The Living World Notes

The Living World

What Is Living?

  • The living world, that we see around us, has a large amount of variety. It includes different types of animals, plants, and microorganisms, that live in varied habitats such as mountains, lakes, oceans, forests etc.
  • A roaring tiger, a flying bird, or a blooming flower, all of them are so different yet they are parts of the same living world! Such an extent of diversity makes us wonder— what is life? Let us find out what it is.
  • What is life? Can it be defined? It is very difficult to define life. Life may be regarded as a property of living things which distinguishes them from non-living things,  Again, scientists have tried to define life on the basis of a  number of characteristic features. These features include metabolism, reproduction, etc.
  • Anything that exhibits these characteristics is called living. There are several unanswered questions about life. For example, how did life originate? Which species was the first to originate on Earth? Hence, there is ample scope for research in this field. The branch of science which studies all the aspects of life is called life science or biology.

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The-Living-World-Aristotle

  • The history of biology traces the study of the living world from ancient to modern times. Biological sciences emerged from traditional disciplines of medicine and J natural history, dating back to the works of Aristotle and Galen in ancient Greece.
  • The term ‘biology’ in its modern f sense appears to have been introduced independently by Thomas Beddoes (in 1799), and Kar Friedrich Burdach (in  Life originated on earth from non-living matter. These non-living matters include the different inorganic elements. Simple organic molecules are formed from these inorganic elements.

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  • For example, when inorganic elements C, H, N, and O combine, they form organic compounds such as amino acids, nucleotides, etc. These were organised into larger molecules like proteins, nucleic acids, etc. Gradually, through these molecules, life came into existence. Life originated on earth, probably around 3.8 to 4 billion years ago.
  • The origin of life had been discussed in Rig Veda, Manu Samhita, Agni Purana, and other Indian texts about 3000-3500 years ago. Charak, the principal contributor to the ancient science of ayurveda, tried to define life.
  • Attempts to define life were also made by great European scholars like Aristotle and Darwin. In general, life is considered a unique and complex organization of molecules, ions, and even cells. It manifests itself through various characteristics such as growth, development, metabolism, and reproduction.

The Living World Notes

Characteristics of living organisms:

The major characteristics of living organisms are as follows—

Growth and development: 

  1. Growth is a; fundamental characteristic of all living organisms. A single-celled zygote grows into a multicellular organism, Multicellular organisms show growth through repeated cell divisions.
  2. Growth can be characterized by increase in mass and number of cells in an organism. Cell increases in mass due to the aggregation of protoplasmic I material in them.
  3. In plants, growth takes place throughout their life, but in animals, it occurs up to a certain age.
  4. Development is a post-zygotic phase in  which an organism expresses itself through cell division, cell differentiation, and cell rearrangement.
  5. Non-living things do not show growth and development. They may also increase in size due to the deposition of materials on their surfaces. This is known as accretion. This is seen in case of sand dunes in desert. They grow in size due to the deposition of sand.

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 Differences between growth and development:

The Living World Difference Between Growth And Development

Shape and size:

  1. The shape and size of the organisms are determined by their genes.
  2. Organisms are either macroscopic or microscopic. Macroscopic organisms can be observed through the naked eye. In nature, microscopic- organisms can be observed only through a microscope. Microscopic organisms are much more in number than the macroscopic ones.
  3. A young organism grows into an adult. An adult organism does not show non-living things, on the other hand, remains same in shape and size.
  4. Giant sequoia (redwood tree) is the world’s largest tree and the largest living organism by volume. They grow to an average height of 50-95m and are over 17 m in diameter.

Reproduction:

  1. Reproduction is the process by which an organism produces its young ones. It is one of the unique characteristics of the living organisms. By this process, an organism can transmit hereditary traits or characters to its subsequent generations.
  2. Reproduction may be vegetative, asexual and sexual.
  3. Vegetative reproduction is usually found in plants.
  4. Most plants and animals reproduce through asexual or sexual methods.
  5. Non-living things cannot reproduce.

