Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Balbharti Maharashtra State Board Class 10 English Solutions My English Coursebook Chapter 4.6 A Brave Heart Dedicated to Science and Humanity Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

My English Coursebook Standard Ten Guide Chapter 4.6 A Brave Heart Dedicated to Science and Humanity Textbook Questions and Answers

Warming up:
Chit-chat

Question 1.
Have you ever participated in or visited a science exhibition?
Answer:
Yes, last year I along with my friends participated in ward level science exhibition and selected for district level exhibition.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Question 2.
Did you observe/learn any new things there ? What are those ?
Answer:
Yes, we observed and learnt a lot there.

  1. Everyone was busy in his/her experiment.
  2. Everyone was trying to prove something. We learnt that if we work consistently we can achieve anything in our life.
  3. The pairs or groups of students were working hard together to make the exhibit successful.

Question 3.
“Scientists have to work very hard and make many sacrifices for years to achieve success in their experiments”. Why?
Answer:
Hard work is the key to success. While working on any project they forget everything around them and concentrate on the experiment while doing so they sacrifice their health, hunger, family, likes, hobbies and then only they are able to carry out the undertaken experiment and get the results of it.

Question 4.
What is the difference between a discovery and an invention?
Answer:
Discovery is the process of finding information, a place or an object, especially for the first time. It is an act of finding something which is in existence but not known before. Invention is something that has never been made or processed before. It is something newly designed or created.

1. Discuss in pairs/groups about the precautions one has to take while working in a science laboratory. Write them in the form of Dos and Dont’s:

Question 1.
Discuss in pairs/groups about the precautions one has to take while working in a science laboratory. Write them in the form of Dos and Dont’s:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity 1
Answer:

Dos Don’ts
1. Handle test tubes and glass apparatus with care. 1. Don’t touch chemicals with bare hands.
2. Inform the lab attendant if there is a problem. 2. Don’t eat, drink, chew gum or apply cosmetics in the lab.
3. Wash your hands before you leave the lab for the day. 3. Do not work with chemical until you are sure of their safe handling.
4. Wear appropriate safety attire when in the laboratory-safety goggles, lab aprons, etc. 4. Do not use the phone or computer with gloves on your hands.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

2. Study the underlined Predicate in each of the sentences below and note the difference:

Question 1.
Study the underlined Predicate in each of the sentences below and note the difference:
Answer:
1. I met Debbie, [(verb (met) + noun (Debbie)]
We study Grammar, [verb (study) + noun (Grammar)]
(The object answers WHAT? or WHO? when asked to the verb.)

2. He become tired, [(verb (become) + complement (tired)]
They are hungry, [(verb (are) + complement (hungry)]
The verbs (became/are) incomplete without the words that follow (tired/hungry.) So ‘tired’ and ‘hungry’ are Complements of the verbs ‘became’ and ‘are’.)

3. She spoke softly.
They live here.
He turned early.
(The verbs in the Predicates are followed by Adverbials since they indicate
How?/Where/When the action in the verb takes place.)
Now Say whether the Predicate in the following sentence contain Object or Complement or Adverbial after the verb:
Answer:
(a) He looked upwards. Complement
(b) My brother is injured. Complement
(c) We scored a goal. Object
(d) We beat the opponents. Object
(e) She answered perfectly. Adverbial
(f) The guests arrived early. Adverbial
(g) I shall be happy. Complement
(h) You wrote the address. Object

English Workshop:

1. Write one sentence each, why the following years were landmarks in the lives of Pierre and Marie Curie:

Question 1.
Write one sentence each, why the following years were landmarks in the lives of Pierre and Marie Curie: (Answers are directly given.)
Answer:

  1. 1895 – Marie and Pierre got married.
  2. 1902 – Radium was discovered.
  3. 1903 – The Curies along with Henry Becquerel were awarded the Nobel Prize in Physics for the discovery of Radium and Polonium.
  4. 1906 – Pierre was knocked down and killed by a horse-drawn wagon.
  5. 1911 – Marie was awarded the Nobel Prize for the second time for Chemistry.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

2. Complete the web diagram with the qualities of Madame Curie:

Question 1.
Complete the web diagram with the qualities of Madame Curie:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity 2
Answer:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity 3

3. Name the following:

Question (a)
Two discoverers of New lands.
1. ……………………..
2. …………………….
Answer:
1. Columbus
2. Vasco da Gama

Question (b)
Two conquerers of the peak of Mt. Everest.
1. …………………..
2. ………………….
Answer:
1. Hillary
2. Tenzing

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Question (c)
Original name of Madame Curie.
…………………………………………….
Answer:
Marja Sklowdaska

Question (d)
Her place of birth.
………………………………………………
Answer:
Warsaw, in Poland

Question (e)
Subjects Marie majored in, at the University of Paris.
………………………………………………
Answer:
Physics and Mathematics.

Question (f)
Nationality of her husband Pierre:
………………………………………………
Answer:
French.

Question (g)
Scientist who discovered the properties of uranium-
Answer:
Henri Becquerel

Question (h)
Two radioactive elements discovered by Curies-
Answer:
Polonium and Radium.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

4. Read the text carefully and match the incidents occured in Madam Curie’s life given in table A with the years in table B.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity 4

5. Think and answer in your own words.

Question (a)
What common characteristics did Marie and Pierre share, as a student?
Answer:
When Marie went to Sorbonne the University of Paris she spent her days in poverty. Her friend Pierre was brilliant but a poor scientist. Due to lack of money they began to work in a shabby laboratory. Being students they were brilliant scholars but poverty was an obstacle in their way.

Question (b)
Which of the two scientists was greater than the other? Say why.
Answer:
I think Madame Curie was greater than Pierre Curie because she wanted to study in Paris but the poverty at home did not allow her to continue. But with her own efforts and confidence she could fulfil her childhood dream. Even in that condition she got married to a person who was poor. And even after his death, she did not deter and continued her work and was awarded the Nobel Prize. It shows that she was really a great human being and a scientist.

Question (c)
Why was the gift of a ton of pitchblende, a great stroke of luck to the Curies?
Answer:
Being very expensive the Curies couldn’t buy pitchblende on a large scale but when their admirer, the emperor of Austria gifted them the most precious gift of a ton of pitchblende they could do their experiment on a large scale and find out what they wanted to achieve. So it was a great stroke of luck to the Curies.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Question (d)
Besides the uses of Radium and Polonium mentioned in the text, in what other way do you think, it is used in the field of medicine?
Answer:
Radium is used to produce radon gas which in turn, is used to treat cancer. Radiation given off by radium is sometimes used also to study the composition of metals, plastics and other material.

The alpha rays emitted by Polonium can be used to eliminate static electricity. Polonium is also used in anti-static brushes to eliminate dust on photographic film.

Question (e)
What proves that Marie Curie was a true lover of humanity and not of wealth?
Answer:
Marie Curie could have patented her discovery and earned a lot of money. But she believed in working for Science and whatever she discovered belonged to the people and not to her so in real sense she thought about humanity not money. She was a true lover of humanity and not of wealth.

6. Find words/phrases from the text that are synonyms of the word ‘brave’:

Question 1.
Find words/phrases from the text that are synonyms of the word ‘brave’:
Answer:
courageous indomitable powerful

7. Read and understand the following words. Find out/search for proverbs/thoughts/quotes/ slogans that are related to each of them:

Question 1.
Read and understand the following words. Find out/search for proverbs/thoughts/quotes/ slogans that are related to each of them:
Answer:
(a) Courage: Courage is the complement of fear. One with courage to laugh is master of the world. Courage is a kingdom without a crown.
(b) Strength of character: Knowledge will give you power, but weakness of attitude becomes weakness of character.
(c) Determination: A dream doesn’t become reality through magic, it takes sweat, determination and hardwork. The difference between the impossible and the possible lies in a man’s determination.
(d) Hard-work: Success is the result of perfection, hardwork, learning from failure, loyalty and persistence. There is simply no substitute for hardwork when it comes to achieving success.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

8. Match the branches of Science with what fields they study:

Question 1.
Match the branches of Science with what fields they study:

’A’ B’
(a) Ecology (i) Study of Atmosphere
(b) Geology (ii) Study of Plant-life
(c) Meteorology (iii) Study of the Universe
(d) Anatomy (iv) Study of living organisms and environment
(e) Botany (v) Study of human mind and emotions
(f) Psychology (vi) Study of structure and functions of the body
(g) Cosmology (vii) Study of solid earth and rocks

Answer:
(a) Ecology – Study of living organisms and environment
(b) Geology – Study of solid earth and rocks
(c) Meteorology – Study of Atmosphere
(d) Anatomy – Study of structure and functions of the body
(e) Botany – Study of Plant-life
(f) Psychology – Study of human mind and emotions
(g) Cosmology – Study of the Universe

9. Imagine that your school needs some Scientific equipment, apparatus and formulae charts for the laboratory. Being the in-charge of Science Committee, write a letter to the Headmaster of your school suggesting him some Scientific equipment and apparatus required for Standard X Science practicals. Take help of the information given below to complete your letter.

Question 1.
Imagine that your school needs some Scientific equipment, apparatus and formulae charts for the laboratory. Being the in-charge of Science Committee, write a letter to the Headmaster of your school suggesting him some Scientific equipment and apparatus required for Standard X Science practicals. Take help of the information given below to complete your letter.
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity 5
Answer:
Sanjay Sharma,
In-charge of Science Committee,
New English High School,
Akola-444 001.
1 January, 2020.

To,
The Headmaster,
New English High School,
Akola.
Subject: Requirement of scientific equipment, apparatus and formulae charts for the laboratory.

Respected sir,

With regard to learning Science content, processes and views of the nature of Science experiments play significant role in Science. As per the new syllabus, new subjects and some experiments are included in revised Science coursebooks. Experiments give immense pleasure of self-learning and observations.

The equipment and apparatus that we have been using in our Science laboratory are outdated and rarely of any use. If we bring some new, definitely it will arose interest in learning Science and doing some experiments on our own. Being in-charge of Science Committee, I request you on behalf of all students and teachers to bring following equipment and apparatus in our Science laboratory.

  1. 2 compound microscope
  2. 2 Hand lens
  3. 6 Beakers
  4. 3 Beakers Tongs
  5. 3 Test Tube holder
  6. Thermometer
  7. Safety goggles-6
  8. 6 funnels
  9. 1 microwave oven
  10. 2 pipette
  11. Graduated Cylinder
  12. Vernier Calliper
  13. Meters
  14. Compass
  15. Stop clock

You are well aware that due to lack of equipment our Science teachers can’t conduct some experiments in the lab.

Hope you will understand our difficulty and buy the listed equipment and apparatus as early as possible so that the subject of Science can be well understood by doing various experiments in the laboratory.
Anticipating your valuable help in this respect.

Thanking you,

Yours obediently,
Sanjay Sharma
In-charge of Science Committee.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

10. ‘There is no short-cut to success’. Expand this maxim with a suitable introduction, body, examples and conclusion. Write it in your notebook in about 20 lines.

Question 1.
‘There is no short-cut to success’. Expand this maxim with a suitable introduction, body, examples and conclusion. Write it in your notebook in about 20 lines.
Answer:
‘There is no Short-cut to Success.’
It is often said that success requires hardwork and great diligence as well as commitment. This makes success a long-term achievement. Everybody thinks that there are some easier ways to live life, no need to work hard for livelihood but the truth is that easier ways are always the toughest ways. Most of us forget that for reaching success, the road is not strewn with roses. Success is an achievement by constant diligence and most of us lack the essential quality, diligence. The magic of hardwork and perseverance cannot be underestimated.

