Bhagya

MP Board Class 10th Science Solutions Chapter 4 Carbon and Its Compounds

MP Board Class 10th Science Chapter 4 Intext Questions

Intext Questions Page No. 61

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO₂?
Answer:

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur? (Hint: The eight atoms of sulphur are joined together in the form of a ring).
Answer:

Intext Questions Page No. 68,69

Question 1.
How many structural isomers can you draw for pentane?
Answer:
Three structural isomers are possible for pentane.

Question 2.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:
The two features of carbon that give rise to a large number of compounds are as follows:

1. Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to large molecules. This property is called catenation.
2. Since carbon has a valency of four, it is capable of bonding with four other atoms of carbon or atoms of some other mono-valent element.

Question 3.
What will be the formula and electron dot structure of cyclopentane?
Answer:
The formula for cyclopentane is C₅H₁₀. Its electron dot structure is given below:

Question 4.
Draw the structures for the following compounds:
(i) Ethanoic acid
(ii) Bromopentane
(iii) Butanone
(iv) Hexanal
Are structural isomers possible for bromo-pentane?
Answer:

Yes, there are many structural isomers possible for bromo-pentane. Among them, the structures of the three isomers are given.

Question 5.
How would you name the following compounds?

Answer:
(i) Bromoethane
(ii) Methanal (formaldehyde)
(iii) Hexyne.

Intext Questions Page No. 71

Question 1.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer:
Addition reaction means adding oxygen. Adding ethanol to potassium permanganate, we get ethanoic acid. Hence this reaction is called oxidation reaction.

Since in this reaction one oxygen is added to ethanol, hence it is an oxidation reaction.

Question 2.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?
Answer:
If a mixture of oxygen and ethyne is burnt, then ethyne burns completely producing a blue flame. The oxygen ethyne flame is extremely hot and produces a very high temperature which is used for welding metals. A mixture of ethyne and air is not used for welding because the burning of ethyne in air produces a sooty flame due to incomplete combination which is not enough to melt metals for welding.

Intext Questions Page No. 74

Question 1.
How would you distinguish experimentally between an alcohol and a carboxylic acid?
Answer:

1. We can distinguish between an alcohol and a carboxylic acid on the basis of their reaction with carbonates and hydrogen carbonates. The acid reacts with carbonate and hydrogen carbonate to evolve CO₂ gas that turns lime-water milky.
2. Metal carbonate/Metal hydrogen carbonate + Carboxylic acid → Salt + Water + Carbon dioxide.
3. In the litmus test, alcohol shows no change in colour whereas carboxylic acid turns blue litmus red.
   With sodium metal, alcohol gives effervescence but carboxylic acid does not give it. Alcohols, on the other hand, do not react with carbonates and hydrogen carbonates.

Question 2.
What are oxidising agents?
Answer:
some substances are capable of adding oxygen to others. These substances are known as oxidising agents.

Intext Questions Page No. 76

Question 1.
Would you be able to check if the water is hard by using a detergent?
Answer:
Margents are remaining effective in hardwater. Because of this reason, we can check if water is hard by using a detergent.

Question 2.
People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer:
The soap molecules form structures called micelles in water, where one end of the molecules is towards the oil droplet while the ionic end-faces outside. This forms emulsion in water and we can wash our clothes clean.

MP Board Class 10th Science Chapter 4 Ncert Textbook Exercises

Question 1.
Ethane, with the molecular formula C₂H₆ has –
(a) 6 covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds
Answer:
(b) 7 covalent bonds.

Question 2.
Butanone is a four-carbon compound with the functional group.
(a) Carboxylic acid
(b) Aldehyde
(c) Ketone
(d) Alcohol
Answer:
(c) Ketone.

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that.
(a) The food is not cooked completely.
(b) The fuel is not burning completely.
(c) The fuel is wet.
(d) The fuel is burning.
Answer:
(b) the fuel is not burning completely

Question 4.
Explain the nature of the covalent bond using the bond formation in CH₃Cl.
Answer:
The structure of CH₃Cl is given below:

Carbon has four valence electrons. It shares one electron each with three hydrogen atoms and one electron with chlorine. The bond between C and Cl atoms is covalent but due to higher value of electro-negativity of Cl, the C-Cl bond is polar in nature.

Question 5.
Draw the electron dot structures for:
(a) Ethanoic acid
(b) H₂S
(c) Propanone
(d) F₂

Question 6.
What is a homologous series? Explain with an example?
Answer:
A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain is called a homologous series.
Eg: CH₄ and C₂H₆ – These differ by a CH₂ unit.
C₂H₆ and C₃H₈ – these differ by a CH₂ unit.

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Answer:
Ethanol and Ethanoic acid can be differentiated on the basis of their following properties by:

1. Ethanol is a liquid at room temperature with a pleasant smell. Ethanoic acid has a melting point of 17°C. Since it is below the room temperature so, it freezes during winter. Moreover, ethanoic acid has a smell like vinegar.
2. Ethanol does not react with metal carbonates while, ethanoic acid reacts with metal carbonates to form a salt, water and carbon dioxide.
   For example:
   2CH₃COOH + Na₂CO₃ → 2CH₃COONa + CO₂ +H₂O
3. Ethanol does not react with NaOH while ethanoic acid reacts with NaOH to form sodium ethanoate and water.
   For example,
   CH₃COOH + NaOH → CH₃COONa + H₂O
4. Ethanol is oxidized to give ethanoic acid in the presence of acidified KMnO₄ while no reaction takes place with ethanoic acid in the presence of acidified KMnO₄.

Difference in physical properties:

Ethanol	Ethanoic acid
This is in liquid form at room temperature. Its melting point is 156° K.	Its melting point is 290K and hence it often freezes during winter in cold climates.
Difference in chemical properties
Ethanol will not react with metallic carbo­nates.	Ethanoic acid reacts with carbo­nates and Hydrogen carbonate and forms salts, carbon dioxide and water.

Question 8.
Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Answer:
Soaps are molecules in which the two ends have differing properties, one is hydrophilic that is, it interacts with water, while the other end is hydrophobic, that is it interacts with hydrocarbons. When soap is at the surface of water, the hydrophobic tail of soap will not be soluble in water and the soap will align along the surface of water with the ionic end in water and the hydrocarbon tail protruding out of water. Thus, clusters of molecules in which the hydrophobic tails are in the interior of the cluster and the ionic ends are on the surface of the cluster. This formation is called a micelle. Soap in the form of a micelle is able to clean. A micelle will not be formed in other solvents such as ethanol.

Question 9.
Why are carbon and its compounds used as fuels for most applications?
Answer:
Carbon and its compounds give large amount of heat on combustion due to the high percentage of carbon and hydrogen. Carbon compounds used as fuel have optimum ignition temperature with high calorific values and are easy to handle. Their combustion can be controlled. Therefore, carbon and its compounds are used as fuels.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
When soap reacts with water, calcium and magnesium salts are formed which causes hardness for water. Ionic ends of soap interacts with water while the carbon chain interacts with oil. The soap molecules, thus form structures called micelles where one end of the molecules is towards the oil droplet while the ionic-end faces outside. This forms an emulsion in water.

Question 11.
What change will you observe if you test soap with litmus paper (red and blue)?
Answer:
Since soap is basic in nature, it will turn red litmus blue. However, the colour of the blue litmus will remain blue.

Question 12.
What is hydrogenation? What is its industrial application?
Answer:
Unsaturated Hydrocarbons react with Hydrogen, in presence of catalysts such as palledium or Nickel and forms saturated Hydrocarbons. This is called Hydrogenation of oils.
This process is useful in hydrogenation of oils derived from plants.

Question 13.
Which of the following hydrocarbons undergo addition reactions:
C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄.
Answer:
Unsaturated hydrocarbons undergo addition reactions. Being unsaturated hydrocarbons, C₃H₆ and C₂H₂ undergo addition reactions.

Question 14.
Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
Answer:
Saturated Hydrocarbons will not react with Bromine, but unsaturated hydrocarbons change the colour of Bromine.

Question 15.
Explain the mechanism of the cleaning action of soaps.
Answer:
The dirt present on clothes is organic in nature and insoluble in water. Therefore, it cannot be removed by washing with water only. When soap is dissolved in water, its hydrophobic ends attach themselves to the dirt and remove it from the cloth. Then, the molecules of soap arrange themselves in micelle formation and trap the dirt at the centre of the cluster. These micelles remain suspended in the water. Hence, the dust particles are easily rinsed away by water.

