Class 10

In this article, we will share MP Board Class 10th Social Science Book Solutions Chapter 19 Consumer Awareness Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Social Science Solutions Chapter 19 Consumer Awareness

MP Board Class 10th Social Science Chapter 19 Text book Exercises

Objective Type Questions

Question 1.
Multiple Choice Questions
(Choose the correct answer from the following)

Question (i)
When was the Consumer Protection Act enforced –
(a) 1986
(b) 1996
(c) 1968
(d) None of above
Answer:
(a) 1986

Question (ii)
The meaning of consumer awareness is –
(a) Alertness towards self – rights
(b) Alertness towards self – duties
(c) Alertness towards both, self – right and duties
(d) None of above.
Answer:
(c) Alertness towards both, self – right and duties

Question (iii)
Consumer awareness is necessary for –
(a) Protection from exploitation
(b) High standard of living
(c) To check harmful consumption
(d) All of the above.
Answer:
(d) All of the above.

Question (iv)
Producer can be arbitrary regarding quality and price of commodity in –
(a) Competition
(b) Monopoly
(c) Agro products
(d) None of the above.
Answer:
(d) None of the above.

Question (v)
Agmark is a mark of safety for –
(a) Jewellery
(b) Agriculture Products
(c) Woollen clothes
(d) Electronic equipments.
Answer:
(d) Electronic equipments.

Question 2.
Fill in the blanks:

1. Electronic equipments are labelled with ………….. mark.
2. Consumers have right to ………….. against adulterated food.
3. The World Consumers Day is celebrated on ……………
4. To show scarcity of goods despite of its sufficient supply is known as ……………
5. Agmark is labelled on …………… related products.

Answer:

1. Hallmarks
2. Safety
3. 24 December
4. Black marketing
5. Agriculture.

Question 2.
Macth the coloumn:

Answer:

1. (a)
2. (c)
3. (d)
4. (e)
5. (b)

MP Board Class 10th Social Science Chapter 19 Very Short Answer Type Questions

Question 1.
What is buyer of commodity or service called?
Answer:
A consumer or purchaser of the commodity is called buyer.

Question 2.
What is meant by consumer’s exploitation?
Answer:
The process by which a consumer is misleaded and adulterated with Someone is called consumer’s exploitation.

Question 3.
State any two kinds of consumer exploitation?
Answer:
Adulteration, overcharging, under weighing and misleading.

Question 4.
What does the selling of a ticket of cinema in an increased price than its fixed price called?
Answer:
Misleading the consumer creating artificial crisis of tickets.

Question 5.
What is the consequence of having limited information about a commodity?
Answer:
Due to limited information about a commodity a consumer is trapped and exploited.

Question 6.
What is monopoly?
Answer:
The prodution or sale of a commodity is done by only a single person or institution, the process is called monopoly.

Question 7.
When is the National Consumers Day celebrated?
Answer:
To bring awareness towards the right of cosumers through awareness camps on the occassion of National Consumer Day celebrated on 24 December every year.

MP Board Class 10th Social Science Chapter 19 Short Answer Type Questions

Question 1.
Illustrate with examples the meaning of consumer awareness?
Answer:
Consumer awareness is needed to protect the consumer from the exploitation. It safe guards their interest and removes their ignorance, unawareness and literacy about the goods. Following are the main facts classify the need of making consumers ware:

1. To achieve maximum satisfaction
2. Protection against exploitation of producers
3. Control over consumption of harmful goods
4. Motivation for saving
5. Construction of healthy society.

Question 2.
How is a consumer awareness helpful in keeping a check on anarchy and harmful consumption? (MP Board 2013)
Answer:
Having full of the knowledge regarding commodity, e.g., quality, advertisement and expiry date a consumer can keep aware himself.

Question 3.
What is ISI?
Answer:
It is a mark which standardise the quality of industrial and consumer goods.

Question 4.
What is Consumer Protection Act?
Answer:
After that in 1986, the Consumer Protection Act has been passed by the Government of India. This act is famous as ‘Copra’. The main objective of this act is to decide the complaints of the consumers immediately and to make legal proceeding easy. A three tier judiciary system has been established under Copra at district, state and national levels to resolve the disputes of consumers. The court at district level hears the cases related to the claims up to Rs. 20 lakh. The claims from Rs. 20 Lakh to Rs. 1 crore are heard in the state level courts. The courts at national level hears the cases with the claims of more than. Rs. 1 crore.

MP Board Class 10th Social Science Chapter 19 Long Answer Type Questions

Question 1.
Describe the necessity and importance of consumer awamess? (MP Board 2009, 2012)
Answer:
The following needs of consumer awareness are:

1. To achieve maximum satisfaction:
The income of every individual is limited. He wants to buy maximum goods and services with his income. He gets full satisfaction only by this behaviour. Therefore it is necessary that he should get tire goods which are measured appropriately and he should not be cheated in any way. For this he should be made aWare.

2. Protection against exploitation of producers:
Producers and sellers exploit the consumers in many ways as underweighing, taking more price than the market price, selling duplicate goods etc. Big companies through their advertisment also mislead the consumers.

3. Control over consumption of harmful goods:
There are several such goods available in market which cause harm to some consumers. For example, we can take goods like cigarette, tobacco, liquor etc. The consumer education and awareness motivate people not to purchase such goods, which is very beneficial for them.

4. Motivation for saving:
The awareness controls people from wastage of money and extravagancy and inspire them to take right decision Such consumers are not by attraction of sale, concession, free gifts, attractive paking etc., due to which people can use their income in a right way and can save more.

5. Knowledge regarding solution to problems:
The consumer class is cheated due to illiteracy, innocence and lack of information. Therefore it becomes necessary that the information about their rights should be provided to them so that they cannot be cheated by producers and sellers. Through consumer awareness they are also made known to the proceedings of laws so that they can solve their problems.

6. Construction of healthy society:
Every member of the society is a consumer. So, if the consumer is aware and rationale, then complete society becomes healthy and alert towards their rights. In such a situation it is not possible for producers and sellers to cheat or deceive them.

Question 2.
How do the producer and traders exploit consumers?
Or
Describe the main types of consumer exploitation? (MP Board 2009, 2013)
Answer:
In this age of capitalism and globalisation the main objective of each producer is to maximise his profit. In each and every possible way the producer are trying to increase the sale of their products. Therefore in fulfilment of their aim they forget the side of consumers and start exploiting them. For example, overcharging, under weighing, selling of adulterated and poor quality goods, misleading the consumers by giving false advertisement etc.

