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Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 1 Heredity and Evolution Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 1 Heredity and Evolution

Question 1.
Complete the following diagram.

Answer:

Question 2.
Read the following statements and justify same in your own words with the help of suituble examples.
a. Human evolution began approximately 7 crore years ago.
Answer:

• Approximately around 7 crore years back the ice age began on the earth. In such conditions, dinosaurs became extinct. The evolution and diversity of mammals started during this time. Due to change in climate the forest cover also declined rapidly.
• Ancestors of monkey-like animals were Lemur like animals which evolved during this time period.
• The tails of these monkey-like creatures started vanishing very gradually around 4 crore years ago.
• The body and brain both increased in volume forming first ape like animals. The monkey like ancestors gave rise to two evolutionary links to apes and human like animals.
• Later, the human evolution took place by changes in the brain volume, the ability to walk upright, excessive use of hand for manipulations.
• This journey of human evolution began 7 crore years ago. But the true wise and intelligent man arose around 50,000 years ago.

b. Geographical and reproductive isolation of organisms gradually leads to speciation.
Answer:

• Every species survives in specific geographical conditions. The requirements of food and habitat, is specific for each species. Their reproductive ability and period is also different.
• Therefore, the individuals from one species cannot reproduce with individuals from other species.
• When they are separated by a distance or geographical barriers they are said to be isolated geographically.
• When they cannot reproduce with each other, they are said to be isolated reproductively.
• The ancestor species of both these subspecies may be the same but due to isolation over a very long-time duration, there is genetic variation between the two. Therefore, the isolation leads to speciation.

c. Study of fossils is an important aspect of study of evolution.
Answer:
Answer:

• Fossils offer palaeontological evidence for the evolutionary process.
• Due to some natural calamities the organisms get buried during ancient times.
• The impressions and remnants of such organisms remain preserved underground. The hot lava also traps some organisms or their impressions. All such formations form fossils.
• Study of fossils help the researcher to understand the characteristics of the organisms that existed in the past.
• Carbon dating method also helps in finding out exact age of the fossil. According to the structure of earth’s crust the fossils are obtained at specific depths.
• The oldest ones are obtained at the depth while the relatively recent ones occupy the upper surface. Thus fossils of invertebrates were seen in very old Palaeozoic era. Later were seen fossils of Pisces, Amphibia and Reptilia. The Mesozoic era was dominated by reptiles while Coenozoic era showed presence of mammals.
• In this way, study of fossils unfold the evolutionary secrets.

d. There is evidences of fatal Science among chordates.
[Please read the above question as: Among different chordates there are embryological evidences.]
Answer:

• Very young embryos of fish, amphibians, reptiles, birds and mammals show quite similar structure in the early stages.
• As the further growth takes place, they acquire different patterns.
• The initial similarity between the vertebrate embryos is an evidence that during evolution, there was a common ancestor for all the vertebrate classes.
• This is called embryological evidence for vertebrate evolution.

Question 3.
complete the statements by choosing correct options from bracket.
(Genes, Mutation, Translocation, Transcription, Gradual development, Appendix)
a. The causality behind the sudden changes was understood due to ………… principle of Hugo de Vries.
Answer:
Mutation

b. The proof for the fact that protein synthesis occurs through ……….. was given by George Beadle and Edward Tatum.
Answer:
Genes

c. Transfer of information from molecule of DNA to mRNA is called as …………… process.
Answer:
Transcription

d. Evolution means ………….
Answer:
Gradual development

e. Vestigial organ ……….. present in human body is proof of evolution.
Answer:
Appendix

Question 4.
Write short notes based upon the information known to you.
a. Lamarckism.
Answer:
(1) Lamarckism consists of two theories which were proposed by Jean Baptiste Lamarck. These are as follows: (a) Use and disuse of the organs (b) Inheritance of acquired characters.
(2) In theory of use and disuse of organs, Lamarck says : The characters of organs develop because specific activities that the organisms perform. If such organ is not used it gets degenerated. Thus the morphological changes take place due to activities or inactivity of a particular organism.
(3) To emphasise this theory, he quoted following examples. Due to constant extension of neck to eat foliage from the top of the trees, giraffe’s neck became long. Similarly blacksmith has strong arms due to constant work. Flightless ostrich and emu did not fly and hence their wings became useless. Aquatic birds like swan and duck made their feet suitable for swimming by living in water. Snake lost limbs as it tried burrowing mode.
(4) Such acquired characters are passed from one parental generation to the offspring. This is called inheritance of acquired characters.
(5) The theory of inheritance of acquired characters is not accepted as such transmission of acquired character does not take place. Only genetic characters are transmitted.

b. Darwin’s theory of natural selection.
Answer:

• Charles Darwin proposed the theory of natural selection after making many observations on different specimens. He published a concept ‘Survival of the fittest’.
• Darwin explains this concept as follows: All the organisms reproduce prolifically. Therefore, there is always a competition for food, mate, etc. Only adaptations for sustaining this struggle.
• Natural selection plays important role by selecting only those organisms which are fit to live. Those that do not have better adaptations, perish. Selected sustaining organisms then perform reproduction and form new species in a very long period of time.
• Darwin published his views in the book titled ‘Origin of Species’.

c. Embryology.
Answer:

• Embryology is the study of developing embryos.
• These embryos in their initial stages are very similar to each other.
• These similarities decrease later in the development.
• This similarity in initial stages indicate that these vertebrates have originated from a common ancestor.
• In evolutionary science, comparative study of embryos of various vertebrates provide evidence for evolution.

d. Evolution.
Answer:

• The sequential changes in the groups of living organisms that take place very gradually is called evolution.
• Evolution is also described as the formation of new species due to natural selection.
• The process of evolution takes millions of years for development and speciation of different organisms.
• Changes in stars and planets in space and the changes in biosphere occurring on the Earth are all included under study of evolution.
• Due to evolution organisms become fit, biodiversity is increased, and new species are created.
• Different scientists have put forth theories to explain the process of evolution. Among these Charles Darwin’s theory of natural selection and speciation is accepted worldwide.

e. Connecting link.
Answer:
Some living organisms possess some characters in them which are the distinctive features of different groups or phyla. Such individuals connect these two groups by sharing the characters of both and hence they are known as connective links.

Examples: (1) Peripatus: Peripatus is the connecting link between Annelida and Arthropoda. It shows characters of both animal phyla. Like annelid worm, it shows segmented body, thin cuticle and parapodia. Like an arthropod, it shows open circulatory system and tracheal system for respiration.
(2) Duck Billed platypus: This is a connecting link between reptiles and mammals. Like reptiles it lays eggs but like mammals it has mammary glands and hairy skin.
(3) Lung fish: Lung fish is a connecting link between fishes and amphibians. Though a fish, it shows lungs for respiration as in amphibian animals.
(4) Connecting links indicate the direction and hierarchy of evolution.

Question 5.
Define heredity. Explain the mechanism of hereditary changes.
Answer:
(1) Heredity: Heredity is the process by which the biological characters from parental generation are transmitted to the next generation through genes.

(2) The mechanism of hereditary changes:

• Mutation: Sudden change in the parental DNA can cause mutations. This results into changes in the hereditary characters.
• At the time of meiosis, the crossing over takes place. This creates new recombination of the genetic information. Therefore, the haploid gametes produced carry changed hereditary characters.

Question 6.
Define vestigial organs. Write names of some vestigial organs in human body and write the names of those animals in whom same organs are functional.
Answer:

• Vestigial organs are degenerated or underdeveloped organs of organisms which do not perform any function.
• According to the principle of natural selection, such organs are on the verge of disappearance. But it takes many millions of years for its complete vanishing.
• The vestigial organs in one animal may be of use but to other kind of the animal as they still perform regular functions.
• Appendix is vestigial for humans, it does not perform any function but in ruminant animals it is concerned with digestion.
• Ear muscles are vestigial for us but in monkeys and cattle they are functional.
• Names of vestigial organs in human body-Appendix, tail-bone or coccyx, wisdom teeth and body hair.

Question 7.
Answer the following questions.
a. How are the hereditary changes responsible for evolution?
Answer:
Hereditary characters are transmitted from parental generation to the offspring. These characters are maintained through inheritance. But the genes which are beneficial for the organisms in helping them to adapt to the environment are transmitted to the next generations in a greater proportion. This happens due to natural selection.

The process of evolution happens at a very slow pace. The favourable genes are preserved in the species as they bring about better survival of the individuals. Such individual reproduces more efficiently and evolve. The individuals with unfavourable genes are not selected by nature and are thus removed from the population through natural death. The fuel for evolution is thus truly supplied by the hereditary changes.

b. Explain the process of formation of complex proteins.
Answer:
The proteins are synthesised in following steps, viz. transcription, translation and translocation. Protein synthesis takes place according to the sequence of nucleotides present on the DNA molecule with the help of RNA molecules. This is known as central dogma of protein synthesis.

1. Transcription: In the process of transcription, mRNA is produced as per the nucleotide sequence on the DNA. For this the two strands DNA are separated. Only one strand participates in the formation of mRNA. The sequence of nucleotides which is complementary to that of present on DNA is copied on mRNA. Instead of thymine present in DNA, uracil is added on the mRNA. Transcription takes place in nucleus but the mRNA leaves nucleus, carries the genetic code and enters the cytoplasm. This genetic code is always in triplet form arid hence is known as triplet codon. The code for each amino acid always consists of three nucleotides.

2. Translation: Each mRNA may carry thousands of codons. But each codon is specific for only one amino acid. The tRNA molecule brings the required amino acid as per the code present on mRNA. There is anticodon on each tRNA which is complementary to the codon on mRNA. This process is known as translation.

3. Translocation: In translocation, the ribosome keeps on moving from one end of mRNA molecule to other end by distance of one triplet codon. While this process is taking place, rRNA, helps in joining the amino acids together by peptide bonds. The peptide chains later come together to form complex protein molecules.

c. Explain the theory of evolution and mention the proof supporting it.
Answer:
1. Theory of evolution:

• According to the theory of evolution, first living material was in the form of protoplasm which was formed in ocean.
• Gradually, it gave rise to unicellular organisms. Changes took place in these unicellular organisms which made them evolve into larger and more complex organisms.
• All evolutionary changes were very slow and gradual taking about 300 crore years to happen.
• Different types of organisms were developed as the changes and development that occurred in living organisms wefts all round and multi-dimensional.
• Hence, this overall process of evolution is called organizational and progressive.
• Variety of plants and animals developed from the ancestors having different structural and functional organization during the process of evolution.

2. Proof here means evidences of evolution.
These evidences are as follows:

• Morphological evidences
• Anatomical evidences
• Vestigial organs
• Palaentological evidences
• Connecting links
• Embryological evidences.

d. Explain with suitable examples importance of anatomical evidences in evolution. (July 2019)
Answer:

• There are similarities in the structure and anatomy of different animal groups. E.g. human hand, forelimb of bull, patagium of bat and flipper of whale are all similar in their internal anatomy. There is similarity in the bones and joints of all these specimens.
• External morphology does not show any similarity. Use of each of the organ is also different in different animals. Structurally, they may not be related.
• However, the similarities in the anatomy is an evidence that they may have a common ancestor.
• In this way, the anatomical evidence throws light on the process of evolution.

e. Define fossil. Explain importance of fossils as proof of evolution.
Answer:

• Fossils offer palaeontological evidence for the evolutionary process.
• Due to some natural calamities the organisms get buried during ancient times.
• The impressions and remnants of such organisms remain preserved underground. The hot lava also traps some organisms or their impressions. All such formations form fossils.
• Study of fossils help the researcher to understand the characteristics of the organisms that existed in the.past.
• Carbon dating method also helps in finding out exact age of the fossil. According to the structure of earth’s crust the fossils are obtained at specific depths.
• The oldest ones are obtained at the depth while the relatively recent ones occupy the upper surface. Thus fossils of invertebrates were seen in very old Palaeozoic era. Later were seen fossils of Pisces, Amphibia and Reptilia. The Mesozoic era was dominated by reptiles while Coenozoic era showed presence of mammals.
• In this way, study of fossils unfold the evolutionary secrets.

f. Write evolutionary history of modern man.
Answer:
(1) Ancestors of humans developed from animals which resembled lemur like animals.
(2) Around seven crore years ago, monkey-like animals evolved from some of these lemur like animals.
(3) Then after about 4 crore years ago, in Africa the tails of these monkey like creatures very gradually disappeared.
(4) Simultaneously, there was enlargement in their body and brain volume too. The hands also improved and were provided with opposable thumb. In this way, ape-like animals were evolved.
(5) These ape-like animals independently gave rise to two lines of evolution, one giving rise to apes like gibbon and orangutan in the South and North-East Asia and gorilla and chimpanzee which stayed in Africa around 2.5 crores of years ago.
(6) The other line of evolution gave rise to human like animals around 2 crore years ago.
(7) The climate became dry and this resulted into reduction of forest cover. This made arboreal apes to descend on the land and start terrestrial mode.
(8) Due to this, there were changes in the pefvic
girdle and vertebral column. The hands were also freed from locomotion and thus they became more manipulative.
(9) Later, journey of hominoid species started from around 2 crores years ago. The first record of human like animal is ‘Ramapithecus’ ape from East Africa.
(10) Ramapithecus → Australopithecus → Neanderthal man → Cro-Magnon are the important steps in human evolution.
(11) Neanderthal man was said to be the first wise man. The increasing growth of brain made man more and more intelligent and thinking animal.
(12) Later, more than biological evolution, it was cultural evolution, when man started agriculture, animal , rearing. There was development of civilizations, arts and science etc. About 200 years ago there were industrial inventions and thus man now rules the earth.

Project:

Project 1.
Make a presentation on human evolution using various computer softwares and arrange a group disscussion over it in the class room.

Project 2.
Read the book – ‘Pruthvivur Manus Uparcich’ written by Late Dr. Sureshchandra Nadkarni and note your opinion on evolution.

Can you recall? (Text Book Page No. 1)

Question 1.
Which component of the cellular nucleus of living organisms carries hereditary characters?
Answer:
The chromosomes made up of nucleic acids and proteins, present in the nucleus of the cell are the components that carry hereditary characters in living organisms.

Question 2.
What do we call the process of transfer of physical and mental characters from parents to the progeny?
Answer:
The process of transfer of physical and mental characters from parents to the progeny is called inheritance or heredity.

Question 3.
Which are the components of the DNA molecule?
Answer:
DNA molecule is made up of two helical strands consisting of deoxyribose sugar, phosphoric acid and pairs of nitrogenous bases. These three together is called a nucleotide.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
Darwin has published a book titled …………..
(a) Natural selection
(b) Mutation
(c) Fall of a sparrow
(d) Origin of species
Answer:
(d) Origin of species

Question 2.
The …………. man evolved about 50 thousand years ago.
(a) Cro Magnon
(b) Neanderthal
(c) Java man
(d) Ramapithecus
Answer:
(a) Cro Magnon

Question 3.
About 10 thousand years ago, ………….. started to practise agriculture.
(a) Gorilla
(b) wise man
(c) Ramapithecus
(d) Australopithecus
Answer:
(b) wise man

Question 4.
………………. can be considered as the first example of wise-man.
(a) Australopithecus
(b) Ramapithecus
(c) Cro Magnon
(d) Neanderthal man
Answer:
(d) Neanderthal man

Question 5.
………. is a connecting link between Annelida and Arthropoda. (March 2019)
(a) Duck-billed platypus
(b) Peripatus
(c) Lung fish
(d) Whale
Answer:
(b) Peripatus

Question 6.
………… years ago human brain was sufficiently evolved to call him wise man.
(a) 50,000
(b) 30,000
(c) 20,000
(d) 10,000
Answer:
(a) 50,000

Question 7.
The process by which the gene in the nucleotide suddenly changes its position is called ………. (Board’s Model Activity Sheet)
(a) translation
(b) translocation
(c) mutation
(d) transcription
Answer:
(c) mutation

Question 8.
…………. is not the vestigial organ in the human body. (Board’s Model Activity Sheet)
(a) appendix
(b) Coccyx
(c) Canine
(d) Wisdom teeth
Answer:
(c) Canine

Write whether the following statements are true or false with proper justification for your answer:

Question 1.
It takes thousands of years for a useful structure to disappear.
Answer:
False. (The useful structures of the body do not disappear. The functioning of the body is easier due to such organs. It takes thousands of years for a functionless organ to disappear.)

Question 2.
Dr. Har Govind Khorana was awarded Nobel prize for his invention and publication in the journal Radio carbon.
Answer:
False. (Willard Libby was awarded Nobel prize for his invention and publication in the journal Radio carbon.)

Question 3.
Mesozoic era was dominated by variety of mammals.
Answer:
False. (Mesozoic era dominated by variety of reptiles.)

Question 4.
It seems that invertebrates have been slowly originated from vertebrates.
Answer:
False. (Vertebrates have been slowly originated from invertebrates in course of evolution. The primitive type of organisms always give rise to complex life forms. The invertebrates from Palaeozoic era gradually gave rise to vertebrates.)

Question 5.
The decaying process of C-12 occurs continuously from the dead remains of living organisms.
Answer:
False. (The decaying process of C-14 occurs continuously from the dead remains of living organisms. C-12 is not radioactive and hence it does not show decaying process.)

Question 6.
The theory of natural selection which mentions ‘Survival of fittest’ is given by Lamarck.
Answer:
False. (The theory of natural selection which mentions ‘Survival of fittest’ is given by Darwin.)

Question 7.
Changes acquired during life time are transferred to next generation.
Answer:
False. (Changes acquired during life time are not heritable. They are not transferred to next generation. Only the genes are transferred to the next generation.)

Question 8.
Each species grows in specific geographical conditions and has specific food, habitat, reproductive ability and period.
Answer:
True. (Each species has specifically evolved characters due to evolution and speciation.)

Question 9.
Humans walking with upright posture were confined to Africa only during prehistoric period.
Answer:
False. (Humans walking upright existed in Africa and China, Indonesia of Asian continent too.)

Question 10.
Industrial society was established about 200 years ago.
Answer:
True. (After the development and specialization of human brain, he started indulging in science and technology. Before; this period the idea of industrialization was not existing.)

Match the columns:

Question 1.

Answer:
(1) Johann Gregor Mendel – Pioneer of the modern genetics.
(2) Hugo de Vries – Mutational theory.

Question 2.

Answer:
(1) Walter, Sutton – Chromosomes of grasshopper.
(2) Mclyn McCarthy – DNA is genetic material.

Question 3.

Answer:
(1) Morphological evidences – Shape and venation of leaf.
(2) Anatomical evidences – Human hand and fore limb of bull.

Question 4.

Answer:
(1) Palaeontological evidences – Remnants and impressions.
(2) Connecting links – Duck billed Platypus and Peripatus.

Find the odd one out:

Question 1.
Transcription, Translation, Translocation, Mutation
Answer:
Mutation. (All others are stages of protein synthesis.)

Question 2.
Bones of the hands, structure of nostrils, position of eyes, structure of ear pinnae
Answer:
Bones of the hands. (All the others are morphological evidences.)

Question 3.
Venation, Shape of seeds, Leaf petiole, Leaf shape
Answer:
Shape of seeds. (All the others are morphological evidences in plants.)

Question 4.
Human hand, wing of cockroach, forelimb of bull, flipper of whale
Answer:
Wing of cockroach. (All others are anatomical evidences, they are homologous organs.)

Identify the correlation between the first two words and suggest the suitable words in the fourth place:

Question 1.
mRNA : Transcription :: tRNA :…………
Answer:
Translation

Question 2.
Peripatus : Connecting link :: Appendix :……….
Answer:
Vestigial organs

Question 3.
Open circulatory system : Arthropods :: Thin cuticle and parapodia :………..
Answer:
Annelida

Question 4.
Between Annelida and Arthropoda : Peripatus ::……….: Lungfish
Answer:
Pisces/Fish and Amphibia

Question 5.
Theory of natural selection : Charles Robert Darwin :: Theory of inheritance of acquired characters :…………
Answer:
Jean Baptiste Lamarck

Question 6.
Survival of fittest : Darwin :: Acquired characters :……….
Answer:
Lamark

Question 7.
Wisdom teeth : Vestigial organs :: Lungfish :………..
Answer:
Connecting link.

Define the following:

Question 1.
Heredity.
Answer:
The transfer of biological characters from one generation to another through genes is called heredity.

Question 2.
Transcription.
Answer:

Question 3.
Translation.
Answer:
The process of bringing tRNA possessing anticodon that is complementary to the codon on mRNA for protein synthesis is called translation.

Question 4.
Translocation.
Answer:
The process of movement of the ribosome from one end of mRNA to other end by the distance of one triplet codon is called translocation.

Question 5.
Mutation.
Answer:
Sudden and drastic change that occurs in the genetic material is called mutation.

Question 6.
Species.
Answer:
The group of organisms that cap produce fertile individuals through natural reproduction is called a species.

Name the following:

Question 1.
Three Scientists who proved that except viruses, all living organisms have DNA as genetic material.
Answer:
Oswald Avery, Mclyn McCarthy and Colin MacLeod.

Question 2.
Genetic disorder caused due to mutation:
Answer:
Sickle cell anaemia.

Question 3.
Fish that can breathe with help of lungs:
Answer:
Lung fish.

Question 4.
Vestigial organs in human beings:
Answer:
Appendix, tail-bone or coccyx, wisdom teeth and body hair.

Question 5.
Important stages in the journey of human evolution:
Answer:

• Animals like Lemur
• Egyptopithecus
• Dryopithecus
• Ramapithecus
• Australopithecus
• Skilled Human
• Homo erectus i.e. Man with erect posture
• Neanderthal man
• Cro-Magnon man.

Distinguish between the following:

Question 1.
Transcription and Translation.
Answer:
Transcription:

1. In the process of transcription, the sequence of nucleotides present on the DNA molecule is copied
   and carried to the cytoplasm by mRNA.
2. The process of transcription takes place in nucleus.
3. During transcription, RNA is produced from DNA.
4. Only mRNA takes part in transcription.

Translation:

1. In the process of translation, the specific amino acids are picked up according to the codons brought by mRNA.
2. The process of translation takes place in ribosomes located in cytoplasm.
3. During translation, proteins are produced with the help of RNA.
4. mRNA, tRNA and rRNA take part in translation.

