Author name: Prasanna

Devotional Paths to the Divine Class 7 Questions and Answers Provided helps you to answer complex Questions too easily. You can use them while preparing for board exams and all of them are given by subject experts. Reading NCERT Solutions for Class 7 Social Science History Chapter 8 Devotional Paths to the Divine familiarizes you with the kind of questions appearing in the board exams. Students are advised to read these solutions on a regular basis to score well.

Devotional Paths to the Divine Class 7 Questions and Answers History Chapter 8

Make your learning experience enjoyable by preparing from the quick links available on this page. Use the Class 7 SST History Chapter 8 NCERT Solutions and get to know different concepts involved. All the Solutions are covered as per the latest syllabus guidelines. Knowing the NCERT Class 7 History Chapter 8 Questions and Answers helps students to attempt the exam with confidence.

Class 7 History Chapter 8 NCERT Textbook Questions and Answers

Let’s Recall
1. Match the following:

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

Question 2.
Fill in the blanks:
(i) Shankara was an advocate of _________ .
(ii) Ramanuja was influenced by the _________ .
(iii) ______, _______, and _______ were advocates of Virashaivism.
(iv) _______ was an important centre of the Bhakti tradition in Maharashtra.
(v) Mirabai was married into the royal family of ________ .
(vi) Narsi Mehta was a famous saint belong from the state ________ .
(vii) Jalaluddin Rumi was a great Sufi poet from ________ .
(viii) Bijak is collection related to ________ .
(ix) Martin Luther was a ________ .
(x) Khalsa Panth was formed by ________ .

Answer:
(i) Advaita
(ii) Alvars
(iii) Basavanna, Allama Prabhu, Akkamahadevi
(iv) Pandharpur
(v) Mewar
(vi) Gujarat
(vii) Iran
(viii) Kabir Das
(ix) German professor
(x) Guru Gobind Singh

Question 3.
Describe the beliefs and practices of the Nathpanthis, Siddhas and Yogis.
Answer:
Nathpanthis, Siddhas and Yogis that emerged during this period criticized the ritual and other aspects of conventional religion and the social order, using simple, logical arguments. They advocated renunciation of the world. To achieve the salvation they advocated intense training of the mind and body through practices like yogasanas, breathing exercises and meditation. It became popular among low’ castes.

Question 4.
What were the major ideas expressed by Kabir? How did he express these?
Answer:
Kabir’s ideas are found in the form of sakhis and pads. Kabir’s verses are also in the Guru Granth Sahib, Panch Vani and Bijak which is collection of Kabir’s verses. Kabir’s teachings openly ridiculed all forms of external worship of both Brahmanical Hinduism and Islam, the pre-eminence of the priestly classes and the caste system.

He expressed his ideas in the language called ‘sadhukkadi’ or ‘khadibhasha’ and it is a combination of words of many languages, which were spoken and widely understood by the ordinary people.

Let’s Understand

Question 5.
What were the major beliefs and practices of the Sufis?
Answer:
Major beliefs and practices of the Sufis were:

1. Sufis were Muslim mystics. They laid emphasis on love and devotion to God.
2. They totally followed Shariat;
3. They rejected idol worship;
4. They prefer collective prayer called Namaz;
5. They developed elaborate methods of training using chanting the name of almighty God;
6. Sufis composed poems expressing their feelings and a rich literature in prose, including anecdotes and fables, developed around them etc.

Question 6.
Why do you think many teachers rejected prevalent religious beliefs and practices?
Answer:
Many teachers rejected prevalent religious beliefs and practices because these were not good for the society. Many beliefs or traditions was discriminated among men and women, rich and poor, higher caste and lower caste, brahmanas and non-brahmanas. Apart from this, they believed that all human beings are equal.

Question 7.
What were the major teachings of Baba Guru Nanak?
Answer:
The major teachings of Baba Guru Nanak were:

1. Worship of one God
2. He believed that caste, creed or gender was irrelevant for attaining liberation. Thus, he emphasized on equality
3. To use nam, dan and isnan for the essence as right worship (nam), welfare of others (dan) and purity of conduct (isnan)
4. To pursue active life with a strong sense of social commitment
5. He gave importance to honest living and helping others.

Let’s Discuss

Question 8.
For either the Virashaivas or the sants of Maharashtra, discuss their attitude towards caste.
Answer:
Virashaiva movement was initiated by Basavanna and his companions Allama Prabhu and
Akkamahadevi in Karnataka in mid-12th century. They argued strongly for equality of all human beings, opposed Brahmanical ideas on caste and treatment of women. They were also against all forms of ritual and idol worship.

The most important saints of Maharashtra were Jananeshwar, Namdev, Eknath and Tukaram as well as women like Sakhubai who belonged to the “untouchable” Mahar caste. These saints rejected all forms of ritualism, outward display of piety and social differences based on birth.

Question 9.
Why do you think ordinary people preserved the memory of Mirabai?
Answer:
A unique feature of most of the saints is that their works were composed in regional languages and could be sung. She also became immensely popular and her bhajans were sung orally from generation to generation. Usually the poor, the most deprived communities and women transmitted these songs, often adding their own experiences.

Therefore, we can say that the language of Mirabai’s bhajans and their theme of love with Lord Krishana, encouraged the ordinary people to preserve the memory.

Let’s Do

Question 10.
Find out whether in your neighbourhood there are any dargahs, gurudwaras or temples associated with saints of the bhakti tradition in your neighbourhood. Visit any one of these and describe what you see and hear.
Answer:
Students discuss with class teacher and can write the answer.

Question 11.
For any of the saint-poets whose compositions have been included in this chapter, find out more about their works, noting down other poems. Find out whether these are sung, how they are sung, and what the poets wrote about.
Answer:
Students discuss with class teacher and can write the answer.

Question 12.
There are several saint-poets whose names have been mentioned but their works have not been included in the chapter. Find out more about the language in which they composed, whether their compositions were sung, and what their compositions were about.
Answer:
Students discuss with class teacher and can write the answer.

Hope the data shared above regarding the NCERT Class 7 Social Science History Chapter 8 Devotional Paths to the Divine PDF has aided in your exam preparation. If you ever need any assistance you can always reach us and our team will guide you at the soonest possibility.

Tribes, Nomads and Settled Communities Class 7 Questions and Answers Provided helps you to answer complex Questions too easily. You can use them while preparing for board exams and all of them are given by subject experts. Reading NCERT Solutions for Class 7 Social Science History Chapter 7 Tribes, Nomads and Settled Communities familiarizes you with the kind of questions appearing in the board exams. Students are advised to read these solutions on a regular basis to score well.

Tribes, Nomads and Settled Communities Class 7 Questions and Answers History Chapter 7

Make your learning experience enjoyable by preparing from the quick links available on this page. Use the Class 7 SST History Chapter 7 NCERT Solutions and get to know different concepts involved. All the Solutions are covered as per the latest syllabus guidelines. Knowing the NCERT Class 7 History Chapter 7 Questions and Answers helps students to attempt the exam with confidence.

Let Recall

Question 1.
Match the following

Answer:
1. (b)
2. (c)
3. (f)
4. (a)
5. (e)
6. (d)

Question 2.
Fill in the blank.
(a) The new castes emerging within varnas were called _____________ .
(b) ___________ were historical works written by the Ahoms.
(c) The ____________ mentions that Garha Katanga had 70,000 villages.
(d) As tribal states became bigger and stronger, they gave land grants to ______ and ______ .