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Metabolism:

  1. Metabolism is an important characteristic of living organisms.
  2. The body of each living organism is constituted of different chemical compounds. A large number of chemical reactions take place in the living cells. These reactions are known as biochemical reactions.
  3. All such chemical reactions, that occur in an organism, are collectively known as metabolism. This process helps to generate energy within a living body.
  4. Sun is the ultimate source of energy on earth. Green plants can harvest solar energy and convert it into chemical energy, in the form of food through a process called photosynthesis. Herbivorous animals or herbivores feed on these green plants and acquire energy. Carnivorous animals or carnivores get their energy by feeding on the herbivores.
  5. This energy is also required for other processes like transport, growth, etc., in living organisms.
  6. Metabolism is of two types—anabolism and catabolism.
  7. Through anabolism, the living cell synthesises complex organic biomolecules from simple inorganic or organic raw materials at the expense of energy. Consequently, the dry weight of the cell or organism (i.e., weight after removal of all of its water content) increases. These anabolic processes include processes like photosynthesis, protein synthesis from amino acids, nucleic acid synthesis, etc.
  8. Through catabolism, large and complex molecules break up into simpler components, mainly to liberate energy. As a consequence, the dry weight of the cell or organism decreases. An important example of catabolism is respiration, in which food matters break up to release energy, C02 and H20.

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The Living World Metabolism Anabolism And Catabolism

Differences between anabolism and catabolism:

The Living World Differences between Anabolism And Catabolism

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Organisation

Cellular organisation:

  1. Every living body is made up of microscopic structures called cells. It is the structural and functional unit of. a living organism.
  2. Each cell contains a jelly-like substance called protoplasm.
  3. The protoplasm of a cell contains a globular body called the nucleus, that regulates all activities of the cell. The part of the protoplasm surrounding the nucleus is the cytoplasm. It contains cell organelles.
  4. The protoplasm of a cell is enclosed by a membrane known as the cell membrane.
  5. The activity of a cell depends on the structure and function of the protoplasm.
  6. The cell as well as protoplasm is absent in non-living things.

Structural organisation:

  1. The body of an organism may be unicellular to multicellular.
  2. In unicellular organisms, all biological functions are performed by a single cell.
  3. In multicellular organisms, a cell is the structural and functional unit. Groups of cells are organised to form tissues. Tissues are further organised into an organ and several organs collectively constitute an organ system.
  4. An external stimulus and the response to it, both occur in the form of a signal. These signals pass through each part of the body. There is a system within the body, that changes one form of signal to the other. This is called a signal transducing system.

The Living World Body Organisation Multi Celluar Organism

Consciousness:

  1. Consciousness refers to the ability to sense the external environment and to respond to environmental stimuli.
  2. There are continuous interactions between an organism and its environment. The environment sends different external stimuli to the organism, who in turn responds to those stimuli.
  3. The ability to generate a response to such stimuli is called sensitivity.
  4. The degree of response varies from one organism to another.
  5. It is observed in both plants and animals. E.g., Mimosa pudica is a seismonastic (sensitive to touch) plant. In this plant, the leaves droop in response to wind, vibration or touch. Rapid closing of eyes in response to high light intensity is another example of sensitivity.
  6. Non-living things neither show sensitivity nor respond to stimuli.

The Living World Sensitivity Of Mimosa Pudica

Apart from these major features, living organisms also have many other features that differentiate them from non-living things. These are as follows—

Genetic material: Every living organism contains deoxyribonucleic acid (DNA) as the genetic material. The genetic material determines the characteristics and functioning of the organism. It also contains units of heredity called genes. It also transmits these characteristics from one generation to the next. Hence is called the hereditary material.

The Living World RXA And DNA

In some organisms such as viruses, RNA (Ribonucleic acid) is present instead of DNA, as the genetic material. No such genetic material is present in a non-living object.

Mutation:

  1. Any sudden alteration in the structure of DNA that brings about permanent, heritable change(s) in the characteristics of the organism, is called mutation.
  2. Mutations act as raw materials of evolution.

Coordination:

  1. All the physiological events in a living body are correlated and all the organs cooperate with each other for proper functioning of the body. Due to this correlation, living organisms maintain a balance in all their activities.
  2. This leads to continuous exchange of information between internal and external environments of organisms.
  3. In plants, coordination is maintained through different hormones.
  4. In animals, coordination takes place through nerves, hormones, etc.

Homeostasis: 

  1. Homeostasis refers to the ability of maintaining a stable internal condition despite of changing external conditions.
  2. Examples of homeostasis are — in plants, opening, and closing of stomata regulate the process of transpiration and entry of C02; In humans, regulation of blood pressure, heart beat, concentration of components of blood, levels of Ca2+ in body, etc.

Adaptation:

  1. Living organisms modify themselves according to environmental changes and requirements. This is known as adaptation. This results in morphological, physiological, biochemical, and molecular changes within the organisms.
  2. These changes are generally inherited over generations, for survival under adverse conditions.
  3. Adaptation is the inherent characteristic of all living organisms.
  4. Without adaptation, no species can survive in this changing world.