Question 2.
Ape and Human.
Answer:
Ape:

1. Brain of the apes is smaller in size.
2. Ape cannot walk upright.
3. Ape is less intelligent as compared to human.
4. Apes are arboreal in their habitat and they spend more time on the trees.
5. The forelimbs of ape are longer than the hind limbs.

Human:

1. Brain of humans is larger in size.
2. Humans can walk upright.
3. Human is considered to be the most intelligent animal.
4. Humans are terrestrial in their habitat. They cannot stay on the trees.
5. The forelimbs of humans are shorter than the hind limbs.

Give scientific reasons:

Question 1.
Some of the characters of parents are seen in their offspring.
Answer:

• The parental genes are transferred to their progeny through male and female gametes.
• These genes carry hereditary characters.
• Since they are transmitted from the parents to their offspring, one can see the parental characters in their offspring.

Question 2.
Darwin’s work on evolution has been a milestone.
Answer:
(1) Darwin has proposed two very important theories of evolution, viz. Theory of natural selection and Theory of origin of species.
(2) The evolution has taken place on the earth for last many crores of years.
(3) The exact nature and process of these evolutionary changes become clear after studying Darwinism. (4) The observations made by Darwin at that time are now tested according to the modern development in science and are found to be correct. Thus, his work is said to be a milestone.

Question 3.
Peripatus is said to be a connecting link between Annelida and Arthropoda.
Answer:

• Peripatus shows segmented body, thin cuticle, and parapodia-like organs.
• These characters are typical of Annelids.
• Similarly, it also shows tracheal respiration and open circulatory system which is a characteristic feature of Arthropods.
• Since Peripatus shares both these characters, it is said to be a connecting link between j Annelida and Arthropoda.

Question 4.
Vertebrates have been slowly originated from invertebrates.
Answer:

• When the carbon dating method was used to assess the age of fossils, it was understood that invertebrates were present on the earth much before the vertebrates.
• The fossils of invertebrates are present in lower layers of earth’s strata.
• They were seen in Palaeozoic era of geological time period. Vertebrates dominated during Coenozoic era.
• Their fossils are seen in the upper strata of the earth’s crust.
• The structural complexity also increased in vertebrates. All these facts indicate that Vertebrates have slowly originated from invertebrates.

Question 5.
During human evolution the hands became available for use.
Answer:

• During human evolution, the climate of earth started becoming dry.
• This resulted in loss of forest cover.
• The apes which were arboreal on the trees thus descended and started walking on land.
• The lumbar bones underwent change and the apes started walking upright on the grasslands.
• The vertebral column also underwent change. Due to upright posture the forelimbs were freed from locomotion.
• The legs started bearing the weight of the body and the hands became available for use.

Read the following statements and justify the same in your own words with the help of suitable examples:

Question 1.
Geographical and reproductive isolation of organisms gradually leads to speciation.
Answer:

• Every species survives in specific geographical conditions. The requirements of food and habitat, is specific for each species. Their reproductive ability and period is also different.
• Therefore, the individuals from one species cannot reproduce with individuals from other species.
• When they are separated by a distance or geographical barriers they are said to be isolated geographically.
• When they cannot reproduce with each other, they are said to be isolated reproductively.
• The ancestor species of both these subspecies may be the same but due to isolation over a very long-time duration, there is genetic variation between the two. Therefore, the isolation leads to speciation.

Question 2.
Study of fossils is an important aspect of study of evolution.
Answer:
(1) Fossils offer palaeontological evidence for the evolutionary process.
(2) Due to some natural calamities the organisms get buried during ancient times.
(3) The impressions and remnants of such organisms remain preserved underground. The hot lava also traps some organisms or their impressions. All such formations form fossils.
(4) Study of fossils help the researcher to understand the characteristics of the organisms that existed in the.past.
(5) Carbon dating method also helps in finding out exact age of the fossil. According to the structure of earth’s crust the fossils are obtained at specific depths.
(6) The oldest ones are obtained at the depth while the relatively recent ones occupy the upper surface. Thus fossils of invertebrates were seen in very old Palaeozoic era. Later were seen fossils of Pisces, Amphibia and Reptilia. The Mesozoic era was dominated by reptiles while Coenozoic era showed presence of mammals.
(7) In this way, study of fossils unfold the evolutionary secrets.

Question 3.
There is evidences of fatal Science among chordates.
[Please read the above question as: Among different chordates there are embryological evidences.]
Answer:

• Very young embryos of fish, amphibians, reptiles, birds and mammals show quite similar structure in the early stages.
• As the further growth takes place, they acquire different patterns.
• The initial similarity between the vertebrate embryos is an evidence that during evolution, there was a common ancestor for all the vertebrate classes.
• This is called embryological evidence for vertebrate evolution.

Question 4.
Human evolution began approximately 7 crore years ago.
Answer:

• Approximately around 7 crore years back the ice age began on the earth. In such conditions, dinosaurs became extinct. The evolution and diversity of mammals started during this time. Due to change in climate the forest cover also declined rapidly.
• Ancestors of monkey-like animals were Lemur like animals which evolved during this time period.
• The tails of these monkey-like creatures started vanishing very gradually around 4 crore years ago.
• The body and brain both increased in volume forming first ape like animals. The monkey like ancestors gave rise to two evolutionary links to apes and human like animals.
• Later, the human evolution took place by changes in the brain volume, the ability to walk upright, excessive use of hand for manipulations.
• This journey of human evolution began 7 crore years ago. But the true wise and intelligent man arose around 50,000 years ago.

Answer the following questions:

Question 1.
Answer the following questions: (March 2019)
(a) What do you mean by central dogma?
Answer:
Information about protein synthesis is present in DNA. As per this information, proteins are produced by DNA through RNA molecules. This is called central dogma.

(b) What is transcription?
Answer:
The process of synthesis of mRNA as per the nucleotide sequence present in DNA is called transcription. The nucleotide sequence on mRNA is complimentary to that of the single DNA strand used in synthesis. Instead of thymine, mRNA possesses uracil.

(c) What is meant by triplet codon?
Answer:
The code for each amino acids always consists of three nucleotides which is known as triplet codon.

Question 2.
Which animal is called a connecting link between Reptiles and Mammals? (Board’s Model Activity Sheet)
Answer:
Duck billed platypus is called a connecting link between Reptiles and Mammals.

Question 3.
In which way is science of heredity useful these days?
Answer:
The science of heredity is useful in the following ways:

• For diagnosis of hereditary disorders.
• For treatment of hereditary disorders
• For prevention of hereditary disorders
• For production of hybrid varieties of animals and plants
• For using microbes in the industrial processes.

Question 4.
What is meant by carbon dating method?
Answer:
(1) Carbon dating method is technique used for determining the age of fossils.
(2) After the death of the organisms, their consumption of carbon stops. But right from that moment the decaying process of C-14 occurs continuously.
(3) This results in change in the ratio between C-14 and C-12. C-12 is not radioactive as C-14.
(4) Thus the time passed since the death of a plant or animal is calculated by measuring the radioactivity of C-14 and ratio of C-14 to C-12 present in their body.
(5) The points noted during carbon dating are:

• The period after the organism has been dead.
• The activity of C-14 in the dead organism.
• Ratio between C-14 and C-12.

Question 5.
Answer the following questions:
(a) Describe briefly the Darwin’s theory of natural selection.
Answer:
Charles Darwin (1809-1882) proposed the theory of natural selection.
Theory of natural selection: ‘The survival of fittest’, i.e., organisms which are fit for survival, evolve while those that are not, perish. The natural selection thus acts to produce new species.

(b) What were the objections raised against Darwinism?
Answer:
Objections raised against Darwinism:

1. There are other factors too for evolution and just not the Natural Selection.
2. Arrival of useful and useless modifications were not explained by Darwin, though he said about the survival of the fittest.
3. He has not given any explanation about slow changes and abrupt changes occurring during evolution.

(c) Which book was published by Darwin to explain this theory? (Board’s Model Activity Sheet)
Answer:
Charles Darwin wrote the book ‘Origin of Species’.

Question 6.
What were the objections raised against Darwinism?
Answer:
Some of the main objections raised against Darwinism are as follows:

• There are other factors too for evolution and just not the Natural Selection.
• Arrival of useful and useless modifications were not explained by Darwin, though he said about the survival of the fittest.
• He has not given any explanation about slow changes and abrupt changes occurring during evolution.

Question 7.
Answer the following questions:
(a) Explain in brief-Lamarck’s principle of ‘use or disuse of organs’.
Answer:
The theory of use and disuse of organs says that the morphological characters of organism develop because of specific activities that the organisms perform. If some organ is not used it gets degenerated. If excessively, used, it develops. Thus, the morphological changes take place due to activities or non-working of a particular body parts in an organism.

(b) Give two examples.
Answer:
Due to constant extension of neck to eat foliage from the top of the trees, giraffe’s neck became long. Similarly, blacksmith has strong arms due to constant work. The flightless ostrich and emu did not fly and hence their wings became useless. Aquatic birds like swan and duck made their feet suitable for swimming by living in water. Snake lost limbs as it tried burrowing mode.

(c) What are acquired characters?
Answer:
Acquired characters are those characters which are obtained during the life time by any organism and passed on to next generations.

Write short notes:
(OR)
Write short notes based upon the information known to you:

Question 1.
Theory of evolution.
Answer:

• According to the theory of evolution, first living material was in the form of protoplasm which was formed in ocean.
• Gradually, it gave rise to unicellular organisms. Changes took place in these unicellular organisms which made them evolve into larger and more complex organisms.
• All evolutionary changes were very slow and gradual taking about 300 crore years to happen.
• Different types of organisms were developed as the changes and development that occurred in living organisms was all round and multi-dimensional.
• Hence, this overall process of evolution is called organizational and progressive.
• Variety of plants and animals developed from the ancestors having different structural and functional organization during the process of evolution.

By choosing appropriate words given in the bracket, complete the paragraph:

Question 1.
(translation, anticodon, tRNA, mRNA, amino acids, triplet codon, transcription, DNA)
The …….. formed in nucleus comes in cytoplasm. It brings in the coded message from DNA. The message contains the codes for amino acids. The code for each amino acid consists of three nucleotides. It is called as ‘………..’. Each mRNA is made up of thousands of triplet codons. As per the message on mRNA, ……… are supplied by the ………. For this purpose, tRNA has ‘…………’ having complementary sequence to the codon on mRNA. This is called ‘………..’.
Answer:
The mRNA formed in nucleus comes in cytoplasm. It brings in the coded message from DNA. The message contains the codes for amino acids. The code for each amino acid consists of three nucleotides. It is called as ‘triplet codon’. Each mRNA is made up of thousands of triplet codons. As per the message on mRNA, amino acids are supplied by the tRNA. For this purpose, tRNA has ‘anticodon’ having complementary sequence to the codon on mRNA. This is called ‘translation’.

Question 2.
(Cultural, agriculture, fire, brain, Cro-Magnon, Homo sapiens, Neanderthal)
Evolution of upright man continued in the direction of developing its ………. for the period of about 1 lakh years and meanwhile he discovered the ………. Brain of man, 50 thousand years ago had been sufficiently evolved to the extent that it could be considered as member of the species ………… Neanderthal man can be considered as the first example of wise-man. The ……….. man evolved about 50 thousand years ago and afterwards, this evolution had been faster than the earlier. About 10 thousand years ago, wise-man started to practise the ………. It started to rear the cattle-herds and established the cities. ………..development took place later.
Answer:
Evolution of upright man continued in the direction of developing its brain for the period of about 1 lakh years and meanwhile he discovered the fire. Brain of man 50 thousand years ago had been sufficiently evolved to the extent that it could be considered as member of the species Homo sapiens. Neanderthal man can be considered as the first example of wise-man. The Cro-Magnon man eyolved about 50 thousand years ago and afterwards, this evolution had been faster than the earlier. About 10 thousand years ago, wise-man started-to practise the agriculture. It started to rear the cattle-herds and established the cities. Cultural development took place later.

Read the paragraph and answer the questions given below:

With the help of RNA, the genes present in the form of DNA participate in the functioning of cell and thereby control the structure and functioning of the body. Information about protein synthesis is stored in the DNA and synthesis of appropriate proteins as per requirement is necessary for body. These proteins are synthesized by DNA through the RNA. This is called ‘Central Dogma’. mRNA is produced as per the sequence of nucleotides on DNA. Only one of the two strands of DNA is used in this process. The sequence of nucleotides in mRNA being produced is always complementary to the DNA strand used for synthesis. Besides, there is uracil in RNA instead of thymine of DNA. This process of RNA synthesis is called ‘transcription’.

Questions and Answers:

Question 1.
Which part of the cell control the structure and functioning of the body?
Answer:
Genes present in the form of DNA along with RNA control the structure and functioning of the body.

Question 2.
How is a specific protein synthesised in the cell?
Answer:
The information of protein synthesis is stored in the DNA which is utilised as per the requirement of the body. Later the proteins are synthesised by DNA through the RNA.

Question3.
What is the similarity between mRNA and DNA?
Answer:
The sequence of nucleotides on DNA is copied on mRNA. The nucleotide sequence on mRNA is thus complementary to DNA.

Question 4.
Give one difference between RNA and DNA.
Answer:
RNA has uracil instead of thymine which is present in DNA.

Question 5.
Define central dogma.
Answer:
Central dogma is the concept that proteins are synthesised by DNA through the RNA.

Diagram-based questions:

Question 1.
Observe the figure 1.3 of transcription given on page 9 in this chapter and answer the following questions:

(1) What is the sequence of nucleotides present on one strand of the DNA?
Answer:
A T G C A A T T

(2) According to the above sequence on DNA, what will be the transcribed sequence on the mRNA molecule?
Answer:
U A C G U U A A

(3) Which enzyme is taking part in the above process of transcription?
Answer:
RNA polymerase takes part in the process of transcription.

Question 2.
Observe the figure 1.5 of translation and translocation, given on page 9 this chapter and answer the following questions:

(1) Which is the initiation codon? Where is it present?
Answer:
AUG is the initiation codon, which is present on the mRNA.

(2) What are the types of RNA present inside the ribosome? Which triplet codon is present on it?
Answer:
There are two molecules of tRNA present inside the ribosome. The triplet codons present on them are UAC and AAG respectively.

(3) Which genetic code is present on mRNA that is leaving the nucleus? What must be the sequence on the DNA to have such code on mRNA?
Answer:
The mRNA that leaves the nucleus has genetic code: A U G U U C A A A
The genetic code on DNA therefore should be as follows: T A C A A G T T T

Question 3.
Observe the figure 1.6 given on page 10 from this chapter. Answer the following question based on your observations:

What is the significance of this figure from the viewpoint of evolution? Explain in brief.
Answer:
In the figure, the process of mutation is shown. The original nucleotide sequence of TGC is replaced by new mutated sequence GAT. The change in the nucleotide sequence will change the DNA.

This will result in the change in genes and then changing the hereditary characters. Due to such change in genes, the evolution proceeds. The mutation so formed can be minor or major. The greater the impact of the change, the evolution takes place rapidly. The mutation thereby produce recombinations leading to diversity.

Question 4.
Observe the picture and answer the following questions:

(1) Which evidence of evolution is shown in the picture?
Answer:
Embryological evidences of evolution are shown in this picture.

(2) What can be proven with this proof?
Answer:
The similarities in the initial embryonic stages of different vertebrates shows that there was a common origin of all of them. Thus embryological evidences prove that there was common vertebrate ancestor.

(3) Give one more example of evidence of evolution.
Answer:
Palaeontological evidences such as vestigial organs and connecting links are another examples of evolutionary evidences.

Question 5.
Which concept/theory do you remember after seeing this picture of Giraffes? Describe it in brief.

Answer:

• The picture is based on the Lamarck’s principle of ‘use and disuse of organs’.
• The morphological characters of organism develop because of specific activities that the organisms perform.
• If some organ is not used it gets degenerated. If excessively used, it develops further.
• Thus, the morphological changes take place due to activities or non-working of a particular body parts in an organism. Due to constant extension of neck to eat foliage from the top of the trees, giraffe’s neck became long.

Activity-based Questions:

Try this: (Text Book Page No. 4)

Observe the above images and note the similarities between given animal images and plant images.
Answer:
The above pictures of the animals show similarities such as structure of mouth, position of eyes, structure of nostrils and ear pinnae and body fur. In pictures of plants there are similarities in characters like leaf shape, leaf venation, leaf petiole, etc.
These above morphological evidences show that there may be a common ancestor for all of the species shown.

Observe and Discuss:

Question 1.
Observe the pictures given below. (Text Book Page No. 5)

Answer:
(1) Fossils offer palaeontological evidence for the evolutionary process.
(2) Due to some natural calamities the organisms get buried during ancient times.
(3) The impressions and remnants of such organisms remain preserved underground. The hot lava also traps some organisms or their impressions. All such formations form fossils.
(4) Study of fossils help the researcher to understand the characteristics of the organisms that existed in the.past.
(5) Carbon dating method also helps in finding out exact age of the fossil. According to the structure of earth’s crust the fossils are obtained at specific depths.
(6) The oldest ones are obtained at the depth while the relatively recent ones occupy the upper surface. Thus fossils of invertebrates were seen in very old Palaeozoic era. Later were seen fossils of Pisces, Amphibia and Reptilia. The Mesozoic era was dominated by reptiles while Coenozoic era showed presence of mammals.
(7) In this way, study of fossils unfold the evolutionary secrets.

Question 2.
Observe the pictures given and discuss the characters observed. (Text Book Page No. 6)

Answer:
Some living organisms possess some characters in them which are the distinctive features of different groups or phyla. Such individuals connect these two groups by sharing the characters of both and hence they are known as connective links.

Examples: (1) Peripatus: Peripatus is the connecting link between Annelida and Arthropoda. It shows characters of both animal phyla. Like annelid worm, it shows segmented body, thin cuticle and parapodia. Like an arthropod, it shows open circulatory system and tracheal system for respiration.
(2) Duck Billed platypus: This is a connecting link between reptiles and mammals. Like reptiles it lays eggs but like mammals it has mammary glands and hairy skin.
(3) Lung fish: Lung fish is a connecting link between fishes and amphibians. Though a fish, it shows lungs for respiration as in amphibian animals.
(4) Connecting links indicate the direction and hierarchy of evolution.

Project: (Do it your self)

Project 1.
Internet is my friend: (Text Book Page No. 3)
Collect the information from the internet about Big-Bang theory related with the formation of stars and planets and present it in your class.

Project 2.
Use of ICT: (Text Book Page No. 4)
Collect the information of geological dating and present it in the classroom.

Project 3.
Use of ICT: (Text Book Page No. 5)
Find how the vestigial organs in certain animals are functional in others. Present the information in your class and send it to others.

Project 4.
Internet is my friend: (Text Book Page No. 8)
Collect the pictures and information of various species of monkeys from internet.

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 4 Environmental management Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 4 Environmental management

Question 1.
Reorganize the following food chain. Describe the ecosystem to which it belongs.
Grasshopper – Snake – Paddy field – Eagle – Frog.
Answer:

• Correct food chain: Paddy field → Grasshopper → Frog → Snake → Eagle.
• Such food chain is seen in the terrestrial ecosystem. There are many biotic factors in the terrestrial ecosystem, such as insects, birds, mammals etc.
• The above example mentions about paddy field, so it must be in vicinity of coastal lands. There is water logging in the paddy fields. Therefore, it offers a habitat to the frogs.
• In the above example, paddy fields are producers in the ecosystem. The primary consumer is grasshopper. Secondary consumer is frog, tertiary consumer is snake and the apex consumer is eagle. On every trophic level the bacteria, fungi and some scavenging worms can act as the decomposers.
• In this ecosystem. the solar energy is transferred from the paddy crops to eagle in a step wise food
  chain.

Question 2.
Explain the statement – ‘We have got this Earth planet on lease from our future generations and not as an ancestral property from our ancestors.’
Answer:

• The earth was inhabited by older generations before us. We have replaced them.
• But during their life time, they have created hazardous impact on the earth. The industrialization, the quest for more and more natural resources, wars fought, the construction activities such as dams, roads and bridges, extensive deforestation, etc. were their thoughtless activities.
• All these activities were for development of mankind. But most of them have destroyed the delicate balance between the producers and different levels of consumers.
• Due to ever increasing population of human beings there is shortage of food, clothing and shelter. To procure these basic needs, we have exploited many natural resources causing destruction of the earth’s natural ecosystem.
• Now it is our turn to protect the earth as on the same planet the next generations have to survive. We have to hand over the ecosystems of the earth which are in perfect balance to the new generations.
• The future generations need a good quality of air, water and land along with all other living organisms.
• Due to problems like climate change, global warming, pollution, droughts, etc. the environment is impacted, thus in order to keep sustainability of earth, we have to remember that the earth has not been obtained only as ancestral property but we have to save it for future generations.

Question 3.
Write short notes.
a. Environmental conservation.
Answer:
Due to natural and man-made causes, there are many environmental problems on the earth. These problems affect the existence of various living organisms. In order to save these organisms and maintain the environmental balance, there is need for environmental conservation. If this is not done then there will not be any quality of life for the resident humans. For environmental conservation, the Government has formulated acts and rules. UN has established UNEP for the conservation programs.

The people’s participation in the conservation movement is essential. From school age, the environmental values are inculcated in the young minds. Conservation of environment is the social responsibility of everyone. Judicial use of natural resources conservation also way of environmental conservation.

b. Chipko Movement of Bishnoi.
Answer:
Chipko Movement of Bishnoi or Bishnoi Andolan:
Khejarli or Khejadli is a village in Rajasthan, where Bishnoi community is located. The name of the town is derived from Khejri trees.

The first event of Chipko Movement took place in Khejadli village in 1730 AD. In this village 363 Bishnois, led by Amrita Devi sacrificed their lives for protecting the trees of Khejri trees, which trees are considered as sacred by Bishnoi.

Amrita Devi said, “if a tree is saved even at the cost of one’s head, it’s worth it”. She was killed with the axes that were brought to chop off the trees. The three young daughters Asu, Ratna and Bhagubai also sacrificed their lives for trees.