Answer:
(a) jatis
(b) Buranji
(c) Akbar Nama
(d) temples and Brahmanas

Question 3.
State whether true or false.
(a) Tribal societies had rich oral traditions.
(b) There were no tribal communities in the north-western part of the subcontinent.
(c) The chaurasi in Gond states contained several cities.
(d) The Bhils lived in the north-eastern part of the subcontinent.

Answer:
(a) True
(b) False
(c) False
(d) False

Question 4.
What kinds of exchanges took place between nomadic pastoralists and settled agriculturists?
Answer:
The nomadic pastoralists exchanged the things from agriculturists. They provided them wool, ghee etc., and grain, cloth, utensils and other products were taken from them.

Let’s Understand

Question 5.
How was the administration of the Ahom state-organized?
Answer:
Administration of the Ahom state was organized by the forced labour. Paiks were the people who were forced to work for the state. Each village had to send a number of paiks by rotation. In such a way administration of the Ahom state was organized.

Question 6.
What changes took place in varna-based society?
Answer:
In varna-based society following changes took place:
(i) Smaller castes or jatis emerged within varnas due to the growth of economy and the needs of society;
(ii) Many tribes and social groups were taken to caste-based society and were given jatis as per their status;
(iii) Various artisAnswer: such as smiths, carpenter and masons were also given separate jatis by the Brahmanas;
(iv) Hence, jatis rather than varna, became the basis for organizing society.

Question 7.
How did tribal societies change after being organized into a state?
Answer:
The emergence of large states changed the nature of tribal societies. The Gond society and the Ahom society are such examples which expressed that the tribal societies change after being organized into a state.
(i) Brahmanas received land grants from the Gond rajas and became more influential and, as Gond chiefs wished to be recognized as Rajputs, they changed their names such as Aman Das, the Gond Raja changed his name as Sangram Singh.
(ii) The Ahom Kings granted to the temple and Brahmanas and thus influence of Brahmanas increased. In the reign of Sib Singh, Hinduism became the predominant religion. But the Ahom kings did not completely give up their traditional beliefs after adopting Hinduism.

Let’s Discuss

Question 8.
Were the Banjaras important for the economy?
Answer:
Yes, the banjaras were important for the economy as they were the most important nomads. They played an important role in trAnswer:porting grain to the city markets by the Delhi SultAnswer:. The grains were carried in bullocks from different areas and sold in towns or on the places where it had a great demand. They were also used for trAnswer:porting the food grains for Mughal army during military campaigns. Thus, banjaras were very useful for both the purposes i.e. for economy and administration.

Question 9.
In what ways was the history of the Gonds different from that of the Ahoms? Were there any similarities?
History of the Gonds was different from that of the Ahoms in these ways:
(i) The Gonds lived in a vast forested region called Gondwana whereas the Ahoms in the Brahmaputra valley migrated from Burma (Myanmar)
(ii) Raja or Rai were the different clAnswer: of Gonds while Ahoms created new state by suppressing the old political system of the bhuiy.
(iii) The Ahoms established a large state while the Gonds could not establish a big or large state. Similarities between Gonds and Ahoms: Many sources expressed that both were influenced with the Brahmanas and they had been given an high honour in the state. Apart from this, both the king of Gonds and Ahoms granted them land as prizes on many occasions.

Let’s do

Question 10.
Plot the location of the tribes mentioned in this chapter on a map. For any two, discuss whether their mode of livelihood was suited to the geography and the environment of the area where they lived.
Answer:
The location of the tribes can be seen in the map. They live in different parts of India.

Most of the tribes had their home in the forests or adjoining to the forested areas. Their livelihood was closely associated with the forest. They were engaged in the activities of cattle rearing, shifting cultivation, forest produces and some time in the settled agriculture. Thus, their mode of livelihood was suited to the geography and the environment of the area where they lived.

Question 11.
Find out about present-day government policies towards tribal populations and organize a discussion about these.
Answer:
Since independence of India, our government is continuously making efforts for betterment of tribes through the policies in various fields. Govt, of India has made a separate Ministry as well as State Govts, also established Departments for their welfare. TRIFED, New Delhi is an organization under Govt, of India which purchases their products on a good price and sells them into the open market. Govt, organizations, as well as various NGOs, are also doing work for preserving their unique culture and art.

Question 12.
Find out more about present-day nomadic pastoral groups in the subcontinent. What animals do they keep? Which are the areas frequented by these groups?
Answer:
Present-day various nomadic pastoral groups are found in the subcontinent. Tribes usually thrived in the forests, hills and desert areas of the subcontinents. In the western Himalayas-Gaddi shepherds, in the Jammu and Kashmir-Gujjar Bakarwals, in the Uttar Pradesh, Rajasthan and Madhya Pradesh-Banjaras are living.

These nomadic groups keep different type of cattle as per their convenience. They reared sheep, goat, buffalo, bulls, camels etc. They frequently visit their nearby areas where they sell their animals and its produces. During the visit, they also purchase goods for their need.

Hope the data shared above regarding the NCERT Class 7 Social Science History Chapter 7 Tribes, Nomads and Settled Communities PDF has aided in your exam preparation. If you ever need any assistance you can always reach us and our team will guide you at the soonest possibility.

Towns, Traders and Craftpersons Class 7 Questions and Answers Provided helps you to answer complex Questions too easily. You can use them while preparing for board exams and all of them are given by subject experts. Reading NCERT Solutions for Class 7 Social Science History Chapter 6 Towns, Traders and Craftpersons familiarizes you with the kind of questions appearing in the board exams. Students are advised to read these solutions on a regular basis to score well.

Towns, Traders and Craftspersons Class 7 Questions and Answers History Chapter 6

Make your learning experience enjoyable by preparing from the quick links available on this page. Use the Class 7 SST History Chapter 6 NCERT Solutions and get to know different concepts involved. All the Solutions are covered as per the latest syllabus guidelines. Knowing the NCERT Class 7 History Chapter 6 Questions and Answers helps students to attempt the exam with confidence.

Class 7 History Chapter 6 NCERT Textbook Questions and Answers

Let’s Recall

Question 1.
Fill in the blanks
(a) The Rajarajeshvara temple was built in __________ .
(b) Ajmer is associated with the Sufi saint __________ .
(c) Hampi was the capital of the __________
(d) The Dutch established a settlement at _____________ in Andhra Pradesh.

Answer:
(a) early 11^th century
(b) Khwaja Muinuddin Chishti
(c) Vijayanagara
(d) Masulipatnam

Question 2.
State whether true or false:
(a) We know the name of the architect of the Rajarajeshvara temple from an inscription.
(b) Merchants preferred to travel individually rather than in caravans.
(c) Kabul was a major centre for trade in elephants.
(d) Surat was an important trading port on the Bay of Bengal.

Answer:
(a) True
(b) False
(c) False
(d) False

Question 3.
How was water supplied to the city of Thanjavur?
Answer:
Wells and tanks were formed for water supply to the city of Thanjavur.

Question 4.
Who lived in the “Black Towns” in cities such as Madras?
Answer:
Merchants, artisans, traders and craftspersons lived in the “Black Towns”.

Let’s Understand

Question 5.
Why do you think towns grew around temples?

Answer:
Towns grew around temples because of large number of devotees visited from far places for various rituals and large number of priests, workers, artisans, and traders etc., settled near temple to fulfil, the need of the people. Thus, temples often became central to the economy and society. Gradually, many streets developed in a form of market and around these types of the temples grew towns which were called temple towns.