Most of the time it happens that we start our efforts but soon lose heart half-way and give up our efforts and thus fail to reach the pinnacle of success. If we have the confidence in our efforts, there is nothing that can deter us from achieving. Sitting in front of Television and watching the shopping network channel, fascinating over the new-miracle making abs machine which promises you the six- pack abs only in 5 minutes of daily exercises, you think that product is heaven sent, so you grab your phone book the order and start dreaming about your perfect body. Two months later, that wonder abs machine is at the back of your master closet gathering dust. As you can see, you will have to do the work when reaching a goal and the success that comes with it.

You have to understand that you might have to make sacrifices during your journey and you need to put in the hours required in order to get where you want to reach. Those who sit pretty and just complain about the success of others are the ones who do not do much in life. Those who make continuous efforts for a long time in the right direction rarely lose in the struggle of life. So one important password for all of us to follow is that there is no short cut to success.

Project:

The life and work and inventions of Marie Curie are given in detail in the text. Go to your library or search on internet at least five famous scientists who have given great inventions to the world and write about them or prepare a chart showing the information about them.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Language Study:

1. Pick out the Prefixes and Suffixes from the following words and find the root word:

Question 1.
Pick out the Prefixes and Suffixes from the following words and find the root word:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity 6
Answer:

Word Prefix Suffix Root Word
1. unattainable un able attain
2. indomitable in able
3. disheartening dis ing hearten
4. incalculable in able calculate
5. disconsolate dis ate console
6. ensuring en ing sure

2. Say whether the Predicates in the sentences below contain Objects/Complements/Adverbials:

Question (A)
Say whether the Predicates in the sentences below contain Objects/Complements/Adverbials: (Answers are directly given.)
Answer:
(a) Madame Curie discovered radium. – Object
(b) Pierre was knocked down. – Complement
(c) Pitchblende was expensive. – Complement
(d) The couple took a flat. – Object
(e) They moved cautiously, success came finally. – Adverbial

Question (B)
Pick out the Conjunctions in the following sentences and say whether they are Subordinators or Coordinators.
Answer:
(a) There are women who show extreme courage. – who – subordinator
(b) Pitchblende was an extremely expensive substance, so they could not afford to buy. – so-subordinator
(c) After her daily household work, Marie settled down to studv. – After-subordinator
(d) Marie wanted to study in Paris but her father could not afford it. – but-coordinator
(e) History is full of chapters that tell of extraordinary people! – that-subordinator

Live English!

Question (a)
What is Blogging?
Answer:
The term blog is short for web log. It is an online public diary which lists each diary entry in reverse order so that new diary entries are placed on the top of the page, and older entries are placed below. Each entry is called a post.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Question (b)
Who Blogs?
Answer:
Anyone can create a blog, if they have a basic understanding of the way the software works. There are privately owned blogs on individual webpages, which require a little knowledge about creating webpages, but there are also blog creation services which allow you to quickly and easily set up your own blog. You can add posts as often as you like, and can say almost anything that you want. Blogs are created for a lot of different reasons, and by many different people.

Question (c)
What Are Blogs About?
Answer:
There are many topics that a blog may contain, depending on who is updating it and why. Many blogs provide news or comment in response to a particular subject, like local news, politics or even hobbies. Food blogs are popular for people who want to share recipes, and many political activists and writers have blogs which are used to respond to current political topics. Many blogs also function as personal online diaries.

Question (d)
Blog Classification
Answer:
A blog is usually textual, based on news and commentary, although there are other types of blogs which are gaining popularity. Artlogs are blogs which focus on art, Photoblogs focus on photography, Sketch- blogs focus on sketching, and so on. There are also a few specialised types of blogs, including Vblogging or video blogging, and Podcasting or audio blogging.

Question (e)
Creating a Blog on the Internet
Answer:
If you have a good understanding of webpage design and development, you may be able to create your own blog on a personal website. Otherwise, your best option is to choose a website that does the setting up for you, so that all you have to do is join, customise a little and begin to write. There are many blog websites like www.livejournal.com, www.wordpress.com, www.blogger.com or www.blogs.myspace.com which will allow you to quickly and easily create your own blog. They also give you the option to set controls so that you can choose to allow only friends or only yourself to read them.

Question (f)
Blog Popularity
Answer:
Blogs are becoming more and more popular for many reasons. This is because they offer a wealth of information on a lot of different topics. They also drive traffic to websites by attracting new visitors with interesting blog column commentary. Blogs are a great tool because they allow anyone to express themselves on the Internet. If you are looking for a way to express yourself online, or to share information regarding a favourite subject of yours, beginning your own blog may be the best option for you because everyone has an opinion on the Internet, and soon everyone will have their own personal weblog too.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Now with the help of your teacher create your own Blog on one of the following given topics.

1. Your experiences and recently celebrated birthday party.
2. Your expectations from the school.
3. Describe the surroundings of your residence.
4. Describe a cultural programme/campaign/project conducted at your school.
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity 7

My English Coursebook 10th Digest Chapter 4.6 A Brave Heart Dedicated to Science and Humanity Additional Important Questions and Answers

Simple Factual Activities:

Question 1.
Write the importance of the years. (Answers are directly given.)
Answer:
1. 1867-Marie Curie was born.
2. 1895-Marie and Pierre got married.

Question 2.
State whether the following statements are True or False:
Answer:

  1. Pitchblende is a black, very hard and cheap substance – False
  2. Men could see many substances through the powerful rays of Uranium. – True
  3. The emperor of Austria gifted a ton of pitchblende to the Curies – True
  4. The Curies sacrificed all their luxuries of life to save money to buy pitchblende. – True

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Name the following:

Question 1.
An admirer of the Curies.
Answer:
The emperor of Austria.

Question 2.
A black, very hard and expensive substance-
Answer:
Pitchblende.

Complex Factual Activities:

Question 1.
What qualities do great achievers possess?
Answer:
Great achievers possess qualities like commitment, courage, dedication and singleness of
purpose in their effort.

Question 2.
What information do you get about Marie’s early life?
Answer:
Marie was born in a poor family. Her childhood dream was to study Science in Paris but her father could not afford the expenses for this. So she worked as a governess and saved a little money and went to Sorbonne, the University of Paris to study science. Thus she worked hard in her childhood to fulfil her dream.

Question 3.
What are the signs of poverty suffered by the young couple? ( wt WT oTFlcft?)
Answer:
The young couple was very poor indeed. They started their journey from a shabby laboratory. They took a flat in Paris which contained hardly any furniture. Marie used to work and settle down to study in laboratory in a wooden shed near their flat. It had a leaky skylight and an earthen floor. All these things are the signs of poverty suffered by the young couple.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Question 4.
Complete the following sentences:
(Answers are directly given and underlined.)
Answer:

  1. Madame Curie dedicated her life to the cause of Science and to the welfare of humanity.
  2. The qualities that great achievers possess are extraordinary courage, determination and singleness of purpose.
  3. Marie Curie could not study Science in Paris and fulfil her childhood dream because her father could not afford the expenses of her studies.
  4. Marie and Pierre set up a laboratory in a wooden shed near their flat.

Question 5.
What proves Marie’s strong will power?
Answer:
When the couple could not buy pitchblende which was an expensive substance they sacrificed all the luxuries of life to save money. They lived in extreme poverty. They could not buy costly food and warm clothes for extremely cold Parisian winter. They could not sleep due to lack of warmth. Marie refused to take rest whenever her husband begged her to give up the struggle. All the above incidents prove that Marie had a strong will power to achieve the aim of her life.

Question 6.
Why were the two new substances named Polonium and Radium?
Answer:
When Pierre and Marie found two new substances in their new discovery, she called one Polonium in honour of her country Poland and being the most powerful of the radioactive elements another was called “Radium”.

Question 7.
Which gift did she receive from the emperor of Austria? Why was it the most precious for them?
Answer:
Pitchblende is a very expensive substance which Pierre and Marie could not afford to buy in large quantity for their experiments. The emperor of Austria was an admirer of the Curies. So he gifted them a ton of pitchblende so that the Curies could carry out their experiments easily.

Question 8.
How is radium used in the medical field?
Answer:
The benefits of radium in the world of medicine are incalculable. It has been used with great effect in treatment of cancer. The bacteria of such diseases as typhus, cholera and anthrax can also be killed by radium so it was a great gift in the medical field.

Question 9.
How did Pierre meet with an early death?
Answer:
In 1906 Pierre was knocked down and killed by a horse – drawn wagon and thus met with an early death leaving Marie disconsolate.

Question 10.
What makes Marie Curie an exceptional scientist?
Answer:
When Marie and Perrie discovered Radium and Polonium they could have patented their discovery and become rich but Marie refused to do so and gave it free to the world and the world of medicine. She also believed that Radium belongs to the people not to her. She worked selflessly without expecting money. Her humanity and selflessness make Marie Curie an exceptional scientist.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Question 11.
Describe how Curies first sighted Radium. What was its colour?
Answer:
After continuing their work for more than four years on a September night, the Curies after a day’s tiresome work, went to the laboratory to have another look at the hundreds of small bowls into which they had poured filtered pitchblende. While moving forward in the dark laboratory they found rays of soft, bluish purple light coming from the small glass-covered bowls. Thus they sighted the Radium for the first time. Its colour was bluish purple.

Activity-based on Vocabulary:

Question 1.
Find out the words from the passage which mean:

  1. unbeatable
  2. discouraging
  3. hardly
  4. illiteracy

Answer:

  1. indomitable
  2. disheartening
  3. scarcely
  4. ignorance

Question 2.
Find out antonyms for the following from the passage: (Answers are directly given.)
Answer:

  1. economical × expensive
  2. poverty × luxury
  3. chill × warmth
  4. worthless × precious

Question 3.
Write down the describing words used for the following nouns:

  1. work
  2. pitchblende
  3. colour
  4. desire.

Answer:

  1. tiresome work
  2. filtered pitchblende
  3. beautiful colour
  4. compulsive desire

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Activities based on Contextual Grammar:

Question 1.
Her father could not afford the expenses of her education in Paris. (Make it affirmative.)
Answer:
Her father was unable to afford the expenses of her education in Paris.

Question 2.
Her life in the University was a disheartening experience in poverty and hunger. (Make it negative.)
Answer:
Her life in the University was not an encouraging/heartening experience in poverty and hunger.

Question 3.
Marie took a job as a governess and saved a little money. (Rewrite using ‘by’ + ing.)
Answer:
Marie saved a little money by taking a job as a governess.

Question 4.
Marie Curie’s childhood dream was to study science in Paris, but her father could not afford the j expense for this. (Rewrite using’Though’.)
Answer:
Though Marie Curie’s childhood dream was to study science in Paris, her father could not afford ; the expense for this.

Change the degree:

Question 1.
Pitchblende was the most precious gift the Curies had received.
Answer:
(a) Positive – No other gift the Curies had, received was as precious as pitchblende.
(b) Comparative – Pitchblende was more precious gift than any other gifts the Curies had received.