It has been observed that adaptations over long periods of time, could cause an ancestral species to give rise to several descendant species. This may occur if one population is fragmented into several subpopulations and each inhabiting different habitats.

  • In cases of geographically isolated populations, one species can diverge rapidly into multiple species through adaptation when new resources are available due to environmental changes. This is called adaptive radiation. This may be demonstrated through the famous example of adaptive radiation as seen with Darwin’s finches.
  • In this case, it was observed that several new species of birds (finches) originated from a common ancestor Darwin collected specimens of these birds during his visit to the Galapagos Islands, 900 km off the Pacific coast of South America, in 1835.
  • These species of birds (more commonly called finches) differed on the basis of ecology, song, and morphology, specifically the nature of the beaks.
  • These birds remain scattered throughout the islands, which signified that populations of the same species became isolated geographically and evolved separately. One major difference among the different species within the same island is the shape and size of the beak.
  • Due to adaptation to different food and resources, the finches developed different kinds of beaks. This difference in the beaks, brought about by adaptive radiation helps several species of finches to survive in the same environment, without competition or lack of resources. This further increases the biodiversity in the respective region.

The Living World Evolution Of Finch As Observed By Darwin

Life cycle:

  1. The series of changes that take place in an organism from its birth to death is called its life cycle. Birth, growth, development, reproduction, senescence or ageing and death are the changes involved in a life cycle
  2. The time period from birth to death of an organism is its life span. It varies among

The Living World Life Cycle Of An Animal

organisms of different species from a few hours to several hundred or several thousand of years.

The Living World Life Cycle Of Plant

Ageing and death

  1. After reaching the adult stage, the body of an organism gradually begins to grow old. This is known as senescence or ageing.
  2. It continues up to a certain age after which the organism dies.
  • Morphogenesis: It is the process of development of different germ layers, organs, and parts of an organism. This includes formation of morula from zygote. The morula grows into a blastula. The blastula gradually transforms into gastrula which ultimately gives rise to a tiny organism.
  • Healing and repair: All living organisms have the ability to heal their wounds. In other words, they have the ability to repair the injured tissues. For example, wounds on our skin get healed by growth of new skin. The wounded parts of the plants also get healed by activities of a tissue called cambium.
  • Excretion: Along with useful products, several waste products are also produced by the process of metabolism. Removal of these waste products from the body is known as excretion.
  • Self-regulation: A regulatory system operates within every living organism. It controls the functioning (such as metabolism, excretion, and reproduction) of the body. Non-living organisms do not have any such self-regulatory mechanism.
  • Movement: Movement includes motion of any part of the body and that of the whole body. Movement of the whole body from one place to another is called locomotion.
  • Animals show both movement and locomotion while plants show only movement. Movements that occur due to internal forces are called autonomic movements. Movements that occur due to external stimuli are called paratonic movements.

Evolution: It is the gradual modification of one species over successive generations, into new species. For example, primitive apes have undergone evolution over numerous generations, to give rise to modern man.

Concept Check:

  1. Who is known as the ‘Father of Biology’?
  2. Name two ancient texts in which ‘Origin of Life1 had been discussed about 3000-3500 years ago.
  3. What is life? What is the relation between growth and life?
  4. Explain loss in protoplasmic dry weight, with an example.
  5. Name the ways by which sensitivity is helpful to man.
  6. What is homeostasis?
  7. Reproduction is intimately related to life.1—explain.

Diversity In The Living World

  • We find a large variety of living organisms around us. Some of them are large, while some are small. Some of them are so tiny that they cannot be seen through the naked eye. The variability among the organisms found on earth is known as biodiversity.
  • About 1.7-1.8 million species of organisms have been found on earth till now and several more are yet to be discovered.
  • Biodiversity is the healthy and effective basis of the ecosystem. Biodiversity exists due to the difference in genes among the individuals of a species, and that between different species. Biodiversity deals with the degree of nature’s variety in the biosphere.
  • Definition: Biodiversity may be defined as a large diversification of the biological world due to the diversification of gene pool, species, and ecology.
  • In other words, the diverse types of organisms j present as a whole, in a region at a particular time, is called the biodiversity of that region at that time.
  • Thus, biodiversity may be observed at three levels as follows— 0 genetic diversity,  species diversity, ecological or community diversity.