Question 6.
How important were craftspersons for the building and maintenance of temples?
Answer:
Craftspersons were important for the building and maintenance of temples because of:

1. An artisan community, Viswakarma consisting of goldsmiths, bronzesmiths, blacksmiths, masons and carpenters were essential for building and maintenance of temples in any form.
2. Weavers such as the Saliyar or Kaikkolars were prosperous communities and they made donations to temples.
3. The craftspersons of Bidar were skilled in the inlay work of metals such as copper and silver. They, also played an important role in making the designs in the temples.

Question 7.
Why did people from distant lands visit Surat?

Answer:
People from distant lands visited Surat because of:
(i) It was the emporium of western trade during the Mughal period

• It was the gateway for trade with West Asia via the Gulf of Ormuz;
• Many Muslim pilgrims boarded on ships from Surat for the Mecca;
• Surat had many factories and warehouses;
• There were also several retail and wholesale shops selling cotton textiles;
• Surat was famous for its gold lace borders and had a market in West Asia, Africa and Europe etc.

Question 8.
In what ways was craft production in cities like Calcutta different from that in cities like Thanjavur?

Answer:
The craft production in Calcutta (now called Kolkata) was in the form of cotton textiles, jute textiles and silk textiles while in Thanjavur craft production was in the form of inlay work on metals such as copper and silver.

Let’s Discuss

Question 9.
Compare any one of the cities described in this chapter with a town or a village with which you are familiar. Do you notice any similarities or differences?

Answer:
Students discuss with the teacher and can write the answer.

Question 10.
What were the problems encountered by merchants? Do you think some of these problems persist today?

Answer:
There were many problems encountered by merchants. Due to the long way, a long time was taken in the journey. They had cash or kind with them and hence often threats of loosing the things remained.

I think today there is no such problem. Now plastic money in shape of Debit and Credit cards can be carried easily without any threat. Apart from the plastic money many more options are available such as Demand Draft, online transfer of money etc. The trade articles can be booked by train or transported through roads.

Let’s Do

Question 11.
Find out more about the architecture of either Thanjavur or Hampi, and prepare a scrapbook illustrating temples and other buildings from these cities.
Answer:
Students discuss with the teacher and write the answer.

Question 12.
Find out about any present-day pilgrimage centre. Why do you think people go there? What do they do there? Are there any shops in the area? If so, what is bought and sold there?Answer:
Students discuss with the teacher and write the answer.

Hope the data shared above regarding the NCERT Class 7 Social Science History Chapter 6 Towns, Traders and Craftpersons PDF has aided in your exam preparation. If you ever need any assistance you can always reach us and our team will guide you at the soonest possibility.

The Making of the National Movement 1870s-1947 Class 8 Questions and Answers Provided helps you to answer complex Questions too easily. You can use them while preparing for board exams and all of them are given by subject experts. Reading NCERT Solutions for Class 8 Social Science History Chapter 11 The Making of the National Movement 1870s-1947 familiarizes you with the kind of questions appearing in the board exams. Students are advised to read these solutions on a regular basis to score well.

The Making of the National Movement 1870s-1947 Class 8 Questions and Answers History Chapter 11

Make your learning experience enjoyable by preparing from the quick links available on this page. Use the Class 8 SST History Chapter 11 NCERT Solutions and get to know different concepts involved. All the Solutions are covered as per the latest syllabus guidelines. Knowing the NCERT Class 8 History Chapter 11 Questions and Answers helps students to attempt the exam with confidence.

History Class 8 Chapter 11 NCERT Textbook Questions and Answers

Let’s Recall

Question 1.
Why were people dissatisfied with British rule in the 1870s and 1880s?
Answer:
The dissatisfaction among people with the British rule in the 1870’s to 1880’s had several reason:
(i) The arms act was passed in 1878 disallowing Indians from possessing arms.

(ii) In 1878 only, the Vernacular Press Act was also enacted in an effort to silence those who were critical of the government. This act allowed the government to confiscate the assets of newspapers including their printing presses if the newspapers published anything that was found “objectionable”.

(iii) In 1883, there was a furore over the attempt by the government to introduce the Ilbert Bill. The bill provided for the trial of British or European persons by Indians, and sought equality between British and Indian judges in the country. But when white opposition forced the government to withdraw the Bill, Indians were enraged as the event highlighted the racial attitudes of the British in India.

Question 2.
Who did the Indian National Congress wish to speak for?
Answer:
In the Indian National Congress always wished to speak for the voice of Indians. As the Indian National Congress was composed of the representatives, not of any one class or community of India, but of all the different communities of India, so they always wished to speak for Indians whether belonging to any community.

Question 3.
What economic impact did the First World War have on India?
Answer:
The First World War not only altered the political situation but also the economical situation in India and the economic impact created by it was as follows:
(i) It led to huge rise in the defence expenditure of the Government of India.

• Government in turn increased taxes on the individual’s income and business profits.

(ii) Increased military expenditure and the demands for war supplies led to a sharp rise in prices which had created great difficulties for the common people.

(iii) Business groups reaped fabulous profits from the war.

• As the war created a demand for industrial goods and caused a decline of imports from other countries into India it resulted in the expansion of the Indian industries during the war.
• Indian business group began to demand greater opportunities for development.

Question 4.
What did the Muslim League resolution of 1940 ask for?
Answer:
The Muslim League Resolution of 1940 asked for “Independent States” for Muslims in the North-western and western and eastern states of the country. The resolution did not mention partition or Pakistan.

Let’s Discuss

Question 5.
Who were the Moderates? How did they propose to struggle against British rule?
Answer:
Moderates were leaders who were against the adoption of any violence for convincing the British. They were the ideal followers of non-violence. The moderates proposed to struggle against the British rule by adopting non-violence methods as Moderate leaders wanted to develop public awareness about the unjust nature of the British rule.

They published newspapers, wrote articles, and showed how British rule was leading to economic ruin of the country. They criticised British rule in their speeches and sent representatives to different parts of the country to mobilise public opinion. They felt that the British had respect for the ideas of freedom and justice so they would accept the just demands of Indians. All attempts by the moderates were made to make the government aware of the feeling of Indians.

Question 6.
How was Hie politics of the Radicals within the Congress different from that of the Moderates?
Answer:
The Radicals within the Congress were the ones who started raising questions about the political style of the Congress by the 1890’s. The Radicals started to explore more objectives and methods. They criticised the Moderates for their “politics of prayers” and emphasized the importance of self-reliance and constructive work. The Radicals argued that people must rely on their own strengths not on the “good” intentions of the government, people must fight for Swaraj.

Question 7.
Discuss Hie various forms that Hie Non-Cooperation Movement took in different parts of India. How did the people understand Gandhiji?
Answer:
Non-cooperation movement took place in different parts of India and in various forms. In many cases, people resisted British rule non-violently, and in either case people linked their movements to local grievances. People also took Gandhiji in different way and linked their movements to local grievances.

• In Kheda of Gujarat, Patidar Peasants organised non-violent campaigns against the high land revenue demand of the British.
• In coastal Andhra and interior Tamil Nadu, liquor shops were picketed.
• In Guntur district of Andhra Pradesh, tribals and poor peasants staged a number of “forest Satyagrahas”. Believing about Gandhiji that Gandhiji would get their taxes reduced and have the forest regulations abolished.
• In many forests, peasants proclaimed swaraj and believed that “Gandhi Raj” was about to be established.
• In Sind, Muslim traders and peasants were very enthusiastic about the Khilafat call.
• The Khilafat – Non-cooperation alliance gave enormous communal unity and strength to the national movement in Bengal too.
• In Punjab, the Akali agitation of the Sikh sought to remove corrupt mahants supported by the British from their gurudwaras. The movement got closely associated with Non-Cooperation movement. All the above incidents show how people thought about Gandhiji. They took Gandhiji at their Messiah as someone who could help them overcome their misery and poverty.