Question 2.
Pitchblende is one of the most expensive substances.
Answer:
(a) Positive – Very few substances are as expensive as pitchblende.
(b) Comparative – Pitchblende is more expensive substance than most other substances.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Change the voice:

Question 1.
In 1906, Pierre was knocked down by a horse – drawn wagon. (Begin the sentence with ‘A horse-drawn wagon’.)
Answer:
A horse-drawn wagon knocked down Pierre in 1906.

Question 2.
In 1911, Marie was awarded the Nobel Prize. (Begin the sentence with ‘They’)
Answer:
They awarded Marie the Nobel Prize in 1911.

Personal Response:

Question 1.
Describe your favourite scientist.
Answer:
C.V. Raman who won the Nobel Prize for Physics in 1930 is my favourite scientist. He was the first Asian and first non-white to receive any Nobel Prize in Science. He was the first to investigate the harmonic nature of the sound of the Indian drums such as the tabla and the Mrudangam. He discovered that, when light traverses a transparent material, some of the deflected light changes in wavelength.

Question 2.
What qualities, do you think, are important to get success in our life?
Answer:
There are some key components of success without which we cannot succeed in our life. First our mindset-without a positive mindset the rest of these components won’t have as much power as they should. Staying focused on goal is also important. Besides it, dedication, discipline and determination are important factors to get success in our life.

Question 3.
Why are awards and prizes given to the people in different fields?
Answer:
An award or a prize is something given to a person, in recognition of his or her excellence in a certain field. It may also simply be a public acknowledgement of excellence. It boosts the j recipient’s confidence. It also encourages and inspires him/her to do better in his/her life. It also inspires other people in different fields to achieve something great.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Language Study:

1. Coordinators: Coordinators are words that join grammatically, equal units together. E.g., and, but, or, nor are the main Coordinators.

2. Subordinators: Words whose function is to establish an unequal grammatical relationship are called subordinators. that, for, who, whether, if, because, so. Dependent clauses Principal clause that, for, who, whether, if, because, so subordinate conjunctions (subordinators)

Do as directed:

Question 1.
Complete the words by using correct letters:
Answer:

  1. bre _ d – bread
  2. b i _ t h – birth
  3. la _ ge – large
  4. p r i _ e – prize

Question 2.
Put these words in alphabetical order:
1. lifeless, hostile, endurance, progress
2. commitment, characterise, courage, costly
Answer:
1. endurance, hostile, lifeless, progress
2. characterise, commitment, costly, courage

Question 3.
Punctuate the following sentences:
1. do you remember the day when you told me that you wanted radium to have a beautiful colour marie said to her husband
2. she replied i am working for science radium belongs to the people not to me
Answer:
1. “Do you remember the day when you told me that you wanted radium to have a beautiful colour?” Marie said to her husband.
2. She replied, “I am working for science. Radium belongs to the people, not to me.”

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Question 4.
Make four words (minimum 3 letters each) using the letters in the given word: ‘ incalculable’
Answer:

  1. ill
  2. call
  3. able
  4. cube

Question 5.
Write related words as shown in the example:
(Answer is directly given and underlined.)
Answer:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity 8

Question 6.
Complete the word-chain of verbs. Add four words, each beginning with the last letter of the previous word!
eat → t……….., ……………, ………….., …………….
Answer:
eat → take , extend , develop , paddle.

1. Attempt any one:

Question (a)
Make a meaningful sentence of your own using the phrase: ‘to fight against ’
Answer:
Kailash Satyarthi has been fighting against child labour for the last several years.

OR

Question (b)
Add a clause to the following sentence to expand it meaningfully:
The history is the subject ………………. .
Answer:
The history is the subject which gives the narration of episodes happened in the past.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

2. Attempt any one:

Question (a)
Add prefix or suffix to make new words :
1. ignore
2. dedicate
Answer:
1. ignorance
2. dedication

OR

Question (b)
Make a meaningful sentence using any one of the following words :
1. ignore
2. dedicate
Answer:
1. Don’t ignore your parent’s advice.
2. Dedicate your life for poor; the father told his son.

Live English:

Blogging:

  1. Terms – (Blogging)
  2. Blog – Web log which is an online public diary.
  3. Purpose – Anybody can create blog for any reason.
  4. About – Function as personal online diaries
  5. Types – Artlogs – Focus on art
  6. Photologs – Focus on photography
  7. Sketch – Focus on sketching
    blogging – Video blogging
  8. Podcasting – Audio blogging
  9. Creating – You can create your own blog on a personal website or any other websites like www.livejournal.com, www.wordpress.com, www.blogger. corn, www.blogs.rny-space.com
  10. Popularity – Blogs allow anyone to express themselves on the Internet and it becomes popular if people like it.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.6 A Brave Heart Dedicated to Science and Humanity

Steps of creating blogs:

Starting a blog is the best way to share your Ideas and expertise online.
Follow the given steps to create your blog.

  • Choose your blogging platform. (word press)
  • Choose a Domain name. (www.yourblog.com)
  • Sign up with Web Host and install Word Press (www.bluehost.com is good hosting provider)
  • Sign up with a web hosting.
  • Choose the hosting plan.
  • Choose the domain name you want.
  • Complete your registration.
  • Install wordpress with one-click installation.
  • Log into your new blog.
  • Customizing and choosing a blog theme.
  • Adding Post and Pages.
  • Get Traffic and monetize your blog.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Balbharti Maharashtra State Board Class 10 English Solutions My English Coursebook Chapter 4.5 Joan of Arc Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

My English Coursebook Standard Ten Guide Chapter 4.5 Joan of Arc Textbook Questions and Answers

Warming up:
Chit-chat

1. Discuss in groups whether you would like to join the Armed Forces. Yes/No. Say Why? Why not? Each one should give 2 to 3 reasons for their responses.

Question 1.
Discuss in groups whether you would like to join the Armed Forces. Yes/No. Say Why? Why not? Each one should give 2 to 3 reasons for their responses.
Answer:
(a) Yes, I would like to join the Armed Forces because ……………..

  1. It is my passion to serve my country by joining Armed forces.
  2. It gives an opportunity to serve at different places and interact with different people everyday.
  3. It keeps us physically and mentally fit.

(b) No, I wouldn’t like to join the Armed Forces because …………………..

  1. I am not physically and mentally prepared for this field.
  2. I have decided another career option for myself.
  3. I don’t like to leave my parents and live away from them.
  4. I am afraid of wars and killing and disaster that comes with them. I hate them all.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question 2.
Also discuss what areas of work are open for women in Armed forces in India.
Answer:
All wings of the Indian Armed Forces allow women in combat roles (Junior rank) and combat supervisory roles (Officers).

  1. They can work in administrative sections in the Armed Forces.
  2. They can work in the medical field as a doctor, nurse or any other related post.
  3. They can join the paramilitary forces of India.

1. Discuss in groups/pairs and make lists of the weapons used in the old times and in the present times:

Question 1.
Discuss in groups/pairs and make lists of the weapons used in the old times and in the present times:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc 1
Answer:

Weapons used in the Past Weapons used nowadays
Sword
Axe
Spear
Shield
Dagger and knives
Rifles, Machine guns,
Tanks, Grenade
Aircraft Submarines
Chemical weapons (gas, etc.)
Biological weapons (germs, etc.)

2. Imagine that you are the captain of your school Kabaddi/Football team. Your final match is against a very strong team. Prepare a short pep-talk of about 60 to 80 words that you would give as a Captain, to encourage your team.

Question 1.
Imagine that you are the captain of your school Kabaddi/Football team. Your final match is against a very strong team. Prepare a short pep-talk of about 60 to 80 words that you would give as a Captain, to encourage your team.
Answer:
My dear teammates,

I have heard about a good quotation-winning or losing a game is not important but playing with spirit is important. We get good and useful experiences as participants. Success is not always about winning but being happy with the participation and playing to fullest ability.

We know that strength of our team is each individual member and the strength of each member is the team. The player that fights and works the hardest will always come out on the top. If you find out the strength in you and push forward to work hard for yourself and for the team, I am sure you can do a lot.

Leave your negativity and get in the habit of the doing things the right way in the game. Blaming others for not reaching your goal is pointless. It is your goal and you are the only one who can achieve it. I know champions are those who work to the point of exhaustion when no one else is watching.

Use your calibre and be a champion and success will be yours.
All the best!

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

3. When different Prepositions are added to the same action verb, meaning of the phrase thus formed changes.

Question 1.
When different Prepositions are added to the same action verb, meaning of the phrase thus formed changes.
Examples:

  1. call out – announce
  2. call at – visit
  3. call far – summon
  4. call up – make a telephone call

Try to guess the meanings of the underlined phrases and write them down:

Question (a)
He promised me to look into the matter.
Answer:
to inspect carefully, to investigate.

Question (b)
He wanted to look for his lost book.
Answer:
to search for.

Question (c)
An epidemic of cholera broke-out in the village.
Answer:
widely spread.

Question (d)
The thieves broke into the apartment.
Answer:
entered by force.

Question (e)
She has to carry out her duty regularly.
Answer:
discharge/perform

Question (f)
You must carry on trying for success.
Answer:
continue

English Workshop:

1. Read the extract from G. B. Shaw’s play on Joan of Arc and fill in the Tree diagram:

Question 1.
Read the extract from G. B. Shaw’s play on Joan of Arc and fill in the Tree diagram: (Answer is directly given.)
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc 2
Answer:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc 3

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

2. Pick out from the play 2 lines each that prove the following:

Question 1.
Pick out from the play 2 lines each that prove the following:
Joan of Arc
Her confidence
1. …………………..
2. …………………..
Her courage
1. …………………..
2. …………………..
Her determination
1. …………………..
2. …………………..
Answer:

1. Her confidence 1. The Dauphin will give me, all I need, to free Orleans.
2. Three men will be enough for you to send with me.
2. Her courage 1. You wouldn’t see me. But here I am.
2. I can find a soldier’s armour that will fit me well enough.
3. Her determination 1. There will be no trouble for you, Squire.
2. I will teach them all to fight for France.

3. Match the comparisons as given in the extract:

Question 1.
Match the comparisons as given in the extract:

A B
1. as easy as (a) like sheep
2. as mad as (b) like a rat in the corner
3. drive the soldiers (c) a bit of a miracle
4. The Dauphin is (d) as she is
5. Joan of Arc is (e) the steward
(f) chasing a cow

Answer:

  1. as easy as – chasing a cow
  2. drive the soldiers – like sheep
  3. The Dauphin is – like a rat in a corner
  4. Joan of Arc – a bit of miracle
  5. as mad as – she is

4. Say Why?

Question (a)
Joan wanted to meet the Captain Squire? (%MT wq iN^dl ift?)
Answer:
Captain Squire was to give her a horse, armour and some soldiers and send her to the Dauphin. He thought that she wouldn’t dare to meet him. But being courageous she came to meet the Captain and showed her confidence. She wanted the Captain to give her his order to go to Dauphin, the prince of France.

Question (b)
Joan did not ask for many soldiers from the Captain Squire?
Answer:
Joan did not ask for many soldiers from the Captain Squire because she was fully confident about fighting against the English with a few warriors who would join her in the battle. She was also sure that the Dauphin would give her soldiers and all that she needed to free Orleans.

Question (c)
Poulengey, Jack and Dick offered to accompany Joan?
Answer:
Being kind and a gentleman Jack would go willingly with Joan and even she managed Poulengey as he was sure that she was like a miracle. They had promised to go with Joan because they had faith in her valour. Every one was as mad as she was for freedom of their country; so they had offered to accompany Joan.

