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Sources of Energy Chapter Wise Important Questions Class 10 Science

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Sources  of Energy Chapter Wise Important Questions Class 10 Science

PREVIOUS YEARS QUESTIONS

Very Short Answer Type Questions [1 Mark] – Year 2015

1.Name any two fossil fuels.
Ans. A good fuel is the one which

  •  has high calorific value,
  •  is non-polluting,
  •  is easy to transport and easily available,
  •  has moderate ignition temperature.

2.Mention the purpose of blackening the interior of a solar cooker.
Answer. The purpose of blackening the interior of a solar cooker is that the black surface absorbs more heat radiations of incident solar energy (about 98%) as compare to white or other light coloured surface

Short Answer Type Question[ll] [3 Marks] – Year 2015

3.Why do people oppose the construction of Tehri Dam on the river Ganga and Sardar sarover project on the river Narmada.  (Or)
Mention three disadvantage of producing hydroelectricity by constructing the dams.(Or)
List any three ways in which construction of dams for production of electricity adversely affects the environment of that place.
Answer. 

  1. Large area is required to build the dam that  result rehabilitation of displaced people.
  2. Large eco-systems are destroyed when submerged under the water in dams.
  3. The vegetation which is submerged rots under anaerobic conditions and gives rise to large amounts of methane gas which leads to a green-house effect.

4. Reena’s father works with a company that builds dams. Once he took her to the site where a dam was being built. Reena saw a group of people sitting there and shouting slogans against the building of dam. She talked the group of people and asked them about their problems and then discussed it with her father. She then tried to convince the people to talk with the authorities and come to an amicable solution. The discussion was successful. Reena’s father was very proud of his daughter. Now answer the following questions:
(i) Why was Reena’s father proud of his daughter?
(ii)Why was the group of people protesting against the building of dam?
(iii) Reena’s contribution to the peaceful resolution of the conflict proved to boon for many citizens of the country. How?
Answer.
(i) Because she convinced the people by explaining them the advantage of constructing dam and come with a solution of rehabilitation of displaced peoples. So the contribution to the peaceful resolution of the conflict make Reena’s father proud of his daughter.
(ii)The group of people protesting against the building of dam because of following disadvantages
(a) A large area of agricultural land and human habitation are sacrificed as it get submerged at the dam site.
(b) Large ecosystems are destroyed when submerged under water.
(c) The decomposition of vegetation under anaerobic condition produces methane gas which is also a greenhouse gas.
(iii)(a) The citizens are not affected by flood as it can be controlled by storing the water behind the dam.
(b) Farmers get benefited as water for irrigation is available for the whole year.
(c) Low cost of production of electricity gives monetary gain which is a boon to citizens.

5.Bio gas is an excellent fuel. Justify the statement by giving two reasons. Mention the main constituents of bio gas along with its percentage.
Answer. 
(i) It burns without smoke and leave no residue therefore causes no atmospheric pollution.
(ii)Its heating capacity is high, i.e. it has high calorific value. Main constituents of bio gas: The composition of bio gas varies depending upon the nature of organic matter feeding in’the digester and advanced waste treatment technology. The typical composition of bio gas is
Methane                                50 – 75%
Carbon dioxide                     25 – 50%
Nitrogen                                   0 – 10%
Hydrogen                                 0 – 1%
Hydrogen sulphide                0 – 3%

6.(a) Define tidal energy.
(b) Explain how the tidal energy is harnessed and write one limitation of the use of tidal energy.
Answer.
(a) Tidal energy: The energy produced by the surge of ocean water during high and low tides due to difference in sea-levels is called tidal energy. The high and low tides occur due to the gravitational pull of the moon. This causes enormous movement of water.
(b) Tidal energy is harnessed by constructing a dam near the shores. During the high tides water flows into the dam and during the low tides, water flows out. This flowing water rotates the turbine, present at the opening of the dam and produces electricity.

7. Define process of nuclear fission. Writ the steps involved in generating electricity in a nuclear reactor.
Answer. Nuclear Fission: The process in which a heavy nucleus (such as uranium, plutonium or thorium) is broken into two nearly equal fragments when bombarded with low-energy neutrons and a tremendous amount of energy is released. This process is called nuclear fission.
Steps involved in generating electricity:

  1. The fuel rods full of uranium pellets are placed in a nuclear reactor chamber.
  2. Low-energy neutrons are bombarded on uranium fuel rod.
  3.  A self-sustaining fission chain reaction starts that releases energy at a controlled rate.
  4. With this heat the reactor converts water to steam at a high temperature and pressure.
  5. This high temperature and pressure steam spins generator turbines producing electricity.
  6. The steam cools back into water, which can then be used over again.

Long Answer Type Question [5 Marks] – Year 2015

8.Nikhil and Neha went to a remote village in Kerala to meet their grandmother. They were surprised to see a biogas plant in Mr. Nair’s house in the neighbourhood. There were plenty of livestocks and the household used cooking gas from the plant. Also their farms had rich vegetation. They contacted sarpanch of the village and convinced him to set up a biogas plant for village community.
(a) Mention the values displayed by Mr. Nair, Nikhil and Neha.
(b) Explain the possible arguments given by Nikhil and Neha to the sarpanch to convince him to set up community biogas plant.
Answer.
(a) Mr Nair: Higher degree of general awareness, helping nature.
Nikhil and Neha: Knowledge sharing.
• helping nature.
• concern for community advantage.
(b) Possible arguments given by Nikhil and Neha in favour to set up community biogas plant, to the sarpanch on

  1. Economical source of energy.
  2.  Cleanliness of the whole village.
  3.  Use of slurry left behind as mannar which is rich in nitrogen and phosphorous.
  4. It reduces air pollution and greenhouse effect.
  5.  It raises the standard of living.
  6. It is beneficial to entire village community. Basically they discuss and explain the advantages of community bio gas plant for the’entire village.

9.Solar cooker takes more time as compared to the LPG to boil potato or rice, yet Kunal uses solar cooker for such type of cooking:
(i) Why does Kunal uses Solar cooker instead of LPG? Give reasons for your answer.
(ii) Name the phenomenon which is responsible for obtaining high temperature in solar cooker.
Answer.
(a)(i) There are no energy losses while cooking on the solar cooker as the food gets cooked in a controlled environment whereas cooking on the LPG leads to maximum percentage of energy loss.
(ii) Temperature controlled cooking retains the nutrient value of the food.
(iii) Using the solar cooker, LPG can be served which result reduction in the emission of  CO2.
(iv) It saves a lot of prestigious time which is normally consumed for cooking purpose.
(b) Greenhouse effect.

10. Manoj’s father works in a nuclear power reactor. He asked his father to take him for a visit to nuclear power reactor. But his father was not keen to take him there. Now answer the following questions:
(i) What could be the possible reason of Manoj’s father not taking him to the nuclear power reactor?
(ii) Write one advantages and one disadvantage of setting up a nuclear power reactor. .
Answer.
(a) (i) Safety concerns.
(ii) Health concerns.
(b) Advantage: From a small amount of nuclear fuel, a large amount of energy is released in a nuclear power reactor.
Disadvantage: There is risk of harmful radiation leakage from nuclear waste.

Very Short Answer Type Questions [1 Mark] – Year 2014

11. Name the part of a biogas plant where reactions take place in the absence of oxygen.
Answer. Digester chamber.

12. Name the kind of energy possessed by wind and the device used to harness it.
Answer. Kinetic energy, wind mill.

13. List two non-conventional sources of energy.
Answer. Geothermal, solar, biomass, water, wind are the non-conventional sources of energy, (any two)

14. A black surface absorbs more heat radiations as compared to a white or a
reflecting surface under identical conditions. List two solar devices which make use of this property in their design.
Answer. Solar cooker, solar water heater.

15. Name any two elements that are used in fabricating solar cells.
Answer. Germanium, Silicon.

16. Why a solar cooker painted black form outside?
Answer. Black surface absorbs more heat as compared to white or reflecting surface under identical conditions.

Short Answer Type Question [I] [2 Marks] – Year 2014

17. Define fuel. List any two characteristics that you would look for in a good fuel.
Answer. A substance that produces useful energy when it burn or undergoes a chemical f or nuclear reaction. The fuel such as coal, wood, oil, or gas provides energy when burned. A good fuel is the one which produces a huge amount of heat on burning. It does not produce a lot of smoke and is easily available.

Short Answer Type Question[ll] [3 Marks] – Year 2014

18 State any three reasons to justify that LPG is considered as an ideal fuel.
Answer. LPG is considered as an ideal fuel because

  1.  It is easy to store, handle and transport.
  2.  It produces large amount of heat on burning.
  3.  It does not leave any residue on burning,

19 Sate any three advantages of charcoal over wood.
Answer. Advantages of charcoal over wood

  1.  It has higher calorific value, i.e. higher heat generating efficiency.
  2. Charcoal does not produce smoke on burning so it is a clean fuel.
  3.  It is easier to transport and ready to use in a convenient dry and broken-up form.

20. Ramesh is a student of standard X. He organized many activities in his school to convey the students about the various advantages and disadvantages of using renewable and non-renewable sources of energy. Many students of the school took part and concluded about the best choices of energy sources on this basis.
Answer the following questions:
(i) What activities Ramesh might have had assigned for the students?
(ii) Name any two renewable sources of energy.
(in) Which two values are reflected in Ramesh’s thought and action?
Answer. 
(i) Activities that Ramesh have had assigned for the students are:
(a) Hands on projects related to renewable and non-renewable sources of energy.
(b) Photograph exhibitions on the topic “use of fossil fuel and its impact on the environment.”
(c) Essay writing competition on a “renewable energy technology that could potentially be used for community.”
(d) Group discussion on advantages and disadvantages of using renewable and non-renewable sources of energy for selection of best source of energy.
(ii) Solar energy, hydro energy.
(iii) Ramesh’s thought: Higher degree of general awareness proper use of renewable source of energy.
Ramesh’s action: Concern for community advantage and good management.

21. Megha asked her mother to install a solar water heater on the roof top. Her mother asked about the need of such installation. Megha convinced her mother and succeeded in setting up a solar water heater on her roof top.
(i) State one advantage of installing the solar water heater to Megha’s family and to the society.
(ii) What qualities of Megha are reflected in her action?
(iii) State one limitation in using solar water heater.
Answer.
(i) Advantage of solar water heater (any one)
(a) Once solar water heater installed, the family and society never need to pay for heating up the water again.
(b) Megha’s family and society make their contribution to the nation to curb the emission of green house gases.
(ii) General awareness, concern for community advantage and proper use of ‘ renewable source energy.
(iii) (a) It does not work at night as well as on cloudy day.
(b) Relatively low heating system efficiency of passive solar water.

22. Mariyam made a solar cooker on her own. She used a white metallic box with a lid. She put the food in the solar cooker for heating and closed the lid. But she did not get good results. She then consulted her teacher to know her mistakes. Her teacher pointed out her mistakes, which she later rectified.
(a) According to you, which two mistakes were made by Mariyam and what
did she do to rectify them?
(b) Which qualities of Mariyam are rectified in her action.
answer.
(a) The possible mistakes were made by Mariyam are:

  •  The outer and interior of solar cooker might not have been blackened.
  • Instead of using glass lid she might have been used plastic cover.
  •  The solar cooker might not have been insulated properly.
  •  The cooker utensil used for cooking might not have been blackened.

(b) • Gaining practical knowledge of what she have learned in theory. She wanted to adopt the eco-friendly life style.

23. Aditya suggests his family to install a solar water heater at their residence. But
some of the family members were in favour of installing an electric geyser.
Ultimately the family got water heater installed.
(a) Which according to you was correct? Give two reasons in support of your answer.
(b) Mention two possible changes in the thinking of the family members
because of the arguments Given by Aditya.
answer. (a) The decision of installing the solar water heater was correct this is due to
(i) Their system use solar energy which is a renewable source of energy and free.
(ii)It will reduce our dependence on fossil fuel, improving on our energy security and reduce the country import bill for fuel.
(b) Two possible changes in the thinking of the family members because of the arguments given by Aditya are:
(i) We should conserve the non-renewable fuels for application for which renewable source of energy is not easily available.
(ii)Solar water heating is good investment and cost effective. It is one of the safe and efficient way to deliver hot water free of cost.

24. Traffic jams, outside the school gate was a common sight since most of the students came on their own cars. This became a topic for discussion on every P.T.A meeting. On one such P.T.A meeting, the principle pointed out the examples of four of their teachers who were carpooling for the past several
years. She asked the parents to adopt this method to sort out the problem.
(i) List two values shown by the teachers mentioned by the principal.
(i) Explain two advantages that will occur if more parents emulate the examples of these teachers.
answer.
(a) The two values shown by the teachers mentioned by the principle are:
(i) Eco-friendly life style.
(ii) Co-ordination and friendship.
(b) (i) Reduction in air pollution and traffic congestion: Car pooling reduces the emission of CO2 into the atmosphere as every car pooling participant takes another car off the road.
(ii) Save money: Car pooling saves money by using less fuel. It reduces the cost involved in repetitive or long distance journey.

25. On returning home, Neha, a IXth std. student noticed that her 6 years old brother Neresh, watching T.V in the afternoon with all the lights and fans ‘on’. She noticed that the windows were closed and curtains were drawn, which made the room dark, so, Naresh had put on the lights. She calmly opened the windows, drew the curtain aside, which illuminated and aerated the room. Then she made Naresh put “off” the lights and made him understand the reason behind her action. (Assume that they are getting electric supply from Thermal Power Plant)
(a) List the two value exhibited by Neha.
(b) Explain how she tried to give same values to her brother.
answer.(a) The values exhibited by Neha are:
(i) Responsible citizen in the conservation of energy resources.
(ii) Adopt fuel and money saving technique in her life.
(b) Neha tried to give the same values to her brother Naresh by opening the windows, draw the curtain aside, which illuminated and aerated the room. This way she shows the practice of adopting renewable sources of energy like solar energy and conserve the fossil fuel like coal.

Very Short Answer Type Questions [1 Mark] – Year 2013

26. What is acid rain? 
answer. Acid rain: The rain containing the acidic oxides such as oxides of carbon, nitrogen and sulphur.

27.Write the name of the substance whose vapours are used to run the turbine of the generator of ocean thermal energy plant.
answer. Ammonia.

28. Mention the minimum temperature difference required between surface water and water at a depth of upto 2 km in an ocean thermal energy plant.
answer. 20 °C or 293 K in trapping geothermal energy

Short Answer Type Questions [II] [3 Marks] – Year 2013

29. Distinguish between renewable and non-renewable sources of energy. Also give an example of each of these sources.
answer.
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30. State the principle of working of ocean thermal energy conversion plant.
Explain how the plant works? Write one essential condition for it to operate
properly.
answer. Principle of working of OTEC: The water at the surface of the ocean is warmer
than the water at deeper depths. This temperature difference can be used by
Ocean Thermal Energy Conversion (OTEC) systems to generate electricity.
Working of OTEC:

  • In OTEC plant, the energy of warm surface water is used to convert low i boiling point liquid ammonia into gaseous state.
  •  The vapour of ammonia at high pressure is used to spin the turbines of generators converting the Ocean thermal energy to electricity.
  •  The used vapour pass through the condenser where cold water, pumped from the deeper parts of ocean condenses ammonia vapour back into a liquid.
  • This process is repeated again and again ,to get continuous production of electricity.
    Essential condition for it to operate properly: The temperature difference between the warmer water at the surface and colder water at depths up to 2 km should be 293 K (20°C) or more.

Very Short Answer Type Questions [1 Mark] – Year 2012

31.Name one fuel used in nuclear reactor.
answer. Uranium-235.

32.Name any two elements that are used in fabricating solar cells.
answer. Germanium, Silicon.

33.Name the reaction responsible for large energy production in the sun.
answer. Nuclear fusion.

Short Answer Type Questions [II] [3 Marks] – Year 2012

34.Mention why is it not possible to make use of solar cells to meet all our energy needs? State three reasons to support your answer. Also mention three uses of solar cells.
answer. It is not possible to make use of solar cells to meet all our energy needs because:
(i) of limited availability of special grade semiconducting materials such as silicon and germanium.
(ii)solar cells have lower efficiency as they depend entirely on intensity of solar radiation.
(iii) the process of manufacturing of solar cells is very expensive, silver used for interconnection of cells in the panel further adds to the cost.
Uses of solar cells:
(i) They provide electric power to satellites and space probes.
(ii) They provide electric power to off-shore drilling platforms and light houses.
(iii) TV relay stations or wireless transmission systems located in remote areas use solar panels to get electric power.

35.Out of two elements A and B with mass number 2 and 235 respectively, which one is suitable for making ,
(a) a nuclear reactor,
(b) a hydrogen bomb?
Name the nuclear reaction involved in each case. Write one difference between the two types of nuclear reactions.
answer.
(a) For a nuclear reactor— element ‘B’ with mass number 235 is suitable as a fuel in a nuclear reactor.
(b) For a hydrogen bomb— element ‘A’ with mass number 2 is suitable for making the hydrogen bomb.
The nuclear reaction involved in
(a) nuclear reactor is nuclear fission reaction and
(b) nuclear bomb is nuclear fusion reaction.
Difference between fission and fusion:
Fusion reaction releases much greater energy with non-radioactive products than the energy released in fission reaction with radioactive products.

Very Short Answer Type Questions [1 Mark] – Year 2011

36. List two nutrients that the slurry left behind in the biogas plant contain.
answer. Nitrogen and phosphorous.

37. Biogas is also known as gobar gas. Justify.
answer. Starting material for biogas is mainly cow dung. So, it is also known as gobar gas.

38. List two practical uses of biogas in rural areas.
answer. Practical uses of biogas in rural area are:

  1.  It is an excellent fuel which burns without smoke with high heating capacity.
  2.  It is also used for lighting.

Short Answer Type Questions[l] [2 Marks] – Year 2011

39. Bio gas is considered to be a boon to the farmers. Give reasons.
answer.

  1. It is the source of excellent manure, rich in nitrogen and phosphorous which
    can be obtained from the biogas plant in addition to biogas.
  2.  It provides the safe, efficient and profitable disposal method for bio-waste and sewage material.

40. What are the advantages of nuclear energy?
answer. Advantages of nuclear energy are:

  1.  A small quantity of nuclear fuel is needed to produce a large amount of useful energy.
  2.  Nuclear power plant produces less atmospheric pollution than thermal power plants, if the nuclear fission reaction is performed properly.
  3.  Small amount of nuclear fuel can run a nuclear power plant over a long period of time. There is no need of inserting the nuclear fuel in the nuclear reactor again and again in a short period as in case of thermal power plant.

41. What is a solar cell panel? Mention any three of its applications.
answer. A large number of solar cells combined in an arrangement to obtain large electrical power is called solar cell panel.
Applications of solar cell panel are:
It provides the electric power for the:

  1. working of artificial satellites stationed in outer space,
  2.  running of irrigation water pumps by the farmers in rural areas,
  3.  street lighting in remote areas.

42. Out of two solar cookers, one was covered with a plane glass slab and the other was left open. Which of the two solar cookers will be more efficient and why?
answer. The solar cooker which was covered with a plane glass slab would be more efficient. The glass lid allows the heat radiation from sun to enter the solar cooker but does not allow the reflected heat radiation to escape or go outside the box. Thus, heat trapped inside the box increases the temperature. Glass lid also reduces heat loss due to reflection.

Short Answer Type Questions [II] [3 Marks] Year 2011

43.List any three hazards of nuclear waste. How does the disposal of nuclear waste pose a problem for the plant and animal life?
answer. Hazards of nuclear wastes are:

  1.  Nuclear waste contains radioactive substances which emit harmful nuclear radiations.
  2. There is a high risk of environmental contamination.
  3. It is highly toxic.
    Effect of nuclear waste on plant and animal life is:
    The radiations emitted from the nuclear waste penetrate deep inside the human or animal body where they can damage biological cells thereby initiate cancer or causes genetic disease.
    Increased mortality of plants, soil invertebrates and mammals and reproductive losses in plants and animals have also been observed.

44. (a) Charcoal is a better fuel than wood. Why? .
(b) How does biogas plant help to reduce the problem of pollution?
answer.
(a) Charcoal is considered to be a better fuel than wood because:

  1.  It burns without flames.
  2. It is comparatively smokeless.
  3. It has higher calorific value, i.e. higher heat generating efficiency than wood.

(b) Biogas plant helps to reduce the problem of pollution in the following ways.

  1.  It provides better sanitation due to safe disposal of bio-waste and sewage material.
  2. Biogas obtained from this plant produces less smoke on burning. (Hi) The residue left can be used as a manure which can be used as an alternative
    of fertilizers. Thus, it prevents soil and water from degradation.

45. (a) What is geothermal energy?
(b) What are the advantages of wind energy?
answer.
(a) The heat energy obtained from the molten rocks formed in the deeper hot regions inside the earth are called geothermal energy.
(b) Advantages of wind energy are:

  1.  It is an inexhaustible source of energy.
  2. It does not cause any environmental pollution.
  3.  It is available at free of cost.

46. (a) How does construction of dams across the river get linked with production of greenhouse gases?
(b) How do technological inputs improve the efficiency of biomass fuels?
answer.
 (a) A vast variety of plants get submerged in water, rot under anaerobic conditions and produce large amount of greenhouse gases such as methane.
(b) Traditional uses of biomass fuels are not only efficient but they also produce a lot of pollutants which are hazardous to health. Therefore, technological inputs are necessary to improve the efficiency of these fuels and make them environment friendly. With the help of technology, smokeless chulhas and biogas plants have been designed.

Short Answer Type Questions [2 Marks] -Year 2010

47. Mention any four limitations in harnessing wind energy on a large scale.
answer. Limitations in harnessing wind energy are:

  1.  Speed of wind is not available at all time and at all places.
  2. To establish the wind energy farm, a large area of land is needed.
  3.  Speed of wind should be higher than 15 km/h to harness the wind energy.
  4. Construction of windmill and its installation is very expensive.

48. What happens when wood is burnt in a limited supply of oxygen? Name the residue left behind after the reaction and state two advantages of using this residue as a fuel over wood.
answer. When wood is burnt in a limited supply of oxygen, volatile materials present in it get removed and cooled to get wood tar and wood gas.
The black residue left behind aftqr the reaction is known as charcoal. Advantages of using charcoal as a fuel over wood are:

  1.  Burning of charcoal does not produce smoke. On the other hand, wood produces a lot of smoke on burning.
  2. For a given quantity, charcoal produces ,more heat than wood.

49. Name four gases commonly present in biogas. State two advantages of using this gas over fossil fuels.
answer.  Methane, carbon dioxide, hydrogen and hydrogen sulphide.
Advantages of using biogas over fossil fuels are:

  1.  Biogas burns without smoke, leaves no residue unlike coal.
  2. Biogas is cheaper as compared to fossil fuels.

50. How are the wastes produced in nuclear power plants different from those produced in a thermal power plants ? What happens to the waste of a nuclear power plant?
answer.The waste obtained from nuclear power plants is highly radioactive in nature which emits harmful radiations, whereas waste produced in a thermal power plant is non-radioactive.
Management of nuclear waste is given as follows:

  1.  Some products of nuclear waste are buried in sealed steel/lead containers for a long term storage, buried under the ground or dumped in vacated coal mines.
  2.  Other waste products transforms into less harmful products or to products with a shorter half life.

