.
A random variable X has the following probability distribution.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q8
Determine (i) k, (ii) P(X < 3), (iii) P(X > 6), (iv) P(0 < X < 3).
Solution:
(i) It is a p.m.f. of r.v. X
Σp(x) = 1
p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
9k + 10k2 = 1
10k2 + 9k – 1 = 0
10k2 +10k – k – 1 = 0
∴ 10k(k + 1) – 1(k + 1) = 0
∴ (10k – 1) (k + 1) = 0
∴ 10k – 1 = 0r k + 1 = 0
∴ k = \(\frac{1}{10}\) or k = -1
k = -1 is not accepted, p(x) ≥ 0, ∀ x ∈ R
∴ k = \(\frac{1}{10}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

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

(iii) P(X > 6) = p(7)
= 7k2 + k
= \(7\left(\frac{1}{10}\right)^{2}+\frac{1}{10}\)
= \(\frac{7}{100}+\frac{1}{10}\)
= \(\frac{17}{100}\)

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

Question 9.
The following is the c.d.f. of a r.v. X.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q9
Find the probability distribution of X and P(-1 ≤ X ≤ 2).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q9.1
P(-1 ≤ X ≤ 2) = p(-1) + p(0) + p(1) + p(2)
= 0.2 + 0.15 + 0.10 + 0.10
= 0.55

Question 10.
Find the expected value and variance of the r.v. X if its probability distribution is as follows.
(i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(i)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(i).1

(ii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(ii)
Solution:
E(X) = Σx . p(x)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(ii).1

(iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(iii)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(iii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(iii).2

(iv)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(iv)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q10(iv).1
= 1.25
S.D. of X = σx = √Var(X)
= √1.25
= 1.118

Question 11.
A player tosses two coins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears, and ₹ 2 if no head appears. Find the expected value and variance of the winning amount.
Solution:
S : Two fair coin are tossed
S = {HH, HT, TT, TH}
n(S) = 4
∴ Range of X = {0, 1, 2}
∴ Let Y = amount received corresponds to values of X
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q11
Expected winning amount
E(Y) = Σpy = \(\frac{22}{4}\) = ₹ 5.5
V(Y) = Σpy2 – (Σpy)2
= \(\frac{154}{4}\) – (5.5)2
= 38.5 – 30.25
= ₹ 8.25

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 12.
Let the p.m.f. of the r.v. X be
\(p(x)= \begin{cases}\frac{3-x}{10} & \text { for } x=-1,0,1,2 \\ 0 & \text { otherwise }\end{cases}\)
Calculate E(X) and Var(X).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q12
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q12.1

Question 13.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\(f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
We know that
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q13
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q13.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q13.2

Question 14.
The p.d.f. of the r.v. X is given by
\(f(x)= \begin{cases}\frac{1}{2 a} & \text { for } 0<x<2 a \\ 0 & \text { otherwise }\end{cases}\)
Show that P(X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q14

Question 15.
Determine k if
\(f(x)= \begin{cases}k e^{-\theta x} & \text { for } 0 \leq x<\infty, \theta>0 \\ 0 & \text { otherwise }\end{cases}\)
is the p.d.f. of the r.v. X. Also find P(X > \(\frac{1}{\theta}\)). Find M if P(0 < X < M) = \(\frac{1}{2}\)
Solution:
We know that
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q15
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q15.1

Question 16.
The p.d.f. of the r.v. X is given by
\(f_{x}(x)=\left\{\begin{array}{l}
\frac{k}{\sqrt{x}}, 0<x<4 \\
0, \text { otherwise }
\end{array}\right.\)
Determine k, c.d.f. of X and hence find P(X ≤ 2) and P(X ≥ 1).
Solution:
We know that
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q16
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q16.1

Question 17.
Let X denote the reaction temperature (in °C) of a certain chemical process. Let X be a continuous r.v. with p.d.f.
\(f(x)= \begin{cases}\frac{1}{10}, & -5 \leq x \leq 5 \\ 0, & \text { otherwise }\end{cases}\)
Compute P(X < 0).
Solution:
Given p.d.f. is f(x) = \(\frac{1}{10}\), for -5 ≤ x ≤ 5
Let its c.d.f. F(x) be given by
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 1 Q17

Part – II

Question 1.
Let X ~ B(10, 0.2). Find (i) P(X = 1) (ii) P(X ≥ 1) (iii) P(X ≤ 8)
Solution:
X ~ B(10, 0.2)
n = 10, p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
(i) P(X = 1) = 10C1 (0.2)1 (0.8)9 = 0.2684

(ii) P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)
= 1 – 10C0 (0.2)0 (0.8)10
= 1 – 0.1074
= 0.8926

(iii) P(X ≤ 8) = 1 – P(x > 1)
= 1 – [p(9) + p(10)]
= 1 – [10C9 (0.2)9 (0.8)1 + 10C10 (0.2)10]
= 1 – 0.00000041984
= 0.9999

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

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

(ii) E(X) = 5, V(X) = 2.5
∴ np = 5, ∴ npq = 2.5
∴ 5q = 2.5
∴ q = \(\frac{2.5}{5}\) = 0.5, p = 1 – 0.5 = 0.5
But np = 5
∴ n(0.5) = 5
∴ n = 10

Question 3.
If a fair coin is tossed 4 times, find the probability that it shows (i) 3 heads, (ii) head in the first 2 tosses, and tail in the last 2 tosses.
Solution:
n : No. of times a coin is tossed
∴ n = 4
X : No. of heads
P : Probability of getting heads
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 2 Q3

Question 4.
The probability that a bomb will hit the target is 0.8. Find the probability that, out of 5 bombs, exactly 2 will miss the target.
Solution:
X : No. of bombs miss the target
p : Probability that bomb miss the target
∴ q = 0.8
∴ p = 1 – q = 1 – 0.8 = 0.2
n = No. of bombs = 5
∴ X ~ B(5, 0.2)
∴ p(x) = nCx px qn-x
P(X = 2) = 5C2 (0.2)2 (0.8)5-2
= 10 × 0.04 × (0.8)3
= 10 × 0.04 × 0.512
= 0.4 × 0.512
= 0.2048

