Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6

Maharashtra State Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6

Question 1.
The age group and number of persons, who donated blood in a blood donation camp is given below.
Draw a pie diagram from it.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 1
Solution:
Total number of persons = 80 + 60 + 35 + 25 = 200
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 2

Question 2.
The marks obtained by a student in different subjects are shown. Draw a pie diagram showing the information.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 3
Solution:
Total marks obtained = 50 + 70 + 80 + 90 + 60 + 50 = 400
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 4

Question 3.
In a tree plantation programme, the number of trees planted by students of different classes is given in the following table. Draw a pie diagram showing the information.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 5
Solution:
Total number of trees planted = 40 + 50 + 75 + 50 + 70 + 75 = 360
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 6

Question 4.
The following table shows the percentages of demands for different fruits registered with a fruit vendor. Show the information by a pie diagram.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 7
Solution:
Total percentage = 30 + 15 + 25 + 20 + 10 = 100%
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 8

Question 5.
The pie diagram in the given figure shows the proportions of different workers in a town. Answer the following questions with its help.
i. If the total workers is 10,000, how many of them are in the field of construction?
ii. How many workers are working in the administration?
iii. What is the percentage of workers in production?
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 9
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 10
∴ There are 2000 workers working in the field of construction.

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Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 11
∴ There are 1000 workers working in the administration.

Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 12
∴ 25% of workers are working in the production field.

Question 6.
The annual investments of a family are shown in the given pie diagram. Answer the following questions based on it.
i. If the investment in shares is ? 2000, find the total investment.
ii. How much amount is deposited in bank?
iii. How much more money is invested in immovable property than in mutual fund?
iv. How much amount is invested in post?
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 13
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 14
The total investment is ₹ 12000.

ii. Central angle for deposit in bank (θ) = 90°
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 15
∴ The amount deposited in bank is ₹ 3000.

iii. Difference in central angle for immovable property and mutual fund (θ) = 120° – 60° = 60°
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 16
∴ ₹ 2000 more is invested in immovable property than in mutual fund.

iv. Central angle for post (θ) = 30°
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.6 17
∴ The amount invested in post is ₹ 1000.

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Practice Set 1.1

Maharashtra State Board Class 9 Maths Solutions Chapter 1 Sets Practice Set 1.1

Question 1.
Write the following sets in roster form.
i. Set of even natural numbers
ii. Set of even prime numbers from 1 to 50
iii. Set of negative integers
iv. Seven basic sounds of a sargam (sur)
Answer:
i. A = { 2, 4, 6, 8,….}
ii. 2 is the only even prime number
∴ B = { 2 }
iii. C = {-1, -2, -3,….}
iv. D = {sa, re, ga, ma, pa, dha, ni}

Question 2.
Write the following symbolic statements in words.
i. \(\frac { 4 }{ 3 }\) ∈ Q
ii. -2 ∉ N
iii. P = {p | p is an odd number}
Answer:
i. \(\frac { 4 }{ 3 }\) is an element of set Q.
ii. -2 is not an element of set N.
iii. Set P is a set of all p’s such that p is an odd number.

Question 3.
Write any two sets by listing method and by rule method.
Answer:
i. A is a set of even natural numbers less than 10.
Listing method: A = {2, 4, 6, 8}
Rule method: A = {x | x = 2n, n e N, n < 5}

ii. B is a set of letters of the word ‘SCIENCE’. Listing method : B = {S, C, I, E, N}
Rule method: B = {x \ x is a letter of the word ‘SCIENCE’}

Question 4.
Write the following sets using listing method.
i. All months in the Indian solar year.
ii. Letters in the word ‘COMPLEMENT’.
iii. Set of human sensory organs.
iv. Set of prime numbers from 1 to 20.
v. Names of continents of the world.
Answer:
i. A = {Chaitra, Vaishakh, Jyestha, Aashadha, Shravana, Bhadrapada, Ashwina, Kartika, Margashirsha, Paush, Magha, Falguna}
ii. X = {C, O, M, P, L, E, N, T}
iii. Y = {Nose, Ears, Eyes, Tongue, Skin}
iv. Z = {2, 3, 5, 7, 11, 13, 17, 19}
v. E = {Asia, Africa, Europe, Australia, Antarctica, South America, North America}

Question 5.
Write the following sets using rule method.
i. A = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
ii. B= {6, 12, 18,24, 30,36,42,48}
iii. C = {S, M, I, L, E}
iv. D = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
v. X = {a, e, t}
Answer:
i. A = {x | v = n², n e N, n < 10}
ii. B = {x j x = 6n, n e N, n < 9}
iii. C = {y j y is a letter of the word ‘SMILE’} [Other possible words: ‘SLIME’, ‘MILES’, ‘MISSILE’ etc.]
iv. D = {z | z is a day of the week}
v. X = {y | y is a letter of the word ‘eat’}
[Other possible words: ‘tea’ or ‘ate’]

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Question 1.
Fill in the blanks given in the following table. (Textbook pg. no. 3)
Answer:
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Practice Set 1.1 1

Maharashtra Board Class 9 Maths Solutions

Maharashtra Board Class 9 Maths Solutions Chapter 3 Polynomials Practice Set 3.2

Maharashtra State Board Class 9 Maths Solutions Chapter 3 Polynomials Practice Set 3.2

Question 1.
Use the given letters to write the answer.
i. There are ‘a’ trees in the village Lat. If the number of trees increases every year by ’b‘. then how many trees will there be after ‘x’ years?
ii. For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?
iii. The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Solution:
i. Number of trees in the village Lat = a
Number of trees increasing each year = b
∴ Number of trees after x years = a + bx
∴ There will be a + bx trees in the village Lat after x years.

ii. Total rows = x
Number of students in each row = y
∴ Total students = Total rows × Number of students in each row
= x × y
= xy
∴ There are in all xy students for the parade.

iii. Digit in units place = n
Digit in tens place = m
∴ The two digit number = 10 x digit in tens place + digit in units place
= 10m + n
∴ The polynomial representing the two digit number is 10m + n.