Short Answer Type Questions [2 Marks] -Year 2009

51. In a solar cooker, the following arrangements are made. Write one function of each arrangement.
(a) The box is made of insulating material such as plastic or wood.
(b) The inner walls of the box are painted black.
(c) The box is covered with a transparent glass sheet.
(d) A plane mirror is hinged at an angle at the top of the box.
answer. 
(a) To avoid loss of heat from solar cooker to the surroundings.
(b) Black surface absorbs more heat radiations of incident energy.
(c) Transparent glass sheet does not allow the reflected heat radiation to go outside the box.
(d) To increase the amount of solar energy incident on the transparent glass sheet.

52. Describe how hydro energy can be converted into electrical energy. Write any
two limitations of hydro energy.
answer. Conversion of hydro energy into electrical energy

  • High rise dams are constructed on the river to obstruct the flow of water to collect it at a suitable height. The stored water has a lot of potential energy.
  • The water from a suitable height is allowed to fall on the blades of a turbine located at the bottom of a dam through a pipe.
  • Kinetic energy of flowing water rotates the turbine rapidly. Rotation of turbine helps the armature coil of generator to rotate rapidly in the magnetic field. Thus, hydroelectricity is generated.
    Limitations of hydro energy:
    (i) All river-sites are not suitable for construction of dams.
    (ii) Large ecosystems are destroyed when submerged under the water in dam.

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Carbon and its Compounds Chapter Wise Important Questions Class 10 Science

Carbon and its Compounds Chapter Wise Important Questions Class 10 Science

Important QuestionsNCERT SolutionsSample PapersNCERT Exemplar Solutions

VERY SHORT ANSWER TYPE QUESTION [1 Mark] – Year 2007

Question.1 Give the names of the following functional groups:
(i) —OH (ii) —COOH
Ansswer. (i) Alcohol group (ii) Carboxylic acid group

VERY SHORT ANSWER TYPE QUESTION [1 Mark] – Year 2008

Question.2 What is the difference in the molecular formula of any two consecutive members of a homologous series of organic compounds?
Answer. —CH2— is the difference in the molecular formula of any two consecutive members of a homologous series of organic compounds.

SHORT ANSWER TYPE QUESTIONS[I] [2 Marks] – Year 2008

Question.3
(a) Give a chemical test to distinguish between saturated and unsaturated  hydrocarbons.
(b) (i) Name the products formed when ethanol burns in air. ‘
(ii) What two forms of energy are liberated on burning alcohol?
(c) Why is the reaction between methane and chlorine considered a substitution
reaction?
Answer.
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Question.4
(a) Why are covalent compounds generally poor conductors of electricity?
(b) Name the following compound:
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(c) Name the gas evolved when ethanoic acid is added to sodium carbonate. How would you prove the presence of this gas?
Answer.
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VERY SHORT ANSWER TYPE QUESTIONS [1 Mark] – Year 2009

Question.5 Name the carbon compound which on heating with excess of concentrated sulphuric acid at 443 K gives ethene.
Answer.
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Question.6 What is meant by a sturated hydrocarbon?
Answer. Those hydrocarbons in w hich valency of carbon is satisfied by single bonds only are called sturated hydrocarbons.

Question.7 Name the compound formed when ethanol is warmed with ethanoic acid in the
presence of a few drops of cone.H2S04
Answer.
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Question.8 Draw the structure of CH3COOH molecule.
Answer.
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Question.9 Draw the structure of ethanol molecule.
Answer.
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Question.10 What happens when a small piece of sodium is dropped into ethanol? 
Answer. Hydrogen gas will be evolved.
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SHORT ANSWER TYPE QUESTIONS[II] [3 Marks] – Year 2009

Question.11 What is meant by a functional group in an organic compound? Name the functional group present in
(i) CH3CH2OH
(ii) CH3COOH
(b) State one point of difference between soap and synthetic detergent.
Answer.
(a) Functional group is an atom or group of atoms or reactive part of compound, which determines chemical properties of compounds.
(i) —OH (Alcohol)
(ii) —COOH (Carboxylic acid)
(b) Soaps do not work well with hard water, detergents work well with hard water.

Question.12 Give reasons for the following observations:
(a) The element carbon forms a very large number of compounds.
(b) Air holes of a gas burner have to be adjusted when the heated vessels get blackened by the flame.
(c) Use of synthetic detergents causes pollution of water.
Answer.
(a) Carbon forms large number of compounds since carbon is small in size and can form stable covalent bonds (catenation) and it shows tetravalency.
(b) Air holes of gas burner are made open (adjusted) so that air can pass through, which is needed for complete combustion, so that heated vessels do not get blackened.
(c) Some synthetic detergents are non-biodegradable, therefore, cause pollution of water.

Question.13 What is ethanoic acid? Write the formula of the functional group present in this acid.
What special name is given to its 5 – 8% solution in water? How does ethanoic acid react with sodium carbonate? Write a chemical equation of the reaction and common name of the salt produced.
Answer.
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VERY SHORT ANSWER TYPE QUESTIONS[1 Mark] – Year 2010

Question.14 State two characteristic features of carbon which when put together give rise to large
number of carbon compounds.
Answer. (i) Catenation (ii) Tetravalency of carbon

Question.15 Write the structural formula of chloroethane.
Answer.
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SHORT ANSWER TYPE QUESTIONS[I] [2 Marks] – Year 2010

Question.16 Write the name and molecular formula of an organic compound having its name suffixed with ‘-ol and having two carbon atoms in the molecule. With the help of a
balanced chemical equation indicate what happens when it is heated with excess of r cone.H2S04.
Answer.
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Question.17 Write the names and molecular formula of two organic compounds having functional r group suffixed as ‘-oic acid’. With the help of a balanced chemical equation and explain what happens when any one of them reacts with sodium hydroxide.
Answer.
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Question.18 What is a homologous series? Which two of the following organic compounds belong
to the same homologous? 
CH3 ,C2H6, C2H6O, C2H6O2,CH4O     
Answer.      
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SHORT ANSWER TYPE QUEST10NS[II] [3 Marks] – Year 2010

Question.19 Name the functional group of organic compounds that can be hydrogenated. With the help of suitable example explain the process of hydrogenation mentioning the conditions of the reaction and any one change in physical property with the formation of the product. Name any one natural source of organic compounds that are hydrogenated.
Answer.
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When unsaturated hydrocarbons are heated with hydrogen in the presence of nickel as catalyst, saturated hydrocarbons are formed. If the starting unsaturated hydrocarbons are liquids, they will change into solids. Vegetable oils are hydrogenated to form vegetable ghee. Plants are natural sources of vegetable oils which can be hydrogenated.

Question.20 An ester has the molecular formula C4H8O2. Write its structural formula. What happens when this ester is heated in the presence of sodium hydroxide solution? Write the balanced chemical equation for the reaction and name the products. What is a saponification reaction?
Answer.
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Question.21 An organic compound ‘A’ is an essential constituent of wine and beer. Oxidation of ‘A’ yields an organic acid ‘B’ which is present in vinegar. Name the compounds ‘A’ and ‘B’ and write their structural formula. What happens when ‘A’ and ‘B’ react in the presence of an acid catalyst? Write the chemical equation for the reaction.
Answer.
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Question.22 What is ethanol? State its two properties. What happens when it is heated with excess of cone. H2SO4 at 443 K? What role does cone. H2SO4 play in this reaction? Write chemical equation of the reaction involved and the structural formula of the main product formed.
Answer.
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Question.23 With the help of balanced chemical equations explain what happens when ethanol is heated with (i) alkaline solution of potassium permanganate, (ii) excess concentrated sulphuric acid at 443 K. Mention any two uses of ethanol. 
Answer.
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VERY SHORT ANSWER TYPE QUESTIONS [1 Mark] – Year 2011

Question.24 How many covalent bonds are there in a molecule of ethane (C2H6)?
Answer. There are 7 covalent bonds in a molecule of ethane.

Question.25 Write the electron dot structure of ethene molecule (C2H4).
Answer.
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Question.26 Write the electron dot structure of ethane molecule (C2H6).
Answer.
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Question.27 Draw the structure of butanone molecule, CH3COC2H5.
Answer.
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Question.28 Draw the structure of the hexanal molecule, C5H11CHO.
Answer.
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Question.29 Butanone is a four carbon per molecule compound. Name the functional group present in it.
Answer. Ketone

Question.30 Name the following compound:
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Answer. 1-Hexyne is IUPAC name of the compound.

SHORT ANSWER TYPE QUESTION[II] [3 Marks] – Year 2011

Question.31 What is an ‘esterification’ reaction? Describe an activity to show esterification.
Answer.
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Question.32 Out of HCI and CH3COOH, which one is a weak acid and why? Describe an activity to
support your answer.
Answer. Acetic acid ( CH3COOH) is a weaker acid because it does not dissociate completely into its ions in aqueous solution. .
Activity: Add zinc metal in HCI and CH3COOH respectively. The hydrogen gas will be evolved faster in HCI and slowly in CH3COOH. It shows acetic acid is a weak acid.
Alternative Method:
If we use pH paper, the colour of pH paper will be dark red in HCI and light red in CH3COOH which shows HCI is a strong acid and CH3COOH is a weak acid.

Question.33Write chemical equations for what happens when
(i) sodium metal is added to ethanoic acid.
(ii) solid sodium carbonate is added to ethanoic acid.
(iii) ethanoic acid reacts with a dilute solution of sodium hydroxide.
Answer.
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Question.34 Describe two examples of different oxidations of ethanol. Name the products obtained in each case.
Answer.
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Question.35  Write a chemical equation in each case to represent the following types of chemical reactions of organic compounds:
(i) Oxidation reactions
(ii) Addition reactions
(iii) Substitution reactions
Answer.
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LONG ANSWER TYPE QUESTIONS [5 Marks] – Year 2011

Question.36 (a) In tabular form, differentiate between ethanol and ethanoic acid under the following heads:
(i) Physical state (ii) Taste
(iii) NaHCO3 test (iV) Ester test
(b) Write a chemical reaction to show the dehydration of ethanol.
Answer. 
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Question.37 (a) State two properties of carbon which lead to a very large number of carbon compounds.
(b) Why does micelle formation take place when soap is added to water? Why are micelles not formed when soap is added to ethanol?
Answer.
(a) (i)-Catenation (ii) Tetravalency
(b) It is because large number of molecular ions of soaps get aggregated and form colloidal solution. Soap has hydrophobic tail (hydrocarbon) which dissolves in hydrocarbon part and hydrophilic part dissolves in water. Ethanol is non-polar solvent therefore micelles are not formed because hydrocarbon part gets attracted towards ethanol and ionic end will not dissolve in alcohol.

Question.38 Explain isomerism. State any four characteristics of isomers. Draw the structures of possible isomers of butane, C4H10
Answer. Isomerism is a phenomenon due to which some compounds have same molecular formula but different structural formulae.
Characteristics:
(i) They differ in structural formula.
(ii) They differ in melting point.
(iii) They differ in boiling point.
(iv) They differ in solubility in same solvent.
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Question.39 Give reasons for the following:
(i)Element carbon forms compounds mainly by covalent bonding.
(ii)Diamond has a high melting point.
(iii)Graphite is a good conductor of electricity.
(iv)Acetylene bums with a sooty flame.
(v)Kerosene does not decolourise bromine water while cooking oils do.
Answer.
(i) It is because carbon has four valence electrons, it cannot gain or lose four electrons because high energy is needed. It can only share four electrons.
(ii) It is due to strong covalent bonds and compact structure of diamond.
(iii) It is due to presence of free electrons in graphite because each carbon is linked to three more carbon atoms.
(iv) It is due to high percentage of carbon, it burns with sooty or smoky flame.
(v) Kerosene oil is mixture of saturated hydrocarbons therefore does not decolourise bromine water.

VERY SHORT ANSWER TYPE QUESTIONS [1 Mark] – Year 2012

Question.40 Name the functional group present in each of the following organic compounds:
(i) C2H5CI
(ii) C2H5OH
Answer.
(i) (—Cl) Halogen (Chloro)
(ii) (—OH) Alcohol

Question.41 Name the functional group present in each of the following compounds:
(i) HCOOH
(ii) C2H5CHO
Answer.
(i) —COOH (Carboxylic acid)
(ii) —CHO (Aldehyde)

Question.42 Name the functional group present in each of the following organic compounds:
(i)CH3COCH3
(ii) C2H5COOH
Answer.
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Question.43 Write the name and formula of the second member of the carbon compounds having functional group —OH.
Answer.
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Question.44 Write the name and formula of the first member of the carbon compounds having functional group —CHO.
Answer.
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Question.45 Write the name and formula of the first member of the carbon compounds having functional group —COOH.
Answer.
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Question.46 Write the name and formula of the 2nd member of the series of carbon compounds whose general formula is CnH2n+1OH
Answer. Ethanol, C2H5OH or CH3CH2OH

Question.47 Write the name and formula of the 2nd member of the series of carbon compounds whose general formula is CnH2n.
Answer.
C3H6, H2C=CH—CH3
Propene is second member of series whose general formula is CnH2n.

SHORT ANSWER TYPE QUESTIONS[II] [3 Marks] – Year 2012

Question.48 What are isomers? Draw the structures of two isomers of butane, C4H10. Why can’t we have isomers of first three members of alkane series?
Answer. Those compounds, which have same molecular formula but different structural formulae are called isomers.
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In first three members of alkane series, branching is not possible. Therefore, we cannot have isomers.

Question.49 Define homologous series of organic compounds. List its two characteristics. Write the name and formula of the first member of the series of alkenes.
Answer. The series of organic compounds having same functional group and similar chemical properties is called homologous series.
Each member differs from successive member by —CH2— group. The difference in molecular weight between two successive members is 14 u.
Characteristics:
(i) It has same general formula, from which, all members can be derived.
(ii) They have similar chemical properties.
C2H4, CH2=CH2, Ethene is first member of alkene series.

Question.50 Complete the following equations:
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Answer.
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Question.51 Why homologous series of carbon compounds are so called? Write chemical formula of two consecutive members of a homologous series and state the part of these compounds that determines their
(i) physical properties, and (ii) chemical properties.
Answer. The series consists of members of same family with similar physical and chemical properties, therefore, called homologous series
(i) CH3OH, and (ii) CH3CH2OH are two consecutive members of homologous series.
Alkyl group —CH3 and —CH3CH2 part determines physical properties. Functional group —OH determines chemical properties of the compounds.

Question.52 Name the oxidising agent used for the conversion of ethanol to ethanoic acid. Distinguish between ethanol and ethanoic acid on the basis of (i) litmus test, (ii) reaction with sodium hydrogencarbonate.
Answer. Alkaline potassium permanganate or Acidified potassium dichromate.
(i)Ethanol will not affect litmus paper. Ethanoic acid will turn blue litmus ‘ paper red.
(ii) Ethanol will not react with sodium hydrogen carbonate. Ethanoic acid will give brisk effervescence due to colourless, odourless carbon dioxide gas.

Question.53 Distinguish between esterification and saponification reactions of organic compounds with the help of the chemical equation for each. What is the use of (i) esters and (ii) saponification process?
Answer.
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LONG ANSWER TYPE QUESTIONS [5 Marks] – Year 2012

Question.54 What is the difference between the chemical composition of soaps and detergents? State in brief the action of soaps in removing an oily spot from a shirt. Why are soaps not considered suitable for washing where water is hard?
Answer. oaps are sodium or potassium salts of fatty acids having — COONa group. Detergents are
sodium or potassium salts of sulphonic acids having — SO3Na and — SO4Na group. Cleansing action of soap: Soap molecules consist of a large hydrocarbon tail which is hydrophobic (water-hating or water repelling) with a negatively charged head which is hydrophilic (water-loving) as shown in figure.
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When a soap is dissolved in water, the molecules associate together as clusters called micelles in which water molecules, being polar in nature, surround the ions and the hydrocarbon part of the molecule attracts grease, oil and dirt.
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Question.55 List in tabular form three physical and two chemical properties on the basis of which ethanol and ethanoic acid can be differentiated
Answer.
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Question.56 What are the hydrocorbons write the name and general formula of (i) sturated hydrocarbons, (ii) unsaturated hydrocarbons, and draw the structure of one hydrocarbon of each type. How can an unsaturated hydrocarbon be made saturated?
Answer.
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Question.57 What are detergents chemically? List two merits and two demerits of using detergents for cleansing. State the reason for the suitability of detergents for washing, even in the case of water having calcium and magnesium ions.
Answer. Detergents chemically are sodium or potassium salts of sulphonic acid of benzene or alkene.
Merits:
(i) They work well with hard water.
(ii) They are more effective than soaps.
Demerits:
(i) They are expensive.
(ii) Some of them having branching are non-biodegradable, therefore create water pollution.
Detergents are suitable for hard water having  Mg2+ and Ca2+ ions because they do not form insoluble salts with Mg2+ and Ca2+ ions.

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Heredity and Evolution Chapter Wise Important Questions Class 10 Science

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Heredity  and Evolution Chapter Wise Important Questions Class 10 Science

VERY SHORT ANSWER TYPE QUESTION [1 Mark] -Year 2007

1. What is a gene? [Delhi]
Answer. Gene is the unit of inheritance. Gene is the part of a chromosome which controls the appearance of a set of hereditary characteristics.

SHORT ANSWER TYPE QUESTIONS [I] [2 Marks] -Year 2007

2. What is meant by analogous organs? Taking a suitable example, explain how they support the theory of Organic Evolution. [Delhi]
Answer. Analogous organs are those organs which I have different basic structural design and developmental origin but have similar appearance and perform similar functions.
Example: The wings of birds and bats look similar but have different design in their structure. They have a common function of flying but their origins are not common. So, birds and bats are not closely related.

3. What is a sex chromosome? [Foreign]
Answer. Sex chromosome is a chromosome that operates in the sex-determining mechanism of a species. Many animals have two different types of sex chromosomes. For example, in human there is a large X chromosome and a much smaller Y chromosome.

SHORT ANSWER TYPE QUESTIONS [II] [3 Marks] -Year 2007

4.Define ‘evolution’. Describe Darwin’s theory of evolution. [All India]
Answer. Evolution is the sequence of gradual changes which take place in the primitive organisms over millions of years and new species are produced. Since, the evolution is of the living organisms, so it is called ‘Organic Evolution’.
Darwin’s theory of Evolution: Charles Robert Darwin gave the theory of evolution in his famous book, ‘The Origin of Species’. The theory of evolution proposed by Darwin is known as ‘The Theory of Natural Selection’. It is also called ‘Darwinism’.
According to Darwin’s theory of evolution:

  1. There is natural variation within any population and some individuals have more favourable variations than others.
  2. Population remains fairly constant even though all species produce a large number of off springs.
  3. This is due to ‘competition’ or struggle for existence between same and different species.
  4. The struggle for survival within population eliminates the unfit individuals and those with ‘favourable variations’ survive and pass on these variations to their progeny to continue. This is called natural selection.
  5. The favourable variations are accumulated over a long time period leading to the origin of a new species.

SHORT ANSWER TYPE QUESTIONS [I] [2 Marks] -Year 2008

5.”The sex of the children is determined by what they inherit from their father and not their mother.” Justify. [Delhi(C)]
Answer. It is because a child who inherits an X chromosome from her father will be a girl and one who inherits a Y chromosome from his father will be a boy. But all children inherit a X chromosome from their mother regardless of whether they are boys or girls.

6. Define variation in relation to a species. Why is variation beneficial to the species? [Delhi]
Answer. Variation refers to the differences in the characters or traits among the individuals of a species.
Variations are beneficial to the species because:

  1. They enable the organisms to adapt themselves in changing environment.
  2. Variations form the basis of heredity.
  3. They form the raw materials for evolution and development of new species.

7. Describe briefly four ways in which individuals with a particular trait may increase in a population. [Foreign]
Answer. The four ways in which individuals with a particular trait may increase in a population are as follows:

  1.  Sexual reproduction results into variations.
  2. The individuals with special traits survive the attack of their predators and multiply while the other will perish.
  3.  Genetic drift provides diversity without any adaptation.
  4. Variations in the species may lead to increased survival of the individuals.

8. What are fossils? What do they tell us about the process of evolution? [All India(C): All India 2011]
Answer. The remains of dead plants and animals which were buried under the rocks millions of years ago are called fossils.
Fossils tell us about the process of evolution. The fossils of different organisms have some features similar to one species while some features are similar to the other species. In this way, they show the link between two species. They tell us that one species evolves from the other.

SHORT ANSWER TYPE QUESTIONS [II] [3 Marks] -Year 2009

9. Distinguish between acquired and inherited traits by giving one example of each. Why are traits acquired during the lifetime of an individual not inherited? [Delhi(C)]
Answer. Acquired trait is a particular characteristic that is developed during the lifetime of an individual. Such characteristics are not genetically controlled and cannot be passed on to the next generation. Example: Loss of weight due to starvation.
Inherited trait is the transmission of particular characteristics from parents to their off springs, generation to generation. Such traits are genetically determined characteristic that distinguishes a person.
Example: Colour of skin.
The acquired traits cannot change the DNA of the germ cells. Therefore, the acquired traits cannot be inherited over generations during the lifetime of an individual.

10. The human beings who look so different from each other in terms of colour, size and looks are said to belong to the same species. Why? Justify your answer. [All India(C)]
Answer. Human beings are said to belong to the same species because of the following reasons:

  1. DNA studies.
  2. Number of chromosome is same,
  3. All have a common ancestor.
  4. They interbreed among themselves to produce fertile young ones of their own kind.

SHORT ANSWER TYPE QUESTIONS [I] [2 Marks] -Year 2010

11. Give one example each of characters that are inherited and the ones that are acquired in humans. Mention the difference between the inherited and the acquired characters. [Delhi]
Answer. Eye colour or hair colour of a person is an example of inherited character whereas, body weight is an example of acquired character.
The basic difference between inherited and acquired character is that inherited character is passed on from parent to offspring and acquired characters are acquired by an individual during his lifetime depending upon his lifestyle.

SHORT ANSWER TYPE QUESTIONS [II] [3 Marks] -Year 2010

12. Explain the mechanism of sex determination in humans. [All India]
Or
With the help of a flow chart explain in brief how the sex of a newborn is genetically determined in human beings. Which of the two parents, the mother or the father, is responsible for determination of sex of a child? [Foreign]
Answer. Mechanism of Sex Determination in Human Beings:
In human beings, the sex of the individual is genetically determined.

  • Sex determination is the process by which sex of a new born individual can be determined.
  •  Human beings have 1 unpaired sex chromosome. Sex chromosome of male is XY and of female is XX.
  •  Sex of a child depends on what happens at fertilisation.
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    Thus, father is responsible for the determination of the sex of a child.

13. With the help of suitable examples explain natural selection. [Foreign]
Answer. Let us assume that there exist a group of red beetles in some green bushes. Due to variation during sexual reproduction one green beetle evolved among them. This green beetle can pass the colour to its progeny which are green beetles.
Crows cannot see green beetles in green bushes and therefore, cannot eat them. The progeny of green beetles are not eaten while the progeny of red beetles continued to be eaten. As a result there are more and more green beetles in the beetle population. The progeny of green beetles increase due to natural selection which gives survival advantage.

SHORT ANSWER TYPE QUESTIONS [I] [2 Marks] -Year 2011

14. How is the equal genetic contribution of male and female parents ensured in the progeny? [Delhi]
Answer. During sexual reproduction, a female gamete or egg cell fuses with a male gamete or sperm cell which are haploid to form zygote. Zygote is diploid which contains 23 chromosomes from mother and 23 from father. In this way, an equal genetic contribution of male and female parents is ensured in the progeny.

