Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Balbharti Maharashtra State Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues Important Questions and Answers.

Maharashtra State Board 12th Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Complete the following statements by selecting the appropriate option.

Question 1.
The Rio de Janeiro Earth Summit (1992) focused on
(a) Environment and development
(b) Nuclear non-proliferation
(c) International trade
(d) Gender issues
Answer:
(a) Environment and development

Question 2.
The 2002 Earth Summit recognised as the most important goal at all level.
(a) poverty
(b) socio-economic develop
(c) sustainable development
(d) gender issues
Answer:
(c) sustainable development

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 3.
The 2030 Agenda for sustainable Development was adopted in the year
(a) 2015
(b) 2002
(c) 1992
(d) 2019
Answer:
(a) 2015

Question 4.
The Second UN Decade for poverty eradication was in
(a) 1997-2006
(b) 1971-1980
(c) 2008-2017
(d) 1992-2003
Answer:
(c) 2008-2017

Question 5.
The Revolution aimed to make India self-sufficient in food grains.
(a) Import substitution
(b) Green
(c) White
(d) Infrastructural
Answer:
(b) Green

Question 6.
The is a scheme introduced in 1978-79 to create opportunities for self-employment in
the rural sector.
(a) MGNREGA
(b) JRY
(c) IRDP
(d) KVY
Answer:
(c) IRDP

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Complete the following sentences by selecting the appropriate reason.

Question 1.
Green Revolution was introduced in India in the 1960’s to ………………
(a) make India self sufficient in food grains.
(b) generate employment and tackle poverty.
(c) to combine economic growth with social justice.
Answer:
(a) make India self sufficient in food grain

Question 2.
The Department of Women and Child Development was set up as a separate Ministry in 2006 to …………..
(a) facilitate health and education programmes for women.
(b) provided political representation to women
(c) empower women to live with dignity and contribute as equal partners in development.
Answer:
(c) make India self sufficient in food grains.

State the appropriate concept for the given statements.

Question 1.
Introduction of high yielding variety of seeds and irrigation methods.
Answer:
Green Revolution

Question 2.
Think tank of the Government of India, providing both directional and policy inputs.
Answer:
NITI Aayog

Question 3.
The responsible interaction with the environment to avoid depletion or degradation of natural resources.
Answer:
Sustainable development

Question 4.
Deprivation of common necessities that determine the quality of life.
Answer:
Poverty

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 5.
A view that free market economy promotes development.
Answer:
Capitalism

Question 6.
Blueprint for prosperity for people and planet adopted by the UN in 2015.
Answer:
2030 Agenda for Sustainable Development

Question 7.
Agency set up in 1950 to formulate a plan for effective and balanced utilisation of India’s resources.
Answer:
Planning Commission

Question 8.
India’s policy towards poverty and development.
Answer:
Growth with Social Justice and Equity

Question 9.
The process by which oppressed persons gain control over their own lives.
Answer:
Empowerment

Question 10.
Socially constructed characteristics of women and men such as behaviour, norms, roles and relationships.
Answer:
Gender

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Find the odd word.

Question 2.
Pollution, Deforestation, Poverty, Water Scarcity.
Answer:
Poverty (not a direct environment concern)

Question 2.
Brundtland Commission, Rio+20 Summit, Johannesburg Earth Summit, United Nations Development Programme (UNDP).
Answer:
UNDP (not a commission/ summit associated with sustainable development)

Expand the following abbreviations.

(1) UNCED
(2) ECOSOC
(3) UNDP
(4) IRDP
(5) JGSY
(6) MPI
Answer:
(1) UNCED – United Nations Conference on Environment and Development
(2) ECOSOC – Economic and Social Council of the UN.
(3) UNDP – United Nations Development Programme
(4) IRDP – Integrated Rural Development Programme
(5) JGSY – Jawahar Gram Samridhi Yojana
(6) MPI – Multidimensional Poverty Index

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Complete the concept maps.

Question 1.
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 1
Answer:
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 2

Question 2.
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 3
Answer:
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 4

Question 3.
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 5
Answer:
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 6

State whether the following statements are true or false with reason.

Question 1.
Rio Earth Summit (1992) was a landmark in environment issues.
Answer:
This statement is True.

  1. Rio de Janeiro Earth Summit (1992) sought to raise public awareness on the need to integrate environment and development.
  2. It’s objective was to create a partnership between developing and more industrialized nations to ensure a healthy future for the planet.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 2.
The perception of poverty has undergone a change in recent times.
Answer:
This statement is True.

  1. The traditional perception of poverty focused on the deprivation of the basic necessities of life such as food, shelter, etc.
  2. The new perception of poverty as expressed by the ECOSOC, considers poverty as lack of basic capacity to participate effectively in society.
  3. It is perceived as a denial of choices, opportunities and human dignity.

Question 3.
The 2030 Agenda for Sustainable Development (2015) is a significant document.
Answer:
This statement is True.

  1. The 2030 Agenda for Sustainable Development (2015) adopted by the UN, provides a blueprint for prosperity of the people and planet for now and in the future
  2. It mentions seventeen Sustainable Development Goals for action by all countries in a global partnership for e.g., No Poverty, Climate action, Gender equality, etc.

Question 4.
The economic reforms initiated in India after 1991 had far-reaching implications.
Answer:
This statement is True.

  1. After 1991, India introduced the policies of economic liberalisation and privatisation of the public sector and brought an increase in economic and industrial growth rate.
  2. These reforms changed the role of the State to focus more on social development such as in heath care, sanitation, education, etc.

Explain the correlation between the following.

Question 1.
Department of Women and Child Development – Women’s empowerment.
Answer:
The Department of Women and Child Development, Government of India came into existence as separate ministry in 2006. Before this, it was a department under HRD ministry: The Department of Women and Child Development was constituted with the intention of addressing gaps in State action for women and children and to create gender equitable and child-centered legislations, policies and programmes.

It aims to achieve the empowerment of women i.e., to live in dignity and to contribute as equal partners in development in an environment free from discrimination, injustice and violence. The Ministry has prepared the National Policy for the Empowerment of Women (2001). The Draft of the National Policy for women (2016) aims at women’s empowerment by following a socially inclusive rights-based approach.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Express your opinion of the following.

Question 1.
There is a need to protect the environment.
Answer:
I agree with this statement.
The environment refers to all living and non-living things that make up our surroundings. Many environmental problems result due to unplanned management, human activities and technological development that interfere with the environment. Protection of environment refers to conservation and preservation of environment by reducing pollution, soil erosion, global warming, deforestation etc.

Main environmental concerns hence are climate change, water scarcity, pollution, loss of biodiversity, etc. Environmental degradation and depletion of resources will result in spread of pests and vector diseases, extinction of species, natural disasters like floods; acid rain, melting of glaciers, etc. Life on this earth cannot exist unless we restore environmental balance.

Question 2.
Poor political representation of women is a major concern.
Answer:
I agree with this statement.
Poor political representation of women is a significant gender concern in India. The first Lok Sabha had only 24 women of total 489 members i.e., 5%, and the 17th Lok Sabha (2019) has 78 women (14%). This is the highest representation of women in Lok Sabha till date. The Women’s Reservation Bill which proposes to reserve 33% of all seats in Lok Sabha and in State Assemblies for women is pending in the Lok Sabha. This bill was first introduced in Parliament in 1996, by the United Front government of Prime Minister H.D. Deve Gowda.

As per the bill, seats will be reserved for women on a rotational basis. The UPA-I government (2004-2009) again introduced the bill in May 2008 and it was passed in May 2010 by the Rajya Sabha. Currently, the bill is still pending in Lok Sabha. However, there exists 33% reservation for women in rural and urban local bodies (as per 73rd and 74th Amendment Acts).

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Answer the following question in 80 to 100 words.

Question 1.
What is the Green Revolution?
Answer:
Green Revolution refers to a large increase in crop production that is achieved by the use of high yielding crop varieties, pesticides, artificial fertilizers, machines and better management of agricultural resources. Norman Borlaug, an agricultural scientist of USA is considered as ‘Father of the Green Revolution’ for which he was awarded the Nobel Peace Prize (1970). In India, M.S. Swaminathan is credited with introducing the Green Revolution in 1965 especially in Punjab, Haryana and Uttar Pradesh.

This was necessitated due to famine and acute food shortages due to low productivity in food production. Green Revolution was mainly in wheat, maize and gram production. During it’s early years, the Green Revolution yielded great economic prosperity leading to significant increases in agricultural output and farmers incomes. However, this did not benefit marginal farmers who could not afford expensive inputs.

Question 2.
What is empowerment? Suggest measures for empowerment of women.
Answer:
Empowerment refers to the process by which oppressed or marginalized sections of the population become stronger, especially in controlling their own life and claiming their own rights. This may happen socially, economically, politically or nationally. The approach to women’s issues has progressed from ‘welfare’ to ‘development’ to ‘empowerment’.

Some of the measures for empowerment of women include-

  1. Political empowerment – giving women better representation in legislative bodies and in the decision making process.
  2. Economic empowerment – such as strengthening women’s access to property inheritance and land rights, skill training, work opportunities, micro-credit, increased right to economic resources and power.
  3. Social empowerment – through literacy/education, training and raising awareness, freedom from domestic violence, access to information, health services and sanitation.

In India, schemes to empower women include Beti Bachao Beti Padhao Yojana (2015), Support to Training and Employment Programme for Women (STEP), Mahila Shakti Kendras (2017) etc.

Question 3.
Describe international efforts to deal with environmental concerns.
Answer:
There have been many international efforts to deal with environmental concerns as well as about the relationship between economic development and environmental degradation.
1. UN Conference on Human Environment (Stockholm, 1972) The purpose was to encourage and provide guidelines for protection of the environment.

2. World Commission on Environment and Development (1983) also called Brundtland commission put forward the concept of sustainable growth.

3. Rio de Janeiro Earth Summit (1992) sought to raise public awareness on the need to integrate environment and development and to create a partnership between developing and more industrialized nations to ensure a healthy future for the planet.

4. Earth Summit at Johannesburg (2002) recognized sustainable development as the most important goal for institutions at the national, regional and international levels

5. UN Conference on sustainable development (Rio-2012) is also called Rio Earth Summit or Rio+20. Environmental sustainability is defined as responsible interaction with the environment to avoid depletion or degradation of natural resources and allow for long¬term environmental quality. It’s goal is to conserve natural resources, to reduce pollution, to develop alternate sources of power, etc.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 4.
Elaborate on the aspects of development in India.
Answer:
The purpose of development is to ensure welfare of the people and is associated with economic growth. The traditional approaches to development focus on the predominant role of the State in promoting economic growth (Socialist) or role of free market economy (Capitalist). Since the 1990s, the alternative view of development that focuses on both material and non-material aspects has become acceptable. The focus today is development based on equity, participation, empowerment and sustainability.