Question 8.
Why did Gandhiji choose to break the salt law?
Answer:
Gandhiji choose to break salt law because as per this law the state had a monopoly on the manufacture and sale on salt. In view of Gandhiji along with other nationalists it was sinful to tax salt since it is such an essential item to our food. The Salt March related the general desire of freedom to a specific grievance shared by everybody and thus did not divide the rich and the poor.

Question 9.
Discuss those developments of the 1937-47 period that led to the creation of Pakistan.
Answer:
The provincial elections of 1937 seemed to have convinced the League that Muslims were a minority,
and they would always have to play second fiddle in any democratic structure. It feared that Muslims may even go rejection. The Congress’s unrepresented to the league’s desired movement in the United Provinces in 1937 also annoyed the League.

Failure of Congress in mobilising the Muslim masses in the 1930’s allowed the League to widen its social support. It sought to enlarge its support in the early 1940’s when most Congress leaders were in jail. At the end of the war in 1945, the British opened negotiation between the Congress, the League and themselves for the independence of India. The talks failed because the League saw itself as the sole spokesperson of India’s Muslims. The Congress could not accept this claim since a large member of Muslims still were in its support.

Again the elections to the provinces were held in 1946. The Congress did well in “General Constituencies” but the League’s success in the seats reserved for Muslims was spectacular. It persisted with the demand for “Pakistan”. In March 1946 the British cabinet sent a 3 member mission to Delhi to examine this demand and to suggest a suitable political framework for a free India. This mission suggested that India should remain united and constitute itself as a loose confederation with some autonomy for the Muslim-majority areas. But neither Congress nor the Muslim League agreed to the proposal. This led to inevitability of partition.

After the failure of the Cabinet Mission, the Muslim League decided on mass agitation for winning its Pakistan demand. 16th August 1946 was announced as “Direct Action Day”. On this day riots broke out in Calcutta, lasting several days and resulting in death of 1000’s of people. By the 1947 the violence spread to different parts of northern India. With numerous killings, fleeing and various untold brutalities of riots, a new country Pakistan was born. The joy of country’s Independence came up mixed with pain and violence of partition.

Hope the data shared above regarding the NCERT Class 8 Social Science History Chapter 11 The Making of the National Movement 1870s-1947 PDF has aided in your exam preparation. If you ever need any assistance you can always reach us and our team will guide you at the soonest possibility.

Here you will find Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

Extra Questions for Class 10 Maths Quadratic Equations with Answers Solutions

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers

Quadratic Equations Class 10 Extra Questions Objective Type

Question 1.
The solution of quadratic equation x² – x – 2 = 0 are:
(a) 1, -2
(b) -1, -2
(c) -1, 2
(d) 1, 2.
Answer:
(c) -1, 2

Question 2.
The zero of the polynomial x² + 2x – 3 are:
(a) 1, -3
(b) -1, 3
(c) -1, -3
(d) 1, 3
Answer:
(a) 1, -3

Question 3.
The degree of the polynomial x³ – x + 7 is:
(a) 1
(b) 2
(c) 3
(d) none of these.
Answer:
(c) 3

Question 4.
The zero of the polynomial p(x) = x² + 1 are:
(a) real
(b) not real
(c) (a) and (b) both
(d) none of these.
Answer:
(b) not real

Question 5.
(i) Every quadratic polynomial can have at the most:
(a) one zero
(b) two zeroes
(c) three zeroes
(d) none of these.
Answer:
(b) two zeroes

(ii) From the equation 4√5 x² + 7x – 3√5 = 0, the value of x will be:
(a)
(b)
(c)
(d)
Answer:
(a)

Question 6.
If one root of quadratic equation x² + kx + 3 = 0 is 1, then the value of k will be:
(a) 1
(b) -3
(c) -4
(d) -5.
Answer:
(c) -4

Question 7.
If one root of quadratic equation 2x² + px – 4 = 0 is 2, then the value of p will be:
(a) -3
(b) -2
(c) 2
(d) 3.
Answer:
(b) -2

Question 8.
If one root of quadratic equation ax² + bx + c = 0 is 1, then:
(a) a = 1
(b) b = 1
(c) c = 1
(d) a + b + c = 0
Answer:
(d) a + b + c = 0

Question 9.
If 2x² + 1 = 33, then the value of x will be:
(a) ±2
(b) ±3
(c) ±4
(d) ±1.
Answer:
(c) ±4

Question 10.
One root of quadratic equation x² + 3x – 10 = 0:
(a) -2
(b) +2
(c) 0
(d) 1
Answer:
(b) +2

Question 11.
If = \(\frac {x}{6}\) = \(\frac {6}{x}\), then the value of x will be:
(a) 6
(b) -6
(c) 6
(d) none of these.
Answer:
(c) 6

Question 12.
Discriminant of ax² + bx + c = 0) is:
(a)
(b)
(c)
(d) none of these.
Answer:
(d) none of these.

Question 13.
A quadratic equation ax² + bx + c = 0, a ≠ 0 has equal roots if b² – 4ac is:
(a) equal to 0
(b) ≥ 0
(c) ≤ 0
(d) none of these.
Answer:
(a) equal to 0

Question 14.
The roots of ax² + bx + c = 0, a 50 are real and unequal, if b² – 4ac is:
(a) equal to 0
(b) ≥ 0
(c) ≤ 0
(d) none of these.
Answer:
(b) ≥ 0

Question 15.
The roots of the equation 2x² – 8x + c = 0 are eaual, the value of c is:
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 16.
If the discriminant (D) of a quadratic equation ax² + bx + c = 0, a ≠ 0 is greater than zero, the roots are:
(a) real and unequal
(b) real and equal
(c) not real
(d) none of these.
Answer:
(a) real and unequal

Question 17.
An expression in α and β is called symmetrical expression if by interchanging α and β, the expression is:
(a) changed
(b) not changed
(c) may be (a) and (b)
(d) none of these.
Answer:
(b) not changed

Question 18.
If x² + 5bx + 16 = 0 has no real roots, then:
(a) b > \(\frac {8}{5}\)
(b) b < \(\frac {-8}{5}\)
(c) \(\frac {-8}{5}\) < b < \(\frac {8}{5}\)
(d) none of these.
Answer:
(c) \(\frac {-8}{5}\) < b < \(\frac {8}{5}\)

Question 19.
If the roots of 5x² – px + 1 = 0 are real and distinct, then:
(a) p > √15
(b) p > – 2√5
(c) -2√5 < p < 2√5
(d) p > 2√5 or p < – 2√5.
Answer:
(d) p > 2√5 or p < – 2√5

Question 20.
The sum of the roots of the quadratic equation 3x² + 4x = 0 is:
(a) 0
(b) –\(\frac {3}{4}\)
(c) –\(\frac {4}{3}\)
(d) \(\frac {4}{3}\)
Answer:
(c) –\(\frac {4}{3}\)

Question 21.
The product of the roots of quadratic equation 3x² – 4x = 0) is:
(a) 0
(b) –\(\frac {3}{4}\)
(c) –\(\frac {4}{3}\)
(d) –\(\frac {4}{3}\)
Answer:
(a) 0