Question 2.
Add the given polynomials.
i. x3 – 2x2 – 9; 5x3 + 2x + 9
ii. -7m4+ 5m3 + √2 ; 5m4 – 3m3 + 2m2 + 3m – 6
iii. 2y2 + 7y + 5; 3y + 9; 3y2 – 4y – 3
Solution:
i. (x3 – 2x2 – 9) + (5x3 + 2x + 9)
= x3 – 2x2 – 9 + 5x3 + 2x + 9
= x3 + 5x3 – 2x2 + 2x – 9 + 9
= 6x3 – 2x2 + 2x

ii. (-7m4 + 5m3 + √2 ) + (5m4 – 3m3 + 2m2 + 3m – 6)
= -7m4 + 5m3 + √2 + 5m4 – 3m3 + 2m2 + 3m – 6
= -7m4 + 5m4 + 5m3 – 3m3 + 2m2 + 3m +√2 – 6
= -2m4 + 2m3 + 2m2 + 3m + √2 – 6

iii. (2y2 + 7y + 5) + (3y + 9) + (3y2 – 4y – 3)
= 2y2 + 7y + 5 + 3y + 9 + 3y2 – 4y – 3
= 2y2 + 3y2 + 7y + 3y – 4y + 5 + 9 – 3
= 5y2 + 6y + 11

Question 3.
Subtract the second polynomial from the first.
i. x2 – 9x + √3 ; – 19x + √3 + 7x2
ii. 2ab2 + 3a2b – 4ab; 3ab – 8ab2 + 2a2b
Solution:
i. x2 – 9x + √3 -(- 19x + √3 + 7x2)
= x2 – 9x + √3 + 19x – √ 3 – 7x2
= x2 – 7x29x + 19x + √3 – √3
= – 6x2 + 10x

ii. (2ab2 + 3a2b – 4ab) – (3ab – 8ab2 + 2a2b)
= 2ab2 + 3a2b – 4ab – 3ab + 8ab2 – 2a2b
= 2ab2 + 8ab2 + 3a2b – 2a2b 4ab – 3ab
= 10ab2 + a2b – 7ab

Question 4.
Multiply the given polynomials.
i. 2x; x2 – 2x – 1
ii. x5 – 1; x3 + 2x2 + 2
iii. 2y +1; y2 – 2y + 3y
Solution:
i. (2x) x (x2 – 2x – 1) = 2x3 – 4x2 – 2x

ii. (x5 – 1) × (x3 + 2x2 + 2)
= x5 (x3 + 2x2 + 2) -1(x3 + 2x2 + 2)
= x8 + 2x7 + 2x5 – x3 – 2x2 – 2

iii. (2y + 1) × (y2 – 2y3 + 3y)
= 2y(y2 – 2y3 + 3y) + 1(y2 – 2y3 + 3y)
= 2y3 – 4y4 + 6y2 + y2 – 2y3 + 3y
= -4y4 + 2y3 – 2y3 + 6y2 + y2 + 3y
= -4y4 + 7y2 + 3y

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Question 5.
Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor x Quotient + Remainder’.
i. x3 – 64; x – 4
ii. 5x5 + 4x4 – 3x3 + 2x2 + 2 ; x2 – x
Solution:
i. x3 – 64 = x3 + 0x2 + 0x – 64
Maharashtra Board Class 9 Maths Solutions Chapter 3 Polynomials Practice Set 3.2 1
∴ Quotient = x2 + 4x + 16, Remainder = 0
Now, Dividend = Divisor x Quotient + Remainder
∴ x3 – 64 = (x – 4)(x2 + 4x + 16) + 0

ii. 5x5 + 4x4 – 3x3 + 2x2 + 2 = 5x5 + 4x4 – 3x3 + 2x + 0x + 2
Maharashtra Board Class 9 Maths Solutions Chapter 3 Polynomials Practice Set 3.2 2
∴ Quotient = 5x3 + 9x2 + 6x + 8,
Remainder = 8x + 2
Now, Dividend = Divisor x Quotient + Remainder
∴ 5x5 + 4x4 – 3x3 + 2x2 + 2 = (x2 – x)(5x3 + 9x2 + 6x + 8) + (8x + 2)

Question 6.
Write down the information in the form of algebraic expression and simplify.
There is a rectangular farm with length (2a2 + 3b2) metre and breadth (a2 + b2) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a2 – b2) metre. What is the area of the remaining part of the farm? [4 Marks]
Solution:
Length of the rectangular farm = (2a2 + 3b2) m
Breadth of the rectangular farm = (a2 + b2) m
Area of the farm = length x breadth = (2a2 + 3b2) x (a2 + b2)
= 2a2(a2 + b2) + 3b2(a2 + b2)
= 2a2 + 2a2b2 + 3a2b2 + 3b4
= (2a4 + 5a2b2 + 3b4) sq. m … (i)
The farmer used a square shaped plot of the farm to build a house.
Side of the square shaped plot = (a2 – b2) m
∴ Area of the plot = (side)2
= (a2 – b2)2
= (a4 – 2a2b2 + b4) sq m… .(ii)

∴ Area of the remaining farm = Area of the farm – Area of the plot
= (2a4 + 5a2b2 + 3b4) – (a4 – 2a2b2 + b4) … [From (i) and (ii)]
= 2a4 + 5a2b2 + 3b4 – a4 + 2a2b2 – b4
= 2a4 – a4 + 5a2b2 + 2a2b2 + 3b4 – b4
= a4 + 7a2b2 + 2b4
∴ The area of the remaining farm is (a4 + 7a2b2 + 2b4) sq. m.

Maharashtra Board Class 9 Maths Solutions

Silk Road Extra Questions and Answers Class 11 English Hornbill

Silk Road Extra Questions and Answers Class 11 English Hornbill

Here we are providing Silk Road Extra Questions and Answers Class 11 English Hornbill, Extra Questions for Class 11 English was designed by subject expert teachers.

Silk Road Extra Questions and Answers Class 11 English Hornbill

Silk Road Extra Questions and Answers Short Answer Type

Question 1.
Why was the author disappointed with Darchen?
Answer:
Darchen was shabby. The narrator was sick in Darchen. He passed an extremely uncomfortable night because of his breathing problem. Next day when he felt better, he disliked Darchen less.