  1. Modernisation of the economy – During the Nehruvian period, the focus was on land reforms, irrigation and large-scale industrialisation to provide employment and increase productivity.
  2. Self – reliance – This included ‘import substitution’ strategy to develop indigenous industry.
  3. Socialist pattern of society – It was based on promotion of the Welfare State, employment generation and importance ,to the public sector. The Planning Commission (March 1950 to 31st December 2014) was created to chalk out Five Year Plans for development.

“Growth with Social Justice and Equity’ has been India’s policy towards poverty and development. Its development strategy has evolved over the years. In the early stages, government played a dominant role through the public sector.

Question 5.
Describe the 2030 Agenda for Sustainable Development adopted in 2015 by the UN.
Answer:
In 2015, the United Nations adopted the ‘2030 Agenda for Sustainable Development. It provides a blueprint for peace and prosperity for people and the planet for now and into the future. The Agenda mentions 17 Sustainable Development Goals (SDGs) for action by all developed and developing countries in a global partnership.

The Sustainable Development Goals are:

  1. No Poverty
  2. Zero hunger
  3. Good health and well being
  4. Quality
  5. Gender equality
  6. Clean water and sanitation
  7. Affordable and clean energy
  8. Decent work and economic growth
  9. Industry, innovation and infrastructure
  10. Reduced inequalities
  11. Sustainable cities and communities
  12. Responsible consumption and production
  13. Climate action
  14. Life below water
  15. Life on land
  16. Peace, justice and strong institutions
  17. Partnership goals

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Answer the following question with reference to the given points.

Question 1.
Explain condition of women in India.
(a) Economic inequality
(b) Trafficking and exploitation
(c) Literacy rate
(d) Political representation
Answer:
All over the world, women have to face serious problems such as discrimination and violence, under representation in economic and political decision-making, inequalities in education, health care, landholdings and workplaces and even in human rights.

Some of the important gender issues:
(a) Economic inequality – Participation of women in the labour market is about 28% only. There is also income inequality due to wide gender wage gaps as well as inequality of opportunities to women e.g., unequal access to education and finance. In many countries, women are either not encouraged to work or employed in the informal sector. Women continue to be underrepresented in high-level, highly paid positions and experience gender discrimination and sexual harassment in the workplace. Women experience high levels of poverty, unemployment and other economic hardships.

(b) Trafficking and exploitation – Women are often exploited and denied basic rights. They are trafficked for purpose of sexual and economic exploitation particularly prostitution, ‘sold’ as brides, subjected to sexual crimes, forced labour, street begging, etc. Trafficking in women means that they are deceived or forced/sold, physically confined, abused with no access to protection or health care.

(c) Literacy rate – low literacy among women is acute in India. As per 2011 census report, female literacy rate is only 65.46% while for males it is 82.14%. The reasons for the low female literacy are-

  1. traditional patriarchal notions that do not consider female education as economically advantageous
  2. Schools in rural areas are not easily accessible and travel may not safe or easy.
  3. Benefits of schemes like R.T.E have not reached many rural females.
  4. Social evils like child marriage, female foeticide, dowry, etc.

(d) Political Representation – Poor political representation of women is a significant gender concern in India. The First Lok Sabha had only 24 women of total 489 members i.e., 5% and the 17th Lok Sabha (2019) has 78 women (14%). This is the highest representation of women in Lok Sabha till date. The Women’s Reservation Bill which proposes to reserve 33% of all seats in Lok Sabha and in State Assemblies for women is pending in the Lok Sabha. However, there exists 33% reservation for women in rural and urban local bodies.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 2.
Describe the following poverty alleviation programmes in India.
(a) Integrated Rural Development Programme (IRDP)
(b) Jawahar Gram Samridhi Yojana (JGSY)
(c) Pradhan Mantri Gramin Awaas Yojana (PMGAY)
(d) Mahatma Gandhi National Rural Employment Guarantee Act. (MGNREGA)
(e) Saansad Adarsh Gram Yojana.
Answer:
The poverty alleviation programmes in India have been designed for both rural areas and urban areas. Most of the programmes are designed to target the rural poverty as prevalence of poverty is high in rural areas.
(a) Integrated Rural Development Programme (IRDP) – It was first introduced in 1978¬79. The main objective of IRDP is to create sustainable opportunities for self-employment in the rural sector.

(b) Jawahar Gram Samridhi Yojana (JGSY) – It is the revised version of the Jawahar Rozgar Yojana (JRY). It was started in 1999. The main aim of this programme was development infrastructure like roads to connect the villages to different areas and other social, education (schools) and infrastructure like hospitals in the rural areas.

(c) Pradhan Mantri Gramin Awaas Yojana – It was started in 1985. This scheme aimed at creating housing for everyone.

(d) Mahatma Gandhi National Rural Employment Guarantee Act. (MGNREGA) – It was launched in 2005. It provides 100 days assured employment every year to every rural household.

(e) Saansad Adarsh Gram Yojana – It was started in 2014. According to this yojana, each Member of Parliament will take the responsibility of developing three villages by 2019. The idea is to make India’s villages to be fully developed with physical and institutional infrastructure.

Question 3.
Discuss the following humanitarian issues.
(a) Environmental degradation
(b) Poverty
(c) Gender issues
Answer:
Humanitarian issues refer to concerns about issues affecting human welfare with the aim of alleviation of sufferings. Examples of humanitarian issues include natural disasters, famine, epidemics, armed conflicts, poverty, inequalities and injustice, environmental problems, etc.

(a) Environmental degradation – Indiscriminate exploitation of natural resources had resulted in environmental depletion and degradation. This results in significant environmental concerns such as deforestation, soil erosion,loss of biodiversity, climate change, endangerment of species, water scarcity, acid rain, pollution, spread of diseases etc. The relationship between economic development and environmental degradation has been considered at various international conferences like Rio Earth Summit, etc.

Environmental degradation has serious consequences for humans, animals, plants as well as the ecological balance. The causes of degradation include resource depletion due to overuse, landfills and illegal dumping, waste production, overpopulation and consumption behaviour, etc.

(b) Poverty – The traditional perception of poverty is where people are unable to provide for their basic necessities of life. It is the deprivation of common necessities that determine the quality of life including food, clothing, shelter and safe drinking water. The alternative view of poverty focuses on both material and non-material aspects i.e., focus on community ties, values and availability of common resources, participatory decision making and political and economic decentralisation.

The purpose of development is to ensure welfare of the people and is associated with economic growth. The traditional approaches to development focus on the predominant role of the State in promoting economic growth (socialist) or role of free market economy (capitalist). Since the 1990s, the alternate view of development that focuses on both material and non material aspects has become acceptable. The focus today is development based on equity, participation, empowerment and sustainability.

(c) Gender issues – There is a distinct link between poverty and unsatisfactory conditions of women. In most parts of the world women are poor due to cultural norms, unequal power equations between men and women. In the 1970s, women empowerment came to be accepted as an important philosophy to bring about gender justice. The UN ‘Decade for Women’ began in 1976 to-

  1. link women’s issues with developmental issues
  2. promote equal rights and opportunities for women across the globe.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 4.
Discuss the vision of the development process in India.
(a) Modernisation of the economy
(b) Self-reliance
(c) Socialist pattern of society
Answer:
The purpose of development is to ensure welfare of the people and is associated with economic growth. The traditional approaches to development focus on the predominant role of the State in promoting economic growth (Socialist) or role of free market economy (Capitalist). Since the 1990s, the alternative view of development that focuses on both material and non material aspects has become acceptable. The focus today is development based on equity, participation, empowerment and sustainability. India’s vision of development had three aspects:

(a) Modernisation of the economy – This implied industrialization to provide employment to a growing labour force and to increase productivity During the Nehruvian period, the focus was on large scale industrialization and factory production. Land reforms and irrigation were looked at as the means to achieve agricultural growth and productivity.

(b) Self-reliance – This included ‘import substitution’ strategy to develop indigenous industry. Aid taken from Soviet Union, France, etc. was utilised mainly in the public sector and for infrastructure.

(c) Socialist pattern of society – The approach of planning for development was based on promotion of the welfare State, employment generation and importance to the public sector. The Planning Commission (March 1950 31st December 2014) was created to chalk out Five Year Plans for development.

“Growth with Social Justice and Equity” has been India’s policy towards poverty and development. Its development strategy has evolved over the years. In the early stages, government played a dominant role through the public sector.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Balbharti Maharashtra State Board Class 12 Psychology Important Questions Chapter 5 Emotions Important Questions and Answers.

Maharashtra State Board 12th Psychology Important Questions Chapter 5 Emotions

Choose the correct option and complete the following statements.

Question 1.
According to the theory, we experience physiological arousal and feelings simultaneously and independently.
(a) James-Lange
(b) Cannon-Bard
(c) Schachter-Singer
Answer:
(b) Cannon-Bard

Question 2.
According to Ekman, there are basic emotions.
(a) 2
(b) 5
(c) 6
Answer:
(c) 6

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 3.
is an emotion that is in response to some threat.
(a) Sadness
(b) Fear
(c) Disgust
Answer:
(b) Fear

Question 4.
Plutchik’s model explains that there are primary emotions.
(a) 8
(b) 6
(c) 12
Answer:
(b) 6

Question 5.
is not a primary/ basic emotion according to Plutchik.
(a) Trust
(b) Anticipation
(c) Jealousy
Answer:
(c) Jealousy

Question 6.
Emotions are controlled by the ……………….. in the brain.
(a) cerebral cortex
(b) limbic system
(c) RAS
Answer:
(b) limbic system

Question 7.
Aniket shows creativity and has meaningful relationships. He has …………. emotional well-being.
(a) high
(b) low
(c) moderate
Answer:
(a) high

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

State whether the following statements are true or false.

Question 1.
“I feel sad because I cry”, This is the premise of the Schachter – Singer theory.
Answer:
False

Question 2.
Pride and guilt are considered as higher cognitive emotions.
Answer:
True

Question 3.
The fight or flight reaction is a reaction to sadness.
Answer:
False

Question 4.
Low serotonin levels are associated with clinical depression.
Answer:
True

Answer the following in one sentence each.

Question 1.
What is the premise of the facial feedback hypothesis?
Answer:
According to the facial feedback hypothesis, our facial expressions provide feedback to our brain about our emotions.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 2.
What are complex emotions?
Answer:
Complex emotions are emotions which result form the combination of basic emotions for e.g., surprise and sadness lead to disappointment.

Question 3.
According to Ekman, which are the basic emotions?
Answer:
According to Ekman, there are six basic (universal) types of emotions, i.e., happiness, sadness, anger, fear, surprise, disgust.

Question 4.
When does a person experience disgust?
Answer:
A person usually experiences disgust as a response to some unwanted stimulus.

Question 5.
According to Plutchik, how do emotions result?
Answer:
According to Plutchik, emotions result due to evolving over a period of time for increasing our chances of survival.