Question 22.
If one root of x² + kx + 3 = 0 is 1, then the value of k will be:
(a) -4
(b) -3
(c) 1
(d) 5.
Answer:
(b) -3

Question 23.
If the sum of the roots of the equation 3x² + (2k + 1)x – (k + 5) = 0 be equal to their product, the value of k is:
(a) 5
(b) 4
(c) 3
(d) 2
Answer:
(b) 4

Question 24.
The sum of the roots of quadratic equation 5 – 7x + 3x² = 0 is:
(a) +\(\frac {7}{5}\)
(b) –\(\frac {7}{5}\)
(c) –\(\frac {7}{3}\)
(d) +\(\frac {7}{3}\)
Answer:
(d) +\(\frac {7}{3}\)

Question 25.
If one root of quadratic equation x³ – 3x + 2 = 0 is 2, then the second root is:
(a) 3
(b) -1
(c) 1
(d) 2
Answer:
(c) 1

Quadratic Equations Class 10 Extra Questions Very Short Answer Type

Question 1.
If one root quadratic equation x² + 2x – p = 0 is -2, then find the value of p.
Solution.
Given that – 2 is a root of given quadratic equation x² + 2x – p = 0
∴ (-2)² + 2 (-2) – p = 0
⇒ 4 – 4 – p = 0
∴ p = 0
Hence, p = 0

Question 2.
If x² – 5 = 0, then find the value of x:
Solution.
x² – \(\frac {1}{9}\) = 0
x = ±\(\frac {1}{3}\)

Question 3.
If x = 12 and y = 5, then find the value of 9x² + 49y² – 42xy.
Solution.
We have:
9x² + 49y² – 42xy
= (3x)² + (7y)² – 2.3x.7y
= (3x – 7y)²
= (3.12 – 7.5)²
= (36 – 35)² = 1

Question 4.
Solve the following quadratic equation for x:
4x² + 46x – (a² – 12) = 0
Solution.
4x² + 4bx – a² + b² = 0
⇒ (2x)² + 2.2xb + (b)²-(a)² = 0
⇒ (2x + b)²-(a)² = 0
⇒ (2x + b + a) (2x + b – a) = 0

Question 5.
Find the roots of the quadratic equation:
3x² – 2 √6 x + 2 = 0
Solution.
Given quadratic equation
3x² – 2√6 x + 2 = 0
= (√3x)² – 2√3x × √2 +(√2)² = 0
(√3x – √2)² = 0
∴ roots are \(\frac{\sqrt{2}}{\sqrt{3}}\) or \(\frac{\sqrt{2}}{\sqrt{3}}\)

Question 6.
Find the equation whose roots are b – 2a and b + 2a.
Solution.
α = b – 2a
β = b + 2a
Sum of roots = α + β = 6-2a +b + 2a = 2b.
Product of roots = αß = (b – 2a)(b + 2a)
= b² – (2a)²
= b² – 4a²
The required quadratic equation is
∴ x² – (sum of roots) x + product of roots = 0
∴ x² – (2b)x + (b² – 4a²) = 0
∴ x² – 2bx + b² – 4a² = 0

Question 7.
If α and β are the roots of the equation x² – x – 2 = 0, find that quadratic equation whose roots are (2α + 1) and (2β + 1).
Solution.
We have
x² – x – 2 = 0

The required equation is is

Question 8.
If – 5 is a root of quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots. Find the value of k.
Solution.
Given – 5 is a root of quadratic equation 2x² + px – 15 = 0
∴ 2(-5)² + p(-5) – 15 = 0
50 – 5p – 15 = 0
5p = 35 ⇒ p = 7
Putting the value of p = 7 in given (ii) quadratic equation.
7x² + 7x + k = 0
as the roots of this equation are equal
∴ b² = 4ac
⇒ (7)² = 4 × 7 × k
∴ k = \(\frac {7}{4}\)

Question 9.
If aand Bare the roots of the equation ax² + bx + b = 0, then prove that:

Solution.
Since, a and ß are the roots of the equation
ax² + bx + b = 0

Proved.

Question 10.
The length and breadth of a room are 15 m and 12 m. There is a verandah surrounding the room and area of this is 90m². Find the width of verandah.
Solution.
Suppose the width of Verandah is x metre.
Length of verandah = (2x + 15) m.
Breadth of verandah = (2x + 12) m.
Area of verandah = (2x + 15) (2x + 12) – 15 × 12 = 90

∴ 4x² + 24x + 30x + 180 – 180 – 90 = 0
∴ 4x² + 54x – 90 = 0
∴ 2x² + 27x – 45 = 0
∴ 2x² + 30x – 3x – 45 = 0
∴ 2x(x + 15) – 3(x + 15) = 0
(x + 15) (2x – 3) = 0
x + 15 = 0
or 2x – 3 = 0
2x = 3
x = 3/2
The width of verandah = 3/2 m.

Question 11.
The square of a positive integer is greater than 11 times the integer by 26. Find the positive integer.
Solution.
Let the positive integer be x then according to the given condition,
x² = 11x + 26
⇒ x² – 11x – 26 = 0
⇒ x² – 13x + 2x – 26 = 0
⇒ x(x – 13) + 2(x – 13) = 0
⇒ (x – 13)(x + 2) = 0
∴ x = -2 or x = 13
Neglecting the negative value of x, we have x = 13.
Hence, the integer is 13.

Question 12.
A dealer sells an article for ₹ 24 and gains as much per cent as the cost price of the article. Find the cost price of the article.
Solution.
Let cost price of article be ₹ x. Then, gain = x%.

⇒ x² + 100x – 2400 = 0
⇒ (x + 120)(x – 20) = 0
⇒ x = -120 or x = 20
⇒  x = 20
[∴ Cost Price can never be negative]
Hence, the cost price of the article is ₹ 20.

Quadratic Equations Class 10 Extra Questions Short Answer Type

Question 1.
The area of a right angle triangle is 30cm². Find the lengths of the base if its height is 7 cm more then base.
Solution.
Let the length of base of triangle = x cm
∴ length of its height = (x + 7) cm

Area of right angle ∆ = \(\frac {1}{2}\) × base × height
⇒ 30 = \(\frac {1}{2}\)x.(x + 7)
⇒ x² + 7x – 60 = 0
⇒ x² + (12 – 5)x – 60 = 0
⇒ x² + 12x – 5x – 60 = 0
⇒ x(x +12) – 5(x + 12) = 0
⇒ (x + 12) (x – 5) = 0
∴ x = 5 or – 12
as length is never negative
∴ length of base = 5 cm.

Question 2.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution.
Given that, hypotenuse of right triangle = 13 cm
Let the base of the right triangle = x
According to the question,
Altitude of the triangle = 7 cm
Less than its base = (x – 7) cm

By Pythagoras theorem, we have
In ∆ABC AC² = BC² + Ab²
⇒ (13)² = x² + (x – 7)²
⇒ 169 = x² + x² – 14x + 49
[∵ (a – b)² = a² – 2ab + b²]
⇒ 2x² – 14x – 120 = 0
⇒ x² – 7x – 60 = 0
x² – (12x – 5x) – 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12) (x + 5) = 0
Now, x – 12 = 0 x = 12
and x + 5 = 0 = x = -5
Since, altitude of the triangle cannot be negative, hence x ≠ -5
Hence, base of the triangle = 12 cm
and altitude of the triangle = 12 – 7 = 5 cm.