Genetic diversity:

  • Members of any plant or animal species differ widely in their genetic constitution. It is due to heredity that every individual gets a set of specific characteristics. Thus, for example, within a human population, each one is different from others in that population as well as from all other organisms.
  • This genetic diversity is essential for a population of any species. This genetic variety in wild species constitutes the ‘gene pool’ from which modern-day crops and domestic animals have evolved over the course of thousands of years.
  • Today new variety of plants with more productivity are being developed by using wild relatives of the existing crop plants. This process is also adapted to develop new breeds of domestic animals with desirable and improved features.
  • However, If the number of individuals is reduced in a  species, the genetic diversity is reduced.

Species diversity:

  • Species is a basic unit of classification. It is defined as a  group of similar organisms that mate and produce progeny of the same type with further variations thus, sharing a common lineage.
  • Various species of plants and animals present in a region constitute its species diversity. This diversity is seen both in natural ecosystems and in agricultural ecosystems.
  • Some areas are richer in species diversity than others. Natural undisturbed tropical forests have much greater richness of species than plantations. At present scientists have been able to identify and categorize about 1.75 million species on earth.
  • However, many new species are being continuously identified, especially in the group of flowering plants and insects.
  • Areas that are rich in species diversity are called ‘hotspots’ of diversity. India is among the world’s 15 nations that are exceptionally rich in species diversity.

Ecological or community diversity:

  • There is a large variety of different ecosystems on earth. Each has its own distinctive interlinked species based on the differences in the habitat.
  • Ecosystem diversity can be described for a specific geographical region, or a political entity such as a country, a state, or a district.
  • Distinctive ecosystems include landscapes such as forests, grasslands, deserts, mountains, etc., and aquatic ecosystems such as rivers, lakes, and the sea. Ecosystems are mostly natural in wilder areas.
  • If natural ecosystems are overused or misused, their productivity eventually decreases and they are then said to be degraded. India is exceptionally rich in ecological diversity.

This kind of diversity is again observed at three levels – Alpha (α), Beta (β), Gamma (γ) diversity.

Alpha diversity:

It is the biodiversity within a particular area, community, or ecosystem. It is usually expressed by the number of species (i.e., species richness) in that ecosystem. This can be measured by counting the number of taxa (distinct groups of organisms such as families, genera, and species) within the ecosystem.

Beta diversity:

  • Beta diversity is a measure of biodiversity which is done by comparing the species diversity between ecosystems or habitats. This comparison involves the number of taxa unique to each of the ecosystems.
  • It is the rate of change in species composition across habitats or among communities. It gives a quantitative measure of diversity of communities that experience changing environments.

Gamma diversity:

It refers to the total species richness over a large area or region. It is a measure of the overall diversity for the different ecosystems within a region.

It is the sum of the diversity of component ecosystems and the diversity between component ecosystems. Gamma diversity can be expressed in terms of the species richness of component communities as follows

γ= SI + S2 – c

Where, SI = the total number of species recorded in the first community, S2 = the total number of species recorded in the second community, c = the number of species common to both communities.

Significance of biodiversity:

The present-day biodiversity on earth is the outcome of nearly 3.5 billion years of evolution. In the period before the appearance of humans, the earth supported more biodiversity than any other period in geological history. However, since the dominance of mankind, biodiversity showed begun a rapid decline, with one species after another suffering extinction.

Biodiversity should be conserved for the following important reasons—

Ecological value: Every species carries out a particular function within an ecosystem. Some of them can capture and store energy. Some of them produce food and others decompose organic material.

  • There are species that help to recycle water and nutrients throughout the ecosystem, fix atmospheric gases, or help to regulate climate.
  • Ecosystems with immense biodiversity provide different sources of food. These also provide services without which we can not survive.
  • These include decomposition of wastes, purification of the air and water, stabilization and moderation of the climate, reducing chances of flood, drought, and other environmental disasters.
  • More diverse an ecosystem, the better it can withstand environmental stress and become more productive.

Economic value: For mankind, biodiversity is first a resource for daily life. Crop diversity is very important for mankind as a food source and is also called agrobiodiversity. Biodiversity is also viewed as a reservoir of resources like food, pharmaceuticals, and other necessary products. Thus resource shortages may be related to the loss of biodiversity.

The Living World Cinchona Plant

Some of the economically important commodities that biodiversity supplies to humans are as follows.