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 5.
The probability that a lamp in the classroom will burn is 0.3. 3 lamps are fitted in the classroom. The classroom is unusable if the number of lamps burning in it is less than 2. Find the probability that the classroom can not be used on a random occasion.
Solution:
X : No. of lamps not burning
p : Probability that the lamp is not burning
∴ q = 0.3
∴ p = 1 – q = 1 – 0.3 = 0.7
n = No. of lamps fitted = 3
∴ X ~ B(3, 0.7)
∴ p(x) = nCx px qn-x
P(classroom cannot be used)
P(X < 2) = p(0) + p(1)
= 3C0 (0.7)0 (0.3)3-0 + 3C1 (0.7)1 (0.3)3-1
= 1 × 1 × (0.3)3 + 3 × 0.7 × (0.3)2
= (0.3)2 [0.3 + 3 × 0.7]
= 0.09 [0.3 + 2.1]
= 0.09 [2.4]
= 0.216

Question 6.
A large chain retailer purchases an electric device from the manufacturer. The manufacturer indicates that the defective rate of the device is 10%. The inspector of the retailer randomly selects 4 items from a shipment. Find the probability that the inspector finds at most one defective item in the 4 selected items.
Solution:
X : No. of defective items
n : No. of items selected = 4
p : Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
P(At most one defective item)
P(X ≤ 1) = p(0) + p(1)
= 4C0 (0.1)0 (0.9)4-0 + 4C1 (0.1)1 (0.9)4-1
= 1 × 1 × (0.9)4 + 4 × 0.1 × (0.9)3
= (0.9)3 [0.9 + 4 × 0.1]
= (0.9)3 × [0.9 + 0.4]
= 0.729 × 1.3
= 0.9477

Question 7.
The probability that a component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 components tested survive.
Solution:
p = 0.6, q = 1 – 0.6 = 0.4, n = 4
x = 2
∴ p(x) = nCx px qn-x
P(X = 2) = 4C2 (0.6)2 (0.4)2 = 0.3456

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 8.
An examination consists of 5 multiple choice questions, in each of which the candidate has to decide which one of 4 suggested answers is correct. A completely unprepared student guesses each answer randomly. Find the probability that this student gets 4 or more correct answers.
Solution:
n : No. of multiple-choice questions
∴ n = 5
X : No. of correct answers
p : Probability of getting correct answer
∵ There are 4 options out of which one is correct
∴ p = \(\frac{1}{4}\)
∴ q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∵ X ~ B(5, \(\frac{1}{4}\))
∴ p(x) = nCx px qn-x
P(Four or more correct answers)
P(X ≥ 4) = p(4) + p(5)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 2 Q8

Question 9.
The probability that a machine will produce all bolts in a production run with in the specification is 0.9. A sample of 3 machines is taken at random. Calculate the probability that all machines will produce all bolts in a production run within the specification.
Solution:
n : No. of samples selected
∴ n = 3
X : No. of bolts produce by machines
p : Probability of getting bolts
∴ p = 0.9
∴ q = 1 – p = 1 – 0.9 = 0.1
∴ X ~ B(3, 0.9)
∴ p(x) = nCx px qn-x
P(Machine will produce all bolts)
P(X = 3) = 3C3 (0.9)3 (0.1)3-3
= 1 × (0.9)3 × (0.1)0
= 1 × (0.9)3 × 1
= (0.9)3
= 0.729

Question 10.
A computer installation has 3 terminals. The probability that anyone terminal requires attention during a week is 0.1, independent of other terminals. Find the probabilities that (i) 0 (ii) 1 terminal requires attention during a week.
Solution:
n : No. of terminals
∴ n = 3
X : No. of terminals need attention
p : Probability of getting terminals need attention
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
∵ X ~ B(3, 0.1)
∴ p(x) = nCx px qn-x
(i) P(No attention)
∴ P(X = 0) = 3C0 × (0.1)0 (0.9)3-1
= 1 × 1 × (0.9)3
= 0.729

(ii) P(One terminal need attention)
∴ P(X = 1) = 3C1 (0.1)1 (0.9)3-1
= 3 × 0.1 × (0.9)2
= 0.3 × 0.81
= 0.243

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 11.
In a large school, 80% of the students like mathematics. A visitor asks each of 4 students, selected at random, whether they like mathematics, (i) Calculate the probabilities of obtaining an answer yes from all of the selected students, (ii) Find the probability that the visitor obtains the answer yes from at least 3 students.
Solution:
X : No. of students like mathematics
p: Probability that students like mathematics
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
n : No. of students selected
∴ n = 4
∵ X ~ B(4, 0.8)
∴ p(x) = nCx px qn-x
(i) P(All students like mathematics)
∴ P(X = 4) = 4C4 (0.8)4 (0.2)4-4
= 1 × (0.8)4 × (0.2)0
= 1 × (0.8)4 × 1
= 0.4096

(ii) P(Atleast 3 students like mathematics)
∴ P(X ≥ 3) = p(3) + p(4)
= 4C3 (0.8)3 (0.2)4-3 + 0.4096
= 4 × (0.8)3 (0.2)1 + 0.4096
= 0.8 × (0.8)3 + 0.4096
= (0.8)4 × 0.4096
= 0.4096 + 0.4096
= 0.8192

Question 12.
It is observed that it rains on 10 days out of 30 days. Find the probability that
(i) it rains on exactly 3 days of a week.
(ii) it rains at most 2 days a week.
Solution:
X : No. of days it rains in a week
p : Probability that it rains
∴ p = \(\frac{10}{30}=\frac{1}{3}\)
∴ q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
n : No. of days in a week
∴ n = 7
∴ X ~ B(7, \(\frac{1}{3}\))
(i) P(Rains on Exactly 3 days of a week)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 2 Q12

(ii) P(Rains on at most 2 days of a week)
∴ P(X ≤ 2) = p(0) + p(1) + p(2)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 2 Q12.1

Question 13.
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.
Solution:
X : Follows Possion Distribution
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8 IV Part 2 Q13
∴ m = 1
∴ Mean = m = Variance of X = 1

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

Question 14.
If X has Poisson distribution with parameter m, such that
\(\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}\)
find probabilities P(X = 1) and P(X = 2), when X follows Poisson distribution with m = 2 and P(X = 0) = 0.1353.
Solution:
Given that the random variable X follows the Poisson distribution with parameter m = 2
i.e. X ~ P(2)
Its p.m.f. is satisfying the given equation.
\(\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}\)
When x = 0,
\(\frac{\mathrm{P}(\mathrm{X}=1)}{\mathrm{P}(\mathrm{X}=0)}=\frac{2}{0+1}\)
P(X = 1) = 2P(X = 0)
= 2(0.1353)
= 0.2706
When x = 1,
\(\frac{\mathrm{P}(\mathrm{X}=2)}{\mathrm{P}(\mathrm{X}=1)}=\frac{2}{1+1}\)
P(X = 2) = P(X = 1) = 0.2706