Question 6.
Why is Plutchik’s model important?
Answer:
Plutchik’s model is important from the perspective of emotional literacy, i.e., understanding emotional levels, complexity and change as well as appropriate emotional labelling.

Question 7.
What does emotional well-being mean?
Answer:
Emotional well-being means managing both positive and negative emotions, so that we can lead a healthy and productive life.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 8.
What does anger management mean?
Answer:
Anger management is an intervention programme to prevent anger from turning into a habit or obstacle by creating awareness of and responsibility for our emotions.

Explain the concepts in 25 – 30 words each.

Question 1.
Emotions
Answer:
The word emotion is derived from the latin word ‘emovere’, which means to stir up or to move. An emotion refers to an involuntary, aroused state of an organism involving physical, cognitive and behavioural components. It is described as a combination of bodily arousal, e.g., increased heartrate, thoughts and feelings, i.e. emotional tone and expressive behaviour i.e. facial expression.

Question 2.
Emotional well-being.
Answer:
Emotional well-being means managing our emotions, both positive and negative ones, so that we can lead a healthy and productive life. Persons who have high emotional well-being experience benefits such as-

  1. better able to deal with stress
  2. better self-regulation
  3. increased productivity in tasks undertaken
  4. increased creativity and openness to new experiences
  5. life satisfaction due to meaningful activities and relationship.

Question 3.
Emotional Abuse.
Answer:
Emotional abuse is any kind of abuse that is emotional rather than physical in nature. It occurs when one person subjects or exposes another person to intentionally harmful behaviour that may result in anxiety, depression and psychological trauma in the victim.
The types of emotional abuse may be-

  1. verbal abuse such as blaming, insulting, labeling, threatening, swearing, etc.
  2. nonverbal abuse such as ignoring, rejection, bullying, spying, etc.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 4.
Anger Management.
Answer:
Anger management is an intervention programme to prevent anger from turning into a habit or obstacle. It enables the person to create awareness of and responsibility for his/her emotions. Anger management involves two aspects (i) managing one’s own anger (ii) learning to respond effectively to anger in others. The three R’s in anger management are Relax, Reassess and Respond.

Answer the following questions in 35 – 40 words each.

Question 1.
What are the techniques to deal with emotional abuse?
Answer:
Emotional abuse is any kind of abuse that is emotional rather than physical in nature. It occurs when one person subjects or exposes another person to intentionally harmful behaviour that may result in anxiety, depression and psychological trauma for the victim.
The techniques to deal with emotional abuse-

  1. Accept that emotional abuse is not because of you, i.e., don’t justify the actions of the abuser.
  2. Respond assertively to the abuser but seek distance from him/her.
  3. Give yourself time to heal.
  4. Prioritize your self-care, eating right, exercise, etc.
  5. Mobilise support from family and friends. If needed, seek professional help.

Question 2.
What are the benefits of emotional well-being?
Answer:
Emotional well-being means managing our emotions, both positive and negative ones, so that we can lead a healthy and productive life. Persons who have high emotional well-being experience benefits such as-

  1. better able to deal with stress
  2. better self-regulation
  3. increased productivity in tasks undertaken
  4. increased creativity and openness to new experiences
  5. life satisfaction due to meaningful activities and relationship.

Write short notes on the following in 50 – 60 words each.

Question 1
Characteristics of emotions
Answer:
The word emotion is derived from the latin word ‘emovere’, which means to stir up or to move. An emotion refers to an involuntary, aroused state of an organism involving physical, cognitive and behavioural components. It is described as a combination of bodily arousal, e.g., increased heartrate, thoughts and feelings, i.e. emotional tone and expressive behaviour i.e. facial expression.
Some characteristics of emotions are-

  1. Emotions may be positive, e.g., joy or negative, e.g., anger.
  2. Emotions may occur for a brief period or may be long lasting.
  3. Emotions may be important for our survival, e.g., fear or for our psychological well-being, e.g., love.
  4. Emotions differ in intensity in expression for e.g., annoyance-anger-rage.
  5. Complex emotions (higher cognitive level emotions) result from the combination of basic emotions, for e.g., surprise and sadness lead to disappointment.
  6. According to Ekman, there are six basic (universal) types of emotions, i.e., happiness, sadness, anger, fear, surprise, disgust.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 2.
Physiological changes during emotions.
Answer:
Physiological changes refer to automatic reactions that take place in our body in response to some stimulus, for e.g., if you saw a snake, the brain at the cognitive level perceived the stimulus as dangerous. This leads to physiological arousal such as dilated pupils, increased heart rate, increased pulse rate, sweating. At the emotional level you experience fear. The Autonomic Nervous System and Glandular system signal the pituitary gland which activates the adrenal glands to secrete the cortisol hormone. This triggers “the fight or flight” response. At the behavioural level there is an action plan such as running away or calling for help, etc.
Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions 1

Answer the following questions 80 – 100 words each.

Question 1.
Theories of emotions.
Points:
(i) James-Lange theory
(ii) Cannon-Bard theory
(iii) Schachter-Singer theory
Answer:
The word emotion is derived from the latin word ‘emovere’, which means to stir up or to move. An emotion refers to an involuntary, aroused state of an organism involving physical, cognitive and behavioural components. It is described as a combination of bodily arousal, e.g., increased heartrate, thoughts and feelings, i.e. emotional tone and expressive behaviour, i.e., facial expression.
The main theories of emotions are-
(i) James-Lange theory – It was one of the earliest theories to explain emotion. According to William James and Carl Lange, physiological arousal instigates the experience of emotion. It proposes that each specific emotion is attached to a different pattern of physiological arousal. For e.g., we feel sad because we cry. The sequence of events in emotional experience is Emotion Stimulus – Physiological Response Pattern → Affective Experience.

(ii) Cannon-Bard theory – According to Walter Cannon and Philip Bard, we may experience the same physiological arousal but emotions can be different, for e.g., we don’t cry only when we are sad but we also cry when we are angry or extremely happy. We experience physiological arousal and feelings at the same time and independently. For e.g., seeing a man with a gun prompts the feeling of fear as well as a racing heartbeat.

(iii) Schachter and Singer’s Two Factor theory – According to Stanley Schachter and Jerome Singer, emotion is based on two factors, i.e., physiological arousal and cognitive label, for e.g., an environmental stimuli (growling dog) elicits a physiological response (increased heart rate). We cognitively label this response (fear).
Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions 2

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 2.
Emotional well-being.
Points:
(i) Meaning
(ii) Aspects
(iii) Benefits
Answer:
The word emotion is derived from the latin word ‘emovere’, which means to stir up or to move. An emotion refers to an involuntary, aroused state of an organism involving physical, cognitive and behavioural components. It is described as a combination of bodily arousal, e.g., increased heart rate, thoughts and feelings, i.e. emotional tone and expressive behaviour i.e. facial expression.

(i) Meaning – Emotional well-being means managing our emotions, both positive and negative ones, so that we can lead a healthy and productive life. Emotional well-being is not easily observable. It can be guaged on the basis of how a person rationally handles a situation involving some stressors or leading to negative emotions. Emotional well-being refers to understanding and managing one’s emotions without getting overwhelmed by negative emotions but instead encouraging positive emotions.

(ii) Aspects – The aspects of emotional well-being are at three levels viz. physical, emotional and social.

  1. Physical level, i.e., well balanced diet, exercise.
  2. Emotional level, i.e., practise mindfulness, raising levels of motivation and optimism.
  3. Social level, i.e., engaging in prosocial behaviour, meaningful relationships.

(iii) Benefits – Persons who have high emotional well-being experience benefits such as-

  1. Coping with stress – It helps to deal with stress using healthy methods such as exercise, social support, etc.
  2. Better self-regulation – It enables the person to label how they feel and accept negative emotions life fear, anger, etc.
  3. Increases productivity in tasks undertaken – The ability to focus is enhanced, the person feels positive and energized.
  4. Increases creativity – The person indulges in divergent thinking, shows curiosity is open to new experiences.
  5. Life satisfaction – The individual is able to have meaningful interactions and relationships, show empathy, altruism and engage in activities like volunteer work.

Maharashtra Board Miscellaneous Problems Set 1 Class 7 Maths Solutions

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 1 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 1 Class 7 Maths Solutions

Question 1.
Solve the following:
i. (-16) × (-5)
ii. (72) ÷ (-12)
iii. (-24) × (2)
iv. 125 ÷ 5
v. (-104) ÷ (-13)
vi. 25 × (-4)
Solution:
i. (-16) × (-5) = 80

ii. 72 ÷ (-12) = \(\frac { 72 }{ -12 }\)
= \(\frac{1}{(-1)} \times \frac{72}{12}\)
(-1) × 12
= -6

iii. (-24) × 2 = -48

iv. 125 ÷ 5 = \(\frac { 125 }{ 5 }\)
= 25

v. (-104) ÷ (-13) = \(\frac { -104 }{ -13 }\)
= \(\frac { 104 }{ 13 }\)
= 8

vi. 25 × (-4) = -100

Question 2.
Find the prime factors of the following numbers and find their LCM and HCF:
i. 75,135
ii. 114,76
iii. 153,187
iv. 32,24,48
Solution:
i. 75 = 3 × 25
= 3 × 5 × 5
135 = 3 × 45
= 3 × 3 × 15
= 3 × 3 × 3 × 5
∴ HCF of 75 and 135 = 3 × 5
= 15
LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3
= 675

ii. 114 = 2 × 57
= 2 × 3 × 19
76 = 2 × 38
= 2 × 2 × 19
∴ HCF of 114 and 76 = 2 × 19
= 38
LCM of 114 and 76 = 2 × 19 × 3 × 2
= 228

iii. 153 = 3 × 51
= 3 × 3 × 17
187 = 11 × 17
∴ HCF of 153 and 187 = 17
LCM of 153 and 187 = 17 × 3 × 3 × 11
= 1683

iv. 32 = 2 × 16
= 2 × 2 × 8
= 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2
24 = 2 × 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
48 = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
∴ HCF of 32, 24 and 48 = 2 × 2 × 2
= 8
LCM of 32,24 and 48 = 2 × 2 × 2 × 2 × 2 × 3
= 96

Question 3.
Simplify:
i. \(\frac { 322 }{ 391 }\)
ii. \(\frac { 247 }{ 209 }\)
iii. \(\frac { 117 }{ 156 }\)
Solution:
i. \(\frac { 322 }{ 391 }\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 1

ii. \(\frac { 247 }{ 209 }\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 2

iii. \(\frac { 117 }{ 156 }\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 3

Question 4.
i. 784
ii. 225
iii. 1296
iv. 2025
v. 256
Solution:
i. 784
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 4
∴ 784 = 2 × 2 × 2 × 2 × 7 × 7
∴ √784 = 2 × 2 × 7
= 28

ii. 225
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 5
∴ 225 = 3 × 3 × 5 × 5
∴ √225 = 3 × 5
= 15

iii. 1296
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 6
∴ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ √1296 = 2 × 2 × 3 × 3
= 36

iv. 2025
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 7
∴ 2025 = 3 × 3 × 3 × 3 × 5 × 5
∴ √2025 = 3 × 3 × 5
= 45

v. 256
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 8
∴ 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ √256 = 2 × 2 × 2 × 2
= 16

Question 5.
There are four polling booths for a certain election. The numbers of men and women who cast their vote at each booth is given in the table below. Draw a joint bar graph for this data.