Silk Road Extra Questions and Answers

Question 2.
How did the author and his companions cross the first snow blockage on their way to Mount Kailash?
Answer:
Snow was so steep that they could not go around it. They had to go over it. The danger was that they could slip. They flung handfuls of dirt and covered the snow completely with soil. The narrator and Daniel got off the vehicle to lighten the load and Tsetan drove the vehicle over the snow.

Silk Road Short Answer Type Questions and Answers

Question 3.
Comment on the sensitive behaviour of hill folk.
Answer:
In the story ‘Silk road’, the Hill folks are very simple, naive and unsophisticated. They are polite and courteous towards all tourists. They actually take good care of all their tourists. They know the fact that they win their bread because of these tourists. They were very hospitable and God-fearing.

Question 4.
How does the author recount his experience at the Darchen Medical College?
Answer:
The doctor at the Darchen Medical College did not wear the traditional white coat of a doctor. He observed the author and diagnosed his problem as the effect of cold and high altitude. He gave him brown powders and pellets to be taken with hot water. The author benefitted with this treatment.

Question 5.
How was the author’s experience at Hor a stark contrast to earlier accounts of the place?
Answer:
The author was disappointed and rather depressed on arrival at Hor. Previous visitors had been overwhelmed by the beauty of Mansarovar Lake, but the author found Hor shabby and dirty.

Question 6.
How can the presence of salt flats in Tibet be explained?
Answer:
Salt flats are vestiges of Tethys ocean which bordered Tibet before the great continental collision.

Question 7.
Where is the town of Hor situated? Describe the town.
Answer:
Hor is located on the eastwest highway when one travels from Lhasa to Kashmir. It is a grim desolate place littered with accumulated refuse. There is no vegetation in this town. It is located on the shore of Lake Mansarovar. On the whole, it has badly painted concrete buildings.

Question 8.
What is the importance of Lake Mansarovar?
Answer:
Lake Mansarovar holds sacred importance for both the Hindus and the Buddhists. It is a source of four great Indian rivers i.e., the Indus, the Brahmaputra, the Ganges and the Sutlej. It is an extremely beautiful lake. Its first site often move the tourists into tears.

Question 9.
How did the author suffer at Darchen?
Answer:
Due to extreme cold winds at Hor, on reaching Darchen the author suffered from cold and a blocked nose. He gasped for oxygen, could breathe through one nostril only. He could not sleep at night. He sat up and felt better.

Question 10.
Who was Norbu? How was he different from the local people?
Answer:
Norbu has a Tibetan working in Beijing at the Chinese Academy. He was different from other Tibetans as he was wearing a windcheater and metal rimmed spectacles of a Western style and he spoke English fluently.

Question 11.
Why was the narrator relieved on meeting Norbu?
Answer:
The narrator was quite relieved on meeting Norbu firstly because he was all alone at Darchen. He found a companion in Norbu. He could speak English fluently. He was educated. He didn’t believe in doing Kora on foot in the conventional manner. Both of them decided to hire yaks. In every aspect, Norbu appeared to be an ideal companion for the narrator.

Question 12.
Where was the narrator going? Through what kind of terrain would he have to pass?
Answer:
The narrator was journeying towards Mount Kailash and the Mansarovar Lake where they he had to pass through several high mountain passes. He had to journey through a lot of snow and vast open plains.

Question 13.
Did the narrator encounter any wildlife in the course of his journey?
Answer:
Yes, on the course of his journey, the narrator came across gazelles and herd of wild asses.

Question 14.
What have you learnt about the Tibetan mastiff from the essay?
Answer:
After reading the essay, we found that Tibetan mastiffs are too violent and ferocious with big heads. They are black in colour with their red collars. They attack like bullets fired from guns. They possess massive jaws. Their bark is furious. These mastiffs are so fearless that they can also attack cars and jeeps.

Question 15.
How did the narrator and Tsetan negotiate the hurdle of the swathe of snow?
Answer:
The snow was so steep that they could not go around it. They had to go over it. The danger was that they could slip. They flung handfuls of dirt and covered the snow completely with soil. The narrator and Daniel got off the vehicle to lighten the load and Tsetan drove the vehicle over the snow.

Question 16.
What problems did the narrator and his team experience due to low atmospheric pressure?
Answer:
Due to low atmospheric pressure, the narrator and his team felt their heads going heavy. The low pressure also caused the fuel to expand, making it extremely difficult for them to carry forward onto their journey.

Question 17.
Why has the article been titled ‘Silk Road’?
Answer:
The article has been titled ‘Silk Road’ because the narrator travelled along the old Silk Route in the Himalayas that touches Tibet to reach Mansarovar.

Silk Road Extra Questions and Answers Long Answer Type

Question 1.
Describe the difficulties and disillusionment faced by Nick Middleton during his journey to Mount Kailash.
Answer:
Nick Middleton, the narrator had to journey through the difficult terrain to reach Mount Kailash. The path was totally snow covered, the snow was so steep that they could not go around it. They somehow went over it. The danger was that they could slip. They flung handfuls of dirt and covered the snow completely with soil.

The narrator and Daniel got off the vehicle to lighten the load while Tsetan drove the vehicle over the snow. Due to low atmospheric pressure, he got a headache. The fuel of vehicle also expanded due to the low pressure, which would prove perilous for them. In Darchen, the narrator also suffered from blocked sinuses that resulted in extreme cold and breathing problems. The town Hor was shabby, dirty and very depressing. Overall, the journey of the author was adventurous.

Question 2.
What was the purpose of the narrator’s journey? What route did he take to reach his destination?
Answer:
The author had set up on a religious pilgrimage to perform Kora to Mount Kailash and the Mansarovar Lake. He travelled through the old Silk Route in the Himalayas that touches Tibet to reach Mansarovar. One has to pass through several high mountain passes covered with snow and vast open plains in order to reach this destination. Enroute the narrator crossed Hor and Darchen. The weather at Hor was extremely chilly. The narrator found Hor to be depressing and shabby. Darchen was another dirty place where the narrator suffered from extreme cold and had to take medical help.

Question 3.
What physical discomfort did the narrator experience in Darchen? How did he find relief?
Answer:
While in Darchen, it became extremely tough for the narrator as he got extreme cold due to blocked sinuses. He was gasping for oxygen. He could breathe through only one nostril. He could not sleep, felt utter restlessness. He had fear that he might die if he would sleep. So, he sat up the whole night. Next morning, Tsetan took him for the medical help to the Darchen Medical College. He was given a five-day course of Tibetan medicine which gave him a lot of relief and comfort. After undergoing the treatment, he felt somewhat better. Now, he could sleep soundly.