Today a consumer faces difficulty in selecting the right commodity, due to expansion of commercial activities and availability of varieties of goods and services in the economy. The companies give attractive advertisement and information also to attract consumers but many times it is not proper and complete. Therefore the consumers are cheated and exploited.

The following factors are more responsible to exploit consumers:

1. Inferior quality
2. High Prices
3. Adulteration and Implirity
4. False or incomplete information
5. Artificial scarcify, black marketing and hoarding
6. Wrong measurement.

Inferior quality:
Sometimes the purchased goods are of lower quality, for example medicine sold after its expiry date, some times electronic devices either burn itself or burn the consumers through electric current. In this way the consumer is cheated and exploited.

Adulteration and impurity:
Adulteration means mixing of some cheaper material in a quality commodity. This causes harm to health of consumers. White stone chips mixed with rice, colour in spices, Khesari in tuar pulse, mixing of harmful substance in other costly items as ghee, oil etc. So as to earn more profit. Sometimes people develop disability of hands and legs due to consumption of adulterated food.

False or Incomplete Information:
Many times producers and sellers provide wrong and incomplete information to the customers. Due to this, the customers are trapped by purchasing wrong items and their money gets wasted. The correct and complete information about the goods is not given to the consumers regarding the price, quality, expiry date, impact on environment, conditions of purchasing the goods. Therefore after purchasing the goods the consumer becomes harassed.

Artificial scarcity, black marketing and hoarding:
Some times at the time of festivals etc., the traders in order to earn undue profit create an artificial scarcity of goods through hoarding and then by black marketing, they charge higher prices. Sometimes the traders say that a particular commodity is not available. But afterwards on repeated request of the customers they provide goods on double price. Kerosene oil, sugar etc. are such goods for which the sellers show scarcity and take high prices.

Question 3.
Why are the consumers exploited? Discuss any four reasons?
Or
Why are the consumers exploited? Explain the causes. (MP Board 2009, 2011)
Answer:
The following four reasons are as follows:

1. Lack of knowledge:
The main reason for exploitation of consumers is lack of knowledge. Several consumers do not have the knowledge about the price, quality, services related to commodities. Therefore, they simply trust the facts told to them by the sellers and purchase goods, thus become trapped and are exploited.

2. Limited information:
In this age of globalisation the market is full of avariety of products. Producers are free to produce. There are no fixed laws for quality and fixation of prices. There is lack of proper and correct information regarding different aspects of goods j such as price, quality, structure, terms of their use, laws of purchasing etc. Therefore the consumers selection becomes wrong and they have to bear economic loss.

3. Monopoly:
Monopoly means the right on the production and distribution of a commodity of a producer or a group of producers. In the state of monopoly producers have arbitrary behaviour regarding the prices and the quality and availability of the goods. As a result, they succeed in exploiting consumers.

4. Consumers indifference towards the market:
There is a large group of such consumers who behave indifferently regarding purchasing. There are some consumers who think as what to do, everything is alright, leave it, why to take cash memo, whatever good the shopkeeper gives is good, things should be cheap, good and durable, what is the need of the standardisation marks like ISI and AGMARK etc. The producers take full advantage of this indifferent behaviour of consumers and succeed in exploiting them.

Question 4.
What is meaning of standardistation? State standardisation of different products. (MP Board 2009, 2010)
Answer:
Maintaining quality, grades, size and constituents of the product is called standardisation. Standarsation of products is an important measures taken by the government to protect the consumer from lack of quality and varying standards of goods. In India, these, standards have been achieved through Bureau of Indian standard (BIS), which is earlier known as the Indian Standard Institution (ISI), has the responsibility of laying down the standards for industrial and consumer goods on a scientific basis and certifying the goods that met the prescribed quality and standards.

Government try to maintain the standard of things in a number of ways:

1. Every producer is liable to be prosecuted.
2. For certain articles, an ISI mark is a must to ensure its high quality and genuinity.
3. As far as food items are concerned, it is essential to indicate, its weight on each and every packet.
4. The producers of medicines have to print the date of manufacture as well as the date of expiry on the medicine.

Question 5.
What measures will you take so that you are not cheated in the market as a consumer? Discuss also the legal measures of protection.
Answer:
Following are the main measures to protect oneself from exploitation as consumer:
1. Consumer’s education:
Consumer’s education and awareness is the most important measure for the solution of exploitation. The government has made several laws for the protection of consumers. But it is observed that they are not known to general public. Therefore proper education of consumers right should be provided to them.

2. Purchase of standardised goods:
Several types of goods are available in the market. But for the safety against exploitation consumers should always buy standardised goods. Goods which are marketed as ISI, AGMARK and HALLMARK are considered standardised goods.

3. Taking of cash memo or receipt:
To take cash memo along with purchase of a good is very important. Due to this the legal proceeding can be done in case of the quality of good is inferior or does not work properly before the given period of time.

4. Not to be lured by advertisements:
Big companies publicise their products by attractive advertisements through television or other media. The advertisements are shown in such a manner that it has a psychological effect on consumers and so they become desperate to purchase the goods. But the reality is this that the consumer should be alert and aware of advertisements. Before the purchasing of goods the consumer should completely verify its quality, price and quantity etc.

5. To complain collectively:
A consumer alone cannot do anything against the producers and seller, but if complaint is lodged collectively then it has more effect.

6. To verify the expiry date:
Whenever we purchase a medicine, we should always check the expiry date. After this date the medicine has no effect and there is also a possibility of its bad effect. The similar is the case with tinned food. Therefore, it is essential to check the expiry date before purchase.

Question 6.
What are the rights of consumers? Why are these rights provided to them?
Or
Write is brief the rights of consumers. (MP Board 2009)
Answer:
Consumers have the right to buy good commodities and services from the market. The protection of law has been provided to him so that a producer or seller cannot cheat him in anyway. Generally a consumer has got the following rights:

1. Right to safety:
This is essential for producers that they should obey the rules related to the safety of consumers. The reason is if the producers do not obey these rules, then the consumer may have to bear a big risk. For example, in pressure cooker there is a safety valve which if faulty can lead to a fatal accident. The manufacturers of the safety valves should fix a high quality for it. If manufacturers do not do this, then the consumers can take help of consumers’ law.

2. Right to be informed:
When we purchase any product we see that some special informations are written on the packet. Such as Batch number of the commodity, manufacturing date, expiry date, address of manufacturing company of the good etc., when we purchase any medicine, then we get the direction regarding its side effect and dangers. When we purchase clothes, then we should have the wasting directions.