Polling Booths Navodaya Vidyalaya Vidyaniketan School City High School Eklavya School
Women 500 520 680 800
Men 440 640 760 600

Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 9

Question 6.
Simplify the expressions:
i. 45 ÷ 5 + 120 × 4 – 12
ii. (38 – 8) × 2 ÷ 5 + 13
iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
Solution:
i. 45 ÷ 5 + 120 × 4 – 12
= 9 + 80 – 12
= 89 – 12
= 77

ii. (38 – 8) × 2 ÷ 5 + 13
= 30 × 2 ÷ 5 + 13
= 60 ÷ 5 + 13
= 12 + 13
= 25

iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
\(\frac{5}{3}+\frac{4}{7} \times \frac{21}{32}\)
\(\frac{5}{3}+\frac{3}{8}=\frac{40}{24}+\frac{9}{24}\)
\(\frac{49}{24}\)

iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
= 3 × {4[90 – 5] + 2}
= 3 × {4 × 85 + 2}
= 3 × (340 + 2)
= 3 × 342
= 1026

Question 7.
Solve:
i. \(\frac{5}{12}+\frac{7}{16}\)
ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)
iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
Solution:
i. \(\frac{5}{12}+\frac{7}{16}\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 10

ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 11

iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
= 4 × (-2)
= -8

iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
= \(\frac{7}{4} \times \frac{9}{5}\)
= \(\frac { 63 }{ 20 }\)

Question 8.
Construct ∆ABC such that m∠A = 55°, m∠B = and l(AB) = 5.9 cm.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 12

Question 9.
Construct ∆XYZ such that, l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 13

Question 10.
Construct ∆PQR such that, m∠P = 80°, m∠Q = 70°, l(QR) = 5.7 cm.
Ans:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° …. (Sum of the measures of the angles of a triangle is 180°)
∴ 80 + 70 + m∠R = 180
∴ 150 + m∠R = 180
∴ m∠R = 180 – 150
∴ m∠R = 30°
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 14

Question 11.
Construct ∆EFG from the given measures. l(FG) = 5 cm, m∠EFG = 90°, l(EG) = 7 cm.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 15

Question 12.
In ∆LMN, l(LM) = 6.2 cm, m∠LMN = 60°, l(MN) 4 cm. Construct ∆LMN.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 16

Question 13.
Find the measures of the complementary angles of the following angles:
i. 35°
ii. a°
iii. 22°
iv. (40 – x)°
Solution:
i. Let the measure of the complementary
angle be x°.
35 + x = 90
∴35 + x-35 = 90 – 35
….(Subtracting 35 from both sides)
∴x = 55
∴The complementary angle of 35° is 55°.

ii. Let the measure of the complementary angle be x°.
a + x = 90
∴a + x – a = 90 – a
….(Subtracting a from both sides)
∴x = (90 – a)
∴The complementary angle of a° is (90 – a)°.

iii. Let the measure of the complementary angle be x°.
22 + x = 90
∴22 + x – 22 = 90 – 22
….(Subtracting 22 from both sides)
∴x = 68
∴The complementary angle of 22° is 68°.

iv. Let the measure of the complementary angle be a°.
40 – x + a = 90
∴40 – x + a – 40 + x = 90 – 40 + x
….(Subtracting 40 and adding x on both sides)
∴a = (50 + x)
∴The complementary angle of (40 – x)° is (50 + x)°.

Question 14.
Find the measures of the supplements of the following angles:
i. 111°
ii. 47°
iii. 180°
iv. (90 – x)°
Solution:
i. Let the measure of the supplementary
angle be x°.
111 + x = 180
∴ 111 + x – 111 = 180 – 111
…..(Subtracting 111 from both sides)
∴ x = 69
∴ The supplementary angle of 111° is 69°.

ii. Let the measure of the supplementary angle be x°.
47 + x = 180
∴47 + x – 47 = 180 – 47
….(Subtracting 47 from both sides)
∴x = 133
∴The supplementary angle of 47° is 133°.

iii. Let the measure of the supplementary angle be x°.
180 + x = 180
∴180 + x – 180 = 180 – 180
….(Subtracting 180 from both sides)
∴x = 0
∴The supplementary angle of 180° is 0°.

iv. Let the measure of the supplementary angle be a°.
90 – x + a = 180
∴90 – x + a – 90 + x = 180 – 90+ x
….(Subtracting 90 and adding x on both sides)
∴a = 180 – 90 + x
∴a = (90 + x)
∴The supplementary angle of (90 – x)° is (90 + x)°.

Question 15.
Construct the following figures:
i. A pair of adjacent angles
ii. Two supplementary angles which are not adjacent angles.
iii. A pair of adjacent complementary angles.
Solution:
i.
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 17

ii.
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 18

iii.
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 19

Question 16.
In ∆PQR the measures of ∠P and ∠Q are equal and m∠PRQ = 70°, Find the measures of the following angles.
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 20

  1. m∠PRT
  2. m∠P
  3. m∠Q

Solution:
Here, ∠PRQ and ∠PRT are angles in a linear pair.
m∠PRQ + m∠PRT = 180°
∴70 + m∠PRT = 180
∴m∠PRT = 180 – 70
∴m∠PRT = 110°
Now, ∠PRT is the exterior angle of ∆PQR.
∴m∠P + m∠Q = m∠PRT
∴m∠P + m∠P = m∠PRT ….(The measures of ∠P and ∠Q is same)
∴2m∠P = 110
∴m∠P = \(\frac { 110 }{ 2 }\)
∴m∠P = 55°
∴m∠Q =

Question 17.
Simplify
i. 54 × 53
ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)
iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)
iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)
Solution:
Simplify
i. 54 × 53
= 54+3
= 57

ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 21

iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 22

iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 23

Question 18.
Find the value:
i. 1716  ÷ 1716
ii. 10-3
iii. (2³)²
iv. 46 × 4-4
Solution:
i. 1716  ÷ 1716
= 170
= 1

ii. 10-3
= \(\frac{1}{10^{3}}\)
= \(\frac{1}{1000}\)

iii. (2³)²
= 23×2
= 26
= 2 × 2 × 2 × 2 × 2 × 2
= 64

iv. 46 × 4-4
= 46+(-4)
= 42
= 4 × 4
= 16

Question 19.
Solve:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
ii. (3x + 2y) (7x – 8y)
iii. (7m – 5n) – (-4n – 11m)
iv. (11m – 12n + 3p) – (9m + 7n – 8p)
Solution:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
= (6a + 2a) + (-5b + 15b) + (-8c – 5c)
= 8a + 10b – 13c

ii. (3x + 2y) (7x – 8y)
= 3x × (7x – 8y) + 2yx (7x – 8y)
= 21x² – 24xy + 14xy – 16y²
= 21x² – 10xy – 16y²

iii. (7m – 5n) – (-4n – 11m)
= 7m – 5n + 4n + 11m
= (7m + 11m) + (-5n + 4n)
= 18m – n

iv. (11m – 12n + 3p) – (9m + 7n – 8p)
= 11m – 12n + 3p – 9m – 7n + 8p
= (11m – 9m) + (-12n – 7n) + (3p + 8p)
= 2m – 19n + 11p

Question 20.
Solve the following equations:
i 4(x + 12) = 8
ii. 3y + 4 = 5y – 6
Solution:
i. 4(x + 12) = 8
∴4x + 48 = 8
∴4x + 48 – 48 = 8 – 48
….(Subtracting 48 from both sides)
∴ 4x = -40
∴ x = \(\frac { -40 }{ 4 }\)
∴ x = -10

ii. 3y + 4 = 5y – 6
∴ 3y + 4 + 6 = 5y – 6 + 6
….(Adding 6 on both sides)
∴ 3y + 10 = 5y
∴ 3y + 10 – 3y = 5y – 3y
….(Subtracting 3y from both sides)
∴ 10 = 2y
∴ 2y = 10
∴ y = \(\frac { 10 }{ 2 }\)
∴ y = 5

Maharashtra Board Miscellaneous Problems Set 2 Class 7 Maths Solutions

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 2 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 2 Class 7 Maths Solutions

Question 1.
Angela deposited Rs 15000 in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to Rs 5400. For how many years had she deposited the amount?
Solution:
Here, P = Rs 15000, R = 9 p.c.p.a., I = Rs 5400
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 2 1
∴ T = 4
∴ Angela had deposited the amount for 4 years.

Question 2.
Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?
Solution:
Let us suppose that 8 men require x days to tar the road.
Number of days required by 10 men to tar the road = 4
The number of men and the number of days required to tar the road are in inverse proportion.
∴ 8 × x = 10 x 4
∴ \(x=\frac{10 \times 4}{8}\)
∴ x = 5
∴ 8 men will require 5 days to tar the road.

Question 3.
Nasruddin and Mahesh invested Rs 40,000 and Rs 60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?
Solution:
Total amount invested = Rs 40,000 + Rs 60,000
= Rs 1,00,000
Profit earned = 30%
∴ Total profit = 30% of 1,00,000
= \(\frac { 30 }{ 100 }\) × 100000
= Rs 30000
Proportion of investment = 40000 : 60000
= 2:3 …. (Dividing by 20000)
Let Nasruddin’s profit be Rs 2x and Mahesh’s profit be Rs 3x.
∴ 2x + 3x = 30000
∴ 5x = 30000
∴ x = \(\frac { 30000 }{ 5 }\).
∴ x = 6000
∴ Nasruddin’s profit = 2x = 2 × 6000 = Rs 12000
Mahesh’s profit = 3x = 3 × 6000 = Rs 18000
∴ The profits of Nasruddin and Mahesh are Rs 12000 and Rs 18000 respectively.

Question 4.
The diameter of a circle is 5.6 cm. Find its circumference.
Solution:
Diameter of the circle (d) = 5.6 cm
Circumference = πd
= \(\frac{22}{7} \times 5.6\)
= \(\frac{22}{7} \times \frac{56}{10}\)
= 17.6 cm
∴ The circumference of the circle is 17.6 cm.