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 1.
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e-1 = 0.3678.
Solution:
∵ m = 1
∵ X follows Poisson Distribution
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Q1
= e-m × 1 + e-m × 1
= e-1 + e-1
= 2 × e-1
= 2 × 0.3678
= 0.7356

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 2.
If X ~ P(\(\frac{1}{2}\)), then find P(X = 3) given e-0.5 = 0.6065.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Q2

Question 3.
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e-3 = 0.0497
Solution:
∵ X follows Poisson Distribution
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Q3

Question 4.
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives (i) only two complaints on a given day, (ii) at most two complaints on a given day. Use e-4 = 0.0183.
Solution:
∵ m = 1
∵ X ~ P(m = 4)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
X = No. of complaints recieved
(i) P(Only two complaints on a given day)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Q4

(ii) P(Atmost two complaints on a given day)
P(X ≤ 2) = p(0) + p(1) + p(2)
= \(\frac{e^{-4} \times 4^{0}}{0 !}+\frac{e^{-4} \times 4^{1}}{1 !}\) + 0.1464
= e-4 + e-4 × 4 + 0.1464
= e-4 [1 + 4] + 0.1464
= 0.0183 × 5 + 0.1464
= 0.0915 + 0.1464
= 0.2379

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 5.
A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that
(i) no car is used on a given day.
(ii) some demand is refused on a given day, given e-1.5 = 0.2231.
Solution:
Let X = No. of demands for a car on any day
∴ No. of cars hired
n = 2
m = 1.5
∵ X ~ P(m = 1.5)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Q5

Question 6.
Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has (i) no defect, (ii) at least one defect. Use e-1 = 0.3678.
Solution:
∵ X = No. of defects on a plywood sheet
∵ m = -1
∵ X ~ P(m = -1)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(No defect)
P(X = 0) = \(\frac{e^{-1} \times 1^{0}}{0 !}\)
= e-1
= 0.3678

(ii) P(At least one defect)
P(X ≥ 1) = 1 – P(X < 1)
= 1 – p(0)
= 1 – 0.3678
= 0.6322

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 7.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has
(i) exactly 5 rats
(ii) more than 5 rats
(iii) between 5 and 7 rats, inclusive. Given e-5 = 0.0067.
Solution:
X = No. of rats
∵ m = 5
∴ X ~ P(m = 5)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(Exactly five rats)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Q7

(ii) P(More than five rats)
P(X > 5) = 1 – P(X ≤ 5)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Q7.1

(iii) P(between 5 and 7 rats, inclusive)
P(5 ≤ x ≤ 7) = p(5) + p(6) + p(7)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Q7.2
= 0.0067 × 3125 × 0.02
= 0.0067 × 62.5
= 0.42

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 1.
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of (i) 2 successes (ii) at least 3 successes (iii) at most 2 successes.
Solution:
X: Getting an odd no.
p: Probability of getting an odd no.
A die is thrown 4 times
∴ n = 4
∵ p = \(\frac{3}{6}=\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∵ X ~ B(3, \(\frac{1}{2}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(Two Successes)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q1
(ii) P(Atleast 3 Successes)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q1.1
(iii) P(Atmost 2 Successes)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q1.2

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 2.
A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
n: No. of times die is thrown = 3
X: No. of doublets
p: Probability of getting doublets
Getting a doublet means, same no. is obtained on 2 throws of a die
There are 36 outcomes
No. of doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q2

Question 3.
There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Solution:
n: No of sample items = 4
X: No of defective items
p: Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
X ~ B(4, 0.1)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} \mathrm{q}^{n-x}\)
P(Not include more than 1 defective)
P(X ≤ 1) = p(0) + p(1)
= 4C0 (0.1)0 (0.9)4 + 4C1 (0.1)1 (0.9)4-1
= 1 × 1 × (0.9)4 + 4 × 0.1 × (0.9)3
= (0.9)3 [0.9 + 0.4]
= (0.9)3 × 1.3
= 0.977

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that (i) all the five cards are spades, (ii) only 3 cards are spades, (iii) none is a spade.
Solution:
X: No. of spade cards
Number of cards drawn
∴ n = 5
p: Probability of getting spade card
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q4
(i) P(All five cards are spades)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q4.1
(ii) P(Only 3 cards are spades)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q4.2
(iii) P(None is a spade)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q4.3

Question 5.
The probability that a bulb produced by a factory will use fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.
Solution:
X : No. of bulbs fuse after 200 days of use
p : Probability of getting fuse bulbs
No. of bulbs in a sample
∴ n = 5
∴ p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
∵ X ~ B(5, 0.2)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(X = 0) = 5C0 (0.2)0 (0.8)5-0
= 1 × 1 × (0.8)5
= (0.8)5

(ii) P(X ≤ 1) = p(0) + p(1)
= 5C0 (0.2)0 (0.8)5-0 + 5C1 (0.2)1 (0.8)5-1
= 1 × 1 × (0.8)5 + 5 × 0.2 × (0.8)4
= (0.8)4 [0.8 + 1]
= 1.8 × (0.8)4

(iii) P(X > 1) = 1 – [p(0) + p(1)]
= 1 – 1.8 × (0.8)4

(iv) P(X ≥ 1) = 1 – p(0)
= 1 – (0.8)5

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 6.
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Solution:
X : No. of balls drawn marked with the digit 0
n : No. of balls drawn
∴ n = 4
p : Probability of balls marked with 0.
∴ p = \(\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
p(x) = \({ }^{n} C_{x} p^{x} q^{n-x}\)
P(None of the ball is marked with digit 0)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q6

Question 7.
In a multiple-choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Solution:
n: No. of Questions
∴ n = 5
X: No. of correct answers by guessing
p: Probability of getting correct answers
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q7

Question 8.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
X : No. of sixes in 6 throws
n : No. of times dice thrown
∴ n = 6
p : Probability of getting six
∴ p = \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
∵ X ~ B(6, \(\frac{1}{6}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
P(At most 2 sixes)
P(X ≤ 2) = p(0) + p(1) + p(2)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3 Q8