15. What evidence do we have for the origin of life from inanimate matter? [All India]
Answer. Stanley L. Miller and Harold C. Urey provided evidence regarding origin of life from inanimate matter. They assembled an atmosphere similar to that existed on early earth. The atmosphere had molecules like ammonia, methane and hydrogen sulphide, but no oxygen and kept over water at temperature just below 100°C. Sparks were passed through the mixture of gases. At the end of a week, 15% carbon from methane had been converted to simple compounds of carbon like amino acids which make-up protein molecules. So, life arose afresh on earth.

SHORT ANSWER TYPE QUESTIONS [II] [3 Marks] -Year 2011

16. Explain the terms: (i) Speciation (ii) Natural selection [Delhi]
Answer. (i) Speciation is the evolution of reproductive isolation among once-interbreeding populations, i.e. the development of one or more species from an existing species.
(ii) Natural selection is the process, according to Darwin, which brings about the evolution of new species of animals and plants.

17. Explain with examples how the following are evidences in favour of evolution in organisms. (i) Homologous organs (ii) Analogous organs (iii) Fossils [Delhi]
Answer. (i) Forelimb of human and bird are homologous organs. They have same structural design and developmental origin but they have different functions and appearance. Homologous organs help us to understand that the organism has evolved from a common ancestor. The more common characteristics the two species have, the more closely they are related.
(ii) Analogous organs are those organs design and developmental origin but have similar appearance and perform similar functions.
Example: The wings of birds and bats look similar but have different design in their structure. They have a common function of flying but their origins are not common. So, birds and bats are not closely related.
(iii) Fossils and their study is useful in knowing about the species which are no longer alive. They provide evidence and missing links between two classes. They are helpful in forming a sequence of organisms in the pathway of evolution. Thus, fossils have an importance in deciding evolutionary relationship. Archaeopteryx is a fossil bird. It had feathers, fused bones and beak which are exclusively bird structures. It also had some features which are found in reptiles, e.g. teeth in jaw, claws on free fingers and a long tail. This fossil provides a clue that birds have evolved from reptiles.

18. Give an example of body characteristics used to determine how close two species are in terms of evolution and explain it. [All India]
Answer. Homologous organs help to identify the relationship between organisms. These characteristics in different organisms would be similar because they have inherited from a common ancestor. For example, forelimbs of human and wings of birds show closeness between the two species because the organs have similar basic structural design of limbs though it has been modified to perform different functions.

19. What are homologous organs? Can the wing of a butterfly and the wing of a bat be regarded as homologous? Why? [All India]
Answer. Homologous organs are those organs which have the same basic structural design and developmental origin but have different functions and appearance.
Example: The forelimb of a frog, a lizard, a bird and a man seem to be built from the same basic design of bones, but they perform different functions.
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No, the wing of a butterfly and the wing of a bat cannot be considered homologous organs because they have a common function for flying but their origin and structure are not common. So, they are analogous organs.

SHORT ANSWER TYPE QUESTIONS [II] [3 Marks] -Year 2012

20. What is meant by the term speciation? List four factors which could lead to speciation. [Delhi]
Answer. Speciation is the evolution of reproductive isolation among once interbreeding population.
Factors which can lead to speciation are:

  1. Genetic drift: Over generations, genetic drift may accumulate which leads to speciation.
  2. Natural selection: Natural selection may work differently in different location which may give rise to speciation.
  3.  Severe DNA change.
  4. A variation may occur which does not allow sexual act between two groups.

21. A blue colour flower plant denoted by BB is cross bred with that of white colour flower plant denoted by bb.
(a) State the colour of flower you would expect in their F generation plants.
(b) What must be the percentage of white flower plants in F generation if flowers of Fplants are self-pollinated?
(c) State the expected ratio of the genotypes BB and Bb in the Fprogeny. [Delhi]
Answer.
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22. Distinguish between homologous organs and analogous organs. In which category ; would you place wings of a bird and wings of a bat? Justify your answer giving a suitable reason.[Delhi]
Answer.
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Wings of a bird and wings of a bat are analogous organs as they have different basic structural design but have similar appearance and perform similar functions.

23. Define the term ‘evolution’. “Evolution cannot be equated with progress”. Justify this statement. [Delhi]
Answer. Evolution is the sequence of gradual change which takes place in primitive organisms over millions of years and new organisms are formed. Evolution cannot be equated to progress from lower forms to higher forms. It seems to have given rise to more complex body designs even while the simpler body designs continue to flourish. For example, human beings have not evolved from chimpanzees, but both have common ancestor.

24. If we cross pure-bred tall (dominant) pea plant with pure-bred dwarf (recessive) pea plant we will get pea plants of F1 generation. If we now self-cross the pea plant of F generation, then we obtain pea plants of F generation.
(a) What do the plants of F2 generation look like?
(b) State the ratio of tall plants to dwarf plants in F2 generation.
(c) State the type of plants not found in F2 generation but appeared in F2 generation, mentioning the reason for the same. [All India]
Answer. (a) All plants of F1  generation will be tall plants.
(b) 3:1
(c) Dwarf trait is recessive trait which was not expressed in the F1  generation, the recessive trait gets expressed in the F2 generation after self pollination.

25. How are fossils formed? Describe, in brief, two methods of determining the age of fossils. [All India]
Answer. When organisms die, their bodies decompose due to action of micro organisms. However, sometime the body or at least some parts of the body may be in such an environment that does not let it decompose completely. All such preserved traces of living organisms are called fossils.
The age of fossils can be estimated by the following two methods:

  1.  If we dig into the earth and start finding fossils, it can be assumed that the fossils closer to the surface are more recent to those found in deeper layers.
  2. By detecting the ratios of different isotopes of the same element in the fossil material.

26. State the meaning of inherited traits and acquired traits. Which of the two is not passed on to the next generation? Explain with the help of an example.
Answer. Inherited traits are the characteristics transmitted from parents to their offspring.Acquired traits are characteristics which are developed during the lifetime of an individual.Acquired traits are not passed on to the next generation. For example, if we breed a group of mice, all their progeny will have tails. Now, if the tails of these mice are removed by surgery and allowed to breed, the next generation mice will also have tails. If these tails are also removed and allowed to breed, the progeny of mice will again have tails. Removal of tail by surgery is an acquired trait and do not change the genes of germ cells and hence, are not passed on to the next generation.

27.”An individual cannot pass on to its progeny the experiences of its lifetime.” Justify the statement with the help of an example and also give reason for the same. [Foreign]
Answer. Experience achieved during the lifetime of an individual does not make any change in the gene of the individual.
For example, if a person reads a book on birds, the knowledge he earns by reading the book does not make any change in the gene, hence, this knowledge will not get automatically transmitted to his next generation. Such a trait is called acquired trait.

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Human Eye and Colourful World Chapter Wise Important Questions Class 10 Science

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Human Eye and Colourful World Chapter Wise Important Questions Class 10 Science

LONG ANSWER TYPE QUESTION [5 Marks] – Year 2007

1. A 14-year old student is not able to see clearly the questions written on the blackboard placed at a distance of 5 m from him.
(a) Name the defect of vision he is suffering from.
(b) With the help of labelled ray diagrams show how this defect can be corrected.
(c) Name the type of lens used to correct this defect. [All India]
Answer.
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VERY SHORT ANSWER TYPE QUESTIONS [1 Mark] – Year 2008

2. Why is red colour selected for danger signal lights? [Delhi]
Answer. Wavelength of red colour is more and so, it is least scattered. It can be easily seen through a large distance.

3. Study the diagram given below and answer the questions that follow it:
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(a) Which defect of vision is represented in this case? Give reason for your answer.
(b) What could be the two causes of this defect?
(c) With the help of a diagram show how this defect can be corrected by the use of a suitable lens. [Delhi(C)]
Answer. (a) Hypermetropia as the image is formed beyond the retina.
(b) (i) Due to greater focal length of the , lens and
(ii) As eyeball becomes smaller.
(c) It can be corrected by using a convex lens of suitable focal length as shown below.
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SHORT ANSWER TYPE QUESTIONS [I] [2 Marks] – Year 2009

4. Student sitting at the back bench in a class is not able to see what is written on the blackboard. He however, sees it clearly when sitting on the front seat at an approximate distance of 1.5 m from the blackboard. Draw ray diagrams to illustrate the image formation of the blackboard when he is seated at the (i) back seat (ii) front seat. [All India (C)]
Answer. (i) When student is seated at the back seat.
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(ii) When student is seated at front seat.
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5. What is meant by spectrum of white light? How can we recombine the components of white light after a prism has separated them? Draw a diagram to illustrate it. [Foreign, All India (C)]
Answer. The coloured pattern VIBGYOR formed by a prism by splitting the incident white light is called a spectrum. By having two prisms, inverted to each other, one can recombine the light to get white light again.
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6. Explain why do the planets not twinkle but the stars twinkle. [Delhi (C); Delhi 2011]
Answer. Planets being of larger size can be taken as a collection of large number of point-sized objects/sources of light, which nullify the twinkling effect of each other.
Due to varying conditions of atmosphere, starlight undergoes multiple refraction and its path varying slightly while passing through the atmosphere. Therefore, the apparent position of star fluctuates and amount of light entering the eye changing continuously. The star sometimes appear brighter and some other time, it appears fainter. This causes twinkling of star.

LONG ANSWER TYPE QUESTIONS [5 Marks] – Year 2009

7. (a) What is dispersion of white light? What is the cause of such dispersion? Draw a diagram to show the dispersion of white light by a glass prism.
(b) A glass prism is able to produce a spectrum when white light passes through it but a glass slab does not produce any spectrum. Explain why is it so? [All India]
Answer. (a) The splitting up of white light into its constituent colours is called dispersion. The colour sequence is given by the acronym V I B G Y O R – Violet, Indigo, Blue, Green, Yellow, Orange and Red. This colour pattern is called a spectrum.
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Dispersion takes place because the speed of light of different colours through a glass prism is different and so, refractive index, therefore, each colour bends (refracts) through different angles with respect to incident ray as they pass through a prism. The red colour has maximum speed in glass prism. So, it is least deviated while the violet colour has minimum speed so its deviation is maximum. Thus, the ray of each colour emerges along different paths and becomes distinct.
(b) For dispersion, the two refracting surfaces must be inclined to each other as in case of prism. In rectangular glass slab, the refracting surfaces are parallel to each other. So, dispersion cannot occur. This is due to fact that the rectangular glass slab can be considered as equivalent of two identical prisms in inverted position placed in an inverted position with respect to each other. The deviation and dispersion produced by the second inverted prism is equal and opposite to that produced by the first prism.
Therefore, there will neither be dispersion, nor deviation, Le. second inverted prism recombines the , colour to give a white light parallel to the incident ray again and will undergo only lateral displacement. Hence, rectangular glass slab cannot produce any spectrum.

8. (a) Explain the following terms used in relation to defects in vision and correction provided by them:
(i) Myopia (ii) Astigmatism (iii) Bifocal lenses (iv) Far-sightedness.
(b) Why is the normal eye unable to focus on an object placed within 10 cm from the eye? [All India]
Answer. (a) (i) Myopia: Short-sightedness is caused due to excessive curvature in cornea or elongation of eyeball. Image is formed before of the retina. So, a concave lens is used to correct it.
(ii) Astigmatism: The inability to focus the light in both vertical and horizontal lines is called astigmatism. It is caused due to varying curvature in lens both horizontally and vertically. It is . corrected by using cylindrical lens.
(iii) Bifocal lenses: These are used to correct presbyopic eye. These contain lenses with upper concave and lower convex surface. Presbyopia arises with age.
(iv) Far-sightedness: Hypermetropia or far-sightedness is caused due to greater focal length of eye lens and/or when eyeball becomes smaller. Image is formed beyond the retina and can be corrected using a convex lens.
(b) The focal length of the lens cannot be changed up to an extent that objects nearer than 25 cm can be viewed (its near point is 25 cm). So it is not able to focus for 10 cm.

VERY SHORT ANSWER TYPE QUESTIONS [1 Mark] – Year 2010

9. Name the part of our eyes that helps us to focus near and distant objects in quick succession. [Delhi]
Answer. Ciliary muscles help in changing the focal length of the eye lens.

10. A person is advised to wear spectacles with concave lenses. What type of defect of vision is he suffering from? [All India]
Answer. Myopia or short-sightedness.

11. A person is advised to wear spectacles with convex lenses. What type of defect of vision is he suffering from? [All India]
Answer. Hypermetropia or far-sightedness.

12. Why do different components of white light deviate through different angles when passing through a triangular glass prism? [All India]
Answer. Due to change in refractive index offered by the medium.

13. A person can comfortably read a book but finds it difficult to read the number on a bus parked 5 m away from him. Name the type of defect of vision he is suffering from. Which type of lens should he use in his spectacles to correct his vision? [Foreign]
Answer. Myopia or short-sightedness. Concave lens should be used to correct his vision.

14. What will be the colour of the sky, when it is observed from a place in the absence of any atmosphere? Why? [All India; Delhi 2012}
Answer. Sky appears dark.
Reason: In the absence of atmosphere, there would have been no scattering of sunlight at all.

15. The sky appears dark instead of blue to an astronaut. State its reason. [Delhi; Delhi 2012]
Answer. The sky appears dark to the astronaut as scattering does not take place at very high altitude due to the absence of atmosphere.

SHORT ANSWER TYPE QUESTIONS [I] [2 Marks] – Year 2010

16. What is myopia (near-sightedness)? Draw a ray diagram to show how it can be corrected using a lens. [Delhi]
Answer. Myopia is the inability of an eye in viewing long distant objects. The image in this case is formed before the retina. For every myopic eye, there exists a far point beyond which clear image cannot be seen.
The short-sightedness is corrected by using a concave lens which diverges and shifts the image to the retina.
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17.What is hypermetropia (far-sighted-ness)? Draw a ray diagram to show how this defect can be corrected using a lens. [Delhi]
Answer. Hypermetropia is the inability of an eye in viewing the nearby objects. The image in this case is formed beyond the retina. For a hypermetropic eye, there exists a near point.
It is corrected by using a convex lens, which converges and shifts the image to the retina.
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18. Define the term dispersion of white light. State the colour which bends (i) the least and (ii) the most while passing through a glass prism. [All India; All India 2012]
Answer. The splitting up of white light into its component colours is called dispersion. The colour which bends (i) the least is red, and (ii) the most is violet, while passing / through a glass prism.

VERY SHORT ANSWER TYPE QUESTIONS [1 Mark] – Year 2011

19. What is Tyndall effect? [Delhi]
Answer. The phenomenon of scattering of light by the colloidal particles is called Tyndall effect.

20. Give an example of optical phenomena which occurs in nature due to atmospheric refraction. [Delhi]
Answer. Twinkling of stars.

21. Give an example of a phenomenon where Tyndall effect can be observed. [All India]
Answer. When a fine beam of sunlight enters a room containing suspended particles of dust, the path of the beam of light is visible. It is due to the scattering of light (Tyndall effect).

22. Name the type of particles which acts as a prism in the formation of rainbow in the sky. [All India]
Answer. Water droplets present in the atmosphere.

23. What is the cause of dispersion of white light on passing through a prism? [Foreign]
Answer. The refractive index of the material of a prism is different for different colours of light as different colours have different speeds in the material of a prism. Also, prism has non-parallel surfaces.

24. Name the atmospheric phenomenon due to which the sun can be seen above the horizon about two minutes before actual sunrise. [Foreign]
Answer. Atmospheric refraction.

SHORT ANSWER TYPE QUESTIONS [I] [2 Mark] – Year 2011

25. Give reasons:
(i) The extent of deviation of a ray of light on passing through a glass prism depends on its colour.
(ii) Lights of red colour are used for danger signals. [Foreign]
Answer. (i) Refractive index of a medium is different for different colours of light.
(ii) Due to large wavelength, red colour is least scattered and travel to large distance.

26. A star appears slightly higher (above) than its actual position in the sky. Illustrate it with the help of a labelled diagram. [Delhi; All India 2012]
Answer. The gradual change in the refractive index of different layers of the atmosphere due to the varying conditions of it causes atmospheric refraction. When starlight enters the atmosphere, it gets refracted continuously. The higher level of air acts as a rarer medium while the dense air near the surface of earth acts as a denser medium. So, the atmosphere bends the starlight towards the normal. As a result, the apparent position of star is slightly different from its actual position. Thus, star appears slightly higher (above) than its actual position in the sky.
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27 Draw a ray diagram to show the refraction of light through a glass prism. Mark on it (a) the incident ray. (b) the emergent ray and (c) the angle of deviation. [All India]
Answer. i-incident angle, r – refraction angle, δ – angle of deviation, e – angle of emergence.
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SHORT ANSWER TYPE QUESTIONS [II] [3 Mark] – Year 2011

28. (a) What is meant by the power of accommodation of an eye?
(b) A person with a myopic eye cannot see objects beyond 1.2 m directly. What should be the type of the corrective lens used? What would be its power? [All India]
Answer. (a) The maximum variation in power of the lens so that the far-off and nearby objects are viewed clearly is called power of accommodation.
(b) To correct, an object at infinity has to be brought as an image to 120 cm.
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A concave lens of focal length 120 cm and power – 0.83 D is to be used.

29. (a) What are the values of (i) near point and (ii) far point of vision of a normal adult person?
(b) A student has difficulty in reading the blackboard while sitting in the last row. What could be his defect of vision? Draw a ray diagram to illustrate this defect of vision. [All India]
Answer.
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30. Name the three common defects of vision. What are their causes? Name the type of lens used to correct each of them. [Foreign]
Answer. (i) Myopia:
Cause: Elongation of eyeball,
Type of lens used for correction: Concave lens of suitable power.
(ii) Hypermetropia:
Cause: Shrinking of eyeball,
Type of lens used for correction: Convex lens of suitable power.
(iii) Presbyopia:
Cause: Weakening of ciliary muscles,
Type of lens used for correction: Bifocal lens.

VERY SHORT ANSWER TYPE QUESTIONS [1 Mark] – Year 2012

31. Which phenomenon is responsible for making the path of light visible?[Delhi]
Answer. Tyndall effect.

32. State one function of iris in human eye. [All India]
Answer. Iris controls the size of pupil.

33. State one function of pupil in human eye. [All India]
Answer. Pupil regulates and controls the amount of light entering the eye.

34. State one role of ciliary muscles in the human eye. [All India]
Answer. Ciliary muscles help the eye lens to focus the image of an object on the retina by increasing or decreasing the curvature of eye lens.

35. State one function of the crystalline lens in the human eye. [Foreign]
Answer. Crystalline (eye) lens forms a real and inverted image of the object on the retina.

36. State two properties of the image formed by the eye lens on the retina.
Answer.

  1. Image on the retina is real and inverted.
  2.  Diminished in size.

37. State one function of cornea in human eye.
Answer. Cornea provides the refraction of light rays entering the eye.

38. Why does the sun appear reddish at sunrise?
Answer. At sunrise, the sun looks almost reddish because only red colour ( λb < λr ), which is least scattered is received by our eye and appears to come from the sun. Hence, the appearance of sun at sunrise, near the horizon may look almost reddish.

SHORT ANSWER TYPE QUESTIONS [I] [2 Mark] – Year 2012

39. A star sometimes appears brighter and some other times fainter. What is this effect called? State the reason for this effect. [Delhi]
Answer. This effect is called Twinkling effect.
Reason: It is due to atmospheric refraction of starlight. The atmosphere has varying layers. The gradual change in the refractive index of different layers of atmosphere causes atmospheric refraction and starlight undergoes multiple refraction. So, the fluctuation in the position of star occurs continuously due to the changing amount of light entering the eye. The star sometimes appears brighter and at some other times, it appears fainter. This causes twinkling of star.

40. What is the colour of the clear sky during day time? Give reason for it. [All India]
Answer. Clear sky appears blue.
Reason: When sunlight passes through the atmosphere having the molecules of air and other fine particles, whose size is smaller than the wavelength of visible light, these molecules and particles scatter the blue colour more strongly than the other colours of spectrum as the wavelength of blue colour is more. This scattered blue light enters our eye. So, the colour of sky appears blue to us during day time.

41. What is a spectrum? Why do different coloured rays deviate differently on passing through a glass prism? [All India]
Answer. The band of coloured component of a light beam is called its spectrum. The colour sequence is given by the acronym V I B G Y O R — Violet, Indigo, Blue, Green, Yellow, Orange and Red.
The speed of light of different colours in a medium like glass is different. Varying speeds for different colours lead to different refractive indices for different colours. It has been observed that the refractive index of glass for violet colour is more than that for red colour. All the colours present in white light refract through different angles and hence, emerge out from the prism with different directions and become distinct.

SHORT ANSWER TYPE QUESTION [II] [3 Mark] – Year 2012

42. What is meant by the term ‘power of accommodation’ of human eye? How does it help a person to see nearby as well as distant objects clearly.[Foreign]
Answer. The ability of eye lens to adjust its focal length to form the sharp image of the object at varying distances on the retina is called its power of accommodation.
When we see the nearby object, the ciliary muscles contract, it increases the thickness of eye lens. The eye lens then becomes thicker. As a result, the focal length of eye lens decreases in such a way that the clear sharp image of nearby object is formed on the retina. Thus, the object is seen clearly to us.
When we see the distant object, these muscles becomes relaxed, thus the eye lens becomes thinner, and consequently focal length of the lens increases. Therefore, the parallel rays coming from the distant object are focused on the retina and object is seen clearly to us. Thus, the accommodation power of an eye helps a person to see nearby as well as distant objects clearly.

LONG ANSWER TYPE QUESTION [5 Mark] – Year 2012

43. (a) A student cannot see clearly a chart hanging on a wall placed at a distance 3 m from his eye. Name the defect of vision he is suffering from. Draw a ray diagram to illustrate this defect. List its two possible causes.
(b) Draw a ray diagram to show how this defect may be corrected using a lens of appropriate focal length. [ Delhi]
(c) An eye donation camp is being organised by social workers in your locality. How and why would you help in this cause? [All India]
Answer. (a) Myopia or Short-sightedness is the inability of an eye in viewing long distance objects. The image in this case falls before the retina. For every myopic eye, there exists a far point beyond which clear image cannot be seen. Short-sightedness is caused due to
(i) excessive curvature in cornea. (ii) elongation of eyeball.
(b) The short-sightedness is corrected by using a concave lens which diverges and shifts the image to the retina.
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(c) Reasons for donating eyes:

  1. A person’s corneal blindness can only be elevated by a human donor’s cornea.
  2.  It is opportunity to restore someone’s sight.

Methods for supporting this cause:

  1.  Make it a family tradition of donating the eyes.
  2. Motivate and educating others about eye donation.
  3.  Help removing all the myths . about eye donation.

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How Do Organisms Reproduce?Chapter Wise Important Questions Class 10 Science

How Do Organisms Reproduce?Chapter Wise Important Questions Class 10 Science

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SHORT ANSWER TYPE QUESTIONS [3 Marks] – Year 2007

Question. 1(i) What is fertilisation? Distinguish between external fertilisation and internal
fertilisation.
(ii) What is the site of fertilisation in human beings?
Answer.
(i) Fertilisation is defined as the fusion of a male gamete (sperm) with a female gamete (an ovum or egg) to form a zygote during sexual reproduction.
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(ii) The site of fertilisation in human beings is in the fallopian tube of female reproductive system.

Question.2 Define the terms unisexual and bisexual giving one example of each.
Answer. Unisexual is the plant whose flowers contain either stamens or carpels but not both. Example: Papaya, Watermelon.
Bisexual is the plant whose flowers contain both stamens and carpels. Example: Hibiscus, Mustard.