Question 5.
Expand:
i. (2a – 3b)²
ii. (10 + y)²
iii. \(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}\)
iv. \(\left(y-\frac{3}{y}\right)^{2}\)
Solution:
i. Here, A = 2a and B = 3b
∴ (2a – 3b)² = (2a)² – 2 × 2a × 3b + (3b)²
…. [(A – B)² = A² – 2AB + B²]
= 4a² – 12ab + 9b²

ii. Here, a = 10 and b = y
(10 + y)² = 102 + 2 × 10xy + y²
…. [(a + b)² = a² + 2ab + b²]
= 100 + 20y + y²

iii. Here, a = \(\frac { p }{ 3 }\) and b = \(\frac { q }{ 4 }\)
\(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}=\left(\frac{p}{3}\right)^{2}+2 \times \frac{p}{3} \times \frac{q}{4}+\left(\frac{q}{4}\right)^{2}\)
…. [(a + b)² = a² + 2ab + b²]
\(\frac{p^{2}}{9}+\frac{p q}{6}+\frac{q^{2}}{16}\)

iv. Here, a = y and b = \(\frac { 3 }{ y }\)
\(\left(y-\frac{3}{y}\right)^{2}=y^{2}-2 \times y \times \frac{3}{y}+\left(\frac{3}{y}\right)^{2}\)
…. [(a – b)² = a² – 2ab + b²
= \(y^{2}-6+\frac{9}{y^{2}}\)

Question 6.
Use a formula to multiply:
i. (x – 5)(x + 5)
ii. (2a – 13)(2a + 13)
iii. (4z – 5y)(4z + 5y)
iv. (2t – 5)(2t + 5)
Solution:
i. Here, a = x and b = 5
(x – 5)(x + 5) = (x)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= x² – 25

ii. Here, A = 2a and B = 13
(2a – 13)(2a + 13) = (2a)² – (13)²
…. [(A + B)(A – B) = A² – B²]
= 4a² – 169

iii. Here, a = 4z and b = 5y
(4z – 5y)(4z + 5y) = (4z)² – (5y)²
…. [(a + b)(a – b) = a² – b²]
= 16z² – 25y²

iv. Here, a = 2t and b = 5
(2t – 5)(2t + 5) = (2t)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= 4t² – 25

Question 7.
The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?
Solution:
Diameter of the wheel (d) = 1.05 m
∴ Distance covered in 1 rotation of wheel = Circumference of the wheel
= πd
= \(\frac{22}{7} \times 1.05\)
= 3.3 m
∴ Distance covered in 1000 rotations = 1000 x 3.3 m
= 3300 m
= \(\frac { 3300 }{ 1000 }\) km …[1m = \(\frac { 1 }{ 1000 }\)km]
= 3.3 km
∴ The distance covered by the cart in 1000 rotations of the wheel is 3.3 km.

Question 8.
The area of a rectangular garden of length 40 m, is 1000 sq m. Find the breadth of the garden and its perimeter. The garden is to be enclosed by 3 rounds of fencing, leaving an entrance of 4 m. Find the cost of fencing the garden at a rate of Rs 250 per metre.
Solution:
Length of the rectangular garden = 40 m
Area of the rectangular garden = 1000 sq. m.
∴ length × breadth = 1000
∴ 40 × breadth = 1000
∴ breadth = \(\frac { 1000 }{ 40 }\)
= 25 m
Now, perimeter of the rectangular garden = 2 × (length + breadth)
= 2 (40 + 25)
= 2 × 65
= 130 m
Length of one round of fence = circumference of garden – width of the entrance
= 130 – 4
= 126 m
∴ Total length of fencing = length of one round of wire × number of rounds = 126 × 3
= 378 m
∴ Total cost of fencing = Total length of fencing × cost per metre of fencing
= 378 × 250
= 94500
∴ The cost of fencing the garden is Rs 94500.

Question 9.
From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 2 2
In ∆ABC, ∠B = 90°, and l(BC) = 21, and l(AB) = 20
∴ According to Pythagoras’ theorem,
∴ l(AC)² = l(BC)² + l(AB)²
∴ l(AC)² = 21² + 20²
∴ l(AC)² = 441 + 400
∴ l(AC)² = 841
∴ l(AC)² = 29²
∴ l(AC) = 29
Perimeter of ∆ABC = l(AB) + l(BC) + l(AC)
= 20 + 21 + 29
= 70
∴ The length of hypotenuse AC is 29 units, and the perimeter of ∆ABC is 70 units.

Question 10.
If the edge of a cube is 8 cm long, find its total surface area.
Solution: ,
Total surface area of the cube = 6 × (side)²
= 6 × (8)²
= 6 × 64
= 384 sq. cm
The total surface area of the cube is 384 sq.cm.

Question 11.
Factorize: 365y4z3 – 146y2z4
Solution:
= 365y4z3 – 146y2z4
= 73 (5y4z3 – 2y2z4)
= 73y2 (5y2z3 – 2z4)
= 73y2z3(5y2 – 2z)

Practice Set 1 Class 7 Answers Chapter 1 Geometrical Constructions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 1 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 1 Solutions Maharashtra Board

Std 7 Maths Practice Set 1 Solutions Answers

Question 1.
Draw line segments of the lengths given below and draw their perpendicular bisectors:
i. 5.3 cm
ii. 6.7 cm
iii. 3.8 cm
Solution:
i.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 1
Line AB is the perpendicular bisector of seg PQ.

ii.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 2
Line UV is the perpendicular bisector of seg ST.

iii.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 3
Line ST is the perpendicular bisector of seg LM.

Question 2.
Draw angles of the measures given below and draw their bisectors:
i. 105°
ii. 55°
iii. 90°
Solution:
i. 105°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 4

ii. 55°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 5

iii. 90°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 6

Question 3.
Draw, an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 7
The points of concurrence of the angle bisectors of both the triangles lie in the interior of the triangles.

Question 4.
Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie?
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 8
The point of concurrence of the perpendicular bisectors of the sides of the right angled triangle lies on the hypotenuse.

Question 5.
Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
Solution:
Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle.
The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses.
The shop will be at the point of concurrence of the perpendicular bisectors.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 1 Intext Questions and Activities

Question 1.
Draw a line segment PS of length 4cm and draw its perpendicular bisector. (Textbook pg. no. 1)

  1. How will your verify that CD is the perpendicular bisector? m∠CMS = __°
  2. Is l(PM) = l(SM)?

Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 9

  1. Here, m∠CMS = 90°
  2. Also, l(PM) = l(SM) = 2cm
    ∴ line CD is the perpendicular bisector of seg PS.

Std 7 Maths Digest

Practice Set 24 Class 7 Answers Chapter 5 Operations on Rational Numbers Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 24 Answers Solutions Chapter 5 Operations on Rational Numbers.

Operations on Rational Numbers Class 7 Maths Chapter 5 Practice Set 24 Solutions Maharashtra Board

Std 7 Maths Practice Set 24 Solutions Answers

Question 1.
Write the following rational numbers in decimal form.
i. \(\frac { 13 }{ 4 }\)
ii. \(\frac { -7 }{ 8 }\)
iii. \(7\frac { 3 }{ 5 }\)
iv. \(\frac { 5 }{ 12 }\)
v. \(\frac { 22 }{ 7 }\)
vi. \(\frac { 4 }{ 3 }\)
vii. \(\frac { 7 }{ 9 }\)
Solution:
i. \(\frac { 13 }{ 4 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 1

ii. \(\frac { -7 }{ 8 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 2

iii. \(7\frac { 3 }{ 5 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 3

iv. \(\frac { 5 }{ 12 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 4

v. \(\frac { 22 }{ 7 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 5

vi. \(\frac { 4 }{ 3 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 6

vii. \(\frac { 7 }{ 9 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 7

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 24 Intext Questions and Activities

Question 1.
Without using division, can we tell from the denominator of a fraction, whether the decimal form of the fraction will be a terminating decimal? Find out. (Textbook pg. no. 40)
Solution:
If the prime factorization of the denominator of a fraction has only factors as 2 or 5 or a combination of 2 and 5 then the decimal form of that fractional will be a terminating decimal form.
Consider the fractions \(\frac { 17 }{ 20 }\) and \(\frac { 19 }{ 6 }\)
Now, 20 = 2 x 2 x 5, and 6 = 2 x 3
∴ \(\frac { 17 }{ 20 }\) is terminating decimal form while \(\frac { 19 }{ 6 }\) is recurring decimal form.

Practice Set 4 Class 6 Answers Maths Chapter 3 Integers Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 4 Answers Solutions.

Integers Class 6 Maths Chapter 3 Practice Set 4 Solutions Maharashtra Board

Std 6 Maths Practice Set 4 Solutions Answers

Question 1.
Classify the following numbers as positive numbers and negative numbers.
-5, +4, -2, 7, +26, -49, -37, 19, -25, +8, 5, -4, -12, 27
Solution:

Positive Numbers +4, 7, +26, 19, +8, 5, 27
Negative Numbers -5, -2, -49, -37, -25, -4, -12

Question 2.
Given below are the temperatures in some cities. Write them using the proper signs.

Place Shimla Leh Delhi Nagpur
Temperature 7 °C below 0° 12 °C below 0° 22 °C above 0° 31 °C above 0°

Solution:

Place Shimla Leh Delhi Nagpur
Temperature with proper sign -7 °C -12 °C +22 °C +31 °C

Question 3.
Write the numbers in the following examples using the proper signs.

  1. A submarine is at a depth of 512 meters below sea level.
  2. The height of Mt Everest, the highest peak in the Himalayas, is 8848 meters.
  3. A kite is flying at a distance of 120 meters from the ground.
  4. The tunnel is at a depth of 2 meters under the ground.

Solution:

  1. A submarine is at a depth of -512 meters from sea level.
  2. The height of Mt Everest, the highest peak in the Himalayas is +8848 meters.
  3. A kite is flying at a distance of +120 meters from the ground.
  4. The tunnel is at a depth of -2 meters from the ground.

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 4 Intext Questions and Activities

Question 1.
Take warm water in one beaker, some crushed ice in another and a mixture of salt and crushed ice in a third beaker. Ask your teacher for help in measuring the temperature of the substance in each of the beakers using a thermometer. Note the temperatures. (Textbook pg. no. 13)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 1
Solution:
( Students should attempt this activity on their own)

Question 2.
Look at the picture of the kulfi man. Why do you think he keeps the kulfi moulds in a mixture of salt and ice? (Textbook pg. no. 14)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 2
Solution:
Kulfi man keeps the kulfi moulds in a mixture of salt and ice because such a mixture helps in keeping the kulfi cool for a longer period of time. The kulfi kept in the said mixture relatively takes more time to melt. This mixture is Considered ideal as it has the temperature of -4°C as against the temperature of ice i.e. 0°C.

Question 3.
My class, i.e. Std. VI, is a part of my school. My school is in my town. My town is a part of a taluka. In the same way, the taluka is a part of a district, and the district is a part of Maharashtra State. In the same way, what can you say about these groups of numbers? Textbook pg. no. 15)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 3
Solution:
By observing the above given groups of numbers, we can infer that natural numbers are a part of whole numbers. In turn, whole numbers are a part of integers.