Question 3.
Solve the equation

Solution.
Given,

⇒ (3x + 4)(x + 4) = 4(x + 1)(x + 2)
⇒ 3x² + 16x + 16 = 4(x² + 3x + 2)
⇒ 3x² + 16x + 16 = 4x² + 12x + 8
⇒ x² – 4x – 8 = 0
⇒ Here a =1, b = -4, and c = -8.

Question 4.
A motor boat whose speed in still water is 24 km/h. takes 1 hour more to go 32 km up stream than to return down stream to the same spot. Find the speed of stream.
Solution.
Let the speed of stream be x km/h.
Given that the speed of boat in still water is 24 km/h
∴ Speed of boat in down stream = (24 + x) km/h
and speed of boat in up stream = (24 – x) km/h
Time taken by boat to travel 32 km in down

⇒ 32 × 2x = (24 – x) (24 + x)
⇒ 64x = 576 – x²
⇒ x² + 64x – 576 = 0
⇒ x² + (72 – 8)x – 576 = 0
⇒ x² + 72x – 8x – 576 = 0
⇒ x(x + 72) – 8(x + 72) = 0 (x – 8)(x + 72) = 0
∴ x = 8 or -72
as speed is never negative
∴ x = 8
i.e., speed of stream = 8 km/h.

Question 5.
Solve the quadratic equation

by factorization method
Solution.

⇒ x(a + b + x) = – ab
⇒ ax + bx + x² + ab = 0
⇒ x² + ax + bx + ab = 0
⇒ x(x + a) + b (x + a) = 0
⇒ (x + a) (x + b) = 0
∴ x = -a or -6.

Question 6.
If the equations x² + kx + 64 = 0 and x² – 8x + k = 0 have real roots. Find the positive value of k.
Solution.
Given quadratic equation are
x² + kx + 64 = 0. …(i)
and x² – 8x + k = 0…(ii)
If the equations have real roots, then D ≥ 0 Discriminant of (i) k² – 256 and equation

(ii) 64 – 4k
= k² – 256 ≥ 0 and 64 – 4k ≥ 0
⇒ k² ≥ 256 and 64 ≥ 4k
⇒ k ≥ 16 and 16 ≥ k.
Hence, k = 16.

Question 7.
If the speed of a train is increased by 5 km/H then train takes 1 hour less time to cover a distance of 360 km. Find the speed of the train.
Solution.
Let the speed of train be x km/ Hr.
∴ Time taken by the train to cover 360
km = \(\frac {360}{x}\) Hours
New speed of the train = (x + 5) km/Hr.
∴ New time taken to cover 360 km

⇒ 360 × 5 = x² + 5x
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45) (x – 40) = 0
∴ x = 40 or -45
as speed is never negative
∴ speed of train = 40 km/hour.

Question 8.
Solve the following equation

Solution.
Let = \(\frac{x-2}{x+2}\) = y
∴ Equation reduced to
y + \(\frac {3}{y}\) – 4 = 0
⇒ y² + 3 – 4y = 0
⇒ y² – 4y + 3 = 0
⇒ y² – 3y – y + 3 = 0
⇒ y(y – 3)-1 (y – 3) = 0
⇒ (y – 3)(y – 1) = 0
∴ y = 3 or 1.
Where
y = 3 = \(\frac{x-2}{x+2}\)
3x + 6 = x – 2
⇒ 2x = -8
∴ x = -4
Where
y = 1 = \(\frac{x-2}{x+2}\)
x + 2 x + 2 = x – 2 which is not true
∴ x = -4.

Question 9.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is a. Find his present age.
Solution.
Let the present age of Rehman – x year
∴ Rehman’s age, 3 yr ago = (x – 3) yr
Rehman’s age, after 5 yr = (x + 5) yr
According to the question,

∴ 6x + 6 = x² + 2x – 15
⇒ x² + 2x – 6x – 15 – 6 = 0
x² – 4x − 21 = 0
x² – (7x – 3x) – 21 = 0
⇒ x² – 7x + 3x – 21 = 0
(By factorization method)
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7)(x + 3) = 0
∴ x – 7 = 0 ⇒ x = 7 or x + 3 = 0 ⇒ x = -3
But x +-3 age cannot be negative.
So, present age of Rehman = 7 yr.

Question 10.
The difference of two numbers is 2 and the sum of their square is 34. Find the number.
Solution.
Let Numbers be x and y, such x > y
∴ x – y = 2 …..(i)
and x² + y² = 34 …(ii)
from (i) x = 2 + y
⇒ (2 + y)² + y² = 34
⇒ 2y² + 4y – 30 = 0
⇒ y² + 2y – 15 = 0
⇒ (y + 5) (y – 3) = 0
∴ y = 3 or – 5
where y = 3, x = 2 + 3 = 5
∴ Numbers are 5 and 3.

Question 11.
The speed of boat in still water is 15km/H. The boat completes 30 km distance in going downstream and 30 km up stream in total 4 Hr 30 minute. Find the speed of stream.
Solution.
Let the speed of stream = x km/ Hr.
∴ speed of boat in down stream = (15 + x) km/H
and speed of boat in up stream = (15 – x) km/H
Time taken 30 km in down stream = \(\frac{30}{(15+x)}\) Hrs

⇒ 2 × 30 × 30 = 9(225 – x²)
⇒ 200 = 225 – x²
⇒ x² = 225 – 200
= 25
x = √25 = ± 5
as speed is never negative:. speed to stream = 5 km/Hr.

Question 12.
Determine whether the quadratic equation 9x² + 7x – 2 = 0 has real roots. If so, find roots of quadratic equation.
Solution.
Given, quadratic equation
9x² + 7x – 2 = 0
Here a = 9, b = 7 and c= -2
D = b² – 4ac .
= (7)² – 4 × 9 × -2
= 49 + 72 = 121
∴ roots of quadratic equation are real.
∴ roots are

Question 13.
By solving the equation

get a quadratic equation. Find the nature of roots. Using formula, solve the quadratic equation.
Solution.
Given,

Nature of root, D = b² – 4ac = (4)² – 4 × 1 × 0 = 16 > 0
∴ root of the equation are real.

Question 14.
Some vessels are manufactured in a day in a small industry. On a particular day it is observed that the cost of manufacturing each item was 3 more than twice the number of vessels manufactured on that day. If the total manufacturing cost on that day was 90, then find the number of manufactured vessels and manufacturing cost of each item.
Solution.
Let Number of vessel manufacture in a day = x
∴ Cost of manufacturing = (2x + 3).
According to Question,
x(2x +3) = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x( 2x + 15)-6 (2x + 15) = 0
⇒ (x – 6) (2x + 15) = 0
∴ x = 6 or – \(\frac {15}{2}\)
as number of vessels are not negative
∴ x = 6
Also the manufacturing cost = 2 × 6 + 3 = ₹ 15.

Question 15.
Is the following situation possible ? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution.
Let age of one of the two friends = x yr
Then, age of other friend = (20 – x) yr
(∵ The sum of the ages of two friends is 20 yr)
4 yr ago age of one of the two friends = (x – 4) yr
4 yr ago age of the other friend
= (20 – x – 4) yr = (16 – x) yr
According to the question,
(x – 4)(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112=0
On comparing the above equation with ax² + bx + c = 0, we get
a = 1, b = -20 and c = 112
∴ Discriminant, D = b² – 4ac
= (-20)² – 4 × 1 × 112
= 400 – 448 = 48 < 0
which implies that the real roots are not possible because this condition represents imaginary roots. So, the solution does not exist.