Question 4.
Why was the narrator disappointed to find no pilgrims at Darchen? Was his disappointment dispelled?
Answer:
The narrator had to do the Kora to Kailash and Mansarovar. He was expecting to join the bands of pilgrims. But it was too early in the season when he reached Darchen. He became so disappointed and desolate. He felt lonely and unhappy without any companion. It became difficult for him to pass leisure time. But, soon he got over this disappointment upon meeting Norbu.

He interacted with him and felt too comfortable in his company. Somewhere, they had similar nature and thoughts which made them perfect for each other. The narrator decided to do Kora with him. They decided to hire yaks as both of them did not believe in doing Kora on foot in the conventional way.
The best part was that in spite of being a Tibetan, Norbu could converse well in English. The narrator felt Norbu would be a perfect companion.

Question 5.
Discuss the accounts of exotic places in legends and the reality.
Answer:
Exotic places are those which are seemingly exciting and unusual. At times, they are connected to foreign countries. We have read about numerous exciting places in the legends i.e. in the ancient stories which take us into the ancient times. Often the accounts and descriptions of such places.

may be exaggerated or heightened not due to a desire to deceive, but due to the writer’s own heightened emotional response. If we visit these places, the reality may be quite different. We should not forget that while writing, the writer often dives into a deep world of fantasy and imagination. So, at times, the writer might accentuate a place. He never does this to mislead his readers, it is done out of writer’s feelings and emotions.

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Reflective Essay Topic – Necessity Is The Mother Of Invention

You can find Previous Year Reflective Essay Topics asked in ICSE board exams.

Reflective Essay Topic – Necessity Is The Mother Of Invention

Introduction: Universal truth

  • The journey of man has been a relentless pursuit of satisfying needs.
  • Examples of how need to communicate led to great discoveries
  • Advance made in medical and health care

Conclusion: Adds spice to life and makes it comfortable

‘Necessity is the mother of invention.’ There is much truth in this old adage. Ever since the advent of man in the universe, he has always been in pursuit of something to satisfy his needs. In order to satisfy hunger, he took to hunting wild animals. His desire to eat wholesome food, led him to discover fire and cultivate food grains.

To protect his body from the scourging heat and the cold weather, he covered it with fur of wild animals and subsequently with fabrics, made from cotton or wool. Indeed the journey of mankind has been a voyage of satisfying his different needs and wants.

No sooner has one need been satisfied, than a whole plethora of other needs arise. Thus what may have started as sign language to communicate, with fellow beings, gave rise to communication through language. This subsequently led to the need for recording communication, which was done through the evolution of alphabets.

Even today this need is being addressed by different media like telephone, cellular telephone, satellite phones and the Internet. To further augment the need to stay connected there are numerous networking sites like the Facebook, Whatsapp, Linkedin and numerous others.

Thus the necessity to communicate and stay connected led to the invention of different avenues to fulfill the need. In a similar manner our desire to do something new to meet the diverse needs, has led to stupendous development not only in the field of communication but in every sphere of human endeavour.

Thus in the field of transportation, the bullock carts made way to motor cars, trucks, trains and airplanes. In the field of medicine and health care, diseases that were thought to be incurable have been successfully treated. The discovery of new medicines and drugs have eradicated deadly diseases like plague and small pox.

Incurable diseases like cancer has been successfully treated by Chemotherapy and Radium treatment, while heart and other coronary diseases are treated with surgery. In our daily life we observe the universal truth of this statement, as we find new and more efficient ways of doing things.

Indeed necessity is the mother of inventions. Our ever growing needs and their satisfaction, has made the world a wonderful place to live. Our relentless endeavour to satisfy needs, not only makes us discover something new, but also adds spice to life. Had nature put everything before us on a platter, life would have been a eventless drudgery of routine.

Assignment

The journey of mankind has been a voyage in the pursuit of satisfying his needs and wants. Discuss.

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Selina Concise Biology Class 8 ICSE Solutions Chapter 8 Diseases and First Aid

Selina Concise Biology Class 8 ICSE Solutions Chapter 8 Diseases and First Aid includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 8 exams must go through Selina Concise Biology middle school Class 8 Textbook Solutions for Chapter 8 Diseases and First Aid. Students of Class 8 can avail the Chapter 8 Diseases and First Aid Selina ICSE Solutions for all the exercises here.

Selina Publishers Concise Biology Class 8 ICSE Solutions Chapter 8 Diseases and First Aid

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Selina Concise Biology Class 8 ICSE Solutions

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Weathering the Storm in Ersama Extra Questions and Answers Class 9 English Moments

Weathering the Storm in Ersama Extra Questions and Answers Class 9 English Moments

In this page you can find Weathering the Storm in Ersama Extra Questions and Answers Class 9 English Moments, Extra Questions for Class 9 English will make your practice complete.

Weathering the Storm in Ersama Extra Questions and Answers Class 9 English Moments

Weathering the Storm in Ersama Extra Questions and Answers Short Answer Type

Question 1.
Where had Prashant gone on 27th October 1999? Why?
Answer:
Ersama is a small town in coastal Odisha. Prashant had gone to block headquarters of Ersama on 27th October 1999. He went there to spend the day with a friend, seven years after his mother’s death.

Weathering the Storm in Ersama Extra Questions and Answers

Question 2.
Why was falling of two coconut trees on the roof of Prashant’s friend’s house a blessing in disguise?
Answer:
The falling of two coconut trees on the roof was a blessing in disguise because the tender coconuts from the trees kept the trapped family from starving during the fateful days that followed.

Weathering the Storm in Ersama Short Answer Type Questions and Answers

Question 3.
What was the effect of cyclone on Prashant’s village?
Answer:
Prashant was shocked after he reached his village. Where their house once stood, there were only . remnants. All the ninety-six houses in his village were washed away, Children were left orphaned. There was devastation all around. Eighty-six lives were lost in his village.

Question 4.
What did Prashant do for the orphans?
Answer:
Prashant decided to help the orphans. He brought them together and put up a polythene sheet shelter for them. Women were requested to look after them properly. He also arranged food for them.

