Category: Class 12

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Chemistry Model Question Paper 4 English Medium

Time: 3 Hours
Maximum Marks: 70

Instructions:

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

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Depressing agents used to separate ZnS from PbS is …………..
(a) NaCN
(b) NaCl
(c) NaNO₃
(d) NaNO₂
Answer:
(a) NaCN

Question 2.
The basic structural unit of silicates is
(a)(SiO₃)^2-
(b) (SiO₄)^2-
(c) (SiO)^–
(d) (SiO₄)^4-
Answer:
(d) (SiO₄)^4-

Question 3.
………………… is a pungent smelling gas.
(a) Ammonia
(b) Nitric acid
(c) Fluorite
(d) Sodium chloride
Answer:
(a) Ammonia

Question 4.
Consider the following statement.
(i) All the actinoids are non radioactive.
(ii) Neptunium and other heavier elements are produced by artificial transformation of naturally occurring elements by nuclear reactions.
(iii) Most.of the actinoids have long half lives.
Which of the above statements is/are not correct.
(a) i only
(b) i and ii
(c) ii and iii
(d) i and iii
Answer:
(d) i and iii

Question 5.
How many geometrical isomers are possible for [Pt (Py) (NH₃) (Br) (Cl)]?
(a) 3
(b) 4
(c) 0 3
(d) 15
Answer:
(a) 3

Question 6.
Metal excess defect is possible in……………….
(a) AgCl
(b) AgBr
(c) KCl
(d) Fes
Answer:
(c) KCl

Question 7.
The correct difference between first and second order reactions is that ……………………
(a) A first order reaction can be catalysed; a second order reaction cannot be catalysed.
(b) The half life of a first order reaction does not depend on [A₀]; the half-life of a second order reaction does depend on [A₀].
(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations.
Answer:
(b) The half life of a first order reaction does not depend on [A₀]; the half-life of a second order reaction does depend on [A₀].
Solution:
For a first order reaction

Question 8.
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.24 × 10^-4 mol L^-1 solubility product of Ag₂C₂O₄ is
(a) 2.42 × 10^-8 mol³ L^-3
(b) 2.66 × 10^-12mol³ L^-3
(c) 4.5 × 10^-11mol³ L^-3
(d) 5.619 × 10^-12mol³ L^-3
Answer:
(d) 5.619 × 10^-12mol³ L^-3
Solution:

Question 9.
Assertion(A): Copper Sulphate can be stored in a Zinc vessel.
Reason (R): Zinc is less reactive than Copper.
(a) Both A and R are correct
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are wrong

Question 10.
Fog is colloidal solution of……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas

dispersion medium-gas
dispersed phase-liquid

Question 11.
Match the following Column-I with Column-II using the code given below.
c
Answer:
(a) A – 2,B – 3, C – 4, D – 1

Question 12.

(a) anilinium chloride
(b) O – nitro aniline
(c) benzene diazonium chloride
(d) m – nitro benzoic acid

Answer:
(c) benzene diazonium chloride

Question 13.
Which one of the following is the IUPAC name of CH₃ – CH₂ – CH₂ CN?
(a) Propiono nitrile
(b) Butane cyanide
(c) Isobutyro nitrile
(d) Butane nitrile
Answer:
(d) Butane nitrile

Question 14.
The correct corresponding order of names of four aldoses with configuration given below Respectively is ……………
(a) L-Erythrose, L-Threose, L-Eiythrose, D-Threose
(b) D-Threose, D-Erythrose, L-Threose, L-Erythrose
(c) L-Erythrose, L-Threose, D-Erythrose, D-Threose
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
Answer:
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose

Question 15.
Tranquilisers are substances used for the treatment of ………………
(a) cancer
(b) AIDS.
(c) mental diseases
(d) blood infection
Answer:
(c) mental diseases

Part – II

Answer any six questions. Question No. 22 is compulsory. [6 × 2 = 12]

Question 16.
What are leaching process?
Answer:
This method is based on the solubility of the ore in a suitable solvent and the reactions in aqueous solution. In this method, the crushed ore is allowed to dissolve in a suitable solvent, the metal present in the ore is converted to its soluble salt or complex while the gangue remains insoluble. This process is also called chemical method.

Question 17.
What happens when PCI5 is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.

Question 18.
Write the biologically importance of coordination compounds.
Answer:
Biological important of coordination compounds:

(i) Ared blood corpuscles (RBC) is composed of heme group, which is Fe²⁺– Porphyrin complex, it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.

(ii) Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg²⁺ as central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants converts CO₂ and water into carbohydrates and oxygen.

(iii) Vitamin B₁₂(cyanocobalamine) is the only vitamin consist of metal ion. it is a coordination complex in which the central metal ion is CO⁺ surrounded by Porphyrin like ligand.

(iv) Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion, contains a zinc ion coordinated to the protein.

Question 19.
Mention the factors affecting the reaction rate.
Answer:
The rate of the reaction is affected by the following factors.

• Nature and state of the reactant .
• Concentration of the reactant
• Surface area of the reactant
• Temperature of the reaction
• Presence of a catalyst

Question 20.
What is meant by standard reduction potential? What is its application?
Answer:

• The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
• The greater the E° value means greater is the tendency shown by the species to accept electrons and undergo reduction.
• So higher the E° values, lesser is the tendency to undergo corrosion.

Question 21.
Give two important characteristics of physisorption.
Answer:
Important characteristics of physisorption:

• It is reversible
• It has low heat of adsorption
• It has weak van der Waals forces of attraction with adsorbent.
• It increases with increase in pressure.
• It forms multimolecular layer.

Question 22.
Draw the major product formed when 1-ethoxyprop-l-ene is heated with one equivalent
Answer:

This reaction follows SN¹ mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1-iodo prop-1-ene.

Question 23.
Define Tautomerism. Give example. Why tertiary nitro alkanes do not exhibit tautomerism?
Answer:
(i) Tautomerism is an isomerism in which the isomers change into one another with great ease of shifting of proton so that they exist together in equilibrium.

(ii) Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atom.

Question 24.
Why do soaps not work in hard water?
Answer:
Soaps are sodium or potassium salts of long-chain falty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.

This is the reason why soaps do not work in hard water.

Part – III

Answer any six questions. Question No. 31 is compulsory. [6 × 3 = 18]

Question 25.
Write the application of Iron (Fe).
Answer:

• Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
• Cast iron is used to make pipes, valves and pumps stoves etc.
• Magnets can be made of iron and its alloys and compounds.

An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for making cables, automobiles and aeroplane parts. Chrome steels are used for manufacturing cutting tools and crushing machines.

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono (H₂SO_4.H₂O) and di (H₂SO_4.H₂O) hydrates and the reaction is exothermic.
The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.

Question 27.
Give evidence that [CO(NH₃)gCl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they will give different ions in the solution which can be
tested by adding AgNO₃ solution and BaCl₂ solution. When Cl^– ions are the counter ions, a white precipitate will be obtained with AgNO₃ solution. If SO₄^2- ions are the counter ions, a white precipitate will be obtained with BaCl₂ solution.

Question 28.
For a reaction, X + Y → Product; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall .order of the reaction?
Answer:
z = k [x]^m [y]ⁿ …… (1)
8z = k [4x]^m [y]ⁿ…… (2)
I 6z = k [4x]^m [4y]ⁿ…… (3)
Dividing Eq (2) by Eq (1) we get,

1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,

16 = 4^m . 4ⁿ
16 = 4².4ⁿ
16/16 = 4ⁿ
1 = 4n
∴ n = 0 [Zero order with respect toy]
Overall order of the reaction, ‘
k[x]^m[y]ⁿ
k [x]^1.5 [y]⁰
Order = (1.5 + 0) = 1.5

Question 29.
Identify the conjugate acid base pair for the following reaction in aqueous solution
Answer:

• HF and F^– , HS^– and H₂S are two conjugate acid – base pairs.
• F^– is the conjugate base of the acid HF (or) HF is the conjugate acid of F^– .
• H₂S is the conjugate acid of HS^– (or) HS^– is the conjugate base of H₂S.

(ii)

• HPO₄^2- and PO₄^3- , SO₃^-2 and HSO₃^– are two conjugate acid – base pairs.
• PO₄^3- is the conjugate base of the acid HPO₄^2- (or)
  HPO₄^2- is the conjugate acid of PO₄^3-.
• HSO₃^– is the conjugate acid of SO₃^-2 (or) SO₃^-2 is the conjugate base of HSO₃^–

(iii)

• NH₄⁺ and NH₃, CO₃^2- and HCO₃^– are two conjugate acid – base pairs.
• HCO₃^– is the conjugate of acid CO₃^2- (or) CO₃^2- is the conjugate bases of HCO₃^–
• NH₃ is the conjugate base of NH₄⁺ (or) NH₄⁺ is the conjugate acid of NH₃.

Question 30.
In the following fields, how adsorption is applied?
Answer:
(i) Medicine (ii) Metallurgy (iii) Mordant & Dyes (iv) indicators
Answer:
(i) Medicine: – Drugs cure diseases by adsorption on body tissues.

(ii) Metallurgy:- Sulphide ores are concentrated by a process called froth floation in which lighter ore particles are adsorbed by pine oil.

(iii) Mordants and Dyes:- Most of the dyes are adsorbed on the surface of the fabric. Mordants are the substances used for fixing dyes onto the fabric.

(iv) In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after getting absorbed on precipitate. It is used to indicate the end point of filtration.

Question 31.
Identify A, B, C, and D

Answer:

Question 32.
Draw the structural formula and write the IUPAC name of
(i) N, N-dimethyl aniline (ii) Benzyl amine (iii) N-methyl benzylamine
Answer:

Question 33.
Differentiate soap and detergents?
Soap :

1. Soaps are sodium or potassium salt of long chain fatty acid.
2. Soaps are made from animal (or) plant fats and oils.
3. Soaps have lesser cleansing action.
4. Soaps are bio degradable.
5. Soaps are less effective in hard water.
6. They have a tendency to form a scum in hard water.
7. Example: Sodium palmitate.

Detergent :

1. Detergent is sodium salt of alkyl hydrogen sulphate or alkyl benzene sulphonic acid.
2. Detergents are made from petrochemicals.
3. Detergents have more cleansing action.
4. Detergents are non-bio degradable.
5. Detergents are more effective even in hard water.
6. They do not form scum with hard water.
7. Example: Sodium lauryl sulphate.

Part – IV

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

Question 34.
(a) (i) Explain the role of carbon monoxide in the purification of nickel? (2)
(ii) Describe the structure of diborane. (3)
[OR]
(b) (i) What type of hybridisation occur in 1. BrF₅. 2. BrF₃ (2)
(ii) Most of the transition metals act as catalyst. Justify this statement. (3)
Answer:
(a) (i) During the purification of Nickel by Mond’s process, carbon monoxide (CO) is used to convert impure nickel to nickel carbonyl.
Nickel carbonyl is an unstable compound. Heating to higher temperature decomposes it to give pure nickel.

(ii) In diborane two BH₂ units are linked by two bridged hydrogens. Therefore, it has eight B-H bonds. However, diborane has only 12 valance electrons and are not sufficient to
form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre – two electron bond or 2c-2e bond). The remaining four electrons have to used for the bridged bonds.
1. e. two three centred B-H-B bonds utilise two electrons each. Hence, these bonds are three centre- two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure.

In dibome, the boron is sp³ hybridised. Three of the four sp³ hybridised orbitals contains single electron and the fourth orbital is empty. Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled Is orbital of hydrogen.

[OR]

(b) (i) 1. BrF₅. is a AX₅ type. Therefore is has sp³ d² hybridisation. Hence, BrF₅ molecule has square pyramidal shape.
2. BrF₃ is a AX₃ type. Therefore it has sp³ d hybridisation. Hence, BrF₃ molecule has T-shape.

(ii) Many industrial processes use transition metals or their compounds as catalysts. Transition metal has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using its ‘d’ electrons.

For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its n electrons with an empty d orbital of the catalyst.

Question 35.
(a) (i) Why tetrahedral complexes do not exhibit geometrical isomerism. (2)
(ii) Explain about the importance and application of coordination complexes. (3)
[OR]
(b) Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.
Answer:
(a) (i) In tetrahedral geometry

• AH the four ligands are adjacent or equidistant to one another.
• The relative positions of donor atoms of ligands attached to the central metal atom are same with respect to each other.
• It has plane of symmetry

Therefore, tetrahedral complexes do not exhibit geometrical isomerism.

(ii) 1. Phthalo blue – a bright blue pigment is a complex of copper (II) ion and it is used in printing ink and packaging industry.

2. Purification of Nickel by Mond’s process involves formation of [Ni (CO)₄ ] which yields 99.5% pure on decomposition.

3 . EDTA is used as a chelating ligand for the separation of lanthanides, in softening of hard water and also in removing lead poisoning.

4. Coordination complexes are used in the extraction of silver and gold from their ores
by forming soluble cyano complex. These cyano complexes are reduced by zinc to
yield metals. This process is called Mac – Arthur Forrest cyanide process.

5. Some metal ions are estimated more accurately by complex fonnation. For eg., Ni present in Nickel chloride solution is estimated accurately forming an insoluble, complex called [Ni (DMG)₂].

6. Many of the complexes are used as catalyst in organic and inorganic reactions. For e.g.,
(i) Wilkinson’s Catalyst – [(PPh₃)₃, Rh Cl] is used for hydrogenation of alkenes.
(ii) Ziegler – Natta Catalyst [TiCl₄ + Al (C₂H₅) ]₃ is used in the polymerisation of ethene.

7. In photography, when the developed film is washed with sodium thio sulphate solution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called sodium dithio sulphate argentate (I) which can be removed easily by washing the film with water.

AgBr + 2 Na₂ S₂O₃ → Na₃ [Ag (S₂O₃)₂] + 2NaBr

[OR]

(b) (i) AAAA type of three dimensional packing:
This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical. All spheres of different layers of crystal are perfectly aligned horizontally and also vertically, so that any unit cell of such arrangement as simple cubic structure as shown in fig.

In simple cubic packing, each sphere is in contact with 6 neighbouring spheres – Four in its own layer, one above and one below and hence the coordination number of the sphere in simple cubic arrangement is 6.

(ii) ABABA type of three dimensional packing:
In this arrangement, the spheres in the first layer (A type) are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A as shown in figure. The third layer is a repeat of the first. This pattern ABABAB is repeated throughout the crystal. In this arrangement, each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.

(iii) ABCABC type of three dimensional packing:
In this arrangement (FCC) second layer spheres are arranged at the dips of first layer. Third layer spheres are arranged in a manner such that it cover the octahedral void. Then no longer third layer is similar to first or second layer. Third layer gives different arrangement. Fourth Layer spheres are similar to first layer. If the first, second and third layer are represented as A,B,C then this type of packing gives the arrangement of layers as ABCABC… (Le.,), the first three layers do not resemble first, second and third layers respectively and the sequence is repeated with the addition of more layers.

In this arrangement atoms occupy 74% of the available space and thus has 26% vacant space. The coordination number is 12.

Voids: The empty spaces between the three dimensional layers are known as voids. There are
two types of common voids possible. They are tetrahedral and octahedral voids.

Tetrahedral void:
A void formed by three spheres of a layer in contact with each other and also with a sphere on the top or bottom r Layer a layer is a hole between four spheres. The spheres are arranged at the vertices Layerc of a regular tetrahedron such a hole or void is called tetrahedral void.

Octahedral void:
A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.

The buffer capacity or index is defined as the concentration of acid or base to add in order to modify the pH of an aqueous solution.

Question 36.
(a) (i) The rate constant for a first order reaction is 1.54 × 10^-3s^-1 (2)
Calculate its half life time.
(ii) Explain about protective action of colloid. (3)
[OR]
(b) (i) What is buffer solution? Give an example for an acidic buffer and a basic buffer. (2)
(ii The value of k_spof two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 × 10^-15 and 6 × 10^-17 respectively. Which salt is more soluble? Explain. (3)
Answer:
(a) (i) We know that, t_1/2 = 0.693/ k
t_1/2= O.693/1.54 ×10^-3s ^-1 = 450s

(ii) 1. Lyophobic sols are precipitated readily even with small amount of electrolytes. But they are stabilised by the addition of small amount of lyophillic colloid.

2. A smaLl amount of gelatine sol is added to gold sol to protect gold sol.

3. Gold number is a measure of protecting power of a colloid. Gold number is defined as the number of milligrams of hydrophillic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of ImI of 10% NaCI solution. Smaller the gold number, greater the protective power.

[OR]

(b) (i) 1. Buffer is a solution which consists of a mixture of weak acid and its conjugate base (or) a weak base and its conjugate acid.

2. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases and this ability is called buffer action.

3. Acidic buffer solution : Solution containing acetic acid and sodium acetate.
Basic buffer solution : Solution containing NH₄O and NH₄Cl.

(ii)

Ni(OH)₂ is more soluble than AgCN.

Question 37.
(a) Describe about lead storage battery construction and its uses. (5)
[OR]
(b) (i) What happens when m-cresol is treated with acidic solution of sodium dichromate? (3)
(ii) Formic acid is more stronger than acetic acid. Justify this statement. (2)
(a) Lead storage battery :
1. Anode : Spongy lead
Cathode : Lead plate bearing PbO₂
Electrolyte : 38% by mass of H₂SO₄ with density 1.2 g/ml

2. Oxidation occurs at the anode

Pb_(s) → Pb²⁺ _(aq) + 2e^– …………(1)

The Pb²⁺ ions combine with SO₄^2- to form PbSO₄ precipitate

Pb²⁺_(aq) + SO₄^2- → PbS_4(s) …………….(2)

3. Reduction occurs at the cathode

PbO_2(s) + 4H⁺_(aq) + 2e^– – → Pb²⁺_(aq) + 2H₂O ………….(3)

The Pb²⁺ ions also combines with SO₄^2- ions to form sulphuric acid to form PbSO₄ Precipitate

Pb²⁺_(aq) + SO₄^2-_(aq) → PbSO₄ …………..(4)

4. The overall reaction is,
(1) + (2) + (3) + (4)
pb_(s) + PbO_2(s) + 4H⁺ + 2SO₄^2-_(aq) → 2PbSO_4(s) + 2H₂O(l)

5. The emf of a single cell is about 2V. Usually six such cells are combined in series to produce 12 volts.

6. The emf of the cell depends on the concentration of H₂SO₄. As the cell reaction uses SO₄^2- ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1,8V, the cell has to be recharged.