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 9.
Given that X ~ B(n, p),
(i) if n = 10 and p = 0.4, find E(X) and Var(X).
(ii) if p = 0.6 and E(X) = 6, find n and Var(X).
(iii) if n = 25, E(X) = 10, find p and Var(X).
(iv) if n = 10, E(X) = 8, find Var(X).
Solution:
∵ X ~ B (n, p), E(X) = np, V(X) = npq, q = 1 – p
(i) E(X) = np = 10 × 0.4 = 4
∵ q = 1 – p = 1 – 0.4 = 0.6
V(X) = npq = 10 × 0.4 × 0.6 = 2.4

(ii) ∵ p = 0.6
∴ q = 1 – p = 1 – 0.6 = 0.4
E(X) = np
∴ 6 = n × 0.6
∴ n = 10
∴ V(X) = npq = 10 × 0.6 × 0.4 = 2.4

(iii) E(X) = np
∴ 10 = 25 × p
∴ p = 0.4
∴ q = 1, p = 1 – 0.4 = 0.6
∴ S.D.(X) = √V(X)
= \(\sqrt{n p q}\)
= \(\sqrt{25 \times 0.4 \times 0.6}\)
= √6
= 2.4494

(iv) ∵ E(X) = np
∴ 8 = 10p
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
∵ V(X) = npq = 10 × 0.8 × 0.2 = 1.6

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Question 1.
Check whether each of the following is p.d.f.
(i) \(f(x)= \begin{cases}x & \text { for } 0 \leq x \leq 1 \\ 2-x & \text { for } 1<x \leq 2\end{cases}\)
Solution:
Given function is
f(x) = x, 0 ≤ x ≤ 1
Each f(x) ≥ 0, as x ≥ 0.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q1(i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q1(i).1
∴ The given function is a p.d.f. of x.

(ii) f(x) = 2 for 0 < x < 1
Solution:
Given function is
f(x) = 2 for 0 < x < 1 Each f(x) > 0,
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q1(ii)
∴ The given function is not a p.d.f.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Question 2.
The following is the p.d.f. of a r.v. X.
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
Find (i) P(X < 1.5), (ii) P(1 < X < 2), (iii) P(X > 2)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q2
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q2.1

Question 3.
It is felt that error in measurement of reaction temperature (in Celsius) in an experiment is a continuous r.v. with p.d.f.
\(f(x)= \begin{cases}\frac{x^{3}}{64} & \text { for } 0 \leq x \leq 4 \\ 0 & \text { otherwise }\end{cases}\)
(i) Verify whether f(x) is a p.d.f.
(ii) Find P(0 < X ≤ 1).
(iii) Find the probability that X is between 1 and 3.
Solution:
(i) f(x) is p.d.f. of r.v. X if
(a) f(x) ≥ 0, ∀ x ∈ R
(b) \(\int_{0}^{4} f(x) d x\) = 1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q3.1

Question 4.
Find k, if the following function represents the p.d.f. of a r.v. X.
(i) \(f(x)= \begin{cases}k x & \text { for } 0<x<2 \\ 0 & \text { otherwise }\end{cases}\)
Also find P[\(\frac{1}{4}\) < X < \(\frac{1}{2}\)]
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q4(i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q4(i).1

(ii) \(f(x)= \begin{cases}k x(1-x) & \text { for } 0<x<1 \\ 0 & \text { otherwise }\end{cases}\)
Also find (a) P[\(\frac{1}{4}\) < X < \(\frac{1}{2}\)], (b) P[X < \(\frac{1}{2}\)]
Solution:
We know that
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q4(ii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q4(ii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q4(ii).2

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected student and suppose that X has p.d.f.
\(f(x)= \begin{cases}0.5 x & \text { for } 0 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Calculate (i) P(X ≤ 1), (ii) P(0.5 ≤ X ≤ 1.5), (iii) P(X ≥ 1.5).
Solution:
Given p.d.f. of X is f(x) = 0.5x for 0 ≤ x ≤ 2
∴ Its c.d.f. F(x) is given by
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q5

(i) P(X < 1) = F(1)
= 0.25(1)2
= 0.25

(ii) P(0.5 < X < 1.5) = F(1.5) – F(0.5)
= 0.25(1.5)2 – 0.25(0.5)2
= 0.25[2.25 – 0.25]
= 0.25(2)
= 0.5

(iii) P(X ≥ 1.5) = 1 – P(X ≤ 1.5)
= 1 – F(1.5)
= 1 – 0.25(1.5)2
= 1 – 0.25(2.25)
= 1 – 0.5625
= 0.4375

Question 6.
Suppose X is the waiting time (in minutes) for a bus and its p.d.f. is given by
\(f(x)=\left\{\begin{array}{cl}
\frac{1}{5} & \text { for } 0 \leq x \leq 5 \\
0 & \text { otherwise }
\end{array}\right.\)
Find the probability that (i) waiting time is between 1 and 3 minutes, (ii) waiting time is more than 4 minutes.
Solution:
p.d.f. of r.v. X is given by
f(x) = \(\frac{1}{5}\) for 0 ≤ x ≤ 5
This is a constant function.
(i) Probability that waiting time X is between 1 and 3 minutes
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q6
(ii) Probability that waiting time X is more than 4 minutes
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q6.1

Question 7.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\(f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
Since given f(x) is a p.d.f. of r.v. X
Since -2 ≤ x ≤ 2
∴ x2 ≤ 4
∴ 4 – x2 ≥ 0
∴ k(4 – x2) ≥ 0
∴ k ≥ 0 [∵ f(x) ≥ 0]
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q7
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q7.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q7.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q7.3

Question 8.
Following is the p.d.f. of a continuous r.v. X.
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
(i) Find an expression for the c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7, and 5.
Solution:
The p.d.f. of a continuous r.v. X is
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
(i) c.d.f. of continuous r.v. X is given by
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q8

(ii) F(0.5) = \(\frac{(0.5)^{2}}{16}=\frac{0.25}{16}=\frac{1}{64}\) = 0.015
F(1.7) = \(\frac{(1.7)^{2}}{16}=\frac{2.89}{16}\) = 0.18
For any of x greater than or equal to 4, F(x) = 1
∴ F(5) = 1

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Question 9.
The p.d.f. of a continuous r.v. X is
\(f(x)=\left\{\begin{array}{cl}
\frac{3 x^{2}}{8} & \text { for } 0<x<2 \\
0 & \text { otherwise }
\end{array}\right.\)
Determine the c.d.f. of X and hence find (i) P(X < 1), (ii) P(X < -2), (iii) P(X > 0), (iv) P(1 < X < 2).
Solution:
The p.d.f. of a continuous r.v. X is
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q9
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q9.1