  • Food: Biodiversity provides a variety of food sources like — crops, livestock, forestry, and fishes.
  • Medicine: Several plant species have been used for medicinal purposes since the prehistoric times. For example, quinine (to treat malaria) comes from the bark of the Cinchona plant; morphine (used as pain reliever) comes from the Poppy plant. It is estimated that out of 250,000 known plant species, only 5,000 have been explored so far, for possible medical applications.
  • Commercial products: Biodiversity is a source of different products that have commercial use. For example, fibers (used for making cloth), wood (for making furniture), etc.
  • Other such products are oils, lubricants, perfumes, paper, waxes, rubber, latexes, resins, and cork. All of these can be derived from various plant species. Products of animal origin are wool, silk, fur, leather, lubricants, waxes.
  • Transport: Different animals may also be used as modes of transport. Tourism and recreation: Biodiversity also facilitates tourism and provides a source of recreation as well. Ecotourism, in particular, is a growing outdoor recreational activity.
  • Environmental value: The adverse effects of environmental pollution may be prevented through more and more biological diversification.
  • Ethical value:  Each and every organism has the j full right to survive in the world. This has been accepted in the announcement of United Nations Environment Programme in 1982. Therefore, it is the duty of man to save and conserve all the species in the world.
  • Laws for conservation of biodiversity: Several lav/s have been framed, both nationally as well as,  internationally for conservation of biodiversity. Some famous law for conservation of biodiversity in India are The wildlife (protection) act, 1972. 2) The Forest (conservation) act, 1980. 3) The environment (protection act), 1986. 4) Biological diversity act, 2002.

Taxonomy And Systematics

  • Each organism on earth is known by different local names at different places. This gives rise to confusion. To avoid this, a universal name is given to each organism. This name is used to refer to the organism, irrespective of its local name.
  • This system of providing a universal name to an organism is known as nomenclature. The science of naming, classification, and identification of different organisms by their names and characteristic features is called taxonomy. The term was coined by de Candolle in 1813.
  • Carolus Linnaeus is known as the Father of Taxonomy. Taxonomy is a part of a broader branch of study called systematics. Systematics is defined as the scientific study of organisms regarding their identification, naming, and classification.
  • The term was coined by Linnaeus in 1751. Simpson had defined systematics as— the study of diversity of organisms and all their comparative and evolutionary relationships, based on the studies of anatomy, ecology, physiology, and biochemistry.

Taxonomy And Systematics Identification:

Taxonomy And Systematics Definition: Identification is the process of differentiating one organism from others with the help of its specific features.

Necessity of identification:

  1. An organism should be identified, in order to be placed in the appropriate taxon (unit of classification like genus, species, etc.).
  2. Correct identification is essential because further studies on the organism depend on it.

Procedure of identification:

  • Generally, it is carried out directly by comparing the organism to be identified with other organisms of known identity. It can also be carried out indirectly by comparing the features of the organism with those of similar organisms, as mentioned in books and other scientific literature.
  • Sometimes, a specimen may not resemble any previously identified one. In that case, the specimen is considered as a new one.

Nomenclature:

  • Nomenclature Definition: Nomenclature is defined as the scientific and international system of naming any organism.
  • Nomenclature Necessity: People of a particular place know the local plants and animals by their common or vernacular names. But these names vary from place to place. To overcome this problem, it is essential that each organism should have a universally accepted scientific name. Scientific names do not vary with region and help to identify a particular organism in any place, around the world.
  • Nomenclature Types: All organisms on earth have two names— common name or vernacular name and scientific name. Carolus Linnaeus introduced scientific names of plants and animals in his books, Species Plantarum (1753) and Systema Naturae (1758, 10th Ed.) respectively. He named the plants and animals by laying down certain rules.

Common or vernacular name:

  • Common or vernacular name Definition:  Common name or vernacular name is the name that is based on the local language of a particular place.
  • Example:  The cockroach, which is found everywhere in the world, is known by different names in different places. In Hindi, it is known as Tilachatta and Aarshola in Bengali.
  • Common or vernacular name Advantages:  Common names are easy to pronounce. A layman (person without any scientific knowledge) can identify an organism easily by its common name.
  • Common or vernacular name Disadvantages:  Common names are not the same in every language. So it is difficult to identify an organism in different places by its common name if the local language is not known.

Scientific name

Scientific name Definition: Scientific name is the name given to each organism following the rules of international system of nomenclature.

Example: The scientific name of onion is Allium cepa. Onion is scientifically identified by this name, all over the world.