Question.3 Differentiate between ‘self-pollination’ and ‘cross-pollination’. Describe double fertilisation in plants.
Answer.
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During fertilisation in plants, the following events take place:
(i) One of the male gamete fuses with the female gamete present in the embryo sac.
(ii) The other male gamete fuses with the two polar nuclei in the embryo sac.
The first fusion product gives rise to the zygote while the second one forms the endosperm.
The process of two fusions occurring in the embryo sac is called double fertilisation.

VERY SHORT ANSWER TYPE QUESTION [1 Mark] – Year 2008

Question.4 What is the effect of DNA copying which is not perfectly accurate on the reproduction process?
Answer. DNA copying is not perfectly accurate and the resultant errors are a source of variations in populations of organisms.

SHORT ANSWER TYPE QUESTIONS [2 Marks] – Year 2008

Question.5 What is ‘reproduction’? Mention the importance of DNA copying inreproduction.
Answer. Reproduction is the process of producing new individuals of the same species
by existing organisms of a species, i.e. parents. The importance of DNA copying in reproduction are as follows:
(i) DNA copying is called DNA replication. In this process, one copy each of replicated DNA will be passed to daughter cells.
(ii) Variations may be introduced during DNA copying. This inbuilt tendency for variation during reproduction forms the basis of evolution.

Question.6 “Variations that confer an advantage to
an individual organism only will survive in a population.” Justify.
Answer. It is because the chances of survival depend on the nature of variations and different individuals have different kinds of advantages.
For example, a bacteria that can withstand heat will survive better in a heat wave, i.e. the organisms that are fit in the competitive environment and with great variations will be able to survive and adapt. Thus, more offsprings and population with genetic variations will survive.

Question.7 Name one sexually transmitted disease
each caused due to bacterial infection and viral infection. How can these be prevented?
Answer. Sexually transmitted disease caused due to
(i) Bacterial infection is gonorrhoea, and
(ii) Viral infection is AIDS (Acquired Immune Deficiency Syndrome). These diseases can be prevented by responsible sexual behaviour such as use of condom during intercourse, etc.

Question.8(a) In the human body what is the role of
(i) seminal vesicles, and (ii) prostate gland?
(b) List two functions performed by testis in human beings.
Answer. (a) The role of seminal vesicles and the prostate gland are as follows:
(i) Seminal vesicles produce seminal plasma which is in the form of fluid makes the transport of sperms smooth.
(ii) Prostate gland secretes prostatic fluid that keeps the sperms alive and helps them to swim vigorously.
(b) Two functions performed by testis
in human beings are as follows:
(i) Formation of sperms takes place in testis.
(ii) They secrete the hormone testosterone which regulates the formation of sperms and brings changes in appearance of boys at the time of puberty.

SHORT ANSWER TYPE QUESTK)NS[II] [3 Marks] – Year 2008

Question.9 Illustrate the following with the help of suitable diagrams:
(i) Regeneration in Planaria.
(ii) Budding in Hydra.
Answer.
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Question.10 Illustrate the following with the help of suitable diagrams:
(i) Binary Fission in Amoeba.
(ii)Leaf of Bryophyllum with buds.
Answer.
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Question.11 Illustrate the following with the help of suitable diagrams:
(i) Spore formation in Rhizopus.
(ii)Multiple fission in Plasmodium.
Answer.
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LONG ANSWER TYPE QUESTIONS [5 Marks] – Year 2008

Question.12 (a) Name the parts labelled A, B, C, D and E.
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(b) Where do the following functions occur?
(i) Production of an egg
(ii) Fertilisation
(iii) Implantation of zygote.
(c) What happens to the lining of uterus:
(i) before release of a fertilised egg?
(ii) if no fertilisation occurs?
Answer. (a)
A – Oviduct or Fallopian tube;
B – Ovary;
C – Uterus;
D – Cervix;
E – Vagina.
(b) (i) Ovaries; (ii) Fallopian tube;
(iii) Lining of the uterus.
(c) (i) The lining of uterus becomes
(ii) The lining of uterus slowly breaks and comes out through the vagina as blood and mucous, if no fertilisation occurs.

Question.13(a) Draw a diagram showing
germination of pollen on stigma of a flower.
(b) Label pollen grain, male germ- cells, pollen tube and female germ-cell in the above diagram.
(c) How is zygote formed?
Answer. (a) and (b)
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(c) Zygote is formed when male gamete, Le. sperm fuses with female gamete, i.e. ovum.

VERY SHORT ANSWER TYPE QUESTIONS [1 Mark] – Year 2009

Question.14 Mention the mode of reproduction used by
(a) Amoeba (b) Planaria.
Answer. Mode of reproduction used by
(a) Amoeba is Binary fission.
(b) Planaria is Regeneration.

Question.15 Name the information source of making proteins in the cell. State two basic events in reproduction.
Answer. The DNA in the cell nucleus is the information source of making proteins.
The two basic events in reproduction are:
(i) Creation of a DNA copy,
(ii) Additional cellular apparatus by the cell involved in the process.

Question.16 State the method used for growing rose
plants.
Answer. Artificial methods of vegetative propagation like cutting are used to grow rose plants.

Question.17 State what type of method is used for
growing jasmine plant.
Answer. Artificial methods of vegetative propagation like layering is used for growing jasmine plant.

Question.18 Name the largest cell present in the
human body.
Answer. The largest cell present in the human body is ovum.

SHORT ANSWER TYPE QUESTIONS [I] [2 Marks] – Year 2009

Question.19 What is regeneration? State a reason why a more complex organism cannot give rise to new individuals through this method.
Answer. Regeneration is the ability of a fully differentiated organism to give rise to new individual organisms from its body parts. More complex organisms cannot give rise to new individuals through regeneration because:
(i) their body is highly complicated.
(ii) there are specific organs to do specific functions.
(iii) there is a labour division in the body of complex organisms.
(iv) regeneration is carried out by specialised cells which are not present in complex organisms.

Question.20 What is reproduction? What are its two types? Which one of the two confers new characteristics on the offsprings and how?
Answer. Reproduction is the process of producing new individuals of the same species by existing organisms of a species, z.e. parents. Its two types are: Asexual reproduction and Sexual reproduction.
Sexual reproduction confers new characteristics on the offspring due to variation in DNA copying.

SHORT ANSWER TYPE QUESTIONS][II] [3 Marks] – Year 2009

Question.21(a) Explain the terms:
(i) Implantation (ii) Placenta
(b) What is the average duration of human pregnancy?
Answer. (a)
(i) Implantation: The embedding
of a fertilised mammalian egg (embryo) into the inner thick wall of the uterus (womb) where it will continue its development is called implantation.
(ii) Placenta: It is a complex double-layered spongy vascular tissue in human female formed by the joint activity of maternal and foetal tissues in the wall of uterus that is meant for attachment, nourishment and waste disposal for the foetus.
(b) The average duration of human pregnancy is 40 weeks or 280 days.

Question.22 What are sexually transmitted diseases?
Name four such diseases. Which one of them damages the immune system of human body?
Answer. Sexually Transmitted Diseases (STDs) are the diseases which are spread by sexual contact from an infected person to a healthy person. They are caused by various microorganisms that live in warm and moist environments of the vagina, urethra, anus and mouth.
The four sexually transmitted diseases are:
(i) Gonorrhoea
(ii) Syphilis
(iii) Trichomoniasis
(iv) AIDS (Acquired Immune Deficiency Syndrome).
AIDS damages the immune system of human body.

SHORT ANSWER TYPE QUESTION[I] [2 Marks] – Year 2010

Question.23 List any four reasons for vegetative propagation being practised in the growth of some type of plants.
Answer.
(i) Vegetative propagation is a cheaper, easier and more rapid method of propagation in plants than growing plants from their seeds.
(ii) Better quality of plants can be maintained by this method.
(iii) It results in propagation of those plants which do not produce viable seeds or produce seeds with prolonged period of dormancy.
(iv) The plants generated from vegetative means are more uniform and genetically similar to the parent stock.

SHORT ANSWER TYPE QUESTIONS][II] [3 Marks] – Year 2010

Question.24 Write the full form of DNA. Name the part of the cell where it is located. Explain its role in the process of reproduction of the cell.
Answer. The full form of DNA is deoxyribonucleic acid. It is the genetic material found in the chromosomes, which are present in the nucleus of a cell.
Role of DNA in the process of repro¬duction of the cell:
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DNA plays an important role in the reproduction of a cell. The reproducing cell produces an identical copy of DNA through some cellular mechanism. Since the newly formed copy of DNA lacks an organised cellular structure, the cell gets divided to provide cell cover to the newly formed DNA. Thus, two daughter cells are formed from the single cell as a result of the copying of DNA.

Question.25 Explain vegetative propagation with the help of two examples. List two advantages of vegetative propagation.
Answer. In vegetative propagation, new plants are obtained from the parts of old plants like stems, roots and leaves, without the help of any reproductive organ.
There are two ways of vegetative propagation:
(a) Natural Vegetative Propagation, and
(b) Artificial Vegetative Propagation.
Natural vegetative propagation by leaves: The fleshy leaves of Bryophyllum bear adventitious buds in the notches along the leaf margin.
Grafting: In this method of reproduction, two plants of closely related varieties are joined together so that they live as one plant.

  • The portion of a plant that is grafted on the other plant is called scion, and the plant in which grafting is performed is called the stock.
  • This method is applied to improve variety of fruits like mango, apple, peas, citrus and guava.
    advantages of vegetative propagation are:

(i) Vegetative propagation is a cheaper, easier and more rapid method of propagation in plants than growing plants from their seeds.
(ii) Better quality of the plants can be maintained by this method.

Question.26 List any three differences between pollination and fertilisation.
Answer.
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LONG ANSWER TYPE QUESTIONS [5 Marks] – Year 2010

Question.27 Draw a longitudinal section of a flower and label the following parts:
(i) Part that produces pollen grain.
(ii)Part that transfers male gametes to the female gametes.
(iii) Part that is sticky to trap the pollen grain.
(iv) Part that develops into a fruit.
Answer.
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Question.28 (a) Explain the role of placenta in the
development of human embryo.
(b) Give example of two bacterial and two viral sexually transmitted diseases. Name the most effective contraceptive which prevents spread of such diseases.
Answer.
(a) Role of placenta in the development of human embryo: A special tissue develops between the uterine wall and the embryo (foetus) called placenta, where exchange of nutrients, glucose and oxygen takes place. The developing embryo will also generate waste substances which can be removed by transferring them into the mother’s blood through the placenta. The development of the child inside the mother’s blood takes approximately nine months.
(b) Sexually transmitted diseases (STDs) transmitted by bacteria are:
(i) Gonorrhoea (ii) Syphilis
STDs transmitted by virus:
(i) AIDS (ii) Genital warts
The most effective contraceptive which prevents the spread of these diseases is by the use of mechanical barriers such as physical devices like condoms.

SHORT ANSWER TYPE QUESTIONS [I] [2 Marks] – Year 2011

Question.29 Write any two differences between binary fission and multiple fission in a tabular form as observed in cells of organisms.
Answer.
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Question.30 List any four modes of asexual reproduction.
Answer. Four modes of asexual reproduction are—Binary fission in Amoeba, Fragmentation in Spirogyra, Regeneration in Planaria and Budding in Hydra.

Question.31 Why is DNA copying an essential part of the process of reproduction?
Answer.DNA copying is an essential part of the process of reproduction because:
(i) DNA copying provides cellular apparatus in the daughter cells.
(ii) DNA in daughter cells will be able to control the functioning of daughter cells.
(iii)DNA copies will retain the traits.

SHORT ANSWER TYPE QUESTIONS [II] [3 Marks] – Year 2011

Question.33 What does HIV stand for? Is AIDS an infectious disease? List any four modes of spreading AIDS.
Answer. HIV stands for Human Immunodeficiency Virus.
Yes, AIDS is an infectious disease.
Four modes of spreading AIDS are as follows:
(i) By having sexual contact with an infected person.
(ii) By the transfusion of blood from an infected person.
(iii) Through infected needles used for injection.
(iv) Through the placenta from the mother to child during pregnancy.

Question.34 Expand AIDS. List any four methods of prevention (control) of AIDS.
Answer. AIDS stands for Acquired Immune Deficiency Syndrome.
Four methods of prevention or control of AIDS are as follows:
(i) Use condom during sex.
(ii) Avoid sharing of needles.
(iii) Test blood for AIDS before transfusion.
(iv) Avoid sexual contact with unknown person.

Question.35 (a) List any four reasons for adopting contraceptive methods.
(b) If a woman is using Copper-T, will it help in protecting her from sexually transmitted diseases? Why? 
Answer. (a) Four reasons for adopting contra¬ceptive methods are:
(i) To increase the gap between, two children.
(ii)To prevent unwanted preg¬nancy.
(iii)To prevent transmission of STDs.
(iv)To control population growth. (b) If a woman is using a copper-T,
it will not help in protecting her from sexually transmitted diseases. Copper-T prevents only implantation in the uterus.

Question.36 Explain the following methods of contraception giving one example of each:
(i) Barrier method
(ii) Hormonal imbalance method
(iii) Surgical method.
Answer.
(i) Barrier Method: In this method, physical devices such as condoms, diaphragm and cervical caps are used. These devices prevent the entry of sperm in the female genital tract during copulation, thus acting ‘ as a barrier between them.
(ii) Hormonal Imbalance Method: In
this method, specific drugs are used by females, which are of two types: oral pills and vaginal pills.
Oral pills contain hormones which stop the ovaries from releasing ovum into the fallopian tube. These pills are also called oral contraceptives (OCs) which act by changing the hormonal balance of the body so that eggs are not released and fertilisation cannot occur. The use of Intrauterine Contraceptive Devices (IUCDs) prevents implantation in the uterus. This device is copper-T placed safely inside the uterus by a doctor or nurse.
(iii) Surgical Method: In this method, a small portion of vas deferens in male and the fallopian tube in ‘ female is surgically removed or tied. It is called vasectomy in males and tubectomy in females. In this case, if the vas deferens in male is blocked, sperm transfer will be prevented and if the fallopian tube in the female is blocked, the egg will not be able to reach the uterus, thus fertilisation will not take place.

LONG ANSWER TYPE QUESTIONS [5 Marks] – Year 2011

Question.37 (a) What is fragmentation in organism?
Name a multicell-ular organism which reproduces by this method.
(b) What is regeneration in organism? Describe regeneration in Planaria with the help of a suitable diagram.
Answer. (a) Fragmentation: Multicellular organisms with simple body organisation such as filamentous algae-Spirogyra breaks up into two or more small pieces or fragments upon maturation. These fragments grow into new individuals.
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(b) Regeneration: It is the ability of a fully differentiated organism to give rise to new individual organisms from its body parts.

  •  Small cut or broken parts of the organism body grow or regenerate into separate individuals.
  •  Planaria can be cut into any number of pieces and each piece grows into a complete organism.
  • Regeneration is carried out by specialised cells which proliferate and make large number of cells thus, undergoing changes to become various cell types and tissues.
  • The changes taking place in an organised sequence is development.
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Question.38 With the help of suitable diagrams, explain the various steps of budding in Hydra.
Answer.

  • Hydra reproduces by budding using the regenerative cells.
  • A bud develops as an outgrowth in Hydra due to repeated cell division at one specific site.
  •  When fully matures, the bud detaches itself from the parent body and develops into new indipendent individuals
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Question.39 (a) What is spore formation?
(b) Draw a diagram showing spore formation in Rhizopus.
(c) List two advantages for organisms
to reproduce themselves through spores.
Answer.
(a) When a slice of bread is kept in moist dark place for a few days, spores of Rhizopus present in air settle on the bread to form new fungus plants of Rhizopus.
(b) The Rhizopus consists of fine thread-like projections called hyphae. It has a knob like structure which is involved in reproduction called sporangia, containing spores, that develop into new Rhizopus.
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(c) Two advantages for organisms to reproduce themselves through spores are as follows:
(i) It is a faster mode of reproduc-tion.
(ii) Offsprings produced are identical.

SHORT ANSWER TYPE QUEST10NS[I] [2 Marks] – Year 2012

Question.40 State one genetically different feature between sperms and eggs of humans. What is its consequence?
Answer. The sex chromosome of human male is XY. A sperm of human male carries either an X chromosome or one Y chromosome.
The sex chromosome of human female is XX and hence, the egg always carries the X chromosome.
If a sperm carrying X chromosome fertilises an egg which carries X chromosome, then the’ child born will be a girl. If a sperm
carrying Y chromosome fertilises an egg which carries X chromosome, then the child born will be a boy.

Question.41 List two advantages of vegetative reproduction practised in case of an orange plant.
Answer. Two advantages of practising vegetative reproduction in orange plants are:
(i) The oranges produced are similar in size and shape.
(ii) Many oranges do not produce viable seeds and hence, vegetative method is good alternative.

Question.42 How does growing embryo get nutrition from the mother’s blood?
Answer. The embryo gets nutrition from the mother’s blood with the help of a special tissue called placenta. This is a disc which is embedded in the uterine wall and transfers glucose and oxygen from the mother to the embryo.

Question.43 Define the term puberty. List two changes observed in girls at the time of puberty.
Answer. The period, when the rate of general body growth begins to slow down and reproductive tissues begin to mature, is called puberty.
Two changes observed in girls at the time of puberty are:
(i) The breast size begin to increase, (ii) Menstruation starts.

Question.44 What is meant by asexual reproduction? List its any two different forms.
Answer. Asexual reproduction is the process of producing new organism from a single parent without the involvement of sex cells. Fission and fragmentation are two different forms of asexual reproduction.

Question.45 Name an organism which reproduces by spore formation. List three conditions
favourable for spores to germinate and grow.
Answer. Rhizopus reproduces by spore formation. Conditions favourable for spore formation are:
(i) Cool place, (ii) Moist place and (iii) Dark place.

Question.46 “DNA copies generated during reproduction will be similar but may not be identical to the original.” Justify this statement.
Answer. DNA copies generated will be similar, but may not be identical to the original as some variations are so drastic that new DNA copy cannot work with the cellular apparatus it inherits. Such a newborn cell will simply die. Therefore, there could be many other variations in the DNA copies that would not lead to such a drastic outcome. Thus, the surviving cells are similar but slightly different from each other. This tendency of variation during reproduction is the basis for evolution.

Question.47 List two advantages of practising vegetative propagation in plants. Select two plants raised by this method from the list given below:
Banana, Gram, Pea, Rose, Tomato, Wheat.
Answer. Advantages of vegetative propagation are:

  1. Plants raised by vegetative propa-gation can bear fruits and flowers earlier.
  2.  Plants produced are genetically similar.
    Banana and Rose can be raised by vegetative method.

Question.48 List the parts of human male reproductive system which contribute fluid to the semen. State two advantages semen offers to the sperms.
Answer. Prostate glands and seminal vesicles add fluid in the vas deferens. This makes transportation of sperms easier and also provides nutrition to the sperms.

Question.49 Name the two types of germ-cells
present in human beings. How do they structurally differ from each other? Give two differences.
Answer. The two types of germ-cells present in human beings are sperm and ova. The sperm of human have either X or Y chromosome. The ova always carry X chromosome. The sperm is structurally long with a tail. The ova is round in structure.

SHORT ANSWER TYPE QUESTIONS[II] [3 Marks] – Year 2012

Question.50 List and explain in brief three methods of contraception.
Answer. Methods of contraception are:

  1. Use of condom for penis or for vagina as a mechanical barrier for the sperms to reach the egg.
  2.  Use of oral pills which change the hormonal balance so that eggs are not released.
  3.  Surgical method where either the vas deferens of male is blocked or the fallopian tube of female is blocked.

Question.51 What is AIDS? Which microbe is responsible for AIDS infection? State one mode of transmission of this disease. Explain in brief one measure for the prevention of AIDS.
Answer. AIDS is the Acquired Immune Deficiency Syndrome.
It is caused by a virus called Human Immunodeficiency Virus.
AIDS is transmitted by sexual contact with an infected person.
AIDS can be prevented by avoiding sexual contact with infected person or by using condom during sex.

LONG ANSWER TYPE QUESTIONS [5 Marks] – Year 2012

Question.52 Describe in brief the role of (i) testis (ii) seminal vesicle, (iii) vas deferens, (iv) ureter and (v) prostate gland in human male reproductive system.
Answer.

  1. Testis: Testes are oval shaped primary reproductive organs in men. The function of testes is to produce sperms and male sex hormone testosterone. The scrotum provides optimal temperature for the formation of sperms.
  2.  Seminal vesicle: Seminal vesicles are a pair of thin walled muscular elongated sac which secrete fluid for nourishment of sperms.
  3. vas deferens: The sperms are carried by a long tube called vas deferens to organs called seminal vesicles where the sperms get nourishment and stored.
  4. Ureter It is the tube that carries urine from kidney to the urinary bladder. In humans, there are two ureters, one attached to each kidney.
  5.  Prostate glands: Prostate glands produce a fluid which is released in the urethra along with secretion of seminal vesicles for nourishment and transportation of sperms.

Question.53 Draw a diagram of a human female reproductive system and label the part
(i) that produces egg
(ii) where fusion of egg and sperm take place
(iii) where zygote is implanted
What happens to human egg when it is not fertilised?
Answer.
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If the egg is not fertilised, the thick and nourishing lining of the uterus breaks and comes out through vagina as blood and mucous.

Question.54 State in brief the changes that take place in a fertilised egg (zygote) till birth of the child in the human female reproductive system. What happens to the egg when it is not fertilised?
Answer. The egg gets fertilised in the oviduct. The fertilised egg, the zygote gets implanted in the lining of the uterus and starts dividing. The uterus prepares itself every month to receive and nurture the growing embryo. The lining thickens and is richly supplied with blood to nourish the growing embryo.
The embryo gets nutrition from the mother’s blood with the help of a special tissue called placenta. The development of the child inside the mother’s body takes approximately nine months. On completion of 9 months, the child is born as a result of rhythmic contractions of the muscles in the uterus.
If the egg is not fertilised, the thick and nourishing lining of the uterus breaks and comes out through vagina as blood and mucous.

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Maharashtra Board Class 10 Solutions for Marathi कुमारभारती –हत्तीचा दुष्टान्त

Maharashtra Board  Class 10 Solutions for Marathi कुमारभारती – हत्तीचा दुष्टान्त

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Maharashtra Board SSC Class 10 Solutions for Marathi कुमारभारती

The post Maharashtra Board Class 10 Solutions for Marathi कुमारभारती – हत्तीचा दुष्टान्त appeared first on Learn CBSE.

Maharashtra Board Class 10 Solutions for Marathi कुमारभारती –मीच तुमचा बाजीराव

Maharashtra Board  Class 10 Solutions for Marathi कुमारभारती – मीच तुमचा बाजीराव

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Maharashtra Board SSC Class 10 Solutions for Marathi कुमारभारती

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Maharashtra Board Class 10 Solutions for Marathi कुमारभारती –स्त्रीपुरुष तुलना

Maharashtra Board  Class 10 Solutions for Marathi कुमारभारती – स्त्रीपुरुष तुलना

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Maharashtra Board SSC Class 10 Solutions for Marathi कुमारभारती

The post Maharashtra Board Class 10 Solutions for Marathi कुमारभारती – स्त्रीपुरुष तुलना appeared first on Learn CBSE.


NCERT solutions for class 8 Sanskrit सुभाषितानि

NCERT solutions for class 8 Sanskrit सुभाषितानि

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NCERT solutions for class 8 Sanskrit

The post NCERT solutions for class 8 Sanskrit सुभाषितानि appeared first on Learn CBSE.

