Class 10

Balbharti Maharashtra State Board Class 10 English Solutions Unit 1.6 The Alchemy of Nature Notes, Textbook Exercise Important Questions and Answers.

Class 10 English Chapter 1.6 Question Answer Maharashtra Board

The Alchemy of Nature Poem 10th Std Question Answer

Alchemy Of Nature Class 10 Question 1.
Rearrange the good qualities in each set, so that the first letter of each of the words should make a meaningful word. Join the sets and get a message.

Set 1: U nderstanding/A daptable/Tolerant/N eat/Encouraging Resourceful.
– The word is ……………………………………………………
Set 2: Selfless/Inspiring
– The word is ……………………………………………………
Set 3: Youthful/Modest
– The word is ……………………………………………………
Set 4: Affectionate/Compassionate/Empathetic/Earnest/Honest/Reliable/Trustworthy
– The word is ……………………………………………………
– The message is …………………………………………………… …………………………………………………… ……………………………………………………
Answer:
NATURE
IS
MY
TEACHER
The message Is: NATURE IS MY TEACHER.

Alchemy Of Nature Class 10 Questions And Answers Question 2.
Various aspects of Nature have special features that make them differ from one another.
For example, Birds :- appearance, shape, colour, size, food habits, habitat, sound etc.
Write such special features of each of the following.
Land …………………………………………………… …………………………………………………… ……………………………………………………
Water …………………………………………………… …………………………………………………… ……………………………………………………
Trees …………………………………………………… …………………………………………………… ……………………………………………………
Animals …………………………………………………… …………………………………………………… ……………………………………………………
Insects …………………………………………………… …………………………………………………… ……………………………………………………
Answer:
Land: geographical features, soil colour, terrain, fertility, chemical composition, crops grown.

Water: width and length of various water bodies, chemical composition (sweet water, salt water), colours (according to sand beds), rocks, coral reefs, variety of sea creatures.

Trees: height and shape of leaves, changing (or unchanging) colour during seasons, fruits, flowers, medicinal products, types of soil, climate and habitat required for their existence.

AnImals: shapes, sizes, colours, different habitats, kinds: wild or domestic, place in the food chain.

Insects: shapes, sizes, colours, number of legs, different habitats, soundš, carriers of dIseases.

The Alchemy Of Nature Question Answer Question 3.
Make a list of living creatures in the alphabetical order. You can write more than one beginning with the same letter.
A …………………………………………………… B ……………………………………………………
C …………………………………………………… D ……………………………………………………
E …………………………………………………… F ……………………………………………………
G …………………………………………………… H ……………………………………………………
I …………………………………………………… J ……………………………………………………
K …………………………………………………… L ……………………………………………………
M …………………………………………………… N ……………………………………………………
O …………………………………………………… P ……………………………………………………
Q …………………………………………………… R ……………………………………………………
S …………………………………………………… T ……………………………………………………
U …………………………………………………… V ……………………………………………………
W …………………………………………………… Y ……………………………………………………
Z ……………………………………………………
Answer:
A -alligator, antelope, ant, etc., B ……………. Z.

The Alchemy of Nature Class 10 English Workshop Questions and Answers Maharashtra Board

The Alchemy Of Nature Questions And Answers Question 1.
What things in nature teach us the following :
(a) Nothing is impossible to achieve ……………………………………………………
(b) Problems are not permanent ……………………………………………………
(c) Be humble and adjust ……………………………………………………
(d) Make the best use of time and opportunity ……………………………………………………
(e) Be persistent ……………………………………………………
(f) Many hands make work light ……………………………………………………
(g) Delicate structures are not a sign of weakness ……………………………………………………
Answer:
(a) Nothing is impossible to achieve: ants small bits of grass peeping from cracks In the concrete
(b) Problems are not permanent: trees that are bare in winter
(c) Be humble and adjust: water
(d) Make the best use of time and opportunity: flowers
(e) Be persistent: ants waler
(f) Many bands make light work: ants
(g) Delicate structures are not a sign of weakness: picr_webs

The Alchemy Of Nature English Workshop Question 2.
Read the questions from the lesson. What do they imply?
(a) Are you listening? …………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………………………

(b) What if we too lived our lives, however short, to its fullest? ……………………………………………………
………………………………………………………………………………………………………………………………………………………………

(c) What if we too are consistent, organised, focused . . . ? ……………………………………………………
………………………………………………………………………………………………………………………………………………………………
Answer:
(a) It implies that one must listen.
(b) It Implies that we too should live our lives to… the fullest, however short they may be.
(c) It Implies that we too could do wonders If we were consistent, organised focussed

Question 3.
Go through the lesson again and complete the flow-chart that highlights the life of a ‘hibiscus’ flower.

Answer:

Alchemy Of Nature Class 10 English Workshop Question 4.
Fill in the web.

Answer:

English Workshop 10th The Alchemy Of Nature Question 5.
The writer explains the contrasting features of ‘water’ and ‘rock’ in the lesson. Write all the features of both water and rock in the given table.

Water	Rock
1.	1.
2.	2.
3.	3.
4.	4.

Answer:

Water	Rock
1. gentle	1. hard
2. persistent	2. humble
3. persevering	3. yielding
4. determined	4. adaptable

Alchemy Of Nature Class 10 Solutions Question 6.
The writer has very positively described the different things in nature. Discuss with your partner the special features of each one of them. Add on the list.

Part of Nature	Special feature	Value learned
1. Rainbow	……………………	……………………
2. Caterpillar	……………………	……………………
3. ……………………	……………………	……………………
4. ……………………	……………………	……………………
5. ……………………	……………………	……………………
6. ……………………	……………………	……………………
7. ……………………	……………………	……………………

Answer:

Part of nature	Special feature	Value learnt
1. rainbow	cheerfulness; acceptance.	Even when there are problems in the surroundings, we must be cheerful and spread colour and happiness.
2. caterpillar	patience; acceptance.	There is a bright and beautiful future ahead.
3. hibiscus flower	optimism; cheerfulness	However short life may be, we must live it to the fullest.
4. rocks	obedience to nature; adjustment; humility	Obey the commands of nature; adjust to the situation; be humble
5. bits of grass	optimism; perseverance.	However impossible things may look, there is an opening.
6. bare tree	optimism, faith; conviction	However difficult things are in the present, it will not remain so forever. With conviction we should remind ourselves that this too will pass.
7. water	perseverance; determination; humility.	(i) Even colossal problems can be surmounted if we persist. (ii) Learn to adapt to others without any hint of ego.

Alchemy Of Nature Questions And Answers Question 7.
Think and answer in your own words.
(a) How should you deal with difficulties and problems?
Answer:
When coming across problems In lilt. I turn towards nature for inspiration. I try to understand how the different elements in nature deal with their difficulties and try to solve my own problems in the same way.

(b) ‘An oyster turns a grain of sand into a pearl.’ What can we learn from this example?
Answer:
We learn that there Is a mysterious power or magic In nature that can change things dramatically.

(c) How does nature succeed in its ‘Alchemy’? What can it turn a small person into?
Answer:
Nature succeeds in Its ‘Alchemy’ by changing things in a mysterious way. It can turn a small person Into anything he/she wishes one touches nature and becomes gold oneself.

(d) Which two aspects of nature teach us to accept change and adjust according to the situation?
Answer:
With gentle humility, water changes its form according to the dictates of the sun and the wind. The bare trees wait patiently during the winter months for the arrival of spring, when they get Wesh green leaves. These two aspccts of nature teach us to accept change and adjust according to the situation.