Question 5.
What did Prashant do to divert the attention of the women?
Answer:
Prashant wanted to divert the attention of the women who had lost their family members. He persuaded them to start working in the food-for-work programme started by an NGO. He organised sports activities for children.

Question 6.
Narrate the storm in Ersama in your own words.
Answer:
The storm in Ersama was so fierce that it brought destruction and wreaked havoc everywhere. It brought a devastating flood that took many lives and left people homeless.

Weathering the Storm in Ersama Extra Questions and Answers Long Answer Type

Question 1.
How was Prashant’s village affected by the cyclone? How did he take a lead to help the people? What would have been your reaction in such a condition?
Answer:
Prashant saw that 2,500 people were crowded in a shelter. All the houses in his village were washed away. Eighty-six lives were lost in the village. There was not enough to eat. They had survived on green coconuts which were not adequate. Prashant organised a group of youths and elders. They pressurised the merchant to part with rice. After that the youth task force gathered empty utensils from the shelter.

The children were deputed to lie in the sand with these utensils on their stomach. They succeeded in sending the message to helicopters who threw food packets. As a responsible youth. I would have acted the same way as Prashant. I would lead the movement. I would have organised relief camps and help in arranging food and medicines. I would be an active member of the youth task force.

Question 2.
The true traits of an individual come to the fore in times of emergency. What traits come to one’s notice in Prashant, during the calamity that struck Ersama in coastal Orissa?
Answer:
The storm at Ersama, left a deep impact on the observant Prashant, who realised that he had never before seen such a destruction. His self-determination made him turn a deaf ear to all warnings and set out in search of his family, wading through water across 18 miles. Prashant, a born leader, soon organized the youth and elders of the village for pressurizing the local merchant to release rice stocks.

The innovative youngster made children lie on the sand with utensils on them so as to attract helicopter sorties to drop food for them. An optimist by nature, he got the children playing games of cricket to uplift spirits, and mobilized women for child care activities and men to secure food and materials. A highly cooperative individual he soon joined hands with volunteer NGOs to mend broken lives, and share others’ pains instead of wallowing in his own grief.

Disaster nursing quiz

Question 3.
In adversity, people like Prashant get extraordinary determination and power to fight. Do you agree that adversity or crisis brings people closer?
Answer:
Yes, it is true that some people, like Prashant, get extraordinary powers in adverse conditions. They become heroes. In the story, it was extraordinary courage and strong determination that made Prashant go to his house to seek his family. He covered his eighteen-kilometre long journey with great difficulty. There was a leader in him, although he was a teenager. He had a special quality of taking initiatives.

He helped all and one during the disaster. He organized a team of volunteers to help the victims of super cyclone in Orissa. Adversity or crisis also brings people together. There is a bond of humanity. The victims of cyclone in Ersama helped one another. They worked together during their resettlement. They took care of orphans and widows. Prashant managed to unite them and work for one another.

The Brave Little Bowman Question and Answers

 

Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5

Maharashtra State Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5

Question 1.
Yield of soyabean per acre in quintal in Mukund’s field for 7 years was 10, 7, 5,3, 9, 6, 9. Find the mean of yield per acre.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 1
Mean = 7
The mean of yield per acre is 7 quintals.

Question 2.
Find the median of the observations, 59, 75, 68, 70, 74, 75, 80.
Solution:
Given data in ascending order:
59, 68, 70, 74, 75, 75, 80
∴ Number of observations(n) = 7 (i.e., odd)
∴ Median is the middle most observation
Here, 4th number is at the middle position, which is = 74
∴ The median of the given data is 74.

Question 3.
The marks (out of 100) obtained by 7 students in Mathematics examination are given below. Find the mode for these marks.
99, 100, 95, 100, 100, 60, 90
Solution:
Given data in ascending order:
60, 90, 95, 99, 100, 100, 100
Here, the observation repeated maximum number of times = 100
∴ The mode of the given data is 100.

Question 4.
The monthly salaries in rupees of 30 workers in a factory are given below.
5000, 7000, 3000, 4000, 4000, 3000, 3000,
3000, 8000, 4000, 4000, 9000, 3000, 5000,
5000, 4000, 4000, 3000, 5000, 5000, 6000,
8000, 3000, 3000, 6000, 7000, 7000, 6000,
6000, 4000
From the above data find the mean of monthly salary.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 2
∴ The mean of monthly salary is ₹ 4900.

Question 5.
In a basket there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows:
60, 70, 90, 95, 50, 65, 70, 80, 85, 95.
Find the median of the weights of tomatoes.
Solution:
Given data in ascending order:
50, 60, 65, 70, 70, 80 85, 90, 95, 95
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, 5th and 6th numbers are in the middle position
∴ Median = \(\frac { 70+80 }{ 2 }\)
∴ Median = \(\frac { 150 }{ 2 }\)
∴ The median of the weights of tomatoes is 75 grams.

Question 6.
A hockey player has scored following number of goals in 9 matches: 5, 4, 0, 2, 2, 4, 4, 3,3.
Find the mean, median and mode of the data.
Solution:
i. Given data: 5, 4, 0, 2, 2, 4, 4, 3, 3.
Total number of observations = 9
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 3
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 4
∴ The mean of the given data is 3.

ii. Given data in ascending order:
0,2, 2, 3, 3, 4, 4, 4,5
∴ Number of observations(n) = 9 (i.e., odd)
∴ Median is the middle most observation
Here, the 5th number is at the middle position, which is 3.
∴ The median of the given data is 3.

iii. Given data in ascending order:
0,2, 2, 3, 3, 4, 4, 4,5
Here, the observation repeated maximum number of times = 4
∴ The mode of the given data is 4.

Question 7.
The calculated mean of 50 observations was 80. It was later discovered that observation 19 was recorded by mistake as 91. What Was the correct mean?
Solution:
Here, mean = 80, number of observations = 50
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
∴ The sum of 50 observations = 80 x 50
= 4000
One of the observation was 19. However, by mistake it was recorded as 91.
Sum of observations after correction = sum of 50 observation + correct observation – incorrect observation
= 4000 + 19 – 91
= 3928
∴ Corrected mean
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 5
= 78.56
∴ The corrected mean is 78.56.