Question (d)
French soldiers were always beaten?
Answer:
The French soldiers were always beaten because they used to fight only to save their skins. They used to run away from the battlefield to save their lives. They always cared only for their own lives, and not for their nation.

Question (e)
Captain Squire Robert said, “I wash my hands off it.”-Why did he say so?
Answer:
Captain Squire Robert realised that Joan could inspire anybody. But he was not ready to take any responsibility, because he knew that he was taking a big chance and was not sure about the win. He wanted to stay away if anything went wrong by his decision. He would be responsible only for sending Joan to Dauphin and nothing else. So he said, “I wash myhands off it.”

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

5. Using a dictionary, find the difference between the following pairs of phrases. Make sentences of your own with each of them:

Question 1.
Using a dictionary, find the difference between the following pairs of phrases. Make sentences of your own with each of them:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc 4
Answer:

Phrases Meaning Own Sentences
1. (a) to cut in
(b) to cut out
(a) interrupt
(b) remove/cut
(a) When I was talking with Arjun, Abid cut in our conversation.
(b) The director cut out several scenes when objection was raised by Censor Board.
2. (a) to be held by

(b) to be held up

(a) kept/maintained/will take place
(b) delay/stop/block the movement
(a) The meeting will be held by next Saturday.
(b) The match was held up by heavy rains.
3. (a) to run away
(b) to run for
(a) to leave a place secretely
(b) run for something
(a) He ran away from home when he was only fifteen.
(b) The picnickers ran for shelter when the rain started.
4. (a) to be known as
(b) to be known for
(a) to be best known as
(b) to be famous or known because of something
(a) Mohandas Karamchand Gandhi is known as Mahatma Gandhi to all of us.
(b) Miller is known for his whimsical paintings and sculpture.
5. (a) to go with
(b) to go after
(a) to choose or accept something
(b) to try to catch or stop something
(a) I think we should go with yellow for the drawing room.
(b) You would better go after her and tell her you’re sorry.
6. (a) to put fire into
(b) to put fire out
(a) heat up/put on the fire.
(b) to extinguish a fire.
(a) She spends much of her time in putting fire into others lives.
(b) The firemen were able to put out the fire before too much damage.

6. From an Indian history book or Internet find out information about Indian Women (queens) who led battles.

Question 1.
From an Indian history book or Internet find out information about Indian Women (queens) who led battles. (For example, Rani of Jhansi and Rani Karnawati of Mewad). Write out 3 points of similarity and 3 points of contrast between any one of the above Indian Queens and Joan of Arc. Write in your own words:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc 5

Answer:

Similarities Contrast
(a) young and beautiful (a) Joan of Arc was unmarried.
(b) brave and courageous (b) She was poor being a peasant girl.
(c) fought for their kingdom/nation. (c) Rani of Jhansi fought for her adopted son and her kingdom. Rani Karnawati of Mewad fought for her kingdom but Joan of Arc fought for her nation and led the French army to several victories during the Hundred Years War.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

7. Read the script from:

Question 1.
Read the script:
Answer:
Joan: Good morning, … … … … … … … … … promised to come with me. On Coursebook page 157.
Write a summary of that part of the script (in Indirect Speech) in 15 to 20 lines.
Answer:
Wishing Good morning to Captain Squire, Joan said that he was to give her a horse and armour and some soldiers, and send her to the Dauphin. Robert felt that, that girl was indeed mad so he asked Steward why he had not told him about it. Hearing this Steward told him not to anger her and requested him to give her what she wanted.

After hearing his remark Robert got furious and told Joan that he would send her back to her father with orders to put her under lock and key. But Joan was confident and told him that he thought he would. But it wouldn’t happen that way. She further said that he said he would not see her but there she was. Robert asked her if she was assuming that he would give her what she wanted.

Joan confidently said that he would do so and continued that a horse would cost sixteen francs. It was a big amount of money. But she could save it on the armour. She didn’t need beautiful armour made to her measure, she could find a soldier’s armour that would fit her well enough. She wouldn’t want many soldiers and the Dauphin would give her all, she needed, to free Orleans. After hearing her Robert was shocked and asked if she could free Orleans. She continued with confidence and told him that, it was true and only three men would be enough for him to send with her. Polly and Jack had promised to come with her.

Language Study:

Question (A)
Make the following sentences affirmative without change of meaning.
(a) Negative: I am not so sure, now.
Affirmative: …………………………………..
(b) Negative: He will not be able to stop them.
Affirmative: …………………………………..
(c) Negative: I dont’t remember.
Affirmative: ……………………………………
(d) Negative: I can do no more.
Affirmative: ……………………………………
(e) Negative: Sir, do not anger her.
Affirmative: ……………………………………
(f) Negative: I shall not want many soldiers.
Affirmative: ……………………………………

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question (B)
Fill in the gaps in the table.
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc 6
Answer:

Noun Verb Adjective Adverb
1. success succeed successful successfully
2. inspiration inspire inspirational inspirationally
3. safety safe safe safely
4. belief believe believable believably
5. thought think thoughtful thoughtfully
6. brightness brighten bright brightly
7. courage encourage courageous courageously
8. haste hasten hasty hastily

My English Coursebook 10th Digest Chapter 4.5 Joan of Arc Additional Important Questions and Answers

Simple Factual Activities:

Question 1.
Name the following:
Answer:

  1. Military officer – Captain Robert de Baudricourt
  2. A peasant girl – Joan
  3. The oldest son of the King France – Dauphin
  4. The persons who Joan needed to free Orleans – Squire Jack, John Godsave, Dick the Archer, John of
    Honecourt and Julian

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question 2.
State whether the following statements are True or False :
Answer:

  1. Monsieur de Poulengey and Monsieur de Metz want to go with Joan – True
  2. Chinon is one of the cities in England. – False
  3. Poulengey was sure that only miracle can save them. – True
  4. Joan’s words have put fire into Robert. – False

Question 3.
Complete the following sentences: (Answers are directly given and underlined.)
Answer:

  1. Soldiers called Joan ‘the Maid’.
  2. Joan wanted a soldier’s dress.
  3. The shortest way to save your skin is to run away.
  4. According to Joan, their soldiers
  5. Joan is a person of immense faith.

Complex Factual Activities:

Question 1.
Was the Dauphin fit to be a Prince and heir?
Answer:
No, Dauphin was not at all fit to be a Prince and heir.

Question 2.
What was the Squire’s opinion about miracles?
Answer:
According to the Squires miracles don’t happen nowadays.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question 3.
What did Robert tell the Steward?
Answer:
Robert told the Steward to go with Joan and keep eyes on her.

Question 4.
What did Robert accuse Poulengey of?
Answer:
Robert accused Poulengey that he was as mad as Joan.

Activity-based on Vocabulary:

Question 1.
Find out two adjectives and two adverbs from the passage :
Answer:
1. Adjectives – wrong, mad
2. Adverbs – seriously, obstinately

Question 2.
What shows Joan was a person of immense faith?
Answer:
Robert thought that his soldiers would not be inspired by anything but Joan was very sure that she along with her soldiers would drive the enemy like sheep and there would not be a single English soldier on the soil of France. It shows that Joan was a person of immense faith.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question 3.
What made Robert finally agree to the plan?
Answer:
When Robert realized that if Joan could put fire into Poulengey, she could put it into anybody. She could inspire his soldiers too, so he finally agreed to the plan.

Question 4.
Give one word for the following :
Answer:

  1. A person of high rank – Squire
  2. A stupid person – blockhead
  3. To take for granted – to assume
  4. Military unit consisting of armoured fighting vehicles – armour

Question 5.
Find out synonyms:
Answer:

  1. looting – plundering
  2. seriously- gravely
  3. following- chasing
  4. hayfield – meadow.

Activities based on Contextual Grammar:

Question 1.
He is a very kind gentleman.
(Make it exclamatory.)
Answer:
What a kind gentleman he is!

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question 2.
She is a well-built strong country girl.
(Make it exclamatory.)
Answer:
What a well-built, strong country girl she is!

Make the following sentences affirmative without change of meaning:

Question 1.
Sir, do not anger her.
Answer:
Sir, be calm with her.

Question 2.
I shall not want many soldiers.
Answer:
I shall want few soldiers.

Make it affirmative:

Question 1.
I am not sure, now.
Answer:
I am unsure (doubtful) now.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question 2.
He will not be able to stop them.
Answer:
He will be unable to stop them.

Choose the correct question tag:

Question 1.
Choose the correct question tag:
(haven they?, doesn’t he?, aren’t you?, am I?)

  1. The Steward retreats hastily.
  2. You are as mad as she is,
  3. I am not so sure now,
  4. Jack and Dick have offered to go with her,

Answer:
doesn’t he? aren’t you? am I? haven’t they?

Rewrite as affirmative sentences:

Question 1.
I don’t remember.
Answer:
I forget.

Question 2.
You do not understand Squire.
Answer:
You fail to understand Squire.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question 3.
I can do no more.
Answer:
I am unable to do anything more.

Personal Response:

Question 1.
Describe any brave lady as Joan of Arc, from India who fought for her nation.
Answer:
Lakshmibai, the Rani of Jhansi, was the queen of the Princely state of Jhansi in North India. She was one of the leading figures of the first Indian War of Independence of 1857 and became the symbol of resistance to the British Raj by Indian nationalists.

After the death of her husband, the then Head of the British Government of India, Lord Dalhousie, refused to allow her adopted son to become Raja of Jhansi. British then forcibly took possession of Jhansi. Rani Lakshmibai with Tatya Tope and others fought against the British rule. She sacrificed her life to regain her kingdom.

Question 2.
What is your opinion about Joan?
Answer:
I think Joan was really a courageous and brave lady. She dared to fight against the enemy for her country without caring for her own life and family life. It shows that she was really a patriotic person who was proud of her country and loved her country from the bottom of her heart.

Question 3.
Do you love your country? Why?
Answer:
I love my country from the bottom of my heart. It is the place where I was born, brought up and saw this beautiful world. I owe everything to this country. I always think about the betterment of
my country because I really love my country as it is my birthplace, my motherland.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Do as directed:

Question 1.
Complete the words by using correct letters:

  1. h o _ s e
  2. t h i _ k
  3. m o _ e y
  4. c r a _ y

Answer:

  1. horse
  2. t h i n k
  3. m o n e y
  4. c r a z y

Question 2.
Put the words in alphabetical order:
1. order, trouble, captain, promise.
2. assume, armour, afraid, always.
Answer:
1. captain, order, promise, trouble.
2. afraid, always, armour, assume.

Question 3.
Punctuate the following sentences:
1. joan said i dont think it can be very difficult if god is on your side
2. joan said and the dress i may have a soldiers dress squire
Answer:
1. Joan said, “I don’t think it can be very difficult if God is on your side.”
2. Joan said, “And the dress? I may have a soldier’s dress, Squire?”

Question 4.
Make four words (minimum 3 letters each) using the letters in the word : ‘plundering’
Answer:
plunder, ring, under, pun.

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question 5.
Write related words as shown in the example: (Answers are directly given and underlined.)
Answer:
Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc 7

Question 6.
Complete the word-chain of nouns. Add four words, each beginning with the last letter of previous word:
officer, → r…………., …………., …………., ………….
Answer:
officer, → ration , nation , novel, logo.

1. Attempt any one:

Question (a)
Make a meaningful sentence by using the phrase :’ to be afraid of’
Answer:
Most of the people are afraid of ghosts and darkness.