Ultimate Book of Accountancy Class 12 Part A Solutions – Goodwill

Ultimate Book of Accountancy Class 12 Part A Solutions – Goodwill

You can buy this text book from here: http://amzn.to/2s2YSxN

Practical Problems

Solution 1:
Average Profit = 8,000 + 10,000 + 14,000 + 12,000 = 44,000/4 = 11,000
Goodwill = 11,000 × 2 = 22,000

Solution 2:
Average Profit = 6,000 – 4,000 – 2,000 + 8,000 + 12,000 = 20,000/5 = 4,000
Goodwill = 4,000 × 2.5 = 10,000

Solution 3:
Average Profit = 20,000 + 20,000 + 40,000 + 20,000 = 1,00,000/4 = 25,000
Goodwill = 25,000 × 1.5 = 37,500

Solution 4:
Average Profit = 20,000 + 40,000 + 40,000 + 20,000 + 30,000 = 1,50,000/5 = 30,000
Goodwill = 30,000 × 2 = 60,000

Solution 5:
Average Profit = 50,000 + 60,000 + 70,000 = 1,80,000/3 = 60,000 – 36,000 = 24,000
Goodwill = 24,000 × 2 = 48,000

Solution 6:
Average Profit = 45,000(60,000 – 15,000) + 1,50,000 (1,20,000 + 30,000) + 1,05,000 (1,14,000 – 9,000) = 3,00,000/3 = 1,00,000
Goodwill = 1,00,000 × 2 = 2,00,000

Solution 7:
Average Profit =
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Average Profit = 1,62,000/4 = 40,500
Goodwill = 40,500 × 2 = 81,000

Solution 8:
Calculation of Goodwill :

Profits Weights Product
20,000 1 20,000
25,000 2 50,000
30,000 3 90,000
35,000 4 1,40,000
60,000 5 3,00,000
15 6,00,000

Weighted Average Profit = 6,00,000/15 = 40,000
Goodwill = 40,000 × 2.5 = 1,00,000

Solution 9:
Calculation of Goodwill :

Profits Weights Product
12,000 3 36,000
8,000 1 8,000
10,000 2 20,000
15,000 4 60,000
25,000 5 1,25,000
15 2,49,000

Weighted Average Profit = 2,49,000/15 = 16,600
Goodwill = 16,600 × 2 = 33,200

Solution 10:
Calculation of Goodwill :
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Calculation of weighted average profits:

Year Profit Weight Product
2006 7,700 1 7,700
2007 8,800 2 17,600
2008 11,700 3 35,100
2009 12,310 4 49,240
Total 10 1,09,640

Weighted Average Profit = 1,09,640/10 = 10,964
Goodwill = 10,964 × 3 = 32,892

Solution 11:
Average Profit = 27,000 + 30,000 + 33,000 = 90,000/3 = 30,000
Normal Profit = Capital investment × 10/100 = 90,000 × 10/100 = 9,000
Super Profit = 30,000 – 9,000 = 21,000
Goodwill = 21,000 × 2 = 42,000

Solution 12:
Goodwill (given) = 48,000
Super Profit = 48,000/4 = 12,000
Normal Profit = 1,50,000 × 10/100 = 15,000
Super Profit = Average Profit – Normal Profit
12,000 = ? – 15,000
Average profit = 12,000 + 15,000 = 27,000

Solution 13:
Average Profit = 14,000 + 13,000 + 16,000 + 15,000 + 12,000 = 70,000/5 = 14,000
Normal Profit = 80,000 × 12/100 = 9,600
Super Profit = 14,000 – 9,600 = 4,400
Goodwill = 4,400 × 3 = 13,200

Solution 14:
Average Profit = 60,000 + 70,000 + 86,000 = 2,16,000/3 = 72,000
Remuneration = 72,000 – 12,000 = 60,000
Normal Profit = 4,00,000 × 10/100 = 40,000
Super Profit = 60,000 – 40,000 = 20,000
Goodwill = 20,000 × 2 = 40,000

Solution 15:
Average Profit = 12,200 + 15,000 – 2,000 + 21,000 = 46,200/4 = 11,550
Remuneration = 11,550 – 3,600 = 7,950
Normal Profit = 50,000 × 10/110 = 5,000
Super Profit = 7,950 – 5,000 = 2,950
Goodwill = 2,950 × 3 = 8,850

Solution 16:
Capitalised value of Average Profit = 20,000 × 100/10 = 2,00,000
Goodwill = 2,00,000 – 1,64,000 = 36,000

Solution 17:
Capialised value of Average Profit = 21,600 × 100/10 = 2,16,000
Net Assets = 3,00,000 – 90,000 – 1,20,000 = 1,71,000
Goodwill = 2,16,000 – 1,71,000 = 45,000

Solution 18:
Average Profit = 9,000
Normal Profit = 60,000 × 10/100 = 6,000
Super Profit = 9,000 – 6,000 = 3,000
Goodwill = 3,000 × 100/10 = 30,000

Solution 19:
Average Profit = 12,000
Normal Profit = 30,000 × 20/100 = 6,000
Super Profit = 12,000 – 6,000 = 6,000
Goodwill = 6,000 × 100/20 = 30,000

Solution 20:
Average Profit = 20,000 + 16,000 – 10,000 – 6,000 + 40,000 = 60,000/5 = 12,000
Remuneration = 12,000 – 4,000 = 8,000
Normal Profit = 50,000 × 10/100 = 5,000
Net Assets = 60,000 – 36,000 = 24,000
Case 1:
Goodwill by Average Profits = 8,000 × 2 = 16,000
Case 2:
Goodwill by Super Profit:
Super Profit = Average Profit – Normal Profit
8,000 – 5,000 = 3,000
Goodwill = 3,000 × 2.5 = 7,500
Case 3:
Capitalisation of Average Profit:
Capitalised value = 8,000 × 100/10 = 80,000
Goodwill = Capitalised value – Net Assets
56,000 = 80,000 – 24,000
Case 4:
Capitalisation of Super Profit:
Goodwill = Super Profit × 100/Rate = 3,000 × 100/10 = 30,000

Solution 21:
Average Profit = 72,000
Normal Profit = 1,87,500 + 1,50,000 + 1,12,500 = 4,50,000 × 10/100 = 45,000
Super Profit = 72,000 – 45,000 = 27,000
Goodwill = 27,000 × 100/10 = 2,70,000

Solution 22:
Average profit = 1,20,000 + 1,44,000 + 1,68,000 = 4,32,000/3 = 1,44,000
salary = 1,44,000 – 48,000 = 96,000
Normal Profit = 4,00,000 × 15/100 = 60,000
Super Profit = 96,000 – 60,000 = 36,000
Goodwill = 36,000 × 2 = 72,000
Note: Salary is 48,000 (24,000 each)

Brilliant Problems

Solution 1:
Case 1:
Average Profit = 1,47,600 + 1,48,100 + 1,52,500 = 4,48,200/3 = 1,49,400
Remuneration = 1,49,400 – 12,000 = 1,37,400
Goodwill = 1,37,400 × 3 = 4,12,200
Case 2:
Average profit = 1,37,400
Normal Profit = 7,00,000 × 18/100 = 1,26,000
Super Profit = Average Profit – Normal Profit
11,400 = 1,37,400 – 1,26,000
Goodwill = 11,400 × 3 = 34,200
Case 3:
Capitalisation of Average Profit = 1,37,400 × 100/18 = 7,63,333
Net Assets = 7,54,762 – 31,329 = 7,23,433
Goodwill = 7,63,333 – 7,23,433 = 39,900
Case 4:
Super Profit x 100/Rate = 11,400 × 100/18 = 63,333

Solution 2:
Average Profit = 32,280 + 36,870 + 43,350 = 1,12,500/3 = 37,500 – 1,800 = 35,700
Capital Employed = Total Assets (ignore non-trade investment) – outsiders liabilities = Creditors
2,42,040 – 30,000 – 38,040 = 1,74,000
Normal Profit = 1,74,000 × 10/100 = 17,400
Super Profit = 35,700 – 17,400 = 18,300
Goodwill = 18,300 × 2 = 36,600

Weighted Method:

Profits Weight Products
32,280 – 1,800 1 30,480
36,870 – 1,800 2 70,140
43,350 – 1,800 3 1,24,650
6 2,25,270

Weighted Average Profit = 2,25,270/6 = 37,545
Normal Profit = 17,400
Super Profit = 37,545 – 17,400 = 20,145
Goodwill = 20,145 × 2 = 40,290

Solution 3:
Goodwill (given) = 24,000
Super Profit = 24,000/4 = 6,000
Normal Profit = 75,000 × 20/100 = 15,000
Super Profit = Average Profit – Normal Profit
6,000 = ? – 15,000
Average Profit = 6,000 + 15,000 = 21,000

Solution 4:
Goodwill = 18,000 (given)
Super Profit = 18,000/4 = 4,500
Capital Employed = 15,00,000 – 12,00,000 = 3,00,000
Normal Profit = 3,00,000 × 10/100 = 30,000
Average Profit = 30,000 + 4,500 = 34,500

Solution 5:
Average Profit = 45,000 + 60,000 + 63,000 (75,000 – 12,000) = 1,68,000/3 = 56,000
Normal Profit = 5,00,000 × 10/100 = 50,000
Super Profit = 56,000 – 50,000 = 6,000
Goodwill = 6,000 × 100/10 = 60,000

Solution 6:
Normal Profit = 10,00,000 × 20/100 = 2,00,000
Average Profit = 3,00,000 (given)
Super Profit = 3,00,000 – 2,00,000 = 1,00,000
Goodwill = 1,00,000 × 100/20 = 5,00,000
Goodwill = 1,00,000 × 2 = 2,00,000

Solution 7:
Capital employed = 20,00,000 – 3,60,000 = 16,40,000
Normal Profit = 16,40,000 × 10/100 = 1,64,000
Average Profit = 2,00,000
Super Profit = 2,00,000 – 1,64,000 = 36,000
Goodwill = 36,000 × 100/10 = 3,60,000
Goodwill = 36,000 × 3 = 1,08,000

Solution 8:
Average Profit = 1,56,800 + 1,40,000 + 1,80,000 + 1,20,000 = 5,96,800/4 = 1,49,200 – Remuneration (4,800 × 4) = 1,30,000
Normal Profit = 9,20,000 × 13/100 = 1,19,600
Super Profit = 1,30,000 – 1,19,600 = 10,400
Net Assets = 10,40,000 – 1,60,000 = 8,80,000
Goodwill by capitalisation of Average Profit = 1,30,000 × 100/13 = 10,00,000
10,00,000 – Net Assets 8,80,000 = 1,20,000
Goodwill by capitalising Super Profit = 10,400 × 100/13 = 80,000

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Ultimate Book of Accountancy Class 12 Part A Solutions – Basics of Partnership

Ultimate Book of Accountancy Class 12 Part A Solutions – Basics of Partnership

You can buy this text book from here: http://amzn.to/2s2YSxN

Practical Problems

Solution 1:
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Solution 2:
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Solution 3:
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Solution 4:
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Solution 5:
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Solution 6:
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Solution 7:
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Solution 8:
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Solution 9:
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Solution 10:
Case 1:
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Case 2:
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Case 3:
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Solution 11:
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Working Note:
Profit = 20,100 – 4,500 = 15,600
Out of which 12,000 divided as 50%, 30% and 20% and remaining 3,600 in equal ratio.
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Solution 12:
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Solution 13:
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Solution 14:
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Solution 15:
Case 1: Claim of B is not valid because partnership deed is silent on interest on partner’s loan. So he is entitled only for 6% p.a.
Case 2: Profit sharing ratio will be equal.

Solution 16:
(a) No salary will be given to any partner.
(b) No interest on capital to any partner. Only X will get 6% p.a. interest on loan.
(c) New partner can be admitted with the consent of old partners. In this case Y is objective, so X cannot admit his son as a partner.

Solution 17:
(a) No salary will be paid.
(b) No interest on capital will be given to any partner.
(c) Interest on partner’s loan will be 6% p.a.
(d) Profit sharing ratio will be equal
(e) No interest on drawings will be charged.

Solution 18:
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Solution 19:
Interest on M’s Loan (for 9 months) = 8,000 x 6/100 x 9/12 = 360

Solution 20:
Interest on A’s Loan = 15,000 x 6/100 x 6/12 = 450
Interest on B’s Loan = 15,000 x 6/100 x 6/12 = 450

Solution 21:
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Solution 22:
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Solution 23:
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Solution 24:
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Solution 25:
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Solution 26:
Interest on capital to B
Opening Capital = 8,000 + 3,000 (Drawings) – 1,000 (Profit)
Interest = 10,000 x 5/100 = 500
Note: Total profit was Rs.6,000 during the year out of which Rs.4,000 was kept aside and Rs.2,000 distributed by the partners in equally i.e. 1,000 each.

Solution 27:
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Solution 28:
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Solution 29:
Interest on John’s Capital = 2,00,000 x 5/100 = 10,000
Interest on Smith’s Capital = 2,00,000 x 5/100 = 10,000

Solution 30:
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Solution 31:
Case (a):
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Interest to Hero Rs.6,000 and Zero Rs.3,000 = 6,000 + 3,000 = 9,000 but profit is only 7,500
Ratio of appropriations = 6,000 : 3,000 = 2 : 1
Now : 7,500 x 2/3 = 5,000 and 7,500 x 1/3 = 2,500
Case (b):
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Solution 32:
(a) Partnership deed is silent as to payment to interest on capital so no interest on capital will be calculated and profit will be shared by the partners directly in their profit sharing ratio i.e. 3 : 2.
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(d) No interest on capital, because interest on capital is an appropriation and in case of loss no appropriation to the partners. Loss will be shared by the partners in their profit sharing ratio 3:2.
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Solution 33:
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Solution 34:
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Solution 35:
English = 1,00,000 x 6/100 x 7.5/12 = 3,750
Hindi = 1,00,000 x 6/100 x 4.5/12 = 2,250

Solution 36:
A = 72,000 x 15/100 x 6.5/12 = 5,850
B = 72,000 x 15/100 x 5.5/12 = 4,950
C = 72,000 x 15/100 x 6/12 = 5,400

Solution 37:
Hina = 60,000 x 6/100 x 6.5/12 = 1,950
Rahim = 60,000 x 6/100 x 5.5/12 = 1,650

Solution 38:
Keyboard = 1,80,000 x 10/100 x 5/12 = 7,500
Mouse = 1,80,000 x 10/100 x 4/12 = 6,000
Printer = 1,80,000 x 10/100 x 4.5/12 = 6,750

Solution 39:
A = 90,000 x 8/100 x 3.5/12 = 2,100
B = 90,000 x 8/100 x 2.5/12 = 1,500
C = 90,000 x 8/100 x 3/12 = 1,800

Solution 40:
(i) 45,000 x 10/100 x 6/12 = 2,250
(ii) 45,000 x 10/100 x 6/12 = 2,250
(iii) 8,000 x 10/100 x 6/12 = 400

Solution 41:
Tom = 24,000 x 10/100 = 2,400
Jerry = 16,000 x 10/100 = 1,600

Solution 42:
(i) 60,000 x 10/100 x 6/12 = 3,000
(ii) 50,000 x 10/100 = 5,000
(iii) 20,000 x 10/100 = 2,000
10,00 x 10/100 = 1,000
Total = 2,000 + 1,000 = 3,000

Solution 43:
(a) 12,000 x 10/100 x 6.5/12 = 650
(b) 12,000 x 10/100 x 5.5/12 = 550
(c) 12,000 x 10/100 x 6/12 = 600
(d) 24,000 x 10/100 x 6/12 = 1,200
(e) 12,000 x 10/100 x 6/12 = 600
(f) 12,000 x 10/100 x 7.5/12 = 750
(g) 12,000 x 10/100 x 4.5/12 = 450
(h) 12,000 x 10/100 x 6/12 = 600
(i) 36,000 x 10/100 x 5/12 = 1,500
(j) 72,000 x 10/100 x 3.5/12 = 2,100

Solution 44:
Interest on A’s Loan = 60,000 x 3/5 = 36,000
36,000 x 6/100 x 6/12 = 1,080
Interest on B’s Loan = 60,000 x 2/5 = 24,000
24,000 x 6/100 x 6/12 = 720

Solution 45:
Case 1:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q45-i

Case 2:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q45-ii

Case 3:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q45-iii

Solution 46:
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Solution 47:
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Solution 48:
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Solution 49:
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Solution 50:
Z’s Capital A/c Dr. 135
To X’s Capital A/c 120
To Y’s Capital A/c 15
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q50

Solution 51:
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Solution 52:
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Solution 53:
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Solution 54:
Y’s Capital A/c Dr. 225
To X’s Capital A/c 150
To Z’s Capital A/c 75
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q54

Solution 55:
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Solution 56:
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Solution 57:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q57-ii

Note for teachers: Partners have directly divided profit Rs.1,80,000 in ratio of 3:2:1 i.e. A Rs.90,000; B Rs.60,000 and C Rs.30,000.
But they should have prepared P/L Appropriation Account to provide salary, commission and guaranteed amount etc.
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q57-iii

Note : Deficiency of Rs.8,125 borne by B and C in 3:2 ratio.

Solution 58:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q58

Solution 59:
A’s Capital A/c Dr. 121
To B’s Capital A/c 121
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q59

Solution 60:
Anil’s Capital A/c Dr. 1,100
To Mohan’s Capital A/c 1,100
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q60

Solution 61:
S’s Current A/c Dr. 10,112
To R’s Current A/c 10,112
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q61

Note : Interest on drawings should not be deducted because this is already taken benefit.

Solution 62:
B’s Current A/c Dr. 5,600
To A’s Current A/c 5,600
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q62

Solution 63:
Q’s Capital A/c Dr. 4,000
R’s Capital A/c Dr. 1,000
To P’s Capital A/c 5,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q63

Solution 64:
B’s Capital A/c Dr. 4,000
To A’s Capital A/c 3,000
To C’s Capital A/c 1,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q64

Solution 65:
Anu Dr. 250
To Beena 25
To Ceema 100
To Deepa 125
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q65

Solution 66:
Mohan’s Current A/c Dr. 60
To Ram’s Current A/c 60
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q66

Solution 67:
Z’s Current A/c Dr. 400
To Y’s Current A/c 400
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q67

Solution 68:
S’s Capital A/c Dr. 27,000
To R’s Capital A/c 27,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q68

Solution 69:
Jagdish Dr. 3,000
To Ashish 2,000
To Deepak 1,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q69

Hint: Profits are taken for last 4 years and not for 5 years.

Solution 70:
A’s Capital A/c Dr. 31,666
To B’s Capital A/c 6,333
To C’s Capital A/c 25,333
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q70

Solution 71:
X Dr. 45,000
Y Dr. 5,000
To Z 50,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q71

Solution 72:
Y Dr. 7
Z Dr. 23
To X 30
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q72

Solution 73:
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Working Note: Profit distribution
Robine = 80,000 x 5/10 = 40,000 – 1,000
William = 80,000 x 4/10 = 32,000 – 1,000
Henry = 80,000 x 1/10 = 8,000 + 1,000 from Robin + 1,000 from William
Gauranteed amount to Henry 10,000 but he is getting only 8,000, deficiency of 2,000 will be borne by Robin and William equally.
OR
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q73-ii

Solution 74:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q74-ii

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q74-iii

Solution 75:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q75

Note : Deficiency of Rs.2,000 will be borne by Austin and Alex in 3:2.

Solution 76:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q76

Working Note : Deficiency 500 – 230 = 270
Ram will pay = 270 x 3/5 = 162
Mohan will pay = 270 x 2/5 = 108

Solution 77:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q77-ii

Solution 78:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q78

Solution 79:
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Calculation of new share : 1- 1/5 = 4/5 Remaining share
New share of A = 2/3 x 4/5 = 8/15
New share of B = 1/3 x 4/5 = 4/15
New share of C = 1/5 x 3/3 = 3/15
Deficiency 10,000 will be borne by A.

Solution 80:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q80

Amount due as manager: Salary 10,000 + Commission 4,000 (84,000 x 5/105) = 14,000
Amount due as partner (profit) = 94,000 x 1/5 = 18,800
Deficiency = 18,800 – 14,000 = 4,800
Actual profit after deducting salary and commission was Rs.80,000 i.e.
94,000 – 10,000 Salary – 4,000 Commission = 80,000
Profit to X = 80,000 x ¾ = 60,000 – 4,800 Deficiency = 55,200
Profit to Y = 80,000 x ¼ = 20,000

Solution 81:
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Solution 82:
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Solution 83:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q83-i

Amount payable as partner = 54,300 x 1/6 = 9,050
Deficiency = 9,050 – 8,300 = 750
Actual profit = 54,300 – Salary – Commission = 46,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q83-ii

Solution 84:
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Solution 85:
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Solution 86:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q86-ii

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q86-iii

Solution 87:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q87

Solution 88:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-q88

Brilliant Problems

Solution 1:
Manager’s Commission = 1,10,000 x 10/110 = 10,000
A’s Commission = 1,10,000 – 10,000 = 1,00,000 x 12/100 = 12,000
B’s Commission = 1,00,000 – 12,000 = 88,000 x 10/110 = 8,000

Solution 2:
C’s Capital A/c Dr. 17,000
To A’s Capital A/c 9,000
To B’s Capital A/c 8,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q2-i

Note for teachers: Partners have directly divided profit Rs.1,05,000 in ratio of 1:1:1 i.e. A Rs.35,000; B Rs.35,000 and C Rs.35,000.
But they should have prepared P/L Appropriation Account to provide salary, commission and guaranteed amount etc.
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q2-ii

Note : Deficiency of Rs.6,000 borne by A and C in 3:1 ratio.

Solution 3:
X’s Current A/c Dr. 750
To Y’s Current A/c 150
To Z’s Current A/c 600
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q3

Solution 4:
C’s Capital A/c Dr. 5,000
To A’s Capital A/c 2,600
To B’s Capital A/c 2,400
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q4-i

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q4-ii

Solution 5:
Nisha’s Capital A/c Dr. 55,000
To Preeti’s Capital A/c 55,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q5

Solution 6:
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Solution 7:
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Solution 8:
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Note: No partnership deed so do not calculate interest on drawings and profits will be shared equally.

Solution 9:
E’s Current A/c Dr. 22,200
To F’s Current A/c 1,200
To G’s Current A/c 21,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q9

Solution 10:
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Solution 11:
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Note: No partnership deed, so no salary to partners.