Scientific name Advantages:

  1. A scientific name helps to avoid the confusion that arises due to different local names of the same organism, around the world.
  2. Scientific names are universally accepted.
  3. The rules of scientific naming or nomenclature are the same all over the world.

Method of naming: There are three common methods of nomenclature.

The Living World Method Of Naming

  • Binomial nomenclature:  Latin term bi means two and nomen means name. This type of naming includes two levels—genus and species.
  • Binomial Definition: The scientific system of naming of any organism by using two words, where the first word is the generic name (genus) and the second word is the specific name (species), is known as binomial nomenclature.
  • Example: Mangifera indica, where Mangifera is the genus and indica is the species.
  • Caspar Bauhin (1560-1624) was the first to implement a system of naming, which was the precursor of the binomial system of nomenclature. Later on, this was followed by Carolus Linnaeus who used binomial nomenclature for the plants in his book Species Plantarum (1753) and for the animals in his book Systema Naturae. He is considered the Father of modern taxonomy and binomial nomenclature.

The Living World Bauhin And Linnaeus

Rules for binomial nomenclature:

  • The rules for binomial nomenclature are framed and standardised by some international scientific organisations. These rules are followed all over the world. International rules for naming plants are called International Code of Botanical Nomenclature (ICBN).
  • International rules for naming animals are called International Code of Zoological Nomenclature (ICZN). There are international rules for naming other organisms as well, such as International Code of Nomenclature of Bacteria (ICNB) for naming bacteria, International Code of Viral Nomenclature (ICVN) for naming viruses, etc.

Common recommendations

Some common recommendations for binomial  nomenclature are

  1. Every organism should have one scientific name, that is composed of two Latin words. The first word should be genus and the second is species.
  2. The scientific name should always be typed in italics.  In case it is handwritten, both generic and specific epithets should be underlined separately.
  3. The generic name should always start with capital letter while the specific epithet should start with small letter.
  4. In some cases, the name is further followed by the name of the; author (the taxonomist who first described the species; and named it). If the author’s name is too long, the abbreviated form is used. For example, the complete  scientific name of a human being is Homo sapiens (L.).  Here, T denotes the name of the author Linnaeus.
  5. Organisms of different kingdoms such as animals and plants, should be named differently. Genus names of two different types of organisms belonging to the same kingdom should not be the same. But the same species name can be used with different genus names. For example, Cyriocosmus elegans is the scientific name of a spider species (tarantula) and Centruroides elegans is the: scientific name of a scorpion.

Some information related to nomenclature

  • Homonym: If the names of two different organisms are pronounced and spelled in the same way, then they are known as homonyms.
  • According to international code for biological nomenclature, the older of the two homonyms is accepted while the other name is rejected. For example, the Astragalus rhizanthus Boiss (1843) is a homonym of Astragalus rhizanthus Royle (1835).
  • Toponym: When an organism is named after its habitat or geographical location, then this type is known as toponym. For example, Myiagra caledonica is a bird found in New Caledonia. This bird is sometimes known as the New Caledonian flycatcher.
  • Tautonym: If the genus and species name of any organism are same, then it is an example of tautonym. Example, Gorilla gorilla. This type of name is not permitted in botanical nomenclature but is accepted in zoological nomenclature.

Scientific names of India’s national tree, animal and bird:

The Living World Scientific Names Of Indias National Tree Animal And Bird

Scientific names of some plants:

The Living World Scientific Names Of Some Plants

Scientific names of some animals:

The Living World Scientific Names Of Some Animals

  • Trinomial nomenclature: In this type of naming, scientific names include genus names, species names, and subspecies (subgroups under a species) names.
  • Example: The scientific name of house crow is Corvus splendens. The subspecies of house crow are differentiated using trinomial nomenclature. E.g., Corvus splendens splendens, Corvus splendens insolens and Corvus splendens protegatus. Here splendens, insolens, and protegatus are the subspecies.
  • Polynomial nomenclature: In this type of nomenclature organisms are named by using a series of 1 descriptive words, hence, it is known as polynomial j nomenclature. It was in use before the publication of Systema Naturae by Linnaeus.
  • Example: Caryophyllum saxatilis folis gramineus umbellatis corymbis is the full scientific name of the plant Caryophyllum. The name says that this plant grows on the rocks, its leaves are grass-like, and floral arrangement is umbellate corymb (flowers are arranged 1 to form an umbrella-like shape).

Some information related to nomenclature Classification: Diverse types of living organisms are present on earth. Each year a number of new species are added to the list of identified organisms. Many fossils have also been discovered in recent times. To study the different types of living organisms easily, scientists have classified them into various groups or categories.