Question 16.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth.
Solution.
Let the breadth of the park = x metre
Then, according to the question, Perimeter of a rectangular park = 80 m
= 2(Length + Breadth)=80 m
= Length + Breadth = 40 m
Length = (40 – x) m
∴ Area of a rectangular park = Length × Breadth = (40 – x)x m²
But according to the question,
Area of the rectangular park is 400 m².
∴ (40 – x)x = 400
⇒ x² – 40x + 400 = 0
⇒ x² – 2x × 20 + (20)² = 0
⇒ (x – 20)² = 0
[∵ a² – 2ab + b² = (a – b)²]
⇒ x = 20
Thus, breadth of the park = 20 m
and length of the park = (40 – 20) m = 20 m
So, it is possible to design the rectangular park of perimeter 80 m and area 400 m². But this rectangle will be a square of side 20 m.

Quadratic Equations Class 10 Extra Questions Long Answer Type

Question 1.
Solve:

x ≠ 0.
Solution.
The given equation is:

Therefore, the given equation can be written as

Now, substituting y = x + \(\frac {1}{x}\), we get
4y² + 8y – 45 = 0
⇒ 4y² + 18y – 10y – 45 = 0
⇒ 2y (2y + 9) – 5(2y +9) = 0
⇒ (2y + 9) (2y – 5) = 0

Now, from (ii), we get,
2(x² + 1) = 5x
⇒ 2x² – 5x + 2 = 0
⇒ 2x² – 4x – x + 2 = 0
⇒ 2x(x – 2) – 1(x – 2) = 0
⇒ (x – 2)(2x – 1) = 0
x = 2 and x = \(\frac {1}{2}\)
Hence, the solutions of the given equation are

Question 2.
Solve: 2x⁴ – x³ – 11x² + x + 2 = 0.
Solution.
Given equation is
2x⁴ – x³ – 11x² – x + 2 = 0
Divide both sides by x²,

Now substitute x + \(\frac {1}{x}\) = y
∴ 2y² – y – 15 = 0
⇒ 2y² – 6y + 5y – 15 = 0
⇒ 2y(y – 3) + 5(y – 3) = 0
⇒ (y – 3)(2y + 5) = 0
∴ y = 3 or y = –\(\frac{5}{2}\)
y = 3 = \(\frac{x^{2}+1}{x}\)
When
⇒ x² + 1 = 3x
⇒ x² – 3x + 1 = 0
Here, a = 1, b = -3 and c = 1

⇒ 2x² + 5x + 2 = 0
⇒ 2x² + 4x + x + 2 = 0
⇒ 2x(x + 2) + 1(x + 2) = 0
⇒ (x + 2)(2x + 1) = 0
x = -2, – \(\frac{1}{2}\)
Hence, the solutions are -2, –\(\frac{1}{2}\), \(\frac{3 \pm \sqrt{5}}{2}\).

Question 3.
Solve \(\sqrt{3 x+10}\) + \(\sqrt{6-x}\) = 6.
Solution.
Given equation is:
\(\sqrt{3 x+10}\) + \(\sqrt{6-x}\) = 6.
We must look for solutions which satisfy
3x + 10 ≥ 0 and 6 – x ≥ 0
x ≥ –\(\frac {10}{3}\) and x ≤ 6
i.e., –\(\frac {10}{3}\) ≤ x ≤ 6.
Now, we first transform one of the radicals co the R.H.S.
We have,
\(\sqrt{3 x+10}\) = 6 – \(\sqrt{6-x}\)
Squaring both sides, we get
3x + 10 = 36 + 6 – x – 12\(\sqrt{6-x}\)
or 4x – 32 = -127\(\sqrt{6-x}\)
or 8 – x = 3\(\sqrt{6-x}\)
Again, squaring both sides, we get
64 + x² – 16x = 9 (6 – x)
or 64 + x² – 16x – 54 + 9x = 0
or x² – 7x + 10 = 0
or (x − 2)(x – 5) = 0
∴ x = 2 and x = 5.
which satisfy \(\frac{-10}{3}\) ≤ x ≤ 6.
Hence, the solutions are x = 2 and x = 5.

Question 4.
One person purchased some cloth for ₹ 2250. Had the price of cloth been ₹ 50 per metre higher, then he would have been able to purchase the cloth less by 1.5 metre. Find the measure of the cloth purchased and price per metre.
Solution.
Let the price per metre of cloth be ₹ x
Amount spend = ₹ 2250
∴ Length of cloth purchased
\(\frac{2250}{x}\) = metre.
Now, new price of cloth per metre become = ₹ (x + 50)
∴ Length of cloth purchase
= \(\frac{2250}{x + 50}\) metre.
According to the question,

⇒ 2250 × 50 × 2 = 3(x² + 50x)
⇒ x² + 50x – 75000 = 0
⇒ x² + 300x – 250x – 75000 = 0
⇒ x(x + 300) – 250(x + 300) = 0
⇒ (x + 300)(x – 250) = 0
∴ x = -300 or x = 250
Neglecting the -ve sign value i.e.,
x = -300
Hence, cost of cloth per metre = ₹ 250.
and length of cloth \(\frac{2250}{250}\) = 9 metre.

Question 5.
A two-digit number is such that the product of the digits is 12, when 36 is added to the number, the digit interchange their places. Find the number.
Solution.
Suppose two digit number is 10x + y.
Whose y = unit’s place and x = ten’s place digit.
According two question
xy = 12 …..(i)
and 10x + y + 36 = 10y + x.
∴ 9x – 9y = -36
∴ x – y = -4
x = y – 4
Put x = y – 4 in (i)
(y – 4)y = 12
∴ y² – 4y – 12 = 0
∴ y² – 6y + 2y – 12 = 0
∴ y(y – 6) + 2(y – 6) = 0
∴ (y – 6) (y + 2) = 0
y – 6 = 0 or y + 2 = 0
y = 6 or y = -2
(not acceptable).
∴ Put y = 6 in (i) we get
x (6) = 12
x = 2
The two digit number is
10x + y = 10(2) + 6 = 26.

Question 6.
The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides of the field.
Solution.
Let PQRS be the rectangular field. Let the shorter side QR of the rectangle = x m.

According to the question,
Diagonal of the rectangle PR = 60 m more than the shorter side = (x + 60) m
Side of the rectangle PQ = 30 m more than the shorter side = (x + 30) m
∴ By Pythagoras theorem,
In ∆PQR, PR² = PQ² + QR²
(∵ In rectangle every adjacent side makes an angle 90° to each other)
⇒ (x + 60)² = (x + 30)² + x²
⇒ x² + 120x + 3600 = x² + 60x + 900 + x²
[∵ (a + b)² = a² + 2ab + b²]
⇒ (2x² – x²) + (60x – 120x) + (900 – 3600) = 0
x² – 60x – 2700 = 0
x² – (90x – 30x) – 2700 = 0
x² – 90x + 30x – 2700 = 0
(By factorization method)
⇒ x(x – 90) + 30(x – 90) = 0
⇒ (x – 90)(x + 30) = 0
∴ Either x – 90 = 0 ⇒ x = 90 or x + 30 = 0
⇒ x = – 30 which is not possible because side cannot be negative.
∴ x = 90
So, breadth of the rectangle = 90 m and length of the rectangle = 90 + 30 = 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 required numbers be x and y, where x > y
Difference of squares of two numbers = 180
Given, x² – y² = 180 …(i)
Square of smaller number = 8 × Larger number
⇒ y² = 8x …..(ii)
From Eqs. (i) and (ii), we have
x² – 8x = 180
⇒ x² – 8x – 180 = 0
x² – (18x – 10x) – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
(By factorization method)
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x – 18)(x + 10) = 0
∴ x – 18 = 0 or x + 10 = 0
∴ x = 18 or x = – 10
Now, x = 18
⇒ y² = 8 × 18 = 144
⇒ y = ± 12
⇒ y = 12 or -12 [From EQuestion (ii)]
Again, x = -10
⇒ y² = [8 (-10)] = -80
which is not possible i.e., imaginary value.
Hence, the numbers are (18 and 12) or (18 and -12).