(e) Why does the writer begin by quoting the lines from William Blake’s poem (Auguries of Innocence)?
Answer:
These lines show that it we pause to relleci, there Is much beauty In nature and plenty that we cari learn from it. The write-up expands on the sanie idea, thus reflecting the philosophy of the quoted lines from William Blake.

Question 8.
(A) Pick out from the lesson 10 examples of each.
Concrete Nouns (that you can touch or see)
For example, sand
(1) …………………………………………
(2) …………………………………………
(3) …………………………………………
(4) …………………………………………
(5) …………………………………………
(6) …………………………………………
(7) …………………………………………
(8) …………………………………………
(9) …………………………………………
(10) …………………………………………
Answer:
(1) bird,
(2) ant,
(3) oyster.
(4) pearl.
(5) caterpillar,
(6) butterfly
(7) flower,
(8) ocean,
(9) rock,
(10) water.

Abstract Nouns (that which you cannot touch or see)
For example, infinity
(1) …………………………………………
(2) …………………………………………
(3) …………………………………………
(4) …………………………………………
(5) …………………………………………
(6) …………………………………………
(7) …………………………………………
(8) …………………………………………
(9) …………………………………………
(10) …………………………………………
Answer:
(1) majesty.
(2) conviction.
(3) perseverance,
(4) passion
(5) infinity.
(6) Imagination,
(7) joy,
(8) significance,
(9) experience,
(10) difference.

(B) Underline the verbs in the sentences below and say whether they are Transitive (needing an Object) or Intransitive (need not have an Object).
(1) One can witness and experience the beauty of Heaven.
(2) It leaves me in complete awe.
(3) Nature soothes and nurtures.
(4) It withers completely.
(5) The flower comes to life only for a day.
(6) A rainbow colours the entire sky.
(7) It smiles and dances.
Answer:
(1) One can witness and experience the beauty of Heaven. (T)
(2) It leaves mc In complete awe. (T)
(3) Nature soothes and nurtures. (I)
(4) It withers completely. (I)
(5) The flower comes to life one for a day. (I)
(6) A rainbow colours the entire sky. (T)
(7) It smiles and dances. (I)

Question 9.
(A) Compose about 8 to 10 sets of imaginary dialogues between a bird, a tree and its fruit regarding the effects of environmental changes. Write it in your notebook.
Answer:
Conversation between a bird, a tree and its fruit:
Conversation 1:
Bird: Hi there, tree! How are you doing?
Tree: Not too well. I can’t breathe. There’s too much of dust and smoke here.
Bird: What, here too? I’ve come here to escape just that!
Fruit: Wrong place, birdie. Look at me do I look healthy? I’m not growing. Too many insecticides have been sprayed on me.
Bird: Oh, oh. Maybe I should leave this place too.

Conversation 2:
Tree: Oh, no! Something is happening! My roots are getting loose!
Bird: That’s called erosion. Soil erosion.
Tree: What happens next?
Bird: The next time it rains heavily-boom! Down you will go! And away I will fly.
Tree: Can’t someone help me?
Fruit: What about me?

(B) Prepare a Fact file of any of the following plants/trees, using the points given.
[coconut / neem / basil / cactus / apple]

• Name of Plant/Tree …………………………………………
• Scientific name …………………………………………
• Region and climate …………………………………………
• Features …………………………………………
• Growth …………………………………………
• Size, shape and colour …………………………………………
• Uses …………………………………………
• Any special feature …………………………………………

Answer:
Profile of the Coconut Tree and Fruit
(1) Names: English name – Coconut Sanskrit name – Narikela Hindi name – Nariyal.
(2) Scientific name: cocos nucifera belongs to the family ‘Palmae’ or the palm family (also known as Arecaceae).
(3) Region and Climate: Tropical and sub-tropical coastal regions, especially near sea beaches.
(4) Features:

• Fruit: has a thick fibrous coir over the hard shell inside the kernel colourless liquid;
• leaves: feather-shaped and split into lots of leaflets.

(5) Growth – Size and shape: Coconut trees can: grow from 15 to 30 metres in height in plantations. Coconut fruits are oval in shape. The trunk of the coconut tree is ringed with scars where old leaves have fallen. The top of the trunk is crowned with a rosette of leaves. The leaves can grow up to 7 feet long and can have 250 leaflets.
(6) Uses:

• Coir and leaves: matting, thatching and weaving.
• Hard outer shell about 10 to 15 inches in length used to make articles such as spoons, eating utensils, charcoal, etc.
• Inside of the shell: lined with a white edible layer called the meat used for cooking, or extraction of oil which is used in making soaps or cosmetics also to make chemical, industrial and medicinal products contains coconut water which is very nutritious.
• Husk and leaves: used as material to make a variety of products for furnishing and decorating,

(7) Any special feature: Known in India as ‘kalpavriksha’ or the ‘tree of heaven’ because of its many uses the term coconut is derived from the 16th century Portuguese and Spanish, meaning ‘grinning face’, from the three small holes on the coconut shell that resemble human facial features.

Question 10.
‘Impossible’ itself says ‘I M possible’. Do you agree? Justify your answer by citing something that you have experienced or heard from someone.
Answer:
Yes, I agree. It is possible to do the most , difficult of things providing we have the will and conviction to do so. The example that comes to my mind as justification is that of Sudha Chandran, the j dancer. Though her leg was amputated below the knee, with great grit and determination she restarted dancing, and went on to become a famous dancer ! and actress. This shows that one can do things which seem to be impossible.

Question 11.
You have an environment protection week celebration in your school. You have invited an environmentalist. Your friend interviews him/her about how to save the environment. Frame suitable Interview Questions.
Answer:
Jai interviews Mr. Ali, an environmentalist:
Jai: Good morning, Sir. Welcome to our school. I would like to ask you a few questions for a write-up in our school magazine. My first question: What is your opinion about the concept of having am ‘environment protection week’ celebration?

Mr. Ali: I think it’s a wonderful idea, though I would not call it a ‘celebration’.

Jai: Why is that, sir?

Mr. Ali: Well, you have a celebration when you are happy about something. In our town, the protection of the environment is so poor that I, as an environmentalist, am not at all happy about it.

Jai: Can you give us some tips to protect the environment?

Mr. Ali: Certainly, my boy. First of all, we should grow more trees wherever possible. It should be made mandatory for every factory, office, residential building, etc. to plant trees before starting construction.

Jai: What about the menace of plastic, sir?

Mr. Ali: We cannot eliminate plastic completely however, the thickness of plastic bags and the methods of disposal should be made clear to all. Air, water and noise pollution must be dealt with too.

Jai: Please expand on that, sir.

Mr. Ali: Well, industrial and vehicular pollution must be controlled water bodies must be kept clean. Those who break rules must be penalized. Loudspeakers must be banned during the night hours. Oh, there are lots of things to be done, lots of things.

Jai: Thank you sir, for giving some of your precious time for this interview.

Question 12.
Write a News Report on the ‘Environment Day’ celebrated in your school.
Answer:
Environment Day Celebrations
Nagpur, June 7: ‘World Environment Day’ was celebrated in New Era School with great fanfare on June 5. The main purpose of the celebration was j to spread awareness about the need to protect the: environment and the ways to do it.

The day was flagged off by a tree plantation drive in the locality. Three hundred quick-growing trees, which do not need much water on a daily basis, were planted near the school wall and in the surrounding area. An eminent environmentalist, Mr. A.T. Ali, spoke on the ways to protect the environment. He also judged the ‘Posters and Photographs’ exhibition “and gave away prizes for the best entries. Environmentally- friendly articles, like disposable plates and cups made from bamboo and banana stem, bags made from leaf waste, etc. were on sale. Students gave power-point ) presentations on the threats to the environment. Last i but not least, was the spirited debate on the topic ‘Man: The worst enemy of the Environment’.