3. Right to be choose:
Consumer also have the right to choose goods and services possible at competitive prices. The seller can not force the consumer to buy a certain product.

4. Right to be heard:
The consumer interestes should receive due consideration at appropriate forms. They must be assumed that their complaints and grievances about the goods and services will be heard.

5. Right to seek redressal:
Consumers have the right to seek redressal in case of being cheated and exploited by the producer. The government has set up many consumer courts for the specific purpose.

6. Right to consumer education:
The consumer must be educated about their rights which have been granted by the law to protect their interest.

Question 7.
Describe about consumer’s movement in India?
Answer:
The consumer’s movement was started due to the dissatisfaction of the consumer’s. There were many reasons of consumers dissatisfaction like lack of food stuff, hoarding, black marketing, adulteration etc. In order to deal with such problems for the first time in 1955, Essential goods Act was passed. Through this the efforts were made to control production, distribution, price and supply of essential goods.

After this, in 1976 Weight and Standard Measure Act was passed to systematise the measurement and weight. After that in 1986, the Consumer Protection Act has been passed by the Government of India. This act is famous as ‘Copra’. The Consumer Protection Act has contributed a lot in making the consumer movement extensive in India. These courts along with deciding the disputes of consumers also guide them.

The Department of Consumer Affairs of the Government of India has launched several schemes to make the consumers aware, such as establishment of consumer organisations in every district, and consumer clubs in schools etc. To bring awareness towards the right of consumers, Awareness camps are organised at different levels. In India, 24th December, is celebrated as the ‘National Consumers Day’. As a result of consumer movement in India awareness has increased among consumers sufficiently. Today, there are more than 700 consumer organisations working. But due to the indifferen behaviour of consumers the number of effective and recognised organisations are very less.

MP Board Class 10th Social Science Chapter 19 Additional Important Questions

Objective Type Questions

Question 1.
Multiple Choice Questions
(Choose the correct answer from the following)

Question (i)
National Consumer Day is celebrated on
(a) 20 December
(b) 22 Decemeber
(c) 24 December
(d) 26 Decemeber.
Answer:
(c) 24 December

Question (ii)
The mark of Agmark and ISI is given to
(a) Industrial goods
(b) Agricultural goods
(c) Consumer goods
(d) Both (a) and (c).
Answer:
(d) Both (a) and (c).

Question (iii)
Essential Goods Act was passed in the year of –
(a) 1953
(b) 1954
(c) 1955
(d) 1956.
Answer:
(c) 1955

Question 2.
Fill in the blanks:

1. Golden jewellery are marked by ………………
2. Weights and Standard Measures Act was passed in the year of ………………
3. Consumer ……………… is necessery.
4. The Consumer Protection Act inforced in ……………… (MP Board 2009)

Answer:

1. Hallmark
2. 1976
3. Awareness
4. 1986.

Question 3.
True and False:

1. ISI is related to the right of Consumer Act.
2. Government passed the Right to Information Act in 2005.
3. Producer of a goods is itself the seller of the goods.
4. The mark which standardises the quality of agriculture products is called Agmark. (MP Board 2009, Set D)

Answer:

1. False
2. True
3. True
4. True.

Answer in One – Two Words or One Sentence:

Question 1.
What do you mean by consumer awareness?
Answer:
Consumer awareness means creating awareness of a consumer towards his rights and duties.

Question 2.
Who is consumer?
Answer:
Consumer is a person who buy commodity from the market. We all are consumers.

Question 3.
What is inflation?
Answer:
The sharp and persistent rise in the price level is known as inflation.

Question 4.
Mention two beneficiaries of price rise?
Answer:

1. Debtors
2. Farmers

Question 5.
Mention three sufferers of price rise?
Answer:

1. Salaried people
2. Creditors
3. Middle class.

Question 6.
What do you mean by standardisation of product?
Answer:
Maintaining quality, grade, size and elements of the product uniform is known as standardisation of product.

Question 7.
What is public distribution system?
Answer:
Distribution of essential commodities through government controlled points of distribution is termed as public distribution system.

Question 8.
Mention two acts concerned with the standardisation of product?
Answer:

1. The Agricultural Produce (Grading and Marketing) Act, 1937.
2. The Bureau of Indian Standard Act, 1986.

Question 9.
Name the beneficiaries of price rise.
Answer:

1. Debtors
2. Agriculture
3. Exporters
4. Job seekers
5. Manufactures, wholesalers and retailers

Question 10.
Name the sufferers from price rise?
Answer:

1. Wage earners
2. Salaried people
3. Middle class,
4. Tax payers
5. Creditors and
6. Importers.

Question 11.
Name the father of Consumer Movement?
Answer:
Ralph Nadar a consumer activist is known as the father of consumer movement.

MP Board Class 10th Social Science Chapter 19 Very Short Answer Type Questions

Question 1.
What is high prices?
Answer:
Generally the shopkeepers take more price than the fixed retail price according to their own wish. You might also have experienced that we buy a costly commodity from a shop and the same good is in less price in another shop. If we show him the printed price then he gives us some excuse such as local tax etc.

Question 2.
Why the consumer protection organisations are less effective?
Answer:
As a result of consumer movement in India awareness has increased among consumers sufficiently. Today, there are more than 700 consumer organisations working. But due to the indifferent behaviour of consumers the number of effective and recognised organisations are very less.

Question 3.
Name the consumer courts at the district, state and national levels?
Answer:

1. District Forum.
2. State Consumer Commission
3. National Consumer Commission.

Question 4.
Establish relationship of price rise or fall of goods with their demand?
Answer:
When the price of goods supply tends to fall and their supply tends to increase. When the price of goods falls the demand for goods tends to rise and their supply tends to fall.

MP Board Class 10th Social Science Chapter 19 Short Answer Type Questions

Question 1.
How inflation is stimulant to growth?
Answer:
According to Therwal and Brown A slow and steady rate of inflation provides the most powerful aid to the attainment of a steady rate of economic progress. According to this school of thought deliberate deficit financing through forced savings and channelising them to capital formation will increase output and result in all-round progress. Inflationary financing is justified theoretically on the ground of low rate of voluntary savings in under developed countries, in elastic tax structure and structural rigidities.

Question 2.
Mention factors leading to, consumer exploitation?
Answer:

1. Illiteracy and ignorance of consumers
2. Fatalism of consumers
3. Compromising attitude
4. Unrecorded sales
5. Non – standard product
6. Dishonest manufactures and sellers.