7. Recharge of the cell: During recharge process, the role of anode and cathode is reversed and H₂SO₄ is regenerated.

The above reaction is exactly the reverse of redox reaction which takes place while discharging.

8. Uses: Lead storage battery is used in automobiles, trains, inverters.

[OR]

(b) (i) When m-cresol is treated with acidic solution of sodium dichromate it gives 4-hydroxy benzoic acid.

(ii) The electron releasing groups (+1 groups) increase the relative charge on the carboxylate ion and destabilise it and hence the loss of proton becomes difficult.
+I groups are CH₃, – C₂H₅, – C₃H₇

Question 38.
(i) An aromatic compound ‘A’ on treatment with aqueous ammonia and heating
forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’
of molecular formula C₆H₇N. Write the structures and IUPAC names of compound
A,BandC. (3)
(ii) Lactose act as reducing sugar. Justify this statement. (2)
[OR]
(b) (i) Explain the mechanism of enzyme action? (3)
(ii) Which chemical is responsible for the antiseptic properties of dettol? (2)
Answer:
(a) (i) Step-1: To find out the structure of compounds‘B’and‘C’.
1. Since compound ‘C’ with molecular formula C₆H₇N is formed from compound ‘B’ on treatment with Br₂ + KOH (i.e, ., Hoffmann bromamide reaction). Therefore, compound ‘B’ must be an amide and ‘C’ must be an amine. The only amine having the molecular formula C₆H₇N is C₆H₅NH₂ (i.e., aniline or benzenamine).

2. Since ‘C’ is aniline, therefore, the amide from which it is formed must be benzamide (C₆H₅CONH₂). Thus, compound ‘B’ is benzamide:
The chemical equation showing the conversion of ‘B’ to ‘C’ is

Step-2: To find out the structure of compound ‘A’. Since compound ‘B’ is formed from compound ‘A’ by treatment with aqueous ammonia and heating. Therefore, compound ‘A’ must be benzoic acid or benzenecarboxylic acid.

(ii) Lactose is a disaccharide and contains one galactose unit and one glucose unit.
In lactose, the β-D galactose and β-D glucose are linked by β-1, 4 – glycosidic bond.
The aldehyde carbon is not involved in the glycosidic bond hence it retains its reducing property and is called a reducing sugar.

[OR]

(b) (i) Enzymes are bio catalysts that catalyse a specific bio chemical reaction. They generally activate the reaction by reducing the activation energy by stabilising the transition state.

In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme substrate complex. During this stage the substrate is converted into product and the enzyme becomes free and is ready to bind to another substrate molecule.

(ii) (a) Chloroxylenol and (b) Terpineol are the chemicals responsible for the antiseptic properties of dettol. But among these two, chloroxylenol plays more important role. Chloroxylenol is an antiseptic and disinfectant which is used for skin disinfection and cleaning surgical instruments.

Students can Download Tamil Nadu 12th Physics Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 3 English Medium

General Instructions:

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

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. the correct plot for electrostatic potential due to this spherical shell is ………

Answer:
(b)

Question 2.
The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1Ω. The resistance of the wire will be 2Ω at …………….
(a) 1154 K
(b) 1100 K
(c) 1400K
(d) 1127 K
Answer:

Question 3.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in horns equal to ………..
(a) n²R
(b) \(\frac{R}{n^{2}}\)
(c) \(\frac{R}{n}\)
(d) nR
Answer:
(a) n²R
Hint :
Resistance in parallel combination, R= r/n = r = Rn
Resistance in series combination , R = nr = n²R

Question 4.
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is……………………….
(a) 5 μT
(b) 7 μT
(c) 8 μT
(d) 10 μT
Answer:
(b) 7 μT

Question 5.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor, when the energy is stored equally between the electric and magnetic fields, is……………………….
(a) \(\frac{Q}{2} \)
(b) \(\frac{Q}{\sqrt{3}}\)
(c) \(\frac{Q}{\sqrt{2}}\)
(d) Q
Answer:
(c) \(\frac{Q}{\sqrt{2}}\)

Question 6.
The direction of induced current during electro magnetic induction is given by …………………………………..
(a) Faraday’s law
(b) Lenz’s law
(c) Maxwell’s law
(d)  Ampere’s law
Answer:
(b) Lenz’s law

Question 7.
In an electromagnetic wave in free space the rms value of the electric field is 3 V m . The peak value of the magnetic field is………………………..
(a) 1.414 x 10^-8 T
(b) 1.0 x 10^-8 T
(c) 2.828 x 10^-8 T
(d) 2.0 x 10^-8 T
Answer:
(a) 1.414 x 10^-8 T

Question 8.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
(a) 1m
(b) 5m
(c) 3m
(d) 6m
Answer:
(b) 5m
Hint:
Resolution limit sin θ =

Question 9.
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is ……………..
(a) six
(b) four
(c) five
(d) three
Answer:
(a) six
Hint:
Number of images formed \(n=\frac{360}{\theta}-1=\frac{360}{60}-1=5\)

Question 10.
The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is
{a) 4125 Å
(b) 3750  Å
(c) 6000 Å
(d) 2062.5 Å
Answer:
(b) 3750  Å

Question 11.
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is…..

Answer:
(c)

Question 12.
The speed of the particle, that can take discrete values is proportional to ……………
(a) n^-3/2
(b) n^-3
(c) n^1/2
(d) n
Answer:
(d) n
Hint:
\(P=m v=\frac{n h}{2 a} ; V \propto n\)

Question 13.
Doping semiconductor results in ………………………………
(a) The decrease in mobile charge carriers
(b) The change in chemical properties
(c) The change in the crystal structure
(d) The breaking of the covalent bond
Answer:
(c) The change in the crystal structure

Question 14.
The signal is affected by noise in a communication system …………………….
(a) At the transmitter
(b) At the modulator
(c) In the channel
(d) At the receiver
Answer:
(c) In the channel

Question 15.
The technology used for stopping the brain from processing pain is…………………………..
(a) Precision medicine
(b) Wireless brain sensor
(c) Virtual reality
(d) Radiology
Answer:
(c) Virtual reality

Part – II

Answer any six questions. Question No. 21 is compulsory. [6 x 2 = 12]

Question 16.
Define ‘Electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
\(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
The electric field is a vector quantity and its Sl unit is Newton per Coulomb (NC^-1).

Question 17.
State the principle of potentiometer.
Answer:
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Question 18.
Compute the intensity of magnetisation of the bar magnet whose mass, magnetic moment and density are 200 g, 2 A m² and 8 g cm^-3, respectively.
Answer:

Question 19.
What is displacement current?
Answer:
The displacement current can be defined as the current which comes into play in the region in ‘ which the electric field and the electric flux are changing with time.

Question 20.
What is principle of reversibility?
Answer:
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 21.
Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to. 81.9 x 10¹⁵ (Given: mass of proton is 1836 times that of electron).
Answer:

Question 22.
Give the symbolic representation of alpha decay, beta decay and gamma decay.
Answer:
Alpha decay: The alpha decay process symbolically in the following way
Beta decay: β decay is represented by \(_{ z }^{ A }X\rightarrow _{ z,-1 }^{ A }Y+e^{ + }+v\)
Gamma decay: The gamma decay is given \(\mathrm{z} \mathrm{X}^{*} \rightarrow_{Z}^{\mathrm{A}} \mathrm{X} \)+ gamma(y)rays

Question 23.
Explain the need for a feedback circuit in a transistor oscillator.
Answer:
The circuit used to feedback a portion of the output to the input is called the feedback network. If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases. It is necessary for sustained oscillations.

Question 24.
Give the factors that are responsible for transmission impairments.
Answer:

• Attenuation
• Distortion (Harmonic)
• Noise

Part – III

Answer any six questions. Question No. 32 is compulsory. [6 x 3 = 18]

Question 25.
A water molecule has an electric dipole moment of 6.3 x 10 Cm. A sample contains 10 water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude 3 x 10^s NC^-1. How much work is required to rotate all the water molecules from θ = 0° to 90°?
Answer:
When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) – U(0°)
As we know, U = -pE cos θ, Next we calculate the work done to rotate one water molecule
from θ = 0° to 90°,
For one water molecule, W = – pE cos 90° + pE cos 0° = pE
W = 6.3 x10^-30 x 3 x 10⁵ = 18.9 x 10^-25
For 10²² water molecules, the total work done is W_tot = 18.9x 10 ²⁵ x 10²² = 18.9 x 10 ³ J

Question 26.
What are the properties of an equipotential surface?
Answer:
Properties of equipotential surfaces
(i) The work done to move a charge q between any two points A and B,
W = q (V_B – V_A ). If the points A and B lie on the same equipotential surface, work done is zero because V_B – V_A

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to an equipotential surface.

Question 27.
An electronics hobbyist is building a radio which requires 150 ft in her circuit, but she has only 220 Ω 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?
Answer:
Required effective resistance = 150 ft
Given resistors of resistance, R = 220 ft, R = 79 ft, R = 92 ft
Parallel combination of R₁ and R₂

Question 28.
What is magnetic permeability?
Answer:
Magnetic permeability: The magnetic permeability can be defined as the measure of ability of the material to allow the passage of magnetic field lines through it or measure of the capacity of the substance to take magnetisation or the degree of penetration of magnetic field through the substance.

Question 29.
A circular metal of area 0.03 m² rotates in a uniform magnetic field of 0.4 T. The axis of rotation passes through the centre and perpendicular to its plane and is also parallel to the field. If the disc completes 20 revolutions in one second and the resistance of the disc is 4 Ω calculate the induced emf between the axis and the rim and induced current flowing in the disc.
Answer:
A = 0.03 m²; B = 0.4T; f = 20rps; R = 4Ω
Area covered in 1 sec = Area of the disc x frequency
= 0.03 x 20 = 0.6 m²

Question 30.
Give the characteristics of image formed by a plane mirror.
Answer:

• The image formed by a plane mirror is virtual, erect, and laterally inverted.
• The size of the image is equal to the size of the object.
• The image distance far behind the mirror is equal to the object distance in front of it.
• If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as,\(n=\left(\frac{360}{\theta}-1\right)\)

Question 31.
An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related?
Answer:

Question 32.
Assuming V_CEsat = 0.2 V and β = 50, find the minimum base current (I_B) required to drive the transistor given in the figure to saturation.
Answer:

V_CEsat = 0.2 V and β = 50
V_CE = V_CC -I_C R_C
0.2= 3 – I_c(1 k)

Question 33.
Mention any two advantages and disadvantages of Robotics.
Answer:
(i) Advantages of Robotics:

• The robots are much cheaper than humans.
• Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
• Robots are more precise and error free in performing the task.

(ii) Disadvantages of Robotics:

• Robots have no sense of emotions or conscience.
• They lack empathy and hence create an emotionless workplace.
• If ultimately robots would do all the work, and the humans will just sit and monitor them* health hazards will increase rapidly.

Part – IV

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

Question 34.
(a) Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
Answer:
Expression for electrostatic potential energy of the dipole in a uniform electric field:
Considera dipole placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\) placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\) . A dipole experiences a torque when kept in an uniform electric field E.

This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole. The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is

This work done is equal to the potential energy difference between the angular positions 0 and 0′. U(θ) – (Uθ’) = ΔU = -pE cos θ + pE cos θ’.

If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos 90° = 0. The potential energy stored in the system of dipole kept in the uniform electric field is given by.
U = -pE cos θ = -p-E  ……………. (3)

In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.

The potential energy is maximum when the dipole is aligned anti-parallel (θ = Jt) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

This online reference angle calculator finds the quadrant of a trigonometric angle in a standard position.

[OR]

(b) Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer:
Magnetic field due to long straight conductor carrying current: Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O.
Consider an element of length dl of the wire at a distance l from point O and \(\vec{r} \) be the vector joining the element dl with the point P. Let θ be the angle between \(d \vec{l} \text { and } \vec{r}\). Then, the magnetic field at P due to the element is
\(d \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{Id} l}{4 \pi r^{2}} \sin \theta\)(unit vector perpendicular to \(d \vec{l} \text { and } \vec{r}\)

The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors \(d \vec{l} \text { and } \vec{r}\)
(let it be \(\hat{n}\)). The net magnetic field can be determined by integrating equation with proper limits. \(\overrightarrow{\mathrm{B}}=\int d \overrightarrow{\mathrm{B}}\) From the figure, in a right angle triangle PAO,

This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating \(d \overrightarrow{\mathrm{B}} \) by varying the angle from θ = φ₁ to θ =φ₂ is

For a an infinitely long straight wire, 1 =0 and 2 =, the magnetic field is \(\overrightarrow{\mathrm{B}}=\frac{\mu_{0} I}{2 \pi a} \hat{n} \) ………… (3)
Note that here \(\hat{n} \) represents the unit vector from the point O to P.

Question 35.
(a) Obtain an expression for motional emf from Lorentz force.
Answer:
Motional emf from Lorentz force: Consider a straight conducting rod AB of length l in a uniform magnetic field \(\vec{B}\) which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity \(\vec{v}\) towards right side. When the rod moves, the free electrons present in it also move with same velocity \(\vec{v}\) in B. As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation  \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\)=-
\( e(\vec{v} \times \overrightarrow{\mathrm{B}}) \) …………. (1)

The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field E directed along BA. Due to the electric field \(\overrightarrow{\mathrm{E}}\), the coulomb force starts acting on the free electrons along AB and is given by
\(\overrightarrow{\mathrm{F}}_{\mathrm{E}}=-e \overrightarrow{\mathrm{E}}\) ……….. (2)

The magnitude of the electric field E keeps on increasing as long as accumulation of electrons at the end A continues. The force \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\) also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\) and the coulomb force F balance each other and no further accumulation of free electrons at the end A takes place,
i.e

The potential difference between two ends of the rod is
V= El
V= vBl
Thus the Lorentz force on the free electrons is responsible to maintain this potential difference and hence produces an emf

ε = B lv    ………….. (4)
As this emf is produced due to the movement of the rod, it is often called as motional emf.

[OR]

(b) Explain the Maxwell’s modification of Ampere’s circuital law.
Answer:
Maxwell argued that a changing electric field between the capacitor plates must induce a magnetic field. As currents are the usual sources of magnetic fields, a changing electric field must be associated with a current. Maxwell called this current as the displacement current.

If ‘A’ be the area of the capacitor plates and ‘cf be the charge on the plates at any instant ‘f during the charging process, then the electric field in the gap will be

But \(\frac{d q}{d t}\) is the rate of change of charge on the capacitor plates. It is called displacement current and given by \(\mathrm{I}_{d}=\frac{d q}{d t}=\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}\)

This is the missing term in Ampere’s Circuital Law. The total current must be the sum of the
conduction current I_c Hence, the modified form of the ampere’s law.
\(\oint_{c} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0}\left(\mathrm{I}_{c}+\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}\right)\)

Question 36.
(a) Obtain the equation for resolving power of microscope.
Answer:
Resolving power of microscope: The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object objective but also in resolving two points on the object separated by a small distance d_min  Smaller the value of d_min better will be the resolving power of the microscope.

The radius of central maxima is, \(r_{0}=\frac{1.22 \lambda v}{a}\) . In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is d_min, then the magnification m is, \(m=\frac{r_{o}}{d_{\min }}\)


To further reduce the value of d_min the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.

Such an objective is called oil immersed objective. The term n sin β is called numerical aperture NA.

[OR]

(b) Derive an expression for de Broglie wavelength of electrons.
Answer:
An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
\(\frac{1}{2} m v^{2}=e \mathrm{V}\)
Therefore, the speed v of the electron is \(v=\sqrt{\frac{2 e \mathrm{V}}{m}}\)
Hence, the de Broglie wavelength of the electron is \(\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 e m \mathrm{V}}}\)
Substituting the known values in the above equation, we get

For example, if an electron is accelerated through a potential difference of 100V, then its de Broglie wavelength is 1.227 Å. Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
\(\lambda=\frac{h}{\sqrt{2 m \mathrm{K}}}\)

Question 37.
(a) Explain the variation of average binding energy with the mass number by graph and discuss its features.
Answer:
We can find the average binding energy per nucleon \(\overline{\mathrm{BE}}\).
It is given by

The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus. \(\overline{\mathrm{BE}}\) is plotted against A of all known nuclei.

 

Important inferences from of the average binding energy curve:

• The value of BE rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A= 56 (iron) and then it slowly decreases.
• The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.
• For higher mass numbers, the curve reduces slowly and BE for uranium is about 7.6 MeV. They are unstable and radioactive.  If two light nuclei with A < 28 combine with a nucleus with A < 56, the binding energy per nucleon is more for final nucleus than the initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.
• If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.

[OR]

(b) State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.
Answer:
Laws of Boolean Algebra: The NOT, OR and AND operations are \(\bar{A}\), A + B, A.B are the Boolean operations. The results of these operations can be summarised as:

Complement law:

The complement law can be realised as \(\overline{\mathrm{A}}\) = A

Commutative laws
A + B = B +A
A . B = B . A

Associative laws
A + (B + C) = (A + B) + C
A . (B . C) = (A .B) . C

Distributive laws
A( B + C) = AB + AC
A + BC = (A + B) (A + C)
The above laws are used to simplify complicated expressions and to simplify the logic circuitry.

Question 38.
(a) Modulation helps to reduce the antenna size in wireless communication – Explain. Antenna size:
Answer:
Antenna is used at both transmitter and receiver end. Antenna height is an important parameter to be discussed. The height of the antenna must be a multiple of \(\frac{\lambda}{4}\)
\(h=\frac{\lambda}{4}\) ……………….. (1)
where X is wavelength \(\left(\lambda=\frac{c}{v}\right), c\) c is the velocity of light and v is the frequency of the signal v to be transmitted.