83 Bishnoi villages came together and villagers sacrificed their lives after hearing about Amrita Devi’s sacrifice. Three hundred and sixty-three Bishnois were killed as they opposed the king. After realizing the mistake, the king ordered stoppage of the felling of trees. Honouring the courage of the Bishnoi community, the ruler of Jodhpur, Maharaja Abhay Singh, apologized. He issued a royal decree to protect trees and wild life.

Chipko movement of 20th century in Uttar Pradesh also followed the same pattern of embracing the trees and saving them from cutting.

c. Biodiversity.
Answer:
Biodiversity means the diverse life forms that inhabit any area. Biodiversity is seen due to variety of life forms and different ecosystems that lodge these organisms. In nature there is biodiversity on the three different levels, viz. genetic diversity, species diversity and ecosystem diversity. This means that there is diversity in the individuals belonging to the same specips due to genetic reasons, there is diversity among the different species of organisms and there is also a diversity in the ecosystems that are present in any region.

Due to development of mankind, the biodiversity is threatened. There are special efforts taken to restore the lost and threatened biodiversity. Some of these are establishing sanctuaries, National Parks, biodiversity hotspots and reserves etc. Certain endangered species are protected by carrying out conservation projects.

d. Sacred Groves.
Answer:
Sacred grove is the green patch of the forest which is conserved by local people in the name of God. It does not belong to forest department. It is like a sanctuary that is conserved by the common people and tribals in the area. It is rich in the biodiversity.

It is conserved as there is a faith that God or deity reside in the sacred grove. Hence in local language, they are known as Deorai. Due to this reason, people do not fell the trees. Also hunting of any wild life is not done here. More than 13000 sacred groves have been reported in India. Most of these are in Western Ghats in Maharashtra, Karnataka and Kerala. Also, in remaining parts of India sacred groves are reported. Role of sacred grove is tremendous in conserving the biodiversity.

e. Disaster and its management.
Answer:

• To save human life from disasters. To help them for moving away from the place of disasters by rapid action.
• To supply essential commodities to the affected people. This helps to reduce the gravity of disaster. People are given grains, water and clothes and other basic necessities under this objective.
• To bring back the conditions of affected people to normalcy.
• To rehabilitate the affected and displaced victims.
• To think and execute the protective measures in order to develop capability to face the disasters in
  future.

Question 4.
How will you justify that overcoming the pollution is a powerful way of environmental management?
(OR)
“Solving the problem of pollution is an effective way of environmental management.” Justify the statement.
Answer:
1. Pollution is created only due to human activities. Air, water, soil, noise, radiation, thermal, light, plastic are different types of pollution.
2. All types of pollution affect environment and particularly threatening the survival of living organisms.
3. Pollution must be controlled in order to have good quality of the environment. E.g. When plastic is thrown anywhere, it causes pollution of the land, it clogs the rain water drains, it affects feeding of the animals. Plastic pollution can be completely stopped by us through proper management of plastic waste. By recycling or reusing, we can overcome the plastic pollution. This would be a powerful way of environmental management.
4. Similarly, when we reduce pollution of different types, we automatically help to regain the environmental health.

Question 5.
Which projects will you run in relation to environmental conservatioh? How?
(OR)
Write six strategies implemented by you for conservation of the environment.
Answer:
Initially, assessment of the environmental problems will be done. The nature and severity of these problems will be understood by detailed study of the same. Then the projects can be undertaken to combat these problems.

1. Tree plantation is one such easier project that can be undertaken to conserve environment. The further nurturing of the tree will also be our responsibility. While selecting the tree, the local and sturdy varieties will be selected. Such trees can survive in polluted environment too and even under the pressure of urbanization.

2. Solid waste management is another very important project that should be undertaken by every society, colony or school. Segregation of waste into dry and wet types and then its proper disposal will be taught to all the people in the neighbouring area.

3. To ban the plastic and make people aware about harmful effects of plastic is another very significant project.

4. Fossil fuels are non-renewable and polluting. Therefore, their use should be reduced as far as possible. Therefore, using bicycle, or walking down for shorter distances or using public transport systems are the better alternatives. The awareness drive about these facts will be taken up as a project.

5. To take care of stray animals, provide shelter, feeding endangered birds like sparrows and allowing them to survive with our support is also one of the essential act to conserve other species.

6. Attempts will be made for bringing awareness among minds of everyone. Such small acts can bring about major shift in the attitude of the people. This will certainly help in the environmental conservation.

Question 6.
Answer the following:
a. Write the factors affecting environment.
Answer:

• The biotic and abiotic- factors affect the environment. Among abiotic factors, the physical and chemical factors can alter the conditions of the environment.
• Abiotic factors are either natural or man-made.
• The various interrelationships between different living organisms can also affect environment.
• The natural disasters such as earthquake, forest fires, cyclones, cloud bursting, drought, etc. change the environment.
• The human activities such as deforestation, urbanisation, constructions etc. cause permanent damage to the ecosystems. Due to man-made impact, there can be large scale changes in the environment.

b. Human beings have important place in environment.
Answer:

• Man came last on the earth during evolution of animals. But due to his intelligence, imagination, critical thinking and memory, he made progress in all fields.
• By virtue of these qualities he became the supreme.
• All the natural resources on the earth were very rapidly exploited by man.
• Under the pretext of technology and. development he made degradation of almost all¬natural ecosystems.
• He never obeys the rules of nature.
• Phenomena like pollution, urbanization, industrialization and deforestation are exclusively his creations.
• Hunting and poaching other animals were his contribution to the extinction of many other animals.
• Except man no other organism on the earth can change the ecosystems in such a drastic way. Therefore, it is rightly said that human beings have important place in environment.

c. Write the types and examples of biodiversity.
Answer:
Biodiversity is documented on the following three levels, viz. genetic diversity, species diversity and ecosystem diversity.
1. Genetic Diversity: Diversity seen among the organisms of same species due to genetic differences is called genetic diversity. E.g. The individual human beings are different from each other. No two animals or plants are exactly alike.

2. Species Diversity: The difference between the different species is the species diversity, e.g. All the species of plants, animals and microbes which are seen in any natural environment.

3. Ecosystem Diversity: In one region there may be different ecosystems, such diversity in the ecosystems is called ecosystem diversity. Ecosystems are natural or artificial. Every region shows different types of ecosystems such as aquatic, terrestrial, desert or forest ecosystems. Each ecosystem has its own habitats with resident flora and fauna.

d. How the biodiversity can be conserved?
Answer:
Biodiversity can be conserved by the following ways:

• Protection of the rare species of plants and animals.
• Creating habitats for the animals and plants by establishing National Parkland Sanctuaries.
• Declaration of bioreserves, the areas which are protected through conservation.
• Conservation projects for protecting special species.
• Conservation of all flora and fauna.
• Strict observance of the acts and rules.
• Use of traditional knowledge and maintaining record of traditional knowledge.

e. What do we learn from the story of Jadav Molai Payeng? (Board’s Model Activity Sheet)
Answer:
Jadav Molai Payeng is a common man who was just a simple forest worker. But he has conscience about plants and tree plantations. He single-handedly planted thousands of trees. He converted a barren patch of land into forest which is spread over 1360 acres. For these plantations he continuously worked.

He has shown that a single determined person, can establish a new forest! We understand the values of
hard work, sincerity and devotion to the nature through the story of Jadav Molai Payeng. Even a common man can contribute a lot for the conservation and protection of the environment by learning the story of Payeng.

f. Write the names of biodiversity hot spots.
Answer:

• In entire world, 34 highly sensitive biodiversity spots are reported.
• These hotspots occupied 15.7% area of the Earth.
• However, currently about 86% of the sensitive areas are already destroyed.
• Now about 2.3% area of the Earth still has such sensitive biodiversity spots.
• There are 1,50,000 plant species which are about 50% of the species in the world.
• In India, out of 135 species of animals, 85 species are found in the jungles of eastern region.
• There are about 1,500 endemic plant species in Western Ghats.
• About 50,000 plants species out of the total plants in the world are said to be endemic.

g. Which are the reasons for endangering the many species of plants and animals? How can we save those diversity?
Answer:

• The animals and plants species are endangered majorly due to man-made causes.
• Some natural disasters like earthquakes, climate change, forest fires, drought and cyclones also affect the living organisms due to lack of food and water.
• In man-made causes, hunting and poaching are the main reasons.
• Also animal-human conflicts occur due to invasion of human settlements into the habitats of wild animals.
• Construction of dams, roads, and colonies destroy the habitats of wild life.
• Industrialization, urbanization and population explosion of humans are putting severe pressure on all the existing biodiversity.
• In order to save and protect the biodiversity, many scientists and naturalists come together. A stretch of land is protected by declaring it as the sanctuary or a national park by the Government. Even the locals can protect it as a sacred grove.
• Various acts and rules have been formulated to protect the organisms. The violators of such rules are punished accordingly.

Question 7.
What are the meanings of the following symbols? Write your role accordingly. (July ’19; Board’s Model Activity Sheet)
(OR)
What do these symbols indicate? Explain your opinion about those symbols.

Answer:
1. The first symbol is for giving the message, “Reduce, reuse and recycle”. This is important mantra for the utilization of natural resources.
The second symbol gives the message about ‘Save water’.
The third symbol advocates the use of solar energy.

2. These symbols inculcate the importance of being eco-friendly. The first symbol is essential to maintain the natural resources by reusing and recycling them. As far as possible, one should reduce the excessive use of resources by preventing consumerism.

3. Water problems persist in many major cities and villages. In villages it results in drought like conditions. It also reflects into loss of agricultural produce. Therefore, the message about saving water or to make judicious use of water should be spread far and wide.

4. The solar energy is the renewable energy option which is very easily available in country like India. By using solar energy, we can replace the polluting and exhaustible fuels. Thereby, pollution will also be reduced.

Due to such symbols, important messages about environment conservation reach, us and we can change ourselves into more eeofriendly persons.

Project: (Do it your self)

Project 1.
Make a presentation on pollution of Gangci and Yamuna Rivers and effects of air pollution on Taimuhal.

Can you recall? (Text Book Page No. 36)

Question 1.
What is ecosystem? Which are its different components?
Answer:
In any environment, there are biotic and abiotic components. There are interactions among these components. All such interactions make an ecosystem.

The different components in the ecosystem are as follows:
Abiotic components : Air, water, soil, sunlight, temperature, humidity, etc.
Biotic components: All the types of living organisms, like bacteria, fungi, plants and animals.

Question 2.
Which are the types of consumers? What are the criteria for their classification?
Answer:
Primary consumers, secondary consumers, tertiary consumers or apex consumers are the different types of consumers. These types are according to the trophic level to which they belong.

Question 3.
What may be the relationship between lake and birds on tree?
Answer:
The birds on the tree depend on the aquatic organisms in the lake for their feeding. Birds stay on the trees which are in the vicinity of the lake, so that it is easier for them to capture fishes, frogs, etc. They must also be using the same lake water for drinking.

Question 4.
What is difference between food chain and food web?
Answer:
In every ecosystem, there are always interactions between producers, consumers and decomposers. This sequence of feeding interactions is called food chain. In every food chain there are links between four to five trophic levels constituting the producers, primary consumers, secondary consumers, tertiary consumers, etc. The links of food chain are in linear sequence. But food web is a complex network of many small food chains. In fact, food web is the collection of many small food chains. Thus, when many food chains are interwoven, they form food web.

Think and Answer! (Text Book Page No. 36)

Question 1.
Write the name and category of each of the component shown in picture.
Answer:
By utilizing the solar energy, the green plants perform photosynthesis. Thus, they are producers of the food chain. This food is consumed by the grasshopper. Thus, it is primary consumer. Frog is secondary consumer as its diet consists of insects like grasshopper. Snake is tertiary consumer as it feeds on frogs, while the hawk is apex consumer as it can kill the snake and feed on it. Last picture in the food chain is of fungi which are acting as decomposers. Few bacteria are shown in the picture, act on all the levels and bring about decomposition.

Question 2.
What is necessary to convert this picture into food web? Why?
Answer:
If this food chain has to be converted into a food web, there should be interactions between the different components. Any living organism can be prey to different predators. Moreover, a predator can also be a prey for other. Frog eats different insects. The same frog can be either eaten by snake or by hawk.

Use your brain power. (Text Book Page No. 40)

Question 1.
Why is it said that pollution control is important?
Answer:
The quantity of pollutants and severity of their effects on the ecosystem have to be taken into consideration constantly. The different methods of pollution control have to be used for checking the hazardous effects of pollution on the living organisms. Especially the impact of pollution on health of human beings is assessed from time to time.

The young children and senior citizens are affected to greater extent by the pollution. If the air and water required for the survival of the people is affected, then exercising the pollution control is to be done immediately. Thus, it is said that pollution control is important.

Enlist and discuss (Text Book Page No. 43)

Question 1.
Find the meaning of given symbols in relation to environment conservation. Make a list of other such symbols.
A.
Answer:
This symbol tells us to keep our wastes carefully. The garbage should not be strewn anywhere. But it should be properly managed. Waste if managed properly can be a wealth.
B.
Answer:
This symbol tells us to save electricity. If electricity is carefully used, we can save our natural resources. This message is given through this picture.
C.
Answer:
Use of bicycle means use of green energy. By riding on a bicycle we save on fuel and use our own muscular energy. It is the best eeofriendly, non polluting vehicle.

Observe and fill the information: (Text Book Page No. 8)

Question 1.
Observe the environment around you. Complete the following flow chart.

Answer:

Complete the Chart: (Text Book Page No. 39)

Question 1.
We have studied the air pollution, water pollution and soil pollution in detail in earlier classes. Based on that, complete the following chart.

Answer:

Complete the Chart: (Text Book Page No. 40)

Question 1.
Now a day, we are observing the environmental degradation everywhere. Complete the flow chart given besides with the help of environment.

Answer:

(Text Book Page No. 43)

Question 1.
Where are such sacred groves in Maharashtra? Make a list and visit with your teachers.
Answer:
Sacred groves: Sacred groves form an important landscape feature in the deforested hill ranges of the Western Ghats. The felling of timber and the killing of animals in sacred groves is not allowed by the locals. It is considered as taboo.

In Maharashtra, sacred groves are found in tribal as well as non-tribal areas. The sacred groves in the western part are called Devrai or Devrahati, which means the abode of the gods. In eastern parts it is called Devegudi by the madiya tribal people.

In Maharashtra 2820 Devrais have been documented. Maruti, Vaghoba, Vira, Bhiroba, Khandoba and Shirkai are some deities to which sacred groves are dedicated.

In the sacred groves, the most commonly found plant species are Portia tree, Casuarina, Silk cotton tree, Indian laurel, Indian Elm, Bead tree, Indian butter tree, Turmeric and Japanese ginger. In Maharashtra, sacred groves are maximum in district of Sindhudurg, (More than 1500 out of total 2820) followed by Ratnagiri, then Pune and in district of Satara.

Choose the correct alternative and write that alphabet against the sub-question number:

Question 1.
Paddy fields are frequently attacked by
(a) goats
(b) birds
(c) grasshoppers
(d) monkeys
Answer:
(c) grasshoppers

Question 2.
Basic functional unit to study the ecology is termed as ……………
(a) environment
(b) niche
(c) ecosystem
(d) food chain
Answer:
(c) ecosystem

Question 3.
As per ……….. trading of rare animals has been completely banned.
(a) clause 48A
(b) clause 49B
(c) clause 49A
(d) all the above
Answer:
(c) clause 49A

Question 4.
(4) The jungle in Kokilamukh of Jorhat district of Assam is well known as ………….
(a) Molai jungle
(b) Rhino jungle
(c) Rhino forest
(d) Payang jungle
Answer:
(a) Molai jungle

Question 5.
Maintaining record of ………. knowledge is very necessary.
(a) modern
(b) mythical
(c) vedic
(d) traditional
Answer:
(d) traditional

Question 6.
………… is world’s largest organization engaged in environmental activities.
(a) Greenpeace
(b) Hariyali
(c) B. N. H. S.
(d) I. I. T.
Answer:
(a) Greenpeace

Question 7.
……….. sanctuary of West Bengal is reserved for tigers.
(a) Gir
(b) Sunderban
(c) Molai
(d) Corbett
Answer:
(b) Sunderban

Question 8.
World Biodiversity Day is celebrated on ……… every year.
(a) 22nd April
(b) 5th June
(c) 16th September
(d) 22nd May
Answer:
(d) 22nd May

Question 9.
Out of the total plant species in the entire world, 50,000 are ……………
(a) extinct
(b) endangered
(c) endemic
(d) rare
Answer:
(c) endemic

Question 10.
Giant squirrel is an ………… species.
(a) indeterminate
(b) rare
(c) endemic
(d) endangered
Answer:
(a) indeterminate

Question 11.
In a food chain, autotrophic plants are present at the ……….. level. (March 2019)
(a) tertiary nutrition
(b) secondary nutrition
(c) producer
(d) apex
Answer:
(c) producer

Question 12.
……….. from Manas sanctuary in Assam is under threat. (Board’s Model Activity Sheet)
(a) one horned rhino
(b) Lion
(c) Musk deer
(d) Giant squirrel/Shekru
Answer:
(a) one horned rhino

Write whether the following statements are true or false, giving suitable explanation for the same:

Question 1.
Only abiotic factors play very important role in the ecosystem.
Answer:
False. (Both abiotic and biotic factors play very important role in the ecosystem. Only abiotic factors will not decide the working of an ecosystem.)

Question 2.
Paddy fields are frequently attacked by frogs.
Answer:
False. (Paddy fields are frequently attacked by grasshoppers. Frogs feed on grasshoppers and control the population of these insects that cause destruction of the crops.)

Question 3.
Environmental pollution is necessary and acceptable change in the surrounding environment.
Answer:
False. (Environmental pollution is never acceptable. It is always harmful to the entire ecosystem and thus never necessary.)

Question 4.
X-rays and radiations from atomic energy plants are natural radiations.
Answer:
False. (X-rays are not present in natural radiations. Infra-red and ultra-violet rays are present in natural radiations.)

Question 5.
The person breaching the Environmental Conservation Act is entitled for either one year imprisonment or fine up to ₹ 5 lakh.
Answer:
False. (The person breaching the Environmental conservation Act is fined upto ₹ 1 lakh. He is also entitled to imprisonment for five years.)

Question 6.
Many people come together to establish arnew forest but a single person, if determined can destroy the entire forest!
Answer:
False. (When anything constructive has to be done even a single man can start such action. In case of ‘Molai jungle’, this statement holds true. But when destructive actions are done, many people come together and cause damage.)

Question 7.
There are clusters of thick forests only in Western Ghats of India.
Answer:
False. (Entire India is rich in biodiversity. Just not in Western Ghats but in entire India one can observe the clusters of thick forests and this is mainly due to suitable tropical climate.)

Question 8.
86 highly sensitive biodiversity spots are reported all over the world.
Answer:
False. (As per the latest information and available data, there are 34 highly sensitive biodiversity spots.)

Question 9.
Flow of nutrients in an ecosystem is unidirectional.
Answer:
False. (Flow of energy in an ecosystem is unidirectional. Flow of nutrients is cyclic.)

Match the columns:

Question 1.

Answer:
(1) Physical, chemical and biological factors together form – Environment.
(2) The science of interactions between biotic and abiotic factors – Ecology.

Question 2.

Answer:
(1) Basic functional unit in the environment – Ecosystem.
(2) Different types of living organisms – Biodiversity.

Question 3.

Answer:
(1) Sound Pollution (Control and Prevention) Rule – 2000.
(2) Biomedical Waste (Management and Handling) Rule – 1998.

Question 4.

Answer:
(1) Forest Conservation Act – 1980.
(2) Environmental Conservation Act – 1986.

Question 5.

Answer:
(1) Endangered Species – Lion tailed monkey, lesser florican.
(2) Rare Species – Red panda, Musk deer.

Question 6.

Answer:
(1) Vulnerable Species – Tiger, Lion.
(2) Indeterminate Species – Giant squirrel (Shekhru).

Find the odd one out:

Question 1.
Ash, Carbon dioxide, Lead, Asbestos
Answer:
Carbon dioxide. (All others are solid particulate pollutants.)

Question 2.
Manas sanctuary, Sunderbans sanctuary, The Western Ghats, Tadoba National Park
Answer:
Tadoba National Park. (All others are endangered heritage places of India.)

Question 3.
Lion tailed monkey, White rats, Musk deer, Tiger
Answer:
White rats. (All others are species that are threatened.)

Question 4.
Conservation, Regulation, Pollution, Prohibition
Answer:
Pollution. (All others are ways of environmental protection.)

Question 5.
IPCC, UNEP, IUCN, BNHS
Answer:
BNHS. (All others are international organizations. BNHS is Bombay Natural History Society.)

Find the correlation:

Question 1.
Rare species : Musk deer : : ………… : Lesser florican.
Answer:
Endangered species

Question 2.
Red panda : Rare species : : Giant Squirrel : …………
Answer:
Indeterminate species

Question 3.
Nitrogen, Oxygen : Gaseous cycle : : Soil and Rocks : …………
Answer:
Sedimentary cycle

Question 4.
Manas : One horned Rhino : : Gir : ………..
Answer:
Asiatic lion

Question 5.
Mumbai : Bombay Natural History Society : : TehriGarhwal : ………….
Answer:
Chipko centre.

Answer the following questions in detail:

Question 1.
Answer the following questions:
If frog population in paddy field declines all of a sudden,
(a) What will be the effect on paddy crop?
Answer:
If the population of frog declines, then there will be rise in the population of grasshoppers. The paddy fields will hence be infested with insect pests.

(b) Number of which consumers will decline and which will increase?
Answer:
The food chain if altered, results in imbalance in the ecosystem. ‘Paddy → Grasshoppers → Frog → Snake’, this food chain is natural. When by any reason there is dec1ine in the number of frogs, thus secondary consumer will also decline. Due to this decline, snake which is at tertiary consumer level will also decline. Theprimary consumers i.e. grasshoppers will increase as there is now no check on their population. Due to increase in their population the paddy production will be reduced. Due to reduced number of snakes, rats and other rodents from neighbouring areas would also rise, which are also secondary consumers.

(c) Name the Indian states where paddy is cultivated on a large scale.
Answer:
West Bengal, Uttar Pradesh, Haryana, Punjab, Tamil Nadu, Andhra Pradesh, Bihar, Chhattisgarh, Odisha, Assam and Maharashtra.