Std 6 Maths Digest

Maharashtra Board 8th Class Maths Miscellaneous Exercise 2 Solutions

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Miscellaneous Exercise 2 8th Std Maths Answers Solutions.

Miscellaneous Exercise 2 8th Std Maths Answers

Question 1.
Questions and their alternative answers are given. Choose the correct alternative answer.
i. Find the circumference of a circle whose area is 1386 cm²? [Chapter 15]
(A) 132 cm²
(B) 132 cm
(C) 42 cm
(D) 21 cm²
Solution:
(B) 132 cm

Hint:
i. Area of the circle = πr²
1386 = \(\frac { 22 }{ 7 }\) x r²
r² = 1386 x \(\frac { 7 }{ 22 }\)
= 63 x 7
= 441
r = √441 … [Taking square root of both sides]
= 21 cm
Circumference of the circle = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 21
= 132 cm

ii. The side of a cube is 4 m. If it is doubled, how many times will be the volume of the new cube, as compared with the original cube? [Chapter 16]
(A) Two times
(B) Three times
(C) Four times
(D) Eight times
Solution:
(D) Eight times

Hint:
ii. Original volume = (4)³ = 64 cu.m
New side = 8 m
∴ New volume = (8)² = 512 cu.m
Now, \(\frac{\text { new volume }}{\text { original volume }}=\frac{512}{64}\) = 8
original volume 64
∴ volume of new cube will increase 8 times as compared to the volume of original cube.

Question 2.
Pranalee was practicing for a 100 m running race. She ran 100 m distance 20 times. The time required, in seconds, for each attempt was as follows. [Chapter 11]
18, 17, 17, 16,15, 16, 15, 14,16, 15, 15, 17, 15, 16,15, 17, 16, 15, 14,15
Find the mean of the time taken for running.
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 1
∴ The mean of the time taken for running 100 m race is 15.7 seconds.

Question 3.
∆DEF and ∆LMN are congruent in the correspondence EDF ↔ LMN. Write the pairs of congruent sides and congruent angles in the correspondence. [Chapter 13]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 2
∆EDF ≅ ∆LMN
∴side ED ≅ side LM
side DF ≅ side MN
side EF ≅ side LN
∠E ≅∠L
∠D ≅∠M
∠F ≅∠N

Question 4.
The cost of a machine is Rs 2,50,000. It depreciates at the rate of 4% per annum. Find the cost of the machine after three years. [Chapter 14]
Solution:
Here, P = Cost of the machine = Rs 2,50,000
R = Rate of depreciation = 4%
N = 3 Years
A = Depreciated price of the machine
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 3
∴The cost of the machine after three years will be Rs 2,21,184.

Question 5.
In ☐ABCD, side AB || side DC, seg AE ⊥ seg DC. If l(AB) = 9 cm, l(AE) = 10 cm, A(☐ABCD) = 115 cm² , find l(DC). [Chapter 15]
Solution:
Given, side AB || side DC.
∴ ☐ABCD is a trapezium.
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 4
Given, l(AB) = 9 cm, l(AE) = 10 cm,
A(☐ABCD) = 115 cm²
Area of a trapezium
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
∴ A(☐ABCD) = \(\frac { 1 }{ 2 }\) x [l(AB) + l(DC) x l(AE)]
∴ 115 = \(\frac { 1 }{ 2 }\) x [9 + l(DC)] x 10
∴ \(\frac { 115 \times 2 }{ 10 }\) = 9 + l(DC)
∴ 23 = 9 + l(DC)
∴ l(DC) = 23 – 9
∴ l(DC) = 14cm

Question 6.
The diameter and height of a cylindrical tank is 1.75 m and 3.2 m respectively. How much is the capacity of tank in litre?
[π = \(\frac { 22 }{ 7 }\)] [Chapter 16]
Solution:
Given: For cylindrical tank:
diameter (d) = 1.75 m, height (h) = 3.2 m
To Find: Capacity of tank in litre
diameter (d) = 1.75 m
= 1.75 x 100
….[∵ 1 m = 100cm]
= 175 cm
∴ radius (r) = \(=\frac{\mathrm{d}}{2}=\frac{175}{2}\) cm
h = 3.2 cm
= 3.2 x 100
= 320 cm
Capacity of tank = Volume of the cylindrical tank
= πr²h
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 5
∴ The capacity of the tank is 7700 litre.

Question 7.
The length of a chord of a circle is 16.8 cm, radius is 9.1 cm. Find its distance from the centre. [Chapter 17]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 6
Let CD be the chord of the Circle with centre O.
Draw seg OP ⊥ chord CD
∴l(PD) = \(\frac { 1 }{ 2 }\) l(CD)
…[Perpendicular drawn from the centre of a circle to its chord bisects the chord]
∴l(PD) = \(\frac { 1 }{ 2 }\) x 16.8 …[l(CD) = 16.8cm]
∴l(PD) = 8.4 cm …(i)
∴In ∆OPD, m∠OPD = 90°
∴[l(OD)]² = [l(OP)]² + [l(PD)]² …..[Pythagoras theorem]
∴(9.1)² = [l(OP)]² + (8.4)² … [From (i) and l(OD) = 9.1 cm]
∴(9.1)² – (8.4)² = [l(OP)]²
∴(9.1 + 8.4) (9.1 – 8.4) = [l(OP)]²
…[∵ a² – b² = (a + b) (a – b)]
∴17.5 x (0.7) = [l(OP)]²
∴12.25 = [l(OP)]²
i.e., [l(OP)]² = 12.25
∴l(OP) = √12.25
…[Taking square root of both sides]
∴l(OP) = 3.5 cm
∴The distance of the chord from the centre is 3.5 cm.

Question 8.
The following tables shows the number of male and female workers, under employment guarantee scheme, in villages A, B, C and D.

Villages A B C D
No. of females 150 240 90 140
No. of males 225 160 210 110

i. Show the information by a sub-divided bar-diagram.
ii. Show the information by a percentage bar diagram. [Chapter 11]
Solution:
i.

Villages A B C D
No. of females 150 240 90 140
No. of males 225 160 210 110
Total 375 400 300 250

Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 7

ii.

Villages A B C D
No. of females 150 240 90 140
No. of males 225 160 210 110
Total 375 400 300 250
Percentage of females 40% 60% 30% 56%
Percentage of males 60% 40% 70% 44%

Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 8

Question 9.
Solve the following equations.
i. 17 (x + 4) + 8 (x + 6) = 11 (x + 5) + 15 (x + 3)
ii. \(\frac{3 y}{2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
iii. 5(1 – 2x) = 9(1 -x)
[Chapter 12]
Solution:
i. 17 (x + 4) + 8 (x + 6) = 11 (x + 5) + 15 (x + 3)
∴ 17x + 68 + 8x + 48 = 11x + 55 + 15x + 45
∴ 17x + 8x + 68 + 48 = 11x + 15x + 55 + 45
∴ 25x + 116 = 26x + 100
∴ 25x + 116 – 116 = 26x + 100 – 116
… [Subtracting 116 from both the sides]
∴ 25x = 26x – 16
∴ 25x – 26x = 26x – 16 – 26x
… [Subtracting 26x from both the sides]
∴ -x = -16
∴ \(\frac{-x}{-1}=\frac{-16}{-1}\)
∴ x = 16

ii. \(\frac{3 y}{2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{3 y \times 2}{2 \times 2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{6 y}{4}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{6 y}{4} \times 4+\frac{y+4}{4} \times 4=5 \times 4-\frac{y-2}{4} \times 4\)
……[Multiplying both the sides by 4]
∴ 6y + y + 4 = 20 – (y – 2)
∴ 7y + 4 = 20 – y + 2
∴ 7y + 4 = 22 – y
∴ 7y + 4 – 4 = 22 – y – 4
…..[Subtracting 4 from both the sides]
∴ 7y = 18 – y
∴ 7y + y = 18 – y + y
…[Adding y on both the sides]
∴ 8y = 18
∴ \(\frac{8 y}{8}=\frac{18}{8}\) … [Dividing both the sides by 8]
∴ \(y=\frac { 9 }{ 4 }\)

iii. 5(1 – 2x) = 9(1 – x)
∴ 5 – 10x = 9 – 9x
∴ 5 – 10x – 5 = 9 – 9x – 5
….[Subtracting 5 from both the sides]
∴ -10x = 4 – 9x
∴ -10x + 9x = 4 – 9x + 9x
… [Adding 9x on both the sides]
∴ -x = 4
∴ -x x (- 1) = 4 x (- 1)
… [Multiplying both the sides by – 1]
∴ x = – 4

Question 10.
Complete the activity according to the given steps.
i. Draw rhombus ABCD. Draw diagonal AC.
ii. Show the congruent parts in the figure by identical marks.
iii. State by which, test and in which correspondence ∆ADC and ∆ABC are congruent.
iv. Give reason to show ∠DCA ≅ ∠BCA, and ∠DAC ≅ ∠BAC
v. State which property of a rhombus is revealed from the above steps. [Chapter 13]
Solution:
a.
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 9
b. In ∆ADC and ∆ABC,
side AD ≅ side AB …..[Sides of a rhombus]
side DC ≅ side BC …..[Sides of a rhombus]
side AC ≅ side AC … [Common side]
∆ADC ≅ ∆ABC … [By SSS test]
∠DCA ≅ ∠BCA …[Corresponding angles of congruent triangles]
∠DAC ≅ ∠BAC …[Corresponding angles of congruent triangles]
From the above steps, property of rhombus revealed is ‘diagonal of a rhombus bisect the opposite angles’.

Question 11.
The shape of a farm is a quadrilateral. Measurements taken of the farm, by naming its corners as P, Q, R, S in order are as follows. l(PQ) = 170 m,
l(QR) = 250 m, l(RS) = 100 m, l(PS) = 240 m, l(PR) = 260 m.
Find the area of the field in hectare (1 hectare = 10,000 sq.m). [Chapter 15]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 10
Area of the field = A(∆PQR) + A(∆PSR)
In ∆PQR, a = 170 m, b = 250 m, c = 260 m
Semiperimeter of ∆PQR = s
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 11
Area of the field = A(∆PQR) + A(∆PSR)
= 20400 + 12000
= 32400 sq.m
= \(\frac { 32400 }{ 10000 }\)
…[1 hectare = 10,000 sq.m]
= 3.24 hectares
∴ The area of the field is 3.24 hectares.