Question 3.
Write any five duties of consumer? (MP Board 2009)
Or
Write the duties of consumer? (MP Board 2009)
Answer:
The duties of consumers are:

1. Consumers should look at the quality of goods
2. Consumers should look at the guarantee of the products and services.
3. Consumers should ask for cash memo, wherever possible.
4. Consumers should purchase goods marked with ISI, AGMARK, etc.
5. Consumers should form consumer awareness organisation.
6. Consumers must make complaints for their genuine grievances.

Question 4.
How can a consumer protect himself against exploitation?
Or
Write any five measures to save the consumer from exploitation. (MP Board 2009)
Answer:
A consumer protects himself against exploitation through the following rules of carefulness:

1. To take bill, receipt, guarantee cards etc., and to preserve them carefully.
2. To increase or decrease the consumption according to the supply of commodity.
3. To have die knowledge of consumer protection laws.
4. To discourage the black marketing and smuggling.
5. One must complaint about the actual problems whether the price of product may be too less. Due to this the cheating tendency of the sellers is reduced.

MP Board Class 10th Social Science Chapter 19 Long  Answer Type Questions

Question 1.
Explain briefly sufferers of price rise?
Answer:
Adverse Consequences of Price rise (Sufferers from Price rise):

1. Wage earners:
As wages do not increase in proportion to the rise in price level, wage earners are the most hard hit.

2. Salaried people:
The real purchasing power of money has fallen in India due to inflation. The salaries do not increase in the proportion to a rise in price level. Therefore, the salaried people are loser in inflation.

3. Middle class:
The middle class is the worst sufferer in inflation. Owing to an abnormal rise in the cost of living, they are most exposed to the evil consequences of inflation.

4. Creditors:
Creditors are also losers, because they had lent out, when money was dearer; and is being paid back, when it has become cheaper.

5. Tax payers:
Tax payers are also hard hit during the inflation period. They have to bear the increasing tax burden, which government imposes to meet out its increasing expenditure.

6. Importers:
As importers have to pay more in terms of rupees, so they are losers. Inflation generates widespread unrest. It has threatened political and social stability. This is why, we plan to achieve ‘growth with stability’. Price stability is undoubtedly a basic requirement for overall stability.

Question 2.
Explain briefly factors leading to consumer exploitation?
Answer:
Factors Leading to Consumer Exploitation:
Inspite of a lot of cry about consumer awareness, it is felt that the consumer is being exploited even now. The following factors are responsible for the exploitation of consumers.

1. lliteracy and ignorance of the consumers:
The majority of our population is illiterate and ignorant. They cannot differentiate between right quality and sub – standard quality of goods. They cannot read the names and constituents of the product. Due to these weaknesses they are befooled.

2. Fatalism of consumers:
Many consumers think that whatever is supplied to them at whatever price is the gift of God. The substandard quality of goods was fated to them. No one else is responsible for their misery.

3. Compromising attitude:
Indian consumers are by nature compromising. They are not habituated of making complaints and fight with the sellers. There is no one to safeguard their interest, if they^are not going even to complain about the cheating. In law also tire aggrieved party has to move the court.

4. Unrecorded sales:
Most of the sales remain unrecorded, so no suit can be filed against such sales. In practice, we do not insist on receipts and cash memos. Consequently we are cheated and no one can help us.

5. Non – standard product:
Generally products being used by the majority of people is non – standard. There is no guarantee for non – standard product and the poor consumers are cheated.

6. Dishonest manufacturers and sellers:
The manufacturers and sellers are dishonest both in the manufacturing and supplying product. They manipulate the deal in their favour and cheat the consumers.

7. Lengthy legal process:
The educated and literate consumers also avoid filing suit in the consumer courts, because the suit lasts long. It takes long time and the consumer has to remain in tension about decision of the case.

Question 3.
Make a list of different legislations enacted to protect consumers interest?
Answer:
Legislations Ensuring Consumer Rights: In order to protect consumer’s interest the government has enacted the following legislations:

1. The Agricultural Produce (Grading and Marketing) Act, 1937.
2. Essential Commodities Act, 1955.
3. Prevention and Control of Pollution Act, 1974.
4. Environment Protection Act.
5. The Standards of Weights and Measures Act, 1976.
6. The Prevention of Goods Adulteration Act, 1976.
7. The Drugs and Cosmetics Act, 1946.
8. The Monopolies and Restrictive Trade Practices Act, 1969.
9. The Prevention of Black Marketing and Maintenance of Supplies of Essential Commodities Act, 1980.
10. The Bureau of Indian Standards Act, 1986.
11. Consumer Protection Act, 1986.

Question 4.
Explain briefly, what remedies are available to consumers?
Answer:
Remedies Available to the Consumer:

1. To remove the defect pointed out by the appropriate laboratory from the goods supplied.
2. To replace the defective goods with new goods of similar description which are free from any defect.
3. To return the complainant the price paid for the goods or the charges paid for the services.
4. To pay the aggrieved consumer such amount as may be fixed by the Forum as compensation for any loss or injury suffered by the consumer due to the negligence of the opposite party.
5. To remove the defects or deficiencies in the services supplied.
6. To discontinue the unfair trade practice or the restrictive trade practice or not to repeat them.
7. Not to offer the hazardous goods for sale.
8. To withdraw the hazardous goods from being offered for sale.
9. To provide for adequate costs to parties.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.
Which term of the AP : 121,117,113, ……., is its first negative term? [Hint: Find n for a_n < 0]
Solution:
We have the A.P. having a = 121 and d = 117 – 121 = -4

Thus, the first negative term is 32nd term.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP
Solution:

Question 3.
A ladder has rungs 25 cm apart (see figure).The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \(2 \frac{1}{2}\) m apart, what is the length of the wood required for the rungs? 250
[Hint: Number of rungs = \(\frac{250}{25}+1\)]

Solution:

Length of the 1^st rung (bottom rung) = 45 cm
Length of the 11^th rung (top rung) = 25 cm
Let the length of each successive rung decreases by x cm.
∴ Total length of the rungs = 45 cm + (45 – x) cm + (45 – 2x) cm + ….. + 25 cm
Here, the number 45, (45 – x), (45 – 2x), …., 25 are in an AP such that
First term (a) = 45
and last term (l) = 25
Number of terms, n = 11