Question 10.
If a r.v. X has p.d.f.
\(f(x)= \begin{cases}\frac{c}{x} & \text { for } 1<x<3, c>0 \\ 0 & \text { otherwise }\end{cases}\)
Find c, E(X) and V(X). Also find f(x).
Solution:
The p.d.f. of r.v. X is
f(x) = \(\frac{c}{x}\), 1 < x < 3, c > 0
For p.d.f. of X, we have
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q10
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q10.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q10.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2 Q10.3

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Question 1.
Let X represent the difference between a number of heads and the number of tails obtained when a coin is tossed 6 times. What are the possible values of X?
Solution:
∵ A coin is tossed 6 times
S = {6H and 0T, 5H and 1T, 4H and 2T, 3H and 3T, 2H and 4T, 1H and 5T, 0H and 6T}
X: Difference between no. of heads and no. of tails.
X = 6 – 0 = 6
X = 5 – 1 = 4
X = 4 – 2 = 2
X = 3 – 3 = 0
X = 2 – 4 = -2
X = 1 – 5 = -4
X = 0 – 6 = -6
X = {-6, -4, -2, 0, 2, 4, 6}

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
S : Two bolts are drawn from the Urn
S = {RR, RB, BR, BB}
X : No. of black balls
∴ X = {0, 1, 2}

Question 3.
Determine whether each of the following is a probability distribution. Give reasons for your answer.
(i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q3(i)
Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{2} \mathrm{P}(\mathrm{X}=x)\) = p(0) + p(1) + p(2)
= 0.4 + 0.4 + 0.2
= 1
∴ The function is a p.m.f.

(ii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q3(ii)
Solution:
Here, p(3) = -0.1 < 0
∴ P(X = x) ≯ 0, ∀ x
∴ The function is not a p.m.f.

(iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q3(iii)
Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{2} \mathrm{P}(\mathrm{X}=x)\) = p(0) + p(1) + p(2)
= 0.1 + 0.6 + 0.3
= 1
∴ The function is a p.m.f.

(iv)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q3(iv)
Solution:
Here, P(Z = z) ≥ 0, ∀ z and
\(\sum_{x=-1}^{3} \mathrm{P}(\mathrm{Z}=z)\) = p(-1) + p(0) + p(1) + p(2) + p(3)
= 0.05 + 0 + 0.4 + 0.2 + 0.3
= 0.95
≠ 1
∴ The function is not a p.m.f.

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

(v)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q3(v)
Solution:
Here, P(Y = y) ≥ 0, ∀ y and
\(\sum_{x=-1}^{2} \mathrm{P}(\mathrm{Y}=y)\) = p(-1) + p(0) + p(1)
= 0.1 + 0.6 + 0.2
= 0.9
≠ 1
∴ The function is not a p.m.f.

(vi)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q3(vi)
Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{0} \mathrm{P}(X=x)\) = p(-2) + p(-1) + p(0)
= 0.3 + 0.4 + 0.2
= 0.9
≠ 1
∴ The function is not a p.m.f.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin,
(ii) number of trails in three tosses of a coin,
(iii) number of heads in four tosses of a coin.
Solution:
(i) S: Coin is tossed two times
S = {HH, HT, TH, TT}
n(S) = 4
X: No. of heads
Range of X = {0, 1, 2}
p.m.f. Table
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q4(i)

(ii) S: 3 coin are tossed
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
X: No. of heads
Range of X = {0, 1, 2, 3}
p.m.f. Table
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q4(ii)

(iii) S: Four coin are tossed
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
n(S) = 16
X: No. of heads
Range of X = {0, 1, 2, 3, 4}
p.m.f. Table
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q4(iii)

Question 5.
Find the probability distribution of the number of successes in two tosses of a die if successes are defined as getting a number greater than 4.
Solution:
S = A die is tossed 2 times
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
X = No. getting greater than 4
Range of X = {0, 1, 2}
p(0) = \(\frac{16}{36}=\frac{4}{9}\)
p(1) = \(\frac{16}{36}=\frac{4}{9}\)
p(2) = \(\frac{4}{36}=\frac{1}{9}\)
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q5

Question 6.
A sample of 4 bulbs is drawn at random with replacement from a lot of 30 bulbs which includes 6 defective bulbs. Find the probability distribution of the number of defective bulbs.
Solution:
Total no. of bulbs = 30
No. of defective bulbs = 6
A sample of 4 bulbs are drawn from 30 bulbs.
∴ n(S) = \({ }^{30} \mathrm{C}_{4}\)
∴ No. of non-defective bulbs = 24
Let X = No. of defective bulbs drawn in sample of 4 bulbs.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q6
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q6.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. Find the probability distribution of a number of tails in two tosses.
Solution:
Here, the head is 3 times as likely to occur as the tail.
i.e., If 4 times coin is tossed, 3 times there will be a head and 1 time there will be the tail.
∴ p(H) = \(\frac{3}{4}\) and p(T) = \(\frac{1}{4}\)
Let X : No. of tails in two tosses.
And coin is tossed twice.
∴ X = {0, 1, 2}
For X = 0,
p(0) = p(both heads)
= p(H) × p(H)
= \(\frac{3}{4} \times \frac{3}{4}\)
= \(\frac{9}{16}\)
For X = 1,
p(1) = p(HT or TH)
= p(HT) + p(TH)
= p(H) × p(T) + p(T) × p(H)
= \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}\)
= \(\frac{6}{16}\)
For X = 2,
p(2) = p(both tails)
= p(T) × p(T)
= \(\frac{1}{4} \times \frac{1}{4}\)
= \(\frac{1}{16}\)
The probability distribution of the number of tails in two tosses is
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q7

Question 8.
A random variable X has the following probability distribution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q8
Determine (i) k, (ii) P(X < 3), (iii) P(0 < X < 3), (iv) P(X > 4).
Solution:
(i) It is a p.m.f. of r.v. X
∴ Σp(x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
∴ k + 2k + 2k + 3k + k2 + 2k2 + (7k2 + k) = 1
∴ 10k2 + 9k = 1
∴ 10k2 + 9k – 1 = 0
∴ 10k2 + 10k – k – 1 = 0
∴ 10k(k + 1) – (k + 1) = 0
∴ (10k – 1)(k + 1) = 0
∴ 10k – 1 = 0 or k + 1 = 0
∴ k = \(\frac{1}{10}\) or k = -1
but k = -1 is not accepted
∴ k = \(\frac{1}{2}\) is accepted