Question 12.
In a library, 50% of total number of books is of Marathi. The books of English are \(\frac { 1 }{ 3 }\) of Marathi books. The books on Mathematics are 25% of the English books. The remaining 560 books are of other subjects. What is the total number of books in the library? [Chapter 12]
Solution:
Let the total number of books in the library be x
50% of total number of books is of Marathi.
Number of Marathi books = 50% of x
= \(\frac { 50 }{ 100 }x\)
= \(\frac { x }{ 2 }\)
The books of English are \(\frac { 1 }{ 3 }\) of Marathi books.
Number of books of English = \(\frac{1}{3} \times \frac{x}{2}\)
= \(\frac { x }{ 6 }\)
The books on Mathematics are 25% of the English books.
Number of books of Mathematics
= 25% of \(\frac { x }{ 6 }\)
= \(\frac{25}{100} \times \frac{x}{6}\)
= \(\frac { x }{ 24 }\)
Since, there are 560 books of other subjects, the total number of books in the library are
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 12
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 13
∴ 24x – 17x = 17x + 13440 – 17x
∴ 7x = 13440
∴ \(\frac{7 x}{7}=\frac{13440}{7}\)
∴ x = 1920
∴ The total number of books in the library are 1920.

Question 13.
Divide the polynomial (6x³ + 11x² – 10x – 7) by the binomial (2x + 1). Write the quotient and the remainder. [Chapter 10]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 14
∴ Quotient = 3x² + 4x – 7,
remainder = 0
Explanation:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 15

Maharashtra Board Class 8 Maths Solutions

Maharashtra Board 8th Class Maths Miscellaneous Exercise 1 Solutions

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Miscellaneous Exercise 1 8th Std Maths Answers Solutions.

Miscellaneous Exercise 1 8th Std Maths Answers

Question 1.
Choose the correct alternative answer for each of the following questions.
i. In ₹PQRS, m∠P = m∠R = 108°, m∠Q = m∠S = 72°. State which pair of sides of those given below is parallel. [Chapter 8]
(A) side PQ and side QR
(B) side PQ and side SR
(C) side SR and side SP
(D) side PS and side PQ
Solution:
(B) side PQ and side SR

Hint:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 1
In ₹PQRS,
m∠P + m∠S = 108°+ 72
= 180°
Since, interior angles are supplementary.
∴ side PQ || side SR

ii. Read the following statements and choose the correct alternative from those given below them. [Chapter 8]
a. Diagonals of a rectangle are perpendicular bisectors of each other.
b. Diagonals of a rhombus are perpendicular bisectors of each other.
c. Diagonals of a parallelogram are perpendicular bisectors of each other.
d. Diagonals of a kite bisect each other.
(A) Statements (b) and (c) are true
(B) Only statement (b) is true
(C) Statements (b) and (d) are true
(D) Statements (a), (c) and (d) are true.
Solution:
(B) Only statement (b) is true

iii. If 19³ = 6859, find \(\sqrt[3]{0.006859}\). [Chapter 3]
(A) 1.9
(B) 19
(C) 0.019
(D) 0.19
Solution:
(D) 0.19

Hint:
\(\sqrt[3]{0.006859}\)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 2

Question 2.
Find the cube roots of the following numbers. [Chapter 3]
i. 5832
ii. 4096
Solution:
i. 5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= (2 × 3 × 3) × (2 × 3 × 3) × (2 × 3 × 3)
= (2 × 3 × 3)³
= (18)³
\(\sqrt[3]{5832}=18\)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 3

ii. 4096 = (4 × 4) × (4 × 4) × (4 × 4)
= (4 × 4)
= 16³
\( \sqrt[3]{4096}=16\)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 4

Question 3.
m∝n,n = 15 when m = 25. Hence
i. Find m when n = 87,
ii. Find n when m = 155. [Chapter 7]
Solution:
Given that, m ∝ n
∴ m = kn …(i)
where, k is the constant of variation.
When m = 25, n = 15
∴ Substituting, m = 25 and n = 15 in (i), we get
m = kn
∴ 25 = k × 15
∴ k = \(\frac { 25 }{ 15 }\)
∴ k = \(\frac { 5 }{ 3 }\)
Substituting k = \(\frac { 5 }{ 3 }\) in (i), we get
m = kn
∴ m = \(\frac { 5 }{ 3 }n\) …(ii)

i. When n = 87, m = ?
Substituting n = 87 in (ii), we get
m = \(\frac { 5 }{ 3 }n\)
m = \(\frac { 5 }{ 3 }\) × 87
m = 5 × 29
m = 145

ii. When m = 155, n = ?
∴ Substituting m = 155 in (ii), we get
m = \(\frac { 5 }{ 3 }n\)
∴ 155 = \(\frac { 5 }{ 3 }n\)
∴ \(\frac{155 \times 3}{5}=n\)
∴ n = 31 × 3
∴ n = 93

Question 4.
y varies inversely with x. If y = 30 when x = 12, find [Chapter 7]
i. y when x = 15,
ii. x when y = 18.
Solution:
Given that,
\(y \propto \frac{1}{x}\)
∴ \(y=k \times \frac{1}{x}\)
where, k is the constant of variation.
∴ y × x = k …(i)
When x = 12, y = 30
∴ Substituting, x = 12 and y = 30 in (i), we get
y × x = k
∴ 30 × 12 = k
∴ k = 360
Substituting, k = 360 in (i), we get
y × x = k
∴ y × x = 360 ….(ii)

i. When x = 15,y = ?
∴ Substituting x = 15 in (ii), we get
y × x = 360
∴ y × 15 = 360
∴ y = \(\frac { 360 }{ 15 }\)
∴ y = 24

ii. When y = 18, x = ?
∴ Substituting y = 18 in (ii), we get
y × x = 360
∴18 × x = 360
∴ x = \(\frac { 360 }{ 18 }\)
∴ x = 20

Question 5.
Draw a line l. Draw a line parallel to line l at a distance of 3.5 cm. [Chapter 2]
Solution:
Steps of construction:

  1. Draw a line l and take any two points M and N on the line.
  2. Draw perpendiculars to line l at points M and N.
  3. On the perpendicular lines take points S and T at a distance 3.5 cm from points M and N respectively.
  4. Draw a line through points S and T. Name the line as n.

Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 5
Line n is parallel to line l at a distance of 3.5 cm from it.

Question 6.
Fill in the blanks in the following statement.
The number \((256)^{\frac{5}{7}}\) is __ of __ power of __. [Chapter 3]
Solution:
The number \((256)^{\frac{5}{7}}\) is 7th root of 5th power of 256.

Question 7.
Expand.
i. (5x – 7) (5x – 9)
ii. (2x – 3y)³
iii. \(\left(a+\frac{1}{2}\right)^{3}\) [Chapter 5]
Solution:
i. (5x – 7) (5x – 9)
= (5x)² + (-7 -9) 5x + (-7) × (-9).
…[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 25x² + (-16) × 5x + 63
= 25x² – 80x + 63

ii. Here, a = 2x and b = 3y
(2x – 3y)³
= (2x)³ – 3 (2x)² (3y) + 3 (2x) (3y)² – (3y)³
…[∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8x³ – 3 (4x²) (3y) + 3 (2x) (9y²) – 27y³
= 8x³ – 36x²y + 54xy² – 27p³

iii. Here, A= a and B = \(\frac { 1 }{ 2 }\)
\(\left(a+\frac{1}{2}\right)^{3}=(a)^{3}+3(a)^{2}\left(\frac{1}{2}\right)+3(a)\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}\)
…[(A + B)³ = A³ + 3A²B + 3AB² + B³]
\(=\mathbf{a}^{3}+\frac{3 \mathbf{a}^{2}}{2}+\frac{3 \mathbf{a}}{4}+\frac{1}{8}\)

Question 8.
Draw an obtuse angled triangle. Draw all of its medians and show their point of concurrence. [Chapter 4]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 6
The point of concurrence of the medians PS, RU and QV is G.

Question 9.
Draw ∆ABC such that l(BC) = 5.5 cm, m∠ABC = 90°, l(AB) = 4 cm. Show the orthocentre of the triangle. [Chapter 4]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 7
Here, point B is the orthocentre of ∆ABC.

Question 10.
Identify the variation and solve.
It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr. If the speed of the bus is reduced by 8 km/hr, how much time will it take for the same travel? [Chapter 7]
Solution:
Let, v represent the speed of the bus and t represent the time required to travel from one town to the other.
The speed of the bus varies inversely with the time required to travel from one town to the other.
∴ \(\mathbf{v} \propto \frac{1}{\mathbf{t}}\)
∴ \(\mathbf{v}=\mathbf{k} \times \frac{1}{\mathbf{t}}\)
where, k is the constant of variation.
∴ v × t = k …(i)
It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr.
i.e., when v = 48, t = 5
∴ Substituting v = 48 and t = 5 in (i), we get
v × t = k
∴ 48 × 5 = k
∴ k = 240
Substituting k = 240 in (i), we get
v × t = k
∴ v × t = 240 …(ii)
Since, the speed of the bus is reduced by 8 km/hr,
∴ Speed of the bus in second case (v)
= 48 – 8 = 40 km/hr
∴ When v = 40, t = ?
∴ Substituting v = 40 in (ii), we get
v × t = 240
∴ 40 × t = 240
∴ \(t=\frac { 240 }{ 40 }\)
∴ t = 6
∴ The problem is of inverse variation and the bus would take 6 hours to travel the distance if its speed is reduced by 8 km/hr.

Question 11.
Seg AD and seg BE are medians of ∆ABC and point G is the centroid. If l(AG) = 5 cm, find l(GD). If l(GE) = 2 cm, find l(BE). [Chapter 4]
Solution:
The centroid of a triangle divides each median in the ratio 2:1.
i. Point G is the centroid and seg AD is the median.
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 8

ii. Point G is the centroid and seg BE is the median.
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 9
∴ l(BG) × 1 = 2 × 2
∴ l(BG) = 4 cm
Now, l(BE) = l(BG) + l(GE)
∴ l(BE) = 4 + 2
∴ l(BE) = 6 cm

Question 12.
Convert the following rational numbers into decimal form. [Chapter 1]
i. \(\frac { 8 }{ 13 }\)
ii. \(\frac { 11 }{ 7 }\)
iii. \(\frac { 5 }{ 16 }\)
iv. \(\frac { 7 }{ 9 }\)
Solution:
i. \(\frac { 8 }{ 13 }\)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 10

ii. \(\frac { 11 }{ 7 }\)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 11

iii. \(\frac { 5 }{ 16 }\)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 12

iv. \(\frac { 7 }{ 9 }\)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 13

Question 13.
Factorize.
i. 2y² – 11y + 5
ii. x² – 2x – 80
iii. 3x² – 4x + 1
Solution:
i. 2y² – 11y + 5
= 2y² – 10y – y + 5
= 2y(y – 5) – 1(y – 5)
= (y – 5)(2y – 1)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 14

ii. x² – 2x – 80
= x² – 10x + 8x – 80
= x (x – 10) + 8 (x – 10)
= (x – 10)(x + 8)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 15

iii. 3x² – 4x + 1
= 3x² – 3x – x + 1
= 3x(x – 1) – 1(x – 1)
= (x – 1) (3x – 1)
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 16

Question 14.
The marked price of a T.V. set is Rs 50,000. The shopkeeper sold it at 15% discount. Find the price of it for the customer. [Chapter 9]
Solution:
Here, marked price = Rs 50,000,
discount = 15%
Let the discount percent be x
∴x = 15%
i. Discount
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 17
= 500 × 15
= Rs 7,500

ii. Selling price = Marked price – Discount
= 50,000 – 7,500
= Rs 42,500
∴The price of the T.V. set for the customer is Rs 42,500.