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. [Hint: S_x-1 = S₄₉ – S_x]
Solution:
We have the following consecutive numbers on the houses of a row; 1, 2, 3, 4, 5, ….. , 49.
These numbers are in AP, such that a = 1, d = 2 – 1 = 1, n = 49
Let one of the houses be numbered as x.
∴ Number of houses preceding it = x – 1
Number of houses following it = 49 – x
Now, the sum of the house-numbers preceding

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see fig.) Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the 1st step \(\frac{1}{4} \times \frac{1}{2} \times 50 m^{3}\)]

Solution:


\(d=\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\)
Here, total number of steps n = 15
Total volume of concrete required to build 15 steps is given by the sum of their individual volumes.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x³ + x² – 5x + 2; \(\frac{1}{2}\), 1, -2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Solution:

Again, p(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
⇒ 1 is a zero of p(x).
Also, p(-2) = 2(-2)³ + (-2)² – 5(-2) + 2
= -16 + 4 + 10 + 2 = -16 +16 = 0
= -2 is a zero of p(x).
Now, p(x) = 2x³ + x² – 5x + 2
∴ Comparing it with ax³ + bx² + cx + d, we have a = 2, b = 1, c = -5, and d = 2

Thus, the relationship between the coefficients and the zeroes of p(x) is verified.

(ii) Here, p(x) = x³ – 4x² + 5x – 2
∴ p(2) = (2)³ – 4(2)² + 5(2) – 2
= 8 – 16 + 10 – 2 = 18 – 18 = 0
⇒ 2 is a zero of p(x)
Again p(1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2 = 6 – 6 = 0
⇒ 1 is a zero of p(x)
Now, comparing p(x) = x³ – 4x² + 5x – 2
with ax³ + bx² + cx + d = 0, we have
a = 1, b = -4, c = 5 and d = -2
Also 2, 1 and 1 are the zeroes of p(x).
Let α = 2,
β = 1,
γ = 1
Now, sum of zeroes = α + β + γ
= 2 + 1 + 1 = 4 = -b/a

Thus, the relationship between the zeroes and the coefficients of p(x) is verified.

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:
Let the required cubic polynomial be ax³ + 6x² + cx + d = 0 and its zeroes be α, β and γ.

∴ The requied cubic polynomial is
1x³ + (-2)x² + (-7)x + 14 = 0
= x³ – 2x² – 7x + 14 = 0

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a -b, a, a + b, find a and b.
Solution:
We have p(x) = x³ – 3x² + x + 1
Comparing it with Ax³ + Bx² + Cx + D,
We have A = 1, B = -3, C = 1 and D = 1
∵ It is given (a – b), a and (a + b) are the zeroes of the polynomial.

Question 4.
If two zeroes of the polynomial
x⁴ – 6x³ – 26x²+ 138x – 35 are 2±\(\sqrt{3}\), find other zeroes.
Solution:
Here, p(x) = x⁴ – 6x³ – 26x³ + 138x – 35
∵ Two of the zeroes of p(x) are : 2 ± \(\sqrt{3}\)

(x² – 4x + 1)(x² – 2x – 35) = p(x)
⇒ (x² – 4x + 1) (x² – 7x + 5x – 35) = p(x)
⇒ (x² – 4x + 1) [x(x – 7) + 5(x – 7)] = p(x)
⇒ (x² – 4x + 1)(x – 7)(x + 5) = p(x)
i.e., (x – 7) and (x + 5) are other factors of p(x).
∴ 7 and – 5 are other zeroes of the given polynomial.

Question 5.
If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
Applying the division algorithm to the polynomials x⁴ – 6x³ + 16x² – 25x + 10 and x² – 2x + k, we have

∴ Remainder = (2k – 9)x – k(8 – k) + 10
But the remainder = x + a (Given)
Therefore, comparing them, we have

and a = -k(8 – k) + 10
= -5(8 – 5) + 10
= -5(3) + 10 = -15 + 10 = -5
Thus k = 5 and a = -5

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) \(\frac{x}{2}+\frac{2 y}{3}\) = -1 and \(x-\frac{y}{3}\) = 3
Solution:
(i) Elimination method :
x + y = 5 … (1)
2x – 3y = 4 …. (2)
Multiplying (1) by 3, we get
3x + 3y = 15 …. (3)
Adding (2) and (3) , we get
2x – 3y + 3x + 3y = 19
⇒ 5x = 19 ⇒ x = \(\frac{19}{5}\)
Now, putting x = \(\frac{19}{5}\) in (1) , we get

Substitution Method :
x + y = 5 ⇒ y = 5 – x … (1)
2x – 3y = 4 … (2)
Put y = 5 – x in (2), we get
2x – 3(5 – x)= 4 ⇒ 2x – 15 + 3x = 4

(ii) Elimination method :
3x + 4y = 10 … (1)
2x – 2y = 2 … (2)
Multiplying equation (2) by 2, we have
⇒ 4x – 4y = 4 … (3)
Adding (1)and (3) , we get
3x + 4y + 4x – 4y = 10 + 4
⇒ 7x = 14 ⇒x = \(\frac{14}{7}\) = 2
Putting x = 2 in (1) , we get,
3(2)+ 4y = 10

(iii) Elimination method :
3x – 5y – 4 = 0 … (1)
9x = 2y + 7 ⇒ 9x – 2y – 7 = 0 …. (2)
Multiplying equation (1) by (3) , we get
⇒ 9x – 15y – 12 = 0(3)
Subtracting (2)from (3),
⇒ 9x – 15y – 12 – 9x + 2y + 7 = 0

(iv) Elimination method :

Question 2.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces 1
to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
(i)Let the numerator = x
and the denominator = y

(ii) Let, present age of Nuri = x years
and the present age of Sonu = y years
Age of Nuri = (x – 5) years
Age of Sonu = (y – 5) years
According to question:
Age of Nuri = 3[Age of sonu]
⇒ x – 5 = 3[y – 5] ⇒ x – 5 = 3y – 15
⇒ x – 3y + 10 = 0 … (1)
10 years later :
Age of Nuri = (x + 10) years,
Age of Sonu = (y + 10) years,
According to question :
Age of Nuri = 2[Age of Sonu]
⇒ x + 10 = 2(y + 10) ⇒ x + 10 = 2y + 20
⇒ x – 2y -10 = 0 …. (2)
Subtracting (1) from (2),
x – 2y – 10 – x + 3y – 10 = 0 ⇒ y – 20 = 0 ⇒ y = 20
Putting y = 20 in (1), we get
x – 3(20) + 10 = 0 ⇒ x – 50 = 0 ⇒ x = 50
Thus, x = 50 and y = 20
∴ Age of Nuri = 50 years
and age of Sonu = 20 years