Question 8.
Following 10 observations are arranged in ascending order as follows. 2, 3 , 5 , 9, x + 1, x + 3, 14, 16, 19, 20. If the median of the data is 11, find the value of x.
Solution:
Given data in ascending order :
2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20.
∴ Number if observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
∴ \( \text { Median }=\frac{(x+1)+(x+3)}{2}\)
∴ 11 = \(\frac { 2x+4 }{ 2 }\)
∴ 22 = 2x + 4
∴ 22 – 4 = 2x
∴ 18 = 2x
∴ x = 9

Question 9.
The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observations is 25. Find the 18th observation.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations
= Mean x Total number of observations
The mean of 35 observations is 20
∴ Sum of 35 observations = 20 x 35 = 700 ,..(i)
The mean of first 18 observations is 15
Sum of first 18 observations =15 x 18
= 270 …(ii)
The mean of last 18 observations is 25 Sum of last 18 observations = 25 x 18
= 450 …(iii)
∴ 18th observation = (Sum of first 18 observations + Sum of last 18 observations) – (Sum of 35 observations)
= (270 + 450) – (700) … [From (i), (ii) and (iii)]
= 720 – 700 = 20
The 18th observation is 20.

Question 10.
The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
The mean of 5 observations is 50
Sum of 5 observations = 50 x 5 = 250 …(i)
One observation was removed and mean of remaining data is 45.
Total number of observations after removing one observation = 5 – 1 = 4
Now, mean of 4 observations is 45.
∴ Sum of 4 observations = 45 x 4 = 180 …(ii)
∴ Observation which was removed
= Sum of 5 observations – Sum of 4 observations = 250 – 180 … [From (i) and (ii)]
= 70
∴ The observation which was removed is 70.

Question 11.
There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.
Solution:
Total number of students = 40
Number of boys =15
∴ Number of girls = 40 – 15 = 25
The mean of marks obtained by 15 boys is 33
Here, sum of the marks obtained by boys
= 33 x 15
= 495 …(i)
The mean of marks obtained by 25 girls is 35 Sum of the marks obtained by girls = 35 x 25
= 875 …(ii)
Sum of the marks obtained by boys and girls = 495 + 875 … [From (i) and (ii)]
= 1370
∴ Mean of all the students
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 6
= 34.25
∴ The mean of all the students in the class is 34.25.

Question 12.
The weights of 10 students (in kg) are given below:
40, 35, 42, 43, 37, 35, 37, 37, 42, 37. Find the mode of the data.
Solution:
Given data in ascending order:
35, 35, 37, 37, 37, 37, 40, 42, 42, 43
∴ The observation repeated maximum number of times = 37
∴ Mode of the given data is 37 kg

Question 13.
In the following table, the information is given about the number of families and the siblings in the families less than 14 years of age. Find the mode of the data.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 7
Solution:
Here, the maximum frequency is 25.
Since, Mode = observations having maximum frequency
∴ The mode of the given data is 2.

Question 14.
Find the mode of the following data.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 8
Solution:
Here, the maximum frequency is 9.
Since, Mode = observations having maximum frequency
But, this is the frequency of two observations.
∴ Mode = 35 and 37

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.5 Intext Questions and Activities

Question 1.
The first unit test of 40 marks was conducted for a class of 35 students. The marks obtained by the students were as follows. Find the mean of the marks.
40, 35, 30, 25, 23, 20, 14, 15, 16, 20, 17, 37, 37, 20, 36, 16, 30, 25, 25, 36, 37, 39, 39, 40, 15, 16, 17, 30, 16, 39, 40, 35, 37, 23, 16.
(Textbook pg, no. 123)
Solution:
Here, we can add all observations, but it will be a tedious job. It is easy to make frequency distribution table to calculate mean.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 9
= 27.31 marks (approximately)
∴ The mean of the mark is 27.31.

Maharashtra Board Class 9 Maths Solutions

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Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6

Maharashtra State Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6

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

i. For different types of investments what is the maximum permissible amount under section 80C of income tax ?
(A) ₹ 1,50,000
(B) ₹ 2,50,000
(C) ₹ 1,00,000
(D) ₹ 2,00,000
Answer:
(A) ₹ 1,50,000

ii. A person has earned his income during the financial year 2017-18. Then his assessment year is….
(A) 2016 – 17
(B) 2018 – 19
(C) 2017 – 18
(D) 2015 – 16
Answer:
(B) 2018 – 19

Question 2.
Mr. Shekhar spends 60% of his income. From the balance he donates ₹ 300 to an orphanage. He is then left with ₹ 3,200. What is his income ?
Solution:
Let the income of Shekhar be ₹ x.
Shekhar spends 60% of his income.
∴ Shekhar’s expenditure = 60% of x
∴ Amount remaining with Shekhar = (100 – 60)% of x
= 40% of x
= \(\frac { 1 }{ 2 }\) × x
= 0.4x
From the balance left, he donates ₹ 300 to an orphanage.
∴ Amount left with Shekhar = 0.4x – 300
Now, the amount left with him is ₹ 3200.
∴ 3200 = 0.4x- 300
∴ 0.4x = 3500
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 1
∴ The income of Mr. Shekhar is ₹ 8750.

Question 3.
Mr. Hiralal invested ₹ 2,15,000 in a Mutual Fund. He got ₹ 3,05,000 after 2 years. Mr. Ramniklal invested ₹ 1,40,000 at 8% compound interest for 2 years in a bank. Find out the percent gain of each of them. Whose investment was more profitable ?
Solution:
Mr. Hiralal:
Amount invested by Mr. Hiralal in mutual fund = ₹ 2,15,000
Amount received by Mr. Hiralal = ₹ 3,05,000
∴ Mr. Hiralal’s profit = Amount received – Amount invested
= 305000 – 215000 = ₹ 90000
Mr. Hirala’s percentage of profit
= \(\frac { 90000 }{ 215000 }\) × 100
= 41.86%

Mr. Ramniklal:
P = ₹ 140000, R = 8%, n = 2 years
∴ Compound interest (I)
= A – P
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 2
= 140000 [(1 + 0.08)2 – 1]
= 140000 [ (1.08)2 – 1]
= 140000(1.1664 – 1)
= 140000 x 0.1664
= ₹ 23296
∴ Mr. Ramniklal’s percentage of profit
= \(\frac { 23296 }{ 140000 }\) × 100
= 16.64%
∴ The percentage gains of Mr. Hiralal and Mr. Ramniklal are 41.86% and 16.64% respectively, and hence, Mr. Hiralal’s investment was more profitable.