An example
Let us consider two baseband signals. One signal is modulated and the other is not modulated. The frequency of the original baseband signal is taken as v = 10 kHz while the modulated signal is v = 1 MHz. The height of the antenna required to transmit the original baseband signal of frequency
v = 10kHz is

The height of the antenna required to transmit the modulated signal of frequency v = 1 MHz is

Comparing equations (2) and (3), we can infer that it is practically feasible to construct an antenna of height 75 m while the one with 7.5 km is not possible. It clearly manifests that modulated signals reduce the antenna height and are required for long distance transmission.

[OR]

(b) What are the possible harmful effects of usage of Nanoparticles? Why?
Answer:
Possible harmful effects of usage of Nanoparticles:

The research on the harmful impact of application of nanotechnology is also equally important and fast developing. The major concern here is that the nanoparticles have the dimensions same as that of the biological molecules such as proteins. They may easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.

The adsorbing nature depends on the surface of the nanoparticle. Indeed, it is possible to deliver a drug directly to a specific cell in the body by designing the surface of a nanoparticle so that it adsorbs specifically onto the surface of the target cell.

The interaction with living systems is also affected by the dimensions of the nanoparticles. For instance, nanoparticles of a few nanometers size may reach well inside biomolecules, which is not possible for larger nanoparticles. Nanoparticles can also cross cell membranes.

It is also possible for the inhaled nanoparticles to reach the blood, to reach other sites such as the liver, heart or blood cells. Researchers are trying to understand the response of living organisms to the presence of nanoparticles of varying size, shape, chemical composition and surface characteristics.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 9 Rectification of Errors Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

11th Accountancy Guide Bank Rectification of Errors Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer.

Question 1.
Error of principle arises when ________.
a) There is complete omission of a transaction
b) There is partial omission of a transaction
c) Distinction is not made between capital and revenue items
d) There are wrong postings and wrong castings
Answer:
c) Distinction is not made between capital and revenue items

Question 2.
Errors not affecting the agreement of trial balance are ________.
a) Errors of principle
b) Errors of overcasting
c) Errors of undercasting
d) Errors of partial omission
Answer:
a) Errors of principle

Question 3.
The difference in trial balance is taken to ________.
a) The capital account
b) The trading account
c) The suspense account
d) The profit and loss account
Answer:
c) The suspense account

Question 4.
A transaction not recorded at all is known as an error of ________.
a) Principle
b) Complete omission
c) Partial omission
d) Duplication
Answer:
b) Complete omission

Question 5.
Wages paid for installation of machinery wrongly debited to wages account is ah error of ________.
a) Partial omission
b) Principle
c) Complete omission
d) Duplication
Answer:
b) Principle

Question 6.
Which of the following errors will not affect the trial balance?
a) Wrong balancing of an account
b) Posting an amount in the wrong account but on the correct side
c) Wrong totaling of an account
d) Carried forward wrong amount in a ledger account
Answer:
b) Posting an amount in the wrong account but on the correct side

Question 7.
Goods returned by Senguttuvan were taken into stock, but no entry was passed in the books. While rectifying this error, which of the following accounts should be debited?
a) Senguttuvan account
b) Sales returns account
c) Returns outward account
d) Purchases returns account
Answer:
b) Sales returns account

Question 8.
A credit purchase of furniture from Athiyaman was debited tg purchases account. Which of the following accounts should be debited while rectifying this error?
a) Purchases account
b) Athiyaman account
c) Furniture account
d) None of these
Answer:
c) Furniture account

Question 9.
The total of purchases book was overcast. Which of the following accounts should be debited in the rectifying journal entry?
a) Purchases account
b) Suspense account
c) Creditor account
d) None of the above
Answer:
b) Suspense account

Question 10.
Which of the following errors will be rectified using suspense account?
a) Purchases returns book was undercast by ₹ 100
b) Goods returned by Narendran was not recorded in the books ;
c) Goods returned by Akila ₹ 900 was recorded in the sales returns book as ₹ 90
d) A credit sale of goods to Ravivarman was not entered in the sales book.
Answer:
a) Purchases returns book was undercast by ₹ 100

II. Very Short Answer Type Questions

Question 1.
What is meant by the rectification of errors?
Answer:
Correction of errors in the books of accounts is not done by erasing, rewriting, or striking the figures which are incorrect. Correcting the errors that have occured is called Rectification.

Question 2.
What is meant by the error of principle?
Answer:
The error of principle means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.

Question 3.
What is meant by the error of partial omission?
Answer:
When the accountant has failed to record a part of the transaction, it is known as an error of partial omission. This error usually occurs in posting. This error affects only one account.

Question 4.
What is meant by the error of complete omission?
Answer:
It means the failure to record a transaction in the journal or subsidiary book or failure to post both the aspects in the ledger. This error’ affects two or more accounts.

Question 5.
What are compensating errors?
Answer:
The errors that make up for each other or neutralize each other are known as compensating errors. These errors may occur in related or unrelated accounts. Thus, excess debit or credit in one account may be compensated by excess credit or debit in some other account. These are also known as offsetting errors.

III. Short Answer Questions

Question 1.
Write a note on the error of principle by giving an example.
Answer:
It means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.
Example:
Entering the purchase of an asset in the purchases book. Machinery purchased on credit for ₹ 10,000 by M/s. Anbarasi garments manufacturing company entered in the purchases book.

Question 2.
Write a note on the suspense account.
Answer:
1. When the trial balance does not tally, the amount of difference is placed to the debit (when the total of the credit column is higher than the debit column) or credit (when the total of the debit column is higher than the credit column) to a temporary account known as ‘suspense account

2. Suspense account will remain in the books until the location and rectification of errors.

3. Afterrectifying the errors and posting the rectification entries to the respective ledger accounts, the suspense account appearing in the ledger is to be balanced.

4. If all the errors are located and rectified, the suspense account gets closed.

Question 3.
What are the errors not disclosed by a trial balance?
Answer:
Certain errors will not affect the agreement of trial balance. Though such errors occur in the books of accounts, the total debit and credit balance will be the same. The trial balance will tally. Errors of complete omission, error of principle, compensating error, a wrong entry in the subsidiary books are not disclosed by the trial balance.

Question 4.
What are the errors disclosed by a trial balance?
Answer:
Certain errors affect the agreement of trial balance. If such errors have occurred in the books of accounts, the total of debit and credit balances will not be the same. The trial balance will not tally. The error of partial omission and error of commission affect the agreement of trial balance.
Examples of such errors are as follows:

• Entered in the journal but posted to one account and omitted to be posted to the other.
• Posting an amount to the wrong side of a ledger account
• Posting twice in a ledger account.
• Over-casting or under-casting in a subsidiary book.
• Posting a wrong amount to the wrong side of an amount.
• Errors in carrying forward the page total from one page to the next page of an account or subsidiary book.
• Errors arising in the balancing of an account.
• Omission to post an entry from a subsidiary book.

Question 5.
Write a note on one-sided errors and two-sided errors.
Answer:
(a) Rectification of one-sided errors before preparing trial balance:

• When a one-sided error is detected before preparing the trial balance, no journal entry is required to be passed in the books.
• In such cases, the error can be rectified by giving an explanatory note in the account affected as to whether the concerned account is to be debited or credited.
• Example: Sales book is undercast by ₹ 100.
• In this case, the sales book is undercast by ₹ 100.
• The total of sales book is posted to the credit side of the sales account in the ledger.
• The under casting has resulted in an under-crediting of sales accounts by ₹ 100.
• This is an error of commission.
• The error is only in the sales account. There is short credit in the sales account by ₹ 100. Hence, it is rectified by crediting sales account by ₹ 100.

(b) Rectification of two-sided errors before preparing the trial balance:
1. When a two-sided error is detected before preparing the trial balance, it must be rectified by passing a rectifying journal entry in the journal proper after analyzing the error.
Example: Goods sold to Anand for ₹ 1,000 on credit was not entered in the sales book. The entry will be

Method of deriving the rectifying entries

IV. Exercises

Question 1.
State the account/s affected in each of the following errors
a) Goods purchased on credit from Saranya for ₹ 150 was posted to the debit side of her account.
b) The total of purchases book ₹ 4,500 was posted twice.
Solution:
a) Saranya account should be credit with ₹ 300
b) Purchases account should be credit with ₹ 4,500

Question 2.
State the account/s affected in each of the following errors
a) Goods sold to Vasu on credit for ₹ 1,000 was not recorded in the sales book.
b) The total of the sales book ₹ 2,500 was posted twice.
Solution:
a) Sales account should be credit with ₹ 1,000
b) Sales account should be Debited with ₹ 2,500

Question 3.
Rectify the following errors discovered before the preparation of the trial balance
a) Sales book was undercast by ₹ 100
b) Purchases returns book was overcast by ₹ 200
Solution:
a) Sales account should be Credited with ₹ 100
b) Purchases returns account should be debited with ₹ 200

Question 4.
Rectify the following errors before the preparation of trial balance
a) Returns outward book was undercast by ₹ 2,000
b) Returns inward book total was taken as ₹ 15,000 instead of ₹ 14,000
c) The total of the purchases account was carried forward ₹ 100 less.
Solution:
a) Purchases return account should be Credited with ₹ 2,000
b) Sales returns account should be Debited with ₹ 1,000
c) Purchases accounts should be debited ₹ 100

Question 5.
Rectify the following errors assuming that the trial balance is yet to be prepared
a) Sales book was undercast by ₹ 400
b) Sales returns book was overcast by ₹ 500
c) Purchases book was undercast by ₹ 600
d) Purchases returns book was overcast by ₹ 700
e) Bills receivable book was undercast by ₹ 800
Solution:
a) Sales account should be credited with ₹ 400
b) Sales return accounts should be Credited with ₹ 500
c) Purchases account should be debited with ₹ 600
d) Purchases returns account should be debited with ₹ 700
e) Bills received account should be debited with ₹ 800

Question 6.
Rectify the following errors before preparing trial balance
a) The total of purchases book was carried forward ₹ 90 less.
b) The total of purchases book was carried forward ₹ 180 more.
c) The total of sales book was carried forward ₹ 270 less.
d) The total of sales returns book was carried forward ₹ 360 more.
e) The total of purchase returns book was carried forward ₹ 450 less.
Solution:
a) Purchases account is to be Debited with ₹ 90
b) Purchases account is to be Credited with ₹ 180
c) Sales account is to be Credited with ₹ 270
d) Sales returns account is to be Credited with ₹ 360
e) Purchases returns account is to be Credit with ₹ 450

Question 7.
The following errors were located by the accountant before the preparation of trial balance. Rectify them.
a) The total of the discount column of ₹ 1,100 on the debit side of the cash book was not yet osted.
b) The total of the discount column on the credit side of the cash book was undercast by ₹ 500.
c) Purchased goods from Anbuchelvan on credit for ₹ 700 was posted to the debit side of his account.
d) Sale of goods to Ponmukil on credit for ₹ 78 was posted to her account as ₹ 87.
e) The total sales returns book of ₹ 550 was posted twice.
Solution:
a) Discount account should be debited with ₹ 1,100
b) Discount account should be Credited with ₹ 500
c) Anbuchelvan account should be Credited with ₹ 1,400
d) Ponmuki account should be Credit with ₹ 9
e) Sales return account should be Credit with ₹ 550

Question 8.
The accountant of a firm located the following errors before preparing the trial balance. Rectify them.
a) Machinery purchased for ₹ 3,000 was debited to the purchases account.
b) Interest received ₹ 200 was credited to the commission account.
c) An amount of ₹ 1,000 paid to Tamil selvan as salary was debited to his personal account.
d) Old furniture sold for ₹ 300 was credited to sales account.
e) Goods worth ₹ 800 purchased from Soundarapandian on credit was not recorded in the books of accounts.
Solution:
Journal Entries

Question 9.
Rectify the following errors which were located before preparing the trial balance.
a) Wages paid ₹ 2,000 for the erection of machinery was debited to wages account.
b) Sales returns book was short totaled by ₹ 1,000.
c) Goods purchased for ₹ 200 were posted as ₹ 2,000 to the purchases account.
d) The sales book was overcast by ₹ 1,500.
e) Cash paid to Mukil ₹ 2,800 which was debited to Akhil’s account as ₹ 2,000.
Solution:
Rectification of Errors:

Question 10.
Rectify the following errors which were located at the time of preparing the trial balance
a) The total of the discount column on the debit side of the cash book of ₹ 225 was posted twice.
b) Goods of the value of ₹ 75 returned by Ponnarasan were not posted to his account.
c) Cash received from Yazhini ₹ 1,000 was not posted.
d) Interest received ₹ 300 has not been posted.
e) Rent paid ₹ 100 was posted to rent account as ₹ 10
Solution:
Rectification of Errors

Question 11.
The following errors were located at the time of preparing the trial balance. Rectify them.
a) A personal expense of the proprietor ₹ 200 was debited to the traveling expenses account.
b) Goods of ₹ 400 purchased from Ramesh on credit was wrongly credited to Ganesh’s account.
c) An amount of ₹ 500 paid as salaries to Mathi was debited to his personal account.
d) An amount of ₹ 2,700 paid for the extension of the building was debited to the repairs account.
e) A credit sale of goods of ₹ 700 on credit to Mekala was posted to Krishnan’s account.
Solution:
Rectification of Errors

Question 12.
Rectify the following journal entries.

Solution:
Rectification of Errors

Question 13.
Rectify the following errors discovered after the preparation of the trial balance
a) Rent paid was carried forward to the next page ₹ 500 short.
b) Wages paid were carried forward ₹ 250 excess.
Solution:
a) Rent account should be debited with ₹ 500
b) Wages account should be Credited with ₹ 250

Question 14.
Rectify the following errors after preparation of trial balance
a) Salary paid to Ram ₹ 1,000 was wrongly debited to his personal account.
b) A credit sale of goods to Balu for ₹ 450 was debited to Balan.
Solution:
Rectification of Errors

Question 15.
Pass necessary journal entries to rectify the following errors located after the preparation of trial balance
a) Sales book was undercast by ₹ 1,000.
b) A amount of ₹ 500 paid for wages was wrongly posted to the machinery Account.
Solution:
Rectification of Errors

Question 16.
Give journal entries to rectify the following errors discovered after the preparation of trial balance
a) Purchases book was overcast by ₹ 10,000.
b) Repairs to the furniture of ₹ 500 were debited to the furniture account.
c) A credit sale of goods to Akilnilavan for ₹ 456 was credited to his account as ₹ 654.
Solution:
Rectification of Errors

Question 17.
Rectify the following errors located after the preparation of trial balance
a) Purchases book was undercast by ₹ 900.
b) Sale of old furniture for ₹ 1,000 was credited to sales account.
c) Purchase of goods from Arul for ₹ 1,500 on credit was not recorded in the books.
Solution:
Rectification of Errors

Question 18.
The following errors were located after the preparation of the trial balance. Pass journal entries to rectify them. Assume that there exists a suspense account.
a) The total sales book was undercast by ₹ 350.
b) The total of the discount column on the debit side of the cash book ₹ 420 was not posted.
c) The total of one page of the purchases book of ₹ 5,353 was carried forward to the next page as ₹ 5,533.
d) Salaries ₹ 2,400 were posted as ₹ 24,000.
e) Purchase of goods from Sembiyanmadevi on credit for ₹ 180 was posted to her account as ₹ 1,800
Solution:
Rectification of Errors

Question 19.
Rectify the following errors assuming that the trial balance is already prepared and the difference was placed to suspense account
a) Sales book was undercast by ₹ 250
b) Purchases book was undercast by ₹ 120
c) Sales book was overcast by ₹ 130
d) Bills receivable book was undercast by ₹ 75
e) Purchases book was overcast by ₹ 35.’
Solution:
Rectification of Errors

Question 20.
The following errors were located after the preparation of the trial balance. The difference in trial balance has been taken to the suspense account. Rectify them.
a) The total of purchases book was carried forward ₹ 70 less.
b) The total sales book was carried forward ₹ 340 more.
c) The total of purchases book was carried forward ₹ 150 more.
d) The total of sales book was carried forward ₹ 200 less.
e) The total of purchase returns book was carried forward ₹ 350 less.
Solution:
Rectification of Errors

Question 21.
The following errors were located by the accountant after the preparation of the trial balance. There exists a suspense account. Rectify them.
a) The total of the discount column of ₹ 1,180 on the debit side of the cash book was not posted.
b) Purchase of goods from Arivuchelvan on credit for ₹ 600 was posted to the debit side of his account.
c) The total of the discount column on the credit side of the cash book was undercast by ₹ 400.
d) The total sales returns book of ₹ 570 was posted twice.
e) Sold goods to Mukil on credit for ₹ 87 was posted to her account as ₹ 78.
Solution:
Rectification of Errors

Question 22.
The accountant of a firm located the following errors after preparing the trial balance. Rectify them assuming that there is a suspense account.
a) Machinery purchased for ₹ 3,500 was debited to purchases account.
b) ₹ 1,800 paid to Raina as salary was debited to his personal account.
c) Interest received ₹ 200 was credited to the commission account.
d) Goods worth ₹ 1,800 purchased from Amudhanila on credit was not recorded in the books of accounts,
e) Used furniture sold for ₹ 350 was credited to the sales account.
Solution:
Rectification of Errors

Question 23.
The book-keeper of a firm found that the trial balance was cut by ₹ 922 (excess credit). He placed the amount in the suspense account and subsequently found the following errors
a) The total discount column on the credit side of the cash book ₹ 78 was not posted in the ledger.
b) The total of purchases book was short by ₹ 1,000.
c) A credit sale of goods to Natarajan for ₹ 375 was entered in the sales book as ₹ 735.
d) A credit sale of goods to Mekala for ₹ 700 was entered in the purchases book.
You are required to give rectification entries and prepare a suspense account.
Solution:
Rectification of Errors

Suspense Account

Question 24.
The books of Raman did not agree. The accountant placed the difference of ₹ 1,270 to the debit of the suspense account. Rectify the following errors and prepare the suspense account
a) Goods taken by the proprietor for his personal use ₹ 75 was not entered in the books.
b) A credit sale of goods to Shanmugam for ₹ 430 was credited to his account as ₹ 340.
c) A purchase of goods on credit for ₹ 400 from Vivek was entered in the sales book. However, Vivek’s account was correctly credited.
d) The total of the purchases returns book ₹ 300 was not posted.
Solution:
Rectification of Errors

Suspense Account

11th Accountancy Guide Rectification of Errors Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
The errors can be classified into ……………… types.
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(d) Four

Question 2.
When one or both aspects of a transaction are recorded in the wrong category of an account, this is called ________.
a) Error of Principle
b) Error of Omission
c) Error of Commission
d) Error of original entry
Answer:
a) Error of Principle

Question 3.
The errors that make up for each other or neutralize each other are known as ………………
(a) Errors of commission
(b) Errors of principle
(c) Errors of omission
(d) Compensating errors
Answer:
(d) Compensating errors

Question 4.
₹ 60,000 paid on the extension of the building wrongly debited to Repairs account. This is called the error of ________.
a) Commission
b) Omission
c) Principle
d) None of these
Answer:
c) Principle

Question 5.
Sales book is undercast by ₹ 100, classify the error
(a) Errors of principle
(b) Errors of commission
(c) Errors in casting
(d) Errors of omission
Answer:
(c) Errors in casting

Question 6.
________ are those which cancel themselves out.
a) Error of principle
b) Error of Commission
c) Error of Omission
d) Compensating errors
Answer:
d) Compensating errors

Question 7.
________ errors can be located in the preparation of trial balance.
a) Error of principle
b) Error of Commission
c) Error of Omission
d) Compensating errors
Answer:
b) Error of Commission

Question 8.
An entry of ₹ 75 has been debited to Rajesh’s A/c as ₹ 57 is an error of ________.
a) Error of principle
b) Error of Commission
c) Error of Omission
d) Compensating errors
Answer:
b) Error of Commission

Question 9.
Casting errors are the result of ________.
a) Wrong totaling
b) Wrong Balancing
c) the wrong carry forward
d) None of the above
Answer:
a) Wrong totaling

Question 10.
Suspense account is usually closed when ________.
a) Accounts are finalized
b) After the completion of auditing
c) All the errors are rectified
d) None of the above
Answer:
c) All the errors are rectified

II. Very Short Answer Questions

Question 1.
What is error or omission?
Answer:
The failure of the accountant to record a transaction or an item in the books of accounts is known as an error of omission. It can be a complete omission or partial omission.