(iii) Let the digit at unit’s place = x
and the digit at ten’s place = y
∴ The number = 10y + x
The number obtained by reversing the digits = 10x + y
According to question,
9[The number] = 2 [Number obtained by reversing the digits]
9[10y+ x] = 2[10x + y]
90y + 9x = 20x + 2y
x – 8y = 0 …. (1)
Also sum of the digits = 9
x + y = 9 …. (2)
Subtracting (1) from (2), we have
x + y – x + 8y = 9
⇒ 9y = 9 ⇒ y = 1
putting y = 1 in (2), we get
x + 1 = 9 ⇒ x = 8
Thus, x = 8 and y = 1
∴ The required number = 10y + x = (10 × 1) + 8 = 10 + 8 = 18

(iv) Let the number of 50 rupees notes = x
and the number of 100 rupees notes = y
According to the condition,
Total number of notes = 25
∴ x + y = 25 …. (1)
Also, the value of all the notes = ₹ 2000
∴ 50x + 100y = 2000 ⇒ x + 2y = 40 …. (2)
Subtracting equation (1) from (2), we get
x + 2y – x – y = 40 – 25
⇒ y = 15
Putting y = 15 in (1),
x + 15 = 25
⇒ x = 25 – 15 = 10
Thus, x = 10 and y = 15
∴ Number of 50 rupees notes = 10
and number of 100 rupees notes = 15

(v) Let the fixed charge (for the three days) = ₹ x
and the additional charge for each extra day = ₹ y
According to question,
Charge for 7 days = ₹ 27
⇒ x + 4y = 27
[∵ Extra days = 7 – 3 = 4]
Also, charge of 5 days = ₹ 21
⇒ x + 2y = 21
[∵ Extra days = 5 – 3 = 2]
Subtracting (2) from (1), we get
x + 4y – x – 2y = 27 – 21
\(\Rightarrow \quad 2 y=6 \Rightarrow y=\frac{6}{2}=3\)
Putting y = 3 in (2), we have
x + 2(3) = 21
⇒ x = 21 – 6 = 15
So, x = 15 and y = 3
∴ Fixed charge = ₹ 15
and additional charge per day = ₹ 3

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²
Solution:
(i) Here, dividend p(x) = x³ – 3x² + 5x – 3, and divisor g(x) = x² – 2
∴ We have

Thus, the quotient = (x – 3) and remainder = (7x – 9)

(ii) Here, dividend p(x) = x⁴ – 3x² + 4x + 5 and divisor g(x) = x² + 1 – x = x² – x + 1
∴ We have

Thus, the quotient = (x² + x – 3) and remainder = 8

(iii) Here, dividend, p(x) = x⁴ – 5x + 6 and divisor, g(x) = 2 – x² = – x² + 2
∴ We have

Thus, the quotient = -x² – 2 and remainder = -5x +10

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t² – 3; 2t⁴ + 3t³ -2t² – 9t – 12
(ii) x² + 3x + 1; 3x⁴ + 5x³ – 7x² + 2x +2
(iii) x³ – 3x + 1; x⁵ – 4x³ + x² + 3x + 1
Solution:
(i) Dividing 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3, we have

∵ Remainder = 0
∴ (t² – 3) is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) Dividing 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1, we have

∵ Remainder = 0
∴ x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) Dividing x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1, we get

∵ Remainder = 2, i.e., remainder = 0
∴ x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x +1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)
Solution:
We have p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5.
Given \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are zeroes of p(x).

Thus, the other zeroes of the given polynomial are -1 and -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Solution:
Here, dividend, p(x) = x³ – 3x² + x + 2, divisor = g(x), quotient = (x – 2) and remainder = (-2x + 4)
Since, (Quotient × Divisor) + Remainder = Dividend
∴ [(x – 2) × g(x)] + [(-2x + 4)] = x³ – 3x² + x + 2
⇒ (x – 2) × g(x)
= x³ – 3x² + x + 2 – (-2x + 4)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
= x³ – 3x² + 3x – 2

Thus, the required divisor g(x) = x² – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) p(x) = 3x² – 6x + 27,
g(x) = 3 and q(x) = x² – 2x + 9.
Now, deg p(x) = deg q(x)
r(x) = 0
⇒ p(x) = q(x) × g(x) + r(x)

(ii) p(x) = 2x³ – 2x² + 2x + 3,
g(x) = 2x² – 1, and
r(x) = 3x + 2, deg q(x) = deg r(x)
⇒ p(x) = q(x) × g(x) + r(x)

(iii) p(x) = 2x³ – 4x² + x + 4,
g(x) = 2x² + 1,
q(x) = x – 2 and r(x) = 6,
deg r(x) = 0
⇒ p(x) = q(x) × g(x) + r(x)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

In each of the following, give the justification of the construction also.

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction :
I. Draw a line segment AB = 7.6 cm.
II. Draw a ray AX making an acute angle with AB.
III. Mark 13 = (8 + 5) equal points on AX, such that AX₁ = X₁X₂ = ……….. X₁₂X₁₃.
IV. Join points X₁₃ and B.
V. From point X₅, draw X₅C || X₁₃B, which meets AB at C.
Thus, C divides AB in the ratio 5 : 8 On measuring the two parts, we get AC = 2.9 cm and CB = 4.7 cm.

Justification:
In ∆ABX₁₃ and ∆ACX₅, we have
CX₅ || BX₁₃
∴ \(\frac{A C}{C B}=\frac{A X_{5}}{X_{5} X_{13}}=\frac{5}{8}\) [By Thales theorem]
⇒ AC : CB = 5 : 8.

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
I. Draw a ∆ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.
II. Draw a ray BX making an acute angle ∠CBX.
III. Mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X₁, X₂, X₃ on BX₁ such that BXj = X₁X₂ = X₂X₃.

IV. Join X₃C.
V. Draw a line through X₂ such that it is parallel to X₃C and meets BC at C’.
VI. Draw a line through C parallel to CA which intersect BA at A’.
Thus, ∆A’BC’ is the required similar triangle.
Justification :
By construction, we have X₃C || X₂C’

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
I. Construct a ∆ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.