All in all, the day was a great success, and has certainly made a difference to the way we view our environment.

– Josh Matthew
New Era School.

Question 13.
Just For Laughs! Enjoy!
Divide the class into two groups. On 12 to 15 slips of paper, Group A writes 12 to 15 conditional clauses beginning with ‘If’.
(For example, If I work very hard, ………………………………………………)
Group B writes 12 to 15 main clauses.)
(For example, I would/shall have a pizza.)
Now, one student from Group ‘A’ reads the first conditional clause (possibility) and one student from Group ‘B’ reads the first main clause. It forms crazy sentences, just for laughter and fun. ENJOY!

Question 14.
Pick out the statements that are false and write them correctly:
(1) On the beach. the author found rocks carved and sculpted by the wInd.
(2) The hibiscus flower smiles with the sun and dances with the wind.
(3) Rocks take the shape that (he water commands.
(4) Our problems are big and so are we.
Answer:
Statements (1) and (4) are false. The corrected statements are:
(1) On the beach, the author found rocks carved and sculpted by the water.
(4) Our problems are very big, and we are very small.

Question 15.
ExplaIn how the hibiscus flower makes the most of Its short life span.
Answer:
The hibiscus flower smiles with the sun and dances with the wind. The flower comes to life only for a day yet It makes the most of the day by living its short life in full splendour, with big, bright and tender blooms.

Question 16.
Complete the flowchart that highlights the life of a hibiscus flower:
Answer:
The life of a hibiscus flower

Question 17.
Guess the meaning of the ‘splendour’.
Answer:
Splendour – great beauty which attracts admiration and attention.

Question 18.
Choose the correct ‘not only … but also …….’ form of the sentence:
Natures soothes and nurtures.
(a) Not only nature soothes but nurtures also.
(b) Nature soothes not only but also nurtures.
(c) Nature soothes but also nurtures not only.
(d) Nature not only soothes but also nurtures.
Answer:
(d) Nature not only soothes but also nurtures.

Question 19.
By evening It falls and becomes one with the earth again. (Rewrite using the ‘-tng form of the underlined word.)
Answer:
By evening it falls, becoming one with the earth agaIn.

Question 20.
Who Is stronger – water or rocks? Justify your answer.
Answer:
I think water is stronger. It wears down tue hard rock by its gentle patience, persistence and perseverance

Question 21.
State whether the ¿ilowlng statements are True or False: (The answers are given directly and underlined.)
Answer:
(i) The spider’s webs are delicate as well as weak. ‘ False
(Ii) The teamwork and perseverance of ants were Impressive. True

Question 22.
What teaches us that hard times do not last forever? How?
Answer:
The following things teach us that hard times do not last forever:
(i) bits of grass peeping through small cracks in a concrete pavement and
(ii) the green leaves on a tree In spring The grass had been nearly destroyed by the concrete but had come to life again. The tree had been bare all through the cold winter months, before regaining Its former green majesty.

Question 23.
Choose the sentence In the Past Perfect Tense from the sentences given below:
(a) The ants had organised themselves around the fly.
(b) The ants organized themselves around the fly.
(c) The ants hat’e organized themselves around the Jly.
Answer:
(a) The ants had organised themselves around the fly.

Question 24.
A rainbow colours the entire sky. (Begin the sentence with The entire sky …‘)
Answer:
The entire sky is coloured by a rainbow.

Question 25.
However Impossible things may look, there Is always an opening. (Rewrite beginning with ‘Even if..’.)
Answer:
Even if things, look Impossible. there is always an openIng.

Question 26.
Spider webs are delicate, yet very strong. (Rewrite beginning with ‘Although ….)
Answer:
Although spider webs arc delicate, they are very strong.

Question 27.
(1) PIck out an Infinitive from the lesson and use It In your own sentence.
(2) Punctuate what If we too had lived our lives however short to its fullest
(3) Find out two hidden words from the given word: approaches
(4) Make a meaningful sentence by using the given phrase: set In
(5) Spot the error and rewrite the correct sentence: When I do. It leave me In complete awe.
(6) IdentIfy the type of sentence: However dimcult things are right now, it will not remain so forever.
(7) WrIte the following words In alphabetical order: perseverance, withers, majesty, ‘oysters.
(8) Pick ont the verb from the following that can form both the present and past participle by doubling the last letter, and write the forms: hit. admtt. turn, feed
Answer:
(1) I went to the beach because I wanted to see the ship.
(2) What if we too had lived our lives, however short, to Its fullest?
(3) approaches — approach, perch (cheap, peach, preach)
(4) Many people try to avoid being In Mumbal when the summer sets in.
(5) When I do, It leaves me In complete awe,
(6) Assertive sentence (negative)
(7) majesty, oysters, perseverance, withers
(8) admit — admitting, admitted

Question 28.
Write 4 words related to things in nature.
Answer:
(1) (i) We must plant more trees to attract rain.
(ii) The new plant started production last year.
(2) We had gone to a rocky beach.
(3) We are soothed and nurtured by nature.
(4) Nature: sand, flower, tree, sun, ocean, rock. water, grass

Question 29.
(1) Use the following word as a verb and a noun in two separate sentences: touch
(2) Analyse the sentence: We saw small bits of grass peeping through the small cracks in a concrete pavement.
Answer:
(1) (i) “Can you touch the sky. Papa7 asked the little girl. (verb)
(ii) The old lady would wake UI) at the slightest touch. (noun)
(2) Simple Sentence.

Read More:

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• The Thief’s Story Question Answers
• On Wings of Courage Question Answers
• All the World’s a Stage Question Answers
• Joan of Arc Question Answers
• The Alchemy of Nature Question Answers

Balbharti Maharashtra State Board Class 10 English Solutions Unit 3.4 The Will to Win Notes, Textbook Exercise Important Questions and Answers.

Class 10 English Chapter 3.4 Question Answer Maharashtra Board

The Will to Win Poem 10th Std Question Answer

The Will To Win Poem Appreciation Question 1.
Get into pairs, discuss and tick the most appropriate answer :
(a) You may have lost the match; but
(i) It is important that you start fighting with your opponent.
(ii) It is important to have the will to win.
(iii) It is important that you blame the organizers for the rough ground.
Answer:
(ii) It is important to have the will to win.

(b) Success is always measured by :
(i) Ability to pounce upon at the opponent and fight with full force.
(ii) Match-fixing before the match begins.
(iii) Ability to bounce back after a fall.
Answer:
(c) The ability to bounce back after a fall.

(c) For attaining success; we need to :
(i) Sleep day in and day out and dream about success.
(ii) Scheme out things to make the opponent fall.
(iii) Work hard day and night for it.
Answer:
(c) Work hard day and night for it.

Appreciation Of Poem The Will To Win Question 2.
How can we achieve success in life? Complete the boxes by filling the essential qualities required for achieving success.

Answer:

Appreciation Of The Poem The Will To Win Question 3.
Discuss and write 5 proverbs/quotations related to the importance of having a strong will-power.
(a) ………………………………..
(b) ………………………………..
(c) ………………………………..
(d) ………………………………..
(e) ………………………………..
Answer:
(a) Where there’s a will, there’s a way.
(b) If at first you don’t succeed, try, try again.
(c) The truest wisdom is a resolute determination. -Napoleon Bonaparte
(d) Nothing is impossible. -Napoleon Bonaparte
(e) Determination is the key to success.

The Will To Win Question 4.
Complete the following table.
Make a list of great personalities of present and past who have achieved success in different walks of life. You can take help of your school library or search on the internet.