OR

Question (b)
Add a clause to expand the sentence:
This is the young boy
Answer:
This is the young boy who bagged the National Championship in boxing.

2. Attempt any one:

Question (a)
Add prefix or suffix to make new words.
1. talk
2. except
Answer:
1. talkative
2. exceptional

OR

Maharashtra Board Class 10 My English Coursebook Solutions Chapter 4.5 Joan of Arc

Question (b)
Make a meaningful sentence using any one of the following words:
1. talk
2. except
Answer:
1. We talked on the phone about our problem.
2. Except Mother, everyone went out for dinner.

Maharashtra Board 10th Class Maths Part 1 Problem Set 4B Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 4B Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Problem Set 4B Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Financial Planning Class 10 Problem Set 4b
Question 1.
Write the correct alternative for the following questions.

i. If the Face Value of a share is ₹ 100 and Market value is ₹ 75, then which of the following statement is correct?
(A) The share is at premium of ₹ 175
(B) The share is at discount of ₹ 25
(C) The share is at premium of ₹ 25
(D) The share is at discount of ₹ 75
Answer:
(B)

ii. What is the amount of dividend received per share of face value ₹ 10 if dividend declared is 50%.
(A) ₹ 50
(B) ₹ 5
(C) ₹ 500
(D) ₹ 100
Answer:
Dividend = 10 × \(\frac { 50 }{ 100 } \) = ₹ 5
(B)

iii. The NAV of a unit in mutual fund scheme is ₹ 10.65, then find the amount required to buy 500 such units.
(A) 5325
(B) 5235
(C) 532500
(D) 53250
Answer:
(A)

iv. Rate of GST on brokerage is _______
(A) 5%
(B) 12%
(C) 18%
(D) 28%
Answer:
(C)

v. To find the cost of one share at the time of buying the amount of Brokerage and GST is to be ______ MV of share.
(A) added to
(B) subtracted from
(C) Multiplied with
(D) divided by
Answer:
(A)

Problem Set 4b Algebra Class 10 Question 2. Find the purchase price of a share of FV ₹ 100 if it is at premium of ₹ 30. The brokerage rate is 0.3%.
Solution:
Here, Face Value of share = ₹ 100,
premium = ₹ 30, brokerage = 0.3%
MV = FV + Premium
= 100 + 30
= ₹ 130
Brokerage = 0.3% of MV
= \(\frac { 0.3 }{ 100 } \) × 130 = ₹ 0.39
Purchase price of a share = MV + Brokerage
= 130 + 0.39
= ₹ 130.39
Purchase price of a share is ₹ 130.39.

Question 3.
Prashant bought 50 shares of FV ₹ 100, having MV ₹ 180. Company gave 40% dividend on the shares. Find the rate of return on investment.
Solution:
Here, Number of shares = 50, FV = ₹ 100,
MV = ₹ 180, rate of dividend = 40%
∴ Sum invested = Number of shares × MV
= 50 × 180
= ₹ 9000
Dividend per share = 40% of FV
= \(\frac { 40 }{ 100 } \) × 100
Dividend = ₹ 40
∴ Total dividend on 50 shares = 50 × 40
= ₹ 2000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 1
∴ Rate of return on investment is 22.2%.

Question 4.
Find the amount received when 300 shares of FV ₹ 100, were sold at a discount of ₹ 30.
Solution:
Here, FV = ₹ 100, number of shares = 300,
discount = ₹ 30
MV of 1 share = FV – Discount
= 100 – 30 = ₹ 70
∴ MV of 300 shares = 300 × 70
= ₹ 21,000
∴ Amount received is ₹ 21,000.

Question 5.
Find the number of shares received when ₹ 60,000 was invested in the shares of FV ₹ 100 and MV ₹ 120.
Solution:
Here, FV = ₹ 100, MV = ₹ 120,
Sum invested = ₹ 60,000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 2
∴ Number of shares received were 500.

Question 6.
Smt. Mita Agrawal invested ₹ 10,200 when MV of the share is ₹ 100. She sold 60 shares when the MV was ₹ 125 and sold remaining shares when the MV was ₹ 90. She paid 0.1% brokerage for each trading. Find whether she made profit or loss? and how much?
Solution:
For purchasing shares:
Here, sum invested = ₹ 10,200, MV = ₹ 100
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 3
For selling shares:
60 shares sold at MV of ₹ 125.
∴ MV of 60 shares = 125 × 60
= ₹ 7500
Brokerage = \(\frac { 0.1 }{ 100 } \) × 7500 = ₹ 7.5
∴ Sale value of 60 shares = 7500 – 7.5 = ₹ 7492.5
Now, remaining shares = 102 – 60 = 42
But 42 shares sold at MV of ₹ 90.
∴ MV of 42 shares = 42 × 90 = ₹ 3780
∴ Brokerage = \(\frac { 0.1 }{ 100 } \) × 3780 = ₹ 3.78
∴ Sale value of 42 shares = 3780 – 3.78 = ₹ 3776.22
Total sale value = 7492.5 + 3776.22 = ₹ 11268.72
Since, Purchase value < Sale value
∴ Profit is gained.
∴ Profit = Sale value – Purchase value
= 11268.72 – 10210.2
= ₹ 1058.52
∴ Smt. Mita Agrawal gained a profit of ₹ 1058.52.

Question 7. Market value of shares and dividend declared by the two companies is given below.
Face value is same and it is 7 100 for both the shares. Investment in which company is more profitable?
i. Company A – ₹ 132,12%
ii Company B – ₹ 144,16%
Solution:
For company A:
FV = ₹ 100, MV = ₹ 132,
Rate of dividend = 12%
Dividend = 12% of FV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 4
∴ Rate of return of company B is more.
∴ Investment in company B is more profitable.

Question 8. Shri. Aditya Sanghavi invested ₹ 50,118 in shares of FV ₹ 100, when the market value is ₹ 50. Rate of brokerage is 0.2% and Rate of GST on brokerage is 18%, then How many shares were purchased for ₹ 50,118?
Solution:
Here, FV = ₹ 100, MV = ₹ 50
Purchase value of shares = ₹ 50118,
Rate of brokerage = 0.2%, Rate of GST = 18%
Brokerage = 0.2% of MV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 5
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 6
∴ 1000 shares were purchased for ₹ 50,118.

Question 9. Shri. Batliwala sold shares of ₹ 30,350 and purchased shares of ₹ 69,650 in a day. He paid brokerage at the rate of 0.1% on sale and purchase. 18% GST was charged on brokerage. Find his total expenditure on brokerage and tax.
Solution:
Total amount = sale value + Purchase value
= 30350 + 69650
= ₹ 1,00,000
Rate of Brokerage = 0.1 %
Brokerage = 0.1 % of 1,00,000
= \(\frac { 0.1 }{ 100 } \) × 1,00,000
= ₹ 100
Rate of GST = 18%
∴ GST = 18 % of brokerage
= \(\frac { 18 }{ 100 } \) × 100
∴ GST = ₹ 18
Total expenditure on brokerage and tax
= 100 + 18 = ₹ 118
∴ Total expenditure on brokerage and tax is ₹ 118.

Alternate Method:
Brokerage = 0.1 %, GST = 18%
At the time of selling shares:
Total sale amount of shares = ₹ 30,350
Brokerage = 0.1% of 30,350
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 7
For purchasing shares:
Total purchase amount of shares = ₹ 69,650
Brokerage = 0.1% of 69,650
= \(\frac { 0.1 }{ 100 } \) × 69650
= ₹ 69.65
GST = 18% of 69.65
= \(\frac { 18 }{ 100 } \) × 69.65
= ₹ 12.537
∴ Total expenditure on brokerage and tax = Brokerage and tax on selling + Brokerage and tax on purchasing
= (30.35 + 5.463) + (69.65 + 12.537)
= ₹ 118
∴ Total expenditure on brokerage and tax is ₹ 118.

Question 10. Sint. Aruna Thakkar purchased 100 shares of FV 100 when the MV is ₹ 1200. She paid brokerage at the rate of 0.3% and 18% GST on brokerage. Find the following –
i. Net amount paid for 100 shares.
ii. Brokerage paid on sum invested.
iii. GST paid on brokerage.
iv. Total amount paid for 100 shares.
Solution:
Here, FV = ₹ 100,
Number of shares = 100, MV = ₹ 1200
Brokerage = 0.3%, GST = 18%
i. Sum invested = Number of shares × MV
= 100 × 1200 = ₹ 1,20,000
∴ Net amount paid for 100 shares is ₹ 1,20,000.

ii. Brokerage = 0.3% of sum invested
= \(\frac { 0.3 }{ 100 } \) × 1,20,000 = ₹ 360
∴ Brokerage paid on sum invested is ₹ 360.

iii. GST = 18% of brokerage
= \(\frac { 18 }{ 100 } \) × 360 = ₹ 64.80
∴ GST paid on brokerage is ₹ 64.80.

iv. Total amount paid for 100 shares
= Sum invested + Brokerage + GST
= 1,20,000 + 360 + 64.80
= ₹ 1,20,424.80
∴ Total amount paid for 100 shares is ₹ 1,20,424.80.

Question 11. Smt. Anagha Doshi purchased 22 shares of FV ₹ 100 for Market Value of ₹ 660. Find the sum invested. After taking 20% dividend, she sold all the shares when market value was ₹ 650. She paid 0.1% brokerage for each trading done. Find the percent of profit or loss in the share trading. (Write your answer to the nearest integer)
Solution:
For purchasing shares:
Here, FV = ₹ 100, MV = ₹ 660, Number of shares = 22, rate of brokerage = 0.1%
Sum invested = MV × Number of shares
= 660 × 22
= ₹ 14,520
Brokerage = 0.1 % of sum invested
= \(\frac { 0.1 }{ 100 } \) × 14520 = ₹ 14.52
∴ Amount invested for 22 shares
= Sum invested + Brokerage
= 14520 + 14.52
= ₹ 14534.52
For dividend:
Rate of dividend = 20%
∴ Dividend per share = 20 % of FV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 8
∴ Percentage of profit in the share trading is 1 % (nearest integer).

Alternate Method:
For purchasing share:
Here, FV = ₹ 100, MV = ₹ 660, Number of shares = 22, rate of brokerage = 0.1%
Sum invested = MV × Number of shares
= 660 × 22
= ₹ 14,520
Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 660 = ₹ 0.66
Amount invested for 1 share = 660 + 0.66
= ₹ 660.66
For dividend:
Rate of dividend = 20%
Dividend = 20% of FV = \(\frac { 20 }{ 100 } \) × 100 = ₹ 20
For selling share:
MV = ₹ 650, rate of brokerage = 0.1%
Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 650 = ₹ 0.65 100
Amount received after selling 1 share
= 650 – 0.65 = 649.35
∴ Amount received including divided
= selling price of 1 share + dividend per share
= 649.35 + 20
= ₹ 669.35
Since, income > Amount invested
∴ Profit is gained.
∴ profit = 669.35 – 660.66 = ₹ 8.69
Profit Percentage = \(\frac { 8.69 }{ 660.66 } \) × 100= 1.31%
∴ Percentage of profit in the share trading is 1 % (nearest integer).