Some information related to nomenclature Definition: The system of categorisation of different organisms into different groups on the basis of their significant characteristics and relationship is known as classification.

Need for classification:

  1. Classification is necessary as it makes the study of diverse organisms very much convenient.
  2. It helps to identify any organism.
  3. It helps us to get an idea about the organisms belonging to a particular group, by studying only a few organisms representing that group.
  4. It helps us to understand the relationship among different groups of organisms.
  5. It provides information about organisms belonging to any specific geographical region.
  6. It also helps to study fossils and to identify extinct species.
  7. It also helps to understand the evolutionary trends (nature of evolution) across different groups of organisms.

Three domains of life:

Living organisms are divided into three groups or domains, on the basis of genetic constitution. These are called the Three Domains of Life. The three domains of life are — Archaea, Bacteria, and Eukarya. This is actually a system of biological classification that was propounded by American microbiologist Carl Woese in 1990. This system has been discussed in detail in (Biological Classification).

Concept Check:

  1. Who coined the term ‘taxonomy’?
  2. What is binomial nomenclature? Give example.
  3. Define taxonomy.
  4. Define systematics.
  5. What do you understand by classification? .
  6. What do you understand by identification in terms of biology?

Taxonomic Categories

Classification of organisms involves various grouping levels. These levels are known as categories or taxonomic categories. The system of arranging categories one above the other is known as hierarchy or taxonomic hierarchy. Each level of this hierarchy is a unit of classification and represents a rank. These ranks are commonly known as taxon (plural: Taxa). For example, Division Angiospermae, here Division is the category, and ‘Angiospermae’ is the taxon. Details are given in the following table.

Different categories and taxa:

The Living World Different Categories And Taxa

The Living World Taxonomic Hierarchy

Some definitions related to taxonomic hierarchy

  • Monotypic genus: A genus which includes only one species is known as monotypic genus. For example, sapiens is the only species under the genus Homo.
  • Polytypic genus: The genus which includes more than one species are known as polytypic genus. For example, tigris and leo are two species present under the genus Panthera.
  • Monotypic species: The species which are not divided j into subspecies, varieties, and races, are known as monotypic species. For example, in Ginkgo biloba, biloba is the monotypic species.
  • Polytypic species: The species which are divided into two or more subspecies, varieties, or races are known as polytypic species. For example, there are many subspecies of Panthera tigris. They are— P. tigris altaica (Siberian tiger), P. tigris tigris (Bengal tiger),  P. tigris jacksoni (Malayan tiger), etc.
  • Species: It is the lowest taxonomic category. The j genetically different but morphologically similar groups of organisms, which are capable of interbreeding, are known as species. Each genus may have one or more species. For example, Panthera leo (lion) and Panthera | tigris (tiger) have different specific epithets (leo and j tigris), but same genus (Panthera). Different species j under the same genus have morphological similarities j but are reproductively isolated.
  • Genus: Genus (PI. genera) is the taxonomic category j that comes above the species. It comprises of j morphologically related species. All the species under j the same genus have a common ancestor. A genus may contain one or more species. For example, lion (Panthera leo), tiger (Panthera tigris), jaguar (Panthera onca), and leopard (Panthera pardus) belong to the same genus (Panthera).
  • Family: It is the taxonomic category that comes above genus. A family contains one or more related genera. For example, Felis (genus of cats) and Panthera (genus of tiger, leopard, lion, and jaguar) are under the same family, Felidae. The suffix used for families are different. For plants, the family name ends with -aceae, while that for animals, the family name ends with -idae.
  • Order: It is the taxonomic category that comes above family. An order comprises one or more related families. For example, families Felidae (family of cats) and Canidae (family of dogs) belong to the same order Carnivora (members of both the families are carnivores and have large canine teeth).

Intermediate Categories

  • Hierarchy categories are of two types— obligate and intermediate categories. The categories which are always used to define the taxonomic position of an organism are known as obligate categories.
  • This includes kingdom, phylum or division, class, order, family, genus, species. Some categories are also added in between obligate categories to make the taxonomic position more precise. These categories are known as intermediate categories. For example, sub-division, superclass, etc.

Class: It is the taxonomic category that comes above order. A class comprises one or more related orders. For example, orders Rodentia (rats), Primata (monkeys) and Carnivora (cats, dogs, etc.) belong to the same class Mammalia (have hairs and milk glands).