Question 2.
At what stages the errors can occur?
The following types of errors may occur in various stages:
Answer:

• At the stage of journalizing
• At the stage of posting
• At the stage of balancing
• At the stage of preparing trial balance

Question 3.
What do you mean by errors?
Answer:
Errors mean recording or classifying or summarising the accounting transactions wrongly or omissions to record them by a clerk or an accountant unintentionally.

Question 4.
What is an error or omission?
Answer:
The failure of the accountant to record a transaction or an item in the books of accounts is known as an error of omission. It can be a complete omission or partial omission.

Question 5.
What is an error of commission?
Answer:
When a transaction is incorrectly recorded, it is known as error of commission. It usually occurs due to lack of concentration or carelessness of the accountant.

Question 6.
Write a note on one-sided errors.
Answer:
When a one-sided error is detected before preparing the trial balance, no journal entry is required to be passed in the books. In such cases, the error can be rectified by giving an explanatory note in the account affected as to whether the concerned account is to be debited or credited.

Question 7.
Write a note on two-sided errors.
Answer:
When a two-sided error is detected before preparing the trial balance, it must be rectified by passing a rectifying journal entry in the journal proper after analyzing the error.

III. Short Answer Questions

Question 1.
What are the steps to be followed to locate the errors after the preparation of the trial balance?
Answer:
While preparing trial balance, if it does not tally, it is an indication of the presence of errors in the books of accounts. The difference in the trial balance is transferred to the suspense account and then errors are to be located and rectified.

The following are the steps to be followed to locate errors after preparing a trial balance

1. The totals of debit and credit columns of trial balance are to be checked
2. The balances of various ledger accounts shown in the trial balance are to be checked to ensure whether they are shown in the respective columns (debit or credit).
3. The difference in the trial balance must be halved and compared with the balances of the ledger to verify whether any ledger balance is recorded on the wrong side of the trial balance.
4. The totals of all the subsidiary books are to be checked, especially if the difference is ₹ 1, ₹ 100, etc.
5. If the difference is divisible by ‘9’, the difference may be due to the transposition of figures in the books (Writing ₹ 127 as ₹ 172). Hence, the possibilities of transposition of figures shall be checked.
6. The accounts of all the creditors and debtors are to be verified.
7. The correctness of the balances of various ledger accounts is to be ensured.
8. The correctness of the balances of various ledger accounts is to be ensured.
9. All the amounts carried forward from one page to the next are to be verified. .
10. If the difference still exists, as a final step all the entries in the journals should be verified.

Question 2.
What are the steps to be followed to locate the errors before preparation of trial balance?
Answer:
Errors may be located before preparing the trial balance either spontaneously or by intentional scrutiny of books of accounts.

The following are the steps to be followed to locate errors before preparing trial balance:

1. Scrutiny of entries made in the journal proper
2. Scrutiny of entries made in the subsidiary books
3. Checking the totals of the subsidiary books
4. Scrutiny of postings made to the ledger accounts
5. Scrutiny of balancing of ledger accounts

Question 3.
Spot out the location of errors before preparation of trial balance.
Answer:
Errors may be located before preparing the trial balance either spontaneously or by intentional scrutiny of books of accounts. The following are the steps to be followed to locate errors before preparing trial balance

1. Scrutiny of entries made in the journal proper.
2. Scrutiny of entries made in the subsidiary books.
3. Checking the totals of the subsidiary books
4. Scrutiny of balancing of ledger accounts.
5. Scrutiny of postings made to the ledger accounts

Question 4.
Rectify the following errors

1. Purchases Book is overcast by ₹ 3,500
2. Sales Book is undercast by ₹ 2,000
3. Purchases returns books have been overcast by ₹ 7,600
4. Sales returns book has been undercast by ₹ 500

Solution:
Rectification:

1. Credit Purchases Account with ₹ 3,500
2. Credit Sales Account with ₹ 2,000
3. Debit Purchases Returns account with ₹ 7,600
4. Debit Sales returns account with ₹ 500

Question 5.
Rectify the following errors
1. Sales book undercast by ₹ 5,000
2. Machinery purchased for ₹ 9000 passed through purchases Book.
3. Sales to Ram for ₹ 11,000 debited to his account as ₹ 10,100.
4. Repairs to building ₹ 3,640 debited to buildings account.
Solution:

Question 6.
Rectify the following errors
1. The purchase of machinery from A for ₹ 3,000 has been entered in the purchase daybook.
2. Received ₹ 1,000 from M but credited to N’s account.
3. ₹ 800 paid as wages for the erection of machine has been charged to Repairs account.
4. ₹ 250 received from Ganesh, previously written off, has been credited to Ganesh account
Solution:

Question 7.
Rectify the following Errors
1. Purchase book is overcast by ₹ 6,000
2. Sales book carried forward ₹ 630 instead of ₹ 360
3. Purchase from Sreesha ₹ 5, 000 has been posted to the debit side of her account.
4. Sale of old machinery for ₹ 50,000 has been entered in Sales book.
Solution:

Question 8.
Rectify the following error
1. Salary paid to manager ₹ 8,000 debited to his personal account.
2. Total the discount column in the debit side of the cash book is wrongly cast short. ₹ 540.
3. Total of sales book has been added ₹ 2,400 excess
4. ₹ 230 received in respect of a book debt was posted to the sales account
5. Goods sold for ₹ 3,873 to Raju were returned to us and recorded in the sales book.
Solution:

Question 9.
When a Trial Balance failed to agree, ₹ 37,900 was transferred to the credit of the suspense account. The following errors were discovered. Give journal entries and prepare a suspense account.
1. Sales daybook was undercast by ₹ 40,000
2. Purchase of machinery for ₹ 60,000 was passed through the purchase book.
3. Goods sold to Velu for ₹ 4,500 were posted to his account as ₹ 5,400.
4. Purchase returns book was overcast by ₹ 2,000
5. The total sales book from one page was carried forward to the next page as ₹ 12,000 instead of ₹ 11,000.
Solution:
Rectification of Entries

Suspense Account

Question 10.
In considering the trial balance, a book-keeper finds that the debit total exceeds the credit total by ₹ 3,520. The amount is placed to the credit of a newly opened suspense account. The following mistakes were discovered. Pass the necessary entries for rectifying the mistakes and show the suspense account.
a) Sales daybook was overcast by ₹ 1000
b) A sale of ₹ 500 to Rajesh was wrongly debited to Ramesh
c) General expenses ₹ 180 was posted as ₹ 800.
d) Cash received from Ganesh was debited to his account ₹ 1,500
e) While carrying forward the total of one page of the purchases day book to the next the amount of ₹ 12,350 was entered as ₹ 13,250.
Solution:
Rectification of Entries

Suspense Account

Question 11.
Pass journal entries to rectify the following entries
1. ₹ 4,500 received in respect of a book debt was posted to sales account
2. Defective goods worth ₹ 260 returned to Saran were recorded through the sales returns book.
3. Goods sold for ₹ 950 to Rakesh were returned to us and recorded in the sales book.
4. A purchase of ₹ 2,100 from Banu on the last day of the year was taken into stock, but the invoice was not passed through the purchase book.
Solution:
Rectification of Entries

Question 12.
Show how you will rectify the following entries.
1. A Credit sales of ₹ 650 to Raja were debited to Kaja.
2. A purchase of goods for ₹ 750 from Shaji was debited to his account.
3. An office typewriter purchased for ₹ 6,500 was debited to the Repairs account.
4. A sum of ₹ 3,900 received from a debtor was debited to his account.
5. Purchase of goods for the consumption of proprietor was debited to purchase account ₹ 1,000
Solution:
Rectification of Entries

Question 13.
Write down the rectifying journal entries for the following errors and show the suspense account.
1. The sales returns books have been undercast by ₹ 5,000
2. Goods worth ₹ 1,500 sold to Batiya have been credited to his account.
3. Purchase of furniture ₹ 7,000 has been entered in the purchases account.
4. Cash Rs.4,500 from Aadhira has been posted to his account as ₹ 5,400.
5. A bill received from X for ₹ 4,000 has been posted to the Bills payable account.
Solution:
Rectification of Entries

Suspense Account

IV. Long Question Answers

Question 1.
Explain the steps to be followed to locate errors after preparing a trial balance
Answer:
The following steps are to be followed to locate errors after preparing a trial balance

1. The totals of the debit and credit columns of the trial balance are to be checked.
2. The balances of various ledger accounts shown in the trial balance are to be checked to ensure whether they are shown in the respective columns (debit or credit).
3. The difference in the trial balance must be halved and compared with the balances of the ledger to verify whether any ledger balance is recorded on the wrong side of the trial balance.
4. The totals of all the subsidiary books are to be checked, especially if the difference is ₹1 to ₹ 100 etc.
5. If the difference is divisible by 9 the difference may be due to the transposition of figures in the books. (Writing ₹ 127 as ₹ 172). Hence, the possibilities of transposition of figures shall be checked.
6. The accounts of all the creditors and debtors are to be verified.
7. Postings from the subsidiary books to different accounts in the ledger are to be checked.
8. The correctness of the balances of various ledger accounts is to be ensured.
9. All the amounts carried forward from one page to the next area to be verified.
10. If the differences still exists, as a final step all the entries in the journals should be verified.

Question 2.
Briefly explain the Classification or errors
Answer:

The failure of the accountant to record a transaction or an item in the books of accounts is known as an error of omission. It can be a complete omission or partial omission.
1. Error of complete omission – It means the failure to record a transaction in the journal or subsidiary book or failure to post both the aspects in the ledger. This error affects two or more accounts.

2. Error of partial omission – When the accountant has failed to record a part of the transaction, it is known as an error of partial omission. This error usually occurs in posting. This error affects only one account.

3. Error of commission – When a transaction is incorrectly recorded, it is known as an error of commission. It usually occurs due to lack of concentration or carelessness of the accountant.

Errors of Principle: It means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.

4. Compensating errors – The errors that make up for each other or neutralize each other are known as compensating errors. These errors may occur in related or unrelated accounts. Thus, excess debit or credit in one account may be compensated by excess credit or debit in some other account. These are also known as offsetting errors.

5. Errors disclosed by the trial balance and errors not disclosed by the trial balance – Generally, one-sided errors are revealed by a trial balance. They will cause disagreement of totals of debit balances and credit balances. Two-sided errors are not revealed by a trial balance.

V. Additional Sums

Question 1.
Rectify the following errors
i) Purchases book overcast by ₹ 1,300
ii) Sales book under cast by ₹ 2,500
Solution:
i) Credit – Purchases A/c with ₹ 1,300
ii) Credit – Sales A/c with ₹ 2,500

Question 2.
Rectify the following errors
i) Purchases return book overcast by ₹ 750
ii) Sales return book under cast by ₹ 600
Solution:
i) Debit – Purchases return A/c with ₹ 750
ii) Debit – Sales return A/c with ₹ 600

Question 3.
Rectify the following errors
i) Purchases book is carried forward ₹ 850 Less
ii) Sales book total is carried toward ₹ 2,500 More
Solution:
i) Debit – Purchases A/c with ₹ 850
ii) Debit – Sales A/c with ₹ 2,500

Question 4.
Rectify the following errors
i) A total of ₹ 7,580 in the purchases book has been carried forward as ₹ 8,570
ii) The total of the sales book ₹ 7,500 or page 20 was carried forward to page 21 as ₹ 5,570
iii) Purchases return book was carried forward as ₹ 1,520 instead of ₹ 5,120
Solution:
i) Credit – Purchases A/c with ₹ 990
ii) Credit – Sales A/c with ₹ 1,980
iii) Credit – Purchases return A/c with ₹ 3,600

Question 5.
Rectify the following errors
i) Purchases from Bagavathi for ₹ 4,500 has been posted to the debit side of her account
ii) Sales to Vijay for ₹ 1,520 has been posted to his credit as ₹ 1,250
Solution:
i) Purchases from Bagavathi should have been posted to the credit of Bagavathi’s A/c, but it has been . debited. Hence, credit Bagavathi’s A/c with double the amount i.e, ₹ 9,000

ii) Sales to Vijay has to be debited in Vijay’s account but his account is credited with ₹ 1,250. Hence Debit Vijay’s A/c with ₹ 1,250 to ₹ 1,520 i.e, ₹ 2,770

Question 6.
Rectify the following errors
i) Purchases from should for ₹ 750 has been omitted to be posted to the personal A/c
ii) Sales to Khader for ₹ 780 has been posted to his account as ₹ 870
Solution:
i) This is an error of omission. Posting must be to the credit of Shakila’s A/c. Hence, post ₹ 750 to the credit of Skakila’s A/c.

ii) Here Khader’s has been debited with a wrong amount i.e., with an excess amount. To rectify this error, the excess amount must be credited to his account. Hence, Credit Khader’s A/c with ₹ 90.

Question 7.
The following errors were found in the book of pradhu. Give the necessary entries to correct them.
i) Salary of ₹ 10,000 paid to Murali has been debited to his personal account.
ii) ₹ 3,500 paid for a typewriter was charged to the office expenses account.
iii) ₹ 8,000 paid for furniture purchased has been charged to the purchases account.
iv) Repairs made were debited to the building account for ₹ 500
v) An amount of ₹ 5,000 withdrawn by the proprietor for his personal use has been debited to the trade expenses account.
iv) ₹ 2,000 received from Shanthi and Co. has been wrongly entered as from Shakila and Co.
Solution:
In the book of Prathu – Rectification of Errors

Question 8.
Give journal entries to rectify the following errors
i) Purchases of goods from Devi amounting to ₹ 25,000 have been wrongly passed through the sales Book.
ii) Credit sale of goods ₹ 30,000 to Rajan has been wrongly passed through the purchases Book
iii) Sold old furniture for ₹ 3,500 passed through the sales Book.
iv) Paid wages for the construction of Building debited to wages to account ₹ 1,00,000
v) Paid ₹ 10,000 for the installation of machinery debited to wages account.
vi) On 31st December 2003 goods worth ₹ 5,000 were returned by Manjia and were taken into stock on the same date, but no entry was passed in the books.
Solution:
Rectification of Errors

Question 9.
An accountant could not tally the Trial Balance. The difference of ₹ 5,180 was Temporality placed to the credit of suspense account for preparing the final account. The following errors were taken located.
i) Commission of ₹ 500 paid, was posted twice, once to, discount allowed account and once to commission account.
ii) The sales book was undercast by ₹ 1,000
iii) A Credit sales of ₹ 2,780 to Raja though correctly entered in the sale book, was posted wrongly to her account as Rs.3,860
iv) A Credit purchase from Nataraj of ₹ 1,500, though correctly entered in the purchases book, was wrongly debited to his personal account.
v) Discount column of the payments side of the book was wrongly added as ₹ 2,800 instead of ₹ 2,400
You are required to
i) Pass the necessary rectifying entries.
ii) Prepare suspense Account.
Solution:
Rectification of Errors

Suspense Account

Question 10.
Rectify the following Journal Entries
Rectification of Errors

Solution:
Rectification of Errors

Question 11.
Rectify the following Errors
i) ₹ 12,000 paid of salary to cashier Govind, stands debited to his personal account.
ii) An amount of ₹ 5,000 withdrawn by the proprietor for his personal use has been debited to trade Expenses A/c
iii) Cash received from Bala ₹ 300 was credited to Balu.
iv) A credit sale of ₹ 2,000 to Janakiram has been wrongly passed through the purchase book.
Solution:
Rectification of Errors

Question 12.
Rectify the following Errors
i) Repairs made were debited to building account ₹ 5,000
ii) Mahesh networked goods worth ₹ 2,000 no Entry was passed in the Book to this Effects
iii) Purchases of goods from Antony amounting to ₹ 1,500 have been debited to his accounts.
iv) ₹ 5,200 paid for the purchases of the typewriter was charged to the office expenses account.
Solution:
Rectification of Errors

Question 13.
Rectify the following Errors
i) Credit purchases of goods from Madhan of ₹ 300 have been wrongly Entered in the sales book.
ii) ₹ 500 received from Seivan has been credited to Selvi’s account
iii) ₹ 1,000 received as interest was credited to the commission account.
iv) Sales book total ₹ 878 was wrongly totaled as ₹ 788.
v) The total of this discount column, on the debit side of the cash book has been added shortly by ₹ 400.
Solution:
Rectification of Errors

Question 14.
Rectify the following Errors
A Book keeper found his trial balance not balanced, placed the difference amount in the suspense account, and subsequently found the following errors
a) Sales book was overcast by ₹ 1,500
b) ₹ 2900 received from Vani in full settlement of her account of ₹ 3,000 was posted in the cash book but omitted to be entered in her account.
c) The total of the sales book ₹ 12,000 was debited to sales return accounts.
d) ₹ 1,000 received as interest was credited to interest as ₹ 100 Give rectifying entries and show the suspense account.
Solution:
Rectification of Errors

Suspense Account

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 Inverse Trigonometric Functions Ex 4.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.6

Inverse Function Calculator is an online tool that helps find the inverse value for the given function.