Question 2.
What is radioactive pollution? What are its effects?
Answer:
(1) The radiations emitted either through the natural sources or through man-made sources cause radioactive pollution.
(2) The natural radiations is in the form of ultra violet and infrared radiations.
(3) Artificial or man-made radiations are X-rays and radiations from atomic energy plants.
(4) All radiations are highly hazardous for the living organisms. The impact of radiation is also for a very long time.
(5) It has brought about major accidental mishaps at Chernobyl, Windscale, qpd Three Miles Island. These disasters have affected thousands of people.
(6) Some other effects of radiations are as follows – (i) Due to higher radiations of X-rays, cancerous ulceration occurs, (ii) Radiations destroy the body tissues, (iii) Radiations cause mutations and thus genetic changes occur, (iv) There is adverse effect on the vision.

Question 3.
Give one word for “The forest conserved in the name of God.” (Board’s Model Activity Sheet)
Answer:
Deorai.

Give scientific reasons:

Question 1.
Certain scavenging caterpillars, termites and insects found in the dung are important.
Answer:

• Scavenging caterpillars and insects are decomposers. They seem to be Worthless due to filthy surrounding in which they thrive.
• But they carry out most important task of decomposition of complex organic substances into simple inorganic elements.
• This recycling is possible only due to decomposers.
• If they are not present, there will be huge accumulation of garbage. Therefore, these living organisms are important.

Question 2.
Destroying trees is to destroy everything.
Answer:

• When a single huge tree is felled many living organisms which are dependent on it, are exterminated.
• Many insects, fungi, birds, etc. lose their habitat.
• Trees take up carbon dioxide from the atmosphere and release oxygen. These natural cycles are also hindered due to loss of trees.
• Due to trees there is shade, cooler atmosphere and increase in the rainfall. When such trees are destroyed all the components in the ecosystem are destroyed too.

Question 3.
There is no definite information about indeterminate species.
Answer:

• Indeterminate species do not have substantial information about them.
• The organisms belonging to such species appear to be endangered due to their some behavioural habits.
• They are shy and do not come in open so that they can be observed keenly.
• For example, animals like Giant squirrel also do not provide such information.

Question 4.
Tigers from Sunderbans and Rhinos from Manas are under threat.
Answer:

• Manas is in the area-of Assam where there are many dams and Indiscriminate use of water.
• This area is also flood affected. Therefore, rhinos are under threat.
• In Sunderbans, there are also problems such as deforestation, dams, excessive fishing, and dug out trenches.
• All of these cause dangers to the tiger population.

Question 5.
There are clusters of thick forests in the Western Ghats of India.
Answer:

• There are many sacred groves in the region of Western Ghats of India.
• These forests are not conserved by Government Forest Departments but are cared for by the local people, in the name of God.
• Due to such faith in the people, the forests are conserved like sanctuaries.
• Such many clusters are in Western Ghats of Maharashtra.

Question 6.
We can see biodiversity on three levels.
Answer:

1. There is biodiversity in the living organisms belonging to the same genus.
2. This diversity is due to different heredity pattern. This is called genetic biodiversity.
3. The organisms occupying the same area and belonging to the same species also show diversity due to different species. This is species biodiversity.
4. The organisms occupying different ecosystems also show differences, which is called ecosystem biodiversity. Therefore, we observe biodiversity on three different levels.

Questions based on diagrams:

Question 1.
What is shown in the picture? Write name and trophic level of each component.

Answer:
In this picture, food chain having rive trophic levels is shown.
(1) Trophic level 1 = Producers : Green plant.
(2) Trophic level 2 = Primary consumer (Herbivore): Grasshopper.
(3) Trophic level 3 = Secondary consumer (Carnivore): Bird.
(4) Trophic level 4 = Tertiary consumer (Carnivore) : Snake.
(5) Trophic level 5 = Top or Apex consumer (Carnivore) : Owl.

Question 2.
Explain the meaning of following symbols A and B and C.
A.

Answer:
The symbol show types of green energy such as solar energy and wind energy. It also expresses that people
should use such sources of energy for their use.

B.

Answer:
This symbol is giving the message “Save water”. Sustainable use of water is necessary for our future.

C.

answer:
The symbols of WWF and BNHS are shown here. BNHS stands for Bombay Natural History Society. This institute works for the conservation and documentation of flora and fauna.

WWF means World Wild Life Fund. Also known as World Wide Life Fund. This International Institute is looking after the welfare of wildlife through different conservation projects. WWF symbol shows Panda while BNHS symbol has Giant Hornbill.

Question 3.
(a) Identify the following symbols and state their significance: (March 2019)

Answer:
(i) This symbol is giving the message “Save water”. Sustainable use of water is necessary for our future.
(ii) Use of bicycle means use of green energy. By riding on a bicycle we save on fuel and use our own muscular energy. It is the best ecofriendly, non polluting vehicle.

(b) How can biodiversity be conserved?
Answer:
Biodiversity can be conserved by the following ways:

• Protection of the rare species of plants and animals.
• Creating habitats for the animals and plants by establishing National Park and Sanctuaries.
• Declaration of bio reserves, the areas which are protected through conservation.
• Conservation projects for protecting special species.
• Conservation of all flora and fauna.
• Strict observance of the acts and rules.
• Use of traditional knowledge and maintaining record of traditional knowledge.

Activity based questions:

Question 1.
Questions based on the charts.
Complete the flow chart: (July 2019)

Answer:

Question 2.
Collect more information about locations of these hotspots present in the world. (Textbook page no. 44)
Answer:
Students should collect this information.

Question 3.
Where are such sacred groves in Maharashtra? Make a list and visit with your teachers. (Textbook page no. 43)
Answer:
Sacred groves: Sacred groves form an important landscape feature in the deforested hill ranges of the Western Ghats. The felling of timber and the killing of animals in sacred groves is not allowed by the locals. It is considered as taboo.

In Maharashtra, sacred groves are found in tribal as well as non-tribal areas. The sacred groves in the western part are called Devrai or Devrahati, which means the abode of the gods. In eastern parts it is called Devegudi by the madiya tribal people.

In Maharashtra 2820 Devrais have been documented. Maruti, Vaghoba, Vira, Bhiroba, Khandoba and Shirkai are some deities to which sacred groves are dedicated.

In the sacred groves, the most commonly found plant species are Portia tree, Casuarina, Silk cotton tree, Indian laurel, Indian Elm, Bead tree, Indian butter tree, Turmeric and Japanese ginger. In Maharashtra, sacred groves are maximum in district of Sindhudurg, (More than 1500 out of total 2820) followed by Ratnagiri, then Pune and in district of Satara.

Projects: (Do it your self)

Project 1.
Let’s Discuss: (Text Book Page No. 41)
Collect the information about Chipko Movement and discuss between two groups of your class about its importance in present situation.

Project 2.
Collect more information on the organization of Greenpeace. (Text Book Page No. 43)
Answer:
Students are expected to write this answer to this question.

Project 3.
There should be positive attitude of human being towards the environment for welfare of entire living world. For this purpose, following roles are important. You can be a conservator, organizer, guide, plant-friend, etc. Describe about the role you wish to perform and your plans for that role. (Text Book Page No. 42)

Project 4.
Survey the plants and animals in your area. Maintain a record about their characteristics. (Text Book Page No. 45)
Answer:
Students can conduct such surveys with the help of elders.

Project 5.
Internet is my friend! (Collect the information Textbook page no. 41)
(1) Sound Pollution (Control and Prevention) Rule, 2000.
(2) Biomedical Waste (Management and Handling) Rule, 1998.
(3) E-waste (Management and Handling) Rule, 2011.

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 10 Disaster Management Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 10 Disaster Management

Question 1.
Complete the table.

Answer:

Question 2.
Write notes.
a. Disaster Management Authority.
Answer:
Disaster Management Authority is the body that works at the level of government from national level to village level. This work is basically about management of any disaster and tackling the problems of the affected people. At National level there is National Disaster Management Authority for which the Prime Minister is the chairman. For every state there is State Disaster Management Authority, where the chief minister of every state is the chairman. Under the state level, there are district level units where district collector is responsible for disaster management and implementation of rehabilitation schemes. Below district level authority there are Taluka and then Village Disaster Management Committees.

The Tahsildar is the chairman for Taluka level while the Sarpanch of the village is responsible for management of disasters at village level. Collector of each district helps in planning, coordinating and controlling the implementation of rehabilitation programme and also gives essential instructions and reviews the entire system.

b. Nature of disaster management.
Answer:
Disaster management involves either prevention of disasters (Pre-disaster management) or creating preparedness to face them (Post-disaster management). The action plans are prepared for managing disasters. This is done after studying the different aspects such as preventive measures, rehabilitation and reconstruction plans. The disasters are tackled by executing action plans in the following steps: Preparation, redemption, preparedness, action during actual disaster, response, resurgence and restoration. At every level there are other voluntary organizations and Government meteorological institutions for their help.

c. Mock drill.
Answer:

• Mock drill is the practice to check whether there is preparedness for dealing with the sudden attack of disaster.
• For this purpose, virtual or apparent situations that simulate the disaster are created.
• The reaction time for any type of disaster is checked by such activity. In the presence of trained personnel, the execution of the rescue plans are observed.
• People also understand their responsibilities at the time of actual disaster.
• The experts also check execution of plan designed for disaster redressal.
• By such mock drills, the efficacy of the system can be understood. In future, when actual calamity strikes, there is already preparation for disaster redressal. Therefore, mock drill is useful.

d. Disaster Management Act, 2005.
Answer:
Government of India has made Disaster Management Act in 2005. The affected people are given all necessary help as per this act. With the humanitarian view, people are rehabilitated and helped them to come back to normalcy after the disaster.

As per this Act, National Disaster Response Force has been established. This force consists of 12 divisions in entire India which are attached with Indian Army. The headquarter is located in Delhi, but the action is taken all over the country with the help of army. As per the Act, in Maharashtra National Disaster Response Force is in action through State Reserve Polioe Force. The personnel of this force are trained accordingly, and they take part in the rescue work during different disasters.

Question 3.
Answer the following questions.
a. Explain the role of district disaster control unit after occurrence of any disaster.
Answer:
(1) District control unit looks after the ; disaster management of the district.
(2) It is immediately formed either after the impact of disaster or if warning is given about some upcoming disaster.

District-wise Disaster Control Unit performs following role:

• The review of various aspects of disasters is done.
• Through the disaster control unit there is continuous contact established with various agencies like army, air force, navy, telecommunication department, paramilitary forces, etc. for obtaining help.
• The unit also coordinates with various voluntary organizations for their help in disaster management.

b. Give the reasons for increase in human disasters after the World War-II.
Answer:

• After Second World War, the feelings of peace and brotherhood among the global citizens were lost. The geographic, religious, racial and ethnic differences sprang up tremendously.
• Atrocities that Nazi has performed made deep impact on the minds of people. Terrorism, abduction, robberies and social unrest increased in almost all the countries.
• The financial losses had incurred in the World War II. The misuse of science and technology was done to retrieve these deficits.
• At the end of World War II, the atomic bombs were dropped in Japan. This has created health problems in the entire world.
• Social inequality, economic disparity, racial and religious differences were some adversaries that created unrest in the country.
• Later, the neighbouring nations kept on fighting. The geographical boundaries were changed. People always had feelings of insecurity. The terrorism flourished. All such instances gave rise to man-made disasters.

c. What are the objectives of disaster management?
(OR)
State any four objectives of disaster management. (March ’19)
Answer:
Objectives of disaster management:

• To save human life from disasters. To help them for moving away from the place of disasters by rapid action.
• To supply essential commodities to the affected people. This helps to reduce the gravity of disaster. People are given grains, water and clothes and other basic necessities under this objective.
• To bring back the conditions of affected people to normalcy.
• To rehabilitate the affected and displaced victims.
• To think and execute the protective measures in order to develop capability to face the disasters in future.

d. Why is it essential to get the training of first aid? (July ’19)
Answer:
When there is a disaster, we need to immediately help the victim. Till the medical help arrives, one should be in position to treat the injured and save his or her life. In such cases; knowing first- aid is essential. Such kind of a need may arise in case of our parents, our siblings at home or with friends in school. Those who are injured should be treated at once. If we know about techniques of first aid, we can save such person before the medical help arrives. Therefore, it is essential to get the training of the first aid.

e. Which different methods are used for transportation of patients? Why?
Answer:
For the transportation of patients following methods are used:

• Cradle method: This method is used for children and persons with less weight.
• Carrying piggy back: This method is useful in carrying the unconscious persons.
• Human crutch method: If one leg of the person is injured, then the victim is supported with minimum load on the other leg. This is called human crutch method.
• Pulling or lifting method: For carrying an unconscious person for a short distance this method is used.
• Carrying on four-hand chair: This method is used when the support is needed for a part below waist region.
• Carrying on two-hand chair: Patients that cannot use their hands but can hold their body upright, are carried by such method.
• Stretcher: By making temporary stretcher in case of emergency, the unconscious patient can be moved. Such temporary stretchers are made by using bamboos, blanket, etc.

Question 4.
On the basis of the structure of disaster management authority, form the same for your school.
Answer:

Question 5.
Write down the reasons, effects and remedial measures taken for any two disasters experienced by you.
Answer:
Students are expected to write the answer based on their own experiences.

Question 6.
Which different aspects of disaster management would you check for your school? Why?
Answer:
For the pre-disaster management at school following aspects would be inspected.

• Are the telephones 6f the school working properly?
• Is there a first-aid box in each class?
• Are there any basic medicines in the school?
• Is the team ready for rescue of smaller children from lower classes?
• Has monitor or prefect participated in a mock drill? Does he/she know about first aid?
• Is the contact of parent representative available in emergency situations?
• Is the Medical Officer/Doctor present on the school campus?
• Is there enough drinking water and some dry snacks available in the school?
• Are the staircases and corridors suitable for quick evacuation of the children?

Question 7.
Identify the type of disaster.
a. Terrorism.
Answer:
Man-made, intentional.
Due to the activities of terrorism, many innocent lives are lost. Many are seriously injured. Some become crippled for their entire life. Buildings, monuments, vehicles everything is completely destroyed. There is rift between different religions or sects. The peaceful atmosphere is disturbed. The entire society is under the constant fear of insecurity.

b. Soil erosion.
Answer:
Natural, geophysical, geological.
When the upper fertile layer of soil is lost, it becomes barren. The trees are uprooted. The fertility of the area is lost. The land becomes unsuitable for cultivation or farming. Due to wind, flowing water or grazing animals the naturally occurring soil erosion becomes hazardous for the environment.

c. Hepatitis.
Answer:
Natural, biological, animal-origin.
Hepatitis is a viral disease which spreads through the contaminated food and water. The outburst of epidemic of hepatitis is difficult to control. As in big cities the quality of road side food is often consumed, the spread of hepatitis is. fast. People suffer due to hepatitis.

d. Forest fire.
Answer:
Natural, biological, plant-origin.
Due to heat and wind, the dry grass and the shrubs catch fire in the forests, resulting in forest fires. Such rapidly spreading forest fire can finish the biodiversity within a very short span of time. It is difficult to extinguish the naturally lit forest fires. Many trees and other vegetation, animals and birds along with their habitats are destroyed due to forest fire. The smoke emanating causes the air pollution.

e. Famine.
Answer:
Natural, climatic.
Due to famine there is severe water scarcity. In absence of water, the fields and farms become barren as the crops cannot grow without water. There is shortage of food grains. The cattle dies due to want of water and grass. Local people have to migrate in search of food, water and shelter.

f. Theft.
Answer:
Man-made, intentional.
Theft causes economic loss for the one whose money or valuables are looted. The person who suffers the loss also undergo mental and emotional shock. Sometimes the thief may also cause physical harm. It may cost on life too.

Question 8.
Some symbols are given below. Explain those symbols. Which disasters may occur if those symbols are ignored?

Answer:
The above signs are warning symbols which should never be ignored.
The meaning of each is given below. They are giving warnings about explosive, inflammable, oxidizing, compressed gas, corrosive, toxic, irritant, environmentally hazardous and health hazard.

(1) Explosive: Some materials are explosive. While handling such materials care should be taken. We should not take anything that would cause fire leading to explosion. If explosion occurs, there would be a major disaster causing great loss of life and property. Thus if this sign is seen, great care has to be taken.

(2) Inflammable: Similar to explosive substances, the inflammable materials can also catch fire easily. Therefore, to warn people such sign is given on materials that can cause hazard by burning.

(3) Oxidizing: Some chemical substances are oxidizing. They carry out chemical reactions with a rapid speed. E.g. If potassium permanganate falls on the cloth, it starts the reaction on its C-C bonds. Due to such property of carrying out reactions, the cloths may catch fire. Therefore, oxidizing substances should be handled with care.

(4) Compressed: Compressed substances are filled under pressure in some container. If mishandled, they can come out of the container by bursting it open. This can cause some injuries.

(5) Corrosive: The corrosive substances are very reactive. The mere touch of corrosive substances can cause destruction of skin, eyes, respiratory passages, digestive organs, etc. rapidly. Just touching or smelling of such substances can cause major injury and thus warning sign of corrosive substance should never be ignored.

(6) Toxic: To taste a toxic substance or even to smell it, can lead to death. The packing of these substances are therefore marked as dangerous. They should be avoided as far as possible.

(7) Irritant: When skin or any delicate part of the body comes in contact with the irritant substance, it can cause harmful reaction. Especially, eyes, nasal mucosa and skin are affected by contact with corrosive substances.

(8) Environmentally hazardous: Many sub¬stances cause harm to the environment due to their toxicity. Air, water or soil can be polluted due to such pollutants. When environment is affected, ultimately these hazardous effects come back to human species. Therefore, such substances should be carefully used. Their use should be judicious and controlled.

(9) Health hazard: The substances that can cause hazard to our health should always be distanced from us. Such substances should not be kept in proximity. As far as possible they should be kept away and handled with great care if needed for any work. Materials marked with health hazard can cause severe toxicity.

Question 9.
Explain that why is it said like that?
a. Mock drill is useful.
Answer:

• Mock drill is the practice to check whether there is preparedness for dealing with the sudden attack of disaster.
• For this purpose, virtual or apparent situations that simulate the disaster are created.
• The reaction time for any type of disaster is checked by such activity. In the presence of trained personnel, the execution of the rescue plans are observed.
• People also understand their responsibilities at the time of actual disaster.
• The experts also check execution of plan designed for disaster redressal.
• By such mock drills, the efficacy of the system can be understood. In future, when actual calamity strikes, there is already preparation for disaster redressal. Therefore, mock drill is useful.

b. Effective disaster management makes us well prepared for future.
Answer:

• Disaster can strike any time. The sudden disasters can be man-made with some bad intentions or may be accidental.
• When natural calamity strikes suddenly with a huge impact, large scale devastation of property and general environment degradation occurs along with substantial mortality of people and animals.
• Therefore, it is most appropriate to have the preparedness to reduce the impact of any future disasters.
• We cannot control the onset of the natural disaster, but we can definitely reduce the harsh effects of the disaster by following disaster management plan.

Question 10.
Complete the following chart.

Answer:

Question 11.
Following are the pictures of some disasters. How will be your pre and post-disaster management in case you face any of those disasters?

Answer:
In the pictures given, following disasters are shown:
(1) Two groups of children are fighting with each other.
(2) There is gas leakage from the LPG cylinder.
(3) There is heavy downpour due to cloud bursting which has led to waterlogging in the town.
(4) There is cyclone causing a tornado. (Commonly called a twister)

Projects:

Project 1.
Demonstrate the activities shown on page no. 106 of Std. IX Science and Technology textbook in front of the students of other classes. Make a video clip and send it to others.

Project 2.
Form a group of students from your school to demonstrate the mock drill and demonstrate it in the school

Can you recall? (Text Book Page No. 109)

Question 1.
What is disaster?
Answer:
Disaster is the incidence that occurs suddenly causing heavy damage to life and property. The disaster can be man-made or due to natural reason.

Question 2.
Which disasters have you experienced in your area?
Answer:
On September 2019, there was a heavy downpour in Pune. This disaster has been experienced recently.
On 26th November 2008 there was attack at several places by the Pakistani terrorists. The stories about the deaths and damage caused by this disaster were seen in films and learnt about this from our elders.

Question 3.
What are the effects of that disaster on local and surrounding conditions?
Answer:
Due to a heavy downpour in Pune, there was waterlogging in all the shallow areas. All the transport systems collapsed on that day. Large trees fell down injuring the people. The water logging caused condition like floods. Schools, colleges and offices were shut down. People were caught in troublesome situation.

On 26th November 2008 many innocent people lost their lives. There was tremendous damage caused to some of the important places like Taj Palace Hotel and Chabad house.

Use your brain power: (Text Book Page No. 111)

Question 1.
Depending upon information given on page 111, explain the various effects of the disaster of railway accident.
Answer:
The effects of disastrous railway accident:
The effect will be dependent upon the nature of the accident that has occurred. Whether, it is a collapse of bridge or due to derailment of the train, or due to collision of two moving trains, whether it is due to failure in signaling system, due to land slide or due to obstacle in the tunnel, that has to be understood. The impact of such railway accident will be dependent on the way that accident has occurred. Based on this impact the effects will take place.

Observe/Discuss:
Let’s Discuss: (Text Book Page No. 110)
Question 1.
observe the images on textbook page no. 110. whether the places of disasters are known to you? Discuss the emfects of these disasters on public lire. How people could lave been saved from these disasters? Discuss with your friends in the classroom.
Answer:
Students should discuss the disasters given in the pictures by themselves after collecting the information.

Observe: (Text Book Page No. 114)

Question 1.
Observe the disaster cycle given below and explain each aspect of the disaster of earthquake.
Answer:
The main aspects of disaster cycle to tackle disaster of earthquake are as follows:
(1) Preparation : With the help of seismograph, the warning about forthcoming earthquake can be obtained these days. The intensity of the earthquake is also predicted with the help of technology. If the estimate of the Richter scale is on the higher sides, there would be more preparatory measures taken to tackle the forthcoming problem of earthquake.

(2) Redemption: Once this information is obtained the possible impact of the earthquake on the houses, buildings, people can be studied by the geological experts. The meetings of the Disaster Management Authority will be organized for same.

(3) Preparedness: What the general public should do and what action the reserved forces should take, will be decided in case of actual incidence of the earthquake. The schemes and plans will be made ready in this direction.