Solution 12:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q12-i

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q12-ii

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q12-iii

Solution 13:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q13

Working Note: Interest on capital to Bunty Rs.36,000 and his guaranteed share is 82,000 (including interest on capital). His actual share: 36,000 (interest on capital) + 44,000 profit = 80,000 Deficiency = 82,000 – 80,000 = 2,000

Solution 14:
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General Reserve = 1,84,500 + 600 + 450 – 6,600 – 4,800 = 1,74,150 x 10/100 = 17,415

Solution 15:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q15

Note : No interest on capital to Z because his capital account is showing negative balance.
X’s Commission = Profit 99,000 x 10/100 = 9,900
Y’s Commission = 99,000 – 9,900 = 89,100 x 10/100 = 8,910

Solution 16:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q16-ii

Solution 17:
Z’s Capital A/c Dr. 135
To X’s Capital A/c 120
To Y’s Capital A/c 15
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q17

Solution 18:
SK’s Capital A/c Dr. 48
To VK’s Capital A/c 48
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q18

Solution 19:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q19

Solution 20:
X’s Capital A/c Dr. 3,000
Y’s Capital A/c Dr. 16,500
To Z’s Capital A/c 19,500
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q20-i

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q20-ii

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q20-iii

Solution 21:
A’s Capital A/c Dr.8,500
To B’s Capital A/c 4,500
To C’s Capital A/c 4,500
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q21

Solution 22:
(a) 12,000 x 5/100 = 600
(b) 24,000 x 6/100 = 1,440
(c) 20,000 x 6/100 x 6/12 = 600
(d) 10,000 x 6/100 = 600

Solution 23:
(a) 45,000 x 8/100 x 5/12 = 1,500
(b) 18,000 x 8/100 x 4/12 = 480
(c) 9,000 x 8/100 x 4.5/12 = 270
(d) 6,000 x 8/100 x 3.5/12 = 140
(e) 12,000 x 8/100 x 7/12 = 560

Solution 24:
RK’s Capital A/c Dr. 5,000
RS’s Capital A/c Dr. 5,000
To RT’s Capital A/c 10,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q24

Solution 25:
Rajesh’s Current A/c Dr. 264
To Sunil’s Current A/c 264
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q25

Solution 26:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q26

Solution 27:
Candy’s Current A/c Dr. 18,000
To Ashok’s Current A/c 16,000
To Brijesh’s Current A/c 2,000
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q27

Solution 28:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q28-i

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q28-ii

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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q28-iii

Solution 29:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q29

Note : Commission = 13,00,000 x 30/130 = 3,00,000
Ram = 3,00,000 x 4/10 = 1,20,000
Shyam = 3,00,000 x 3/10 = 90,000
Mohan = 3,00,000 x 2/10 = 60,000
Sohan = 3,00,000 x 1/10 = 30,000

Solution 30:
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Profit to AK = 3,60,000 x 7/12 = 2,10,000 – 3,000 to DK = 2,07,000
Profit to BK = 3,60,000 x 3/12 = 90,000
Profit to DK = 3,60,000 x 2/12 = 60,000 + 3,000 from AK
DK’s share of profit including interest should be = 75,000
Profit 60,000 + 12,000 (interest) = 72,000; Deficiency 3,000 met by AK.

Solution 31:
Vinod’s Current A/c Dr. 600
To Suraj’s Current A/c 600
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q31

Solution 32:
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Working Note: Distribution of Profit :
Vandana = 9,00,000 x 1/8 = 1,12,500
Vikas = 9,00,000 profit – Vandana’s share 1,12,500 = 7,87,500 x 3/5 = 4,72,500
Vivek = 9,00,000 profit – Vandana’s share 1,12,500 = 7,87,500 x 2/5 = 2,92,500
Deficiency in Vandana’s Share = 1,50,000 (guaranteed amount) – 1,12,500 = 37,500
Deficiency cover up by:
Vikas = 37,500 x 2/5 = 15,000
Vivek = 37,500 x 3/5 = 22,500

Solution 33:
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ultimate-book-accountancy-class-12-part-solutions-basics-partnership-bp-q33

The post Ultimate Book of Accountancy Class 12 Part A Solutions – Basics of Partnership appeared first on Learn CBSE.

Ultimate Book of Accountancy Class 12 Part A Solutions – Reconstitution of Partnership

Ultimate Book of Accountancy Class 12 Part A Solutions – Reconstitution of Partnership

You can buy this text book from here: http://amzn.to/2s2YSxN

Practical Problems

Solution 1:
Old Share = 1 : 1 : 1
New Share = 5 : 3 : 2
Calculation of Sacrifice/Gain Share = Old share – new share
X = 1/3 – 5/10 = 5/30 Gain
Y = 1/3 – 3/10 = 1/30 Sacrifice
Z = 1/3 – 2/10 = 4/30 Sacrifice

Solution 2:
Old Share = 3 : 2
New Share = 5 : 3
Calculation of Sacrifice/Gain Share = Old share – new share
A = 3/5 – 5/8 = 1/40 Gain
B = 2/5 – 3/8 = 1/40 Sacrifice

Solution 3:
Old Share = 4 : 3 : 2
New Share = 1 : 1 : 1
Calculation of Sacrifice/Gain Share = Old share – new share
A =4/9 – 1/3 = 1/9 sacrifice
B = 3/9 – 1/3 = No sacrifice no gain
C = 2/9 -1/3 = 1/9 Gain

Solution 4:
Old Share = 3 : 2
New Share = 1 : 1
Calculation of Sacrifice/Gain Share = Old share – new share
A = 3/5 – A = 1/10 sacrifice
B = 2/5 – A = 1/10 Gain

B’s Capital A/c Dr. 3,000
To A’s Capital A/c   3,000

Share of B = 30,000 x 1/10 = 3,000 and A’s Share = 30,000 x 1/10 = 3,000

Solution 5:
Old Share = 5 : 3 : 2
New Share = 1 : 1 : 1
Calculation of Sacrifice/Gain Share = Old share – new share
Gupta = 5/10 – 1/3 = 5/30 sacrifice
Verma = 3/10 – 1/3 = 1/30 Gain
Sharma = 2/10 – 1/3 = 4/30 Gain

Verma’s Capital A/c 4,000
Sharma’s Capital A/c 16,000
To Gupta’s Capital   A/c 20,000

Solution 6:
Old Share = 4/9 : 1/3 : 2/9 OR 4 : 3 : 2
New Share = 1 : 1 : 1
Calculation of Sacrifice/Gain Share = Old share – new share
A = 4/9 – 1/3 = 1/9 sacrifice
B = 3/9 -1/3 = No sacrifice no gain
C = 2/9 -1/3 = 1/9 Gain

C’s Capital A/c Dr. 1,200
To A’s Capital A/c   1,200

Solution 7:
Old Share = 3 : 2 : 1
New Share = 5 : 3 : 2
Calculation of Sacrifice/Gain Share = Old share – new share
Akshita = 3/6 – 5/10 = No sacrifice no gain
Bakshi = 2/6 – 3/10 = 1/30 sacrifice
Chandra = 1/6 – 2/10 = 1/30 Gain

Chanda’s Capital A/c Dr. 2,000
To Bakshi   2,000
Note: 60,000 x 1/30 = 2,000

Solution 8:
Old Share = 1 : 1
New Share = 5 : 3
Calculation of Sacrifice/Gain Share = Old share – new share
Lal = ½ – 5/8 = 1/8 gain
Bal = ½ – 3/8 = 1/8 sacrifice

Lal’s Capital A/c Dr. 500
To Bal’s Capital A/c   500

Solution 9:
Old Share = 3 : 3 : 2
New Share = 1 : 1 : 1
Calculation of Sacrifice/Gain Share = Old share – new share
GK = 3/8 – 1/3 = 1/24 sacrifice
SK = 3/8 – 1/3 = 1/24 sacrifice
MK = 2/8 – 1/3 = 2/24 Gain

Goodwill = 3,000 + 3,600 + 2,400 = 9,000/3 = 3,000 x 2 = 6,000

MK’s Capital A/c Dr. 500
To GK’s Capital A/c   250
To SK’s Capital A/c   250

Solution 10:
Old Share = 5 : 3
New Share = 1 : 1
Calculation of Sacrifice/Gain Share = Old share – new share
Ramdev = 5/8 – 1/2 = 1/8 sacrifice
Shyamdev = 3/8 – 1/2 = 1/8 gain

P/L A/c Dr. 2,000
To Ramdev   1,250
To Shyamdev   750

General Reserve Dr. 1,000
To Ramdev   625
To Shyamdev   375

Solution 11:
Old Share = 1 : 1 : 1
New Share = 3 : 2 : 1
Calculation of Sacrifice/Gain Share = Old share – new share
Lagan = 1/3 – 3/6 = 1/6 gain
Magan = 1/3 – 2/6 = No sacrifice no gain
Chagan = 1/3 – 1/6 = 1/6 sacrifice

When they distribute General Reserve
General Reserve Dr. 21,000
To Lagan 7,000
To Magan 7,000
To Chagan 7,000

When they do not distribute General Reserve
Lagan Dr. 3,500
To Chagan 3,500

Lagan = 21,000 x 1/6 = 3,500
Chagan = 21,000 x 1/6 = 3,500

Solution 12:
Old Share = 5 : 4 : 2
New Share = 4 : 3 : 2
Calculation of Sacrifice/Gain Share = Old share – new share
SK = 5/11 – 4/9 = 1/99 sacrifice
VK = 4/11 – 3/9 = 3/99 sacrifice
MK = 2/11 – 2/9 = 4/99 Gain
34,900 – 5,200 = 29,700

MK’s Gain = 29,700 x 4/99 = 1,200
SK’s Sacrifice = 29,700 x 1/99 = 300
VK’s Sacrifice = 29,700 x 3/99 = 900

MK Dr. 1,200 To SK 300
To VK 900

Solution 13:
Old Share = 3 : 2
New Share = 1 : 1 given
Calculation of Sacrifice/Gain Share = Old share – new share
Ram = 3/5 – A = 1/10 sacrifice
Rahim = 2/5 – A = 1/10 gain
Increase in fixed assets = 5,000
Increase in debtors = 250
Total = 5,250

Rahim’s Capital A/c Dr. 525
To Ram’s Capital A/c 525

Solution 14:
Old Share = 3 : 2
New Share = 1 : 1
Calculation of Sacrifice/Gain Share = Old share – new share
A = 3/5 – A = 1/10 sacrifice
B = 2/5 – A = 1/10 gain
Revaluation Account
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Solution 15:
Old Share = 2:2:1
New Share = 1:1:1
MM’s Sacrifice = 2/5 – 1/3 = 1/15
SS’s Sacrifice = 2/5 – 1/3 = 1/15
JJ’s Gain = 1/5 – 1/3 = 2/15

JJ’s Capital A/c Dr. 400
To MM’s Capital A/c 200
To SS’s Capital A/c 200
(Being goodwill adjusted)

Fixed Assets increase = 7,500
Creditors decrease = 250
Total = 7,750

JJ’s Capital A/c Dr. 1,033
To MM’s Capital A/c 517
To SS’s Capital A/c 576

Retained Earnings Dr. 2,000
To MM 800
To SS 800
To JJ 400
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ultimate-book-of-accountancy-class-12-part-a-solutions-reconstitution-of-partnership-q15-ii

Solution 16:
Old Share = 3 : 3 : 2
New Share = 1 : 1 : 1
Sacrifice/Gain
A = 3/8 – 1/3 = 1/24 sacrifice
B = 3/8 – 1/3 = 1/24 sacrifice
C = 2/8 – 1/3 = 2/24 gain

C’s Capital A/c Dr. 3,750
To A’s Capital A/c 1,875
To B’s Capital A/c 1,875
(Being goodwill adjusted)

Goodwill = 10,000 x 4.5 = 45,000 x 2/24 = 3,750
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Total Capital = 74,375 + 74,375 + 46,250 = 1,95,000 (now adjusted it in 1 : 1 : 1)
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ultimate-book-of-accountancy-class-12-part-a-solutions-reconstitution-of-partnership-q16-iv

Brilliant Problems

Solution 1:
Profit is given for 5 years but we need profit of 4 years only.
2,00,000 – 1,00,000 + 3,00,000 + 5,00,000 = 9,00,000
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Ram Dr. 1,50,000
To Shyam   30,000
To Mohan    1,20,000

Solution 2:
Analytical Table
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ultimate-book-of-accountancy-class-12-part-a-solutions-reconstitution-of-partnership-bp-q2

Vinay Dr. 1,35,000
To Suraj    1,35,000

Solution 3:
Already distributed = 39,000 + 54,000 – 18,000 + 75,000 = 1,50,000
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ultimate-book-of-accountancy-class-12-part-a-solutions-reconstitution-of-partnership-bp-q3

C’s Capital A/c Dr. 16,000
To A   10,000
To B   6,000

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Ultimate Book of Accountancy Class 12 Part A Solutions – Admission of Partner

Ultimate Book of Accountancy Class 12 Part A Solutions – Admission of Partner

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Practical Problems

Solution 1:
Let the profit be = 1
Share of new partner = 1/10
Remaining profit = 1 – 1/10 = 9/10
William new share = 3/4 x 9/10 = 27/40
Andrew = 1/4 x 9/10 = 9/40
Colin = 1/10 x 4/4 = 4/40

Solution 2:
Green’s Share = 1/10
He acquires this share equally from Black & White i.e. 1/10 x A = 1/20
Black’s new share = 5/9 – 1/20 = 91/180
White’s new share = 4/9 – 1/20 = 71/180
Green’s new share = 1/10 x 18/18 = 18/180

Solution 3:
Old share = 2:1
New share = 3:1:1
Sacrifice of Keyboard = 2/3 – 3/5 = 1/15
Monitor = 1/3 – 1/5 = 2/15

Solution 4:
Let the profit be = 1
Share of new partner = 1/5
Remaining profit = 1 – 1/5 = 4/5
New share of English = 2/3 x 4/5 = 8/15
French = 1/3 x 4/5 = 4/15
History = 1/5 x 3/3 = 3/15
Sacrifice of English = 3/5 – 8/15 = 1/15
French = 2/5 – 4/15 = 2/15

Solution 5:
C’s share 1/3rd which will be sacrificed by A and B in 1:2
1/3 x 1/3 = 1/9 and 1/3 x 2/3 = 2/9
A’s new share = 1/3 – 1/9 = 2/9
B’s new share = 2/3 – 2/9 = 4/9
C’s Share = 1/3 x 3/3 = 3/9

Solution 6:
L’s new share = 7/10 – 2/7 = 29/70
M’s new share = 3/10 – 1/7 = 11/70
N’s new share = 3/7 x 10/10 = 30/70

Solution 7:
A’s new share = 3/5 x A = 3/20 3/5 – 3/20 = 9/20
B’s new share = 2/5 x 1/5 = 2/25 2/5 – 2/25 = 8/25
C’s new share = 3/20 x 2/25 = 23/100 9/20 : 8/25 : 23/100 OR 45:32:23

Solution 8:
Vinod’s new share = 3/5 x 1/3 = 3/15
3/5 – 3/15 = 6/15
Shubh’s new share = 2/5 x 1/4 = 2/20
2/5 – 2/20 = 6/20
Swami’s share = 3/15 (given by Vinod)
Chandel’s Share = 2/20 (given by Shubh)
New share = 24:18:12:6 OR 4:3:2:1

Solution 9:
Let the profit be = 1
Share of new partner = 1/4
Remaining profit = 1 – 1/4 = 3/4
A’s new share = 1/2 x 3/4 = 3/8
B’s new share = 1/2 x 3/4 = 3/8
C’s new share = 1/4 x 2/2 = 2/8
A’s Sacrifice = 3/5 – 3/8 = 9/40
B’s Sacrifice = 2/5 – 3/8 = 1/40

Solution 10:
X’s new share = 7/12 – 1/12 = 6/12
Y’s new share = 5/12 – 1/6 = 3/12
Z’s share = 1/12 + 1/6 = 3/12
New Share = 2:1:1
Sacrificing Share of X = 7/12 – 2/4 = 1/12
Y = 5/12 – 1/4 = 2/12

Solution 11:
Old share of A and C = 1:1
New Share = 5:4:3
A’s Sacrifice = 1/2 – 5/12 = 1/12
C’s Sacrifice = 1/2 – 4/12 = 2/12

Solution 12:
Let the profit be = 1
Share of Vinod and Sunita = 1/6 + 1/6 = 1/3
Remaining profit = 1 – 1/3 = 2/3
Simran’s new share = 3/5 x 2/3 = 6/15
Kuku’s new share = 2/5 x 2/3 = 4/15
New share = 6/15:4/15:1/6:1/6 OR 12:8:5:5

Solution 13:
A’s new share = 4/5 x 1/4 = 4/20
4/5 – 4/20 = 12/20
B’s new share = 1/5 x 1/2 = 1/10
1/5 – 1/10 = 1/10
C’s share = 4/20 + 1/10 = 6/20
12/20: 1/10: 6/20 OR 12:2:6 OR 6:1:3
Sacrifice of A = 4/5 – 6/10 = 2/10 and B = 1/5 – 1/10 = 1/10

Solution 14:
Old Share 7 : 3
A’s new share = 7/10 – 1/7 = 39/70
B’s new share = 3/10 x 1/3 = 1/10
Now 3/10 – 1/10 = 2/10
C’s share = 1/7 + 1/10 = 17/70
A = 39/70 B 2/10 i.e. 14/70 just to make the denominator same C 17/70
Sacrifice = A 7/10 – 39/70 = 10/70
B 3/10 – 14/70 = 7/70

Solution 15:
Old Share = 2:2 i.e. 1:1
L’s new share = 1/2 – 1/4 = 1/4
M’s new share = 1/2 – 1/8 = 3/8
N’s share = 1/4 + 1/8 = 3/8
New Share = 2 : 3 : 3
Note : L’s share is 1/4 but it is converted to 2/8 just to make the denominator same

Solution 16:
X’s new share = 3/5 x 1/2 = 3/10
= 3/5 – 3/10 = 3/10
Y’s new share = 2/5 x 1/2 = 2/10
= 2/5 – 2/10 = 2/10
Z’s new share = 3/10 + 2/10 = 5/10
X’s Sacrifice = 3/5 – 3/10 = 3/10
Y’s Sacrifice = 2/5 – 2/10 = 2/10

Solution 17:
Mohan’s new share = 5/9 – 1/9 = 4/9
Rohan 4/9 Sohan 1/9 New share = 4:4:1

Solution 18:
A’s new share = 5/8 x 1/20 = 1/32
5/8 – 1/32 = 19/32
B’s new share = 3/8 x 1/24 = 1/64
3/8 – 1/64 = 23/64
C’s share = 1/32 + 1/64 = 3/64
New share = 19/32 : 23:64: 3/64 OR 38: 23:3

Solution 19:
Old share = 4:1
New share = 1:1:1 given
Edwin’s Sacrifice = 4/5 – 1/3 = 7/15
Charles’s Gain = 1/5 – 1/3 = 2/15

Solution 20:
E’s Share 20% OR 1/5
Let the profit be = 1
Share of new partner = 1/5
Remaining profit = 1 – 1/5 = 4/5
A’s new share = 3/10 x 4/5 = 12/50
B’s new share = 4/10 x 4/5 = 16/50
C’s new share = 2/10 x 4/5 = 8/50
D’s new share = 1/10 x 4/5 = 4/50
E’s new share = 1/5 x 10/10 = 10/50
New share = 12:16:8:4:10 OR 6:8:4:2:5

Solution 21:
Let the profit be = 1
Share of new partner = 1/4
Remaining profit = 1 – 1/4 = 3/4
A’s new share = 3/5 x 3/4 = 9/20
B’s new share = 2/5 x 3/4 = 6/20
C = 1/4 x 5/5 = 5/20
New Share = 9:6:5
Now D enters and his share is 1/5
Let the profit be = 1
Share of new partner = 1/5
Remaining profit = 1 – 1/5 = 4/5
New share of A = 9/20 x 4/5 = 36/100
B = 6/20 x 4/5 = 24/100
C = 5/20 x 4/5 = 20/100
D = 1/5 x 20/20 = 20/100
New Share = 36:24:20:20 OR 9:6:5:5

Solution 22:
Anita’s new share = 9/14 – 1/12 = 47/84
Tina’s new share = 5/14 – 1/24 = 53/168
Chetan’s share = 1/8 x 21/21 = 21/168
New Share = 47/84: 53:168: 21:168 OR 94:53:21

Solution 23:
A’s new share = 3/5 x 1/6 = 3/30
3/5 – 3/30 = 15/30
B’s new share = 2/5 x A = 2/20
2/5 – 2/20 = 6/20
C’s Share = 3/30 + 2/20 = 12/60
New Share = 15/30: 6/20: 12:60 OR 30:18:12 OR 5:3:2
Sacrificing Share: A = 3/5 – 5/10 = 1/10
B = 2/5 – 3/10 = 1/10

Solution 24:
Let the profit be = 1
Share of new partner = 1/6
Remaining profit = 1 – 1/6 = 5/6
Simon’s new share = 5/8 x 5/6 = 25/48
Murray’s new share = 3/8 x 5/6 = 15/48
George’s Share = 1/6 x 8/8 = 8/48
No journal entries for premium paid privately.

Solution 25:
No entry for premium paid privately

Blair Dr. 4,000
Brent Dr. 3,000
To Goodwill 7,000
(Being existing goodwill written off)

Profit will be distributed by the old partners in 4:3.
P/L A/c Dr. 21,000
To Blair 12,000
To Brent 9,000

Solution 26:
Let the profit be = 1
Share of new partner = 1/4
Remaining profit = 1 – 1/4 = 3/4
Paul’s new share = 3/5 x 3/4 = 9/20
Peter’s new share = 2/5 x 3/4 = 6/20
Roy = 1/4 x 5/5 = 5/20
Sacrificing Ratio = Paul = 3/5 – 9/20 = 3/20 and Peter = 2/5 – 6/20 = 2/20

Cash A/c Dr. 20,000
To Premium for goodwill 20,000
(Being premium brought by new partner)

Premium for goodwill Dr. 20,000
To Paul 12,000
To Peter 8,000
(Being premium adjusted)

Solution 27:
Old Share = 7:3
New Share = 5:3:2
Sacrificing share = Ronald = 7/10 – 5/10 = 2/10
Sam = 3/10 – 3/10 = Nil

Cash A/c Dr. 20,000
To Premium for goodwill 20,000
(Being premium brought by new partner)

Premium for goodwill Dr. 20,000
To Ronald 20,000
(Being premium adjusted)

Solution 28:
X’s new share = 5/9 – 2/9 = 3/9
Y’s new share = 4/9 – 1/9 = 3/9
Z’s share = 1/3 x 3/3 = 3/9
Sacrifice = X = 5/9 – 1/3 = 2/9
Y = 4/9 – 1/3 = 1/9

Cash A/c Dr. 1,500
To Premium for goodwill 1,500
(Being premium brought by new partner)

Premium for goodwill Dr. 1,500
To X 1,000
To Y 500
(Being premium adjusted)

Solution 29:
Old share = 5:3:2
New share = 3:2:2:1
Sacrifice/Gain
A = 5/10 – 3/8 = 5/40
B = 3/10 – 2/8 = 2/40
C = 2/10 – 2/8 = 2/40

Bank A/c Dr. 2,05,000
To D’s Capital 1,75,000
To Premium for goodwill 30,000
(Being premium and capital brought by new partner)

Premium for goodwill Dr. 30,000
C’s Capital A/c 12,000 (2,40,000 2/40)
To A’s Capital A/c 30,000
To B’s Capital A/c 12,000
(Being premium adjusted)

Solution 30:
Bat’s new share = 3/5 x 1/2 = 3/10
3/5 – 3/10 = 3/10
Ball’s new share = 2/5 x 1/2 = 2/10
2/5 – 2/10 = 2/10
Helmet’s new share = 3/10 + 2/10 =5/10
Sacrifice = Bat = 3/5 – 3/10 = 3/10 and Ball = 2/5 – 2/10 = 2/10

Cash A/c Dr. 30,000
To Premium for goodwill 30,000
(Being premium brought by new partner)

Premium for goodwill Dr. 30,000
To Bat 18,000
To Ball 12,000
(Being premium adjusted)

Solution 31:
Let the profit be = 1
New partner’s share = 1/4
Remaining share = 1 – 1/4 = 3/4
Albert’s new share = 1/2 x 3/4 = 3/8
Barry’s new share = 1/2 x 3/4 = 3/8
Craig = 1/4 x 2/2 = 2/8
Sacrifice:
Albert = 3/5 – 3/8 = 9/40
Barry = 2/5 – 3/8 = 1/40

Cash A/c Dr. 40,000
To Premium for goodwill 40,000
(Being premium brought by new partner)