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

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

(iv) P(X > 4) = p(5) + p(6) + p(7)
= k2 + 2k2 + (7k2 + k)
= 10k2 + k
= \(10\left(\frac{1}{10}\right)^{2}+\frac{1}{10}\)
= \(\frac{2}{10}\)
= \(\frac{1}{5}\)

Question 9.
Find expected value and variance of X using the following p.m.f.
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q9
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q9.1
E(X) = Σxp = -0.05
V(X) = Σx2p – (Σxp)2
= 2.25 – (-0.05)2
= 2.25 – 0.0025
= 2.2475

Question 10.
Find expected value and variance of X, the number on the uppermost face of a fair die.
Solution:
S : A fair die is thrown
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
X: No obtained on uppermost face of die
Range of X = {1, 2, 3, 4, 5, 6}
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q10
E(X) = Σxp = \(\frac{21}{6}=\frac{7}{2}\) = 3.5
V(X) = Σx2p – (Σxp)2
= \(\frac{91}{6}\) – (3.5)2
= 15.17 – 12.25
= 2.92

Question 11.
Find the mean of the number of heads in three tosses of a fair coin.
Solution:
S : A coin is tossed 3 times
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Range of X = {0, 1, 2, 3}
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q11
∴ Mean = E(X) = Σxp = \(\frac{12}{8}=\frac{3}{2}\) = 1.5

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
S : Two dice are thrown
S = {(1, 1), (1, 2), (1, 3), ……, (6, 6)}
n(S) = 36
Range of X = {0, 1, 2}
First 6 positive integers are 1, 2, 3, 4, 5, 6
X = Larger two numbers selected
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q12
∴ E(X) = Σxp = \(\frac{12}{36}=\frac{1}{3}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:
First 6 positive integers are 1, 2, 3, 4, 5, 6
X : The larger of the selected two numbers
S = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
n(S) = 30
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q13
E(X) = Σxp = \(\frac{140}{30}=\frac{14}{3}\) = 4.67

Question 14.
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance of X.
Solution:
S : Two fair dice are rolled
S = {(1, 1), (1, 2), (1, 4), ……, (6, 6)}
n(S) = 36
X : Sum of the two numbers.
Range of X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q14
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q14.1
V(X) = Σx2p – (Σxp)2
= \(\frac{1952}{36}-\left(\frac{252}{36}\right)^{2}\)
= 54.22 – (7)2
= 5.22
SD(X) = √V(X) = √5.22 = 2.28

Question 15.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. If X denotes the age of a randomly selected student, find the probability distribution of X. Find the mean and variance of X.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q15
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q15.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Question 16.
70% of the member’s favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and V(X).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1 Q16
E(X) = Σxp = 0.7
V(X) = Σx2p – (Σxp)2
= 0.7 – (0.7)2
= 0.7 – 0.49
= 0.21

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

(I) Choose the correct alternative.

Question 1.
In sequencing, an optimal path is one that minimizes ___________
(a) Elapsed time
(b) Idle time
(c) Both (a) and (b)
(d) Ready time
Answer:
(c) Both (a) and (b)

Question 2.
If job A to D have processing times as 5, 6, 8, 4 on first machine and 4, 7, 9, 10 on second machine then the optimal sequence is:
(a) CDAB
(b) DBCA
(c) BCDA
(d) ABCD
Answer:
(b) DBCA

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 3.
The objective of sequence problem is
(a) to find the order in which jobs are to be made
(b) to find the time required for the completing all the job on hand
(c) to find the sequence in which jobs on hand are to be processed to minimize the total time required for processing the jobs
(d) to maximize the cost
Answer:
(c) to find the sequence in which jobs on hand are to be processed to minimize the total time required for processing the jobs

Question 4.
If there are n jobs and m machines, then there will be ___________ sequences of doing the jobs.
(a) mn
(b) m(n!)
(c) nm
(d) (n!)m
Answer:
(d) (n!)m

Question 5.
The Assignment Problem is solved by
(a) Simple method
(b) Hungarian method
(c) Vector method
(d) Graphical method
Answer:
(b) Hungarian method

Question 6.
In solving 2 machine and n jobs sequencing problem, the following assumption is wrong
(a) No passing is allowed
(b) Processing times are known
(c) Handling times is negligible
(d) The time of passing depends on the order of machining
Answer:
(d) The time of passing depends on the order of machining

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 7.
To use the Hungarian method, a profit maximization assignments problem requires
(a) Converting all profit to opportunity losses
(b) A dummy person or job
(c) Matrix expansion
(d) Finding the maximum number of lines to cover all the zeros in the reduced matrix
Answer:
(a) Converting all profits to opportunity losses

Question 8.
Using the Hungarian method the optimal assignment obtained for the following assignment problem to minimize the total cost is:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 I Q8
(a) 1 – C, 2 – B, 3 – D, 4 – A
(b) 1 – B, 2 – C, 3 – A, 4 – D
(c) 1 – A, 2 – B, 3 – C, 4 – D
(d) 1 – D, 2 – A, 3 – B, 4 – C
Answer:
(a) 1 – C, 2 – B, 3 – D, 4 – A

Question 9.
The assignment problem is said to be unbalanced if
(a) Number of rows is greater than the number of columns
(b) Number of rows is lesser than number of columns
(c) Number of rows is equal to the number of columns
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 10.
The assignment problem is said to be balanced if
(a) Number of rows is greater than the number of columns
(b) Number of rows is lesser than number of columns
(c) Number of rows is equal to the number of columns
(d) If the entry of rows is zero
Answer:
(c) Number of rows is equal to number of columns

Question 11.
The assignment problem is said to be balanced if it is a
(a) Square matrix
(b) Rectangular matrix
(c) Unit matrix
(d) Triangular matrix
Answer:
(a) Square matrix

Question 12.
In an assignment problem if the number of rows is greater than the number of columns then
(a) Dummy column is added
(b) Dummy row is added
(c) Row with cost 1 is added
(d) Column with cost 1 is added
Answer:
(a) Dummy column is added

Question 13.
In a 3 machine and 5 jobs problem, the least of processing times on machines A, B, and C are 5, 1 and 3 hours and the highest processing times are 9, 5 and 7 respectively, then it can be converted to a 2 machine problem if the order of the machines is:
(a) B – A – C
(b) A – B – C
(c) C – B – A
(d) Any order
Answer:
(b) A – B – C

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 14.
The objective of an assignment problem is to assign
(a) Number of jobs to equal number of persons at maximum cost
(b) Number of jobs to equal number of persons at minimum cost
(c) Only the maximize cost
(d) Only to minimize cost
Answer:
(b) Number of jobs to equal number of persons at minimum cost

(II) Fill in the blanks.