Choose the most suitable answer from the given four alternatives:

Question 1.
The value of sin^-1(cos x), 0 ≤ x ≤ π is
(a) π – x
(b) x – \(\frac {π}{2}\)
(c) \(\frac {π}{2}\) – x
(d) x – π
Solution:
(c) \(\frac {π}{2}\) – x
Hint:
sin^-1(cos x) = sin^-1(sin(\(\frac {π}{2}\) – x)) = \(\frac {π}{2}\) – x

Question 2.
If sin^-1 x + sin^-1 y = \(\frac {2π}{3}\); then cos^-1 x + cos^-1 y is equal to
(a) \(\frac {2π}{3}\)
(b) \(\frac {π}{3}\)
(c) \(\frac {π}{6}\)
(d) π
Solution:
(b) \(\frac {π}{3}\)
Hint:
sin^-1x + cos^-1x + cos^-1y + sin^-1y = \(\frac {π}{2}\) + \(\frac {π}{2}\) = π
\(\frac {2π}{3}\) + cos^-1x + cos^-1y = π
cos^-1x + cos^-1y = π – \(\frac {2π}{3}\) = \(\frac {3π-2π}{3}\) = \(\frac {π}{3}\)

Question 3.
sin^-1\(\frac {3}{5}\) – cos^-1\(\frac {12}{13}\) + sec^-1\(\frac {5}{3}\) – cosec^-1\(\frac {13}{12}\) is equal to
(a) 2π
(b) π
(c) 0
(d) tan^-1\(\frac {12}{65}\)
Solution:
(c) 0
Hint:

Question 4.
If sin^-1 x = 2sin^-1 α has a solution, then
(a) |α| ≤ \(\frac {1}{√2}\)
(b) |α| ≥ \(\frac {1}{√2}\)
(c) |α| < \(\frac {1}{√2}\)
(d) |α| > \(\frac {1}{√2}\)
Solution:
(a) |α| ≤ \(\frac {1}{√2}\)
Hint:
If sin^-1 x = 2sin^-1 α has a solution then
–\(\frac {π}{2}\) ≤ 2sin^-1α ≤ \(\frac {π}{2}\)
–\(\frac {π}{4}\) ≤ sin^-1α ≤ \(\frac {π}{4}\)
sin(\(\frac {-π}{4}\)) ≤ α ≤ sin\(\frac {π}{4}\)
–\(\frac {1}{√2}\) ≤ α ≤ \(\frac {1}{√2}\)
-|α| ≤ \(\frac {1}{√2}\)

Question 5.
sin^-1(cos x) = \(\frac {π}{2}\) – x is valid for
(a) -π ≤ x ≤ 0
(b) 0 ≤ x ≤ π
(c) –\(\frac {π}{2}\) ≤ x ≤ \(\frac {π}{2}\)
(d) –\(\frac {π}{4}\) ≤ x ≤ \(\frac {3π}{4}\)
Solution:
(b) 0 ≤ x ≤ π
Hint:
sin^-1 (cosx) = \(\frac {π}{2}\) – x is valid for
cos x = sin (\(\frac {π}{2}\) – x)
cos x ∈ [0, π]
∴ 0 ≤ x ≤ π

Question 6.
If sin^-1 x + sin^-1 y + sin^-1 z = \(\frac {3π}{2}\), the value of show that x²⁰¹⁷ + y²⁰¹⁸ + z²⁰¹⁹ – \(\frac {9}{x^{101}+y^{101}+z^{101}}\) is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:

Question 7.
If cot^-1 x = \(\frac {2π}{5}\) for some x ∈ R, the value of tan^-1 x is
(a) –\(\frac {π}{10}\)
(b) \(\frac {π}{5}\)
(c) \(\frac {π}{10}\)
(d) –\(\frac {π}{5}\)
Solution:
(c) \(\frac {π}{10}\)
Hint:
tan^-1 x + cos^-1 \(\frac {π}{2}\)
tan^-1x = \(\frac {π}{2}\) – cot^-1 x = \(\frac {π}{2}\) – \(\frac {2π}{5}\)
= \(\frac {5π-4π}{10}\) = \(\frac {π}{10}\)

Question 8.
The domain of the function defined by f(x) = sin^-1 \(\sqrt {x-1}\) is
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Solution:
(a) [1, 2]
Hint:
f(x) = sin^-1 \(\sqrt {x-1}\)
\(\sqrt {x-1}\) ≥ 0
-1 ≤ \(\sqrt {x-1}\) ≤ 1
∴ 0 ≤ \(\sqrt {x-1}\) ≤ 1
0 ≤ x – 1 ≤ 1
1 ≤ x ≤ 2
x ∈ [1, 2]

Question 9.
If x = \(\frac {1}{5}\) the value of cos(cos^-1x + 2sin^-1x) is
(a) –\(\sqrt{\frac {24}{25}}\)
(b) \(\sqrt{\frac {24}{25}}\)
(c) \(\frac {1}{5}\)
(d) –\(\frac {1}{5}\)
Solution:
(d) –\(\frac {1}{5}\)
Hint:
cos[cos^-1x + sin^-1x + sin^-1x] = cos(\(\frac {π}{2}\) + sin^-1x)
= -sin(sin^-1x)
[∵ cos(90+θ) = -sin θ]
= -x = –\(\frac {1}{5}\)

Question 10.
tan^-1(\(\frac {1}{4}\)) + tan^-1(\(\frac {2}{9}\)) is equal to
(a) \(\frac {1}{2}\)cos^-1(\(\frac {3}{5}\))
(b) \(\frac {1}{2}\)sins^-1(\(\frac {3}{5}\))
(c) \(\frac {1}{2}\)tan^-1(\(\frac {3}{5}\))
(d) tan^-1(\(\frac {1}{2}\))
Solution:
(d) tan^-1(\(\frac {1}{2}\))
Hint:

Question 11.
If the function f(x) = sin^-1(x² – 3), then x belongs to
(a) [-1, 1]
(b) [√2, 2]
(c) [-2, -√2]∪[√2, 2]
(d) [-2, -√2]
Solution:
(c) [-2, -√2]∪[√2, 2]
Hint:
-1 ≤ x² – 3 ≤ 1
-1 + 3 ≤ x² ≤ 1 + 3
⇒ 2 ≤ x² ≤ 4
±√2 ≤ x ≤ ± 2
[-2, -√2]∪[√2, 2]

Question 12.
If cot^-1 2 and cot^-1 3 are two angles of a triangle, then the third angle is
(a) \(\frac {π}{4}\)
(b) \(\frac {3π}{4}\)
(c) \(\frac {π}{6}\)
(d) \(\frac {π}{3}\)
Solution:
(b) \(\frac {3π}{4}\)
Hint:
A + B + C = π (triangle)
cot^-1 2 + cot^-1 3 + C = π

Question 13.
sin^-1(tan\(\frac {π}{4}\)) – sin^-1(\(\sqrt{\frac {3}{x}}\)) = \(\frac {π}{6}\). Then x is root of the equation
(a) x² – x – 6 = 0
(b) x² – x – 12 = 0
(c) x² + x – 12 = 0
(d) x² + x – 6 = 0
Solution:
(b) x² – x – 12 = 0
Hint:

Question 14.
sin^-1(2 cos²x – 1) + cos^-1(1 – 2 sin²x) =
(a) \(\frac {π}{2}\)
(b) \(\frac {π}{3}\)
(c) \(\frac {π}{4}\)
(d) \(\frac {π}{6}\)
Solution:
(a) \(\frac {π}{2}\)
Hint:
sin^-1(2 cos² x – 1) + cos^-1(1 – 2 sin²x)
= sin^-1 (2 cos² x – 1) + cos^-1 (1 – sin² x – sin² x)
= sin^-1(2 cos² x – 1) + cos^-1(cos² x – (1 – cos²x))
= sin^-1(2 cos² x – 1) + cos^-1(cos² x – 1 + cos²x)
= sin^-1(2 cos² x – 1) + cos^-1(2 cos² x – 1)
= \(\frac {π}{2}\) [∵ sin^-1 x + cos^-1 x = \(\frac {π}{2}\)]

Question 15.
If cot^-1(\(\sqrt {sinα}\)) + tan^-1(\(\sqrt {sinα}\)) = u, then cos 2u is equal to
(a) tan²α
(b) 0
(c) -1
(d) tan 2α
Solution:
(c) -1
Hint:
cot^-1 x + tan^-1 x = \(\frac {π}{2}\)
∴ u = \(\frac {π}{2}\)
cos 2u = cos 2(\(\frac {π}{2}\)) = cos π = -1

Question 16.
If |x| ≤ 1, then 2 tan^-1 x – sin^-1\(\frac {2x}{1+x²}\) is equal to
(a) tan^-1x
(b) sin^-1x
(c) 0
(d) π
Solution:
(c) 0
Hint:
sin^-1\(\frac {2x}{1+x²}\) = 2 tan^-1x
∴ 2 tan^-1 x – 2 tan^-1 x = 0

Question 17.
The equation tan^-1 x – cot^-1 x = tan^-1(\(\frac {1}{√3}\)) has
(a) no solution
(b) unique solution
(c) two solutions
(d) infinite number of solutions
Solution:
(b) unique solution
Hint:
tan^-1 x – cot^-1 x = tan^-1(\(\frac {1}{√3}\)) …….. (1)
tan^-1 x – cot^-1 x = \(\frac {π}{2}\) ……… (2)
Add 1 and 2
2 tan^-1 x = \(\frac {π}{6}\) + \(\frac {π}{2}\) = \(\frac {2π}{3}\)
tan^-1 x = \(\frac {π}{3}\)
x = √3 which is uniqe solution.

Question 18.
If sin^-1 x + cot^-1(\(\frac {1}{2}\)) = \(\frac {π}{2}\), then x is equal to
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{√5}\)
(c) \(\frac {2}{√5}\)
(d) \(\frac {√3}{2}\)
Solution:
(b) \(\frac {1}{√5}\)
Hint:

Question 19.
If sin^-1 \(\frac {x}{5}\) + cosec^-1\(\frac {5}{4}\) = \(\frac {π}{2}\), then the value of x is
(a) 4
(b) 5
(c) 2
(d) 3
Solution:
(d) 3
Hint:

Question 20.
sin(tan^-1 x), |x| < 1 is equal to
(a) \(\frac {x}{\sqrt{1-x^2}}\)
(b) \(\frac {1}{\sqrt{1-x^2}}\)
(c) \(\frac {1}{\sqrt{1+x^2}}\)
(d) \(\frac {x}{\sqrt{1+x^2}}\)
Solution:
(d) \(\frac {x}{\sqrt{1+x^2}}\)
Hint:
tan a = x
W.K.T 1 + tan² a = sec² a
1 + x² = sec² a
sec a = \(\sqrt{1+x^2}\)
\(\frac {1}{cosa}\) = \(\sqrt{1+x^2}\)
cos a= \(\frac {1}{\sqrt{1+x^2}}\)
sin a = \(\sqrt{1-cos^2a}\) = \(\sqrt{1-\frac {1}{1+x^2}}\)
\(\sqrt{\frac{1+x^2 -1}{1+x^2}}\) = \(\frac {x}{\sqrt{1+x^2}}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 8 Bank Reconciliation Statement Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

11th Accountancy Guide Bank Reconciliation Statement Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer.

Question 1.
A bank reconciliation statement is prepared by ________.
a) Bank
b) Business
c) Debtor to the business
d) Creditor to the business
Answer:
b) Business

Question 2.
A bank reconciliation statement is prepared with the help of ________.
a) Bank statement
b) Cash book
c) Bank statement and bank column of the cash book
d) Petty cash book
Answer:
c) Bank statement and bank column of the cash book

Question 3.
Debit balance in the bank column of the cash book means ________.
a) Credit balance as per bank statement
b) Debit balance as per bank statement
c) Overdraft as per cash book
d) None of the above
Answer:
a) Credit balance as per bank statement

Question 4.
A bank statement is a copy of ________.
a) Cash column of the cash book
b) Bank column of the cash book
c) A customer’s account in the bank’s book
d) Cheques issued by the business
Answer:
c) A customer’s account in the bank’s book

Question 5.
A bank reconciliation statement is prepared to know the causes for the difference between:
a) The balance as per the cash column of the cash book and bank column of the cashbook
b) The balance as per the cash column of the cash book and bank statement
c) The balance as per the bank column of the cash book and the bank statement
d) The balance as per petty cash book and the cash book
Answer:
c) The balance as per the bank column of the cash book and the bank statement

Question 6.
When money is withdrawn from bank, the bank ________.
a) Credits customer’s account
b) Debits customer’s account
c) Debits and credits customer’s account
d) None of these
Answer:
b) Debits customer’s account

Question 7.
Which of the following is not the salient feature of bank reconciliation statement?
a) Any undue delay in the clearance of cheques will be shown up by the reconciliation
b) Reconciliation statement will discourage the accountant of the bank from embezzlement
c) It helps in finding the actual position of the bank balance
d) Reconciliation statement is prepared only at the end of the accounting period
Answer:
d) Reconciliation statement is prepared only at the end of the accounting period

Question 8.
Balance as per cash book is ₹ 2,000. Bank charge of ₹ 50 debited by the bank is not yet shown in the cash book. What is the bank statement balance now?
a) 1,950 credit balance
b) 1,950 debit balance
c) 2,050 debit balance
d) 2,050 credit balance
Answer:
a) 1,950 credit balance

Question 9.
Balance as per bank statement is 1, 000. Cheque deposited, but not yet credited by the bank is 2, 000. What is the balance as per bank column of the cash book?
a) 3,000 overdraft
b) 3,000 favourable
c) 1,000 overdraft
d) 1,000 favourable
Answer:
b) 3,000 favourable

Question 10.
Which one of the following is not a timing difference?
a) Cheque deposited but not yet credited
b) Cheque issued but not yet presented for payment
c) Amount directly paid into the bank
d) Wrong debit in the cash book
Answer:
d) Wrong debit in the cash book

II. Very Short Answer Questions

Question 1.
What is meant by bank overdraft?
Answer:
It is not possible to have unfavourable cash balance in the cash book. But, it is possible to have unfavourable balance in the bank account. When the business is not having sufficient money in its bank account, it can borrow money from the bank. As a result of this, the amount is overdrawn from the bank.

Question 2.
What is a bank reconciliation statement?
Answer:
The bank reconciliation statement is a statement that reconciles the balance as per the bank column of the cash book with the balance as per the bank statement by giving the reasons for such difference along with the amount. The internal record of the business (bank column of cash) can be reconciled with the external record (bank statement).

Question 3.
State any two causes of disagreement between the balance as per bank column of cash book and bank statement.
Answer:

1. Cheques issued but not yet presented for payment.
2. Cheques deposited into the bank but not yet credited.

Question 4.
Give any two expenses which may be paid by the banker as per standing instruction.
Answer:

1. Bank Charges
2. Interest

Question 5.
Substitute the following statements with one word/phrase
(a) A copy of the customer’s account issued by the bank.
(b) Debit balance as per bank statement.
(c) Statement showing the causes of disagreement between the balance as per cash book and balance as per bank statement.
Answer:
(a) Pass book
(b) Pass book favourable
(c) (1) Timing difference, (2) Errors in recording

Question 6.
Do you agree with the following statements? Write “yes” if you agree, and write “no” if you Disagree
Answer:

1. The bank reconciliation statement is prepared by the banker. – Yes
2. Adjusting the cash book before preparing the bank reconciliation statement is compulsory. – No
3. Credit balance as per bank statement is an overdraft. – No
4. Bank charges debited by the bank increases the balance as per bank statement. – No
5. Bank reconciliation statement is prepared to identify the causes of differences between balance as per bank column of the cash book and balance as per cash column of the cash book. – Yes

III. Short Answer Questions

Question 1.
Give any three reasons for preparing a bank reconciliation statement.
Answer:

1. To identify the reasons for the difference between the bank balance as per the cash book and bank balance as per bank statement.
2. To identify the delay in the clearance of cheques.
3. To ascertain the correct balance of the bank column of the cash book.
4. To discourage the accountants of the business as well as the bank from misusing funds.

Question 2.
What is meant by the term “cheque not yet presented?”
Answer:

1. When the cheques are issued by the business, it is immediately entered on the credit side of the cash book by the business.
2. This may not be entered in the bank statement on the same day.
3. It will be entered in the bank statement only after it is presented with the bank.

Question 3.
Explain why does money deposited into the bank appears on the debit side of the cash book, but on the credit side of the bank statement?
Answer:
When the cheques are deposited into the bank, the amount is debited in the cash book on the same day. But, these may not be shown in the bank pass book on the same day because these will be entered in the bank statement only after the collection of the cheques.