II. Draw a ray BX such that ∠CBX is an acute angle.
III. Mark 7 points of X₁, X₂, X₃, X₄, X₅, X₆ and X₇ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ – X₄X₅ = X₅X₆ = X₆X₇
IV Join X₅ to C.
V. Draw a line through X₇ intersecting BC (produced) at C’ such that X₅C || X₇C’
VI. Draw a line through C’ parallel to CA to intersect BA (produced) at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ Using AA similarity, ∆ABC ~ ∆A’BC’

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :
I. Draw BC = 8 cm
II. Draw the perpendicular bisector of BC which intersects BC at D.

III. Mark a point A on the above perpendicular such that DA = 4 cm.
IV. Join AB and AC.
Thus, ∆ABC is the required isosceles triangle.
V. Now, draw a ray BX such that ∠CBX is an acute angle.
VI. On BX, mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X₁, X₂ and X₃ such that BX₁ = X₁X₂ = X₂X₃
VII. Join X₂C.
VIII. Draw a line through X₃ parallel to X₂C and intersecting BC (extended) to C’.
IX. Draw a line through C’ parallel to CA intersecting BA (extended) to A’, thus, ∆A’BC’ is the required triangle.
Justification:
We have C’A’ || CA [By construction]
∴ Using AA similarity, ∆A’BC’ ~ ∆ABC

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution:
Steps of construction :
I. Construct a ∆ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
II. Draw a ray BX such that ∠CBX is an acute angle.

Mark four points [greater of 3 and 4 in \(\frac{3}{4}\)] X₁, X₂, X₃, X₄ on BX such that 4
BX₁ = X₁X₂ = X₂X₃ = X₃X₄
IV. Join X₄C and draw a line through X₃ parallel to X₄C to intersect BC at C’.
V. Also draw another line through C’ and parallel to CA to intersect BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
In ∆BX₄C we have
X₄C || X₃C’ [By construction]

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°.Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC.
Solution:
Steps of Construction :
I. Construct a AABC such that BC = 7 cm, ∠B = 45°, ∠A = 105° and ∠C = 30°
II. Draw a ray BX making an acute angle ∠CBX with BC.

III. On BX, mark four points [greater of 4 and 3 in \(\frac{4}{3}\) ] X₁, X₂, X₃ and X₄ such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄.
IV. Join X₃C.
V. Draw a line through X₄ parallel to X₃C intersecting BC(extended) at C’.
VI. Draw a line through C parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle. Justification:
By construction, we have
C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
I. Construct the right triangle ABC such that ∠B = 90°, BC = 4 cm and BA = 3 cm.
II. Draw a ray BX such that an acute angle ∠CBX is formed.
III. Mark 5 points X₁, X₂, X₃, X₄ and X₅ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅.
IV. Join X₃C.
V. Draw a line through X₅ parallel to X₃C, intersecting the extended line segment BC at C’.
VI. Draw another line through C’ parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]

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 15.2 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 4 Quadratic Equations Ex 4.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² – 7x + 3 = 0
(ii) 2x² + x – 4 = 0
(iii) 4x² + 4\(\sqrt{3}\)x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) We have, 2x² – 7x + 3 = 0
Dividing both sides by 2, we get

(ii) We have 2x² + x – 4 = 0
Divide both sides by 2, we get

(iv) We have, 2x² + x + 4 = 0
Dividing both sides by 2, we get

Since, the square of a number cannot be negative.
∴ There is no real value of x satisfying the given equation.

Question 2.
Find the roots of the following quadratic equations, using the quadratic formula:
(i) 2x² – 7x + 3 = 0
(ii) 2x² + x – 4 = 0
(iii) 4x² + 4\(\sqrt{3}\)x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) Comparing the given equation with ax² + bx + c = 0, we get a = 2, b = -7, c = 3
∴ b² – 4ac = (-7)² – 4(2)(3) = 49 – 24 = 25 > 0
Since b² – 4ac > 0, therefore the given equation has real roots, which are given by

Taking negative sign, x = \(\frac{7-5}{4}=\frac{2}{4}=\frac{1}{2}\)
Thus, the roots of the given equation are 3 and \(\frac{1}{2}\).

(ii) Comparing the given equation with ax² + bx + c = 0, we get a = 2, b = 1, c = – 4
b² – 4ac = (1)² – 4(2)(-4) = 1 + 32 = 33 > 0
Since b² – 4ac > 0, therefore the given equation has real roots, which are given by

(iii) Comparing the given equation with
ax² + bx + c = 0, we get
a = 4, b= 473, c = 3
b² – 4ac = (473)² – 4(4)(3)
= (16 × 3) – 48 = 48 – 48 = 0
Since b² – 4ac = 0, therefore the given equation has real and equal roots, which are given by

(iv) Comparing the given equation with ax² + bx + c = 0, we get a = 2, b = 1, c = 4
b² – 4ac = (1 )² – 4(2)(4) = 1 – 32 = -31 < 0 Since
b² – 4ac < 0, therefore the given equation does not have real roots.

Question 3.
Find the roots of the following equations:

Solution:


Taking negative sign, x = \(\frac{3-1}{2}\)
Thus, the required roots of the given equation are 2 and 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:
Let the present age of Rehman be x years.
3 years ago, Rehman’s age = (x – 3) years
5 years later, Rehman’s age = (x + 5) years
According to the condition,

⇒ 3(2x + 2) = x² + 2x – 15
⇒ 6x + 6 = x² + 2x – 15
⇒ x² + 2x – 6x – 15 – 6 = 0
⇒ x² – 4x – 21 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0,
we get a = 1, b = -4, c = -21
b² – 4ac = (-4)² – 4(1)(-21) = 16 + 84 = 100

Since, age cannot be negative.
∴ x ≠ -3 ⇒ x = 7
So, the present age of Rehman = 7 years

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics = x
∴ Marks in English = 30 – x
According to the condition,
(x + 2) × [(30 – x) – 3] = 210
⇒ (x + 2) × (30 – x – 3) = 210
⇒ (x + 2)(- x + 27) = 210
⇒ -x² + 25x + 54 = 210
⇒ -x² + 25x + 54 – 210 = 0
⇒ -x² + 25x – 156 = 0
⇒ x² – 25x + 156 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0, we get a = 1,b = -25, c = 156

When x = 12, then 30 – x = 30 – 12 = 18 Thus, marks in Mathematics = 13, marks in English = 17 or
Marks in Mathematics = 12, Marks in English = 18.