Question 1.
An assignment problem is said to be unbalanced when ___________
Answer:
the number of rows is not equal to the number of columns

Question 2.
When the number of rows is equal to the Number of columns then the problem is said to be ___________ assignment problem.
Answer:
balanced

Question 3.
For solving assignment problem the matrix should be a ___________
Answer:
square matrix

Question 4.
If the given matrix is not a ___________ matrix, the assignment problem is called an unbalanced problem.
Answer:
square

Question 5.
A dummy row(s) or column(s) with the cost elements as ___________ the matrix of an unbalanced assignment problem as a square matrix.
Answer:
zero

Question 6.
The time interval between starting the first job and completing the last, job including the idle time (if any) in a particular order by the given set of machines is called ___________
Answer:
Total elapsed time

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 7.
The time for which a machine j does not have a job to process to the start of job i is called ___________
Answer:
Idle time

Question 8.
The maximization assignment problem is transformed to minimization problem by subtracting each entry in the table from the ___________ value in the table.
Answer:
maximum

Question 9.
When the assignment problem has more than one solution, then it is ___________ optimal solution.
Answer:
multiple

Question 10.
The time required for printing four books A, B, C, and D is 5, 8, 10, and 7 hours. While its data entry requires 7, 4, 3, and 6 hrs respectively. The sequence that minimizes total elapsed time is ___________
Answer:
A – D – B – C

(III) State whether each of the following is True or False.

Question 1.
One machine – one job is not an assumption in solving sequencing problems.
Answer:
False

Question 2.
If there are two least processing times for machine A and machine B, priority is given for the processing time which has the lowest time of the adjacent machine.
Answer:
True

Question 3.
To convert the assignment problem into a maximization problem, the smallest element in the matrix is deducted from all other elements.
Answer:
False

Question 4.
The Hungarian method operates on the principle of matrix reduction, whereby the cost table is reduced to a set of opportunity costs.
Answer:
True

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 5.
In a sequencing problem, the processing times are dependent on the order of processing the jobs on machines.
Answer:
False

Question 6.
The optimal assignment is made in the Hungarian method to cells in the reduced matrix that contain a Zero.
Answer:
True

Question 7.
Using the Hungarian method, the optimal solution to an assignment problem is fund when the minimum number of lines required to cover the zero cells in the reduced matrix equals the number of people.
Answer:
True

Question 8.
In an assignment problem, if a number of columns are greater than the number of rows, then a dummy column is added.
Answer:
False

Question 9.
The purpose of a dummy row or column in an assignment problem is to obtain a balance between a total number of activities and a total number of resources.
Answer:
True

Question 10.
One of the assumptions made while sequencing n jobs on 2 machines is: two jobs must be loaded at a time on any machine.
Answer:
False

(IV) Solve the following problems.

Part – I

Question 1.
A plant manager has four subordinates, and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. This estimate of the times each man would take to perform each task is given in the effectiveness matrix below.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q1
How should the tasks be allocated, one to a man, as to minimize the total man-hours?
Solution:
The hr matrix is given by
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q1.1
Subtracting row minimum from all values in that row we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q1.2
Subtracting column minimum from all values in that column we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q1.3
The minimum no. of lines covering ail the zeros (4) is equal to the order of the matrix (4)
∴ The assignment is possible.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q1.4
The assignment is
A → I, B → III, C → II, D → IV
For the minimum hrs. take the corresponding value from the hr matrix.
Minimum hrs = 7 + 3 + 18 + 9 = 37 hrs

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 2.
A dairy plant has five milk tankers, I, II, III, IV & V. These milk tankers are to be used on five delivery routes A, B, C, D & E. The distances (in kms) between the dairy plant and the delivery routes are given in the following distance matrix.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q2
How should the milk tankers be assigned to the chilling centre so as to minimize the distance travelled?
Solution:
The distance matrix is given by
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q2.1
Subtracting row minimum from all values in that row we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q2.2
Subtracting column minimum from each value in that column we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q2.3
The number of lines covering all the zeros (3) is less than the order of the matrix (5) so the assignment is not possible. The modification is required.
The minimum uncovered value (15) is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same. We get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q2.4
The minimum lines covering all the zeros (4) are less than the order of the matrix (5) so the assignment is not possible. The modification is required the minimum uncovered value (5) is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q2.5
The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) So assignment is possible.
The assignment is
A → II, B → III, C → V, D → I, E → IV
Total minimum distance is = 120 + 120 + 175 + 40 + 70 = 525 kms.

Question 3.
Solve the following assignment problem to maximize sales:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q3
Solution:
As it is a maximization problem so we need to convert it into a minimization problem.
Subtracting all the values from the maximum value (19) we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q3.1
Also, it is an unbalanced problem so we need to add a dummy row (E) with all values zero, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q3.2
Subtracting row minimum from all values in that row we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q3.3
Subtracting column minimum from all values in that column we get the same matrix
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q3.4
The minimum number of lines covering all the zero (4) is less than the order of the matrix (5) So assignment is not possible. The modification is required. The minimum uncovered value (2) is subtracted from the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q3.5
The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so the assignment is not possible. The modification is required. The minimum uncovered value (1) is subtracted from the uncovered value and added to the values at the intersection. The values on the lines remain the same. We get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q3.6
The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q3.7
The assignment is
A → V, B → II, C → IV, D → III, E → I
No salesman goes to I as E is a dummy row.
For the maximum value take the corresponding values from the original matrix.
We get Maximum value = 15 + 19 + 14 + 17 + 0 = 65 units

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 4.
The estimated sales (tons) per month in four different cities by five different managers are given below:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q4
Find out the assignment of managers to cities in order to maximize sales.
Solution:
This is a maximizing problem. To convert it into minimizing problem subtract all the values of the matrix from the maximum (largest) value (39) we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q4.1
Also as it is an unbalanced problem so we have to add a dummy column (T) with all the values as zero. We get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q4.2
Subtracting row minimum from all values in that row we get the same matrix
Subtracting column minimum from all values in that column we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q4.3
The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so assignments are not possible. The modification is required. The minimum uncovered value (1) is subtracted from the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q4.4
The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q4.5
So I → S, II → T, III → Q, IV → P, V → R.
As T is dummy manager II is not given any city.
To find the maximum sales we take the corresponding value from the original matrix
Total maximum sales = 35 + 39 + 36 + 35 = 145 tons