Premium for goodwill Dr. 40,000
To Albert 36,000
To Barry 4,000
(Being premium adjusted)

Solution 32:
Old Share = 3:2
New Share = 4:3:1
A’s Sacrifice = 3/5 – 4/8 = 4/40
B’s Sacrifice = 2/5 – 3/8 = 1/40

Cash A/c Dr. 4,000
To Premium for goodwill 4,000
(Being premium brought by new partner)

Premium for goodwill Dr. 4,000
To A’s Current A/c 3,200
To B’s Current A/c 800
(Being premium adjusted)

Solution 33:
Old Share = 3:2:1
New Share = 2:2:2:1
Sacrifice/Gain
Mark = 3/6 – 2/7 = 9/42
Murray = 2/6 – 2/7 = 2/42
Fox = 1/6 – 2/7 = 5/42 Gain

Bank/Cash A/c Dr. 6,90,000
To Hero’s Capital 6,00,000
To Premium 90,000
(Being capital and premium brought by new partner)

Premium Dr. 90,000 Fox’s Capital A/c 75,000
To Mark’s Capital A/c 1,35,000
To Murray’s Capital A/c 30,000

90,000 + 75,000 in 9:2

Total goodwill = 90,000 x 7/1 = 6,30,000
Fox’s Share = 6,30,000 x 5/42 = 75,000

Solution 34:
Old Share = 5:2
New Share = 2:3:3
Sacrifice/Gain
A = 5/7 – 2/8 = 26/56
B = 2/7 – 3/8 = 5/56

Bank/Cash A/c Dr. 5,10,000
To Premium 2,10,000
To C’s Capital A/c 3,00,000

Premium Dr. 2,10,000
B’s Capital A/c 50,000
To A’s Capital A/c 2,60,000

Total capital of firm = 2,10,000 x 8/3 = 5,60,000
B’s share = 5,60,000 x 5/56 = 50,000

Solution 35:
Old Share = 3:1
New Share = 1:1:1
Sacrifice/Gain
A = 3/4 – 1/3 = 5/12
B = 1/4 – 1/3 = 1/12 Gain

Bank/Cash A/c Dr. 1,60,000
To C’s Capital A/c 1,00,000
To Premium 60,000

Premium Dr. 60,000
B’s Capital A/c 15,000
To A’s Capital A/c 75,000

Total goodwill of the firm = 60,000 x 3/1 = 1,80,000
B’s Share = 1,80,000 x 1/12 = 15,000

Solution 36:
Old Share = 3:2
New Share = 5:3:2
Sacrifice:
A = 3/5 – 5/10 = 1/10
B = 2/5 – 3/10 = 1/10

Bank/Cash A/c Dr. 40,000
To C’s Capital A/c 30,000
To Premium 10,000

Premium Dr. 10,000
To A’s Capital A/c 5,000
To B’s Capital A/c 5,000

A’s Capital A/c Dr. 1,800
B’s Capital A/c Dr. 1,200
To Goodwill 3,000

Solution 37:
Old Share = 3:2
C’s share 3/10 which he acquires in 2:1
i.e. 3/10 x 2/3 = 6/30
3/10 x1/3 = 3/30
A’s new share = 3/5 – 6/30 = 12/30
B’s new share = 2/5 – 3/30 = 9/30
C’s Share = 3/10 x 3/3 = 9/30
New Share = 12:9:9 OR 4:3:3

Sacrifice = A = 3/5 – 4/10 = 2/10
B = 2/5 – 3/10 = 1/10

Bank/Cash A/c Dr. 3,000
To Premium fog goodwill 3,000

Premium Dr. 3,000
To A’s Capital A/c 2,000
To B’s Capital A/c 1,000

A’s Capital A/c Dr. 1,200
B’s Capital A/c Dr. 8000
To Goodwill 2,000

Solution 38:
Old Share = 3:2
New Share = 3:3:2
Sacrifice = A = 3/5 – 3/8 = 9/40
B = 2/5 – 3/8 = 1/40

Bank/Cash A/c Dr. 1,000
To Premium 1,000

Premium Dr. 1,000
To A’s Capital A/c 900
To B’s Capital A/c 100

A’s Capital A/c Dr. 600
B’s Capital A/c Dr. 400
To Goodwill 1,000

Solution 39:
Old Share = 3:1
New Share = 1:1:1
Sacrifice/Gain
Rich = 3/4 – 1/3 = 5/12 sacrifice
Poor = 1/4 – 1/3 = 1/12 Gain

Cash A/c Dr. 25,000
Furniture Dr. 45,000
Machinery Dr. 50,000
To Premium 1,20,000

Premium Dr. 1,20,000
Poor’s Capital A/c 30,000
To Rich 1,50,000

Poor’s share = 1,20,000 x 3/1 = 3,60,000 x 1/12 = 30,000

Solution 40:
Old Share = 3:2
New Share = 5:5:3
Sacrifice/Gain
Koyal = 3/5 – 5/13 = 14/65
Goyal = 2/5 – 5/13 = 1/65

Stock Dr. 20,000
Debtors Dr. 30,000
Land Dr. 50,000
Plant & Mach. Dr. 30,000
To Premium 60,000
To Royal’s Capital A/c 70,000

Premium Dr. 60,000
To Koyal 56,000
To Goyal 4,000

Solution 41:
David’s share = 3/13
Remaining Share = 1 – 3/13 = 10/13
Jack’s new share = 3/5 x 10/13 = 30/65
James new share = 2/5 x 10/13 = 20/65

Furniture 45,000
Machinery 35,000
Stock 30,000
Debtors 20,000
To David’s Capital A/c 70,000
To Premium 60,000

Premium Dr. 60,000
To Jack 36,000
To James 24,000

Solution 42:
Let the profit be = 1
New partner’s share = 1/10
Remaining profit = 1 – 1/10 = 9/10
Tom’s new share = 3/4 x 9/10 = 27/40
Josh’s new share = 1/4 x 9/10 = 9/40
Max’s new share = 1/10 x 4/4 = 4/40
Sacrifice = Tom = 3/4 – 27/40 = 3/40
Josh = 1/4 – 9/40 = 1/40

Stock 40,000
Land & Building 3,00,000
Plant & Mach. 2,60,000
To Max’s Capital A/c 4,00,000
To Premium (Bal. fig.) 2,00,000

Premium Dr. 2,00,000
To Tom’s Capital A/c 1,50,000
To Josh’s Capital A/c 50,000

Solution 43:
Old Share = 7:3
New Share = 5:2:3
Sacrifice/Gain
Thomas = 7/10 – 5/10 = 2/10
Daniel = 3/10 – 2/10 = 1/10

Bank/Cash Dr. 12,000
To Premium for goodwill 12,000

Premium Dr. 12,000
To Thomas 8,000
To Daniel 4,000

Thomas 800
Daniel 400
To Cash 1,200

Solution 44:
Old Share = 3:2
New Share = 3:3:2
Sacrificing Ratio = Vinayak = 3/5 – 3/8 = 9/40
Sahee = 2/5 – 3/8 = 1/40

Cash A/c Dr. 20,000
To Premium 20,000

Premium Dr. 20,000
To Vinayak 18,000
To Sahee 2,000

Vinayak Dr. 18,000
Sahee Dr. 2,000
To Cash 20,000

Vinayak Dr. 2,400
Sahee 1,600
To Goodwill 4,000

Solution 45:
Old Share = 7.5
New Share = 1:1:2
Sacrifice = Lagan = 7/12 – 1/4 = 4/12 and Mayank 5/12 – 1/4 = 2/12

Cash A/c Dr. 16,000
To Premium 6,000
To Abhit 10,000

Premium Dr. 6,000
To Lagan 4,000
To Mayank 2,000

Lagan Dr. 4,000
Mayank Dr. 2,000
To Cash 6,000

Solution 46:
A’s new share = 5/8 – 2/10 = 17/40
B’s new share = 3/8 – 1/10 = 11/40
C’s share = 3/10 x 4/4 = 12/40
Sacrifice = A = 5/8 – 17/40 = 8/40 and B = 3/8 – 11/40 = 4/40

Cash A/c Dr. 3,000
To Premium 3,000

Premium 3,000
To A 2,000
To B 1,000

C’s Current A/c 4,800
To A 3,200
To B 1,600

Solution 47:
Let the profit be = 1
New partner’s share = 1/4
Remaining profit = 1 – 1/4 = 3/4
A’s new share = 1/2 x 3/4 = 3/8
B’s new share = 1/2 x 3/4 = 3/8
C’s new share = 1/4 x 2/2 = 2/8
Sacrifice: A = 1/2 – 3/8 = 1/8
B = 1/2 – 3/8 = 1/8

Cash A/c Dr. 1,000
To Premium for goodwill 1,000

Premium for goodwill Dr. 1,000
To A 500
To B 500

Solution 48:
Old Share = 3:2
New Share = 4:3:2 given
Sacrifice: Louis = 3/5 – 4/9 = 7/45
Jack = 2/5 – 3/9 = 3/45

Cash A/c Dr. 45,000
To Billy’s Capital A/c 40,000
To Premium for goodwill 5,000

Premium for goodwill Dr. 5,000
To Louis’s Capital A/c 3,500
To Jack’s Capital A/c 1,500

Billy’s Current A/c Dr. 3,000
To Louis’s Capital A/c 2,100
To Jack’s Capital A/c 900

Solution 49:
Old Share = 5:3
New Share = 4:3:2

Sacrifice: X = 5/8 – 4/9 = 13/72
Y = 3/8 -3/9 = 3/72

Cash A/c Dr. 1,00,000
To Z’s Capital A/c 1,00,000

Z’s Current A/c Dr. 40,000
To X’s Capital A/c 32,500
To Y’s Capital A/c 7,500

X’s Capital A/c Dr. 1,50,000
Y’s Capital A/c Dr. 90,000
To Goodwill 2,40,000

Solution 50:
Let the profit be = 1
New partner’s share = 1/4
Remaining Share = 1 – 1/4 = 3/4
A’s new share = 3/5 x 3/4 = 9/20
B’s new share = 2/5 x 3/4 = 6/20
C’s Share = 1/4 x 5/5 = 5/20

Sacrifice: A = 3/5 – 9/20 = 3/20
B = 2/5 – 6/20 = 2/20

C’s Current A/c Dr. 5,250
To A’s Capital A/c 3,150
To B’s Capital A/c 2,100

Solution 51:
Let the profit be = 1
New partner’s share = 1/5
Remaining Share = 1 – 1/5 = 4/5
Pandey’s new share = 1/4 x 4/5 = 4/20
Vinayak’s new share = 3/4 x 4/5 = 12/20
Raghu’s Share = 1/5 x 4/4 = 4/20
New Share = 1:3:1
Sacrifice: Pandey = 1/4 – 1/5 = 1/20
Vinayak = 3/4 – 12/20 = 3/20

Raghu’s Current A/c Dr. 20,000
To Pandey’s Capital A/c 5,000
To Raghu’s Capital A/c 15,000

Solution 52:
Old Share = 5:3
New Share = 4:3:2
Sacrifice: William = 5/8 – 4/9 = 13/72
Roy = 3/8 – 3/9 = 3/72

Cash A/c Dr. 1,00,000
To Edward’s Capital A/c 1,00,000

Edward’s Current A/c Dr. 40,000
To William’s Capital A/c 32,500
To Roy’s Capital A/c 7,500

William’s Capital A/c Dr. 1,50,000
Roy’s Capital A/c Dr. 90,000
To Goodwill 2,40,000

Solution 53:
X’s new share = 5/9 – 2/9 = 3/9
Y’s new share = 4/9 – 1/9 = 3/9
Z’s share = 1/3 x 3/3 = 3/9
New Share = 1:1:1
Sacrifice: X = 5/9 – 3/9 = 2/9
Y = 4/9 – 3/9 = 1/9

Cash A/c Dr. 2,00,000
To Z’s Capital A/c 2,00,000

Z’s Current A/c Dr. 60,000
To X’s Capital A/c 40,000
To Y’s Capital A/c 20,000

Solution 54:
Let the profit be = 1
New partner’s share = 1/4
Remaining Share = 1 – 1/4 = 3/4
A’s new share = 3/5 x 3/4 = 9/20
B’s new share = 2/5 x 3/4 = 6/20
C’s Share = 1/4 x 5/5 = 5/20

Sacrifice: A = 3/5 – 9/20 = 3/20
B = 2/5 – 6/20 = 2/20

C’s Current A/c Dr. 21,000
To A’s Capital A/c 12,600
To B’s Capital A/c 8,400

Solution 55:
Let the profit be = 1
New partner’s share = 1/5
Remaining Share = 1 – 1/5 = 4/5
A’s new share = 3/5 x 4/5 = 12/25
B’s new share = 2/5 x 4/5 = 8/25
C’s Share = 1/5 x 5/5 = 5/25
Sacrifice: A = 3/5 – 12/25 = 3/25
B = 2/5 – 8/25 = 2/25

Hidden Goodwill = Total Capital 1,80,000 x 5/1 = 9,00,000
Combined Capital = 2,40,000 + 1,50,000 + 1,80,000 = 5,70,000
Goodwill 3,30,000 = Total Capital 9,00,000 – Combined Capital 5,70,000

C’s share = 3,30,000 x 1/5 = 66,000

Cash A/c Dr. 2,46,000
To Premium for goodwill 66,000
To C’s Capital A/c 1,80,000

Premium for goodwill Dr. 66,000
To A 39,600
To B 26,400

Solution 56:
Old Share = 1:1
Bhandari’s Share = 1/5
Let the Profit be = 1
Remaining share = 1 – 1/5 = 4/5
Sushant’s new share = 1/2 x 4/5 = 4/10
Ranauq’s new share = 1/2 x 4/5 = 4/10
Bhandari = 1/5 x 2/2 = 2/10
New Share = 2:2:1
Sacrifice: Sushant = 1/2 – 2/5 = 1/10
Ranauq = 1/4 x 2/5 = 1/10
Sacrifice 1:1

Hidden Goodwill = Total Capital = 16,000 x 5/1 = 80,000
Combined capital = 26,000 + 18,000 + 16,000 = 60,000
Goodwill = 80,000 – 60,000 = 20,000

Bhandari’s share = 20,000 x 1/5= 4,000

Cash A/c Dr. 20,000
To Bhandari’s Capital A/c 16,000
To Premium of goodwill 4,000

Premium for goodwill Dr. 4,000
To Sushant’s Capital A/c 2,000
To Ranauq’s Capital A/c 2,000

Solution 57:
Arjun’s share = 1/5 which he acquires equally from Parag and Digvijay
i.e. 1/5 x 1/2 = 1/10
Parag’s new share = 7/10 – 1/10 = 6/10
Digvijay = 3/10 – 1/10 = 2/10
Arjun = 1/5 x 2/2 = 2/10
New Share = 6:2:2 OR 3:1:1
Sacrifice: Parag = 7/10 – 3/5 = 1/10
Digvijay = 3/10 – 1/5 = 1/10
Sacrifice = 1:1

Hidden Goodwill = Total Capital = 25,000 x 5/1 = 1,25,000
Combined capital = 40,000 + 30,000 + 25,000 = 95,000
Goodwill = 1,25,000 – 95,000 = 30,000 and Arjun’s share 30,000 x 1/5 = 6,000

Cash A/c Dr. 31,000
To Arjun’s Capital A/c 25,000
To Premium for goodwill 6,000

Premium for goodwill Dr. 6,000
To Parag’s Capital A/c 3,000
To Digvijay’s Capital A/c 3,000

Solution 58:
Let the profit be = 1
New partner’s share = A
Remaining profit = 1 -1/4 = 3/4
A’s new share = 1/2 x 3/4 = 3/8
B’s new share = 1/2 x 3/4 = 3/8
C = 1/4 x 2/2 = 2/8
Sacrifice = A = 1/2 – 3/8 = 1/8
B = 1/2 – 3/8 = 1/8

Hidden Goodwill = Total Capital = 1,40,000 x 4/1 = 5,60,000
Combined Capital = 1,20,000 + 2,40,000 + 1,40,000 = 5,00,000
Goodwill = 5,60,000 – 5,00,000 = 60,000 C’s Share = 60,000 x 1/4 = 15,000

Cash A/c Dr. 1,55,000
To Premium for goodwill 15,000
To C’s Capital A/c 1,40,000

Premium for goodwill Dr. 15,000
To A’s Capital A/c 7,500
To B’s Capital A/c 7,500

Solution 59:
General Reserve Dr. 32,000
To Uday’s Capital A/c 20,000
To Bharat’s Capital A/c 12,000

Uday’s Capital A/c 30,000
Bharat’s Capital A/c 18,000
To P/L A/c 48,000

Solution 60:
Old share = 4:1
New Share = 4:2:3
Sacrifice = Ram = 4/5 – 4/9 = 16/25
Mohan = 1/5 – 2/9 = 1/45 Gain

General Reserve Dr. 50,000 (divided in old ratio)
To Ram’s Capital A/c 40,000
To Mohan’s Capital A/c 10,000

Ram’s Capital A/c Dr. 12,800
Mohan’s Capital A/c Dr. 3,200
To P/L A/c 16,000

Cash A/c Dr. 30,000
To Premium for goodwill 30,000

Premium for goodwill Dr. 30,000
Mohan’s Capital A/c 2,000
To Ram’s Capital A/c 32,000

Total goodwill of firm = 30,000 x 3/1 = 90,000

Solution 61:
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Solution 62:
CASE – 1: Hidden Goodwill
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Hidden Goodwill = Total Capital 10,000 x 5/1 = 50,000
Combined capital = 5,000 + 2,500 + 10,000 = 17,500
Goodwill = 50,000 – 17,500 = 32,500
C’s Share = 32,500 x 1/5 = 6,500
CASE – 2
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ultimate-book-accountancy-class-12-part-solutions-admission-partner-q62-ii

Goodwill = Total Capital 50,000 – Combined capital 27,500 = 22,500
C’s Share = 22,500 x 1/5 = 4,500

Solution 63:
Old Share = 5:3
Henry’s share = 1/5
Let the profit be = 1
Remaining share = 1 – 1/5 = 4/5
Anthony’s share = 5/8 x 4/5 = 20/40
Justin’s share = 3/8 – 4/5 = 12/40
Henry = 1/5 x 8/8 = 8/40
New Share = 20:12:8 OR 5:3:2
Sacrifice = Anthony = 5/8 – 5/10 = 5/40
Justin = 3/8 – 3/10 = 3/40
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Solution 64:
New Share = 12:8:5
Sacrifice = 3:2
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Solution 65:
New Share = 9:6:5
Sacrifice = 3:2
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Solution 66:
New Share = 2:1:1
Sacrifice = 2:1
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Solution 67:
New Share = 3:2:1
Sacrifice = 3:2
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Solution 68:
New Ratio = 21:14:5
Sacrifice = 3:2
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Solution 69:
New Ratio 10 : 15 : 25 : 10 i.e. 2 : 3 : 5 : 2
Sacrifice = 2 : 3 : 5
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ultimate-book-accountancy-class-12-part-solutions-admission-partner-q69-iii

Solution 70:
New Ratio = 12 : 8 : 5
Sacrifice = 3:2
Hidden Goodwill = Total Capital = 30,000 x 5/1 = 1,50,000
Combined Capital = 30,000 + 20,000 + 30,000 = 80,000
Hidden Goodwill = 1,50,000 – 80,000 = 70,000
Pulkit’s Share = 70,000 x 1/5 = 14,000
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Solution 71:
New Share = 5:3:2
Sacrifice = 1:1
Total capital 1,00,000 x 5/1 = 5,00,000
Combined capital = 1,47,900 + 73,600 + 1,00,000 = 3,21,500
Hidden Goodwill = 5,00,000 – 3,21,500 = 1,78,500
Paul’s share = 1,78,500 x 2/10 = 35,700
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Solution 72:
CASE – 1
New Share = 4 : 3 : 1
Sacrifice = 4 : 1
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Total capital = 5,000 x 8/1 = 40,000
Combined capital = 8,500 + 6,500 + 5,000 = 20,000
Hidden goodwill = Total capital 40,000 – combined capital 20,000 = 20,000
C’s share = 20,000 x 1/8 = 2,500
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CASE – 2
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Total Capital = 5,000 x 8/1 = 40,000
Combined capital = 10,000 + 7,500 + 5,000 = 22,500
Hidden goodwill = 40,000 – 22,500 = 17,500
C’s share = 17,500 x 1/8 = 2,188

Cash A/c Dr. 7,188
To C’s Capital A/c 5,000
To Premium for goodwill 2,188

Premium for goodwill Dr. 2,188
To A’s Current A/c 1,750
To B’s Current A/c 438
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ultimate-book-accountancy-class-12-part-solutions-admission-partner-q72-vi

Solution 73:
New Share = 3:3:2
Sacrifice = Only M is sacrificing
Goodwill = Super Profit 20,000 – 12,000 = 8,000 x 4 = 32,000
N’s Share = 32,000 x 1/4 = 8,000
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Solution 74:
New Share = 2:1:1
Sacrifice = 2:1
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Solution 75:
New Share = 3:1:1
Sacrifice = 3:1
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Solution 76:
New Share = 3:1:1
Sacrifice = 3:1
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Solution 77:
New Share = 12:8:5
Sacrifice = 3:2
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Solution 78:
New Ratio = 7 : 5 : 3 and Sacrifice = 19 : 5
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Solution 79:
New Share = 3:2:1
Sacrifice = Only A is sacrificing
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Solution 80:
New Ratio = 3 : 3: 2
Sacrifice = 9 : 1
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Solution 81:
New Share = 3:1:1 and Sacrifice = 3:1
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ultimate-book-accountancy-class-12-part-solutions-admission-partner-q81-iv

Solution 82:
New Share = 3:1:1 and Sacrifice 3:1
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Solution 83:
New Share = 6:5:3:2 and sacrifice 6:5:3
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Total capital = 17,700 x 1/8 = 1,41,600
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Solution 84:
New Share = 5:3:2 and sacrifice = 1:1
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Solution 85:
New Share = 9:6:5
Sacrifice = 3:2
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Solution 86:
New Share = 9:6:3:2 and Sacrifice = 3:2:1
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Solution 87:
New share = 9:3:4 and 3:1
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Solution 88:
New share = 9:6:5 and sacrifice = 3:2
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Solution 89:
New Share = 3:2:1 and Sacrifice = 3:2
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Total capital = 20,000 x 6/1 = 1,20,000
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Solution 90:
New Share = 5:3:4 and sacrifice = 5:3
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Total capital = 10,000 x 3/1 = 30,000
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Solution 91:
New Share = 4:3:1 and Sacrifice 4:3
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Total capital = 1,68,000 + 1,26,000 = 2,94,000 x 8/7 = 3,36,000 x 1/8 = 42,000
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Solution 92:
New Share = 6:4:5 and Sacrifice 3:2
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Solution 93:
New Share 3:2:1 and Sacrifice 3:2
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Solution 94:
New share = 9:6:5 and Sacrifice 1:1
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Solution 95:
New share = 3:3:2
Sacrifice = 1:1

Cash A/c Dr. 6,000
To Anna’s Capital A/c 6,000
(Being Capital brought by Anna)

Anna’s Current A/c Dr. 1,500
To Garry’s Capital A/c 750
To Marry’s Capital A/c 750
(Being premium for goodwill adjusted)
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Solution 96:
New Share 12 : 8 : 5 and Sacrifice 3:2
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Solution 97:
New Share 3:3:2 and sacrifice 9:1
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Solution 98:
New Share 9:6:5 and sacrifice 3:2
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Solution 99:
New share = 3:3:2 and sacrifice 9:1
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Solution 100:
New Share = 1:1:1 and sacrifice 4:1
Capital employed = 3,00,000 given
Normal profit = 3,00,000 x 10/100 = 30,000
Super profit = Average Profit – Normal profit
15,000 = 45,000 – 30,000
Goodwill = 15,000 x 2 = 30,000 and Peter’s share = 30,000 x 1/3 = 10,000
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The post Ultimate Book of Accountancy Class 12 Part A Solutions – Admission of Partner appeared first on Learn CBSE.