(4) Impact of Earthquake: In case of disaster of earthquake, people will be helped to safety. The trapped people will be rescued. First aid and other necessary help will be provided. The data about the losses and the intensity of this disaster will be noted and reported for the further process.

(5) Response: In this phase the response of the people as well as the action of Government can be well studied. The response should be quick and positive. The maximum lives and property should be saved by such responses. The disaster of earthquake should be managed with positivity and through help given to the sufferers.

(6) Resurgence: Earthquakes can destroy the entire households or even entire community. Such homeless people should be given the place to stay. Resurgence is important phase for the national welfare. If the citizens of India are cared for, the nation too will progress.

(7) Restoration: The earthquake victims should be settled by providing them with new settlements. Sometimes, entire village is to be settled. E.g. In Latur or Kutch, there twas very large scale devastation. But Government of India as well as some NGOs helped to reconstruct the houses. In such earthquake-prone areas, houses are built in specific pattern to withstand any possible future calamities.

Observe: (Text Book Page No. 117)

Question 1.
Give the reference of following pictures and explain importance of each of those in disaster management. Which are other such activities ?

Answer:
The actions shown in the above pictures are as follows:
(a) The patient is made to lie on the stretcher. He must be unconscious and injured to greater extent, so that he cannot move by himself.
(b) The patient is helped by giving artificial respiration. Probably the victim is suffocated and needs oxygen supply.
(c) and (d) The patient is being picked up. Most probably the patient is unconscious.
The unconscious person who cannot move by himself is carried by these two methods. In method ‘c’, the weight of the patient is less and hence he can be lifted as shown in the picture. In picture ‘d’ the victim has to be lifted in other way, may be due to his greater body weight.
(e) The patient is carried on the back as in ‘carrying piggy back’ position. He too is unconscious and needs to be shifted for medical treatment.
(f) The patient in this picture is carried by ‘human crutch method’. When victim’s one leg is injured, he cannot walk without support. Hence, he needs to be carried in such a way.

In all the above methods, the injured person at the time of disaster is transported to hospital or dispensary for further medical help. The primary first-aid is given to the victim. Now the volunteer is taking him for further treatment. Such rescue activities depend upon the type of disaster and the extent of the injury. Hence the methods will be of different nature.

Let’s Think: (Text Book Page No. 111)

Question 1.
What will be the effect on yourself and surrounding, if any accident-like disaster occurs during the sports on playground or in school?
Answer:
When in school, there is an accident, first of all we get scared. But with caring help of the teacher, we will give the first aid to the injured friends by using first aid kit. If the injury is serious, we will take him to the medical centre of the school. While playing or during sports event, children flock around and make unnecessary crowding. In case of such accident, first of all the crowd will have to be dispersed. If there is major disaster, one should not fumble but manage the disaster in a wise way with the help of teachers.

Let’s Think: (Text Book Page No. 112)

Question 1.
Explain the nature and scope of the disaster of flood with the help of six points given on text book page no. 112.
Answer:
The nature and the scope of the disaster of flood can be described according to the six points:
(1) Pre-disaster phase: Due to Indian Meteorological Department the warning predictions are received before any climatic disaster strikes. If the scope of the flood is predicted to be high, then the people who may be affected by the calamity are relocated to a safer area.

(2) Warning phase: In the warning phase the Government warns the general public about the forthcoming disaster of floods through mass media like radio, television, newspapers, etc. In recent times, even the cellphone messages are sent to people for warning them. The people living in coastal areas will be worst affected and hence such people are given greater care and they are immediately made to leave their houses. They are taken to the safe places.

(3) Emergency phase: When the flood waters actually start rising up, the low-lying areas are submerged. Houses, roads and shops everything goes under water. The rescue operations are carried out by army men from National Disaster Rescue Force. They take every possible effort to rescue the trapped people. The emergency continues till the water does not recede. Later after the water starts receding, people who had been taken to places on heights, start coming back. During this phase, search, rescue operations, medical treatment, and first aid are all the aspects on which the attention is focussed.

(4) Rehabilitation phase: The people affected due to floods are given emotional and financial support. The fields, farms, houses or cattle-shed are under water. Such people are given transient accommodations. Many cattle and other animals die by drowning. Their rotting carcasses have to be disposed as soon as possible because the decaying process spreads epidemics of diseases. People are given vaccinations to protect them from diseases of such kind. Special care of young children and senior citizens is taken during this period. Supply of food and drinking water is also very important task during this time.

(5) Recovery phase: During recovery phase, the life comes gradually back to normalcy. The removal of dead, decaying material and the debris is the first priority. The water connections and electricity is restored back. Various NGOs or Government organizations provide help of various kind to the affected people. This help is to be distributed to those who are in real need. This work is also done by Disaster Management Department.

(6) Reconstruction phase: The houses and building that collapse due to floods are built back. Agricultural activities start again. Roads and water supply is once again normalized. Schools and colleges start once again. Thus, the once flood affected area comes back to routine functioning again.

Let’s Think: (Text Book Page No. 117)

Question 1.
Following are some pictures of disasters. Which precautions would you take during those disasters?

Answer:
The pictures shown above are showing earthquake, fire and snake bite respectively.
In the above disasters, the initial precautions to be taken are as follows:
(1) Earthquake: In case of earthquake, one should immediately come out of house and stand in the open ground. If this is not possible, one has to go below table or any other cover. During collapse of the building, there should not be a head injury. This precaution is basically for prevention of dangerous injuries and saving our life. Switch off the power supply. If in journey, stay inside the vehicle.

(2) Fire: First and foremost is to save ourselves from fire. Then one can help others in rescue operations. Help others to extinguish fire. Call the fire department for immediate action.

(3) Snake bite: Many a times the biting snake can be non-venomous too. But the victim is psychologically affected too. The tourniquet should be tied in the region above the snake bite. The rope, piece of cloth or even handkerchief can be used for this purpose, so that the venom, if any should not rise and reach vital organs. The wound should be made near the bite-wound so that the blood will ooze out and some venom can automatically flow out. Though these are first-aid measures, the victim should be rushed to a qualified doctor for an injection of antivenin.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
Disasters definitely affect the ………… of the nation.
(a) people
(b) economy
(c) security
(d) employment
Answer:
(b) economy

Question 2.
If local ………… is not strong enough, citizens become confused.
(a) leadership
(b) women
(c) politicians
(d) cattle
Answer:
(a) leadership

Question 3.
…………… problems arise diming the disaster.
(a) Local
(b) Global
(c) Administrative
(d) Private
Answer:
(c) Administrative

Question 4.
Stinking pollution caused due to decomposing corpses of humans and other animal is ………… disaster.
(a) environmental
(b) health
(c) necessary
(d) effective
Answer:
(a) environmental

Question 5.
After the subsidence of any type of disaster, rehabilitation work is started in ………… phase.
(a) later
(b) transitional
(c) terminal
(d) ultimate
Answer:
(b) transitional

Question 6.
…………. phase is highly complicated phase.
(a) Reconstruction
(b) Recycling
(c) Reuse
(d) Redevelopment
Answer:
(a) Reconstruction

Question 7.
There had been a huge ……….. in the village Malin, Tal. Ambegaon in 2014.
(a) earthquake
(b) storm
(c) landslide
(d) change
Answer:
(c) landslide

Question 8.
The atomic energy plant at Chernobyl was used only for generating ………….
(a) electricity
(b) solar power
(c) atomic energy
(d) agriculture
Answer:
(a) electricity

Question 9.
Supply of essential ………….. to the affected people can reduce the effect of the disaster.
(a) food
(b) water
(c) commodities
(d) money
Answer:
(c) commodities

Question 10.
Keeping …………. ready is a practice to check the preparedness of facing the disaster.
(a) First aid
(b) Mock drill
(c) Ambulance
(d) Fire brigade
Answer:
(b) Mock drill

Question 11.
Which of the following is man-made disaster. (March, July ’19)
(a) Earthquake
(b) Flood
(c) Meteor
(d) Leakage of toxic gases
Answer:
(d) Leakage of toxic gases

Question 12.
What should be done if gas cylinder at your house catches fire?
(a) Water should be sprinkled
(b) Sand, soil should be put on it
(c) Cylinder should be covered with wet blanket
(d) one should run away
Answer:
(c) Cylinder should be covered with wet blanket

Which type of disaster is described in the following statements:

Question 1.
On 26th July 2005, entire suburban Mumbai was waterlogged.
Answer:
Cloudbursting and severe downpour

Question 2.
Elephants in the Bandipur forest started running helter and skelter due to smoke.
Answer:
Forest fires

Question 3.
Many innocent people died in the bomb blast that occurred on 11th July 2006 in local trains.
Answer:
Bomb explosion-Terrorism

Question 4.
In Kutch, suddenly many school children were buried under the rubble.
Answer:
Earthquake

Question 5.
Because of lack of crops, people from Vidarbha are migrating to other regions.
Answer:
Dry famine

Question 6.
The huge waves in Chennai engulfed many human lives in December 2004.
Answer:
Tsunami.

Find the correlation:

Question 1.
Earthquake in recent times : Gujarat, Latur : : Devastating floods in 2018 : ………….
Answer:
Earthquake in recent times : Gujarat, Latur : : Devastating floods in 2018 : Kerala/Assam

Question 2.
Toxic gas leakage: Accidental disaster : : war : …………..
Answer:
Toxic gas leakage: Accidental disaster : : war : Intentional

Question 3.
Sun spots : Atmospheric type of disaster : : Salinization : …………..
Answer:
Sun spots : Atmospheric type of disaster : : Salinization : Geological type of disaster

Question 4.
Pre-disaster management : Preparation and warning : : Post-disaster management : ………..
Answer:
Pre-disaster management : Preparation and warning : : Post-disaster management : Resurgence and restoration.

Match the columns:

Question 1.

Answer:
(1) Earthquake and volcano – Geological.
(2) Snowfall and snowstorms – Climatic.

Question 2.

Answer:
(1) Aquatic weeds – Plant origin
(2) Attack by locusts (insects) – Animal origin.

Question 3.

Answer:
(1) Atomic tests – Unintentional
(2) Terrorism – Intentional.

Question 4.

Answer:
(1) Contamination of water – Environmental
(2) Collapsing of transport system – Administrative.

Question 5.

Answer:
(1) Spread of epidemics – Medical
(2) Shortage of funds – Economical.

Question 6.

Answer:
(1) Rift due to religions – Social
(2) Citizens getting confused – Political.

Identify the type of disaster and describe the effects of the same in brief:

Question 1.
Accident at Chernobyl.
Answer:
Man-made, unintentional. At Chernobyl in Russia there was the atomic energy plant, where disastrous accident took place. The radiations emitted through the reactors caused tremendous radiation pollution. These hazardous effects are even seen today.

Give reasons:

Question 1.
When there are riots, the cities, towns or villages show signs of tense atmosphere.
Answer:

• During riots, there is financial loss for all the rioting groups.
• The belongings, houses, shops, etc. are destroyed or damaged.
• Property is looted. There is no guarantee of safety and security for anyone.
• Women and children suffer the most as they are easily victimized. Therefore, when there are riots, the cities, towns or villages show signs of tense atmosphere.

Answer the following questions in detail:

Question 1.
which are the disaters that make Impact for longer duration? Give examples.
Answer:
Those disasters that make the impact for long duration and those disasters, whose after-elfbcts are either severe are long term disasters. Their severity increases with thme. Such disasters are famine, various problems of growth of crop, strikes of workers, rising levels of oceans, desertification, etc.

Question 2.
What types of disaster are the following? Explain their impacts.
(a) Floods (b) War. (Board’s Model Activity Sheet)
Answer:
(a) Flood is geophysical climatic disaster.
(b) War is man-made intentional disaster.

Impact of flood : The low-lying and the coastal areas are seen to be submerged. The entire region is waterlogged.
Impact of war: Tremendous destruction causing loss. Many lives are lost. The costs of all the items rise due to war conditions. Entire nation faces insecurity.

Question 3.
Explain in brief the sensitive issues of general public about disaster.
(OR)
Which are the three aspects of disaster tjiat are important for common citizens?
Answer:
The phase of emergency, transitional phase and reconstruction phase are the three phases of disaster that are important for common citizens.

(1) Phase of emergency: If timely and rapid action is taken during this phase, maximum lives can be saved. Search and rescue operations, medical help, first aid, restoring communication services, removing the people from affected area are done during this phase. The gravity of disaster can be estimated during this phase.

(2) Transitional Phase: The disaster subsides and then the work of transitional phase starts. The main concern is rehabilitation work for the affected and displaced people. This work includes clearing of debris, restoring water supply, repairing roads, etc. to bring normalcy in public life. Help from different voluntary and Government institute is taken to offer the monetary provision and essential commodities to affected victims. Permanent means of livelihood is given to the people to reduce their mental and emotional stress. The victims are truly rehabilitated.

(3) Reconstruction Phase: Reconstruction phase is a highly complicated phase which actually overlaps with transition stage. Help is offered to people to reconstruct their buildings. Other facilities like roads and water supply are restored. Farming practices are restarted. It is a very gradual phase that makes the victims to completely rehabilitate.

Question 4.
What are the objectives of mock drill?
Answer:
Objectives of Mock Drill:

• To evaluate the response of the people to the disaster.
• To improve the coordination between various departments of disaster control.
• To identify one’s own abilities if disaster approaches.
• To improve the ability to quick response to disaster and taking rapid action.
• To check the competency of the planned actions.
• To identify the possible errors and risks while dealing with disasters.

Question 5.
Write down the names of international organizations that work for disaster management.
Answer:
Following international organizations work for disaster management.

• United Nations Disaster Relief Organization
• United Nations Centre for Human Settlements
• Asian Disaster Reduction Centre.
• Asian Disaster Preparedness Centre.
• World Health Organization.
• United Nations Educational, Scientific and Cultural Organization.

Question 6.
The building in which you are residing ( has caught fire on the ground floor. What necessary rescue steps will you take? (July ’19)
Answer:

• We shall call out for help.
• We shall immediately call fire brigade.
• We shall try to extinguish fire with the help of other people.
• We will give first aid to people who are injured, fill the medical help arrives.
• We will cover our nose and mouth with moist cloth to prevent suffocation.

Write short notes:

Question 1.
Pre-disaster management.
Answer:
The management measures taken before onset of a disaster is called pre-disaster management.
In pre-disaster management, complete preparation and planning to face any type of disaster is done. For this purpose, following steps are taken.

• Identifying the areas where the disaster can strike. Such disaster-prone areas are to be thoroughly studied.
• Through predictive intensity maps and hazard maps, the information is collected about the intensity of disaster and probable sites of disasters respectively.
• Special training for disaster management is given to the concerned people.
• The mass awareness is created about disaster management through training programmes, mass media and internet, etc.

Question 2.
Post-disaster management.
Answer:
The management measures taken after the striking of a disaster is called post-disaster management.
Following steps are taken during post-disaster management:

• Helping the victims of disasters by giving all possible help needed for their survival.
• Local people are trained to take part in the disaster management so that affected people can be saved rapidly.
• Establishing the help centres that could provide all the necessary help. Such centres will be different in case of different disasters.
• Collection and categorization of the material received from control centre for helping the victims. Distributing the same and reviewing the measures continuously.
• Disaster rescue programmes are mainly focused.

Some symbols are given below. Explain those symbols. Which disasters may occur if those symbols are ignored?

Question 1.
Write what the signs indicate:

Answer:

• Figure A indicates inflammable substances. They can catch fire if they come in contact with oxygen-rich air.
• Figure B indicates corrosive substances which can cause damage to tissues of skin, eyes and other delicate organs etc.
• Both the symbols are warning signs for people to keep away or handle carefully such substances.

Question 2.
What does the symbols below indicate? Write in brief. (Board’s Model Activity Sheet)

Answer:
(1) Symbol ‘A’ indicates Irritant. When skin or any delicate part of the body comes in contact with the irritant substance, it can cause harmful reaction. Especially, eyes, nasal mucosa and skin are affected by contact with corrosive substances.
(2) Symbol ‘B’ indicates toxic substance. To taste a toxic substance or even to smell it, can lead to death. The packing of these substances are therefore marked as dangerous. They should be avoided as far as possible.

Complete the paragraph by choosing the appropriate words given in the brackets:

(Capability, Rehabilitation, Commodities, Human, objectives, normalcy, amusements)
The ………. of disaster management comprise of the following aspects …………. life is saved from the disasters. People are helped to move away from the place of disasters. They are given essential ……….. by the government so that the gravity of disaster is reduced. The disaster conditions are brought back to ………… of the affected and displaced victims is done. Moreover, protective measures for future are also planned to develop ………… among the people to face any possible disasters in future.
Answer:
The objectives of disaster management, comprise of the following aspects. Human life is saved from the disasters. People are helped to move away from the place of disasters. They are given essential commodities by the government so that the gravity of disaster is reduced. The disaster conditions are brought back to normalcy. Rehabilitation of the affected and displaced victims is done. Moreover, protective measures for future are also planned to develop capability among the people to face any possible disasters in future.

Paragraph based questions:

1. Read the paragraph and answer the questions given below:
Disasters can be properly classified into three categories, viz. natural disasters, technological disasters and man-made disasters. The forces that cause natural disasters cannot be controlled. Moreover, they are becoming more frequent in the current years due to phenomena of climate change. On and off incidences of cyclones, cloud bursting, floods, etc. am creating havoc in the lives of people. Technological disasters are due to improper and callous behaviour at the different processes carried out in technical establishments. Man-made disasters are conflicts arising due to different religions, regions and-terrorism.

Questions and Answers:

Question 1.
What are three broad areas of disasters?
Answer:
Natural disasters, technological disasters and manmade disasters are three broad areas of disasters.

Question 2.
Which disasters cannot be controlled? Why?
Answer:
Natural disasters cannot be controlled as they are due to natural phenomena beyond the human power to stop them.

Question 3.
Which type of disasters were very common in Western Maharashtra in recent times? Why?
Answer:
Cloud bursting and floods were very common in Western Maharashtra caused due to climate change.

Question 4.
Give any one example of technological disaster that shook the entire India.
Answer:
Bhopal gas tragedy that occurred in 1984 was a worst disaster that shook the entire India.

Question 5.
Which types of disasters can be controlled in order to lead happy, peaceful and secured life? How?
Answer:
We have to control manmade disasters such as wars, riots, terrorism, etc. by having peaceful negotiations, respect for each human being and feeling of brotherhood among all.

Questions based on tables and charts:

Question 1.
Complete the chart: (Text Book Page No. 111)
Different problems occur with disasters. In the concept map different effects are mentioned. Read it and fill the blank places.
Answer:
(Answers are given directly in bold.)

Question 2.
Complete the chart: (Text Book Page No. 117)
Complete the chart as per the objectives of the first aidr:t
Answer:
(Answers are given directly in bold.)

Activity based questions:

Question 1.
Observe the images ‘A’ and ‘B’ and answer the following questions.

(i) Which disasters are shown in the images ?
Answer:
Image A is showing damage due to earthquake. Image B is showing house on fire.

(ii) Which primary precautions will you take in case of disaster shown in ‘A’?
Answer:
In the above disasters, the initial precautions to be taken are as follows :
Earthquake: In case of earthquake, one should immediately come out of house and stand in the open ground. If this is not possible, one has to go below table or any other cover. During collapse of the building, there should not be a head injury. This precaution is basically for prevention of dangerous injuries and saving our life. Switch off the power supply. If in journey, stay inside the vehicle.

(iii) Which type of first-aid is offered to the injured people in disaster ‘B’?
Answer:
First aid given to burn victim:

• The person who is injured by fire should be dotised with cold water on his/her body. This will extinguish fire and give some relief caused due to inflammation. Do not break the blisters. Give water to drink.
• Cover the burnt part by wet and moist cloth. Wash the wounds with antiseptic solution.
• If the person is severely burnt, transfer him/her immediately to hospital.

Question 2.
Correct the following diagram:

Answer:

Projects:

Project 1.
Can you tell? ( Textbook page no. 118)
Whether there had been mock drill by fire fighters under the disaster management scheme in your school? Which techniques did you see during the drill?

Project 2.
Try this: (Textbook page no. 115)
Which factors will you consider while designing the pre-disaster management plan for your school/home? Prepare a survey report with the help of your teacher.

Project 3.
Get information:
(1) Visit the district collector or Taluka Tehasildar office and collect the information about disaster management. (Textbook page no. 115)
(2) Meet the medical officer/doctor from your village and collect information about providing the first aid. (Textbook page no. 118)

Project 4.
Internet is my friend:
(1) Search for the video clips of disasters. Discuss in your class about effects of disasters and remedies over it. (Textbook page no. 110)
(2) Find out more about the activities of international organizations that work for disaster management.
(Textbook page no. 116)
1. United Nations Disaster Relief Organization.
2. United Nations Centre for Human Settlements.
3. Asian Disaster Reduction Centre.
4. Asian Disaster Preparedness Centre.
5. World Health Organization.
6. United Nations Educational, Scientific and Cultural Organization.

Balbharti Maharashtra State Board Class 8 History Solutions Chapter 2 Europe and India Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 8 History Solutions Chapter 2 Europe and India

Class 8 History Chapter 2 Europe and India Textbook Questions and Answers

1. Rewrite the statements by choosing the appropriate options:

Question 1.
In 1453, the city of was conquered by Ottoman Turks.
(a) Venice
(b) Constantinople
(c) Rome
(d) Paris
Answer:
(b) Constantinople

Question 2.
The Industrial Revolution began in
(a) England
(b) France
(c) Italy
(d) Portugal
Answer:
(a) England

Question 3.
…….. tried to put restrictions on the illegal trade of British.
(a) Siraj-ud-Daulah
(b) Mir Kasim
(c) Mir Jafar
(d) Shah Alam
Answer:
(b) Mir Kasim

2. Explain the following concepts:

Question 1.
Colonialism:
Answer:
1. Exploitation of one country by another and making it into a colony is called Colonialism.
2. On the basis of economic and military strength one country occupies a region of another country and establishes its political supremacy.
3. Many European nations established their colonies in various parts of the continents of America, Asia, Africa and Australia.
4. They grabbed the power by using force to exploit the colonies systematically. This gave rise to Imperialism.