MP Board Class 10th Science Chapter 4 Additional Questions

MP Board Class 10th Science Chapter 4 Multiple Choice Questions

Question 1.
Which of the following is a three-carbon compound?
(a) Ethene
(b) Ethane
(c) Propane
(d) Acetylene
Answer:
(c) Propane

Question 2.
Which one of the following is an unsaturated hydrocarbon?
(a) Acetylene
(b) Butane
(c) Propane
(d) Decane
Answer:
(a) Acetylene

Question 3.
Two neighbours of homologous series differ by:
(a) -CH
(b) -CH₂
(c) -CH₃
(d) -CH₄
Answer:
(b) -CH₂

Question 4.
General formula of alkanes is –
(a) C_nH_2n+2
(b) C_nH_2n
(c) C_nH_2n-2
(d) C_nH_n
Answer:
(a) C_nH_2n+2

Question 5.
Which of the following represents alkynes?
(a) -C – C-
(b) -C = C-
(c) -C ≡ C-
(d) None of these
Answer:
(c) -C ≡ C-

Question 6.
Which of the following represents ketones?
(a) -C = O
(b) -OH
(c) -CHO
(d) COOH
Answer:
(a) -C = O

Question 7.
Which of the following is not an aliphatic hydrocarbon?
(a) ethene
(b) ethane
(c) propyne
(d) benzene
Answer:
(d) benzene

Question 8.
Complete combustion of a hydrocarbon gives:
(a) CO + H₂O
(b) CO₂ + H₂O
(c) CO + H₂
(d) CO₂ + H₂
Answer:
(b) CO₂ + H₂O

Question 9.
Which is NOT correct for isomers of a compound?
(a) They differ in physical properties.
(b) They differ in chemical properties.
(c) They have the same molecular formula.
(d) They have the same structural formula.
Answer:
(d) They have the same structural formula.

Question 10.
Buckminsterfullerene is an example of ………….. of carbon.
(a) an isomer
(b) an isotope
(c) an allotrope
(d) a functional group
Answer:
(c) an allotrope

Question 11.
Who prepared urea for the first time by heating ammonium cyanate?
(a) Wohler
(b) Lavoisier
(c) Fuller
(d) Haber
Answer:
(a) Wohler

Question 12.
Hexanone is a four-carbon compound with the functional group:
(a) Carboxylic acid
(b) Aldehyde
(c) Ketone
(d) Alcohol
Answer:
(c) Ketone

Question 13.
Major constituent of LPG is ………….
(a) Ethene
(b) Butane
(c) Propane
(d) Pentane
Answer:
(b) Butane

Question 14.
The gas used in welding and cutting metals is:
(a) Ethyne
(b) Ethene
(c) Ethane
(d) Propane
Answer:
(a) Ethyne

Question 15.
How is carbon atoms arranged in Buckminster fullerenes?
(a) Triangle shape
(b) Hexagonal array
(c) Football shape
(d) None
Answer:
(c) Football shape

Question 16.
Vinegar is a solution of –
(a) 40%-45% acetic acid.
(b) 90%-95% acetic acid.
(c) 5-20% acetic acid and water.
(d) 35-40% acetic acid and water.
Answer:
(c) 5-20% acetic acid and water.

Question 17.
How many covalent bonds are there in Bromoethane?
(a) 4
(b) 6
(c) 10
(d) 7
Answer:
(d) 7

Question 18.
Which functional group is present in propane?
(a) Aldehyde
(b) No group
(c) Ketone
(d) Alcohol
Answer:
(b) No group

Question 19.
Which compound/molecule is being presented by the following formula: H: C:: C: H
(a) Ethane
(b) Ethene
(c) Ethyne
(d) None of these
Answer:
(b) Ethene

Question 20.
Next homologous to C₂H₅OH will be:
(a) CH₄
(b) C₂H₆
(c) C₃H₅
(d) C₃H₇OH
Answer:
(d) C₃H₇OH

Question 21.
When we burn naphthalene it produces:
(a) Smoky flame
(b) Non-sooty flame
(c) Colourless flame
(d) No flame
Answer:
(a) Smoky flame

Question 22.
Bunsen burner is used for:
(a) making food.
(b) study flames type.
(c) low heating work.
(d) all the above.
Answer:
(c) low heating work.

Question 23.
See the figure carefully.

Choose the suitable name of isomer:
(a) Neo-pentane
(b) n-pentane
(c) Iso-pentane
(d) All
Answer:
(c) Iso-pentane

Question 24.
Which of the following is a structure of ethanoic acid?

Answer:

Question 25.
What is the name of CH₃-CH₂-Br? Choose from the following:
(a) Hex-1-one
(b) Hexanal
(c) Ethanoic acid
(d) None
Answer:
(d) None

Question 26.
What happens on the litmus test of soap?
(a) No change
(b) Red litmus turns blue
(c) Red litmus turn purple
(d) Red litmus turn green
Answer:
(b) Red litmus turns blue

MP Board Class 10th Science Chapter 4 Very Short Answer Type Questions

Question 1.
Name two groups which can have the same general formula.
Answer:
Both alkenes and cyclo-alkanes can be represented by the same general formula.

Question 2.
Which group of compounds have general formula C₂H_2n?
Answer:
The general formula C_nH_2n represents alkenes group of compounds.

Question 3.
What is the common name and IUPAC name for CH₃COCH₃?
Answer:
Acetone is the common name and propanone is the IUPAC name for CH₃COCH₃.

Question 4.
Do isomers always show same chemical properties?
Answer:
No, isomers always do not show the same chemical properties.

Question 5.
What is the common name and formula for ethanol?
Answer:
Alcohol, CH₃CH₂OH.

Question 6.
What are the products of complete combustion of a hydrocarbon?
Answer:
Carbon dioxide and water.

Question 7.
What is next homologue of C₃H₇OH is called?
Answer:
The next homologue of C₃H₇OH is called butanol C₄H₉OH.

Question 8.
What are isomers?
Answer:
The compounds which have the same molecular formula but different structures and chemical properties are called isomers.

Question 9.
Which one are more reactive unsaturated hydrocarbons or saturated hydrocarbons? Give reason.
Answer:
Unsaturated hydrocarbons: The Presence of double and triple covalent bonds make them more reactive.

Question 10.
Discuss the general nature of covalent compounds in water.
Answer:
Generally, they are insoluble in water.

Question 11.
What type of hydrocarbons takes part in an addition reaction?
Answer:
Unsaturated hydrocarbons.

Question 12.
Which carboxylic acid freezes during winter or under cold climate conditions?
Answer:
Acetic acid and hence, known as glacial acetic acid.

Question 13.
What is the difference in molecular masses of any two successive homologous alkanes?
Answer:
14 units.

Question 14.
What is the molecular formula of the alcohol which can be derived from propane?
Answer:
Alcohol obtained from propane is propanol -1 and the molecular formula is C₃H₇OH.

Question 15.
Give the names of the functional groups: (CBSE 2007)

1. -CHO
2. -COOH

Answer:

1. Aldehydic group.
2. Carboxylic acid group.

Question 16.
Give the names of the following functional groups: (CBSE 2007)

1. -OH
2. -CO

Answer:

1. Alcoholic.
2. Ketonic.

MP Board Class 10th Science Chapter 4 Short Answer Type Questions

Question 1.
What is meant by the term functional group?
Answer:
An atom or a group of atoms, which makes a carbon compound reactive and decides its properties, is called a functional group.
For example aldehyde, ketone etc.

Question 2.
Which R functional groups always occur at the terminal position of a carbon chain?
Answer:
Aldehydic Group, R-CHO (R is the alkyl group),
Carboxyl Group, R-COOH (R is the alkyl group)

Question 3.
Why a candle flame burns yellow, while a highly-oxygenated gas fuel flame burns blue?
Answer:
The most important factor determining the colour of the flame is oxygen supply and the extent of fuel-oxygen, which determines the rate of combustion and thus, the temperature and reaction paths, thereby producing different colour hues. In case of a candle, it is incomplete combustion and the flame temperature is not high. This gives a yellow flame, while a highly-oxygenated gas (e.g., ethyne) flame burns blue because of complete combustion raising a very high temperature.

Question 4.
Why is the reaction between methane and chlorine considered a substitution reaction? (CBSE 2008)
Answer:
Methane reacts with chlorine in the presence of sunlight to form chloromethane and hydrogen chloride. Since chlorine substitutes or replaces hydrogen of methane to form chloromethane, it is considered as substitution reaction.
CH₄ + Cl₂ → CH₃Cl + HCl
With the excess of chlorine, four hydrogen atoms of methane are replaced by chlorine atoms to form carbon tetrachloride (CCl₄).

Question 5.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Being tetravalent carbon atom, it is neither capable of losing all of its four valence electrons nor it can easily accept four electrons to complete its octet. Both of these are requirements of ionic bond formation and are energetically less favourable. Carbon completes its octet by sharing electrons and hence, covalent bonding is preferred.

Question 6.
What do you mean by Octane rating?
Answer:
Gasoline is rated on a scale known as octane rating, which is based on the way they burn in an engine. The higher the octane rating, the greater the percentage of complex-structured hydrocarbons that are present in the mixture, the more uniformly the gasoline burns, and the less knocking there is in the automobile engine. Thus, a gasoline rated 92 octane will burn more smoothly than one rated 87 octanes.

Question 7.
What is covalent bonding?
Answer:
The chemical bonding that takes place due to the mutual sharing of electron pairs of two or more atoms of different elements is called covalent bonding. By mutual sharing of electron pairs, atom attains noble gas configuration, e.g., hydrogen molecule (H₂), the two H-atoms combine by covalent bonding (H-H).