Politics	Social work	Sports	Music

Answer:

Politics	Sports	Social Work	Music
Abraham Lincoln Franklin Roosevelt George Washington Winston Churchill Woodrow Wilson Sardar Vallabhbhai Patel	Michael Jordan Allyson Felix Shekhar Naik Pele H. Boniface Prabhu Mahendra Singh Dhoni	Medha Patkar Helen Keller Emmeline Pankhurst Florence Nightingale	Ravindra Jain Cher (singer) Beethoven Mozart Tansen

The Will to Win Class 10 English Workshop Questions and Answers Maharashtra Board

The Will To Win Appreciation Question 1.
(A) In order to achieve success the poet wants us to
(a) ………………………………..
(b) ………………………………..
(c) ………………………………..
Answer:
(a) go out and fight for it
(b) work day and night for it
(c) give up time, peace and sleep for ft.

(B) Write as many phrases as you can using ‘enough’ and use them in your sentences.
Example : good enough
Answer:
(1) hard enough : If you work hard enough, you will be rewarded.
(2) bold enough : He was not bold enough to realise his ambitions.
(3) fast enough : He was not fast enough to understand the joke at his expense.
(4) strong enough : The little bird was still not strong enough to fly away on its own.
(5) reasonable enough : The price seemed reasonable enough; yet I hesitated to buy It.

(C) What does the word ‘Scheme’ mean here ? Choose the correct alternative from the following.
(a) Housing colony
(b) Goal in life
(c) Rhyme pattern
(d) Plan of action
Answer:
(d) plan of action

(D) Do you think the line ‘Give up your time and your peace and your sleep for it’ means that one should be ‘desperate’ or restless’ to achieve one’s goal. Explain your opinion, in your notebook.
Answer:
No. If you are ‘desperate’, you might take unnecessary risks to achieve your goal. If you are ‘restless’, you will have no peace of mind. What the poet means by losing ‘time’. ‘peace’ and sleep’ means spending many hours working hard to gain one’s objectives and to continually keep on thinking about it until it is achicvcd.

(E) Following lines are given to you. Find their appropriate meanings after discussing with your partner.
(a) To go out and fight for it.
(b) If you gladly sweat for, fret for and plan for it.
(c) Lose all your terror of opposition for it.
(d) With all your capacity, strength and sagacity.
Answer:
(a) to try one’s best and struggle hard in order to achieve something.
(d) making full use of all your capabilities, power and wisdom.
(c) sweat for it, fret for and plan for it and lose all your terror of the opposition for it
(d) to try one’s best and struggle hard in order to achieve something.

(F) Read the poem again and complete the web showing all those things that can turn one away from ones efforts towards a goal.

Answer:

(G) Find the lines from the poem which mean-
(a) become crazy for the goal
(b) toil hard happily
(c) get rid of all fears
(d) make efforts continuously
(e) extremely poor condition
Answer:
(a) if you are mad enough for it
(b) gladly sweat
(c) work day and night
(d) If neither cold poverty, famish or gaunt,

Question 2.
The poet has mentioned some hurdles in the poem that keep us away from achieving our goal in our life. Discuss with your partner and make a list of all the hurdles mentioned in the poem.

• sickness

Answer:

• or sickness or pain

Appreciation Of The Will To Win Question 3.
The poem explicitly describes some strengths and weaknesses with the help of some words and phrases. The poet wants us to possess all the strengths and keep away from all the weaknesses. Make a list of all the words and phrases showing Strengths in table A and Weaknesses in table B. One is done for you.

A Strengths	B Weaknesses
Work day and night	Sickness

Answer:

A (Strengths)	B (Weaknesses)
capacity, strength, sagacity faith, hope, confidence doggedness, grimness the help given by God	cold poverty famish gaunt sickness or pain of body and brain

A Will To Win Question 4.
Listen to the poem carefully and state whether the following statements are true or false. Correct the false statements.
(a) If you want a thing you should not give up your sleep.
(b) You should be afraid of your opposition.
(c) Cold or poverty cannot keep you away from achieving your goal.
(d) You can achieve your goal with the help of God.
(e) Life will not seem useless and worthless without achieving your goal.
Answer:
(a) False
(b) False
(c) True
(d) True
(e) False

The Will To Win Poem Question 5.
Form pairs and complete the web with suitable responses. Tell the class what all things the poet wants us to do to win.

Answer:

The Will To Win Speech Question 6.
Match the phrases in table A with lines of the poem given in table B.

Phrases	Lines
(1)    Toil hard (2)    Get rid of all (3)    Extremely poor condition (4)    Need desperately	(a)     If you want a thing bad enough… (b)    If neither cold or poverty, famished (c)    To work day and night for it. (d)    Lose all your terror of the opposition for it.

Answer:

‘A’ (Phrases)	‘B,’ (Lines)
(1) Toil hard	(b)    If neither cold or poverty, famished
(2) Get rid of all	(c)    To work day and night for it.
(3) Extremely poor condition….	(d)    Lose all your terror of the opposition for it.
(4) Need desperately	(a)     If you want a thing bad enough

Will To Win Question 7.
Every stanza begins with word ‘if’. How does it add to the effectiveness of the poem?
Answer:
Without the word ‘if’, the poem would remain as a simple statement of rules for success. By beginning each stanza with the word ‘if’, first of all, the poet implies that every human being has the choice of wanting to be successful in life. This is a more effective way of expression. Also by using the word ‘if’, he puts forward the conditions that will determine success in any sphere.

Will To Win Poem Appreciation Question 8.
(A) In poetry, when words/ideas are arranged in an ascending order of importance, the figure of speech used is called ‘Climax’.
For example, Man should work for his family, his country, but most of all for God.
Pick out two examples of ‘Climax’ from the poem.
Answer:
(a) to go out and fight for it, work day and night for it, give up your time and your peace and your sleep for it
(b) … sweat for it, fret for and plan for it and lose all your terror of the opposition for it …

(B) When some words, in the line of the poem, express the same idea in different ways, the figure of speech used is ‘Tautology’.
For example, . . . happy and joyful.
. . . motionless and still.
Pick out two examples of ‘Tautology’ from the poem.
Answer:
… neither cold poverty, famish
The words ‘poverty’ and ‘famish’ imply almost the same human physical condition.

(C) Pick out one example of the following Figures of Speech.
(1) Antithesis : ………………………
(2) Alliteration : ………………………
(3) Repetition : ………………………
Answer:
(1) Work day and night for it.
(2) … of body and brain
(3) … or sickness or pain

Will To Win Poem Question 9.
Work in group and prepare and present a speech on ‘How to Achieve Success.’
Answer:
How to Achieve Success
Friends,
I, Ajai Chitnis, do not hold with those who say, ‘Success is only for the privileged few!’ In my opinion, any and everyone can achieve success. All that is required is belief in one’s potential and a few rules of thumb in order to be a winner.

First of all you must have a goal. In order to reach that goal, you must have the necessary education and skills. For this you have to study or train. Then you have to look for opportunities. Life presents choices. You have to make the right choice. Having made the right choice, you must work hard to edge towards your goal. Only after much labour, sacrifice, determination and effort will you finally achieve success. This is my opinion on ‘How to achieve success’.

Thank you.

Question 10.
Read the poem again and write an appreciation of the poem ‘The Will to Win’ in a paragraph format.
Answer:
Point Format (for understanding)
The title of the poem : ‘The Will To Win’
The poet : Berton Braley
Rhyme scheme : No specific rhyme scheme, rhymes are used randomly
Figures of speech : Repetition, Climax, Tautology, Antithesis, etc.
The central ideatheme : What we should do and what we should avoid in order to achieve success.