Question 2.
Imperialism:
Answer:

1. Imperialism means a powerful country controls other country by establishing its overall domination.
2. The objective of imperialism is to establish more and more colonies.
3. European countries with strong military power and imperialist aspirations enslaved the countries in Asia and Africa.
4. Imperialism gained momentum after industrial revolution and came to an end in twentieth century.

Question 3.
Age of Renaissance:
Answer:
(1) Renaissance literally means rebirth or revival. In the latter phase of medieval Europe, reformation, religious reform movement and geographical discoveries gained momentum.
(2) It was a comprehensive movement which touched all aspects of human life. The Greek and Roman traditions in art, architecture and philosophy were revived in the Age of Renaissance.
(3) Humanitarianism got great impetus.
(4) Renaissance inspired all round progress and ushered a new era in the history of world.
(5) This developments took place from 13th century to 16th century in Europe. This period is known as the Age of Renaissance.

Question 4.
Capitalism:
Answer:
1. Many traders participated in the competition for trade with Asia after the discovery of the new sea routes to the east.
2. As it was not possible to carry out trade single-handedly, it was decided by the traders to pool their resources to raise capital for trade.
3. Many trading companies like the East India Company were formed.
4. Trade with Oriental countries was very profitable and responsible for economic prosperity.
5. The ruling powers gave protection and trade concessions to these companies.
6. This led to the accumulation of wealth in Europe. This wealth was used as capital for trade and commerce which gave rise to capitalism in Europe.

3. Explain the following statements with reasons:

Question 1.
Siraj-ud-Daulah was defeated in the Battle of Plassey.
Answer:
1. The officers of the British East India Company misused the trade concessions and built fortification around the factory in Kolkata.
2. Siraj-ud-Daulah captured the factory at Kolkata.
3. This created discontent in England.
4. Robert Clive diplomatically bribed Mir Jafar, Commander in Chief of Nawab’s army and promised to make him Nawab of Bengal.
5. So. the Army of Siraj-ud-Daulah under the command of Mir Kasim did not join the battle against the British at Plassey in 1757 AD.
6. Thus, the British won the battle of Plassey by treachery and deceit.

Question 2.
The European countries felt the necessity of finding new sea routes to Asia.
Answer:

1. The Ottoman Turks captured Constantinople. the capital of Byzantine Empire. in 1453 AD.
2. All land trade routes joining Asia and Europe passed through this city.
3. It stopped the flow of valuable goods from the countries in the east. Therefore, the European countries felt the necessity of finding new sea routes to Asia.

Question 3.
European rulers started giving military protection and trade concessions to the trading companies.
Answer:
1. With the discovery of new sea routes to the east, trade with oriental countries became very profitable. Thus, European trade prospered.
2. Competition for trade began among many European nations.
3. The traders pooled their resources and formed trading companies like the East India Company.
4. As this trade brought economic prosperity, the European rulers started giving military protection and trade concessions to these trading companies.

4. Complete the following table.

Question 1.

Answer:

Leonardo Da Vinci :

1. A famous personality of the Renaissance period who was well versed in different branches of science and arts.
2. He had mastery over varied subjects such as sculpture, architecture, mathematics, engineering, music, astronomy, etc.
3. His paintings ‘Monalisa’ and ‘The Last Supper’ became immortal.

Let us Know:

Project:

Collect information and pictures about the work of famous painters, writers, scientists during renaissance period with the help of reference books and internet. Present a project on it in the classroom.

Class 8 History Chapter 2 Europe and India Additional Important Questions and Answers

Rewrite the statements by choosing the appropriate options:

Question 1.
American colonies organised the army under the leadership of and won against England.
(a) Thomas Jefferson
(b) Robert Clive
(c) George Washington
(d) Johannes Gutenberg
Answer:
(c) George Washington

Question 2.
The British East India Company established its factory at
(a) Kolkata
(b) Cochin
(c) Mumbai
(d) Surat
Answer:
(d) Surat

Question 3.
Siraj-ud-Daulah was defeated because diplomatically bribed Mir Jafar.
(a) Robert Clive
(b) Lord Wellesley
(c) Lord Dalhousie
(d) Lord Cornwallis
Answer:
(a) Robert Clive

Question 4.
Tipu Sultan died in the battle of in 1799.
(a) Mysore
(b) Bengaluru
(c) Srirangapatna
(d) Buxar
Answer:
(c) Srirangapatna

Question 5.
The chieftain of rose against the British.
(a) Multan
(b) Bengal
(c) Carnatic
(d) Jhansi
Answer:
(a) Multan

Question 6.
was the first Portuguese sailor who landed on the western coast of India.
(a) Christopher Columbus
(b) Bartolomeu Días
(c) Vasco-da- Gama
(d) Leonardo da Vinci
Answer:
(c) Vasco-da- Gama

Question 7.
England established colonies on the eastern coast of America.
(a) ten
(b) thirteen
(c) eight
(d) seventeen
Answer:
(b) thirteen

Name the following:

Question 1.
Invented printing press
Answer:
Johannes Gutenberg

Question 2.
Empire of Ottoman Turks
Answer:
Byzantine

Question 3.
The Bill which laid limitations on powers of the King in England.
Answer:
Bill of Rights

Question 4.
Mughal Emperor who gave permission to Britishers to establish factory at Surat.
Answer:
Emperor Jahangir

Question 5.
Came to be known as ‘World Factory’.
Answer:
England.

Answer the following in one sentence:

Question 1.
Which period in the history of Europe is known as Period of Renaissance?
Answer:
In the history of Europe, the last phase of medieval period i.e. 13th to century in Europe is known as the Period of Renaissance.

Question 2.
Name the developments that laid the foundation of modern era.
Answer:
The Renaissance, the Reformation and the geographical discoveries laid the foundation of the modern era.

Question 3.
State the significance of the invention of printing press.
Answer:
It became possible to take new ideas, new concepts and knowledge to all the sections of society due to the invention of the printing press.

Question 4.
What was the effect of the Reformation movement in the area of religion?
Answer:
Individual freedom and rationalism gained importance in the area of religion due to the Reformation movement.

Question 5.
State the principles that the French Revolution gave to the world.
Answer:
The French Revolution gave the principles of liberty equality and fraternity to the world.

Question 6.
Why did the British establish East India Company?
Answer:
The British established the East India Company to carry out trade in India.

Question 7.
What was the outcome of the American War of Independence?
Answer:
Due to American War of Independence, a new nation known as the United States of America with a federal government, written constitution and based on the principles of democracy was formed.

Do as Directed: 

1. Complete the concept Map:

Question 1.

Answer:

Question 2.

Answer:

2. Arrange the following events in chronological order on timeline:
1. Battle of Srirangapatna
2. Second Anglo-Sikh War
3. Battle of Buxar
4. Battle of Plassey
Answer:

Write short notes:

Question 1.
Leonardo-da-Vinci:
Answer:

1. Leonardo-da-Vinci is regarded as an all round personality of the Renaissance period.
2. He was well versed with the different branches of science and art.
3. He had mastery over varied subjects such as sculpture, architecture, mathematics, engineering, music, astronomy, etc.
4. But he became world-famous as a painter.
5. His paintings ‘Monalisa’ and ‘The Last Supper’ became immortal.

Question 2.
Carnatic Wars :
Answer:

1. There was competition among the European nationals to secure monopoly of trading rights in India.
2. Due to this three Cainatic wars were fought between England and France.
3. England defeated France in the third Carnatic War.
4. Alter this, there was no strong European competitor for British East India Company in India.

Question 3.
The Battle of Buxar:
Answer:

1. Mir Kasim was replaced by Mir Jafar when he tried to put restrictions on illegal practices of the British.
2. Mir Kasim, the deposed Nawab of Bengal, Shuja-ud-Daulah, the Nawab of Ayodhya and Mughal Emperor Shah Alam formed an affiance.
3. They undertook a campaign to restrain the activities of the British in Bengal.
4. They were defeated by the British in the Battle of Buxar in 1764 in Bihar.
5. The Treaty of Allahabad was signed after the Battle of Buxar. The British secured the right to collect revenue from Bengal province by this treaty.

Explain the following statements with reasons:

Question 1.
The period between 18th and 19th century came to be known as ‘Age of Revolutions’.
Answer:
1. The British Parliament passed The Bill of Rights in 1689. It established sovereignty of Parliament and curtailed powers of the King.
2. This gave momentum to the development of Parliamentary Democracy in England.
3.  The thirteen British colonies in America won the War of Independence in 1783.
4. These thirteen colonies established the first federal republic, known as the United States of America.
5. The people of France revolted against the unjust and uncontrolled monarchy and feudalism. They established a Republic.
6. The Industrial Revolution in Europe in the latter part of the 18th century brought revolutionary changes in Europe.
7. These changes had long-lasting effect on the entire world.
8. All these revolutionary events took place during 18th and 19th century period. Hence it is known as the ‘Age of Revolutions’.

Question 2.
England was described as World Factory.
Answer:

1. There were many revolutionary changes in the latter part of the 18th century in Europe.
2. Many technological innovations in Europe gave enormous outputs
3. Goods were produced with the help of machines running on steam and later on electricity. It was the dawn of the age of machines.
4. Though the industrial revolution started in England it gradually spread to other ports of the western world.
5. During this period. England enjoyed industrial prosperity and was described as World Factory.

Question 3.
The British captured Sindh in 1843.
Answer:

1. The British were afraid of Russian aggression on India via Afghanistan.
2. They decided to bring Afghanistan under their control to ensure the safety of their Indian Empire.
3. The province of Sindh was located at the Northwest frontier.
4. All routes to Afghanistan passed through Sindh.
5. Realising its strategic importance, the British captured Sindh in 1843.

Answer the following questions in 25 to 30 words:

Question 1.
Write about the Religious Reform Movement.
Answer:

1. The independent intellectuals attacked the old religious ideologies and beliefs of Roman Catholic Church.
2. The Christian priests used to give undue importance to religious rites and practices by taking advantage of the ignorance of the people.
3. They robbed people in the name of religion.
4. The movement which started against this is known as Religious Reform Movement.

Question 2.
What is Intellectual Revolution?
Answer:

1. Due to Renaissance, the European society stepped out of past ignorance and blind faith.
2. The pre-determined customs and traditions came to be seen from a critical point of view.
3. These changes are addressed as Intellectual Revolution.

Question 3.
State the significance of French Revolution.
Answer:

1. The French Revolution was the first event in world history in which the might of the people’s power was witnessed.
2. It had put an end to uncontrolled and unjust monarchy and the feudalism.
3. France became a Republican state. It gave the world principles of liberty, equality and fraternity.
4. The spread of democratic form of government began with the French Revolution.

Question 4.
What were the changes brought about by the Industrial Revolution in the system of production?
Answer:

1. The latter part of 18th century saw many technological innovations in Europe.
2. The production started with the help of machines running on steam.
3. The big factories replaced cottage industries.
4. Handlooms were replaced by machines.
5. New modes of transport like railways and steamer were available.

Question 5.
How did the conflict between Siraj- ud-Daulah and the British start?
Answer:

1. Siraj-ud-Daulah became the Nawab of Bengal in 1756.
2. The British traders misused trade concessions and fortified their factories in Kolkata without seeking permission from the Nawab.
3. So, Siraj-ud-Daulah captured the factory. This was the beginning of conflict between Siraj-ud-Daulah and the British.

Question 6.
Why were the Sikhs defeated in the First Anglo-Sikh War?
Answer:

1. Ranjitsingh, the ruler of Punjab was succeeded by his minor son Duleep Singh.
2. Queen Jindan assumed the reins of administration on behalf of her minor son.
3. Taking advantage of the inability of Queen Jindan to control the officers, the British lured some of them to their side.
4. The Sikhs attacked the British under the impression that the British would invade Punjab.
5. Thus, the Sikhs were defeated in the first Anglo-Sikh war.

Question 7.
What brought an end to the Sikh rule in Punjab in 1849?
Answer:

1. After the First Anglo-Sikh War Duleep Singh remained on throne, but the power rested in the hands of the British.
2. The Sikhs could not accept the growing impact of the British over Punjab.
3. The growing influence of the British led to the revolt by Mulraj, the chieftain of Multan.
4. Thousands of Sikh soldiers participated in this war.
5. The British defeated the Sikhs in the Second Anglo-Sikh war and annexed the whole of Punjab in 1849.

Answer the following questions in detail: (4 marks each)

Question 1.
Give detailed account of Renaissance.
Answer:
The foundation of Renaissance period was laid between 13th century and 16th century in Europe.

1. It revived the Greek and Roman traditions.
2. The changes were seen in the field of art, architecture and philosophy in Europe.
3. Humanitarianism gained great momentum.
4. There was a change in the outlook of treating each other as a human being.
5. Man became the centre of all ideologies instead of religion.
6. The sentiments and emotions of human being found expressions in the art and literature.
7. The reform movement could be found in science and different art forms.
8. Literature was produced in local languages for the better understanding of the people.
9. Renaissance was a comprehensive movement which pervaded all sectors of human life.

Question 2.
Give information about American War of Independence.
Answer:
1. After the discovery of American continent, the imperialist European powers took control of different regions of America and established their colonies.
2. England established 13 colonies on the East coast of America.
3. Initially, England kept nominal domination but later British Parliament laid oppressive restrictions and taxes on the colonies.
4. This created discontent among the people and the colonies revolted.
5. George Washington led the American army in the war against England.
6. The colonies won the war and established the first federal democratic republic, known as the United States of America.
7. USA became the first country in the world to have a written constitution and was based on the principles of Democracy.

Question 3.
What according to you are the advantages of Religious Reform Movement?
Answer:
With changing times we find changes in every field. Religious reforms are part of this process. I feel the following are the advantages of Religious Reforms Movement.

1. Superstitions give way to rational thinking.
2. Priest cannot take advantage of people’s ignorance.
3. People will not indulge in unnecessary rituals.
4. They will engage themselves in constructive work which will help them eventually.
5. Some people take advantage in the name of religion and exploit others. Religious Reforms will stop such malpractices.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Miscellaneous Exercise 8

Solve the following system of inequalities graphically.

Question 1.
x ≥ 3, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 3.
2x + y ≥ 6, 3x + 4y < 12
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 4.
x + y ≥ 4, 2x – y ≤ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 5.
2x – y ≥1, x – 2y ≤ -1
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 6.
x + y ≤ 6, x + y ≥ 4
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 8.
x + y ≤ 9, y > x, x ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 10.
3x + 4y ≤ 60, x +3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 14.
3x + 2y ≤ 150, x + 4y ≥ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 15.
x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.3

Find the graphical solution for the following system of linear inequations.

Question 1.
x – y ≤ 0, 2x – y ≥ -2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 2.
2x + 3y ≥ 12, -x + y ≤ 3, x ≤ 4, y ≥ 3
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 3.
3x + 2y ≤ 1800, 2x + 7y ≤ 1400
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 4.
0 ≤ x ≤ 350, 0 ≤ y ≤ 150
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 5.
\(\frac{x}{60}+\frac{y}{90}\) ≤ 1, \(\frac{x}{120}+\frac{y}{75}\) ≤ 1, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 6.
3x + 2y ≤ 24, 3x + y ≥ 15, x ≥ 4
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.2

Question 1.
Solve the following inequations graphically in a two-dimensional plane
(i) x ≤ -4
Solution:
Given, inequation is x ≤ -4
∴ corresponding equation is x = -4
It is a line parallel to Y-axis passing through the point A(-4, 0)
Origin test:
Substituting x = 0 in inequation, we get
0 ≤ -4 which is false.
∴ Points on the origin side of the line do not satisfy the inequation.
So the points on the non-origin side of the line and points on the line satisfy the inequation
∴ all the points on the line and left of it satisfy the given inequation.
The shaded portion represents the solution set.

(ii) y ≥ 3
Solution:
Given, inequation is y ≥ 3
∴ corresponding equation is y = 3
It is a line parallel to X-axis passing through point A(0, 3)
Origin test:
Substituting y = 0 in inequation, we get
0 ≥ 3 which is false.
∴ Points on the origin side of the line do not satisfy the inequation
∴ Points on the non-origin side of the line satisfy the inequation.
∴ all the points on the line and above it satisfy the given inequation.
The shaded portion represents the solution set.

(iii) y ≤ -2x
Solution:
Given, inequation is y ≤ -2x
∴ corresponding equation is y = -2x
It is a line passing through origin O(0, 0).
To draw the line, we need one more point.
To find another point on the line, we can take any value of x,
say, x = 2.
∴ substituting x = 2 in y = -2x, we get
y = -2(2)
∴ y = -4
∴ another point on the line is A(2, -4)
Now, the origin test is not possible as the origin lies on the line y = -2x
So, choose a point which does not lie on the line say, (2, 1)
∴ substituting x = 2, y = 1 in inequation, we get
1 ≤ -2(2)
∴ 1 ≤ -4 which is false.
∴ the points on the side of the line y = -2x, where (2, 1) lies do not satisfy the inequation.
∴ all the points on the line y = -2x and on the opposite side of the line where (2, 1) lies, satisfy the inequation
The shaded portion represents the solution set.

(iv) y – 5x ≥ 0
Solution:
Given, inequation is y – 5x ≥ 0
∴ corresponding equation is y – 5x = 0
It is a line passing through the point O(0, 0)
To draw the line, we need one more point.
To find another point on the line,
we can take any value of x, say, x = 1.
Substituting x = 1 in y – 5x = 0, we get
y – 5(1) = 0
∴ y = 5
∴ Another point on the line is A(1, 5)
Now origin test is not possible as the origin lies on the line y = 5x
∴ choose a point that does not lie on the line, say (3, 2).
∴ substituting x = 3, y = 2 in inequation, we get
2 – 5(3) ≥ 0
∴ 2 – 10 ≥ 0
∴ -8 ≥ 0 which is false.
∴ the points on the side of line y = 5x where (3, 1) lies do not satisfy the inequation.
∴ the points on the line y = 5x and on the opposite of the line where (3, 2) lies, satisfy the inequation.
The shaded portion represents the solution set.

(v) x – y ≥ 0
Solution:
Given, inequation is x – y ≥ 0
∴ Corresponding equation is x – y = 0
It is a line passing through origin O(0, 0)
To draw the line we need one more point.
To find another point on the line, we can take any value of x,
Say, x = 2.
∴ substituting x = 2 in x – y = 0, we get
2 – y = 0
∴ y = 2
∴ another point on the line is A(2, 2)
Now origin test is not possible as the origin lies on the line y = x
∴ choose a point which not lie on the line say (3, 1)
∴ substituting x = 3, y = 1 in inequation, we get
3 – 1 ≥ 0
∴ 2 ≥ 0 which is true.
∴ all the points on line x – y = 0 and the points on the side where (3, 1) lies satisfy the inequation
The shaded portion represents the solution set.

(vi) 2x – y ≤ -2
Solution:
Given, inequation is 2x – y ≤ -2
∴ corresponding equation is 2x – y = -2
∴ \(\frac{2 x}{-2}-\frac{y}{-2}=\frac{-2}{-2}\)
∴ \(\frac{x}{-1}+\frac{y}{2}=1\)
∴ intersection of line with X-axis is A(-1, 0),
intersection of line with Y-axis is B(0, 2)
Origin test:
Substituting x = 0, y = 0 in the given inequation, we get
2(0) – (0) ≤ -2
∴ 0 ≤ -2
which is false.
∴ Points on the origin side of the line do not satisfy the inequation.
∴ Points on the non-origin side of the line satisfy the inequation
∴ all the points on the line and above it satisfy the given inequation.
The shaded portion represents the solution set.

(vii) 4x + 5y ≤ 40
Solution:
Given, inequation is 4x + 5y ≤ 40
∴ Corresponding equation is 4x + 5y = 40
∴ \(\frac{4 x}{40}+\frac{5 y}{40}=\frac{40}{40}\)
∴ \(\frac{x}{10}+\frac{y}{8}=1\)
∴ Intersection of line with X-axis is A(10, 0)
Intersection of line with Y-axis is B(0, 8)
Origin test:
Substituting x = 0, y = 0 in the inequation, we get
4(0) + 5(0) ≤ 40
∴ 0 ≤ 40 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the given inequation.
The shaded portion represents the solution set.

(viii) \(\left(\frac{1}{4}\right) x+\left(\frac{1}{2}\right) y\) ≤ 1
Solution:
Given, inequation is \(\left(\frac{1}{4}\right) x+\left(\frac{1}{2}\right) y\) ≤ 1
∴ corresponding equation is \(\frac{x}{4}+\frac{y}{2}\) = 1
∴ intersection of line with X-axis is A(4, 0),
intersection of line with Y-axis is B(0, 2)
Origin test:
Substituting x = 0, y = 0 in the given inequation, we get
\(\frac{1}{4}(0)+\frac{1}{2}(0)\) ≤ 1
∴ 0 ≤ 1 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the given inequation.
The shaded portion represents the solution set.

Question 2.
Mr. Rajesh has ₹ 1,800 to spend on fruits for the meeting. Grapes cost ₹ 150 per kg. and peaches cost ₹ 200 per kg. Formulate and solve it graphically.
Solution:
Let x and y be the number of kgs. of grapes and peaches bought.
The cost of grapes is ₹ 150/- per kg, cost of peaches is ₹ 200/- per kg.
∴ cost of v kg of grapes is ₹ 150x
and the cost of y kg of peaches is ₹ 200y.
Mr. Rajesh has ₹ 1800 to spend on fruits.
∴ the total cost of grapes and peaches must be less than or equal to ₹ 1800.
∴ required inequation is 150x + 200y ≤ 1800
i.e., 3x + 4y ≤ 36 ……(i)
Since the number of kg of grapes and peaches can not be negative
∴ x ≥ 0, y ≥ 0
Now, corresponding equation is 3x + 4y = 36
∴ \(\frac{3 x}{36}+\frac{4 y}{36}=\frac{36}{36}\)
∴ \(\frac{x}{12}+\frac{y}{9}=1\)
∴ the intersection of the line with the X-axis is A(12, 0)
the intersection of the line with the Y-axis is B(0, 9)
Origin test:
Substituting x = 0, y = 0 in inequation, we get
3(0) + 4(0) ≤ 36
∴ 0 ≤ 36 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the inequation.
Also, x ≥ 0, y ≥ 0
∴ the solution set is the points on the sides of the triangle OAB and in the interior of ∆OAB.
∴ the shaded portion represents the solution set.