Question 8.
What are hydrocarbons? Give examples.
Answer:
Compounds of carbon and hydrogen are called hydrocarbons. Methane, ethane, butane, ethyne, propane, benzene, petroleum products – all are examples of hydrocarbons.

Question 9.
What are saturated hydrocarbons? (CBSE 2011)
Answer:
The hydrocarbons in which valency of carbon is satisfied by a single covalent bond are called saturated hydrocarbons. Alkanes like methane (CH₄), ethane(C₂H₆), propane (C₃H₈) etc. are examples of saturated hydrocarbons. Saturated hydrocarbons will generally give a clean flame.

Question 10.
Why do ionic compounds have high melting points? (HOTS)
Answer:
Ions have strong electrostatic forces of attraction among them forming ionic compounds. It requires a lot of energy to break these ionic bonds or forces. That’s why ionic bonds have high melting points.

Question 11.
What are homologous series? (HOTS)
Answer:
Homologous series are:

1. Compounds with the same formula.
2. Belong to the same functional group.
3. Have general methods of separation.
4. Have similar chemical properties.

Show similar gradation of physical properties, e.g., boiling points of alcohol increase with an increase in their molecular weights. Similarly, solubility decreases with increase in molecular weights.

Question 12.
What is a heteroatom? What is the heteroatom in the alcohol functional group? (HOTS)
Answer:
In a hydrocarbon chain, one or more hydrogen atoms can be replaced by other atoms according to their valencies. The element wh replaces hydrogen in the chain is called a heteroatom, e.g., in alcohol (-OH) functional group, oxygen is the heteroatom.

MP Board Class 10th Science Chapter 4 Long Answer Type Questions

Question 1.
Distinguish between saturated and unsaturated hydrocarbons by the way of their burning in air and bromine test inferences.
Answer:
1. Saturated compounds are burnt in air, to give a clear (blue) flame but the burning of unsaturated compounds (alkenes and alkynes) give a sooty (yellowish) flame because saturated compounds contain comparatively less percentage of carbon which is completely oxidized by the oxygen present in the air.

On the other hand, the percentage of carbon in unsaturated compounds is more and it requires more oxygen to get completely oxidized that is not fulfilled by air. So, due to incomplete oxidation, they burn with a sooty flame.

2. Bromine-water test: Br₂ water is a brown coloured liquid:

1. Unsaturated hydrocarbons give addition reaction with Br₂. So, the colour of Br₂ water gets decolourised.
   
2. Saturated hydrocarbons do not react with Br₂ water, so the colour of B₂-water does not get decolourised.

Question 2.
Two compounds A and B react with each other in the presence of a dehydrating agent to produce an ester. Both react with Na to evolve hydrogen gas. On reaction with Na₂CO_3, only A evolves CO₂. Identify the functional groups present in A and B giving the reason for your answer.
Answer:
Compound A contains -COOH group while compound B contains -OH group. Since, carboxylic acids and alcohols react with each other to form an ester, out of A and B, one is an alcohol and the other is a carboxylic acid. This is further strengthened by the reaction of both with Na to evolve hydrogen gas. Only carboxylic acids react with Na₂CO₃ to evolve CO₂, A contains -COOH group while B contains -OH group.

Question 3.
An organic compound ‘X’ is widely used as a preservative in pickles and has a molecular formula C₂H₂O₂. This compound reacts with ethanol to form a sweet-smelling compound ‘Y’.

1. Identify the compound ‘X’.
2. Write the chemical equation for its reaction with ethanol to form compound ‘Y’.
3. How can we get compound ‘X’ back from ‘Y’?
4. Name the process and write a corresponding chemical equation.
5. Which gas is produced when compound ‘X’ reacts with washing soda? Write the chemical equation.
6. Answer:
7. Compound X is ethanoic acid which gives and ester (Y) when reacts with ethanol.
8. CH₃COOH + CH₃CH₂OH → CH₃COOC₂H₅.
9. Esters give back alcohol and carboxylic acid in the presence of acid or base.
10. Saponification reaction: CH₃COOC₂H₅ + NaOH → C₂H₅OH + CH₃COOH + Na.
11. CO₂ gas is released,
    CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂.

Question 4.
“Saturated hydrocarbons burn with a blue flame while unsaturated hydrocarbons burn with a sooty flame.” Why?
Answer:
Saturated hydrocarbons have only C-C and C-H single bonds and thus, contain the maximum possible number of hydrogen atoms per carbon atom. With sufficient oxygen, saturated hydrocarbons burn completely and give blue flame,
CH₄ + 2O₂ → CO₂ + 2H₂O
Unsaturated hydrocarbons contain a carbon-carbon double bond (C=C) or triple bond (C=C). Hence, they contain less number of hydrogen than carbon. Unsaturated hydrocarbons undergo incomplete combustion and give yellow flame along with black sooty carbon.
C₂H₄ + O₂ → CO₂ + 2H₂O + C(s)

Question 5.
What makes some molecular formula compound different? (HOTS)
Answer:
The arrangement makes them different compounds with identical molecular formula but different structures are called structural isomers. Organic compounds show a great level of isomerism. Isomers may be structural (due to difference in the arrangement of C atoms forming chain) or stereo (due to arrangement of bonds in a chain). With the increase in the number of carbon atoms in molecular formula, it leads to an increase in the number of isomers.
For example:

MP Board Class 10th Science Chapter 4 Textbook Activities

Class 10 Science Activity 4.1 Page No. 58

1. Make a list of ten things you have used or consumed since the morning.
2. Compile this list with the lists made by your classmates and then sort the items into the following table.
3. If there are items which are made up of more than one material, put them into both the relevant columns.
   
4. (C) Indicates carbon. Most substances contain carbon in it.

Class 10 Science Activity 4.2 Page No. 67

1. Calculate the difference in the formulae and molecular masses for
   (a) CH₃OH and C₂H₅OH.
   (b) C₂H₅OH and C₃H₇OH.
   (c) C₃H₇OH and C₄H₉OH.
2. Is there any similarity between these three?
3. Arrange these alcohols in the order of increasing carbon atoms to get a family. Can we call this family a homologous series?
4. Generate the homologous series for compounds containing up to four. carbons for the other functional groups given in the above table.

Difference: 70 – 60 = 14U
All three groups given above are homologous.

Class 10 Science Activity 4.3 Page No. 69

Caution:

1. This Activity needs the teacher’s assistance.
2. Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatula and burn them.
3. Observe the nature of the flame and note whether smoke is produced.
4. Place a metal plate above the flame. Is there a deposition on the plate in case of any of the compounds?

Observations:

Class 10 Science Activity 4.4 Page No. 69

1. Light a bunsen burner and adjust the air hole at the base to get different types of flames/presence of smoke.
2. When do you get a yellow, sooty flame?
3. When do you get a blue flame?

Observations:

1. Yellow, Sooty flame is formed – when the hole is closed.
2. A blue flame is observed – when the hole is open.

Class 10 Science Activity 4.5 Page No. 70

1. Take about 3 ml of ethanol in a test tube and warm it gently in a water bath.
2. Add a 5% solution of alkaline potassium permanganate drop by drop to this solution.
3. Does the colour of potassium permanganate persist when it is added initially?
4. Why does the colour of potassium permanganate not disappear when excess is added?

Observations:
Doing the above activities we found that potassium permanganate act here as oxidising agents only and their colour do not change at,

Class 10 Science Activity 4.6 Page No. 72

Teacher’s demonstration:

1. Drop a small piece of sodium, about the size of a couple of grains of rice, into ethanol (absolute alcohol).
2. What do you observe?
3. How will you test the gas evolved?

Observations:
Sodium is an inflammable substance hence, it should be handled very carefully. When we place it in alcohol, hydrogen gas is evolved and sodium ethoxide is formed,
2Na + 2CH₃CH₂OH → 2CH₃CH₂ONa⁺ + H₂ ↑

Class 10 Science Activity 4.7 Page No. 73

1. Compare the pH of dilute acetic acid and dilute hydrochloric acid using both litmus paper and universal indicator.
2. Are both acids indicated by the litmus test?
3. Does the universal indicator show them as equally strong acids?

Observations:
The litmus test and pH test show the acidity and alkalinity of substance or chemical:

Class 10 Science Activity 4.8 Page No. 73

1. Take 1 ml ethanol (absolute alcohol) and 1 ml glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
2. Warm in a water-bath for at least five inutes as shown in Figure.
3. Pour into a beaker containing 20-50 ml of water and smell the resulting mixture.
   

Observations:
When acetic acid reacts with alcohol a new compound with an ester functional group is formed. It has fruit like smell. This reaction is called esterification reaction.

Class 10 Science Activity 4.9 Page No. 74

1. Take a spatula full of sodium carbonate in a test tube and add 2 ml of dilute ethanoic acid.
2. What do you observe?
3. Pass the gas produced through freshly prepared lime-water. What do you observe?
4. Can the gas produced by the reaction between ethanoic acid and carbonate be identified by this test?
5. Repeat this Activity with sodium hydrogen carbonate instead of sodium carbonate.