Maharashtra Board 10th Class Maths Part 1 Problem Set 3 Solutions Chapter 3 Arithmetic Progression

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Problem Set 3 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Question 1.
Choose the correct alternative answer for each of the following sub questions.

i. The sequence – 10,- 6,- 2, 2, …
(A) is an A.P. Reason d = – 16
(B) is an A.P. Reason d = 4
(C) is an A.P. Reason d = – 4
(D) is not an A.P.
Answer:
(B)

ii. First four terms of an A.P. are …, whose first term is -2 and common difference is -2.
(A) -2, 0, 2, 4
(B) -2, 4,- 8, 16
(C) -2, -4, -6, -8
(D) -2,-4, -8, -16
Answer:
(C)

iii. What is the sum of the first 30 natural numbers?
(A) 464
(B) 465
(C) 462
(D) 461
Answer:
(B)

iv. For an given A.P. t7 = 4, d = – 4, then a = ………
(A) 6
(B) 7
(C) 20
(D) 28
Answer:
(D)

v. For an given A.P. a = 3.5, d = 0, n = 101, then tn = ….
(A) 0
(B) 3.5
(c) 103.5
(D) 104.5
Answer:
(B)

vi. In an A.P. first two terms are – 3, 4, then 21st term is ….
(A) -143
(B) 143
(C) 137
(D) 17
Answer:
(C)

vii. If for any A.P. d = 5, then t18 – t13 = ….
(A) 5
(B) 20
(C) 25
(D) 30
Answer:
(C)

viii. Sum of first five multiples of 3 is …
(A) 45
(B) 55
(C) 15
(D) 75
Answer:
(A)

ix. 15, 10, 5, … In this A.P. sum of first 10 terms is…
(A) -75
(B) -125
(C) 75
(D) 125
Answer:
(A)

x. In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is 399, then n = ….
(A) 42
(B) 38
(C) 21
(D) 19
Answer:
(B)

Hints:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 1

Question 2.
Find the fourth term from the end in an
A.P.: -11, -8, -5, …, 49.
Solution:
The given A.P. is
-11,-8,-5, ……. 49
Reversing the A.P., we get 49, …, -5, -8, -11
Here, a = 49, d = -11 -(-8) = -11 + 8 = -3
Since, tn = a + (n – 1)d
∴ t4 = 49 + (4 – 1)(-3)
= 49 + (3) (-3)
= 49 – 9
= 40
∴ Fourth term from the end in the given A.P. is 40.
[Note: If an AY. is reversed, then the resulting sequence is also an A.P.]

Question 3.
In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P.
Solution:
For an A.P., let a be the first term and d be the common difference.
t10 = 46, t5 + t7 = 52 …[Given]
Since, tn = a + (n – 1)d
∴ t10 = a + (10 – 1)d
∴ 46 = a + 9d
i. e. a + 9d = 46 …(i)
Also, t5 + t7 = 52
∴ a + (5 – 1)d + a + (7 – 1)d = 52
∴ a + 4d + a + 6d = 52
∴ 2a + 10d = 52
∴ 2 (a + 5d) = 52
∴ a + 5d = \(\frac { 52 }{ 2 } \)
∴ a + 5d = 26 …(ii)
Subtracting equation (ii) from (i), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 2
Substituting d = 5 in equation (ii), we get
a + 5(5) = 26
∴ a + 25 = 26
∴ a = 26 – 25 = 1
t1 = a = 1
t2 = t1 + d = 1 + 5 = 6
t3 = t2 + d = 6 + 5 = 11
t4 = t3 + d = 11 + 5 = 16
The required A.P. is 1,6,11,16,….

Question 4.
The A.P. in which 4th term is -15 and 9th term is -30. Find the sum of the first 10 numbers.
Solution:
t4 = -15, t9 = – 30 …[Given]
Since, tn = a + (n – 1)d
∴ t4 = a + (4 – 1)d
∴ – 15 = a + 3d
i. e. a + 3d = -15 …(i)
Also, t9 = a + (9 – 1)d
∴ -30 = a + 8d
i.e. a + 8d = -30 …(ii)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 3
∴ The sum of the first 10 numbers is -195.

Question 5.
Two given A.P.’s are 9, 7, 5, … and 24, 21, 18, … If nth term of both the .progressions are equal then find the value of n and n,h term.
Solution:
The first A.P. is 9, 7, 5,…
Here, a = 9, d = 7- 9 = -2
∴nth term = a + (n – 1)d
= 9 + (n – 1) (-2)
= 9 – 2n + 2
= 11 – 2n
The second A.P. is 24, 21, 18, …
Here, a = 24, d = 21 – 24 = – 3
∴ nth term = a + (n – 1)d
= 24 + (n – 1) (-3)
= 24 – 3n + 3
= 27 – 3n
Since, the nth terms of the two A.P.’s are equal.
∴ 11 – 2n = 27 – 3n
∴ 3n – 2n = 27 – 11
∴ n = 16
∴ t16 = 9 + (16 – 1) (-2)
= 9 + 15 × (-2)
= 9 – 30
∴ t16 = -21
∴ The values of n and nth term are 16 and -21 respectively.

Question 6.
If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is -3, then find the 10th term.
Solution:
for an A.P., let a be the first term and d be the common difference.
According to the first condition,
t3 + tg = 7
∴ a + (3 – 1) d + a + (8 – 1)d = 7 …[∵ tn = a + (n – 1)d]
∴ a + 2d + a + 7d = 7
∴ 2a + 9d = 7 …(i)
According to the second condition,
t7 + t14 = -3
∴ a + (7 – 1)d + a + (14 – 1 )d = -3
∴ a + 6d + a + 13d = -3
∴ 2a + 19 d = – 3 …(ii)
Subtracting equation (i) from (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 4

Question 7.
In an A.P. the first term is -5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
Solution:
Let the number of terms in the A.P. be n and the common difference be d.
Then, a = -5, tn = 45, Sn = 120
Since, tn = a + (n – 1)d
∴ 45 = -5 + (n – 1)d
∴ 45 + 5 = (n – 1)d
∴ (n – 1)d = 50 …(i)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 5
Substituting n = 6 in equation (i), we get
(6 – 1)d = 50
∴ 5d = 50
∴ d = \(\frac { 50 }{ 5 } \) = 10
∴ There are 6 terms in the A.P. and the common difference is 10.

Alternate Method:
Let the number of terms in the A.P. be n.
Then, t1 = a = -5, tn = 45, Sn = 120
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 6
∴ There are 6 terms in the A.P. and the common difference is 10.

Question 8.
Sum of 1 to n natural numbers is 36, then find the value of n.
Solution:
The natural numbers from 1 to n are
1,2, 3, ……, n.
The above sequence is an A.P.
∴ a = 1, d = 2 – 1 = 1
Sn = 36 …[Given]
Now, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ 36 = \(\frac { n }{ 2 } \) [2(1) + (n – 1)(1)]
∴ 36 = \(\frac { n }{ 2 } \) (2 + n – 1)
∴ 36 × 2 = n (n + 1)
∴ 72 = n (n + 1)
∴ 72 = n2 + n
∴ n2 + n – 72 = 0
∴ n2 + 9n – 8n – 72 = 0
∴ n(n + 9) – 8 (n + 9) = 0
∴ (n + 9) (n – 8) = 0
∴ n + 9 = 0 or n – 8 = 0
∴ n = -9 or n = 8
But, n cannot be negative.
∴ n = 8
∴ The value of n is 8.

Question 9.
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
Solution:
Let the three parts of 207 that are in A.P. be
a – d, a, a + d
According to the first condition,
(a – d) + a + (a + d) = 207
∴ 3a = 207
∴ a = \(\frac { 207 }{ 3 } \)
∴ a = 69 …(i)
According to the second condition,
(a – d) × a = 4623
∴ (69 – d) × 69 = 4623 …[From (i)]
∴ 69 – d = \(\frac { 4623 }{ 69 } \)
∴ d = 69 – 67
∴ d = 2
∴ a – d = 69 – 2 = 67
a = 69
a + d = 69 + 2 = 71
∴ The three parts of 207 that are in A.P. are 67, 69 and 71.

Question 10.
There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
Solution:
Since, there are 37 terms in the A.P.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 7
Substituting d = 4 in equation (i), we get
3a + 54(4) = 225
∴ 3a + 216 = 225
∴ 3a = 225 – 216
∴ 3a = 9
∴ a = \(\frac { 9 }{ 3 } \) = 3
∴The required A. P. is
a, a + d, a + 2d, a + 3d, …., a + (n – 1)d
i.e. 3, 3 + 4,3 + 2 × 4, 3 + 3 × 4,…, 3 + (37 – 1)4
i.e. 3, 7,11,15, …,147

Question 11.
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 8
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 9

Question 12.
If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero, (p ≠ q)
Solution:
For an A.P., let a be the first term and d be the common difference.
The sum of first n terms of an A.P. is given by
Sn = [2a + (n – 1)d]
According to the given condition,
Sp = Sq
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 10
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 11
∴ The sum of the first (p + q) terms is zero

Question 13.
If m times the mth term of an A.P. is equal to n times nth term, then show that the (m + n)th term of the A.P. is zero.
Solution:
According to the given condition,
mtm = ntn
∴ m[a + (m – 1)d] = n[a + (n – 1)d]
∴ ma + md(m – 1) = na + nd(n- 1)
∴ ma + m2d – md = na + n2d – nd
∴ ma + m2d – md – na – n2d + nd = 0
∴ (ma – na) + (m2d – n2d) – (md – nd) = 0
∴ a(m – n) + d(m2 – n2) – d(m – n) = 0
∴ a(m – n) + d(m + n) (m – n) – d(m – n) = 0
∴ (m – n)[a + (m + n – 1) d] = 0
∴ [a+ (m + n – 1)d] = 0 …[Dividing both sides by (m – n)]
∴ t(m+n) = 0
∴ The (m + n)th term of the A.P. is zero.