Paragraph Format
Berton Braley has written this inspirational poem, ‘The Will To Win’.

The poem has no specific rhyme scheme, but rhymes are used randomly all throughout. The most common type is lines ending with the word ‘it’; e.g., ‘for it’, ‘of it’, ‘of it’; ‘without it’, ‘about it’; ‘beget it’, ‘get it’; ‘sweat for if, ‘fret for if and so on. Other examples are ‘capacity’, ‘sagacity’, ‘tenacity’: ‘pain’, ‘brain’.

The main figure of speech used is Repetition, as seen in the abundance of ‘for if phrases used throughout the poem. The other figures of speech are Climax, Tautology, Antithesis, etc.

The poet gives us a formula for sure success. He tells us what we should do and what we should avoid doing in order to achieve success.

It is an inspirational poem. It motivates one to set targets and achieve goals.

Question 11.
Project :
Make a list of Berton Braley’s collection of selected poems. You can take help of your teacher, library or search on internet. Recite Braley’s any one poem in front of the class.

Question 11.
State whether the following statements are True of False. Correct the false statements : 
(a) If you want a thing, you should not give up your sleep.
Corrected statement : If you want a thing, you should willingly give up your sleep.

(b) You should be afraid of your opposition.
Corrected statement : You should not fear the opposition.

(c) Life will not seem useless and worthless without achieving your goal.
Corrected statement : Life will seem utterly useless and worthless without achieving the goal.
Answer:
(a) False
(b) False
(c) False

Question 12.
Find the appropriate meanings of the following lines :
(b) makes you quite mad enough
(c) lose all the terror of God or man for it
Answer:
(b) impels you to do crazy things like taking risks
(c) get rid of all your fears of God or man or opposition.

Question 13.
Match the phrases in table A with lines from the extract given in table :

‘A’ (Phrases)	‘B,’ (Lines)
(1) Worry a lot about it	(a) serious and determined….
(2) Keep you away from….	(b) follow something in a determined way….
(3) Dogged and grim….	(c) Prevent you from….
(4) Simply go after…	(d) Fret for It …

Answer:

‘A’ (Phrases)	‘B,’ (Lines)
(1) Worry a lot about it	(d) Fret for It …
(2) Keep you away from….	(c) Prevent you from….
(3) Dogged and grim….	(a) serious and determined….
(4) Simply go after…	(b) follow something in a determined way….

Question 14.
Give an example of each of the following from the extract :
(a) Tautology.
Answer:
Life seems useless and worthless without it. The words ‘useless’ and ‘worthless’ have almost the same meaning.

(b) get rid of all fears
Answer:
lose all your terror

Question 15.
State whether the following statements are True or False. Correct the false statements :
(a) Cold or poverty cannot keep you away from achieving your goal.
(b) You can achieve your goal with the help of God.
Answer:
(a) True
(b) True

Question 16.
Find the lines from the extract which mean :
(a) make winning possible by using forceful action
Answer:
(a) besiege and beget it

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Balbharti Maharashtra State Board Class 10 English Solutions Unit 4.4 The Height of the Ridiculous Notes, Textbook Exercise Important Questions and Answers.

Class 10 English Chapter 4.4 Question Answer Maharashtra Board

The Height of the Ridiculous Poem 10th Std Question Answer

The Height Of The Ridiculous Appreciation Question 1.
The teacher writes incomplete sentences on the board. He/She asks the students to complete them in their notebooks.
(a) Today, I am happy because ……………………………… .
(b) Today after the class, I wish ……………………………… .
(c) Tomorrow, I feel that ……………………………… .
(d) I want to laugh because ……………………………… .
(e) Today, the class seems to be cheerful about ……………………………… .
Answer:
(a) my grandparents are coming for a holiday.
(b) to eat an ice cream.
(c) I will go for a movie.
(d) I am very happy.
(e) the forthcoming football match.

The Height Of The Ridiculous Question 2.
The teacher writes an incomplete sentence and asks the students to complete it in a funny way.
Answer:
(1) Mother gave me cheese but the cat ate it.
(2) I went to the market and bought an elephant.

Appreciation Of Poem The Height Of Ridiculous Question 3.
Give the words related to:

Syllable
A syllable is a unit of spoken language made up of a single uninterrupted sound formed by a vowel and consonants. For example, single syllable : ant, two syllables – water, three syllables : Inferno.
Answer:

The Height Of The Ridiculous Theme Question 4.
Pick out the word from the given box and write it in the correct columns below.

jump, narrow, cable, live, queen, butter, tree, kitten, van, yellow, dale, happy, night, printer, star, sober, paper, cloud, pearl, within, bike, began, slender.

Here the focus is not on the spellings but the pronunciation of the words.

Words with one syllable	Words with two syllables

Answer:

Words with one syllable	Words with two syllables
jump, live, queen, tree, van, dale, night, star, cloud, pearl, bike	narrow, cable, butter, kitten, yellow, happy, printer, sober, paper, within, began, slender

The Height Of The Ridiculous Notes Question 5.
Count the syllables and circle the appropriate number in the box.

Answer:

The Height Of The Ridiculous Question 6.
Write the names of any five of your friends and mention the number of syllables in each name.

Name	Number of syllables

Answer:

Name	Number of syllables
Rohan	2
Namrata	3
Poonam	2
Jai	1
Nilima	3

The Height of the Ridiculous Class 10 English Workshop Questions and Answers Maharashtra Board

The Height Of The Ridiculous Question 1.
Find out expressions from the poem that indicate funny moments.
For example, I laughed as I would die.
…………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………………
Answer:
(1) was all upon the grin
(2) the grin grew broad
(3) and shot from ear to ear
(4) He read the third; a chuckling noise
(5) The fourth; he broke into a roar
(6) The fifth; his waistband split;
(7) The sixth; he burst five buttons off;
(8) And tumbled in a fit.

Appreciation Of The Poem The Height Of Ridiculous Question 2.
Order of sequence : Arrange the following reactions in their proper order, as per the poem.
(a) His waistband split
(b) The grin grew broad.
(c) Sleepless eye.
(d) Was all upon the grin.
(e) He broke into a roar.
(f) He burst five buttons off.
Answer:
(d) Was all upon the grin
(b) The grin grew bro^d
(e) He broke into a roar
(a) His waistband split
(f) He burst five buttons
(c) Sleepless eye

Height Of Ridiculous Appreciation Question 3.
Form pairs and find out the various rhyming words in the poem and two of your own. Complete the following table.

Words	Rhyming words from the poem	Rhyming words more of your own
ear
within
man
split
way
him
die
mood

Answer:

Words	Rhyming words from the poem	Rhyming words more of your own
Way	Pay	Say, ray
Him	Limb	Dim, rim
Die	I	Fly, shy
Mood	Good	Food, wood
Ear	Hear	fear, dear
Within	Grin	sin, bin
Man	Can	fan, ran
Split	Fit	knit, lit

The Height Of Ridiculous Appreciation Question 4.
Match the lines with the Figures of Speech.

Lines	Figures of Speech
1.   In wondrous merry mood 2.  They were so queer, so very queer. 3.  And saw him peep within 4.  The grin grew broad. 5.  And shot from ear to ear. 6.  He broke into a roar. 7.  Ten days and nights with sleepless eye	Tautology Alliteration Onomatopoeia Repetition Hyperbole Repetition Transferred Epithet

Answer:

Lines	Figures of Speech
1. In wondrous, merry mood	Tautology
2. They were so queer, so very queer	Repetition
3.  And saw him peep within	Repetition
4. The grin grew broad	Alliteration
5. And shot from ear to ear	Hyperbole
6. He broke into a roar –	Onomatopoeia
7. Ten days and nights with sleepless eye	Transferred Epithet

Appreciation Of The Poem The Height Of The Ridiculous Question 5.
Copy any two stanzas of the poem in the lines below. Using a coloured pen underline the stressed syllables in each line and put a stress-mark ( ) over each.
Answer:
I wrote some lines once on a time
In wondrous merry mood,
And thought, as usual, men would say
They were exceeding good.