Observations:
Sodium acetate is produced when we add carbonate or hydrogen carbonate to acetic acid.
2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂
CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂

Class 10 Science Activity 4.10 Page No. 74

1. Take about 10 mL of water each in two test tubes.
2. Add a drop of oil (cooking oil) 10 both the test tubes and table them as A and B.
3. To test tube B add a few drops of soap solution.
4. Now shake both the test tubes vigorously for the same period of time.
5. Can you see the oil and water layers separately in both the test tubes immediately after you stop shaking them?
6. Leave the test tubes undisturbed for some time and observe. Does the oil layer separate out? In which test tube does this happen first?

Observations:
Yes, a layer of oil separates out by reacting with the soap solution. Dirt has an oily nature. It happens first in test tube B.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Take π = \(\frac{22}{7}\), unless stated otheriwise

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Radius of the sphere (r₁) = 4.2 cm
∴ Volume of the sphere = \(\frac{4}{3}\) πr₁³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \mathrm{cm}^{3}\)
Radius of the cylinder (r₂) = 6 cm
Let h be the height of the cylinder.
∴ Volume of the cylinder = πr²h
= \(\frac{22}{7}\) × 6 × 6 × h cm³
Since volume of the metallic sphere = Volume of the cylinder

Hence, the height of the cylinder is 2.744 cm

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radii of the given spheres are
r₁ = 6 cm, r₂ = 8 cm and r₃ = 10 cm
⇒ Volume of the given spheres are

= \(\frac{4}{3} \times \frac{22}{7}\) × [1728] cm³
Let the radius of the new big sphere be R. Volume of the new sphere
= \(\frac{4}{3}\) × π × R₃ = \(\frac{4}{3} \times \frac{22}{7}\) × R₃
Since, the two volumes must be equal.
∴ \(\frac{4}{3} \times \frac{22}{7} \times R^{3}=\frac{4}{3} \times \frac{22}{7} \times 1728\)
⇒ R₃ = 1728 ⇒ R = 12 cm
Thus, the required radius of the resulting sphere is 12 cm.

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Diameter of the cylindrical well = 7 m
⇒ Radius of the cylindrical well (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
∴ Volume = πr²h = \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 20 m³
= 22 × 7 × 5 m³
⇒ Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having length = 22 m, breadth = 14 m
Let h be the height of the platform.
∴ Volume of the platform = 22 × 14 × h m³
Since, the two volumes must be equal
∴ 22 × 14 × h = 22 × 7 × 5
Thus, the required height of the platform is 2.5 m.

Question 4.
A well of diameter 3 m is dug 14m deep. The earth taken out of it has been spread evenly all around it in shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of cylindrical well = 3 m
⇒ Radius of the cylindrical well = \(\frac{3}{2}\) m = 1.5 m
Depth of well (h) = 14 m
∴ Volume of cylindrical well

Let the height of the embankment = H m.
Internal radius of the embankment (r) = 1.5 m.
External radius of the embankment (R)
= (4 + 1.5) m = 5.5 m.
∴ Volume of the embankment
= πR²H – πr²H = πH [R² – r²]
= πH (R + r) (R – r)
= \(\frac{22}{7}\) × H (5.5 + 1.5)(5.5 – 1.5)
= \(\frac{22}{7}\) × H × 7 × 4m³
Since, Volume of the embankment=Volume of the cylindrical well
⇒ \(\frac{22}{7}\) × H × 7 × 4 = 99
⇒ H = 99 × \(\frac{7}{22} \times \frac{1}{7} \times \frac{1}{4} m=\frac{9}{8} m\) = 1.125 m
So, the required height of the embankment = 1.125 m.

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and 6. diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
For the circular cylinder:

Diameter = 12 cm
⇒ Radius (r) = \(\frac{12}{2}\) = 6cm and height (h) = 15 cm
∴ Volume of circular cylinder
= πr²h = \(\frac{12}{2}\) × 6 × 6 × 15 cm³
For conical and hemispherical part of icecream :
Diameter = 6 cm ⇒ radius (R) = 3 cm
Height of conical part (H) = 12 cm

Volume of ice cream cone = (Volume of the conical part) + (Volume of the hemispherical part)

Thus, the required number of cones is 10.

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:
For a circular coin:

Diameter = 1.75 cm
⇒ Radius (r) = \(\frac{175}{200}\) cm
Thickness (h) = 2mm = \(\frac{2}{10}\) cm
∴ Volume of one coin = πr²h = \(\frac{22}{7} \times\left(\frac{175}{200}\right)^{2} \times \frac{2}{10} \mathrm{cm}^{3}\)
For a cuboid:

Length (l) = 10 cm,
Breadth (b) = 5.5 cm
and height (h) = 3.5 cm
∴ Volume = l × b × h = 10 × \(\frac{55}{10} \times \frac{35}{10}\) cm³

Thus, the required number of coins = 400.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
For the cylindrical bucket:
Radius (r) = 18 cm and height (h) = 32 cm
Volume of cylindrical bucket = πr²h
= \(\frac{22}{7}\) × (18)² × 32 cm³
⇒ Volume of the sand = (\(\frac{22}{7}\) × 18 × 18 × 32) cm³
For the conical heap:
Height (H) = 24 cm
Let radius of the base be R.
∴ Volume of conical heap

Thus, the required radius = 36 cm and slant height = \(12 \sqrt{13}\) cm.

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Width of the canal = 6 m,
Depth of the canal = 1.5 m
Length of the water column in 1 hr = 10 km
∴ Length of the water column in 30 minutes
(i.e., \(\frac{1}{2}\)hr) = \(\frac{10}{2}\) km = 5 km = 5000 m
∴ Volume of water flown in \(\frac{1}{2}\) hr
= 6 × 1.5 × 5000 m³ = 6 × \(\frac{15}{10}\) × 5000 m³
= 45000 m³
Since, the above amount (volume) of water is spread in the form of a cuboid of height
8 cm (= \(\frac{8}{100}\) m)
Let the area of the cuboid = a
∴ Volume of the cuboid = Area × Height

= 562500 m² = 56.25 hectares
Thus, the required area is 56.25 hectares.

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Diameter of the pipe = 20 cm
⇒ Radius of the pipe (r) = \(\frac{20}{2}\) cm = 10 cm
Since, the water flows through the pipe at 3 km/hr.
∴ Length of water column per hour(h) = 3 km
= 3 × 1000 m = 3000 × 100 cm = 300000 cm.
Length of water column per hour(h) = 3 km
Volume of water flown in one hour = πr²h
= π × 10² × 300000 cm³ = π × 30000000 cm²
Now, for the cylindrical tank :
Diameter = 10 m
⇒ Radius (R) = \(\frac{10}{2}\) m = 5 × 100 cm = 500 cm
Height (H) = 2 m = 2 × 100 cm = 200 cm
∴ Volume of the cylindrical tank = πR²H
= π × (500)² × 200 cm³
Now, time required to fill the tank

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 15 Probability Ex 15.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1

Question 1.
Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = _________
(ii) The probability of an event that cannot happen is ______.Such an event is called _________.
(iii) The probability of an event that is certain to happen is ______ Such an event is called ______
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
(v) The probability of an event is greater than or equal to _______ and less than or equal to ______.
Solution:
(i) 1 : Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) 0, impossible: The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) 1, certain: The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
(iv) 1: The sum of the probabilities of all the elementary events of an experiment is 1.
(v) 0, 1: The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) It depends on various factors such as whether the car will start or not. So, the probability of car will start does not equal to the probability of car will not start.
∴ The outcomes are not equally likely.
(ii) It depends on the player’s ability. So, probability that the player shot the ball is not the same as the probability that the player misses the shot.
(iii) The outcomes are equally likely as the probability of answer either right or wrong is \(\frac{1}{2}\)
(iv) The outcomes are equally likely as the probability of ‘newly born baby to be either bay or girls’ is \(\frac{1}{2}\) .

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
Since on tossing a coin, the outcomes ‘head’ and ‘tail’ are equally likely, the result of tossing a coin is completely unpredictable and so it is a fairway.

Question 4.
Which of the following cannot be the probability of an event?
(A) \(\frac{2}{3}\)
(B) -1.5
(C) 15%
(D) 0.7
Solution:
Since, the probability of an event cannot be negative.
∴ -1.5 cannot be the probability of an event.

Question 5.
If P(E) = 0.05, what is the probability of ‘not E’?
Solution:
∵ P(E) + P(not E) = 1
∴ 0.05 + P(not E) = 1 ⇒ P(not E) = 0.95
Thus, probability of ‘not E’ = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Since there are only lemon flavoured candies in the bag.
∴ Taking out orange flavoured candy is not possible.
⇒ Probability of taking out an orange flavoured candy = 0.