Ultimate Book of Accountancy Class 12 Part A Solutions – Retirement

Ultimate Book of Accountancy Class 12 Part A Solutions – Retirement

You can buy this text book from here: http://amzn.to/2s2YSxN

Practical Problems

Solution 1:
(a) New share when X retires = 2:1
New share when Y retires = 2:1
New share when Z retires = 1:1

(b) Rohan’s share 5/20 this share is taken by Sandeep and Mukesh equally 5/20 x 1/2 = 5/40 OR 1/8
New share of Sandeep = 7/20 + 1/8 = 19/40
New share of Mukesh = 8/20 + 1/8 = 21/40
New share = 19:21

Solution 2:
Share of Mohan = 7/20 this share is taken by Ramesh and Naresh equally
7/20 x 1/2 = 7/40
New share of Ramesh = 5/20 + 7/40 = 17/40
New share of Naresh = 8/20 + 7/40 = 23/40

Solution 3:
Shivam’s share = 7/20 this share is taken by Satyam and Sundram in 1:2
7/20 x1/3 = 7/60
7/20 x 2/3 = 14/60
Satyam’s new share = 8/20 + 7/60 = 31/60
Sundram’s new share = 5/20 + 14/60 = 29/60

Solution 4:
Great’s share = 6/20 this share is taken by Super and Medium in 1:1
6/20 x 1/2 = 6/40
Super’s share = 7/20 + 6/40 = 20/40
Medium’s share = 7/20 + 6/40 = 20/40
New Share = 1:1

Solution 5:
Old Share 3:2:1
Y’s share = 2/6 this share is taken by X and Y in 3:2
2/6 x 3/5 = 6/30
2/6 x 2/5 = 4/30
X’s new share = 3/6 + 6/30 = 21/30
Z’s new share = 1/6 + 4/30 = 9/30
New share = 7:3

Solution 6:
Old Share = 5:3:2
New share = 6:4
X’s Gain = 6/10 – 5/10 = 1/10
Z’s Gain = 4/10 – 2/10 = 2/10
Gain = 1:2

Solution 7:
Manish’s Gain = 5/8 – 5/10 = 5/40
Naresh’s Gain = 3/8 – 3/10 = 3/40
Gain = 5:3

Solution 8:
Jamwal’s Gain = 1/2 – 15/32 = 1/32
Gupta’s loan = – 9/32 = 7/32
Gain = 1:7

Solution 9:
Old share = 4:1:3
C’s share = 3/8
Surrender to A = 3/8 x 4/9 = 12/72
Surrender to B = 3/8 x 5/9 = 15/72
A’s new share = 4/8 + 12/72 = 48/72
B’s new share = 1/8 + 15/72 = 24/72
48: 24 i.e. 2:1

Solution 10:
Old share = 4:3:2 and new share 5:3
Gain of Verma = 5/8 – 3/9 = 21/72
Gain of Dharma = 3/8 – 2/9 = 11/72

Solution 11:
Old Share = 5:3:2
Gain share = 8,000 : 4,000 = 8:4 OR 2:1 (Gain)

Solution 12:
Old Share = 4:3:2 and new share = 5:3
A’s Gain = 5/8 – 4/9 = 13/72
C’s Gain = 3/8 – 2/9 = 11/72

A’s Capital A/c Dr. 13,000
C’s Capital A/c Dr. 11,000
To B’s Capital A/c   24,000 (72,000 x 3/9)

Solution 13:
Old share = 4:3:2
New Share 4:2 OR 2:1
Azad’s Gain = 2/3 – 4/9 = 2/9
Chandel’s Gain = 1/3 – 2/9 = 1/9
Gain 2:1

Azad’s Capital A/c Dr. 2,000
Chandel’s Capital A/c Dr. 1,000
To Bajaj’s Capital A/c   3,000 (9,000 x 1/3)

Solution 14:
Old Share 3:2:1
New Share = 1:1
Mehra’s Gain = 1/2 – 3/6 = Nil
Prera’s Gain = 1/2 – 1/6 = 2/6

Prera’s Capital A/c Dr. 10,000
To Nehra’s Capital A/c   10,000

Solution 15:
Old Share = 2:2:1:1
New Share = 1:1:1
Ravi’s Gain = 1/3 – 2/6 = Nil
Naresh = 1/3 – 1/6 = 1/6
Yogesh = 1/3 – 1/6 = 1/6

Naresh’s Capital A/c Dr. 30,000
Yogesh’s Capital A/c Dr. 30,000
To Mukesh’s Capital A/c   60,000 (1,80,000 x 2/6)

Solution 16:
Old Share 4:3:1
A retires and his share 4/8 is taken over by B and C in 1:1
4/8 x = 4/16 to B and 4/8 x = 4/16 to C
B’s new share = 3/8 + 4/16 = 10/16
C’s new share = 1/8 + 4/16 = 6/16
Gain 10:6 OR 5:3

B’s Capital A/c Dr. 2,000
C’s Capital A/c Dr. 2,000
To A’s Capital A/c   4,000

Solution 17:
Old Share = 1:1:1 and New Share 3:2
P’s Gain = 3/5 – 1/3 = 4/15
Q’s Gain = 2/5 – 1/3 = 1/15

P’s Capital A/c Dr. 9,600
Q’s Capital A/c Dr. 2,400
To R’s Capital A/c   12,000 (36,000 x 1/3)

Solution 18:
Old Share = 5:3:2 and new share 2:3
Ramesh = 2/5 – 5/10 = 1/10 sacrifice
Suresh = 3/5 – 2/10 = 4/10 Gain

Suresh’s Capital A/c Dr. 48,000
To Ramesh’s Capital A/c   12,000 (1,20,000 x 1/10)
To Naresh’s Capital A/c   36,000 (1,20,000 x 3/10)

Solution 19:
Old Share = 2:1:1 and new share = 4:11
Gain of Tara = 4/15 – 1/5 = 1/15
Ravi = 11/15 – 2/5 = 5/15
Gain 1:5

Tara’s Capital A/c Dr. 6,000
Ravi’s Capital A/c Dr. 30,000
To Chander’s Capital A/c   36,000

Solution 20:
Old Share = 1:2:2 and New share 3:2
Gain of T = 3/5 – 2/5 = 1/5
U = 2/5 – 2/5 = Nil

T’s Capital A/c Dr. 12,000
To S’s Capital A/c   12,000

Solution 21:
Old share = 1:1:1 and New Share = 3:2
Gain of A = 3/5 – 1/3 = 4/15 and B = 2/5 – 1/3 = 1/15

A’s Capital A/c 3,200
B’s Capital A/c 800
To C’s Capital A/c   4,000

A’s Capital A/c 2,000
B’s Capital A/c 2,000
C’s Capital A/c 2,000
To Goodwill   6,000

Solution 22:
Old Share = 5:4:3
B retired and his share 4/12 is taken over by A and C in 1:1 4/12 x 1/2 = 4/24
A’s new share = 5/12 + 4/24 = 14/24
C’s new share = 3/12 + 4/24 = 10/24
New Share = 7:5 and Gain 1:1

Solution 23:
Ratio of 3,000 and 2,0000 is 3:2
Alay’s new share = 3/5 x 3/8 = 9/40
Now 4/8 + 9/40 = 29/40
Chandel’s new share = 2/5 x 3/8 = 6/40
now 1/8 + 6/40 = 11/40
New Share = 29:11

Ajay’s Capital A/c Dr. 3,000
Chandel’s Capital A/c Dr. 2,000
To Bajaj’s Capital A/c   5,000

P/L App. A/c Dr. 8,000
To Ajay’s Capital A/c   5,800
To Chandel’s Capital A/c   2,200

Solution 24:
Old Share = 3:2:1 and new share = 1:2
Gain = X = 1/3 – 3/6 = 1/6
Sacrifice Y = 2/3 – 2/6 = 2/6 OR 1/3 Gain

Y’s Capital A/c Dr. 10,000
To X’s Capital A/c   5,000
To Y’s Capital A/c   5,000

Solution 25:
Old Share = 4:3:2 and new share 5:3
Gain = N = 5/8 – 4/9 = 13/72 and S = 3/8 – 2/9 = 11/72
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Solution 26:
Old Share 3:2:1 and new share 5:1
Gain of Tanya = 5/6 – 3/6 = 2/6 OR 1/3
Bhawna = 1/6 – 1/6 = Nil
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Solution 27:
Old Share = 1/2:1/3:1/6 OR 3:2:1 and new share 3:2 and Gain = 3:7
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Solution 28:
Old Share = 1:1:1 and New Share 1:1 and Gain = 1:1
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Solution 29:
Old Share = 3:2:1
new Share = 3:1 and Gain 3:1
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Solution 30:
Old Share 1/2 : 1/3: 1/6 OR 3:2:1
New Share = 3:2 and Gain 3:2
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Solution 31:
Old Share = 3:2:1
New Share 3:1 and Gain 3:1
Revaluation Account
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Solution 32:
Old Share 2:2:1
New Share 2:1 and Gain 2:1
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Solution 33:
Old Share 3:3:2
New Share 3:2 and Gain 3:2
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Solution 34:
Old Share = 4:3:2
New Share 4:2 OR 2:1 Gain 2:1
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Solution 35:
Old Share 5:3:2 and New Share 3:2 Gain = 1:2
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Solution 36:
Old share = 3:2:1
New Share = 3:1 and Gain share 3:1
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Solution 37:
Old Share = 2:2:1
New share = 5:3 Gain 9:7
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Solution 38:
Old share = 5:3:2
New share 2:1 Gain 5:4
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Solution 39:
Old share 1/2 : 1/3: 1/6 OR 3:2:1
New share 2:1 Gain 2:1
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Solution 40:
Old Share 2:2:1
New Share 2:1 and Gain 2:1
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Solution 41:
Old Share 3:2:1
New Share 2:1 and Gain 2:1
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Solution 42:
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Solution 43:
Old Share 5:3:2
New Share 5:2 Gain 5:2
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Solution 44:
Old Share 3:2:1
New Share 2:1 Gain 2:1
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Solution 45:
Old Share 3:2:1
New Share 3:1 Gain 3:1
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Solution 46:
Old Share 3:2:1
New Share 3:2 and Gain 3:7
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Solution 47:
Old Share = 5:3:2
New share 3:2 Gain 3:2
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Solution 48:
Old Share 5:3:2
New Share 3:2 and Gain 1:2
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Solution 49:
Old Share 1:1:1
New Share 5:3 Gain 7:1
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Solution 50:
Old Share 3:2:1
New share 3:1 and Gain 3:1
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Solution 51:
Old share 1/2 : 3/10: 1/5 OR 5:3:2 Gain 5:1
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Solution 52:
Old Share 3:2:1
New Share 2:1 ……….. only A is Gainer partner and B No sacrifice or gain
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Solution 53:
Old Share 3:2:1
New share 5:4 and Gain 1:2
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Solution 54:
Old Share 5:3:2
New Share 3:2 and Gain 1:2
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Solution 55:
Old Share 1/2: 1/6: 1/3 OR 3:1:2 and Gain 3:1
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Solution 56:
Old share 3:1:1
New Share 3:1 and Gain 3:1
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Solution 57:
Old Share 4:2:2 OR 2:1:1
New Share 2:1 and Gain 2:1
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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-q57-iii

Solution 58:
Old Share 24:18:12 OR 4:3:2
New share 5:3 and Gain 13:11
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Solution 59:
Old share 2:2:1
New share 3:1 Gain 7:1
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Solution 60:
Old share 4:3:2
New share 2:1 Gain 2:1
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Solution 61:
Old Share 4:3:1
Y’s share taken by X and Z in 4:5
X’s Gain = 3/8 x 4/9 = 12/72 and Z = 1/8 x 5/9 = 15/72
X’s new share = 4/8 + 12/72 = 48/72
Z’s new share = 1/8 + 15/72 = 24/72
New Share of X and Z 2:1
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Brilliant Problems

Solution 1:
Old Share 1/2 : 1/3: 1/6 OR 3:2:1
New Share 2:1 and Gain 2:1
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Solution 2:
Old Share 1/2 : 1/3: 1/6 OR 3:2:1
new Share 3:1 and Gain 3:1
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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q2-ii

Solution 3:
Old Share 3:3:2
New Share 3:2 Gain 3:2
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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q3

Solution 4:
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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q4

Solution 5:
Old Share 76:57:38 OR 4:3:2
New Share 3:2 and Gain 3:2
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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q5-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q5-ii

Solution 6:
Old Share = 4:6:4 OR 2:3:2
New Share 3:4 given and Gain only Usha is gainer.
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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q6-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q6-ii

Solution 7:
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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q7

Solution 8:
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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q8-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q8-ii

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ultimate-book-of-accountancy-class-12-part-a-solutions-retirement-bp-q8-iii

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Ultimate Book of Accountancy Class 12 Part A Solutions – Death of Partner

Ultimate Book of Accountancy Class 12 Part A Solutions – Death of Partner

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Practical Problems

Solution 1:
Calculation of Share of Profit of Deceased partner:
28,000 + 32,000 + 36,000 = 96,000/3 = 32,000
32,000 x 6/12 = 16,000 x 1/5 = 3,200

Solution 2:
Percentage of last year’s profit = 5,000/50,000 x 100 = 10%
Three months profit = 15,000 x 10/100 = 1,500
Mukesh’s share = 1,500 x 1/5 = 300

Solution 3:
Percentage of profit last year = 10,000/50,000 x 100 = 20%
Three months Turnover = 25,000 x 3/12 = 6,250
Three months profit = 6,250 x 20/100 = 1,250
Gupta’s share = 1,250 x 1/5 = 250

Solution 4:
Average Profit = 7,000 + 8,000 + 9,000 = 24,000/3 = 8,000
Two months profit = 8,000 x 2/12 = 1,333
Anoop’s Share = 1,333 x 1/3 = 444.44

Solution 5:
Old Share 5:3:2
New Share 1:1 and Gain 2:3
X’s Share of Goodwill = 50,000 x 5/10 = 25,000

Y’s Capital A/c Dr. 10,000
Z’s Capital A/c Dr. 15,000
To X’s Capital A/c 25,000

Solution 6:
Old Share 5:4:3
New Share 3:2 and Gain 11:9
T’s share of goodwill = 2,40,000 x 4/12 = 80,000

Solution 7:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q7

Solution 8:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q8

Note: Share of profit = 30,000 x 2/12 x 2/5 = 2,000
Interest on capital = 80,000 x 12/100 x 2/12 = 1,600
Salary for 2 months = 12,000/12 = 1,000 per month so 1,000 x 2 = 2,000

Solution 9:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q9

Solution 10:
Old Share 4:3:3
New Share 4:3 and Gain 4:3
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q10

Solution 11:
Old Share 6:3:2
new share 3:2 Gain 3:2
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q11

Solution 12:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q12

Solution 13:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q13

Solution 14:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q14

Solution 15:
Old Share 3:2:1
New Share 3:1 and Gain 3:1
Goodwill = 22,000 + 26,000 + 12,000 = 60,000/3 = 20,000 x 3 = 60,000
Bhanu’s share = 60,000 x 2/6 = 20,000
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q15

Share of profit = 12,000 x 3/12 x 2/6 = 1,000

Solution 16:
Goodwill = 40,000 + 40,000 + 30,000 + 40,000 + 50,000 = 2,00,000/5 = 40,000 x 3 = 1,20,000
X’s share of goodwill = 1,20,000 x 3/6 = 60,000
Share of profit = 40,000 + 50,000 = 90,000/2 = 45,000 x 2/12 x 3/6 = 3,750
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q16

Solution 17:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q17

Cash available = 12,000 + 13,250 (investment sold) = 25,250 – 9,450 = 34,700
Share of profit = 1,20,000 x 10/100 x 2/6 = 4,000
Goodwill = 8,200 + 9,000 + 9,800 = 27,000 x 2 = 54,000 – 20% = 43,200
43,200 x 2/6 = 14,400
Interest on capital = 12,000 x 10/100 x 3/12 = 300

Solution 18:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q18

Share of profit = 45,000 x 6/12 x 5/10 = 11,250
Interest on capital = 90,000 x 10/100 x 6/12 = 4,500

Solution 19:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q19

Solution 20:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q20

Solution 21:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q21

Solution 22:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q22

Solution 23:
Old Share 3:2:5
New Share 1:1 given and Gain: only Ram is gainer
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q23-i

First Alternative of preparing Capital Account and Balance Sheet
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q23-ii

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q23-iii

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q23-iv

Second Alternative of preparing Capital Account and Balance Sheet
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q23-v

Solution 24:
Old Share 7:2:1 and New share 3:2 —- Sacrifice of Sonu 1/10 and Tonu’s Gain 3/10
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q24-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q24-ii

First Alternative of preparing Capital Account and Balance Sheet
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q24-iii

Second Alternative of preparing Capital Account and Balance Sheet
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q24-iv

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q24-v

Solution 25:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q25

Solution 26:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q26

Solution 27:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q27-i

Goodwill = Average profit 36,000 x 2.5 = 90,000 x 5/10 = 45,000
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q27-ii

Solution 28:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q28-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q28-ii

Solution 29:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q29-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-q29-ii

Brilliant Problems

Solution 1:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q1-i

Goodwill = Average profit 7,000 x 2.5 = 17,500 x 2/10 = 3,500
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q1-ii

Solution 2:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q2-i

Goodwill = Average profit = 18,000 x 1/3 = 6,000
Manoj’s share of profit = 18,000 x 3/12 x 1/3 = 1,500
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q2-ii

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q2-iii

Solution 3:
Goodwill = Profit of last 4 years 3,20,000 x 3/8 = 1,20,000
Profit credited to Doly in last 4 years = 1,20,000
Half of which is treated as goodwill = 1,20,000 x 1/2 = 60,000
Bad debts recovered (Doly’s share) = 8,000 x 3/8 = 3,000

Solution 4:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q4

Solution 5:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q5

Solution 6:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q6-i

First Alternative of preparing Capital account and balance sheet
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q6-ii

Second Alternative for preparing Partners capital account and Balance Sheet
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q6-iii

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q6-iv

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q6-v

Solution 7:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q7-i

First Alternative to prepare Partners Capital Account and Balance Sheet
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q7-ii

Second Alternative to prepare Partners Capital Account and balance sheet
Sonu is sacrificing 1/10 and Tonu is gainer 3/10
So entry for Monu’s share of profit will not be done with P/L suspense because there is change in new ratio of the continuing partners.
Tonu’s Capital A/c 2,400
To Sonu’s Capital A/c 800
To Monu’s Capital A/c 1,600
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q7-iii

Solution 8:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q8-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q8-ii

Solution 9:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q9-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q9-ii

Solution 10:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q10-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q10-ii

Solution 11:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q11

Solution 12:
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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q12-i

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ultimate-book-of-accountancy-class-12-part-a-solutions-death-of-partner-bp-q12-ii

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Ultimate Book of Accountancy Class 12 Part A Solutions – Dissolution of Partnership Firm

Ultimate Book of Accountancy Class 12 Part A Solutions – Dissolution of Partnership Firm

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Practical Problems

Solution 1:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q1

Solution 2:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q2

Solution 3:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q3

Solution 4:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q4-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q4-ii

Solution 5:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q5

Solution 6:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q6

Solution 7:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q7

Solution 8:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q8-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q8-ii

Solution 9:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q9

Solution 10:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q10

Solution 11:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q11-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q11-ii

Solution 12:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q12-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q12-ii

Solution 13:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q13

Solution 14:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q14-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q14-ii

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q14-iii

Solution 15:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q15-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q15-ii

Solution 16:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q16-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q16-ii

Solution 17:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q17

Solution 18:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q18-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q18-ii

Solution 19:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q19-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q19-ii

Solution 20:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q20

Solution 21:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q21-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q21-ii

Solution 22:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q22-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q22-ii

Solution 23:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q23

Solution 24:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q24-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q24-ii

Solution 25:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q25-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q25-ii

Solution 26:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q26

Solution 27:
Realisation Account
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q27

Solution 28:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q28-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q28-ii

Solution 29:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q29-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q29-ii

Solution 30:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q30

Solution 31:
Realisation Account
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q31-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q31-ii

Solution 32:
Realisation Account
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q32

Solution 33:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q33-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q33-ii

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q33-iii

Solution 34:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q34-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q34-ii

Solution 35:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q35-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-q35-ii

Brilliant Problems

Solution 1:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q1-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q1-ii

Solution 2:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q2-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q2-ii

Solution 3:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q3

Solution 4:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q4

Solution 5:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q5-ii

Solution 6:
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Solution 7:
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Solution 8:
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Solution 9:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q9-i

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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q9-ii

Solution 10:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q10

Solution 11:
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ultimate-book-accountancy-class-12-part-a-solutions-dissolution-partnership-firm-bp-q11

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The post Ultimate Book of Accountancy Class 12 Part A Solutions – Dissolution of Partnership Firm appeared first on Learn CBSE.

RD Sharma Class 7 solutions Chapter-1 Integers Exercise-1.3 Q9

RD Sharma Class 7 solutions Chapter-1 Integers Exercise-1.3 Q10

RD Sharma Class 7 solutions Chapter-1 Integers Exercise-1.3 Q11

RD Sharma Class 9 Solutions Number Systems

RD Sharma Class 9 Solutions Number Systems

Exercise 1.1

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.1 Question 1

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.1 Question 1_1

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.1 Question 3

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.1 Question 4

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.1 Question 4_1

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.1 Question 5

Exercise 1.2

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.2 Question 1 (i) and (ii)

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.2 Question 1 (iii)

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.2 Question 2

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.2 Question 2 (iii)

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.2 Question 2 (iv)

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.2 Question 2 (v)

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.2 Question 2 (vi)

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.2 Question 3

Exercise 1.3

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.3 Question 1

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.3 Question 1_1

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.3 Question 1_2

Solution 2:
(i)
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.3-Q2-i

(ii)
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.3-Q2-ii

Exercise 1.4

Solution 1:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q1

Solution 2:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q2

Solution 3:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-i

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-ii

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-iii

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-iv

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-v

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-vi

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-vii

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-viii

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-ix

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-x

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-xi

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-xii

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q3-xiii

Solution 4:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q4-i

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q4-ii

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q4-iii

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q4-iv

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q4-v

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q4-vi

Solution 5:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q5-i

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q5-ii

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q5-iii

Solution 6:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q6

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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q6-i

Solution 7:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q7

Solution 8:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q8

Solution 9:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q9

Solution 10:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q10

Solution 11:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q11

Solution 12:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q12

Solution 13:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q13

Solution 14:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.4-Q14

Exercise 1.5

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 1

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 2

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 2_1
v
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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 2_2

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 2_3

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3_1

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3_2

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3_3

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3_4

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3_5

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3_6

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3_7

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 3_8

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RD Sharma class 9 maths Solutions chapter 1 Number System Exercise 1.5 Question 4

Exercise 1.6

Solution 1:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.6-Q1

Solution 2:
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RD-Sharma-Class-9-Solutions-Chapter-1-Number-System-Ex1.6-Q2

The post RD Sharma Class 9 Solutions Number Systems appeared first on Learn CBSE.

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