Question 4.
At the start of a year there were ₹ 24,000 in a savings account. After adding ₹ 56,000 to this the entire amount was invested in the bank at 7.5% compound interest. What will be the total amount after 3 years ?
Solution:
Here, P = 24000 + 56000
= ₹ 80000
R = 7.5%, n = 3 years
Total amount after 3 years
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 3
= 80000 (1 + 0.075)3
= 80000 (1.075)3
= 80000 x 1.242297
= 99383.76
∴ The total amount after 3 years is ₹ 99383.76.

Question 5.
Mr. Manohar gave 20% of his income to his elder son and 30% to his younger son. He gave 10% of the balance income as donation to a school. He still had ₹ 1,80,000 for himself. What was Mr. Manohar’s income ?
Solution:
Let the income of Mr. Manohar be ₹ x.
Amount given to elder son = 20% of x
Amount given to younger son = 30% of x
Total amount given to both sons = (20 + 30)% of x = 50% of x
∴ Amount remaining with Mr. Manohar = (100 – 50)% of x
= 50% of x 50
= \(\frac { 50 }{ 100 }\) × 100
= 0.5 x
He gave 10% of the balance income as donation to a school.
Amount donated to school = 10% of 0.5x
= \(\frac { 10 }{ 100 }\) × 0.5x
= 0.05x
∴ Amount remaining with Mr. Manohar after donating to school = 0.5x – 0.05x
= 0.45x
Mr. Manohar still had 1,80,000 for himself after donating to school.
∴ 180000 = 0.45x
∴ \(x=\frac{180000}{0.45}=\frac{180000 \times 100}{0.45 \times 100}=\frac{18000000}{45}=400000\)
∴ The income of Mr. Manoliar is ₹4,00,000.

Question 6.
Kailash used to spend 85% of his income. When his income increased by 36% his expenses also increased by 40% of his earlier expenses. How much percentage of his earning he saves now ?
Solution:
Let the income of Kailash be ₹ x.
Kailash spends 85% of his income.
∴ Kailash’s expenditure = 85% of x
= \(\frac { 85 }{ 100 }\) × x = 0.85 x
Kailash’s income increased by 36%.
∴ Kailash’s new income = x + 36% of x
= x + \(\frac { 36 }{ 100 }\) × x
= x + 0.36x
= 1.36x
Kailash’s expenses increased by 40%.
∴ Kailash’s new expenditure = 0.85x + 40% of 0.85x
= 0.85x + \(\frac { 40 }{ 100 }\) × 0.85 × 100
= 0.85x + 0.4 × 0.85x
= 0.85x (1 + 0.4)
= 0.85x × 1.4
= 1.19x
∴ Kailash’s new saving = Kailash’s new income – Kailash’s new expenditure
= 1.36x – 1.19x
= 0.17x
Percentage of Kailash’s new saving
= \(\frac { 0.17x }{ 1.36x }\) × 100
= 12.5%
∴ Kailash saves 12.5% of his new earning.

Question 7.
Total income of Ramesh, Suresh and Preeti is ₹ 8,07,000. The percentages of their expenses are 75%, 80% and 90% respectively. If the ratio of their savings is 16 : 17 : 12, then find the annual saving of each of them.
Solution:
Let the annual income of Ramesh, Suresh and Preeti be ₹ x, t y and ₹ z respectively.
Total income of Ramesh, Suresh and Preeti = ₹ 8,07,000
∴ x + y + z = 807000 …(i)
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 4
∴ Savings of Ramesh = 25% of x
= ₹ \(\frac { 25x }{ 100 }\) ..(ii)
Savings of Suresh = 20% of y
= ₹\(\frac { 20y }{ 100 }\) …(iii)
Savings of Preeti = 10% of z
= ₹\(\frac { 10z }{ 100 }\) …..(iv)

Ratio of their savings = 16 : 17 : 12
Let the common multiple be k.
Savings of Ramesh = ₹ 16 k … (v)
Savings of Suresh = ₹ 17 k … (vi)
Savings of Preeti = ₹ 12 k .. .(vii)
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 5
∴ z = 120k …(x)
From (i), (viii), (ix) and (x), we get
64k + 85k + 120k = 807000
269k = 807000
k = \(\frac { 807000 }{ 269 }\)
k = 3000
∴ Annual saving of Ramesh = 16k
= 16 x 3000
= ₹ 48,000
Annual saving of Suresh = 17k
= 17 x 3000
= ₹ 51,000
Annual saving of Preeti = 12k
= 12 x 3000
= ₹ 36,000
The annual savings of Ramesh, Suresh and Preeti are ₹ 48,000, ₹ 51,000 and ₹ 36,000 respectively.

Question 8.
Compute the income tax payable by following individuals.
i. Mr. Kadam who is 35 years old and has a taxable income of ₹13,35,000.
ii. Mr. Khan is 65 years of age and his taxable income is ₹4,50,000.
iii. Miss Varsha (Age 26 years) has a taxable income of ₹2,30,000.
Solution:
i. Mr. Kadam is 35 years old and his taxable income is ₹13,35,000.
Mr. Kadam’s income is more than ₹ 10,00,000.
∴ Income tax = ₹1,12,500 + 30% of (taxable income -10,00,000)
= ₹ 1,12,500 + 30% of (13,35,000 – 10,00,000)
= 112500+ \(\frac { 30 }{ 100 }\) x 335000 100
= 112500+ 100500
= ₹ 213000
Education cess = 2% of income tax
= \(\frac { 2 }{ 100 }\) x 213000
= ₹ 4260.
Secondary and Higher Education cess
= 1% of income tax
= \(\frac { 1 }{ 100 }\) x 213000 100
= 2130
Total income tax = Income tax + Education cess + Secondary and higher education cess
= 213000 + 4260 + 2130 = ₹ 2,19,390
∴ Mr. Kadam will have to pay income tax of ₹ 2,19,390.