(ii) Probability of taking out a lemon flavoured candy = 1.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let the probability of 2 students having same birthday = P(SB)
And the probability of 2 students not having the same birthday = P(NSB)
∴ P(SB) + P(NSB) = 1
⇒ P(SB) + 0.992 = 1 ⇒ P(SB) = 1 – 0.992 = 0.008

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution:
Total number of balls = 3 + 5 = 8
∴ umber of possible outcomes = 8
(i) ∵ There are 3 red balls.
∴ Number of favourable outcomes = 3
∴ P (red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}\)
= \(\frac{3}{8}\)
(ii) Probability of the ball drawn which is not red = 1 – P(red) = \(1-\frac{3}{8}=\frac{8-3}{8}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total number of marbles = 5 + 8 + 4 = 17
∴ Number of all possible outcomes = 17
(i) ∵ Number of red marbles = 5
∴ Number of favourable outcomes = 5
∴ Probability of red marbles, P(red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{5}{17}\)

(ii) Number of white marbles = 8
∴ Probability of white marbles, P(white) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{8}{17}\)
_ Number of favourable outcomes _ 8 Number of all possible outcomes 17

(iii) Number of green marbles = 4 Number of marbles which are not green
= 17-4 = 13
i.e., Favourable outcomes = 13
∴ Probability of marbles ‘not green’, P(not greeen)
\(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50p coin?
(ii) will not be a ₹ 5 coin?
Solution:
Number of coins 50 p = 100, ₹ 1 = 50 ₹ 2 = 20, ₹ 5 = 10
Total number of coins = 100 + 50 + 20 +10 = 180
∴ Total possible outcomes = 180

(i) For a 50 p coin:
Favourable outcomes = 100

(ii) For not a ₹ 5 coin:
Y Number of ₹ 5 coins = 10
∴ Number of ‘not ₹ 5’ coins = 180 – 10 = 170
⇒ Favourable outcomes = 170

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Number of male fishes = 5
Number of female fishes = 8
∴ Total number of fishes = 5 + 8 = 13
⇒ Total number of outcomes = 13
For a male fish:
Number of favourable outcomes = 5

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution:
Total number marked = 8
∴ Total number of possible outcomes = 8
(i) When pointer points at 8:
Number of favourable outcomes = 1

(ii) When pointer points at an odd number:
∵ Odd numbers are 1, 3, 5 and 7
∴ Total odd numbers from 1 to 8 = 4
⇒ Number of favourable outcomes = 4

(iii) When pointer points at a number greater than 2:
∵ The numbers 3, 4, 5, 6, 7 and 8 are greater than 2
∴ Total numbers greater than 2 = 6
⇒ Number of favourable outcomes = 6

(iv) When pointer points at a number less than 9:
∵ The numbers 1, 2, 3, 4, 5, 6, 7 and 8 are less than 9.
∴ Total numbers less than 9 = 8
∴ Number of favourable outcomes = 8

Question 13.
A die is thrown once. Find the probability of getting:
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Since, numbers on a die are 1, 2, 3, 4, 5 and 6.
∴ Total number of possible outcomes = 6
(i) Since 2, 3 and 5 are prime number.
∴ Favourable outcomes = 3

(ii) Since the numbers between 2 and 6 are 3, 4 and 5
∴ Favourable outcomes = 3

(iii) Since 1, 3 and 5 are odd numbers.
⇒ Favourable outcomes = 3

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Solution:
Number of cards in deck = 52
∴ Total number of possible outcomes = 52
(i) ∵ Number of red colour kings = 2
[∵ King of diamond and heart is red]
Number of favourable outcomes = 2

(ii) For a face card:
∵ 4 kings, 4 queens and 4 jacks are face cards
∴ Number of face cards = 12
⇒ Number of favourable outcomes = 12

(iii) Since, cards of diamond and heart are red
∴ There are 2 kings, 2 queens, 2 jacks i.e., 6 cards are red face cards.
∴ Number of favorable outcomes = 6

(iv) Since, there is only 1 jack of hearts.
∴ Number of favourable outcomes = 1

(v) There are 13 spades in a pack of 52 cards.
∴ Number of favourable outcomes = 13

(vi) ∵ There is only one queen of diamond.
∴ Number of favourable outcomes = 1

Question 15.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
We have five cards.
∴ Total number of possible outcomes = 5
(i) ∵ Number of queen = 1
∴ Number of favourable outcomes = 1

(ii) The queen is drawn and put aside.
∴ Only 5 – 1 = 4 cards are left.
∴ Total number of possible outcomes = 4
(a) ∵ There is only one ace.
∴ Number of favourable outcomes = 1

(b) Since, the only queen has been put aside already.
∴ Number of favourable outcomes = 0

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
We have number of good pens = 132 and number of defective pens = 12
∴ Total number of pens = 132 + 12 = 144 = Total possible outcomes
There are 132 good pens.
∴ Number of favourable outcomes = 132

Question 17.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
Since, there are 20 bulbs in the lot.
Total number of possible outcomes = 20
(i) ∵ Number of defective bulbs = 4
∴ Favourable outcomes = 4

(ii) ∵ The bulb drawn above is not included in the lot.
∴ Number of remaining bulbs = 20 – 1 = 19.
⇒ Total number of possible outcomes = 19.
∵ Number of bulbs which are not defective = 19 – 4 = 15
⇒ Number of favourable outcomes = 15

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
We have total number of discs = 90
∴ Total number of possible outcomes = 90
(i) Since the two-digit numbers are 10, 11, 12, ………, 90.
∴ Number of two-digit numbers = 90 – 9 = 81
∴ Number of favourable outcomes = 81

(ii) Perfect square from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
∴ Number of perfect squares = 9
∴ Number of favourable outcomes = 9

(iii) Numbers divisible by 5 from 1 to 90 are 5, 10,15, 20, 25, 30, 35,40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
i. e., There are 18 numbers from (1 to 90) which are divisible by 5.
∴ Numbers of favourable outcomes = 18

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Since there are six faces of the given die and these faces are marked with letters

∴ Total number of letters = 6
∴ Total number of possible outcomes = 6
(i) ∵ Number of faces having the letter A = 2
∴ Number of favourable outcomes = 2

(ii) ∵ Number of faces having the letter D = 1
∴ Number of favourable outcomes = 1

Question 20.
20. Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution:
Here, area of the rectangle = 3m × 2m = 6 m²
And, the area of the circle = πr²

Question 21.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it? (ii) She will not buy it?
Solution:
Total number of ball pens = 144
⇒ Total number of possible outcomes = 144
(i) Since there are 20 defective pens.
∴ Number of good pens = 144 – 20 = 124
⇒ Number of favourable outcomes = 124

(ii) Probability that Nuri will not buy it = 1 – [Probability that she will buy it]
= \(1-\frac{31}{36}=\frac{36-31}{36}=\frac{5}{36}\)

Question 22.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. An event is defined as the sum of the two numbers appearing on the top of the dice.
(i) Complete the following table

(ii) A student argues that’there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\) Do you agree with this argument? Justify your answer.
Solution:
∵ The two dice are thrown together.
∴ Following are the possible outcomes :
(1, 1) ; (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1) ; (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1) ; (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6,.2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ Total number of possible outcomes is 6 × 6 = 36
(i) (a) The sum on two dice is 3 for (1, 2) and (2, 1)
∴ Number of favourable outcomes = 2
⇒ P(3) = \(\frac{2}{36}\)

(b) The sum on two dice is 4 for (1, 3), (2, 2) and (3, 1).
∴ Number of favourable outcomes = 3
⇒ P(4) = \(\frac{3}{36}\)

(c) The sum on two dice is 5 for (1, 4), (2, 3), (3, 2) and (4,1)
∴ Number of favourable outcomes = 4
⇒ P(5) = \(\frac{5}{36}\)

(d) The sum on two dice is 6 for (1, 5), (2, 4), (3, 3), (4, 2) and (5,1)
∴ Number of favourable outcomes = 5
⇒ P(6) = \(\frac{5}{36}\)

(e) The sum on two dice is 7 for (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1)
∴ Number of favourable outcomes = 6
⇒ P(7) = \(\frac{62}{36}\)

(f) The sum on two dice is 9 for (3, 6), (4, 5), (5, 4) and (6, 3)
∴ Number of favourable outcomes = 4
⇒ P(9) = \(\frac{4}{36}\)

(g) The sum on two dice is 10 for (4, 6), (5, 5), (6,4)
∴ Number of favourable outcomes = 3
⇒ P(10) = \(\frac{3}{36}\)

(h) The sum on two dice is 11 for (5, 6) and (6,5)
∴ Number of favourable outcomes = 2
⇒ P(11) = \(\frac{2}{36}\)

Thus, the complete table is as follows:

(ii) No. The number of all possible outcomes is 36 not 11.
∴ The argument is not correct.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Let T denotes the tail and H denotes the head.
∴ All the possible outcomes are:
{H H H, H H T, H T T, T T T, T T H, T H T, T H H, H T H)
∴ Number of all possible outcomes = 8
Let the event that Hanif will lose the game denoted by E.
∴ Favourable events are: {HHT, HTH, THH, THT, TTH, HTT}
⇒ Number of favourable outcomes = 6
∴ P(E) = \(\frac{6}{8}=\frac{3}{4}\)