Question 3.
The Diet of the sick person must contain at least 4000 units of vitamin. Each unit of food F₁ contains 200 units of vitamin, whereas each unit of food F₂ contains 100 units of vitamins. Write an inequation to fulfill a sick person’s requirements and represent the solution set graphically.
Solution:
Let the diet of the sick person contain, x units of food F₁ and y units of food F₂.
Since each unit of food F₁ contains 200 units of vitamins.
∴ x units of food F₁ contain 200x units of vitamins.
Also, each unit of food F₂ contains 100 units of vitamins.
y units of food F₂ contain 100y units of vitamins.
Now, Diet for a sick person must contain at least 4000 units of vitamins.
∴ he must take food F₁ and F₂ in such away that total vitamins must be greater than or equal to 4000.
∴ required inequation is 200x + 100y ≥ 4000
i.e., 2x + y ≥ 40
Also x and y cannot be negative.
∴ x ≥ 0, y ≥ 0
Corresponding equation is 2x + y = 40
∴ \(\frac{2 x}{40}+\frac{y}{40}=\frac{40}{40}\)
∴ \(\frac{x}{20}+\frac{y}{40}=1\)
∴ intersection of line with X-axis is A(20, 0)
intersection of line with Y-axis is B(0, 40)
Origin test:
Substituting x = 0, y = 0 in inequation, we get
2(0) + (0) ≥ 40
∴ 0 ≥ 40 which is false
∴ all the points on the non origin side of the line and points on the line satisfy the inequation.
Also, x ≥ 0, y ≥ 0
∴ the solution set is as shown in the figure.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Linear Inequations Ex 8.1

Question 1.
Write the inequations that represent the interval and state whether the interval is bounded or unbounded:
(i) [-4, \(\frac{7}{3}\)]
Solution:
[-4, \(\frac{7}{3}\)]
Here, x takes values between -4 and \(\frac{7}{3}\) including -4 and \(\frac{7}{3}\)
∴ the required inequation is -4 ≤ x ≤ \(\frac{7}{3}\)
∴ it is a bounded (closed) interval.

(ii) (0, 0.9]
Solution:
(0, 0.9]
Here, x takes values between 0 and 0.9, including 0.9 and excluding 0.
∴ the required inequation is 0 < x ≤ 0.9
∴ it is a bounded (semi-right closed) interval.

(iii) (-∞, ∞)
Solution:
(-∞, ∞)
Here, x takes values between -∞ and ∞
∴ the required inequation is -∞ < x < ∞
∴ it is an unbounded (open) interval.

(iv) [5, ∞)
Solution:
[5, ∞)
Here, x takes values between 5 and ∞ including 5.
∴ the required inequation is 5 ≤ x < ∞
∴ it is an unbounded (semi-left closed) interval.

(v) (-11, -2)
Solution:
(-11, -2)
Here, x takes values between -11 and -2
∴ the required inequation is -11 < x < -2
∴ it is a bounded (open) interval.

(vi) (-∞, 3)
Solution:
(-∞, 3)
Here, x takes values between -∞ and 3
∴ the required inequation is -∞ < x < 3
∴ it is an unbounded (open) interval.

Question 2.
Solve the following inequations
(i) 3x – 36 > 0
Solution:
3x – 36 > 0
Adding 36 both sides, we get
3x – 36 + 36 > 0 + 36
∴ 3x > 36
Dividing both sides by 3, we get
\(\frac{3 x}{3}>\frac{36}{3}\)
∴ x > 12
∴ x takes all real values more than 12
∴ Solution set = (12, ∞)

(ii) 7x – 25 ≤ -4
Solution:
7x – 25 ≤ -4
Adding 25 on both sides, we get
7x – 25 + 25 ≤ -4 + 25
∴ 7x ≤ 21
Dividing both sides by 7, we get
x ≤ 3
∴ x takes all real values less or equal to 3.
∴ Solution Set = (-∞, 3]

(iii) 0 < \(\frac{x-5}{4}\) < 3
Solution:
0 < \(\frac{x-5}{4}\) < 3
0 < x – 5 < 12
Adding 5 on both sides, we get
5 < x < 17
x takes all real values between 5 and 17.
∴ Solution set = (5, 17)

(iv) |7x – 4| < 10
Solution:
|7x – 4| < 10
-10 < 7x – 4 < 10 …….[|x| < k is same as -k < x < k]
Adding 4 on both sides, we get
-6 < 7x < 14
Dividing both sides by 7, we get
\(-\frac{6}{7}<x<\frac{14}{7}\)
∴ \(-\frac{6}{7}\) < x < 2
∴ x takes all real values between \(-\frac{6}{7}\) and 2.
∴ Solution set = (\(-\frac{6}{7}\), 2)

Question 3.
Sketch the graph which represents the solution set for the following inequations:
(i) x > 5
Solution:
x > 5
Here, x takes all real values that are greater than 5.
∴ Solution set represents the unbounded (open) interval (5, ∞)
∴ the required graph of the solution set is as follows:

(ii) x ≥ 5
Solution:
x ≥ 5
Here, x takes all real values that are greater than or equal to 5
∴ Solution set represents the unbounded (semi-left closed) interval [5, ∞)
∴ the required graph of the solution set is as follows:

(iii) x < 3
Solution:
x < 3
Here, x takes all real values that are less than 3.
∴ Solution set represents the unbounded (open) interval (-∞, 3)
∴ the required graph of the solution set is as follows:

(iv) x ≤ 3
Solution:
x ≤ 3
Here, x takes all real values less than and including 3
∴ Solution set represents the unbounded (semi-right closed) interval (-∞, 3]
∴ the required graph of the solution set is as follows:

(v) -4 < x < 3
Solution:
-4 < x < 3
Here, x takes all real values between -4 and 3.
∴ Solution set represents the bounded (open) interval (-4, 3)
∴ the required graph of the solution set is as follows:

(vi) -2 ≤ x < 2.5
Solution:
-2 ≤ x < 2.5
Here, x takes all values between -2 and 2.5 including -2.
∴ Solution set represents the bounded (semi-left closed) interval [-2, 2.5)
∴ the required graph of the solution set is as follows.

(vii) -3 ≤ x ≤ 1
Solution:
-3 ≤ x ≤ 1
Here, x takes all real values between -3 and 1 including -3 and 1
∴ Solution set represents the bounded (closed) interval [-3, 1]
∴ the required graph of the solution set is as follows:

(viii) |x| < 4
Solution:
|x| < 4 ⇒ -4 < x < 4
Here, x takes all real values between -4 and 4.
∴ Solution set represents bounded (open) interval (-4, 4)
∴ the required graph of the solution set is as follows:

(ix) |x| ≥ 3.5
Solution:
|x| ≥ 3.5 ⇒ x ≥ 3.5 or x ≤ -3.5
Here, x takes values greater than or equal to 3.5 or it takes values less than or equal to -3.5
∴ Solution set represents the unbounded (semi-left closed) interval [3.5, ∞) or the unbounded (semi-right closed) interval (-∞, -3.5]
∴ x ∈ (-∞, -3.5] ∪ [3.5, ∞)
∴ the required graph of the solution set is as follows:

Question 4.
Solve the inequations:
(i) 5x + 7 > 4 – 2x
Solution:
5x + 7 > 4 – 2x
Adding 2x on both sides, we get
7x + 7 > 4
Subtracting 7 from both sides, we get
7x > -3
Dividing by 7 on both sides, we get
∴ x > \(-\frac{3}{7}\)
i.e., x takes all real values greater than \(-\frac{3}{7}\)
∴ the solution set is (\(-\frac{3}{7}\), ∞)

(ii) 3x + 1 ≥ 6x – 4
Solution:
3x + 1 ≥ 6x – 4
Subtracting 3x from both sides, we get
1 ≥ 3x – 4
Adding 4 on both sides, we get
5 ≥ 3x
Dividing by 3 on both sides, we get
\(\frac{5}{3}\) ≥ x
i.e., x ≤ \(\frac{5}{3}\)
i.e., x takes all real values less than or equal to \(\frac{5}{3}\).
∴ the solution set is (-∞, \(\frac{5}{3}\)]

(iii) 4 – 2x < 3(3 – x)
Solution:
4 – 2x < 3(3 – x)
∴ 4 – 2x < 9 – 3x
Adding 3x on both sides, we get
4 + x < 9
Subtracting 4 from both sides, we get
x < 5
i.e., x takes all real values less than 5
∴ the solution set is (-∞, 5)

(iv) \(\frac{3}{4}\)x – 6 ≤ x – 7
Solution:
\(\frac{3}{4}\)x – 6 ≤ x – 7
Multiplying by 4 on both sides, we get
3x – 24 ≤ 4x – 28
Subtracting 3x from both sides, we get
-24 ≤ x – 28
Adding 28 on both the sides, we get
∴ 4 ≤ x i.e., x ≥ 4
i.e., x takes all real values greater or equal to 4.
∴ the solution set is [4, ∞)

(v) -8 ≤ -(3x – 5) < 13
Solution:
-8 < -(3x – 5) < 13 Multiplying by -1 throughout (so inequality sign changes) 8 ≥ 3x – 5 > -13
i.e., -13 < 3x – 5 ≤ 8
Adding 5 on both the sides, we get
-8 < 3x ≤ 13
Dividing, by 3 on both sides, we get
∴ \(-\frac{8}{3}\) < x ≤ \(\frac{13}{3}\)
i.e., x takes all real values between \(-\frac{8}{3}\) and \(\frac{13}{3}\) including \(\frac{13}{3}\).
∴ the solution set is \(\left(-\frac{8}{3}, \frac{13}{3}\right]\)

(vi) -1 < 3 – \(\frac{x}{5}\) ≤ 1
Solution:
-1 < 3 – \(\frac{x}{5}\) ≤ 1
Subtracting 3 from both sides, we get
-4 < –\(\frac{x}{5}\) < -2 Multiplying by -1 throughout (so inequality sign changes) ∴ 4 > \(\frac{x}{5}\) > 2
i.e., 2 < \(\frac{x}{5}\) < 4
Multiplying by 5 on both sides, we get
10 < x < 20
i.e., x takes all real values between 10 and 20.
∴ the solution set is (10, 20)

(vii) 2|4 – 5x| ≥ 9
Solution:
2|4 – 5x | ≥ 9
∴ |4 – 5x| ≥ \(\frac{9}{2}\)
∴ 4 – 5x ≥ \(\frac{9}{2}\) or 4 – 5x ≤ –\(\frac{9}{2}\) ……[|x| ≥ a implies x ≤ -a or x ≥ a]
Subtracting 4 from both sides, we get
-5x ≥ \(\frac{1}{2}\) or -5x ≤ \(\frac{-17}{2}\)
Divide by -5 (so inequality sign changes)
∴ x ≤ \(-\frac{1}{10}\) or x ≥ \(\frac{17}{10}\)
∴ x takes all real values less than or equal to \(-\frac{1}{10}\)
or it takes all real values greater or equal to \(\frac{17}{10}\).
∴ the solution set is (-∞, \(-\frac{1}{10}\)] or [\(\frac{17}{10}\), ∞)

(viii) |2x + 7| ≤ 25
Solution:
|2x + 7| < 25
∴ -25 ≤ 2x + 7 ≤ 25 …..[|x| ≤ a implies -a ≤ x ≤ a]
Subtracting 7 from both sides, we get
-32 ≤ 2x ≤ 18
Dividing by 2 on both sides, we get
-16 ≤ x ≤ 9
∴ x can take all real values between -16 and 9 including -16 and 9.
∴ the solution set is [-16, 9]

(ix) 2|x + 3| > 1
Solution:
2|x + 3| > 1
Dividing by 2 on both sides, we get
|x + 3| > \(\frac{1}{2}\)
∴ x + 3 < –\(\frac{1}{2}\) or x + 3 > \(\frac{1}{2}\) …..[|x| > a implies x < -a or x > a]
Subtracting 3 from both sides, we get
x < – 3 – \(\frac{1}{2}\) or x > -3 + \(\frac{1}{2}\)
∴ x < \(\frac{-7}{2}\) or x > \(\frac{-5}{2}\)
∴ x can take all real values less \(\frac{-7}{2}\) or it can take values greater than \(\frac{-5}{2}\).
∴ Solution set is (-∞, \(\frac{-7}{2}\)) ∪ (\(\frac{-7}{2}\), ∞)

(x) \(\frac{x+5}{x-3}\) < 0
Solution:
\(\frac{x+5}{x-3}\) < 0
Since \(\frac{a}{b}\) < 0, when a > 0 and b < 0 or a < 0 and b > 0
∴ either x + 5 > 0 and x – 3 < 0
or x + 5 < 0 and x – 3 > 0
Case I:
x + 5 > 0 and x – 3 < 0 ∴ x > -5 and x < 3
∴ -5 < x < 3
∴ solution set = (-5, 3)
Case II:
x + 5 < 0 and x – 3 > 0
∴ x < -5 and x > 3
which is not possible
∴ solution set = Φ
∴ solution set of the given inequation is (-5, 3)

(xi) \(\frac{x-2}{x+5}\) > 0
Solution:
\(\frac{x-2}{x+5}\) > 0
Since \(\frac{a}{b}\) > 0,
when a > 0 and b > 0 or a < 0 and b < 0 b
∴ either x – 2 > 0 and x + 5 > 0
or x – 2 < 0 and x + 5 < 0 Case I: x – 2 > 0 and x + 5 > 0
∴ x > 2 and x > -5
∴ x > 2
∴ solution set = (2, ∞)
Case II:
x – 2 < 0 and x + 5 < 0
∴ x < 2 and x < -5
∴ x < -5
∴ solution set = (-∞, -5)
∴ the solution set of the given inequation is (-∞, -5) ∪ (2, ∞)

Question 5.
Rajiv obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution:
Let x₁, x₂, x₃ denote the marks in 1st, 2nd and 3rd unit test respectively. Then
\(\frac{x_{1}+x_{2}+x_{3}}{3}\) ≥ 60
∴ \(\frac{70+75+x_{3}}{3}\) ≥ 60
∴ 145 + x₃ ≥ 3(60)
Subtracting 145 from both sides, we get
x₃ ≥ 180 – 145
∴ x₃ ≥ 35
Rajiv must obtain a minimum of 35 marks to maintain an average of at least 60 marks.

Question 6.
To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in the first four examinations are 87, 92, 94, and 95, find the minimum marks that Sunita must obtain in the fifth examination to get a grade ‘A’ in the course.
Solution:
Let x₁, x₂, x₃, x₄, x₅ denote the marks in five examinations. Then
\(\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}\) ≥ 90
∴ \(\frac{87+92+94+95+x_{5}}{5}\) ≥ 90
∴ 368 + x₅ ≥ 450
Subtracting 368 from both sides, we get
∴ x₅ ≥ 82
Sunita must obtain a minimum of 82 marks in the 5th examination to get a grade of A.

Question 7.
Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.
Solution:
Let two consecutive positive integers be 2n – 1, 2n + 1 where n ≥ 1 ∈ Z,
Given that 2n – 1 < 10 and 2n + 1 < 10
∴ 2n < 11 and 2n < 9
∴ 2n < 9
∴ n < \(\frac{9}{2}\) …..(i) Also, (2n – 1) + (2n + 1) > 11
∴ 4n > 11
∴ n > \(\frac{11}{4}\) …….(ii)
From (i) and (ii)
\(\frac{11}{4}<n<\frac{9}{2}\) Since, n is an integer,
∴ n = 3, 4
n = 3 gives 2n – 1 = 5, 2n + 1 = 7
and n = 4 gives 2n – 1 = 7, 2n + 1 = 9
∴ The pairs of positive consecutive integers are (5, 7) and (7, 9).

Question 8.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Solution:
Let 2n, 2n + 2 be two positive consecutive integers where n ≥ 1 ∈ Z.
Given that 2n > 5 and 2n + 2 > 5
∴ n > \(\frac{5}{2}\) and 2n > 3
∴ n > \(\frac{5}{2}\) and n > \(\frac{3}{2}\)
∴ n > \(\frac{5}{2}\) ……(i)
Also (2n) + (2n + 2) < 23
∴ 4n + 2 < 23
∴ 4n < 21
∴ n < \(\frac{21}{4}\) ……(ii)
From (i) and (ii)
\(\frac{5}{2}<n<\frac{21}{4}\) and n is an integer.
∴ n = 3, 4, 5
n = 3 gives 2n = 6, 2n + 2 = 8
n = 4 gives 2n = 8, 2n + 2 = 10
n = 5 gives 2n = 10, 2n + 2 = 12
∴ The pairs of positive even consecutive integers are (6, 8) (8, 10), (10, 12)

Question 9.
The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum integer length of the shortest side.
Solution:
Let the shortest side be x.
Then longest side length = 2x
and third side length = x + 2
Perimeter = x + 2x + x + 2 = 4x +2
Given, perimeter > 166
∴ 4x + 2 > 166
∴ 4x > 164
∴ x > 41
∴ Minimum integer length of shortest side is 42 cm.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Miscellaneous Exercise 7

Question 1.
From a group of 2 men (M₁, M₂) and three women (W₁, W₂, W₃), two persons are selected. Describe the sample space of the experiment. If E is the event in which one man and one woman are selected, then which are the cases favourable to E?
Solution:
Let S be the sample space of the given event.
∴ S = {(M₁, M₂), (M₁, W₁), (M₁, W₂), (M₁, W₃), (M₂, W₁), (M₂, W₂), (M₂, W₃), (W₁, W₂) (W₁, W₃), (W₂, W₃)}
Let E be the event that one man and one woman are selected.
∴ E = {(M₁, W₁), (M₁, W₂), (M₁, W₃), (M₂, W₁), (M₂, W₂), (M₂, W₃)}
Here, the order is not important in which 2 persons are selected e.g. (M₁, M₂) is the same as (M₂, M₁)

Question 2.
Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys and 1 girl and 3 boys. One child is selected at random from each group. What is the chance that the three selected consist of 1 girl and 2 boys?
Solution:

Let G₁, G₂, G₃ denote events for selecting a girl,
and B₁, B₂, B₃ denote events for selecting a boy from 1st, 2nd and 3rd groups respectively.
Then P(G₁) = \(\frac{3}{4}\), P(G₂) = \(\frac{2}{4}\), P(G₃) = \(\frac{1}{4}\)
P(B₁) = \(\frac{1}{4}\), P(B₂) = \(\frac{2}{4}\), P(B₃) = \(\frac{3}{4}\)
Where G₁, G₂, G₃, B₁, B₂ and B₃ are mutually exclusive events.
Let E be the event that 1 girl and 2 boys are selected
∴ E = (G₁ ∩ B₂ ∩ B₃) ∪ (B₁ ∩ G₂ ∩ B₃) ∪ (B₁ ∩ B₂ ∩ G₃)
∴ P(E) = P(G₁ ∩ B₂ ∩ B₃) + P(B₁ ∩ G₂ ∩ B₃) + P(B₁ ∩ B₂ ∩ G₃)

Question 3.
A room has 3 sockets for lamps. From a collection of 10 light bulbs, 6 are defective. A person selects 3 at random and puts them in every socket. What is the probability that the room, will be lit?
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 light bulbs in ¹⁰C₃ ways.
∴ n(S) = ¹⁰C₃
Let A be the event that room is lit.
∴ A’ is the event that the room is not lit.
For A’ the bulbs should be selected from the 6 defective bulbs.
This can be done in ⁶C₃ ways.
∴ n(A’) = ⁶C₃
∴ P(A’) = \(\frac{\mathrm{n}\left(\mathrm{A}^{\prime}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{6} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}\)
∴ P(Room is lit) = 1 – P(Room is not lit)

Question 4.
There are 2 red and 3 black balls in a bag. 3 balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls.
Solution:
There are 2 + 3 = 5 balls in the bag and 3 balls can be drawn out of these in
⁵C₃ = \(\frac{5 \times 4 \times 3}{1 \times 2 \times 3}\) = 10 ways.
∴ n(S) = 10
Let A be the event that 2 balls are red and 1 ball is black
2 red balls can be drawn out of 2 red balls in ²C₂ = 1 way
and 1 black ball can be drawn out of 3 black balls in ³C₁ = 3 ways.
∴ n(A) = ²C₂ × ³C₁ = 1 × 3 = 3
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3}{10}\)
Let B be the event that 1 ball is red and 2 balls are black
1 red ball out of 2 red balls can be drawn in ²C₁ = 2 ways
and 2 black balls out of 3 black balls can be drawn in ³C₂ = \(\frac{3 \times 2}{1 \times 2}\) = 3 ways.
∴ n(B) = ²C₁ × ³C₂ = 2 × 3 = 6
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6}{10}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B) = P(A) + P(B)
= \(\frac{3}{10}+\frac{6}{10}\)
= \(\frac{9}{10}\)

Question 5.
A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product of the numbers is even?
Solution:
Two tickets can be drawn out of 25 tickets in ²⁵C₂ = \(\frac{25 \times 24}{1 \times 2}\) = 300 ways.
∴ n(S) = 300
Let A be the event that product of two numbers is even.
This is possible if both numbers are even, or one number is even and other is odd.
As there are 13 odd numbers and 12 even numbers from 1 to 25.
∴ n(A) = ¹²C₂ + ¹²C₁ × ¹³C₁
= \(\frac{12 \times 11}{1 \times 2}\) + 12 × 13
= 66 + 156
= 222
∴ Required probability = P(A)
= \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}\)
= \(\frac{222}{300}\)
= \(\frac{37}{50}\)

Question 6.
A, B and C are mutually exclusive and exhaustive events associated with the random experiment. Find P(A), given that
P(B) = \(\frac{3}{2}\) P(A) and P(C) = \(\frac{1}{2}\) P(B)
Solution:
P(B) = \(\frac{3}{2}\) P(A) and P(C) = \(\frac{1}{2}\) P(B)
Since A, B, C are mutually exclusive and exhaustive events,

Question 7.
An urn contains four tickets marked with numbers 112, 121, 122, 222, and one ticket is drawn at random. Let A_i (i = 1, 2, 3) be the event that ith digit of the number of the ticket drawn is 1. Discuss the independence of the events A₁, A₂, and A₃.
Solution:
One ticket can be drawn out of 4 tickets in ⁴C₁ = 4 ways.
∴ n(S) = 4
According to the given information,
Let A₁ be the event that 1st digit of the number of tickets is 1
A₂ be the event that the 2nd digit of the number of tickets is 1
A₃ be the event that the 3rd digit of the number of tickets is 1
∴ A₁ = {112, 121, 122}, A₂ = {112}, A₃ = {121}

∴ A₁, A₂, A₃ are not pairwise independent
For mutual independence of events A₁, A₂, A₃
We require to have
P(A₁ ∩ A₂ ∩ A₃) = P(A₁) P(A₂) P(A₃)
and P(A₁) P(A₂) = P(A₁ ∩ A₂),
P(A₂) P(A₃) = P(A₂ ∩ A₃),
P(A₁) P(A₃) = P(A₁ ∩ A₃)
∴ From (iii),
A₁, A₂, A₃ are not mutually independent.