The Height Of The Ridiculous Question 6.
Complete the lines of the poem by choosing proper pairs of rhyming words and make it meaningful.
– We returned home late, one ………………………. ,
In the window, there glowed a ………………………. .
Burglars !! was our very first ………………………. ;
For defence, sticks ‘n stones we ………………………. .
”Let’s grab the loot and ………………………. ,”
was uttered soft, by ………………………..
The door we softly ……………………….,
And then we were truly ………………………..
Oops! Before, outside, we’d ……………………….,
The television had been left ………………………..
(run, shocked, gone, night, sought, on, someone, thought, light, unlocked)
Answer:
We returned home late, one night,
In the window there glowed a light.
Burglars! Was our very first thought,
For defence, sticks ‘n stones we sought.
“Let’s grab the loot and run.”
Was uttered soft, by someone.
The door we softly unlocked.
And then we were truly shocked.
Oops! Before outside we’d gone,
The television had been left on!

The Height Of Ridiculous Poem Appreciation Question 7.
Form goups in your class and together compose a short humorous poem. Use jokes, experiences, etc. and convert it to a poetic form. Write and decorate it on chart-paper and put it up in your class, in turns.

Appreciation Of The Height Of The Ridiculous Question 8.
Go through the poem and write an appreciation of the poem in a paragraph format.
Answer:
Point Format
(for understanding)
The title of the poem: ‘The Height of the Ridiculous’
The poet: Oliver Wendell Holmes
Rhyme scheme: abcb.
Figures of speech: Transferred Epithet, Hyperbole, Onomatopoeia, Tautology, Alliteration, etc.
Theme/Central idea: A funny poem to simply entertain the audience; written for Enjoyment.

Paragraph Format
The poem ‘The Height of the Ridiculous’ is written by Oliver Wendell Holmes.

The rhyme scheme of the poem is abcb. There are many figures of speech, like Hyperbole, Tautology, Onomatopoeia, Alliteration, etc. but the one that stands out is Transferred Epithet. In the line ‘Ten days and nights, with sleepless eye’, the adjective ‘sleepless’ should be for the man and not for the eye.

The poem is a humorous one written for enjoyment, with plenty of funny expressions. The main purpose of the poet is to simply entertain the reader.

Appreciation Of Poem The Height Of The Ridiculous Question 9.
Project :
Reading a poem.
Arrange the poetry reading competition. Select the poem of your choice.

• Read the poem silently.
• Repeat the reading of the poem.
• Focus on the pauses, stresses, intonation etc.
• Pay attention to the proper pronunciations.

Poem Appreciation Of The Height Of Ridiculous Question 10.
Choose the correct alternatives: (The answers are given directly and underlined.)
(1) The poet was in a very …………….. mood when he wrote the lines.
(a) tired
(b) happy
(c) bored
(d) wondering
Answer:
(b) happy

(2) The poet was generally a ……………… man.
(a) humorous
(b) wonderful
(c) serious
(d) good
Answer:
(c) serious

Question 11.
Explain:
(a) the contrast between the poet and his servant.
Answer:
The poet was a thin and slender man while his servant was strong and muscular.

(b) the poet’s reaction when he read the lines.
Answer:
The poet laughed heartily when he read the lines. He laughed so hard he thought he would die.

Question 12.
Find out the expression from the extract that indicates funny moments:
Answer:
‘I laughed as I would die’.

Question 13.
Match the lines with the figures of speech:
Lines – Figures of Speech
(a) A sober man am I – (c) Tautology
(b) To mind a slender man like me – (d) Inversion
Answer:
(a) A sober man am I – Inversion
(b) To mind a slender man like me – Alliteration

Question 14.
Complete the following:
(1) There was a grin on the servant’s face when …………………………
(2) The chuckling noise was heard when ……………………..
(3) When he read the fifth line ………………….
(4) The grin grew from ear to ear when the servant ………………….
Answer:
(1) he read the first line.
(2) the servant read the third line.
(3) his waistband split.
(4) read the second line.

Question 15.
Describe the outcome of this experience on the poet.
Answer:
After this experience, the poet has never dared to write any more funny poems.

Question 16.
Which line suggests that the servant was totally out of control?
Answer:
The line ‘And tumbled into a fit’ suggests that the servant was totally out of control.

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Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 6 Trigonometry.

10th Standard Maths 2 Practice Set 6.1 Chapter 6 Trigonometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 6.1 Chapter 6 Trigonometry Questions With Answers Maharashtra Board

Question 1.
If sin θ = \(\frac { 7 }{ 25 } \), find the values of cos θ and tan θ.
Solution:
sin θ = \(\frac { 7 }{ 25 } \) … [Given]
We know that,
sin² θ + cos² θ = 1

…[Taking square root of both sides] Now, tan θ = \(\frac{\sin \theta}{\cos \theta}\)

Alternate Method:
sin θ = \(\frac { 7 }{ 25 } \) …(i) [Given]
Consider ∆ABC, where ∠ABC 90° and ∠ACB = θ.
sin θ = \(\frac { AB }{ AC } \) … (ii) [By definition]
∴ \(\frac { AB }{ AC } \) = \(\frac { 7 }{ 25 } \) … [From (i) and (ii)]

LetAB = 7k and AC = 25k
In ∆ABC, ∠B = 90°
∴ AB² + BC² = AC² … [Pythagoras theorem]
∴ (7k)² + BC² = (25k)²
∴ 49k² + BC² = 625k²
∴ BC² = 625k² – 49k²
∴ BC² = 576k²
∴ BC = 24k …[Taking square root of both sides]

Question 2.
If tan θ = \(\frac { 3 }{ 4 } \), find the values of sec θ and cos θ.
Solution:

Alternate Method:
tan θ = \(\frac { 3 }{ 4 } \) …(i)[Given]
Consider ∆ABC, where ∠ABC 90° and ∠ACB = θ.
tan θ = \(\frac { AB }{ BC } \) … (ii) [By definition]
∴ \(\frac { AB }{ BC } \) = \(\frac { 3 }{ 4 } \) … [From (i) and (ii)]

Let AB = 3k and BC 4k
In ∆ABC,∠B = 90°
∴ AB² + BC² = AC² …[Pythagoras theorem]
∴ (3k)² + (4k)² = AC²
∴ 9k² + 16k² = AC²
∴ AC² = 25k²
∴ AC = 5k …[Taking square root of both sides]

Question 3.
If cot θ = \(\frac { 40 }{ 9 } \), find the values of cosec θ and sin θ
Solution:

..[Taking square root of both sides]

Alternate Method:
cot θ = \(\frac { 40 }{ 9 } \) ….(i) [Given]
Consider ∆ABC, where ∠ABC = 90° and
∠ACB = θ
cot θ = \(\frac { BC }{ AB } \) …(ii) [By defnition]
∴ \(\frac { BC }{ AB } \) = \(\frac { 40 }{ 9 } \) ….. [From (i) and (ii)]
Let BC = 40k and AB = 9k

In ∆ABC, ∠B = 90°
∴ AB² + BC² = AC² … [Pythagoras theorem]
∴ (9k)² + (40k)² = AC²
∴ 81k² + 1600k² = AC²
∴ AC² = 1681k²
∴ AC = 41k … [Taking square root of both sides]