Question 15.
Rajabhau sold his flat to Vasantrao for Rs 88,00,000 through an agent. The agent charged 2 % commission for both of them. Find how much commission the agent got. [Chapter 9]
Solution:
Here, selling price of the flat = Rs 88,00,000
Rate of commission = 2%
Commission = 2% of selling price
= \(\frac { 2 }{ 100 }\) × 88,00,000
= 2 × 88,000
= Rs 1,76,000
∴ Total commission = Commission from Rajabhau + Commission from Vasantrao
= Rs 1,76,000 + Rs 1,76,000
= Rs 3,52,000
∴ The agent got a commission of Rs 3,52,000.

Question 16.
Draw a parallelogram ABCD such that l(DC) = 5.5 cm, m∠D = 45°, l(AD) = 4 cm. [Chapter 8]
Solution:
Opposite sides of a parallelogram are congruent.
∴ l(AD) = l(BC) = 4 cm and
l(DC) = l(AB) = 5.5 cm
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 18

Question 17.
In the figure, line l || line m and line p || line q. Find the measures of ∠a, ∠b, ∠c and ∠d. [Chapter 2]
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 1 19
Solution:
i. line l|| line m and line p is a transversal.
∴m∠a = 78° …(i) [Corresponding angles]

ii. line p || line q and line m is a transversal.
∴m∠d = m∠a …[Corresponding angles]
∴m∠d = 78° …(ii)[From (i)]

iii. m∠b = m∠d …[Vertically opposite angles]
∴m∠b = 78° …[From (ii)]

iv. line l|| line m and line q is a transversal.
∴m∠c + m∠d = 180° …[Interior angles]
∴m∠c + 78° = 180° … [From (ii)]
∴m∠c =180° – 78°
∴m∠c = 102°
∴m∠a = 78°, m∠b = 78°, m∠c = 102°, m∠d = 78°

Maharashtra Board Class 8 Maths Solutions

Practice Set 1.1 Geometry 9th Standard Maths Part 2 Chapter 1 Basic Concepts in Geometry Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

9th Standard Maths 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 1.
Find the distances with the help of the number line given below.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 1
i. d(B, E)
ii. d (J, J)
iii. d(P, C)
iv. d(J, H)
v. d(K, O)
vi. d(O, E)
vii. d(P, J)
viii. d(Q, B)
Solution:
i. Co-ordinate of the point B is 2.
Co-ordinate of the point E is 5.
Since, 5 > 2
∴ d(B, E) = 5 – 2
∴ d(B, E) = 3

ii. Co-ordinate of the point J is -2.
Co-ordinate of the point A is 1.
Since, 1 > -2
∴ d(J, A) = 1 – (-2)
= 1 + 2
∴ d(J, A) = 3

iii. Co-ordinate of the point P is -4.
Co-ordinate of the point C is 3.
Since, 3 > -4
∴ d(P,C) = 3 – (-4)
= 3 + 4
∴ d(P,C) = 7

iv. Co-ordinate of the point J is -2.
Co-ordinate of the point H is -1.
Since, -1 > -2
∴ d(J,H) = – 1 – (-2)
= -1 + 2
∴ d(J,H) = 1

v. Co-ordinate of the point K is -3.
Co-ordinate of the point O is 0.
Since,0 > -3
∴ d(K, O) = 0 – (-3)
= 0 + 3
∴ d(K, O) = 3

vi. Co-ordinate of the point O is 0.
∴ Co-ordinate of the point E is 5.
Since, 5 > 0
∴ d(O, E) = 5 – 0
∴ d(O, E) = 5

vii. Co-ordinate of the point P is -4.
Co-ordinate of the point J is -2.
Since -2 > -4
∴ d(P, J) = -2 – (-4)
= – 2+ 4
∴ d(P, J) = 2

viii. Co-ordinate of the point Q is -5.
Co-ordinate of the point B is 2.
Since,2 > -5
∴ d(Q,B) = 2 – (-5)
= 2 + 5
∴ d(Q, B) = 7

Question 2.
If the co-ordinate of A is x and that of B is . y, find d(A, B).
i. x = 1, y = 7
ii. x = 6, y = -2
iii. x = -3, y = 7
iv. x = -4, y = -5
v. x = -3, y = -6
vi. x = 4, y = -8
Solution:
i. Co-ordinate of point A is x = 1.
Co-ordinate of point B is y = 7
Since, 7 > 1
∴ d(A, B) = 7 – 1
∴ d(A, B) = 6

ii. Co-ordinate of point A is x = 6.
Co-ordinate of point B is y = -2.
Since, 6 > -2
∴ d(A, B) = 6 – ( -2) = 6 + 2
∴ d(A, B) = 8

iii. Co-ordinate of point A is x = -3.
Co-ordinate of point B is y = 7.
Since, 7 > -3
∴ d(A, B) = 7 – (-3) = 7 + 3
∴ d(A, B) = 10

iv. Co-ordinate of point A is x = -4.
Co-ordinate of point B is y = -5.
Since, -4 > -5
∴ d(A, B) = -4 – (-5)
= -4 + 5
∴ d(A, B) = 1

v. Co-ordinate of point A is x =-3.
Co-ordinate of point B is y = -6.
Since, -3 > -6
∴ d(A, B) = -3 – (-6)
= -3 + 6
∴ d(A, B) = 3

vi. Co-ordinate of point A is x = 4.
Co-ordinate of point B is y = -8.
Since, 4 > -8
∴ d(A, B) = 4 – (-8)
= 4 + 8
∴d(A, B) = 12

Question 3.
From the information given below, find which of the point is between the other two. If the points are not collinear, state so.
i. d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
ii. d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
iii. d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
iv. d(L, M) =11, d(M, N) = 12, d(N, L) = 8
v. d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
vi. d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Solution:
i. Given, d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
d(P, Q) = 10 …(i)
d(P, R) + d(Q, R) = 7 + 3 = 10 .. .(ii)
∴ d(P, Q) = d(P, R) + d(Q, R) …[From (i) and (ii)]
∴ Point R is between the points P and Q
i. e., P – R – Q or Q – R – P.
∴ Points P, R, Q are collinear.

ii. Given, d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
d(R, S) = 8 …(i)
d(S, T) + d(R, T) = 6 + 4 = 10 …(h)
∴ d(R, S) ≠ d(S, T) + d(R, T) … [From (i) and (ii)]
∴ The given points are not collinear.

iii. Given, d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
d(A, B) = 16 …(i)
d(C, A) + d(B, C) = 9 + 7 = 16 …(ii)
∴ d(A, B) = d(C, A) + d(B, C) …[From(i) and (ii)]
∴ Point C is between the points A and B.
i. e., A – C – B or B – C – A.
∴ Points A, C, B are collinear

iv. Given, d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
d(M, N) = 12 …(i)
d(L, M) + d(N, L) = 11 + 8 = 19 …(ii)
∴d(M, N) + d(L, M) + d(N, L) … [From (i) and (ii)]
∴ The given points are not collinear.

v. Given, d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
d(X, Y) = 15 …(i)
d(X,Z) + d(Y, Z) = 8 + 7= 15 …(ii)
∴ d(X, Y) = d(X, Z) + d(Y, Z) …[From (i) and (ii)]
∴ Point Z is between the points X and Y
i. e.,X – Z – Y or Y – Z – X.
∴ Points X, Z, Y are collinear.

vi. Given, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
d(E, F) = 8 …(i)
d(D, E) + d(D, F) = 5 + 6 = 11 …(ii)
∴ d(E, F) ≠ d(D, E) + d(D, F) … [From (i) and (ii)]
∴ The given points are not collinear.

Question 4.
On a number line, points A, B and C are such that d(A, C) = 10, d(C, B) = 8. Find d(A, B) considering all possibilities.
Solution:
Given, d(A, C) = 10, d(C, B) = 8.

Case I: Points A, B, C are such that, A – B – C.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 2
∴ d(A, C) = d(A, B) + d(B, C)
∴ 10 = d(A, B) + 8
∴ d(A, B) = 10 – 8
∴ d(A, B) = 2

Case II: Points A, B, C are such that, A – C – B.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 3
∴ d(A, B) = d(A, C) + d(C, B)
= 10 + 8
∴ d(A, B) = 18

Case III: Points A, B, C are such that, B – A – C.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 4
From the diagram,
d (A, C) > d(B, C)
Which is not possible
∴ Point A is not between B and C.
∴ d(A, B) = 2 or d(A, B) = 18.

Question 5.
Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8, find d(X, Z).
Solution:
Given,d(X, Y) = 17, d(Y, Z) = 8
Case I: Points X, Y, Z are such that, X – Y – Z.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 5
∴ d(X, Z) = d(X, Y) + d(Y, Z)
= 17 + 8
∴ d(X, Z) = 25

Case II: Points X, Y, Z are such that, X – Z – Y.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 6
∴ d(X,Y) = d(X,Z) + d(Z,Y)
∴ 17 = d(X, Z) + 8
∴ d(X, Z) = 17 – 8
∴ d(X, Z) = 9

Case III: Points X, Y, Z are such that, Z – X – Y.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 7
From the diagram,
d(X, Y) > d (Y, Z)
Which is not possible
∴ Point X is not between Z and Y.
∴ d(X, Z) = 25 or d(X, Z) = 9.

Question 6.
Sketch proper figure and write the answers of the following questions. [2 Marks each]
i. If A – B – C and l(AC) = 11,
l(BC) = 6.5, then l(AB) = ?
ii. If R – S – T and l(ST) = 3.7,
l(RS) = 2.5, then l(RT) = ?
iii. If X – Y – Z and l(XZ) = 3√7,
l(XY) = √7, then l(YZ) = ?
Solution:
i. Given, l(AC) =11, l(BC) = 6.5
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 8
l(AC) = l(AB) + l(BC) … [A – B – C]
∴ 11= l(AB) + 6.5
∴ l(AB) = 11 – 6.5
∴ l(AB) = 4.5

ii. Given, l(ST) = 3.7, l(RS) = 2.5
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 9
l(RT) = l(RS) + l(ST) … [R – S – T]
= 2.5 + 3.7
∴ (RT) = 6.2

iii. l(XZ) = 3√7 , l(XY) = √7,
Maharashtra Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1 10
l(XZ) = l(X Y) + l(YZ) … [X – Y – Z]
∴ 3 √7 ⇒ √7 + l(YZ)
∴ l(YZ)= 3√7 – √7
∴ l(YZ) = 2 √7

Question 7.
Which figure is formed by three non-collinear points?
Solution:
Three non-collinear points form a triangle.

Class 9 Maths Digest