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Solution:
Since, throwing a die twice or throwing two dice simultaneously is the same.
∴ All possible outcomes are:
(1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1) ; (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6, 2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ All possible outcomes = 36
(i) Let E be the event that 5 does not come up either time.
∴ Numebr of favourable outcomes = [36 – (5 + 6)] = 25
∴ P(E) = \(\frac{25}{36}\)
(ii) Let N be the event that 5 will come up at least once, then number of favourable outcomes = 5 + 6 = 11
∴ P(N) = \(\frac{11}{36}\)

Question 25.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
(ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{3}\)
Solution:
(i) Given argument is not correct. Because, if two coins are tossed simultaneously then four outcomes are possible (HH, HT, TH, TT). So total outcomes is 4.
∴ The required probability = \(\frac{1}{4}\).
(ii) Given argument is correct.
Since, total numebr of possible outcomes = 6
Odd numbers = 3 and even numbers = 3
So, favourable outcomes = 3 (in both the cases even or odd).
∴ Probability = \(\frac{3}{6}=\frac{1}{2}\)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Use π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
We have,

d₁ = 4 cm
∴ r₁ = \(\frac{d_{1}}{2}\) = 2 cm
and d₂ = 2 cm
r₂ = \(\frac{d_{2}}{2}\) = 1 cm
and h = 14 cm
Volume of the glass

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
We have,
Slant height (l) = 4 cm
Circumference of one end = 2πr₁ = 18 cm
and Circumference of other end = 2πr₂ = 6 cm

⇒ πr₁ = \(\frac{18}{2}\) = 9 cm
and πr₂ = \(\frac{6}{2}\) = 3 cm
∴ Curved surface area of the frustum of the cone
= π(r₁ + r₂) l = (πr₁ + πr₂) l = (9 + 3 ) × 4 cm²
= 12 × 4 cm² = 48 cm².

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:
Here, the radius of the open side (r₁) = 10 cm
The radius of the upper base (r₂) = 4 cm
Slant height (l) = 15 cm
∴ Area of the material required = [Curved surface area of the frustum] + [Area of the top end]
= π(r₁ + r₂)l + πr₂²

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm². (Take π = 3.14)
Solution:
We have, r₁ = 20 cm, r₂ = 8 cm and h = 16 cm



Area of the bottom = πr₂²
= (\(\frac{314}{100}\) × 8 × 8) cm² = 200.96 cm²
∴ Total area of metal required
= 1758.4 cm² + 200.96 cm² = 1959.36 cm²
Cost of metal required for 100 cm² = ₹ 8
∴ Cost of metal required for 1959.36 cm²
= ₹ \(\frac{8}{100}\) × 1959.36 = ₹ 156.75

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{1}{16}\) find the length of the wire
Solution:
Let us consider the frustum DECB of the metallic cone ABC


Thus, the required length of the wire = 7964.44 m

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 10 Circles Ex 10.2

In questions 1 to 3, choose the correct option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
(A): ∵ QT is a tangent to the circle at T and OT is radius

∴ OT⊥QT
Also, OQ = 25 cm and QT = 24 cm
∴ Using Pythagoras theorem, we get
OQ² = QT² + OT²
⇒ OT² = OQ² – QT² = 25² – 24² = 49
⇒ OT = 7
Thus, the required radius is 7 cm.

Question 2.
In figure, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ =110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
(B): TQ and TP are tangents to a circle with centre O and ∠POQ = 110°
∴ OP⊥PT and OQ⊥QT
⇒ ∠OPT = 90° and ∠OQT = 90°
Now, in the quadrilateral TPOQ, we get
∠PTQ + 90° + 110° + 90° = 360° [Angle sum property of a quadrilateral]
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 360° – 290° = 70°

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
(A) : Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.
∴ OA⊥AP and OB⊥BP
⇒ ∠OAP = ∠OBP = 90°

Now, in quadrilateral PAOB, we have
∠BPA + ∠PAO + ∠AOB + ∠OBP = 360°
⇒ 80° + 90° + ∠AOB + 90° = 360°
⇒ 260° + ∠AOB = 360°
⇒ ∠AOB = 360° – 260° ⇒ ∠AOB = 100°
In right ∆OAP and right ∆OBP, we have
OP = OP [Common]
∠OAP = ∠OBP [Each 90°]
OA = OB [Radii of the same circle]
∴ ∆OAP ≅ ∆OBP [By RHS congruency]
⇒ ∠POA = ∠POB [By CPCT]
∴ ∠POA = \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 100° = 50°

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
In the figure, PQ is diameter of the given circle and O is its centre.
Let tangents AB and CD be drawn at the end points of the diameter PQ.
Since, the tangents at a point to a circle is perpendicular to the radius through the point.

∴ PQ⊥AB
⇒ ∠APQ = 90°
And PQ⊥CD
⇒ ∠PQD = 90° ⇒ ∠APQ = ∠PQD
But they form a pair of alternate angles.
∴ AB || CD
Hence, the two tangents are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
In the figure, the centre of the circle is O and tangent AB touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.
Join OP.
Since, tangent at a point to a circle is perpendicular to the radius through that point.

∴ OP⊥AB
⇒ ∠OPB = 90° ……….. (1)
But by construction, PQ⊥AB
⇒ ∠QPB = 90° ………….. (2)
From (1) and (2),
∠QPB = ∠OPB
which is possible only when O and Q coincide. Thus, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
∵ The tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OTA = 90°
Now, in the right ∆OTA, we have
OA² = OT² + AT² [Pythagoras theorem]

⇒ OT² = 5² – 4²
⇒ OT² = (5 – 4)(5 + 4)
⇒ OT² = 1 × 9 = 9 = 3²
⇒ OT = 3
Thus, the radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
In the figure, O is the common centre, of the given concentric circles.
AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.
Since, OP is the radius of the smaller circle.
∴ OP⊥AB ⇒ ∠APO = 90°
Also, radius perpendicular to a chord bisects the chord.

∴ OP bisects AB
⇒ AP = \(\frac{1}{2}\) AB
Now, in right ∆APO,
OA² = AP² + OP²
⇒ 5² = AP² + 3² ⇒ AP² = 5² – 3²
⇒ AP² = 4² ⇒ AP = 4 cm
⇒ \(\frac{1}{2}\) AB = 4 ⇒ AB = 2 × 4 = 8 cm
Hence, the required length of the chord AB is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure).

Prove that AB + CD = AD + BC
Solution:
Since, the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touches the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.
∴ AP = AS, BP = BQ,
DR = DS and CR = CQ
Adding them, we get
(AP + BP) + (CR + RD) = (BQ + QQ) + (DS + SA)
⇒ AB + CD = BC + DA

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.

Prove that ∠AOB = 90°.
Solution:
∵ The tangents drawn to a circle from an external point are equal.
∴ AP = AC ……… (1)
Join OC.
In ∆PAO and ∆CAO, we have
AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC [From (1)]
⇒ ∆PAO ≅ ∆CAO [SSS congruency]
∴ ∠PAO = ∠CAO
⇒ ∠PAC = 2∠CAO …………. (2)
Similarly, ∠CBQ = 2∠CBO ……………… (3)

Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
2∠CAO + 2∠CBO = 180° [From (2) and (3)]
⇒ ∠CAO + ∠CBO = \(\frac{180^{\circ}}{2}\) = 90° …………… (4)
Also, in ∆AOB,
∠BAO + ∠OBA + ∠AOB = 180° [Sum of angles of a triangle]
⇒ ∠CAO + ∠CBO + ∠AOB = 180°
⇒ 90° + ∠AOB = 180° [From (4)]
⇒ ∠AOB = 180° – 90°
⇒ ∠AOB = 90°

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution:
Let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right ∆OAP and right ∆OBP, we have
PA = PB [Tangents to circle from an external point]
OA = OB [Radii of the same circle]
OP = OP [Common]
⇒ ∆OAP ≅ ∆OBP [By SSS congruency]
∴ ∠OPA = ∠OPB [By CPCT]
and ∠AOP = ∠BOP
⇒ ∠APB = 2∠OPA and ∠AOB = 2∠AOP
In right ∆OAP,
∠AOP + ∠OPA + ∠PAO = 180°
⇒ ∠AOP = 180° – 90° – ∠OPA
⇒ ∠AOP = 90° – ∠OPA
⇒ 2∠AOP = 180° – 2∠OPA
⇒ ∠AOB = 180° – ∠APB
⇒ ∠AOB + ∠APB = 180°

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre O.
Since, tangents to a circle from an external point are equal in length

∴ AP = AS
BP = BQ
CR = CQ
DR = DS
On adding, we get
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
But AB = CD [Opposite sides of parallelogram]
and BC = AD
∴ AB + CD = AD + BC ⇒ 2AB = 2BC
⇒ AB = BC
Similarly, AB = DA and DA = CD
Thus, AB = BC = CD = DA
Hence, ABCD is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Solution:
Here ∆ABC circumscribes the circle with centre O. Also, radius = 4 cm
Let AC and AB touches the circle at E and F, respectively and join OE and OF.
∵ The sides BC, CA and AB touches the circle at D, E and F respectively.
∴ BF = BD = 8 cm
[ ∵ Tangents to a circle from an external point are equal]
CD = CE = 6 cm
AF = AE = x cm (say)