Question 4.
What will be the effect of interest charged by the bank, if the balance is an overdraft?
Answer:

1. If the business has taken any loan or overdrawn, interest has to be paid by the business.
2. The entries for bank charges and interest are made in the bank statement.
3. The cash book shows more balance than the bank statement.

Question 5.
State the timing differences in BRS with examples.
Answer:
The timing differences in BRS are:

1. cheques issued but not yet presented for payment
2. cheques deposited into the bank but not yet credited
3. bank charges and interest on loan and overdraft
4. interest and dividends collected by the bank
5. dishonour of cheques and bills
6. amount paid by parties directly into the bank
7. payment made directly by the bank to others
8. bills collected by the bank on behalf of its customer

IV. Exercises

Question 1.
From the following particulars prepare a bank reconciliation statement of Jayakumar as of 31st December 2016.
a) Balance as per cash book ₹ 7,130
b) Cheque deposited but not cleared ₹ 1,000
c) A customer has deposited ₹ 800 into the bank directly
Solution:
Bank reconciliation statement of Jayakumar as of 31st December 2016.

Question 2.
From the following particulars of Kamakshi traders, prepare a bank reconciliation statement as of 31st March 2018.
a) Debit balance as per cash book ₹ 10,500
b) Cheque deposited into bank amounting to ₹ 5,500 credited by bank but entered twice in the cash book
c) Cheques issued and presented for payment amounting to ₹ 7,000 omitted in the cash book
d) Cheque book charges debited by the bank ₹ 200 not recorded in the cash book.
e) Cash of ₹ 1,000 deposited by a customer of the business in cash deposit machine not recorded in the cash book.
Solution:
Bank reconciliation statement of Kamakshi traders as of 31st March 2018.

Question 3.
From the following information, prepare a bank reconciliation statement to find out the bank statement balance as of 31st December 2017.

Solution:
Bank reconciliation statement of Kamakshi traders as on 31st March, 2018.

Question 4.
On 31st March 2017, Anand’s cash book showed a balance of ₹ 1,12,500, Prepare bank reconciliation statement,
a) He had issued cheques amounting to ₹ 23,000 on 28,3.2017, of which cheques amounting to ₹ 9,000 have so far been presented for payment.

b) A cheque for ₹ 6,300 deposited into bank on 27.3.2017, but the bank credited the same only on 5th April 2017.

c) He had also received a cheque for ₹ 12,000 which, although entered by him in the cash book, was not deposited in the bank.

d) Wrong credit given by the bank on 30th March 2017 for ₹ 2,000.

e) On 30th March 2017, a bill already discounted with the bank for ₹ 3,000 was dishonoured, but no entry was made in the cash book.

f) Interest on debentures of ₹ 700 was received by the bank directly.

g) Cash sales of ₹ 4,000 wrongly entered in the bank column of the cash book.
Solution:
Bank reconciliation statement of Mr, Anand as on 31st March, 2017

Question 5.
From the following particulars of Siva and Company, prepare a bank reconciliation statement as of 31st December 2017.
a) Credit balance as per cash book ₹ 12,000
b) A cheque of ₹ 1,200 Issued and presented for payment to the bank, wrongly credited in the cash book
c) Debit side of bank statement was undercast by ₹ 100
Solution:
Bank reconciliation statement as of 31st December 2017.

Question 6.
From the following particulars of Raheem traders, prepare a bank reconciliation statement as of 31st March 2018.
a) Overdraft as per cash book ₹ 2,500
b) Debit side of cash book was under cash by ₹ 700
c) Amount received by the bank through RTGS amounting to ₹ 2,00,000, omitted in the cash book.
d) Two cheques issued for ₹ 1,800 and ₹ 2,000 on 29th March 2018. Only the second cheque is presented for payment.
e) Insurance premium on the car for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book
Solution:
Bank reconciliation statement as of 31st March 2018.

Question 7.
From the following information, prepare a bank reconciliation statement as of 31st December,

Solution:
Bank reconciliation statement as on 31st December, 2017

Question 8.
Prepare bank reconciliation statement from the following data.

Solution:
Bank reconciliation statement

Question 9.
From the following particulars of Veera traders, prepare a bank reconciliation statement as on 31st December, 2017.
a) Credit balance as per bank statement ₹ 6,000
b) Amount received by bank through NEFT for ₹ 3,500, entered twice in the cash book.
c) Cheque dishonoured amounting to ₹ 2,500, not entered in cash book.
Solution:
Bank reconciliation statement as on 31st December, 2017.

Question 10.
Prepare bank reconciliation statement from the following data and find out the balance as per cash book as on 31st March, 2018.

Solution:
Bank reconciliation statement as on 31st March, 2018.

Question 11.
Ascertain the cash book balance from the following particulars as of 31st December 2017
i) Credit balance as per bank statement ₹ 2,500
ii) Bank charges of ₹ 60 have not been entered in the cash book
iii) Cheque deposited on 28th December 2017 for ₹ 1,000 was not yet credited by the bank
iv) Cheque issued on 24th December 2017 for ₹ 700, not yet presented for payment
v) A dividend of ₹ 400 collected by the bank directly but not entered in the cash book
vi) A cheque of ₹ 600 had been dishonoured, but no entry was made in the cash book
vii) Interest on term loan ₹ 1,200 debited by the bank but not accounted in the cash book
viii) No entry had been made in the cash book for a trade subscription of ₹ 500 paid vide banker’s order on 23rd December 2014
Solution:
Bank reconciliation statement as of 31st December 2017.

Question 12.
From the following particulars of Raja traders, prepare a bank reconciliation statement as on 31st January, 2018.
a) Balance as per bank statement ₹ 5,000

b) Cheques amounting to ₹ 800 had been recorded in the cash book as having been deposited into the bank on 25th January 2018, but were entered in the bank statement on 2nd February 2018.

c) Amount received by bank through NEFT amounting to ₹ 3,000, omitted in the cash book.

d) Two cheques issued for ₹ 3,000 and ₹ 2,000 on 29th March 2018. Only the first cheque is presented for payment.

e) Insurance premium on motor vehicles for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book.

f) Credit side of cash book was undercast by ₹ 700

g) Subsidy received directly by the bank from the state government amounting to ₹ 10,000, not entered in the cash book.
Solution:
Bank reconciliation statement of Mr. Raja traders as of 31st January 2018.

Question 13.
From the following particulars of Simon traders, prepare a bank reconciliation statement as on 31st March 2018.
a) Debit balance as per bank statement ₹ 2,500
b) Cheques deposited amounting to ₹ 10,000, not yet credited by the bank.
c) Payment through net banking for ₹ 2,000, omitted in the cash book.
Solution:
Bank reconciliation statement as on 31st March 2018.

Question 14.
From the following particulars, ascertain the cash book balance as on 31st December 2016.
i) Overdraft balance as per bank statement ₹ 1,26,640
ii) Interest on overdraft entered in the bank statement, but not yet recorded in cash book ₹ 3,200
iii) Bank charges entered in the bank statement, but not found in cash book ₹ 600
iv) Cheques issued, but not yet presented for payment ₹ 23,360
v) Cheques deposited into the bank but not yet credited ₹ 43,400
vi) Interest on investment collected by the bank ₹ 24,000
Solution:
Bank reconciliation statement as of 31st December 2016.

Question 15.
From the following particulars of John traders, prepare a bank reconciliation statement as on 31st March, 2018.
a) Bank overdraft as per bank statement ₹ 4,000

b) Cheques amounting to ₹ 2,000 had been recorded in the cash book as having been deposited into the bank on 26th March 2018 but were entered in the bank statement on 4th April 2018.

c) Amount received by the bank through cash deposit machine amounting to ₹ 5,000, omitted in the cash book.

d) Amount of ₹ 3,000 wrongly debited to John traders account by the bank, for which no details are available.

e) Bills for collection credited by the bank till 29th March 2017 amounting to ₹ 4,000, but no advice received by John traders.

f) Electricity charges made through net banking for ₹ 900 were wrongly entered in the cash column of the cash book instead of the bank column.
Solution:
Bank reconciliation statement as of 31st March 2018.

Question 16.
Prepare bank reconciliation statement from the following data.

Solution:
Prepare bank reconciliation statement.

Question 17.
Prepare bank reconciliation statement as of 31st March 2017 from the following extracts of cash book and bank statement.
Cashbook (Bank column only)

Bank statement

Solution:
Bank reconciliation statement as on 31st March, 2017

Question 18.
A trader received his bank statement on 31st December 2017 which showed an overdraft balance of ₹ 12,000. On the same day, his cash book showed a debit balance of ₹ 2,000.
Analyze the following transactions. Choose the possible causes and prepare a bank reconciliation statement to show the causes of differences.
a) Cheque deposited for ₹ 2,000 on 21st December 2017. The bank credited the same on 26th December 2017.

b) Cheque issued for payment on 26th December 2017 amounting to ₹ 2,500, not yet presented until 31st, December 2017.

c) Bank charges amounting to ₹ 200 not yet entered in the cash book.

d) Online payment for ₹ 1,500 entered twice in the cash book.

e) Cheque deposited amounting to ₹ 1,000, but omitted in the cash book. The same cheque was dishonoured by bank, but not yet entered in the cash book.

f) Cheque deposited, not yet credited by bank amounting to ₹ 17,800.
Solution:
Bank reconciliation statement as of 31st December 2017.

11th Accountancy Guide Bank Reconciliation Statement Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
A Bank Reconciliation Statement is prepared with the help of ________.
a) Bank statement and bank column of the cash book
b) Journal
c) Ledger
d) None of the above
Answer:
a) Bank statement and bank column of the cash book

Question 2.
Bank reconciliation statement is ______
a) Part of bank statement
b) Part of the cash book
c) A separate statement
d) A sub-division of journal
Answer:
a) Part of bank statement

Question 3.
Uncollected cheques are also known as ______.
a) Outstanding cheques
b) Uncleared cheques
c) Outstation cheques
d) Both a & c
Answer:
d) Both a & c

Question 4.
When a cheque is not paid by the bank it is called as ______.
a) Honoured
b) Endorsed
c) Dishonoured
d) None of these
Answer:
c) Dishonoured

Question 5.
A bank reconciliation statement is prepared by ______.
a) Banker
b) Accountant of the business
c) Auditors₹
d) None of the above
Answer:
b) Accountant of the business

Question 6.
The cheque which is deposited into the bank but not cleared at the end of a particular year is called ______.
a) Uncredited cheque
b) Unpresented cheque
c) Omitted cheque
d) Dishonoured cheque
Answer:
b) Unpresented cheque

Question 7.
In cash book bank charges recorded in the ______.
a) Credit side
b) Debit Side
c) Both a & b
d) None of the above
Answer:
a) Credit side

Question 8.
An amount of Rs,2G0Q is debited twice in the bank statement. What will the reflect when overdraft as per the cash book is the starting point ______.
a) ₹ 2000 will be deducted
b) ₹ 2000 will be added
c) ₹ 4000 will be deducted
d) ₹ 4000 will be deducted
Answer:
b) ₹ 2000 will be added

Question 9.
If any amount is directly deposited into the bank then ______.
a) Cashbook will show less balance & bank book will show more
b) Cashbook will show more balance & bank book will show less
c) Cashbook will show double balance
d) Bank book will show double
Answer:
a) Cashbook will show less balance & bank book will show more

Question 10.
Which of the following error results in an unadjusted cash book balance?
a) Outstanding cheques
b) Unpresented Cheques
c) deposit in Transit
d) Omission of Bank charges
Answer:
d) Omission of Bank charges

Question 11.
Credit balance in the bank column of the cash book means ______.
a) Credit balance as per bank statement
b) Debit balance as per bank statement
c) Overdraft as per cash book
d) None of the above
Answer:
b) Debit balance as per bank statement

Question 12.
When balance as per Cash Book is the starting point, to ascertain balance as per bank statement interest allowed by Bank is ______.
a) Subtracted
b) added
c) not adjusted
d) None of the above
Answer:
b) added

Question 13.
When balance as per Cash Book is the starting point, to ascertain the balance as per bank statement interest charged by Bank is:
a) Added
b) subtracted
c) not adjusted
d) None of the above
Answer:
b) subtracted

Question 14.
When the balance as per Cash Book is the starting point to ascertain balance as per bank statement, direct deposits by customers are:
a) Added
b) subtracted
c) not adjusted
d) None of the above
Answer:
a) Added

Question 15.
When the balance as per Cash Book is the starting point to ascertain balance as per bank statement, direct payments by the bank are:
a) Added
b) subtracted
c) not adjusted
d) None of the above
Answer:
b) subtracted

Question 16.
______ is not possible to have unfavourable cash balance in the cash book.
a) Bank statement
b) Bank overdraft
c) Cash overdraft
d) Cashbook
Answer:
b) Bank overdraft

Question 17.
Bank overdraft is available only to the ______ holders.
a) Saving Account
b) Fixed Account
c) Joint Account
d) Current Account
Answer:
d) Current Account

Question 18.
______ is simply a copy of the customer’s account in the books of a bank.
a) Cashbook
b) Bank statement
c) Bank Account
d) None of these
Answer:
b) Bank statement

Question 19.
A bank statement is a copy of ______.
a) the cash column of a customer’s cash book
b) the bank column of a customer’s cash book
c) the customer’s account in the bank’s ledger
d) None of these
Answer:
c) the customer’s account in the bank’s ledger

Question 20.
Debit balance in the Cash Book means ______.
a) overdraft as per bank statement
b) credit balance as per bank statement
c) overdraft as per Cash Book
d) None of these
Answer:
b) credit balance as per bank statement

II. Additional Questions & Answers

Question 1.
Differences between bank column of cash book and bank statement.
Answer:
Bank column of cash book:

• It is prepared by a business concern.
• Cash deposits are entered on the debit side.
• Cash withdrawals are entered on the credit side.
• Cheque deposits are debited on the day of the deposit.
• Cheques issued are credited on the day of the issue of the cheque.
• Collections and payments as per standing instructions of the business are entered only after checking with the bank statement.
• It is balanced at the end of a specific period.

Bank statement:

• It is prepared by a bank (banker).
• Cash deposits are entered in the credit column.
• Cash withdrawals are entered in the debit column.
• Cheque deposits are credited only at the time of realization of the cheque.
• Cheques issued by customers are debited by the bank on the date on which the payment is made.
• Collections and payments as per standing instructions of the business are entered in the banker’s book on the date of realization of payment.
• It is balanced after each transaction.

Question 2.
What are the items recorded on the debit side of the bank column of the cash book?
Answer:

1. Cheques deposited but not credited.
2. Credits in the passbook only.
   • Interest credited in the bank statement
   • Dividend and other income
   • • Direct deposit by a party
3. Any error in the cash book/ bank statement has the effect of increasing the balance as per the bank statement.

Question 3.
What are the items recorded on the credit side of the bank column of the Cash Book?
Answer:

1. Cheques deposited but not credited
2. Cheques dishonoured but not entered in the cash book
3. Debits in bank statement only
   • Interest debited
   • Insurance premium, loan instalment, etc., paid as per standing instructions
   • Direct payment by banker
4. Any error in cash book/ bank statement which has the effect of decreasing the balance as per bank statement

Question 4.
What is meant by the term “Cheques deposited into bank but not yet credited?”
Answer:
When the cheques are deposited into bank, the amount is debited in the cash book on the same day. But, these may not be shown in the bank pass book on the same day because these will be entered in the bank statement only after the collection of the cheques.

For example, the balances as per cash book and bank statement are ₹ 20,000 for X & Co. X & Co. receives a cheque on 25th March 2016, from ABC Limited for ₹ 5,000. On the same day, X & Co, debits its cash book with ₹ 5,000.

But bank credits X & Co’s account only when the cheque is collected from ABC Limited’s bank. This shows that is a time gap between depositing the cheque by the customer (X & Co) and collection of cheque by the bank.

Question 5.
What will be the effect of Interest and dividends collected by the bank?
Answer:
The bank may collect dividends on its customer’s investment in shares and also interest on any investment. The entry for this will be made in the bank statement on the date of collection. But the entry is made in the cash book only when the bank statement is received by the customer. Till then, the cash book shows less balance than the bank statement.

Question 6.
What will be the effect of Dishonour of cheques and bills?
Answer:
When the cheque is received from outside parties, it is deposited with the bank and debited in the cash book. If the cheque is dishonoured, the bank cannot collect the amount of such cheque from outside parties’ bank. It is not credited in the bank statement. As a result of this, the two records would differ.

While discounting the bills receivables, in the cash book it is entered in the debit side and in the bank statement it is credited. When the bill is presented by the bank to the drawee of the bill and the payment is not received, the bank debits the same to cancel the credit.

But, credit is made in the cash book only when the customer gets the entries made in the bank statement is received. The bank may also charge some amount for such dishonour.

Question 7.
What will be the effect of Amount paid by parties directly into the bank?
Answer:
Sometimes, debtors or the customers of the business may directly deposit the money into bank account of the business. It may be done by directly visiting the branch of the bank by paying cash (including NEFT, RTGS) or swiping debit or credit or business card or depositing the money in cash deposit machine or transfer through online banking facility.

This will be credited in the banker’s book. But the entry is made in the cash book only when the bank statement is received by the customer. Until then, the cash book shows less balance than bank statement.

Question 8.
What will be the effect of Amount paid directly by the bank to others?
Answer:
Sometimes the bank may be instructed to make payments such as, insurance premium, instalment of loan, etc., as an agent of the customer on behalf of its customer. In all such cases, debit is made in bank statement. But, the entry is made in the cash book only when the bank statement is received by the customer. Till then, the cash book shows more balance than bank statement.

Question 9.
What will be the effect of Bills collected by the bank on behalf of its customers?
Answer:
When goods are sold by the business, the documents may be sent through the bank. When the bank collects the amount, it is credited in bank records. But, the entry is made in the cash book only when the bank statement is received by the business. Till then, the bank statement shows more balance than cash book.