Question 14.
₹ 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 12
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 13

Maharashtra Board 9th Class Maths Part 1 Practice Set 2.5 Solutions Chapter 2 Real Numbers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Practice Set 2.5 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
Find the value.
i. | 15 – 2|
ii. | 4 – 9|
iii. | 7| x | -4|
Solution:
i. |15 – 2| = |13| = 13
ii. |4 – 9| = |-5| = 5
iii. |7| x |- 4| = 7 x 4 = 28

Question 2.
Solve.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 1
Solution:
i. |3x – 5| = 1
∴ 3x – 5 = 1 or 3x – 5 = -1
∴ 3x = 1 + 5 or 3x = -1 + 5
∴ 3x = 6 or 3x = 4
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 2

ii. |7 – 2x| = 5
∴ 7 – 2x = 5 or 7 – 2x = -5
∴ 7 – 5 = 2x or 7 + 5 = 2x
∴ 2x = 2 or 2x = 12
∴ x = \(\frac { 2 }{ 2 }\) or x = \(\frac { 12 }{ 2 }\)
∴ x = 1 or x = 6

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 3
∴ 8 – x = 10 or 8 – x = -10 .. [Multiplying both the sides by 2]
∴ 8 – 10 = x or 8 + 10 = x
∴ x = -2 or x = 18

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 4

Maharashtra Board 10th Class Maths Part 1 Practice Set 5.3 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.3 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
Write sample space ‘S’ and number of sample points n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).

i. One die is rolled,
Event A: Even number on the upper face.
Event B: Odd number on the upper face.
Event C: Prime number on the upper face.

ii. Two dice are rolled simultaneously,
Event A: The sum of the digits on upper faces is a multiple of 6.
Event B: The sum of the digits on the upper faces is minimum 10.
Event C: The same digit on both the upper faces.

iii. Three coins are tossed simultaneously.
Condition for event A: To get at least two heads.
Condition for event B: To get no head.
Condition for event C: To get head on the second coin.

iv. Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A: The number formed is even.
Condition for event B: The number is divisible by 3.
Condition for event C: The number formed is greater than 50.

v. From three men and two women, environment committee of two persons is to be formed.
Condition for event A: There must be at least one woman member.
Condition for event B: One man, one woman committee to be formed.
Condition for event C: There should not be a woman member.

vi. One coin and one die are thrown simultaneously.
Condition for event A: To get head and an odd number.
Condition for event B: To get a head or tail and an even number.
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
Solution:
i. Sample space (S) = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Condition for event A: Even number on the upper face.
∴ A = {2,4,6}
∴ n(A) = 3
Condition for event B: Odd number on the upper face.
∴ B = {1, 3, 5}
∴ n(B) = 3
Condition for event C: Prime number on the upper face.
∴ C = {2, 3, 5}
∴ n(C) = 3

ii. Sample space,
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
∴ n(S) = 36
Condition for event A: The sum of the digits on the upper faces is a multiple of 6.
A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}
∴ n(A) = 6

Condition for event B: The sum of the digits on the upper faces is minimum 10.
B = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 6

Condition for event C: The same digit on both the upper faces.
C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(C) = 6

iii. Sample space,
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ n(S) = 8

Condition for event A: To get at least two heads.
∴ A = {HHT, HTH, THH, HHH}
∴ n(A) = 4

Condition for event B: To get no head.
∴ B = {TTT}
∴ n(B) = 1

Condition for event C: To get head on the second coin.
∴ C = {HHH, HHT, THH, THT}
∴ n(C) = 4

iv. Sample space (S) = {10, 12, 13, 14, 15,
20, 21, 23, 24, 25,
30, 31, 32, 34, 35,
40, 41, 42, 43,
45, 50, 51, 52, 53, 54}
∴ n(S) = 25
Condition for event A: The number formed is even
∴ A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54)
∴ n(A) = 13
Condition for event B: The number formed is divisible by 3.
∴ B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
∴ n(B) = 9
Condition for event C: The number formed is greater than 50.
∴ C = {51,52, 53,54}
∴ n(C) = 4

v. Let the three men be M1, M2, M3 and the two women be W1, W2.
Out of these men and women, a environment committee of two persons is to be formed.
∴ Sample space,
S = {M1M2, M1M3, M1W1, M1W2, M2M3, M2W1, M2W2, M3W1, M3W2, W1W2}
∴ n(S) = 10
Condition for event A: There must be at least one woman member.
∴ A = {M1W1, M1W2, M2W1, M2W2, M3W1, M3W2, W1W2}
∴ n(A) = 7
Condition for event B: One man, one woman committee to be formed.
∴ B = {M1W1, M1W2, M2W1, M2W2, M3W2, M3W2}
∴ n(B) = 6
Condition for event C: There should not be a woman member.
∴ C = {M1M2, M1M3, M2M3}
∴ n(C) = 3

vi. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) = 12
Condition for event A: To get head and an odd number.
∴ A = {(H, 1), (H, 3), (H, 5)}
∴ n(A) = 3
Condition for event B: To get a head or tail and an even number.
∴ B = {(H, 2), (H, 4), (H, 6), (T, 2), (T, 4), (T, 6)}
∴ n(B) = 6
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
The greatest number on the upper face of a die is 6.
∴ Event C is an impossible event.
∴ C = { }
∴ n(C) = 0

Maharashtra Board 10th Class Maths Part 1 Practice Set 5.2 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.2 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
For each of the following experiments write sample space ‘S’ and number of sample Point n(S)
i. One coin and one die are thrown simultaneously.
ii. Two digit numbers are formed using digits 2,3 and 5 without repeating a digit.
Solution:
i. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) =12
ii. Sample space,
S = {23,25,32, 35, 52, 53}
∴ n(S) = 6

Question 2.
The arrow is rotated and it stops randomly on the disc. Find out on which colour it may stop.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 1
Solution:
There are total six colours on the disc.
Sample space,
S = {Red, Orange, Yellow, Blue, Green, Purple}
∴ n(S) = 6
∴ Arrow may stop on any one of the six colours.

Question 3.
In the month of March 2019, find the days on which the date is a multiple of 5. (see the given page of the calendar).
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 2
Solution:
Dates which are multiple of 5:
5,10, 15,20,25,30
∴ S = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}
∴ n(S) = 6
∴ The days on which the date will be a multiple of 5 are Tuesday, Sunday, Friday, Wednesday, Monday and Saturday.

Question 4.
Form a ‘Road safety committee’ of two, from 2 boys (B1 B2) and 2 girls (G1, G2). Complete the following activity to write the sample space.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 3

Question 1.
Sample Space

  • The set of all possible outcomes of a random experiment is called sample space.
  • It is denoted by ‘S’ or ‘Ω’ (omega).
  • Each element of a sample space is called a sample point.
  • The number of elements in the set S is denoted by n(S).
  • If n(S) is finite, then the sample space is called a finite sample space.

Some examples of finite sample space. (Textbook pg. no, 117)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 4 Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 5

Maharashtra Board 10th Class Maths Part 1 Practice Set 5.1 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.1 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
How many possibilities are there in each of the following?
i. Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation. Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.
ii. Any day of a week is to be selected randomly.
iii. Select one card from the pack of 52 cards.
iv. One number from 10 to 20 is written on each card. Select one card randomly.
Solution:
i. Here, 8 sites of Maharashtra are given.
∴ There are 8 possibilities in a random experiment of visiting a site out of 8 sites in Maharashtra.

ii. There are 7 days in a week.
∴ There are 7 possibilities in a random experiment of selecting a day of the week.

iii. There are 52 cards in a pack of cards.
∴ There are 52 possibilities in a random experiment of selecting one card from the pack of 52 cards.

iv. There are 11 cards numbered from 10 to 20.
∴ There are 11 possibilities in a random experiment of selecting one card from the given set of cards.

Question 1.
In which of the following experiments possibility of expected outcome is more? (Textbook pg, no. 116)
i. Getting 1 on the upper face when a die is thrown.
ii. Getting head by tossing a coin.
Solution:
i. On a die there are 6 numbers.
∴ There are 6 possibilities of getting any one number from 1 to 6 on the upper face i.e. \(\frac { 1 }{ 6 } \) is the possibility.

ii. There are two possibilities (H or T) on tossing a coin i.e. \(\frac { 1 }{ 2 } \) possibility.
∴ In the second experiment, the possibility of expected outcome is more.

Question 2.
Throw a die, once. What are the different possibilities of getting dots on the upper face? (Textbook pg. no. 114)
Answer:
There are six different possibilities of getting dots on the upper face. They are
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1

Maharashtra Board 10th Class Maths Part 1 Practice Set 4.4 Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Practice Set 4.4 Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Question 1.
Market value of a share is ₹ 200. If the brokerage rate is 0.3% then find the purchase value of the share.
Solution:
Here, MV = ₹ 200, Brokerage = 0.3%
Brokerage = 0.3% of MV
= \(\frac { 0.3 }{ 100 } \) × 200
= ₹ 0.6
∴ Purchase value of the share = MV + Brokerage
= 200 + 0.6
= ₹ 200.60
∴ Purchase value of the share is ₹ 200.60.

Question 2.
A share is sold for the market value of ₹ 1000. Brokerage is paid at the rate of 0.1%. What is the amount received after the sale?
Solution:
Here, MV = ₹ 1000, Brokerage = 0.1%
∴ Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 1000
∴ Brokerage = ₹ 1
∴ Selling value of the share = MV – Brokerage
= 1000 – 1
= ₹ 999
∴ Amount received after the sale is ₹ 999.

Question 3.
Fill in the blanks given in the contract note of sale-purchase of shares.
(B – buy S – sell)
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 1
Solution:
For buying shares:
Here, Number of shares = 100,
MV of one share = ₹ 45
∴ Total value = 100 × 45
= ₹ 4500
Brokerage= 0.2% of total value 0.2
= \(\frac { 0.2 }{ 100 } \) × 4500
CGST = 9% of brokerage
= \(\frac { 9 }{ 100 } \) × 9 = ₹ 0.81
But, SGST = CGST
∴ SGST = ₹ 0.81
∴ Purchase value of shares
= Total value + Brokerage
= 4500 + 9 + 0.81 + 0.81
= ₹ 4510.62

ii. For selling shares:
Here, Number of shares = 75,
MV of one share = ₹ 200
∴ Total value = 75 × 200
= ₹ 15000
Brokerage = 0.2% of total value
= \(\frac { 0.2 }{ 100 } \) × 15000
= ₹ 30
CGST = 9% of brokerage
= \(\frac { 9 }{ 100 } \) × 30 = ₹ 2.70
But, SGST = CGST
∴ SGST = ₹ 2.70
∴ Selling value of shares = Total value – (Brokerage + CGST + SGST)
= 15000 – (30 + 2.70 + 2.70)
= 15000 – 35.40
= ₹ 14964.60
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 2

Question 4.
Smt. Desai sold shares of face value ₹ 100 when the market value was ₹ 50 and received ₹ 4988.20. She paid brokerage 0.2% and GST on brokerage 18%, then how many shares did she sell?
Solution:
Here, face value of share = ₹ 100,
MV = ₹ 50,
Selling price of shares = ₹ 4988.20,
Rate of brokerage = 0.2%, Rate of GST = 18%
Brokerage = 0.2% of MV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 3

Question 5.
Mr. D’souza purchased 200 shares of FV ₹ 50 at a premium of ₹ 100. He received 50% dividend on the shares. After receiving the dividend he sold 100 shares at a discount of ₹ 10 and remaining shares were sold at a premium of ₹ 75. For each trade he paid the brokerage of ₹ 20. Find whether Mr. D’souza gained or incurred a loss? By how much?
Solution:
For purchasing shares:
Here, FV = ₹ 50, Number of shares = 200,
premium = ₹ 100
MV of 1 share = FV + premium
= 50 + 100
= ₹ 150
∴ MV of 200 shares = 200 × 150 = ₹ 30,000
∴ Mr. D’souza invested amount
= MV of 200 shares + brokerage
= 30,000 + 20
= ₹ 30,020
For selling shares:
Rate of dividend = 50 %, FV = ₹ 50,
brokerage = ₹ 20
Number of shares = 200
Dividend per share = 50% of FV
= \(\frac { 50 }{ 100 } \) × 50
= ₹ 25
∴ Dividend of 200 shares = 200 × 25 = ₹ 5,000
Now, 100 shares are sold at a discount of ₹ 10.
∴ Selling price of 1 share = FV – discount
= 50 – 10
= ₹ 40
∴ Selling price of 100 shares = 100 × 40
= ₹ 4000
∴ Amount obtained by selling 100 shares
= selling price – brokerage
= 4000 – 20
= ₹ 3980
Also, remaining 100 shares are sold at premium of ₹ 75.
∴ selling price of 1 share = FV + premium
= 50 + 75
= ₹ 125
∴ selling price of 100 shares = 100 × 125
= ₹ 12,500
∴ Amount obtained by selling 100 shares
= selling price – brokerage
= 12,500 – 20
= ₹ 12,480
∴ Mr D’souza income = 5000 + 3980 + 12480
= ₹ 21460
Now, Mr D’souza invested amount > income
∴ Mr D’souza incurred a loss.
∴ Loss = amount invested – income
= 30020 – 21460
= ₹ 8560
∴ Mr. D’souza incurred a loss of ₹ 8560.