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side i.e., breadth = x metres

∴ The longer side i.e., length = (x + 30) metres and diagonal = (x + 60) metres
In a rectangle,
(diagonal)² = (breadth)² + (length)²
⇒ (x + 60)² = x² + (x + 30)²
⇒ x² + 120 x + 3600 = x² + x² + 60x + 900
⇒ x² + 120x + 3600 = 2x² + 60x + 900
⇒ 2x² – x² + 60x – 120x + 900 – 3600 = 0
⇒ x² – 60x – 2700 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0, we get a = 1, b = -60, c = -2700
∴ b² – 4ac = (-60)² – 4(1)(-2700)
⇒ b² – 4ac = 3600 + 10800
⇒ b² – 4ac = 14400

Since breadth cannot be negative,
x ≠ -30 ⇒ x = 90
∴ x + 30 = 90 + 30 = 120
Thus, the shorter side is 90 m and the longer side is 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the larger number be x.
Since, (smaller number)² = 8(larger number)
⇒ (smaller number)² = 8
⇒ smaller number = \(\sqrt{8 x}\)
According to the condition,

Thus, the smaller number = 12 or -12
Thus, the two numbers are 18 and 12 or 18 and -12.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:


Thus, speed of the train is 40 km/hr.

Question 9.
Two water taps together can fill a tank in \(9 \frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the smaller tap fills the tank in x hours
∴ The larger tap fills the tank in (x – 10) hours.
Amount of water flowing through both the taps in one hour


[ ∵ Time cannot be negative]
x = 25 ⇒ x – 10 = 25 – 10 = 15
Thus, time taken to fill the tank by the smaller tap alone is 25 hours and by the larger tap alone is 15 hours.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of the passenger train be x km/h.
∴ Average speed of the express train = (x + 11) km/h
Total distance covered = 132 km

Comparing equation (1) with ax² + bx + c = 0, we get a = 1, b = 11, c = -1452

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the side of the smaller square be x m.
⇒ Perimeter of the smaller square = 4x m
∴ Perimeter of the larger square = (4x + 24) m
⇒ Side of the larger square

Area of the smaller square = x² m²
Area of the larger square = (x + 6)² m²
According to the condition,
x² + (x + 6)² = 468
⇒ x² + x² + 12x + 36 = 468
⇒ 2x² + 12x – 432 = 0
⇒ x² + 6x – 216 = 0 …(1)
[Dividing both sides by 2] Comparing equation (1) with ax² + bx + c = 0, we get
a = 1, b = 6, c = -216
b² – 4ac = (6)² – 4(1)(-216) = 36 + 864 = 900

But the length of a square cannot be negative,
∴ x ≠ -18 ⇒ x = 12
Length of the smaller square = 12 m
and the length of the larger square = x + 6
= 12+ 6 = 18 m

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 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 13.2

Unless stated otherwise, take π = \(\frac{22}{7}\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Solution:
Here, r = 1 cm and h = 1 cm.
Volume of the conical part = \(\frac{1}{3}\) πr²h and volume of the hemispherical part = \(\frac{2}{3}\) πr³h

∴ Volume of the solid shape
= \(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr³h = \(\frac{1}{3}\) πr²h[h + 2r]
= \(\frac{1}{3}\) π(1)² [1 + 2(1)] cm³
= (\(\frac{1}{3}\) π × 1 × 3) cm³ = π cm³

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Here, diameter = 3 cm
⇒ Radius (r) = \(\frac{3}{2}\) cm
Total height = 12 cm
Height of a cone (h₁) = 2 cm

∴ Height of both cones = 2 × 2 = 4 cm
⇒ Height of the cylinder (h₂) = (12 – 4) cm = 8 cm.
Now, volume of the cylindrical part = πr²h₂
Volume of both conical parts = 2[\(\frac{1}{3}\) πr²h₁]
∴ Volume of the whole model

Question 3.
A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

Solution:
Since a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.

Diameter = 2.8 cm
⇒ Radius (r) = 1.4 cm
∴ Length of the cylindrical part (h)
= 5 cm – (1.4 + 1.4) cm = 5 cm – 2.8 cm = 2.2 cm
Now, volume of the cylindrical part = πr²h and volume of both the hemispherical ends
= 2(\(\frac{2}{3}\) πr³) = \(\frac{4}{3}\) πr³
∴ Volume of a gulab jamun

Volume of 45 gulab jamuns

Question 4.
A pen stand made up of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).

Solution:
Dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm.
∴ Volume of the cuboid = 15 × 10 × \(\frac{35}{10}\) cm³
= 525 cm³
Since, each depression is conical in shape with base radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of each depression
= \(\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times\left(\frac{5}{10}\right)^{2} \times \frac{14}{10} \mathrm{cm}^{3}=\frac{11}{30} \mathrm{cm}^{3}\)
Since, there are 4 depressions
∴ Total volume of 4 depressions = \(\frac{44}{30}\) cm³
Now, volume of the wood in entire stand = [Volume of the wooden cuboid] – [Volume of 4 depressions]
= 525cm³ – \(\frac{44}{30}\) cm³
= \(\frac{15750-44}{30} \mathrm{cm}^{3}=\frac{15706}{30} \mathrm{cm}^{3}\)
= 523.53 cm³

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Height of the conical vessel (h) = 8 cm
Base radius (R) = 5 cm
Volume of water in conical vessel = \(\frac{1}{3}\) πR²h

Thus, the required number of lead shots = 100

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)
Solution:
Height of the big cylinder (h) = 220 cm
Base radius (r) = \(\frac{24}{2}\) cm = 12cm
∴ Volume of the big cylinder
= πr²h = (π(12)² × 220) cm³
Also, height of smaller cylinder (h₁) = 60 cm

Base radius (r₁) = 8 cm
∴ Volume of the smaller cylinder πr₁²h₁
= (π(8)² × 60) cm³
∴ Volume of iron pole = [Volume of big cylinder] + [Volume of the smaller cylinder]
= (π × 220 × 12² × π × 60 × 8²) cm³
= 3.14[220 × 12 × 12 + 60 × 8 × 8] cm³

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Height of the conical part = 120 cm
Base radius of the conical part = 60 cm
Volume of the conical part
= \(\frac{1}{3}\) πr²h = [\(\frac{1}{3} \times \frac{22}{7}\) × (60)² × 120] cm³
Radius of the hemispherical part = 60 cm.

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Volume of the cylindrical part = πr²h
= (3.14 × 1² × 8) cm³ = \(\frac{314}{100}\) × 8 cm³

∴ Total volume of the glass vessel = [Volume of cylindrical part] + [Volume of spherical part]

⇒ Volume of water in the vessel = 346.51 cm³
Since the child finds the volume as 345 cm³
∴ The child’s answer is not correct.
The correct answer is 346.51 cm³.