Question 10.
Explain the differences arising due to errors in recording the entries.
Answer:
Errors committed in recording the transactions by the business in the cash book:
Sometimes, errors may be committed in the cash book. For example, omission or wrong recording of transaction relating to cheques deposited or issued, wrong balancing, etc. In these cases, obviously, there will be differences between bank balance as per bank statement and bank balance as per cash book.

Errors committed in recording the transactions by the bank:
Sometimes errors may be committed in the banker’s book. For example, omission or wrong recording of transaction relating to cheques deposited and wrong balancing. In these cases, obviously, there will be differences between bank balance as per bank statement and bank balance as per cash book.

Question 11.
What is the need for bank reconciliation statement?
Answer:

1. To identify the reasons for the difference between the bank balance as per the cash book and bank balance as per bank statement.
2. To identify the delay in the clearance of cheques.
3. To ascertain the correct balance of bank column of cash book.
4. To discourage the accountants of the business as well as bank from misusing funds.

Additional Sums:

Question 1.
Prepare bank reconciliation statement of Mr. Bala as on 31.03.2013.
a) Balance as per cash book ₹ 15,000
b) Cheques deposited but not cleared ₹ 1,000
c) Cheques issued but not yet present for payments ₹ 1500 d. Interest allowed by bank ₹ 200
Solution:
Prepare bank reconciliation statement

Question 2.
Prepare bank Reconciliation statement to find out balance as per bank statement on 31st March 2018,
1. Cheques deposited but not yet collected by the bank ₹ 1,000
2. Cheques issued but not yet presented for payment ₹ 2,000
3. Bank Interest Charged ₹ 200
4. Rent paid by bank as per standing Instruction ₹ 400
5. Cash book balances ₹ 600
Solution:
Prepare bank reconciliation statement

Question 3.
Form the following particulars of Ashok and company; prepare a bank reconciliation statement as on 31st March 2018.
a) Credit balance as per cash book ₹ 10,000
b) Cheques issued but not yet presented for payment ₹ 10,000
c) Cheques Deposited but not credited ₹ 9000
d) Rent collected by the bank as per standing Instruction ₹ 500
Solution:
Bank Reconciliation statement Mr. Ashok as on 31st March 2018

Question 4.
From the following information, Prepare bank Reconciliationstatement of Mr. Mohan as on 31st Dec. 2017 to find out the balance as per bank statement.
i) Overdraft as per cash book – 20000
ii) Cheques deposited but not yet credited – 10000
iii) Amount wrongly deposited by bank – 600
iv) Interest on overdraft debited by bank – 2000
v) Cheque issued but not yet present for payment – 2000
vi) Payment received from the customer directly by the bank – 1000
Solution:
Bank Reconciliation statement Mr. Mohan as on 31st Dec. 2018.

Question 5.
Prepare bank Reconciliation statement as on 31st December 2017. From the following in information.
a) Balance as perbank statement (pass book) is ₹ 50,000
b) Cheques deposited into bank amount into ₹ 7,000 were not yet collected.
c) Bank charges of ₹ 600 have not been entered in the cash book.
d) Cheques issued amount to ₹ 18,000 have not been presented by for payment.
Solution:
Bank Reconciliation statement as on 31st Dec. 2017.

Question 6.
From the following information, prepare bank Reconciliationstatement as of Mr. Pugazh as on 31st Dec. 2017.
i) Credit balance as per bank statement ₹ 12,000
ii) Cheques deposited on 28th December 2017 but not yet credited ₹ 4000
iii) Cheques issued for 20000 on 20th December 2017 but not yet presented for payment 6000.
iv) Interest on debentures directly in cash book ₹ 8000
v) Insurance premium on building directly paid by the bank ₹ 2000
vi) Amount wrongly credit by bank ₹ 1,000
Solution:
Bank Reconciliation statement of Mr. Pugazh as on 31st Dec. 2017.

Question 7.
From the following data, as certain the cash book balance as on 31st Dec. 2017.
1) Overdraft balance as per bank statement ₹ 13,000
2) Cheques deposited into the bank but not yet credited ₹ 21,000
3) Wrongly credit by the bank ₹ 1,000
4) Cheques issued, but not yet presented for payment₹ 6,000
5) Bank charges debited by bank ₹ 360
6) Insurance Premium on building directly paid by bank ₹ 200
Solution:
Bank Reconciliation statement on 31st Dec. 2017.

Question 8.
Prepare bank Reconciliation statement as on 31st Dec. 2017, From the following balance of cash book, and bank statement.
Cash book (Bank column)

Bank Statement

Solution:
Bank Reconciliation Statement on 31st Dec, 2017.

Question 9.
On 31st March 2017, the pass book of Mr, A showed a credit balance of Rs.92,500. A comparison of pass book and cash book revealed the following:

Solution:
Bank Reconciliation Statement of Mr. A as on 31st March 2017

Question 10.
The bank overdraft of Rajini on 31.12.2017 as per cash book is 90,000. From the following particulars, prepare bank reconciliation statement:

Solution:
Bank Reconciliation Statement as on 31,12.2017 3,000

Question 11.
Prepare a bank reconciliation statement from the following data as on 31.12.2017.
a) Balance as per cash book ₹ 1,25,500
b) Cheques issued but not presented for payment ₹ 9,000
c) Cheques deposited in bank but not collected ₹ 12,000
d) Bank paid insurance premium ₹ 5,000
e) Direct deposit by a customer ₹ 8,000
f) Interest on investment collected by bank ₹ 2,000
g) Bank charges ₹ 1,000
Solution:
Bank Reconciliation Statement as on 31.12.2017

Question 12.
The pass book of X with his bank shows a debit balance of Rs. 500 on 31.10.2017. On comparison of the pass book with the cash book, it is observed that:
i. Cheques issued by X in October 2017 amounted to ₹ 4, 535 of which cheques amounting to ₹ 3,535 were paid by the bank by 31st October 2017.

ii. X deposited cheques amounting to ₹ 5, 000 on 31st October 2017. These cheques were realised by the bank on 1st November, 2017.

iii. Y a customer of X had directly deposited a sum of ₹ 3, 000 on 24th October 2017 to the credit of X account with the bank. X recorded this receipt on 4th November, 2017.

iv. The bank had debited X’s account with ₹ 1, 520 on 31.1.2017 on account of a dishonoured bill. No entry for the same has been made in the account books.

v. On 31.10.1995 X’s account was credited with ₹ 130 being dividend collected by the bank. On the same day, his account was debited with ₹ 10 being bank charges. Both these entries were recorded by X only on 5th November, 2017.
Prepare the Bank Reconciliation Statement as at 31.10.2017.
Solution:
Bank Reconciliation Statement as on 31.10.2017

Question 13.
From the following particulars ascertain the bank balance as would appear in the pass‘book as on 31st December, 2016.
i. The bank overdraft (Credit balance) as per cash book on 31st December 2016 was ₹ 60,000
ii. Interest on overdraft, six months ending 31st December, amounting to ₹ 2,000 is debited in the pass book.
i. Bank charges for the above period also debited in the pass book which amounted to ₹ 500.
ii. Cheques issued but not presented for payment before 31st December amounted to ₹ 15,000.
iii. Cheques paid into the bank, but not cleared and credited before 31st December were ₹ 25,000.
iv. Interest on government securities collected by the bank and credited in the pass book amounted to ₹ 18,000.
Solution:
Bank Reconciliation Statement as on 31st March 2016

Question 14.
From the following information available from the books and records of X & Co., prepare BRS:

Solution:
Bank Reconciliation Statement for Bank A/c No. I

Bank Reconciliation Statement for Bank A/c No. II

Question 15.
From the following particulars ascertain the bank balance that would appear in the cash book of Son 31.12.2016.
i) The bank overdraft as per pass book on 31.12.2016 ₹ 6,340
ii) Interest on on for the year ending 31.12.2016 ₹ 160 is debited in the pass book
iii) Bank charges of ₹130 for the above period are also debited in the pass book.
iv) Cheques issued but not cashed prior to 31.12.2016 amounted to ₹ 11,168
v) Cheques paid into bank but not cleared before 31.12.2016 were ₹ 2170
vi) Interest on investments collected by the bankers and credited in the pass book, ₹ 1,200
Solution:
Bank Reconciliation Statement as on 31st December 2016

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.4

Choose the correct answer

Question 1.
The transportation problem is said to be unbalanced if
(a) Total supply ≠ Total demand
(b) Total supply = Total demand
(c) m = n
(d) m + n – 1
Solution:
(a) Total supply ≠ Total demand

Question 2.
In a non – degenerate solution number of allocation is
(a) Equal to m + n – 1
(b) Equal to m + n + 1
(c) Not equal to m + n – 1
(d) Not equal to m + n + 1
Solution:
(a) Equal to m + n – 1

Question 3.
In a degenerate solution number of allocations is
(a) Equal to m + n – 1
(b) Not equal to m + n – 1
(c) Less then m + n – 1
(d) Greater then m + n – 1
Solution:
(c) Less then m + n – 1

Question 4.
The Penalty in VAM represents difference between the first
(a) Two largest costs
(b) Largest and Smallest costs
(c) smallest two costs
(d) None of these
Solution:
(c) smallest two costs

Question 5.
Number of basic allocation in any row or column in an assignment problem can be
(a) Exactly one
(b) At least one
(c) At most one
(d) None of these
Solution:
(a) Exactly one

Question 6.
North – West Corner refers to
(a) Top left corner
(b) Top right corner
(c) Bottom right corner
(d) Bottom left corner
Solution:
(a) Top left corner

Question 7.
Solution for transportation problem using method is nearer to an optimal
(a) NWCM
(b) LCM
(c) VAM
(d) Row Minima
Solution:
(c) VAM

Question 8.
In an assignment problem the value of iedsion variable x_ij is
(a) 1
(b) 0
(c) 1 or 0
(d) none of them
Solution:
(c) 1 or 0

Question 9.
If number of sources is not equal to number of destinations, the assignment problem is called
(a) balanced
(b) unsymmetric
(c) symmetric
(d) unbalanced
Solution:
(d) unbalanced

Question 10.
The purpose of a dummy row or column in an assignment problem is to
(a) prevent a solution from becoming degenerate
(b) balance between total activities and total resources
(c) provide a means of representing a dummy problem
(d) None of the above
Solution:
(b) balance between total activities and total resources

Question 11.
The solution for an assignment problem is optimal if
(a) each row and each column has no assignment
(b) each row and each column has atleast one assignment
(c) each row and each column has atmost one assignment
(d) each row and each column has exactly one assignment
Solution:
(d) each row and each column has exactly one assignment

Question 12.
In an assignment problem involving four workers and three jobs, total number of assignments possible are
(a) 4
(b) 3
(c) 7
(d) 12
Solution:
(b) 3

Question 13.
Decision theory is concerned with
(a) analysis of information that is available
(b) decision making under certainty
(c) selecting optimal decisions in sequential problem
(d) All of the above
Solution:
(d) All of the above

Question 14.
A type of decision – making environment is
(a) certainty
(b) uncertainty
(c) risk
(d) all of the above
Solution:
(d) all of the above

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.3

Question 1.
Given the following pay – off matrix (in rupees) for three strategies and two states for nature.

Select a strategy using each of the following rule (i) Maximin (ii) Minimax
Solution:

(i) Max min
Max ( 40, -20, -40) = 40. Since the maximum pay-off is 40, the alternative S, is selected.

(ii) Minimax
min (60, 10, 150) = 10, Since the minimum payoff is 10. the alternative S₂ is selected.

Question 2.
A farmer wants to decide which of three crops he should plant on his 100 – acre farm. The profit from each is dependent on the rainfall during the growing season. The farmer has categorized the amount of rainfall as high medium and low. His estimated profit for each is shown in the table.

In the farmer wishes to plant only crop, decide which should be his best crop using (i) Maximin (ii) Minimax
Solution:

(i) Maximin
Max ( 3500, 4500, 2000) = 4500. Since the maximum pay-off is 4500, the alternative ‘Medium’, is selected.

(ii) Minimax
min (8000, 5000, 5000) = 5000, Since the minimum pay-off is 5000. the alternatives both ‘Medium’ and ‘Low’ are selected.

Question 3.
The research department of Hindustan Ltd. has recommended to pay marketing department to launch a shampoo of three different types. The marketing types of shampoo to be launched under the following T estimated pay-offs for various level of sales.

What will be the marketing manager’s decision if (i) Maximin and (ii) Minimax principle applied?
Solution:

(i) Maximin
Max ( 10, 5, 3) = 10. Since the maximum pay of is 10, “Egg Shampoo”, is selected.

(ii) Minimax
min (30, 40, 55) = 30, Since the minimum pay-off is 30. the alternative “Egg Shampoo” is selected

Question 4.
Following pay – off matrix, which is the optimal decision under of the following rule (i) maximin (ii) minimax

Solution:

(i) Maximin
Max ( 5, 7, 9, 8) = 5. Since the maximum pay-off is 5, the alternative A₁ is selected.

(ii) Minimax
min (14, 11, 11, 13) = 11, Since the minimum pay-off is 11. the alternative alternatives A₂ and A₃ are selected.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.2

Question 1.
What is the Assignment problem?
Solution:
Suppose that we have ‘m1 jobs to be performed on ‘n’ machines. The cost of assigning each job to each machine is C_ij (i = 1, 2, … n and j = 1, 2, … , n). Our objective is to assign the different jobs to the different machines (one job per machine) to minimize the overall cost. This is known as an assignment problem.

Question 2.
Give mathematical form of assignment problem.
Solution:
Consider the problem of assigning n jobs to n machines (one job to one machine). Let C_ij be the cost of assigning i^th job to the j^th machine and x_ij represents the assignment of i^th job to the j^th machine.

x_ij is missing in any cell means that no assignment is made between the pair of job and machine.(i.e) x_ij = 0.
x_ij is present in any cell means that an assignment is made their. In such cases x_ij = 1
The assignment model can written in LPP as follows
Minimize Z = \(\sum_{i=1}^{m}\) \(\sum_{j=1}^{n}\) C_ij x_ij
Subject to the constrains
\(\sum_{i=1}^{n}\) x_ij = 1, j = 1, 2, …. n
\(\sum_{j=1}^{n}\) x_ij = 1, i = 1,2,….n and x_ij =0 (or) 1 for all i, j

Question 3.
What is the difference between Assignment Problem and Transportation Problem?
Solution:

Question 4.
Three jobs A, B, and C one to be assigned to three machines U, V, and W. The processing cost for each job machine combination is shown in the matrix given below. Determine the allocation that minimizes the overall processing cost.
(cost is in Rs per unit)

Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Look for atleast one zero in each row and each column.
Here each and every row and columns having exactly one zero No need for step 2 go to step 3.

Step 3.

Mark the zero by □ Mark other zeros in its column by X.
Since each row and each column contains exactly one assignment, all three machines have been assigned a job.

The Optimal assignment (minimum) cost = 46

Question 5.
A computer centre has got three expert programmers. The centre needs three application programmes to be developed. The head of the computer centre, after studying carefully the programmes to be developed, estimates the computer time in minutes required by the experts to the application programme as follows.

Assign the programmers to the programme in such a way that the total computer time is the least.
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero, mark the zero by □. Mark other zeros in its column by X.

Step: 4
Now examine the columns with exactly one zero marks the zero by □. Mark other zeros in its row by X.

Thus all three assignments have been made. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs 280

Question 6.
A departmental head has four subordinates and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimates of the time man would take to perform each task is given below.

How should the tasks to allocated to subordinates so as to minimize the total man-hours?
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero Mark the zero by □. Mark other zeros in its row by X.

Step 4.
Now examine the columns with exactly one zero. Mark the zero by □ Mark other zeros in its row by X.

Step 5.
Cover all the zeros of table 4 with three lines, since three assignments were made check (✓) row S since it has no assignment.

Step 6.
Develop the new revised tableau. Examine those elements that are not covered by a line in table 5. Take the smallest element. This is 1 (one) our case. By subtracting 1 from the uncovered cells.

[Adding 1 to elements (Q, S, R) that line at the intersection of two lines]

Step 7.
Go to step 3 and repeat the procedure until you arrive at an optimal assignment.

Step 8.
Determine an assignment.

Thus all the four assignment have been made. The optimal assignment schedule and total time is

The optimum time (minimum) = 41 Hrs.

Question 7.
Find the optimal solution for the assignment problem with the following cost matrix.

Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and sub tract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero. Mark the zero by □ Mark other zeros in its column by X

Thus all the four assignments have been made. The optimal assignment schedule and total cost.

The Optimum cost (minimum) = Rs 37

Question 8.
Assign four trucks 1, 2, 3 and 4 to vacant spaces A, B, C, D, E and F so that distance travelled is minimized. The matrix below shows the distance.

Solution:
Since the number of columns is less than the number of rows, the given assignment problem is unbalanced one. To balance it, introduce two dummy columns with all the entries zeros.
The revised assignment problem is

Here only 4 tasks can be assigned to 4 vacant spaces.
Step 1.
It is not necessary, since each row contains zero entry. Go to step 2.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Since each row contains more than one zeros. Go to step 4.

Step 4.
Examine the columns with exactly one zero, mark the zero by □ Mark other zeros in its rows by X.

Step 5.
Here all the four assignments have been made we can assign d₁ for D then we will get d₂ for E.

The optimal assignment schedule and total distance is

∴ The Optimum Distant (minimum) = 12 units

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.1

Question 1.
What is transportation problem?
Solution:
The objective of transportation problem is to determine the amount to be transported from each origin to each destinations such that the total transportation cost is minimized.

Question 2.
Write mathematical form of transportation problem.
Solution:
Let there be m origins and n destinations. Let the amount of supply at th i th origin is a_i. Let the demand at j th destination is b_j.
The cost of transporting one unit of an item from origin i to destination j is C_ij and is known for all combination (i,j). Quantity transported from origin i to destination j be x_ij.