Question 5.
Consider the problem of assigning five operators to five machines. The assignment costs are given in the following table.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q5
Operator A cannot be assigned to machine 3 and operator C cannot be assigned to machine 4. Find the optimal assignment schedule.
Solution:
This is a restricted assignment problem, so we assign a very high cost (oo) to the prohibited cells we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q5.1
Subtracting row minimum from all values in that row we get.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q5.2
Subtracting column minimum from all values in that column we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q5.3
As the minimum number of lines covering all the zeros (4) is equal to the order of the matrix (5) so the assignment is not possible. The modification is required. The minimum uncovered value (2) is subtracted from all the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q5.4
As the minimum number of lines covering all the zeros (5) is equal to the order of the matrix, assignment is the possible
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q5.5
So A → 4, B → 3, C → 2, D → 1, E → 5
For the minimum cost take the corresponding values from the cost matrix we get
Total minimum cost = 3 + 3 + 4 + 3 + 7 = 20 units

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 6.
A chartered accountant’s firm has accepted five new cases. The estimated number of days required by each of their five employees for each case are given below, where-means that the particular employee can not be assigned the particular case. Determine the optimal assignment of cases of the employees so that the total number of days required to complete these five cases will be minimum. Also, find the minimum number of days.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q6
Solution:
This is a restricted assignment problem so we assign a very high cost (∞) to all the prohibited cells. The day matrix becomes
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q6.1
Subtracting row minimum from all values in that row we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q6.2
Subtracting column minimum from all values in that column we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q6.3
The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so the assignment is not possible, The modification is required. The minimum uncovered value (1) is subtracted from all the uncovered values and added to the values at the intersection. The values on the lines remain the same, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q6.4
The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part I Q6.5
So E1 → I, E2 → IV, E3 → II, E4 → V, E5 → III
To find the minimum number of days we take the corresponding values from the day matrix.
Total minimum number of days = 6 + 6 + 6 + 6 + 3 = 27 days

Part – II

Question 1.
A readymade garments manufacture has to process 7 items through two stages of production, namely cutting and sewing. The time taken in hours for each of these items in different stages are given below:
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1
Find the sequence in which these items are to be processed through these stages so as to minimize the total processing time. Also, find the idle time of each machine.
Solution:
Let A = cutting and B = sewing. So we have
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.1
Observe min {A, B} = 2 for item 1 for B.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.2
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.3
Now min {A, B} = 3 for item 3 for A
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.4
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.5
New min {A , B} = 4 for item 4 for A.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.6
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.7
Now min(A, B} = 5 for item 6 for B
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.8
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.9
Now min {A, B} = 6 for item 5 for A and item 2 for B
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.10
Now only 7 is left
∴ The optimal sequence is
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.11
Worktable
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q1.12
Total elapsed time = 46 hrs
Idle time for A (cutting) = 46 – 44 = 2 hrs
Idle time for B (Sewing) = 4 hrs

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 2.
Five jobs must pass through a lathe and a surface grinder, in that order. The processing times in hours are shown below. Determine the optimal sequence of the jobs. Also, find the idle time of each machine.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2
Solution:
Let A = lathe and B = surface grinder. We have
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.1
Observe min {A, B} = 1 for job II for A
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.2
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.3
Now min {A, B} = 2 for job IV for A
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.4
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.5
Now min {A, B} = 3 for job I for B
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.6
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.7
Now min {A, B} = 5 for jobs III and V for A
∴ We have two options
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.8
or
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.9
We take the first one.
Worktable
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q2.10
Total elapsed time = 21 hrs
Idle time for A (lathe) = 21 – 17 = 4 hrs
Idle time for B (surface grinder) = 3 hrs

Question 3.
Find the sequence that minimizes the total elapsed time to complete the following jobs. Each job is processed in order AB.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3
Determine the sequence for the jobs so as to minimize the processing time. Find the total elapsed time and the idle time for both machines.
Solution:
Observe min {A, B} = 3 for job VII on B.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.1
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.2
Now min {A, B} = 4 for job IV on B.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.3
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.4
Now min {A, B} = 5 for job III & V on A. we have two options
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.5
or
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.6
We take the first one
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.7
Now min {A, B} = 5 for job II on A
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.8
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.9
Now min {A, B} = 7 for a job I on B and for job VI on A
∴ The optional sequence is
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.10
Worktable
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q3.11
Total elapsed time = 55 units
Idle time for A = 55 – 52 = 3 units
Idle time for B = 9 units.

Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

Question 4.
A toy manufacturing company has five types of toys. Each toy has to go through three machines A, B, C in the order ABC. The time required in hours for each process is given in the following table.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q4
Solve the problem for minimizing the total elapsed time.
Solution:
Min A = 12, Max B = 12
As min A ≥ max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C, We get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q4.1
Now min {G, H} = 16 for type 3 on G
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q4.2
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q4.3
Min (G, H} = 18 for type 1, 4 & 5 on H
We have more than one option, we take
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q4.4
Now only type 2 is left.
∴ The optional sequence is
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q4.5
Worktable
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q4.6
Total elapsed time = 102 hours
Idle time for A = 102 – 84 = 18 hours
Idle time for B = 54 + (102 – 94) = 62 hours
Idle time for C = 38 hours

Question 5.
A foreman wants to process 4 different jobs on three machines: a shaping machine, a drilling machine, and a tapping, the sequence of operations being shaping-drilling-tapping. Decide the optimal sequence for the four jobs to minimize the total elapsed time. Also, find the total elapsed time and the idle time for every machine.
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q5
Solution:
The time matrix is
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q5.1
Min A = 8, Max B = 8, as min A ≥ max B.
The problem can be converted into a two-machine problem.
Let G and H be two fictitious machines such that
G = A + B and H = B + C we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q5.2
Observe min (G, H} = 12 for job 2 on H
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q5.3
The problem reduces to
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q5.4
Now min {G, H} = 14 for job 3 on G and job 4 on H
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q5.5
Now only job 1 is left.
∴ The optimal sequence is
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q5.6
Worktable
Maharashtra Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 IV Part II Q5.7
Total elapsed time = 74 min
Idle time for A (shapping) = 74 – 62 = 12 min
Idle time for B (Drilling) = 47 + (74 – 70) = 51 min
Idle time for C (trapping) = 31 min