Question 1.
Nalinitai invested ₹ 6024 in the shares of FV ₹ 10 when the Market Value was ₹ 60. She sold all the shares at MV of ₹ 50 after taking 60% dividend. She paid 0.4% brokerage at each stage of transactions. What was the total gain or loss in this transaction? (Textbook pg. no. 106)
Solution:
Rate of GST is not given in the example, so it is not considered.
For Purchased Shares:
FV = ₹ 10, MV = ₹ 60
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 4

Question 2.
In the above example if GST was paid at 18% on brokerage, then the loss is ₹ 451.92. Verify whether you get the same answer. (Textbook pg, no. 107)
Solution:
For Purchased Shares:
FV = ₹ 10, MV = ₹ 60, sum invested = ₹ 6024, brokerage = 0.4 %, GST = 18%
Brokerage per share = \(\frac { 0.4 }{ 100 } \) × 60 = ₹ 0.24 100
GST per share = \(\frac { 18 }{ 100 } \) × 0.24 = ₹ 0.0432
∴ Cost of one share = 60 + 0.24 + 0.0432
= ₹ 60.2832
∴ Cost of 100 shares = 100 × 60.2832 = ₹ 6028.32
For sold shares:
FV = ₹ 10, MV = ₹ 50, brokerage = 0.4 %,
GST = 18%, Number of shares = 100
Brokerage per share = \(\frac { 0.4 }{ 100 } \) × 50 = ₹ 0.20
GST per share = \(\frac { 18 }{ 100 } \) × 0.20 = ₹ 0.036
Selling price per share = 50 – 0.2 – 0.036
= ₹ 49.764
Selling price of 100 shares = 100 × 49.764
= ₹ 4976.4
Dividend received 60 %
∴ Dividend per share = \(\frac { 60 }{ 100 } \) × 10 = ₹ 6
Dividend on 100 shares = 6 × 100 = ₹ 600
∴ Nalinitai’s income = 4976.4 + 600 = ₹ 5576.4
∴ Cost of 100 shares = ₹ 6028.32
∴ Loss = 6028.32 – 5576.4 = ₹ 451.92
∴ Nalinitai’s loss is ₹ 451.92.

Maharashtra Board 10th Class Maths Part 1 Problem Set 4A Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 4A Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Problem Set 4A Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Financial Planning Class 10 Problem Set 4a Question 1.
Write the correct alternative for each of the following.

i. Rate of GST on essential commodities is ______
(A) 5%
(B) 12%
(C) 0%
(D) 18%
Answer:
(C)

ii. The tax levied by the central government for trading within state is ______
(A) IGST
(B) CGST
(C) SGST
(D) UTGST
Answer:
(B)

iii. GST system was introduced in our country from ______
(A) 31st March 2017
(B) 1st April 2017
(C) 1st January 2017
(D) 1st July 2017
Answer:
(D)

iv. The rate of GST on stainless steel utensils is 18%, then the rate of state
GST is ______
(A) 18%
(B) 9%
(C) 36%
(D) 0.9%
Answer:
(B)

v. In the format of GSTIN there are ______ alpha-numerals.
(A) 15
(B) 10
(C) 16
(D) 9
Answer:
(A)

vi. When a registered dealer sells goods to another registered dealer under GST, then this trading is termed as ______
(A) BB
(B) B2B
(C) BC
(D) B2C
Answer:
(B)

10th Class Algebra Problem Set 4a Question 2.
A dealer has given 10% discount on a showpiece of ₹ 25,000. GST of 28% was charged on the discounted price. Find the total amount shown in the tax invoice. What is the amount of CGST and SGST.
Solution:
Printed price of showpiece = ₹ 25,000,
Rate of discount = 10%
∴ Amount of discount = 10% of printed price
= \(\frac { 10 }{ 100 } \) × 25000
= ₹ 2500
∴ Taxable value
= Printed price – discount
= 25,000 – 2500 = ₹ 22,500
Rate of GST = 28%
∴ Rate of CGST = 14% and
Rate of SGST = 14%
CGST = 14% of taxable value
= \(\frac { 14 }{ 100 } \) × 22500
= ₹ 3150
∴ CGST = SGST = ₹ 3150
∴ Total amount of tax invoice
= Taxable value + CGST + SGST
= 22500 + 3150 + 3150
= ₹ 28,800
∴ The total amount shown in the tax invoice is ₹ 28,800, and the amount of CGST and SGST is ₹ 3150 each.

Financial Planning Problem Set 4a Question 3.
A ready-made garment shopkeeper gives 5% discount on the dress of ₹ 1000 and charges 5% GST on the remaining amount, then what is the purchase price of the dress for the customer?
Solution:
Printed price of dress = ₹ 1000
Rate of discount = 5%
∴ Amount of discount = 5% of printed price
= \(\frac { 5 }{ 100 } \) × 1000
= ₹ 50
∴ Taxable value = Printed price – discount
= 1000 – 50
= ₹ 950
Rate of GST = 5%
∴ GST = 5% of taxable value
= \(\frac { 5 }{ 100 } \) × 950
∴ GST = ₹ 47.5
Purchase price of the dress
= Taxable value + GST
= 950 + 47.5 = ₹ 997.50
∴ Purchase price of the dress for the customer is ₹ 997.50.

Question 4.
A trader from Surat, Gujarat sold cotton clothes to a trader in Rajkot, Gujarat. The taxable value of cotton clothes is ₹ 2.5 lacs. What is the amount of GST at 5% paid by the trader in Rajkot?
Solution:
Taxable amount of cotton clothes = ₹ 2.5 lacs,
Rate of GST = 5%
GST = 5% of taxable amount
= \(\frac { 5 }{ 100 } \) × 2,50,000
= ₹ 12500
∴ Trader of Rajkot has to pay GST of ₹ 12,500.

Question 5.
Smt. Malhotra purchased solar panels for the taxable value of ₹ 85,000. She sold them for ₹ 90,000. The rate of GST is 5%. Find the ITC of Smt. Malhotra. What is the amount of GST payable by her?
Solution:
Output tax = 5% of 90000
= \(\frac { 5 }{ 100 } \) × 90000
= ₹ 4500
Input tax = 5% of 85000
= \(\frac { 5 }{ 100 } \) × 85000
= ₹ 4250
ITC = ₹ 4250.
∴ GST payable = Output tax – ITC
= 4500 – 4250
GST payable = ₹ 250
∴ ITC of Smt. Malhotra is ₹ 4250 and amount of GST payable by her is ₹ 250.

Question 6.
A company provided Z-security services for the taxable value of ₹ 64,500. Rate of GST is 18%. Company had paid GST of ₹ 1550 for laundry services and uniforms etc. What is the amount of ITC (input Tax Credit)? Find the amount of CGST and SGST payable by the company.
Solution:
Output tax = 18% of 64500
= \(\frac { 18 }{ 100 } \) × 64500
= ₹ 11610
Input tax = ₹ 1550
GST payable = Output tax – ITC
= 11610 – 1550
∴ GST payable = ₹ 10060
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 1
∴ Amount of ITC is ₹ 1550. Amount of CGST and SGST payable by the company is ₹ 5030 each.

Question 7.
A dealer supplied Walky-Talky set of ₹ 84,000 (with GST) to police control room. Rate of GST is 12%. Find the amount of state and central GST charged by the dealer. Also find the taxable value of the set.
Solution:
Let the amount of GST be ₹ x.
Price of walky talky with GST = ₹ 84,000
Taxable value of walky talky = ₹ (84,000 – x)
Now, GST = 12% of taxable value
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 2
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 3
∴ Amount of state and central GST charged by the dealer is ₹ 4,500 each. Taxable value of the set is ₹ 75,000.

Question 8.
A wholesaler purchased electric goods for the taxable amount of ₹ 1,50,000. He sold it to the retailer for the taxable amount of ₹ 1,80,000. Retailer sold it to the customer for the taxable amount of ₹ 2,20,000. Rate of GST is 18%. Show the computation of GST in tax invoices of sales. Also find the payable CGST and payable SGST for wholesaler and retailer.
Solution:
For Wholesaler:
Output tax = 18% of ₹ 1,80,000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 4
Statement of GST payable at each stage of trading:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 5

Question 9.
Anna Patil (Thane, Maharashtra) supplied vacuum cleaner to a shopkeeper in Vasai (Mumbai) for the taxable value of ₹ 14,000, and GST rate of 28% . Shopkeeper sold it to the customer at the same GST rate for ₹ 16,800 (taxable value). Find the following:
i. Amount of CGST and SGST shown in the tax invoice issued by Anna Patil.
ii. Amount of CGST and SGST charged by the shopkeeper in Vasai.
iii. What is the CGST and SGST payable by shopkeeper in Vasai at the time of filing the return.
Solution:
i. For Anna Patil:
Output tax = 28% of 14,000
= \(\frac { 18 }{ 100 } \) × 14000
= ₹ 3920
∴ CGST = SGST = \(\frac { GST }{ 2 } \)
= \(\frac { 3920 }{ 2 } \)
= ₹ 1960
∴ Amount of CGST and SGST shown in the tax invoice issued by Anna Patil is ₹ 1960 each.

ii. For Shopkeeper in Vasai:
Output tax = 28% of 16,800
= \(\frac { 28 }{ 100 } \) × 16,800
= ₹ 4704
∴ CGST = SGST = \(\frac { GST }{ 2 } \)
= \(\frac { 4704 }{ 2 } \)
= ₹ 2352
∴ Amount of CGST and SGST charged by the shopkeeper in Vasai is ₹ 2352 each.

iii. ITC = ₹ 3920
GST payable by shopkeeper in Vasai
= Output tax – ITC
= 4704 – 3920
= ₹ 784
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 6
∴ CGST and SGST payable by shopkeeper in Vasai at the time of filing the return is ₹ 392 each.

Question 10.
For the given trading chain prepare the tax invoice I, II, III. GST at the rate of 12% was charged for the article supplied.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 7
i. Prepare the statement of GST payable under each head by the wholesaler, distributor and retailer at the time of filing the return to the government.
ii. At the end what amount is paid by the consumer?
iii. Write which of the invoices issued are B2B and B2C.
Solution:
i. For wholesaler:
Output tax = 12% of 5000
= \(\frac { 12 }{ 100 } \) × 5000 = ₹ 600
For Distributor:
Output Tax = 12% of 6000
= \(\frac { 12 }{ 100 } \) × 6000 = ₹ 720
ITC = ₹ 600
∴ GST payable = Output tax – ITC
= 720 – 600
= ₹ 120
For Retailer:
Output tax = 12 % of 6500
= \(\frac { 12 }{ 100 } \) × 6500 = ₹ 780
ITC = ₹ 720
∴ GST payable = Output tax – ITC
= 780 – 720 = ₹ 60
Statement of GST payable at each stage of trading:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 8

ii. ITC for consumer = ₹ 780
∴ Amount paid by consumer
= taxable value + ITC
= 6500 + 780
= ₹ 7280
∴ Amount paid by the consumer is ₹ 7280.

iii. B2B = Wholesaler to Distributor
B2B = Distributor to Retailer
B2C = Retailer to Consumer