Question 8.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 kmh^-1 more than that of the passenger train, find the average speed of the two trains.
Solution.
Let the average speed of the passenger train = x kmh^-1.
Then, the average speed of the express train = (x + 11) kmh^-1
Time taken by passenger train to cover

⇒ x (x + 11) = 132 × 11
⇒ x² + 11x – 1452 = 0
x² + (44 – 33)x – 1452 = 0
⇒ x² + 44x – 33x – 1452 = 0
(By fatorization method)
= x(x + 44) – 33(x + 44) = 0
⇒ (x + 44)(x – 33) = 0
∴ x + 44 = 0
⇒ x = -44 or x – 33 = 0
⇒ x = 33
Since, x ≠ -44 because speed can’t be negative.
Hence, the average speed of the passenger train = 33 km/h
and the average speed of the express train = (33 + 11) km/h = 44 km/h.

Question 9.
Out of a certain number of Saras birds, one-fourth of the number are moving about in lotus plants, \(\frac {1}{9}\)th coupled with \(\frac {1}{4}\)th as well as 7 times the square root of the number move on a hill, 56 birds remain in vakula tree. What is the total number of birds ?
Solution.
Let the total number of birds be x.
∴ Number of birds moving about in lotus plants = \(\frac {x}{4}\)
and the number of birds moving on a hill
⇒ \(\frac {x}{9}\) + \(\frac {x}{4}\) + 7√x
Also, the number of birds in vakula tree = 56
∴ According to the given information, we have

9x + 2x + 126 √x – 18x + 1008 = 0
⇒ -7x + 126 √x + 1008 = 0
⇒ x – 18√x – 144 = 0
Putting √x = y, we get
y² – 18y – 144 = 0
⇒ y² – 24y + 6y – 144 = 0
⇒ y(y – 24) + 6(y – 24) = 0
⇒ (y – 24)(y + 6) = 0
⇒ y = 24 or y = -6.
But y = -6, since √x = y is positive
∴ y = 24
⇒ √x = 24
∴ x = 576.
Hence, the total number of birds is 576.

Question 10.
Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m² ? If so, find its length and breadth.
Solution.
Let breadth of a rectangular mango grove = x metre
∴ Length of a rectangular mango grove = 2x metre(By given condition)
∴ According to the question,
Area of rectangular mango grove = 800 m²
⇒ Length × Breadth = 2x (x) = 800
⇒ 2x² = 800
⇒ x² = 400
⇒ x = ± 20
But (∵ breadth never be negative)
∴ Length 2x = 40 m and Breadth = 20 m.

Rulers and Buildings Class 7 Questions and Answers Provided helps you to answer complex Questions too easily. You can use them while preparing for board exams and all of them are given by subject experts. Reading NCERT Solutions for Class 7 Social Science History Chapter 5 Rulers and Buildings familiarizes you with the kind of questions appearing in the board exams. Students are advised to read these solutions on a regular basis to score well.

Rulers and Buildings Class 7 Questions and Answers History Chapter 5

Make your learning experience enjoyable by preparing from the quick links available on this page. Use the Class 7 SST History Chapter 5 NCERT Solutions and get to know different concepts involved. All the Solutions are covered as per the latest syllabus guidelines. Knowing the NCERT Class 7 History Chapter 5 Questions and Answers helps students to attempt the exam with confidence.

Class 7 History Chapter 5 NCERT Textbook Questions and Answers

Let Recall

Question 1.
How is the “trabeate” principle of architecture different from the “arcuate”?

Question 2.
What is a shikhara?
Answer:
A shikhara is the top most portion of a temple.

Question 3.
What is pietra-dura?
Answer:
Coloured, hard stones placed in depressions carved into marble or sandstone creating beautiful ornate patterns is called ‘pietra-dura’.

Question 4.
What are the elements of a Mughal Chahar bagh garden?
Answer:
Placed within rectangular walls and divided into four quarters by artificial channels are the main elements of a Mughal Chahar bagh.

Let’s Understand

Question 5.
How did a temple communicate the importance of a king?
Answer:
Kings built temples to demonstrate their devotion to God and their power and wealth. Constructing places of worship provided rulers with the chance to proclaim their close relationship with God, especially important in an age of rapid political change. Rulers also offered patronage to the learned and pious, and tried to transform their capitals and cities into great administrative, trade and cultural centres that brought fame to their rule and their realm.

Question 6.
An inscription in Shah Jahan’s diwan-i-khas in Delhi stated: “If there is Paradise on Earth, it is here, it is here, it is here.” How was this image created?
Answer:
Shah Jahan’s Diwan-i-khas (audience hall) was specially constructed in the Red Fort at Delhi. The pedestal on which Shah Jahan’s throne was placed was frequently described as the ‘qibla’ (qibla is the direction faced by Muslims at prayer time).

Since everybody faced towards direction of ‘qibla’ when court was in session. The idea of the king was to represent the God on earth. The construction of Shah Jahan’s audience hall aimed to communicate that the king’s justice would treat the high and the low as equals creating a world where all could live together in harmony.

Question 7.
How did the Mughal court suggest that everyone-the rich and the poor, the powerful and the weak received se equally from the emperor?
Answer:
Mughal court ‘Diwan-i-am’ was suggested that justice was made for all in an equal way. Shah Jahan’s audience hall was designed in a manner so that it could communicate that the king’s justice was equal for everyone whether rich or poor, powerful or weak. It has a great message of harmony. Thus, in the Mughal court there was no difference between the rich and poor, powerful and weak.

Question 8.
What role did the Yamuna play in the layout of the new Mughal city at Shahjahanabad?
Answer:
In the new city of Shahjahanabad that Shah Jahan constructed in Delhi, the imperial place commanded the river-front only. All others had to construct their homes in the city away from the River Yamuna.

Let’s Discuss

Question 9.
The rich and powerful construct large houses today. In what ways were the constructions of kings and their courtiers different in the past?
Answer:
The rich and powerful people construct large houses today too. In the metropolitan cities, people live in the societies and some in the bunglows. But a big change can be seen between the buildings of past constructed by the kings and in present days. In present days, cemented and iron pillars, a single baked brick for partition, iron, cement etc. are being used. While, in the past big courtyards, thick walls, domed-roofs, big gardens, heavy stones pillars were made with the help of limestone cement.

Question 10.
Look at Figure 4. How could that building be constructed faster today?
Answer:
In the figure 4, Rajarajeshwara temple has been shown. In the modern time, we have innovative technology for constructing such types of temples and also have many equipment and crane for cutting and lifting the heavy stones. Therefore, now it not as difficult as it would have been during the period when the temple was built. With the help of new technology buildings can be constructed faster today.

Let’s Do

Question 11.
Find out whether there is a stature of or a memorial to a great person in your village or town. Why was it placed there? What purpose does it serve?
Answer:
Students discuss with the teacher and write the answer.

Question 12.
Visit and describe any park or garden in your neighborhood. In what ways is it similar to or different from the gardens of the Mughals?
Answer:
Students discuss with the teacher and can write the answer.

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