Question 4.
If 5 sec θ – 12 cosec θ = θ, find the values of sec θ, cos θ and sin θ.
Solution:
5 sec θ – 12 cosec θ = 0 …[Given]
∴ 5 sec θ = 12 cosec θ

Question 5.
If tan θ = 1, then find the value of

Solution:
tan θ = 1 … [Given]
We know that, tan 45° = 1
∴ tan θ = tan 45°
∴ θ = 45°

Question 6.
Prove that:
i. \(\frac{\sin ^{2} \theta}{\cos \theta}+\cos \theta=\sec \theta\)
ii. cos² θ (1+ tan² θ) = 1
iii. \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta\)
iv. (sec θ – cos θ) (cot θ + tan θ) tan θ. sec θ
v. cot θ + tan θ cosec θ. sec θ
vi. \(\frac{1}{\sec \theta-\tan \theta}=\sec \theta+\tan \theta\)
vii. sin⁴ θ – cos⁴ θ = 1 – 2 cos² θ
viii. \(\sec \theta+\tan \theta=\frac{\cos \theta}{1-\sin \theta}\)

Proof:
i. L.H.S. = \(\frac{\sin ^{2} \theta}{\cos \theta}+\cos \theta\)

ii. L.H.S. = cos² θ(1 + tan² θ)
= cos² θ sec² θ …[∵ 1 + tan² θ = sec² θ]

= 1
= R.H.S.
∴ cos² θ (1 + tan² θ) = 1

iv. L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ. sec θ

v. L.H.S. = cot θ + tan θ

∴ cot θ + tan θ = cosec θ.sec θ

vii. L.H.S. = sin⁴ θ – cos⁴ θ
= (sin² θ)² – (cos² θ)²
= (sin² θ + cos² θ) (sin² θ – cos² θ)
= (1) (sin² θ – cos² θ) ….[∵ sin² θ + cos² θ = 1]
= sin² θ – cos² θ
= (1 – cos² θ) – cos² θ …[θ sin² θ = 1 – cos² θ]
= 1 – 2 cos² θ
= R.H.S.
∴ sin⁴ θ – cos⁴ θ = 1 – 2 cos² θ

viii. L.H.S. = sec θ + tan θ

xi. L.H.S. = sec⁴ A (1 – sin⁴ A) – 2 tan² A
= sec⁴ A [1² – (sin² A)²] – 2 tan² A
= sec⁴ A (1 – sin²A) (1 + sin² A) – 2 tan² A
= sec⁴ A cos²A (1 + sin² A) – 2 tan²A
[ ∵ sin² θ + cos² θ = 1 ,∵ 1 – sin² θ = cos² θ]

Maharashtra Board Class 10 Maths Chapter 6 Trigonometry Intext Questions and Activities

Question 1.
Fill in the blanks with reference to the figure given below. (Textbook pg. no. 124)

Solution:

Question 2.
Complete the relations in ratios given below. (Textbook pg, no. 124)

Solution:
i. \(\frac{\sin \theta}{\cos \theta}\) = [tan θ]
ii. sin θ = cos (90 – θ)
iii. cos θ = (90 – θ)
iv. tan θ × tan (90 – θ) = 1

Question 3.
Complete the equation. (Textbook pg. no, 124)
sin² θ + cos² θ = [______]
Solution:
sin² θ + cos² θ = [1]

Question 4.
Write the values of the following trigonometric ratios. (Textbook pg. no. 124)

Solution:

Maharashtra Board Class 10 Maths Solutions

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Problem Set 6 Class 10 Answers
• Practice Set 7.1 Class 10 Answers
• Practice Set 7.2 Class 10 Answers
• Practice Set 7.3 Class 10 Answers
• Practice Set 7.4 Class 10 Answers
• Problem Set 7 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 7 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

10th Standard Maths 2 Problem Set 7 Chapter 7 Mensuration Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Problem Set 7 Chapter 7 Mensuration Questions With Answers Maharashtra Board

Problem Set 7 Question 1. Choose the correct alternative answer for each of the following questions.

i. The ratio of circumference and area of a circle is 2 : 7. Find its circumference.
(A) 14 π
(B) \(\frac{7}{\pi}\)
(C) 7π
(D) \(\frac{14}{\pi}\)
Answer:

(A)

ii. If measure of an arc of a circle is 160° and its length is 44 cm, find the circumference of the circle.
(A) 66 cm
(B) 44 cm
(C) 160 cm
(D) 99 cm
Answer:

(D)

iii. Find the perimeter of a sector of a circle if its measure is 90° and radius is 7 cm.
(A) 44 cm
(B) 25 cm
(C) 36 cm
(D) 56 cm
Answer:

(B)

iv. Find the curved surface area of a cone of radius 7 cm and height 24 cm.
(A) 440 cm²
(B) 550 cm²
(C) 330 cm²
(D) 110 cm²
Answer:

(B)

v. The curved surface area of a cylinder is 440 cm² and its radius is 5 cm. Find its height.
(A) \(\frac{44}{\pi}\) cm
(B) 22π cm
(C) 44π cm
(D) \(\frac{22}{\pi}\)
Answer:

(A)

vi. A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone.
(A) 15 cm
(B) 10 cm
(C) 18 cm
(D) 5 cm
Answer:

(A)

vii. Find the volume of a cube of side 0.01 cm.
(A) 1 cm
(B) 0.001 cm³
(C) 0.0001 cm³
(D) 0.000001 cm³
Answer:
Volume of cube = (side)³
= (0.01)³ = 0.000001 cm³
(D)

viii. Find the side of a cube of volume 1 m³
(A) 1 cm
(B) 10 cm
(C) 100 cm
(D) 1000 cm
Answer:
Volume of cube = (side)³
∴ 1 = (side)³
∴ Side = 1 m
= 100 cm
(C)

Problem Set 7 Geometry Class 10 Question 2. A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub? = (π = \(\frac { 22 }{ 7 } \))
Given: For the frustum shaped tub,
height (h) = 21 cm,
radii (r₁) = 20 cm, and (r₂) = 15 cm
To find: Capacity (volume) of the tub.
Solution:
Volume of frustum = \(\frac { 1 }{ 3 } \) πh (r₁² + r₂² + r₁ × r₂)

∴ The capacity of the tub is 20.35 litres.

10th Geometry Problem Set 7 Question 3. Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube?
Given: For the cylindrical tube,
height (h) = 90 cm,
outer radius (R) = 30 cm,
thickness = 2 cm
For the plastic spherical ball,
radius (r₁) = 1 cm
To find: Number of balls melted.
Solution:

Inner radius of tube (r)
= outer radius – thickness of tube
= 30 – 2
= 28 cm
Volume of plastic required for the tube = Outer volume of tube – Inner volume of hollow tube
= πR²h – πr²h
= πh(R² – r²)
= π × 90 (30² – 28²)
= π × 90 (30 + 28) (30 – 28) …[∵ a² – b² = (a + b)(a – b)]
= 90 × 58 × 2π cm³

∴ 7830 plastic balls were melted to make the tube.

Problem Set 7 Geometry Question 4.
A metal parallelopiped of measures 16 cm × 11cm × 10cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively?
Given: For the parallelopiped.,
length (l) = 16 cm, breadth (b) = 11 cm,
height (h) = 10 cm
For the cylindrical coin,
thickness (H) = 2 mm,
diameter (D) 2 cm
To find: Number of coins made.
Solution:
Volume of parallelopiped = l × b × h
= 16 × 11 × 10
= 1760 cm³
Thickness of coin (H) = 2 mm
= 0.2 cm …[∵ 1 cm = 10 mm]
Diameter of coin (D) = 2 cm

∴ 2800 coins were made by melting the parallelopiped.