∴ The sides of the ∆ABC are 14 cm, (x + 6) cm and (x + 8) cm
Perimeter of ∆ABC
= [14 + (x + 6) + (x + 8)] cm
= [14 + 6 + 8 + 2x] cm
= (28 + 2x) cm
⇒ Semi perimeter of ∆ABC,
s = \(\frac{1}{2}\) [28 + 2x] cm = (14 + x) cm
∴ s – a = (14 + x) – (8 + x) = 6
s – b = (14 + x) – (14) = x
s – c = (14 + x) – (6 + x) = 8
where, a = AB, b = BC, c = AC

Squaring both sides, we get
(14 + x)² = (14 + x)3x
⇒ 196 + x² + 28x = 42x + 3x²
⇒ 2x² + 14x – 196 = 0
⇒ x² + 7x – 98 = 0
⇒ (x – 7)(x + 14) = 0
⇒ x – 7 = 0 or x + 14 = 0
⇒ x = 7 or x = -14
But x = -14 is rejected.
∴ x = 7
Thus, AB = 8 + 7 = 15 cm, BC = 8 + 6 = 14 cm and CA = 6 + 7 = 13 cm

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
We have a circle with centre O. A quadrilateral ABCD is such that the sides AB, BC, CD and DA touches the circle at P, Q, R and S respectively.
Join OP, OQ, OR and OS.
We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2
∠3 = ∠4
∠5 = ∠6 and ∠7 = ∠8
Also, the sum of all the angles around a point is 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
∴ 2(∠1 + ∠8 + ∠5 + ∠4) = 360°
⇒ (∠1 + ∠8 + ∠5 + ∠4) = 180° …………. (1)
and 2(∠2 + ∠3 + ∠6 + ∠7) = 360°
⇒ (∠2 + ∠3 + ∠6 + ∠7) = 180° ……………. (2)
Since, ∠2 + ∠3 = ∠AOB, ∠6 + ∠7 = ∠COD, ∠1 + ∠8 = ∠AOD and ∠4 + ∠5 = ∠BOC
∴ From (1) and (2), we have
∠AOD + ∠BOC = 180°
and ∠AOB + ∠COD = 180°

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 15 Probability Ex 15.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution:
Here, the number of all the possible outcomes = 5 × 5 = 25
(i) For both customers visiting on same day:
Favourbale outcomes are (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)
∴ Number of favourable outcomes = 5
∴ Required probability = \(\frac{5}{25}=\frac{1}{5}\)

(ii) For both the customers visiting on consecutive days:
Favourable outcomes are (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Wed., Tue.), (Thu., Wed.), (Fri., Thu.)
∴ Number of favourable outcomes = 8
∴ Required probability = \(\frac{8}{25}\)

(iii) For both the customers visiting on different days:
We have probability for both visiting on same day = \(\frac{1}{5}\)
∴ Probability for both visiting on different days = 1 – [Probability for both visiting on the same day]
= \(1-\left[\frac{1}{5}\right]=\frac{5-1}{5}=\frac{4}{5}\)
∴ The required probability = \(\frac{4}{5}\).

Question 2.
A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score in the two throws:

What ist the probability that the total score is
(i) even?
(ii) 6?
(iii) at least 6?
Solution:
The complete table as follows

∴ Number of all possible outcomes = 36
(i) For total score being even:
Favourable outcomes = 18
[∵ The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8,12]
∴ The required probability = \(\frac{18}{36}=\frac{1}{2}\)

(ii) For total score being 6 :
In list of scores, we have four 6’s.
∴ Favourable outcomes = 4
∴ Required probability = \(\frac{4}{36}=\frac{1}{9}\)

(iii) For toal score being at least 6:
The favourable scores are : 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12
∴ Number of favourable outcomes = 15
∴ Required probability = \(\frac{15}{36}=\frac{5}{12}\)

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Let the number of blue balls in the bag be x.
Total number of balls = x + 5 Number of possible outcomes = (x + 5).
For a blue ball, favourable outcomes = x
Probability of drawing a blue ball = \(\frac{x}{x+5}\)
Similarly, probability of drawing a red ball = \(\frac{5}{x+5}\)
Now, we have \(\frac{x}{x+5}=2\left[\frac{5}{x+5}\right]\)
⇒ \(\frac{x}{x+5}=\frac{10}{x+5}\) ⇒ x = 10
Thus the required number of blue balls = 10.

Question 4.
A box contains 12 balls out of whichxare black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
∵ The total number of balls in the box = 12
∴ Number of possible outcomes = 12
Case – I: For drawing a black ball
Number of favourable outcomes = x
∴ Probability of getting a black ball = \(\frac{x}{12}\)

Case – II: When 6 more black balls are added
Then, the total number of balls = 12 + 6 = 18
⇒ Number of possible outcomes = 18
Now, the number of black balls = (x + 6)
∴ Number of favourable outcomes = (x + 6)
∴ Required probability = \(\frac{x+6}{18}\)
According to the given condition,
\(\frac{x+6}{18}=2\left(\frac{x}{12}\right)\)
⇒ 12 (x + 6) = 36x ⇒ 12x + 72 = 36x
⇒ 36x – 12x = 72 ⇒ 24x = 72
⇒ x = \(\frac{72}{24}\) = 3
Thus, the required value of x is 3.

Question 5.
Ajar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\) Find the number of blue marbles in the jar.
Solution:
There are 24 marbles in the jar.
∴ Number of possible outcomes = 24.
Let there are x blue marbles in the jar.
∴ Number of green marbles = 24 – x
⇒ Favourable outcomes = (24 – x)
∴ Required probability for drawing a green marbles \(\frac{24-x}{24}\)
Now, according to the given condition,
\(\frac{24-x}{24}=\frac{2}{3}\)
⇒ 3(24 – x) = 2 × 24 ⇒ 72 – 3x = 48
⇒ 3x = 72 – 48 ⇒ 3x = 24 ⇒ x = \(\frac{24}{3}\) = 8
Thus, the required number of blue marbles is 8.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
A circle can have an infinite number of tangents.

Question 2.
Fill in the blanks :
(i) A tangent to a circle intersects it in _________ point(s).
(ii) A line intersecting a circle in two points is called a ______.
(iii) A circle can have ______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _______.
Solution:
(i) Exactly one
(ii) Secant
(iii) Two
(iv) Point of contact

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm
Solution:
In right ∆QPO,
OQ² = OP² + PQ²

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
We have the required figure, as shown

Here, l is the given line and a circle with centre O is drawn.
Line n is drawn which is parallel to l and tangent to the circle. Also, m is drawn parallel to line 1 and is a secant to the circle.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Solution:
(i) Let the vertices of the triangle be A(2, 3), B(-1, 0) and C(2, – 4)
Here, x₁ = 2, y₁ = 3
x₂ = -1, y₂ = 0
x₃ = 2, y₃ = -4
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [2{0 – (-4)} + (-1){-4 – (3)} + 2{3 – 0}]
= \(\frac{1}{2}\) [2(0 + 4) + (-1)(-4 – 3) + 2(3)]
= \(\frac{1}{2}\) [8 + 7 + 6] = \(\frac{1}{2}\) [21] = \(\frac{21}{2}\) sq. units

(ii) Let the vertices of the triangle be A(-5, -1), B(3, -5) and C(5, 2)
Here, x₁ = -5, y₁ = -1
x₂ = 3, y₂ = -5
x₃ = 5, y₃ = 2
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [-5{-5 – 2} + 3{2 – (-1)} + 5{-1 – (-5)}]
= \(\frac{1}{2}\) [-5{-7} + 3{2 + 1} + 5{-1 + 5}]
= \(\frac{1}{2}\) [35 + 3(3) + 5(4)]
= \(\frac{1}{2}\) [35 + 9 + 20] = \(\frac{1}{2}\) × 64 = 32 sq. units

Question 2.
In each of the following find the value of’k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5)
Solution:
The given three points will be collinear if the triangle formed by them has zero area.
(i) Let A(7, -2), B(5, 1) and C(3, k) be the vertices of a triangle.
∴ The given points will be collinear, if ar (∆ABC) = 0
or \(\frac{1}{2}\) [7(1 – k) + 5(k + 2) + 3(-2 – 1)] = 0
⇒ 7 – 7k + 5k + 10 + (-6) – 3 = 0
⇒ 17 – 9 + 5k – 7k = 0
⇒ 8 – 2k = 0 ⇒ 2k = 8 ⇒ k = \(\frac{8}{2}\) = 4
The required value of k = 4.
(ii) \(\frac{1}{2}\) [8(- 4 + 5) + k(- 5 -1) + 2(1 + 4)] = 0
⇒ 8 – 6k + 10 = 0 ⇒ 6k = 18 ⇒ k = 3.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let the vertices of the triangle be A(0, -1), B(2, 1) and C(0, 3).
Let D, E and F be the mid-points of the sides BC, CA and AB respectively.
∴ Coordinates of D are


ar(∆DEF) : ar(∆ABC) = 1 : 4.