ii. Mr. Khan is 65 years old and his taxable income is ₹ 4,50,000.
Mr. Khan’s income falls in the slab ₹ 3,00,001 to ₹ 5,00,000.
∴ Income tax
= 5% of (taxable income – 300000)
= 5% of (450000 – 300000)
= \(\frac { 5 }{ 100 }\) x 150000 100
= ₹ 7500
Education cess = 2% of income tax
= \(\frac { 2 }{ 100 }\) x 7500
= ₹ 150
Secondary and Higher Education cess = 1 % of income tax
= \(\frac { 1 }{ 100 }\) x 7500
= 75
Total income tax = Income tax + Education cess + Secondary and higher education cess
= 7500+ 150 + 75
= ₹ 7725
Mr. Khan will have to pay income tax of ₹7725.

iii. Taxable income = ₹2,30,000
age = 26 years
The yearly income of Miss Varsha is less than ₹ 2,50,000.
Hence, Miss Varsha will not have to pay income tax.

Maharashtra Board Class 9 Maths Chapter 6 Financial Planning Problem Set 6 Intext Questions and Activities

Question 1.
With your parent’s help write down the income and expenditure of your family for one week. Make 7 columns for the seven days of the week. Write all expenditure under such heads as provisions, education, medical expenses, travel, clothes and miscellaneous. On the credit side write the amount received for daily expenses, previous balance and any other new income. (Textbook pg. no. 98)

Question 2.
In the holidays, write the accounts for the whole month. (Textbook pg. no. 98)

Question 3.
What is a tax? Which are different types of taxes? Find out more information on following websites
www.incometaxindia.gov.in,
www.mahavat.gov.in
www.gst.gov.in (Textbook pg. no. 99)

Question 4.
Obtain more information about different types of taxes from employees and professionals who pay taxes. (Textbook pg. no. 99)

Question 5.
Obtain information about sections 80C, 80G, 80D of the Income Tax Act. (Textbook pg. no. 103)

Question 6.
Study a PAN card and make a note of all the information it contains. (Textbook pg.no. 103)

Question 7.
Obtain information about all the devices and means used for carrying out cash minus transactions. (Textbook pg, no, 103)

Question 8.
Visit www.incometaxindia.gov.in which is a website of the Government of India. Click on the ‘incometax calculator’ menu. Fill in the form that gets downloaded using an imaginary income and imaginary deductible amounts and try to compute the income tax payable for this income. (Textbook pg.no. 107)
[Students should attempt the above activities on their own.]

Maharashtra Board Class 9 Maths Solutions

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Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3

Maharashtra State Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3

Question 1.
State the order of the surds given below.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 1
Answer:
i. 3, ii. 2, iii. 4, iv. 2, v. 3

Question 2.
State which of the following are surds Justify. [2 Marks each]
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 2
Answer:
i. \(\sqrt [ 3 ]{ 51 }\) is a surd because 51 is a positive rational number, 3 is a positive integer greater than 1 and \(\sqrt [ 3 ]{ 51 }\) is irrational.

ii. \(\sqrt [ 4 ]{ 16 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 3
= 2, which is not an irrational number.

iii. \(\sqrt [ 5 ]{ 81 }\) is a surd because 81 is a positive rational number, 5 is a positive integer greater than 1 and \(\sqrt [ 5 ]{ 81 }\) is irrational.

iv. \(\sqrt { 256 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 4
= 16, which is not an irrational number.

v. \(\sqrt [ 3 ]{ 64 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 5
= 4, which is not an irrational number.

vi. \(\sqrt { \frac { 22 }{ 7 } }\) is a surd because \(\frac { 22 }{ 7 }\) is a positive rational number, 2 is a positive integer greater than 1 and \(\sqrt { \frac { 22 }{ 7 } }\) is irrational.

Question 3.
Classify the given pair of surds into like surds and unlike surds. [2 Marks each]
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 6
Solution:
If the order of the surds and the radicands are same, then the surds are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 7
Here, the order of 2\(\sqrt { 13 }\) and 5\(\sqrt { 13 }\) is same and their radicands are also same.
∴ \(\sqrt { 52 }\) and 5\(\sqrt { 13 }\) are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 8
Here, the order of 2\(\sqrt { 17 }\) and 5\(\sqrt { 3 }\) is same but their radicands are not.
∴ \(\sqrt { 68 }\) and 5\(\sqrt { 3 }\) are unlike surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 9
Here, the order of 12\(\sqrt { 2 }\) and 7\(\sqrt { 2 }\) is same and their radicands are also same.
∴ 4\(\sqrt { 18 }\) and 7\(\sqrt { 2 }\) are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 10
Here, the order of 38\(\sqrt { 3 }\) and 6\(\sqrt { 3 }\) is same and their radicands are also same.
∴ 19\(\sqrt { 12 }\) and 6\(\sqrt { 3 }\) are like surds.

v. 5\(\sqrt { 22 }\), 7\(\sqrt { 33 }\)
Here, the order of 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) is same but their radicands are not.
∴ 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) are unlike surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 11
Here, the order of 5√5 and 5√3 is same but their radicands are not.
∴ 5√5 and √75 are unlike surds.

Question 4.
Simplify the following surds.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 12
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 13

Question 5.
Compare the following pair of surds.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 14
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 15
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 16
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 17

Question 6.
Simplify.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 18
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 19

Question 7.
Multiply and write the answer in the simplest form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 20
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 21

Question 8.
Divide and write form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 22
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 23

Question 9.
Rationalize the denominator.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 24
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 25
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 26

Question 1.
\(\sqrt { 9+16 }\) ? + \(\sqrt { 9 }\) + \(\sqrt { 16 }\) (Texbookpg. no. 28)
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 27

Question 2.
\(\sqrt { 100+36 }\) ? \(\sqrt { 100 }\) + \(\sqrt { 36 }\) (Textbook pg. no. 28)
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 28

Question 3.
Follow the arrows and complete the chart by doing the operations given. (Textbook pg. no. 34)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 29
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 30

Question 4.
There are some real numbers written on a card sheet. Use these numbers and construct two examples each of addition, subtraction, multiplication and division. Solve these examples. (Textbook pg. no. 34)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 31
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 32

Maharashtra Board Class 9 Maths Solutions

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