Mensuration Problem Question 5.  The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of ₹ 10 per sq.m.
Given: For the cylindrical roller,
diameter (d) =120 cm,
length = height (h) = 84 cm
To find: Expenditure of levelling the ground.
Solution:
Diameter of roller (d) = 120 cm

Now, area of ground levelled in one rotation = curved surface area of roller
= 3.168 m²
∴ Area of ground levelled in 200 rotations
= 3.168 × 200 =
633.6 m²
Rate of levelling = ₹ 10 per m²
∴ Expenditure of levelling the ground
= 633.6 × 10 = ₹ 6336
∴ The expenditure of levelling the ground is ₹ 6336.

Question 6.
The diameter and thickness of a hollow metal sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm3. Find the outer surface area and mass of the sphere, [π = 3.14]
Given: For the hollow sphere,
diameter (D) =12 cm, thickness = 0.01 m
density of the metal = 8.88 gm per cm³
To find: i. Outer surface area of the sphere
ii. Mass of the sphere.

Solution:
Diameter of the sphere (D)
= 12 cm
∴ Radius of sphere (R)
= \(\frac { d }{ 2 } \) = \(\frac { 12 }{ 2 } \) = 6 cm
∴ Surface area of sphere = 4πR²
= 4 × 3.14 × 6²
= 452.16 cm²
Thickness of sphere = 0.01 m
= 0.01 × 100 cm …[∵ 1 m = 100 cm]
= 1 cm
∴ Inner radius of the sphere (r)
= Outer radius – thickness of sphere
= 6 – 1 = 5 cm
∴ Volume of hollow sphere
= Volume of outer sphere – Volume of inner sphere

∴ The outer surface area and the mass of the sphere are 452.16 cm² and 3383.19 gm respectively.

Question 7.
A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into a shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone?
Given: For the cylindrical bucket,
diameter (d) = 28 cm, height (h) = 20 cm
For the conical heap of sand,
height (H) = 14 cm
To find: Base area of the cone (πR²).
Solution:
Diameter of the bucket (d) = 28 cm


The base area of the cone is 2640 cm².

Question 8.
The radius of a metallic sphere is 9 cm. It was melted to make a wire of diameter 4 mm. Find the length of the wire.
Given: For metallic sphere,
radius (R) = 9 cm
For the cylindrical wire,
diameter (d) = 4 mm
To find: Length of wire (h).
Solution:

∴ The length of the wire is 243 m.

Question 9.
The area of a sector of a circle of 6 cm radius is 157t sq.cm. Find the measure of the arc and length of the arc corresponding to the sector.
Given: Radius (r) = 6 cm,
area of sector = 15 π cm²
To find: i. Measure of the arc (θ),
ii. Length of the arc (l)
Solution:

∴ The measure of the arc and the length of the arc are 150° and 5π cm respectively.

Question 10.
In the adjoining figure, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion.

(π = 3.14, \(\sqrt { 3 }\) = 1.73)
Given: Radius (r) = PA = 8 cm,
PC = 4 cm
To find: Area of shaded region.
Solution:

Similarly, we can show that, ∠BPC = 60°
∠APB = ∠APC + ∠BPC …[Angle sum property]
∴ θ = 60° + 60° = 120°


Area of shaded region = A(P-ADB) – A(∆APB)
= 66.98 – 27.68
= 39.30 cm²
∴ The area of the shaded region is 39.30 cm².

Question 11.
In the adjoining figure, square ABCD is inscribed in the sector A-PCQ. The radius of sector C-BXD is 20 cm. Complete the following activity to find the area of shaded region.

Solution:
Side of square ABCD
= radius of sector C-BXD = [20] cm
Area of square = (side)² = 20² = 400 cm² ….(i)
Area of shaded region inside the square = Area of square ABCD – Area of sector C-BXD

Radius of bigger sector
= Length of diagonal of square ABCD
= \(\sqrt { 2 }\) × side
= 20 \(\sqrt { 2 }\) cm
Area of the shaded regions outside the square
= Area of sector A-PCQ – Area of square ABCD
= A(A – PCQ) – A(꠸ABCD)

Alternate method:

□ABCD is a square. … [Given]
Side of □ABCD = radius of sector (C-BXD)
= 20 cm
Radius of sector (A-PCQ) = Diagonal
= \(\sqrt { 2 }\) × side
= \(\sqrt { 2 }\) × 20
= 20 \(\sqrt { 2 }\) cm

Now, Area of shaded region
= A(A-PCQ) – A(C-BXD)
= 628 – 314
= 314 cm²
∴ The area of the shaded region is 314 cm².

Question 12.
In the adjoining figure, two circles with centres O and P are touching internally at point A. If BQ = 9, DE = 5, complete the following activity to find the radii of the circles.

Solution:
Let the radius of the bigger circle be R and that of smaller circle be r.
OA, OB, OC and OD are the radii of the bigger circle.
∴ OA = OB = OC = OD = R
PQ = PA = r
OQ + BQ = OB … [B – Q – O]
OQ = OB – BQ = R – 9
OE + DE = OD ….[D – E – O]
OE = OD – DE = [R – 5]
As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle,
OQ × OA = OE × OF
∴ (R – 9) × R = (R – 5) × (R – 5) …[∵ OE = OF]
∴ R² – 9R = R² – 10R + 25
∴ -9R + 10R = 25
∴ R = [25units]
AQ = AB – BQ = 2r ….[B-Q-A]
∴ 2r = 50 – 9 = 41
∴ r = \(\frac { 41 }{ 2 } \) = 20.5 units

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Problem Set 6 Class 10 Answers
• Practice Set 7.1 Class 10 Answers
• Practice Set 7.2 Class 10 Answers
• Practice Set 7.3 Class 10 Answers
• Practice Set 7.4 Class 10 Answers
• Problem Set 7 Class 10 Answers

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.

10th Standard Maths 1 Practice Set 5.3 Chapter 5 Probability Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 5.3 Chapter 5 Probability Questions With Answers Maharashtra Board

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

Class 10 Maths Digest

• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Practice Set 5.4 Class 10 Answers
• Problem Set 5 Class 10 Answers

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.

10th Standard Maths 1 Practice Set 5.2 Chapter 5 Probability Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 5.2 Chapter 5 Probability Questions With Answers Maharashtra Board

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.

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).

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:

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:

Class 10 Maths Digest

• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Practice Set 5.4 Class 10 Answers
• Problem Set 5 Class 10 Answers

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.

10th Standard Maths 1 Practice Set 5.1 Chapter 5 Probability Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 5.1 Chapter 5 Probability Questions With Answers Maharashtra Board

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

Class 10 Maths Digest

• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Practice Set 5.4 Class 10 Answers
• Problem Set 5 Class 10 Answers

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.

10th Standard Maths 1 Practice Set 4.4 Chapter 4 Financial Planning Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 4.4 Chapter 4 Financial Planning Questions With Answers Maharashtra Board

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)

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

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

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

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.

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Practice Set 4.3 Class 10 Answers
• Practice Set 4.4 Class 10 Answers
• Problem Set 4A Class 10 Answers
• Problem Set 4B Class 10 Answers

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.

10th Standard Maths 1 Problem Set 4B Chapter 4 Financial Planning Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Problem Set 4B Chapter 4 Financial Planning Questions With Answers Maharashtra Board

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

∴ 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

∴ 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

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

∴ 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


∴ 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

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

∴ 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).

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Practice Set 4.3 Class 10 Answers
• Practice Set 4.4 Class 10 Answers
• Problem Set 4A Class 10 Answers
• Problem Set 4B Class 10 Answers