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution:
Let A(- 4, – 2), B(- 3, – 5), C(3, – 2) and D(2, 3) be the vertices of the quadrilateral.
Let us join diagonal BD.

Now, ar(∆ABD)

Question 5.
You have studied in class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(A, -6), B(3, -2) and C(5, 2).
Solution:
Here, the vertices of the triangle are A(4, -6), B(3, -2) and C(5, 2).
Let D be the midpoint of BC.
∴ The coordinates of the point D are
\(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\) or ( 4, 0)
AD divides the triangle ABC into two parts i.e., ∆ABD and ∆ACD.

ar(∆ADC) = \(\frac{1}{2}\) [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)]
= \(\frac{1}{2}\) [-8 + 32 – 30] = \(\frac{1}{2}\) [-6] = -3
= 3 sq. units (numerically) ………… (2)
From (1) and (2),
ar(∆ABD) = ar(∆ADC)
Thus, a median divides the triangle into two triangles of equal areas.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and 8(3, 7).
Solution:
Let the required ratio be k : 1 and the point C divide them in the above ratio.
∴ Coordinates of C are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\)
Since, the point C lies on the given line 2x + y – 4 = 0.
∴ We have \(2\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)-4\) = 0
⇒ 2(3k + 2) + (7k – 2) = 4 × (k + 1)
⇒ 6k + 4 + 7k – 4k – 4 – 2 = 0
⇒ (6 + 7 – 4)k + (-2) = 0 ⇒ 9k – 2 = 0
⇒ k = \(\frac{2}{9}\)
The required ratio = k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9

Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
Let the given points be A(x, y), B( 1, 2) and C(7, 0) are collinear.
The points A, B and C will be collinear if area of ∆ABC = 0
⇒ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0 or x + 3y – 7 = 0, which is the required relation between x and y.

Question 3.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
Let the points are A (6, -6), B(3, -7) and C(3, 3)
Let P(x, y) be the centre of the circle. Since, the circle is passing through A, B and C.
∴ AP = BP = CP
Taking AP = BP, we have AP² = BP²
⇒ (x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
⇒ x² – 12x + 36 + y² + 12y + 36 = x² – 6x + 9 + y² + 14y + 49
⇒ – 12x + 6x + 12y – 14y + 72 – 58 = 0
⇒ – 6x – 2y + 14 = 0
⇒ 3x + y – 7 = 0 ……………….. (1)
Taking BP = CP, we have BP² = CP²
⇒ (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
⇒ x² – 6x + 9 + y² + 14y + 49 = x² – 6x + 9 + y² – 6y + 9
⇒ – 6x + 6x + 14y + 6y + 58 -18 = 0
⇒ 20y + 40 = 0
⇒ y = \(\frac{-40}{20}\) = -2
From (1) and (2), 3x – 2 – 7 = 0
⇒ 3x = 9 ⇒ x = 3
i.e., x = 3 and y = -2
∴ The required centre is (3, -2).

Question 4.
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
Let we have a square ABCD such that A(-1, 2) and C(3, 2) are the opposite vertices. Let B(x, y) be an unknown vertex.

Since, all sides of a square are equal.
∴ AB = BC ⇒ AB² = BC2²
⇒ (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²
⇒ x² + 2x + 1 + y² – 4y + 4
⇒ x² – 6x + 9 + y² – 4y + 4
⇒ 2x + 1 = -6x + 9
⇒ 8x = 8 ⇒ x = 1  …………………. (1)
Since, each angle of a square = 90°.
∴ ∆ABC is a right angled triangle.
∴ Using Pythagoras theorem, we have AB² + BC² = AC²
⇒ (x + 1)² + (y – 2)²] + [(x – 3)² + (y – 2)²]
= [(3 + 1)² + (2 – 2)²]
⇒ [x² + 2x + 1 + y² – 4y + 4] + [x² – 6x + 9 + y² – 4y + 4]
= [4² + 0²]
⇒ 2x² + 2y² + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16
⇒ 2x² + 2y² – 4x – 8y + 2 = 0
⇒ x² + y² – 2x – 4y + 1 = 0 …………….. (2)
Substituting the value of x from (1) into (2), we have
1 + y² – 2 – 4y + 1 = 0
⇒ y² – 4y + 2 – 2 = 0
⇒ y² – y = 0
⇒ y(y – 4) = 0
⇒ y = 0 or y = 4
Hence, the required other two vertices are (1, 0) and (1, 4).

Question 5.
The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

Solution:
(i) By taking A as the origin and AD and AB as the coordinate axes. We have P(4, 6), Q( 3, 2) and R( 6, 5) as the vertices of ∆PQR.
(ii) By taking C as the origin and CB and CD as the coordinate axes, then the vertices of ∆PQR are P(-12, – 2), Q(-13, – 6) and R(- 10, – 3)
Case I: When P(4, 6), Q(3, 2) and R(6, 5) are the vertices.
∴ ar(∆PQR) = \(\frac{1}{2}\) [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]
= \(\frac{1}{2}\) [-12 – 3 + 24] = \(\frac{9}{2}\) sq. units
Case II: When P(-12, -2), Q(-13, -6) and R(-10, -3) are the vertices.
∴ ar(∆PQR)
= \(\frac{1}{2}\) [-12(- 6 + 3) + (-13)(- 3 + 2) + (-10)(-2 + 6)]
= \(\frac{1}{2}\) [-12(-3) + (-13)(-1) + (-10) × (4)]
= \(\frac{1}{2}\) [36 + 13 – 40] = \(\frac{9}{2}\) sq. units
Thus, in both cases, the area of ∆PQR is the same.

Question 6.
The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. [Recall “The converse of basis proportionality theorem”, and “theorem of similar triangles taking their areas and corresponding sides”]
Solution:

Question 7.
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC.
(i) The median from A meets BCat D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1
(iv) What do you observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]
(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.
Solution:
We have the vertices of ∆ABC as A (4, 2), B(6, 5) and C(1, 4).
(i) Since AD is a median

∴ Coordinates of D are


Also, CR : RF = 2 : 1 i.e., the point R divides CF in the ratio 2 : 1
∴ Coordinates of R are

(iv) We observe that P, Q and R represent the same point.
(v) Here, we have A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC. Also AD, BE and CF are its medians.
∴ D, E and F are the mid points of BC, CA and AB respectively.
We know, the centroid is a point on a median, dividing it in the ratio 2 : 1.
Considering the median AD, coordinates of

Question 8.
ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Solution:
We have a rectangle whose vertices are A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1).
∵ P is mid-point of AB
∴ Coordinates of P are


We see that PQ = QR = RS = SP i.e., all sides of PQRS are equal.
∴ It can be a square or a rhombus.
But PR ≠ QS i.e., its diagonals are not equal.
∴ PQRS is a rhombus.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:
We can calculate the mean as follows :

∴ Mean = \(\overline{x}=\frac{\sum f_{i} x_{i}}{N}=\frac{162}{20}=8.1\)
Thus, mean number of plants per house is 8.1 Since, values of x_i and f_i are small, so we have used the direct method.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Let the assumed mean, a = 150
∵ Class size, h = 20
∴ \(u_{i}=\frac{x_{i}-a}{h}=\frac{x_{i}-150}{20}\)
∴ We have the following table:

Now, \(\overline{x}\) = a + h × {\(\frac{1}{N}\) Σf_iu_i}
= 150 + 20 × \(\left(\frac{-12}{50}\right)\) = 150 – \(\frac{24}{5}\)
= 150 – 4.8 = 145.20

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Solution:
Let the assumed mean, a = 18
∵ Class size, h = 2
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-18}{2}\)
Now, we have the following table:

⇒ [f + 44] (0) = 2[f – 20]
⇒ 2[f – 20] = 0 ⇒ f = 20
Thus, missing frequency is 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:
Let the assumed mean, a = 75.5
∵ Class size, h = 3
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-75.5}{3}\)
Now, we have the following table:

Thus, the mean heart beats per minute is 75.9.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained a varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Let the assumed mean, a = 57
∴ d_i = x_i – 57
Now, we have the following table:

Thus, the average number of mangoes per box = 57.19. We choose assumed mean method.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:
Let the assumed mean, a = 225
∵ Class size, (h) = 50
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-225}{50}\)
Now, we have the following table:

Thus, the mean daily expenditure on food is ₹ 211.

Question 7.
To find out the concentration of S0₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of S0₂ in the air.
Solution:
Let the assumed mean, a = 0.14
∵ Class size, (h) = 0.04
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-0.14}{0.04}\)
Now, we have the following table:

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:
Using the direct method, we have the following table:

Thus, mean number of days a student remained absent = 12.48.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:
Let the assumed mean, a = 70
∵ Class size, (h) = 10
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-70}{10}\)
Now, we have the following table:

Thus, the mean literacy rate is 69.43%