Question 8.
The odds against a certain event are 5 : 2 and the odds in favour of another independent event are 6 : 5. Find the chance that at least one of the events will happen.
Solution:
Let A and B be two independent events.
Odds against A are 5 : 2
∴ the probability of occurrence of event A is given by
P(A) = \(\frac{2}{5+2}=\frac{2}{7}\)
Odds in favour of B are 6 : 5
∴ the probability of occurrence of event B is given by
P(B) = \(\frac{6}{6+5}=\frac{6}{11}\)
∴ P(at least one event will happen) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) P(B) ……[∵ A and B are independent events]

Question 9.
The odds against a husband who is 55 years old living till he is 75 is 8 : 5 and it is 4 : 3 against his wife who is now 48, living till she is 68. Find the probability that
(i) the couple will be alive 20 years hence
(ii) at least one of them will be alive 20 years hence.
Solution:
Let A be the event that husband would be alive after 20 years.
Odds against A are 8 : 5
∴ the probability of occurrence of event A is given by
P(A) = \(\frac{5}{8+5}=\frac{5}{13}\)
∴ P(A’) = 1 – P(A)
= 1 – \(\frac{5}{13}\)
= \(\frac{8}{13}\)
Let B be the event that wife would be alive after 20 years.
Odds against B are 4 : 3
∴ the probability of occurrence of event B is given by
P(B) = \(\frac{3}{4+3}=\frac{3}{7}\)
∴ P(B’) = 1 – P(B)
= 1 – \(\frac{3}{7}\)
= \(\frac{4}{7}\)
Since A and B are independent events
∴ A’ and B’ are also independent events
(i) Let X be the event that both will be alive after 20 years.
∴ P(X) = (A ∩ B)
∴ P(X) = P(A) . P(B)
= \(\frac{5}{13} \times \frac{3}{7}\)
= \(\frac{15}{91}\)

(ii) Let Y be the event that at least one will be alive after 20 years.
∴ P(Y) = P(at least one would be alive)
= 1 – P(both would not be alive)
= 1 – P(A’ ∩ B’)
= 1 – P(A’). P(B’)
= 1 – \(\frac{8}{13} \times \frac{4}{7}\)
= 1 – \(\frac{32}{91}\)
= \(\frac{59}{91}\)

Question 10.
Two throws are made, the first with 3 dice and the second with 2 dice. The faces of each die are marked with the number 1 to 6. What is the probability that the total in the first throw is not less than 15 and at the same time the total in the second throw is not less than 8?
Solution:
When 3 dice are thrown, then the sample space S₁ has 6 × 6 × 6 = 216 sample points.
∴ n(S₁) = 216
Let A be the event that the sum of the numbers is not less than 15.
∴ A = {(3, 6, 6), (4, 5, 6), (4, 6, 5), (4, 6, 6), (5, 4, 6), (5, 5, 5), (5, 5, 6), (5, 6, 4), (5, 6, 5), (5, 6, 6), (6, 3, 6), (6, 4, 5), (6, 4, 6), (6, 5, 4), (6, 5, 5), (6, 5, 6), (6, 6, 3), (6, 6, 4), (6, 6, 5), (6, 6, 6)}
∴ n(A) = 20
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}\left(\mathrm{S}_{1}\right)}=\frac{20}{216}=\frac{5}{54}\)
When 2 dice are thrown, the sample space S₂ has 6 × 6 = 36 sample points.
∴ n(S₂) = 36
Let B be the event that sum of numbers is not less than 8.
∴ B = {(2, 6), (3, 5), (3,6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 15
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}\left(\mathrm{S}_{2}\right)}=\frac{15}{36}=\frac{5}{12}\)
A ∩ B = Event that the total in the first throw is not less than 15 and at the same time the total in the second throw is not less than 8.
∴ A and B are independent events
∴ P(A ∩ B) = P(A) . P(B)
= \(\frac{5}{54} \times \frac{5}{12}\)
= \(\frac{25}{648}\)

Question 11.
Two-thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting first class is 0.25 and that of a boy getting is 0.28. Find the probability that a student chosen at random will get first class.
Solution:
Let A be the event that student chosen is a boy
B be the event that student chosen is a girl
C be the event that student gets first class
∴ P(A) = \(\frac{2}{3}\), P(B) = \(\frac{1}{3}\)
Probability of student getting first class, given that student is boy
Probability of student getting first class given that student is a girl, is
P(C/A) = 0.28 = \(\frac{28}{100}\)
and P(C/B) = 0.25 = \(\frac{25}{100}\)
∴ Required probability = P((A ∩ C) ∪ (B ∩ C))
Since A ∩ C and B ∩ C are mutually exclusive events
∴ Required probability = P(A ∩ C) + P(B ∩ C)
= P(A) . P(C/A) + P(B) . P(C/B)

Question 12.
A number of two digits is formed using the digits 1, 2, 3,……, 9. What is the probability that the number so chosen is even and less than 60?
Solution:
The number of two digits can be formed from the given 9 digits in 9 × 9 = 81 different ways.
∴ n(S) = 81
Let A be the event that the number is even and less than 60.
Since the number is even, the unit place of two digits can be filled in ⁴P₁ = 4 different ways by any one of the digits 2, 4, 6, 8.
Also the number is less than 60, so tenth place can be filled in ⁵P₁ = 5 different ways by any one of the digits 1, 2, 3, 4, 5.
∴ n(A) = 4 × 5 = 20
∴ Required probability = P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{20}{81}\)

Question 13.
A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls each are made without replacement. Find the probability that the first drawing will give 3 white balls and the second drawing will give 3 red balls.
Solution:
Total number of balls = 8 + 5 = 13.
3 balls can be drawn out of 13 balls in ¹³C₃ ways.
∴ n(S) = ¹³C₃
Let A be the event that all 3 balls drawn are white.
3 white balls can be drawn out of 5 white balls in ⁵C₃ ways.
∴ n(A) = ⁵C₃
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{5} C_{3}}{{ }^{13} C_{3}}=\frac{5 \times 4 \times 3}{13 \times 12 \times 11}=\frac{5}{143}\)
After drawing 3 white balls which are not replaced in the bag, there are 10 balls left in the bag out of which 8 are red balls.
Let B be the event that the second draw of 3 balls are red.
∴ Probability of drawing 3 red balls, given that 3 white balls have been already drawn, is given by
P(B/A) = \(\frac{{ }^{8} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}=\frac{8 \times 7 \times 6}{10 \times 9 \times 8}=\frac{7}{15}\)
∴ Required probability = P(A ∩ B)
= P(A) . P(B/A)
= \(\frac{5}{143} \times \frac{7}{15}\)
= \(\frac{7}{429}\)

Question 14.
The odds against student X solving a business statistics problem are 8 : 6 and the odds in favour of student Y solving the same problem are 14 : 16
(i) What is the chance that the problem will be solved, if they try independently?
(ii) What is the probability that neither solves the problem?
Solution:
(i) Let A be the event that X solves the problem B be the event that Y solves the problem.
Since the odds against student X solving the problem are 8 : 6
∴ Probability of occurrence of event A is given by
P(A) = \(\frac{6}{8+6}=\frac{6}{14}\)
and P(A’) = 1 – P(A)
= 1 – \(\frac{6}{14}\)
= \(\frac{8}{14}\)
Also, the odds in favour of student Y solving the problem are 14 : 16
∴ Probability of occurrence of event B is given by
P(B) = \(\frac{14}{14+16}=\frac{14}{30}\) and
P(B’) = 1 – P(B)
= 1 – \(\frac{14}{30}\)
= \(\frac{16}{30}\)
Now A and B are independent events.
∴ A’ and B’ are independent events.
∴ A’ ∩ B’ = Event that neither solves the problem
= P(A’ ∩ B’)
= P(A’) . P(B’)
= \(\frac{8}{14} \times \frac{16}{30}\)
= \(\frac{32}{105}\)
A ∪ B = the event that the problem is solved
∴ P(problem will be solved) = P(A ∪ B)
= 1 – P(A ∪ B)’
= 1 – P(A’ ∩ B’)
= 1 – \(\frac{32}{105}\)
= 1 – \(\frac{73}{105}\)

(ii) P (neither solves the problem) = P(A’ ∩ B’)
= P(A’) P(B’)
= \(\frac{8}{14} \times \frac{16}{30}\)
= \(\frac{32}{105}\)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Ex 7.4

Question 1.
Two dice are thrown simultaneously, if at least one of the dice shows a number 5, what is the probability that sum of the numbers on two dice is 9?
Solution:
When two dice are thrown simultaneously, the sample space is
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
Let A be the event that at least one die shows number 5.
∴ A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)}
∴ n(A) = 11
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{11}{36}\)
Let B be the event that sum of the numbers on two dice is 9.
∴ B = {(3, 6), (4, 5), (5, 4), (6, 3)}
Also, A ∩ B = {(4, 5), (5, 4)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}=\frac{2}{36}\)
∴ Probability of sum of numbers on two dice is 9, given that one dice shows number 5, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{2}{36}}{\frac{11}{36}}=\frac{2}{11}\)

Question 2.
A pair of dice is thrown. If sum of the numbers is an even number, what is the probability that it is a perfect square?
Solution:
When two dice are thrown simultaneously, the sample space is
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
Let A be the event that sum of the numbers is an even number.
∴ A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
∴ n(A) = 18
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{18}{36}\)
Let B be the event that sum of outcomes is a perfect square.
∴ B = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}
Also, A n B= {(1, 3), (2, 2), (3, 1)}
∴ n(A ∩ B) = 3
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}\)
∴ Probability of sum of the numbers is a perfect square, given that sum of numbers is an even number, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{3}{36}}{\frac{18}{36}}=\frac{3}{18}=\frac{1}{6}\)

Question 3.
A box contains 11 tickets numbered from 1 to 11. Two tickets are drawn at random with replacement. If the sum is even, find the probability that both the numbers are odd.
Solution:
Two tickets can be drawn from 11 tickets with replacement in 11 × 11 = 121 ways.
∴ n(S) = 121
Let A be the event that the sum of two numbers is even.
The event A occurs, if either both the tickets with odd numbers or both the tickets with even numbers are drawn.
There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10) from 1 to 11.
∴ n(A) = 6 × 6 + 5 × 5
= 36 + 25
= 61
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{61}{121}\)
Let B be the event that the numbers tickets drawn are odd
∴ n(B) = 6 × 6 = 36
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}\)
Since 6 odd numbers are common between A and B.
∴ n(A ∩ B) = 6 × 6 = 36
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{36}{121}\)
∴ Probability of both the numbers are odd, given that sum is even, is given by
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{36}{121}}{\frac{61}{121}}=\frac{36}{61}\)

Question 4.
A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as
A: a club card is drawn.
B: an ace card is drawn.
Determine whether events A and B are independent or not.
Solution:
One card can be drawn out of 52 cards in ⁵²C₁ ways.
∴ n(S) = ⁵²C₁
Let A be the event that a club card is drawn.
1 club card out of 13 club cards can be drawn in ¹³C₁ ways.
∴ n(A) = ¹³C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}\)
Let B be the event that an ace card is drawn.
An ace card out of 4 aces can be drawn in ⁴C₁ ways.
∴ n(B) = ⁴C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}\)
Since 1 card is common between A and B
∴ n(A ∩ B) = ¹C₁
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{1} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{1}{52}\) …….(i)
∴ P(A) × P(B) = \(\frac{{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}} \times \frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{13 \times 4}{52 \times 52}=\frac{1}{52}\) …….(ii)
From (i) and (ii), we get
P(A ∩ B) = P(A) × P(B)
∴ A and B are independent events.

Question 5.
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 1/3, 1/4, and 1/5 respectively. If all of them try independently, what is the probability that,
(i) problem is solved?
(ii) problem is not solved?
(iii) exactly two students solve the problem?
Solution:
Let A be the event that student A can solve the problem.
B be the event that student B can solve the problem.
C be the event that student C can solve problem.
∴ P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{4}\), P(C) = \(\frac{1}{5}\)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
P(B’) = 1 – P(B) = 1 – \(\frac{1}{4}\) = \(\frac{5}{4}\)
P(C’) = 1 – P(C) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Since A, B, C are independent events
∴ A’, B’, C’ are also independent events
(i) Let X be the event that problem is solved.
Problem can be solved if at least one of the three students solves the problem.
P(X) = P(at least one student solves the problem)
= 1 – P(no student solved problem)
= 1 – P(A’ ∩ B’ ∩ C’)
= 1 – P(A’) P(B’) P(C’)
= 1 – \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}\)
= 1 – \(\frac{2}{5}\)
= \(\frac{3}{5}\)

(ii) Let Y be the event that problem is not solved
∴ P(Y) = P(A’ ∩ B’ ∩ C’)
= P(A’) P(B’) P(C’)
= \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}\)
= \(\frac{2}{5}\)

(iii) Let Z be the event that exactly two students solve the problem.
∴ P(Z) = P(A ∩ B ∩ C’) ∪ P(A ∩ B’ ∩ C) ∪ P(A’ ∩ B ∩ C)
= P(A) . P(B) . P(C’) + P(A) . P(B’) . P(C) + P(A’) . P(B) . P(C)
= \(\left(\frac{1}{3} \times \frac{1}{4} \times \frac{4}{5}\right)+\left(\frac{1}{3} \times \frac{3}{4} \times \frac{1}{5}\right)+\left(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\right)\)
= \(\frac{4}{60}+\frac{3}{60}+\frac{2}{60}\)
= \(\frac{3}{20}\)

Question 6.
The probability that a 50-year old man will be alive till age 60 is 0.83 and the probability that a 45-year old woman will be alive till age 55 is 0.97. What is the probability that a man whose age is 50 and his wife whose age is 45 will both be alive for the next 10 years?
Solution:
Let A be the event that man will be alive at 60.
∴ P(A) = 0.83
Let B be the event that a woman will be alive at 55.
∴ P(B) = 0.97
A ∩ B = Event that both will be alive.
Also, A and B are independent events
∴ P(both man and his wife will be alive) = P(A ∩ B)
= P(A) . P(B)
= 0.83 × 0.97
= 0.8051

Question 7.
In an examination, 30% of the students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subjects I and subject II. A student is selected at random, what is the probability that the student
(i) has failed in the subject I, if it is known that he is failed in subject II?
(ii) has failed in at least one subject?
(iii) has failed in exactly one subject?
Solution:
Let A be the event that the student failed in Subject I
B be the event that the student failed in Subject II
Then P(A) = 30% = \(\frac{30}{100}\)
P(B) = 20% = \(\frac{20}{100}\)
and P(A ∩ B) = 10% = \(\frac{10}{100}\)
(i) P (student failed in Subject I, given that he has failed in Subject II) = P(A/B)
\(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\left(\frac{10}{100}\right)}{\left(\frac{20}{100}\right)}=\frac{10}{20}=\frac{1}{2}\)

(ii) P(student failed in at least one subject) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{30}{100}+\frac{20}{100}-\frac{10}{100}\)
= 0.40

(iii) P(student failed in exactly one subject) = P(A) + P(B) – 2P(A ∩ B)
= \(\frac{30}{100}+\frac{20}{100}-2\left(\frac{10}{100}\right)\)
= 0.30

Question 8.
One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third gun respectively. Assuming that A, B, and C are independent events and that P(A) = 0.5, P(B) = 0.6, and P(C) = 0.8, then find the probability that at least one hit is registered.
Solution:
A be the event that first gun hits the target
B be the event that second gun hits the target
C be the event that third gun hits the target
P(A) = 0.5, P(B) = 0.6, P(C) = 0.8
∴ P(A’) = 1 – P(A) = 1 – 0.5 = 0.5
∴ P(B’) = 1 – P(B) = 1 – 0.6 = 0.4
∴ P(C’) = 1 – P(C) = 1 – 0.8 = 0.2
Now A, B, C are independent events
∴ A’, B’, C are also independent events.
∴ P (at least one hit is registered)
= 1 – P(no hit is registered)
= 1 – P(A’ ∩ B’ ∩ C’)
= 1 – P(A’) P(B’) P(C’)
= 1 – (0.5) (0.4) (0.2)
= 1 – 0.04
= 0.96

Question 9.
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that
(i) first is white and second is black?
(ii) one is white and the other is black?
Solution:
Total number of balls = 10 + 15 = 25
Let S be an event that two balls are drawn at random without replacement in succession
∴ n(S) = ²⁵C₁ × ²⁴C₁ = 25 × 24
(i) Let A be the event that the first ball is white and the second is black.
First white ball can be drawn from 10 white balls in ¹⁰C₁ ways
and second black ball can be drawn from 15 black balls in ¹⁵C₁ ways.
∴ n(A) = ¹⁰C₁ × ¹⁵C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{10} \mathrm{C}_{1} \times{ }^{15} \mathrm{C}_{1}}{25 \times 24}=\frac{10 \times 15}{25 \times 24}=\frac{1}{4}\)

Question 10.
An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, what is the probability that at least one ball is black?
Solution:
Total number of balls in the urn = 4 + 5 + 6 = 15
Two balls can be drawn without replacement in ¹⁵C₂ = \(\frac{15 \times 14}{1 \times 2}\) = 105 ways
∴ n(S) = 105
Let A be the event that at least one ball is black
i.e., 1 black and 1 non-black or 2 black and 0 non-black.
1 black ball can be drawn out of 4 black balls in ⁴C₁ = 4 ways
and 1 non-black ball can be drawn out of remaining 11 non-black balls in ¹¹C₁ = 11 ways
∴ 1 black and 1 non black ball can be drawn in 4 × 11 = 44 ways
Also, 2 black balls can be drawn from 4 black balls in ⁴C₂ = \(\frac{4 \times 3}{1 \times 2}\) = 6 ways
∴ n(A) = 44 + 6 = 50
∴ Required probability = P(A) = \(\frac{n(A)}{n(S)}=\frac{50}{105}\) = \(\frac{10}{21}\)

Alternate Solution:
Total number of balls = 15
Required probability = 1 – P(neither of two balls is black)
Balls are drawn without replacement
Probability of first non-black ball drawn = \(\frac{11}{15}\)
Probability of second non-black ball drawn = \(\frac{10}{14}\)
Probability of neither of two balls is black = \(\frac{11}{15} \times \frac{10}{14}=\frac{11}{21}\)
Required probability = 1 – \(\frac{11}{21}\) = \(\frac{10}{21}\)

Question 11.
Two balls are drawn from an urn containing 5 green, 3 blue, 7 yellow balls one by one without replacement. What is the probability that at least one ball is blue?
Solution:
Total number of balls in the urn = 5 + 3 + 7 = 15
Out of these 12 are non-blue balls.
Two balls can be drawn from 15 balls without replacement in ¹⁵C₂
= \(\frac{15 \times 14}{1 \times 2}\)
= 105 ways.
∴ n(S) = 105
Let A be the event that at least one ball is blue,
i.e., 1 blue and other non-blue or both are blue.
∴ n(A) = ³C₁ × ¹²C₁ + ³C₂
= 3 × 12 + 3
= 36 + 3
= 39
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{39}{105}=\frac{13}{35}\)

Alternate solution:
Total number of balls in the urn = 15
Required probability = 1 – P(neither of two balls is blue)
Balls are drawn one by one without replacement.
Probability of first non-blue ball drawn = \(\frac{12}{15}\)
Probability of second non-blue ball drawn = \(\frac{11}{14}\)
Probability of neither of two ball is blue = \(\frac{12}{15} \times \frac{11}{14}=\frac{22}{35}\)
∴ Required probability = 1 – \(\frac{22}{35}\) = \(\frac{13}{35}\)

Question 12.
A bag contains 4 blue and 5 green balls. Another bag contains 3 blue and 7 green balls. If one ball ¡s drawn from each bag, what is the Probability that two balls are of the same colour?
Solution:
Let A be the event that a blue ball is drawn from each bag.
Probability of drawing one blue ball out of 4 blue balls where there are a total of 9 balls in the first bag and that of drawing one blue ball out of 3 blue balls where there are a total of 10 balls in the second bag is
P(A) = \(\frac{4}{9} \times \frac{3}{10}\)
Let B be the event that a green ball is drawn from each bag.
Probability of drawing one green ball out of 5 green balls where there are a total of 9 balls in the first bag and that of drawing one green ball out of 7 green balls where there are a total of 10 balls in the second bag is
P(B) = \(\frac{5}{9} \times \frac{7}{10}\)
Since both, the events are mutually exclusive and exhaustive events
∴ P(that both the balls are of the same colour) = P(both are of blue colour) or P(both are of green colour)
= P(A) + P(B)
= \(\frac{4}{9} \times \frac{3}{10}+\frac{5}{9} \times \frac{7}{10}\)
= \(\frac{12}{90}+\frac{35}{90}\)
= \(\frac{47}{90}\)

Question 13.
Two cards are drawn one after the other from a pack of 52 cards with replacement. What is the probability that both the cards are drawn are face cards?
Solution:
Two cards are drawn from a pack of 52 cards with replacement.
∴ n(S) = 52 × 52
Let A be the event that two cards drawn are face cards.
First card from 12 face cards is drawn with replacement in ¹²C₁ = 12 ways
and second face card is drawn from 12 face card in ¹²C₁ = 12 ways after replacement.
∴ n(A) = 12 × 12
∴ P(that both the cards drawn are face cards) = P(A)
= \(\frac{n(A)}{n(S)}=\frac{12 \times 12}{52 \times 52}=\frac{9}{169}\)