The objective is to determine the quantity x_ij to be transported over all routes (i,j) so as to minimize the total transportation cost. The supply limits at the origins and the demand requirements at the destinations must be satisfied.
The above transportation problem can be written in the following tabular form:

Now the linear programming model representing the transportation problem is given by
The objective function is Minimize Z = \(\sum_{\mathbf{i}=\mathbf{1}}^{\mathbf{m}}\), \(\sum_{\mathbf{j}=\mathbf{1}}^{\mathbf{n}}\) c_ij x_ij
Subject to the constraints
\(\sum_{\mathbf{j}=\mathbf{1}}^{\mathbf{n}}\) = x_ij = a_i, i = 1, 2 …….. m (supply constraints)
\(\sum_{\mathbf{i}=\mathbf{1}}^{\mathbf{m}}\) = x_ij = b_j, i = 1, 2 …….. n (demand constraints)
x_ij ≥ 0 for all i, j (non- negative restrictions)

Question 3.
What are feasible solution and non-degenerate solutions to the transportation problem?
Solution:
Feasible Solution: A feasible solution to a transportation problem is a set of non-negative values x_ij (i = 1, 2,.., m, j = 1, 2, …n) that satisfies the constraints.
Non-degenerate basic feasible Solution: If a basic feasible solution to a transportation problem contains exactly m + n – 1 allocation in independent positions, it is called a Non-degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem.

Question 4.
What do you mean by a balanced transportation problem?
Solution:
In a transportation problem if the total supply equals the total demand (Σa_i = Σb_j) then it is said to be balanced transportation problem.

Question 5.
Find an intial basic feasible solution of the following problem using north west corner rule.

Solution:
Here total Supply 19 + 37 + 34 = 90
Total demand = 16 + 18 + 31 + 25 = 90
(i.e) Total supply = Total demand
∴ The given problem is balanced transportation problem
∴ we can final an initial basic feasible solution to due given problem.

From the above table we can choose the cell in the North west corner. Here the cell is (Q₁, D₁). Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column’s exhausted.
(i.e) x₁₁ = min (19, 16) = 16

Now the cell in the north west corner is (O₁, D₂) Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the first column is exhausted.
(i.e) x₁₂ = min (3, 18) = 3

Reduced transportation table is

Now the cell in the north west corner is (O₂, D₂)
x₂₂ = min (37, 15) = 15

Now the cell in the north west corner is (O₂, D₃)
x₂₃ = min (22, 31) = 9

Now the cell in the north west corner is (O₃, D₃)
x₃₃ = min (34, 9) = 9

Thus we have the following allocations

Transportation schedule:
O₁ → D₁, O₁, → D₂, O₂ → D₂; O₂ → D₃ O₃ → D₃; O₃ → D₄
= (16 × 5) + (3 × 3) + (15 × 7) + (22 × 9) + (9 × 7) + (25 × 5)
= 80 + 9 + 105 + 198 + 63 + 125
= 580

Question 6.
Determine an intial basic feasible solution of the following transportation problem by north west corner method.

Solution:
Here total capacity (a_i) = 30 + 40 + 50 = 120
Total demand (b_j) = 35 + 28 + 32 + 25 = 120
(i.e) Total capacity = Total demand
∴ The given problem is balanced transportation.
∴ We can find an initial basic feasible solution to the given problem.

From the above table we can choose the cell in the North west corner. Here the cell is (chennai, Bangalore)
x₁₁ = min (30, 35) = 30
Reduced transportation table is

Now the cell in the North west corner is (Madurai, Bangalore)
x₂₁ = min(40, 5) = 5

From the above table we can choose the cell in the North west corner. Here the cell is (Nasik, Madurai)
x₂₂ = min (35, 28) = 28

From the above table we can choose the cell in the North west corner. Here the cell is (Bhopal, Madurai)
x₂₂ = min (7, 32) = 7

From the above table we can choose the cell in the North west corner. Here the cell is (Trichy, Bhopal)
x₃₃ = min (50, 25) = 25
Reduced transportation table is

x₃₄ = min (25, 25) = 25
Thus we have the following allocations

Transportation Schedule:
Chennai → Bangalore; Madurai → Bangalore;
Madurai → Naisk; Madurai → Bhopal
Trichy → Bhopal; Trichy → Delhi
The total transportation cost =
(30 × 6) + (5 × 5) + (28 × 11) + (7 × 9) + (25 × 7) + (25 × 13)
= 180 + 25 + 308 + 63 + 175 + 325
= 1076

Question 7.
Obtain an initial basic feasible solution to the following transportation problem by using least-cost method.

Solution:
The given transportation table is

Total supply = 25 + 35 + 40 = 100
Total demand = 30 + 25 + 45 = 100
(i.e) Total supply = Total demand
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
The least cos is 4 corresponds to the cell (O₂, D₃)
Allocate min (35, 45) = 35 units to this cell

The least cost corresponds to the cell (O₁, D₃)
Allocate min (25, 10) = 10 units to this cell

The least cost is 6 corresponds to the cell (O₃, D₂)
Allocate min (40, 25) = 25 units to this cell

The least cost is 7 corresponds to the cell (O₃, D₁)
Allocate min (15, 30) = 15

Thus we have the following allocations

Transportation Schedule:
O₁ → D₁; O₁, → D₃; O₂ → D₃; O₃ → DI; O₃ → D₂
Total Transportation cost
= (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)
= 135 + 50 + 140 + 105 + 150
= 580

Question 8.
Explain vogel’s approximation method by obtaining initial feasible solution of the following transportation problem.

Solution:
Here Σ ai = 6 + 1 + 10 = 17
Σ bj = 7 + 5 + 3 + 2 = 17
Σ ai = Σ bj
(i.e) Total supply = Total Demand
∴ The given problem is balanced transportation problem
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns.

Choose the largest difference. Here the largest difference is 6 which corresponds to column D₄
In this column choose the least cost. Here the least cost corresponds to (O₂, D₄)
Allocate min (1, 2) = 1 unit to this cell the reduced transportation table is

choose the largest difference 5 which corresponds to column D₂. Here the least cost corresponds to (O₁, D₂).
Allocate min (6, 5) = 5 units in this cell

Choose the largest difference 5 which corresponds to row O₁. Here the least cost corresponds to (O₁, D₁)
Allocate min (1, 7) = 1 unit in this cell

Choose the largest difference 4 which corresponds to row O₃. Here least cost corresponds to (10, 6) = 6 units in this cell.

Choose the largest difference 6 which corresponds to row O₃. Here the least cost corresponds to (O₃, D₄).
Allocate min (4, 1) = 1

Thus we have the following allocations

Transportation schedule
O₁ → D₁; O₁ → D₂; O₂ → D₄;
O₃ → D₁; O₃ → D₃; O₃ → D₄
Total transportation cost:
= (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102

Question 9.
Consider the following transportation problem.

Determine initital basic feasible solution by VAM
Solution:
Given Transportation problem is

Here Σ ai = Σ bj = 100
∴ The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns

Choose the largest difference. Here the difference is 3 which corresponds to D₂
In this column choose the least cost. Here the least cos corresponds to (O₃, D₂)
Allocate min (20, 40) = 20 units to this cell

Choose the largest difference is 4 which corresponds to column D₃. In this column choose the least cost. Here the least cost corresponds to (O₁, D₃).
Allocate min (30, 20) = 20 units to this cell

Choose the largest different is 3 which corresponds to column D₂. In this column choose the least cost. Here the least cost corresponds to (O₂, D₂)
Allocate min (50, 20) = 20 units to this cell

Choose the largest difference is 2 which corresponds to column D₄. In this column choose the least cost. Here the least cost corresponds to (O₂, D₄).
Allocate min (30, 10) = 10 units to this cell

Allocate min (20, 30) = 20 units to this cell

Here
x₁₁ = 10
x₁₃ = 20
x₂₁ = 20
x₂₂ = 20
x₂₄ = 10
x₃₂ = 20
Transportation schedule
O₁ → D₁; O₁ → D₃; O₂ → D₁;
O₂ → D₂; O₂ → D₄; O₃ → D₂
The transportation cost
= (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2)
= 50 + 60 + 80 + 100 + 40 + 40
= 370

Question 10.
Determine basic feasible solution to the following transportation problem using North west Corner rule.

Solution:
Here total supply = 4 + 8 + 9 = 21
Total demand = 3 + 3 + 4 + 5+ 6 = 21
(i.e) Total supply = Total demand
∴ The given problem is balanced transportation problem.
∴ we can find an initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North west corner. Here the cell is (P, A)
Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column is exhausted.

Form the above table we can choose the cell in North west corner. Here the cell is (P,B)
x = min (1, 3) = 1

From the above table we can choose the cell in north west corner. Here the cell is (Q, B)
x = min (2, 8) = 2

From the above table, we can choose the cell in North west corner. Here the cell is (Q, C)

From the above table, we can choose the cell in North west corner. Here the cell is (Q, D)
x = min (2, 5) = 2

From the above table, we can choose the cell in North west corner. Here the cell is (R, D)
x = min (9, 3) = 3

Thus we have the following table

Transportation Schedule:
P → A; P → B; Q → B; Q → C; Q → D R → D; R → E
Total transportation cost:
= (3 × 2) + (1 × 11) + (2 × 4) + (4 × 7) + (2 × 2) + (3 × 8) + (6 × 2)
= 6 + 11 + 8 + 28 + 4 + 24 + 72
= 153

Question 11.
Find the initial basic feasible solution of the following transportation problem:

Using
(i) North West corner tule
(ii) Least Cost method
(iii) Vogel’s approximation method
Solution:
(i) North west corner rule:
Here the total supply = 10 + 10 + 10 = 30
Total demand = 7 + 12 + 11 = 30
(i.e) Total supply = Total demand
The given problem is balanced transportation problem.
∴ we can find an initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North west corner. Here the cell is (A, I)
x₁₁ = min (7, 10) = 7

From the above table we can choose the cell in the north west corner. Here the cell is (B, I)
x = min (3, 12) = 3

From the above table we can choose the cell in the North west corner. Here the cell is (B, II)
x = min (9, 10) = 9

Here the cell in the North west corner is (C, II)
x = min (11, 1) = 1

Thus we have the following allocations

Transportation schedule:
A → I; B → I; B → II; C → II; C → III
Total transportation cost:
= (7 × 1) + (3 × 0) + (9 × 4) + (1 × 1) + (10 × 5)
= 7 + 0 + 36 + 1 + 50
= 94

(ii) Least cost method:
The given transportation table is

Here Total supply = Total demand = 30
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
The least cost is 0 corresponds to the cell (B, I)
Allocate min (12, 10) = 10 units to this cell

The least cost 1 corresponds to the cell (C, II)
Allocate min (11, 10) = 10 units to this cell

Here the least cost 2 corresponds to the cell (B, III)
Allocate min (2, 10) = 2 units to this cell.

Here the least cost is 5 corresponds to the cell (C, III)

Transportation schedule:
A → III; B → I; B → III; C → II; C → III
Total transportation cost:
= (7 × 6) + (10 × 10) + (2 × 2) + (10 × 1) + (1 × 5)
= 42 + 0 + 4 + 10 + 5
= 61

(iii) Vogel’s approximation method:
Here Σ ai = Σ bj = 30
(i.e) Total supply = Total demand
∴ This given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.


A → I; B → I; B → III; C → I; C → II
Total transportation cost:
= (7 × 1) + (2 × 0) + (10 × 2) + (1 × 3) + (10 × 1)
= 7 + 0 + 20 + 3 + 10
= 40

Question 12.
Obtain an initial basic feasible solution to the following transportation problem by north west corner method.

Solution:
Here the total available = 250 + 300 + 400 = 950
Total Required = 200 + 225 + 275 + 250 = 950
(i.e) Total Available = total required
∴ The given problem is balanced transportation problem.
we can find an initial basic feasible solution to the given problem.
From the above table, we can choose the cell in the North west corner. Here the cell is (A, D).
x₁₁ = min (250, 200) = 200

From the above table we can choose the cell in North west corner. Here the cell is (A, E)
x = min (50, 225) = 50

From the above table, the north west corner cell is (B, E)
x = min (300, 175) = 175

From the above table, the north west corner cell is (B, F)
x = min (125, 275) = 125

Here the north west corner cell is (C, F)
x = min (400, 150) = 150

Transportation schedule:
A → D; A → E; B → E; B → F; C → G
Total Transportation cost:
= (200 × 11) + (50 × 13) + (175 × 18) + (125 × 14) + (150 × 13) + (250 × 10)
= 2200 + 650 + 3150 + 1750 + 1950 + 2500
= 12,200

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Question 1.
The following table summarizes the supply, demand and cost information for four factors S₁, S₂, S₃, S₄ Shipping goods to three warehouses D₁, D₂, D₃.

Find an initial solution by using north west corner rule. What is the total cost for this solution?
Solution:
The given transportation table is

Here total supply = 5 + 8 + 7+14 = 34
Total amount =7 + 9 + 18 = 34
(i.e) Total supply = Total demand.
∴ The given problem is balanced transformation problem.
We can find an initial basic feasible solution to the given problem.
First allocation:



Transportation schedule:
S₁ → D₁; S₂ → D₁; S₂ → D₂;
S₃ → D₂; S₃ → D₃; S₄ → D₃
The transportation cost:
= (5 × 2) + (2 × 3) + (6 × 3) + (3 × 4) + (4 × 7) + (14 × 2)
= 10 + 6 + 18 + 12 + 28 + 28
= 102

Question 2.
Consider the following transportation problem

Determine an initial basic feasible solution using (a) Least cost method (b) Vogel’s approximation method.
Solution:
(a) Least cost method
Given transportation table is

Total Availability = Total Requirement = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:



Here
x₁₂ = 10; x₁₃ = 20; x₂₁ = 30;
x₂₂ = 10; x₂₄ = 10; x₃₂ = 20
Transportation Scheme:
O₁ → D₂; O₁ → D₃; O₂ → D₁;
O₂ → D₂; O₂ → D₄; O₃ → D₂
The transportation cost:
=(10 × 8) + (20 × 3) + (30 × 4) + (10 × 5) + (10 × 4) + (20 × 2)
= 80 + 60 + 120 + 50 + 40 + 40
= 390

(ii) Vogel’s approximation method:
Here Σai = Σbj = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:



Here
x₁₁ = 30; x₁₃ = 20; x₂₂ = 20;
x₂₄ = 10; x₃₂ = 20
Transportation Scheme:
O₁ → D₁; O₂ → D₃; O₂ → D₂;
O₂ → D₄; O₃ → D₂;
Total transportation cost:
= (30 × 5) + (20 × 3) + (20 × 5) + (10 × 4) + (20 × 2)
= 150 + 60 + 100 + 40 + 40
= 390

Question 3.
Determine an initital basic feasible solution to the following transportation problem by using (i) North West Corner rule (ii) least cost method.

Solution:
(i) North West Corner rule:
The given transportation table is

Total supply = 25 + 35 + 40 = 100
Total Requirement = 30 + 25 + 45 + 100
(i.e) Total supply = Total requirement.
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:


Transformation schedule:
S₁ → D₁; S₂ → D₁; S₂ → D₂;
S₃ → D₃; S₃ → D₃;
The transformation cost:
= (25 × 9) + (5 × 6) + (25 × 8) + (5 × 4) + (40 × 9)
= 225 + 30 + 200 + 20 + 360 = 835

(ii) Least cost method
The given transportation table is

(i.e) Total supply = Total requirement = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:


Transportation schedule:
S₁ → D₃; S₂ → D₃; S₃ → D₁
S₃ → D₂;
The transportation cost:
= (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6) + (40 × 9)
= 50 + 140 + 105 + 150 + 360 = 805

Question 4.
Explain Vogel’s approximation method by obtaining initial basic feasible solution of the following transportation problem.

Solution:
The given transportation table is

Here Σai = Σbj =17
(i.e) Total supply = Total Demand
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:



Here
x₁₁ = 1; x₁₂ = 5; x₂₄ = 1;
x₃₁ = 6; x₃₃ = 15;
Transportation schedule:
O₁ → D₁; O₁ → D₂; O₂ → D₄
O₃ → D₁; O₃ → D₃
The transportation cost:
=(1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102

Question 5.
A car hire company has one car at each of five depots a,b,c,d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers (destinations) where the customers are given the following distance matrix.

How should the cars be assigned to the customers so as to minimize the distance travelled?
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 4.
Now Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 5.
Cover all the zeros of table 4 with three lives. Since three assignments were made please note that check [✓] Row C and E which have no assignment.

Step 6.
Develop the new revised tableau. Examine those elements that are not covered by a line in Table 5. Take the smallest element in each row and subtract from the uncovered cells, depots

Step 7.
Go to step 3 and repeat the procedure until you arrive at an optimal assignments.
depots

Step 8.
Determine an assignment

Here all the five assignments have been made. The optimal assignment schedule and total distance is

∴ The optimum Distance (minimum) 575 kms

Question 6.
A natural truck – rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance (in kilometers) between the cities with a surplus and the cities with

How should the truck be dispersed so as to minimize the total distance travelled?
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3.
Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 4.
Examine the Columns with exactly one zero. If there is exactly one zero, mark that zero by □ mark other zeros in its rows by X

Step 5.
Cover all the zeros of table 4 with five lines. Since three assignments were made

Step 6.
Develop the new revised tableau. Examine those elements that are not covered by a line in Table 5. Take the smallest element. This is l(one) in our case. By subtracting 1 from the uncovered cells.

Step 7.
Go to step 3 and repeat the procedure until you arrive at an optimal assignments.

Step 8.
Determine an assignment

Here all the six assignments have been made. The optimal assignment schedule and total distance is

∴The optimum Distance (minimum) = 125 kms

Question 7.
A person wants to invest in one of three alternative investment plans: Stock, Bonds and Debentures. It is assumed that the person wishes to invest all of the funds in a plan. The pay – off matrix based on three potential economic conditions is given in the following table

Solution:

(i) Maximin
Max (3000, 1000, 6000) = 6000. Since the maximum pay of is 6000, the alternative ‘Debentures’, is selected.

(ii) Minimax
Min (10000, 8000, 6000) = 6000, Since the minimum pay-off is 6000. the alternative ‘Debentures’ is selected.