Phylum/Division: The next taxonomic category that comes above class is called phylum (in case of animals) or division (in case of plants). It comprises related classes. For example, phylum Chordata includes classes Amphibia, Reptilia, Aves, etc.

Kingdom: The highest category of taxonomic hierarchy is kingdom. Each kingdom consists of same type of phyla/divisions. For example, all the animals are placed under kingdom Animalia, and all plants under kingdom Plantae.

Examples of hierarchical classification:

The Living World Examples Of Hierarchical Classification

Different suffixes used in different hierarchy levels:

The Living World Different Suffixes Used In Different Hierarchy Levels

Concept of species:

  • Species is the basic unit of taxonomy. It is the lowest unit of taxonomic hierarchy. It includes genetically different but morphologically similar groups of organisms. These organisms are capable of interbreeding.
  • They cannot breed with organisms of other species. This feature of any species is termed reproductive isolation. Each genus may have one or more species under it. For example, Panthera leo (scientific name of lion) and Panthera tigris (scientific name of tiger) have different specified names (leo and tigris), but same genus (Panthera).
  • The characteristic of interbreeding is important for a species. However, a species cannot be delimited solely on its basis.
  • This is due to two reasons—0 In some cases, members of different species may interbreed. However, the progeny produced are generally sterile. For example, mule (male ass + mare) and hinny (horse + female ass) are two such sterile offsprings.
  • In some cases of interspecific mating, fertile offsprings such as tigon (male tiger + female lion) and liger (male lion + female tiger) may be produced. 0 Sexual reproduction is absent in organisms like prokaryotes and protists. Hence, the phenomenon of interbreeding is absent in them. In such cases, morphological, anatomical, cytological and other characteristics are considered for categorizing a species.
  • A species may be further divided into subgroups. These subgroups are known as subspecies or varieties. They have characteristic features that distinguish them from other subspecies.
  • Sexual reproduction is an important point in the definition of a species. It produces offsprings that are different from parents.
  • These variations over a course of time may cause a species to change. Thus, species are really dynamic groups and not static as was considered earlier.

Concept Check:

  1. What is the meaning of the word ‘hierarchy?
  2. Name the seven categories of taxonomic hierarchy.
  3. What do you understand by the word ‘family?
  4. What do you understand by the word ‘order?
  5. What is the meaning of the word ‘taxon?

Taxonomical Aids

  • To study the vast diversity of the biological world, we need to identify and classify the organisms correctly. To do this, we need to carry out both fieldwork and laboratory studies.
  • These are carried out by collection and conservation of biological specimens (dead or living) by proper means.
  • The specimens are conserved for further studies. There are several taxonomical aids that help in these studies. These include botanical gardens, herbaria, museum,s and zoological parks.

Museums

  • Museums Definition: A museum is an institution that preserves and exhibits objects of scientific, artistic, cultural, or historical importance.
  • Museums Types: Museums may be classified according to the nature of the objects they preserve. They may be historical museum, natural history museums, etc. The Museum of natural history mainly preserves plants and animals.
  • Working procedure: Museums provide information about various plants and animals to the researchers and students. The working procedures of museums are discussed below.
  • Keeping a record of the sample: Museums keep a record of the samples which they preserve. The Global Biodiversity Information Facility (GBIF), an international organization, maintains the records of the specimens and samples in museums all over the world.
  • Preservation: Preservation methods vary according to the objects that are to be preserved. Plant or animal specimens are preserved in liquid preservatives in jars. The liquid preservative is a mixture of alcohol and formalin. Some large animals are preserved in stuffed forms. Some insects are preserved by drying and mounting in boxes.
  • Exhibition: Science exhibitions in the museum are unique resources for informal education. These help us to remember facts and develop skills in science. They are places to discover, explore, and get ideas about the natural world.

Importance

  1. Museums preserve specimens of plants and animals from various parts of the world. It provides information about these species.
  2. It also helps in the comparative study of different organisms.
  3. Specimens of newly discovered species may also be preserved in some museums as type specimen.
  4. Museums help to draw attention of the common people towards endangered and extinct species.
  5. Biological museum of schools and colleges help the students to study different plant and animal species.

Some famous museums:

The Living World Some Famous Museums

The Living World Zoological Park

Zoological Parks

  • Zoological Parks Definition: A zoological park or garden is a place where living animals are kept in enclosures for study and exhibition.
  • Zoological Parks Types: According to the nature of the conserved animals and their habitat, zoological gardens or parks are of several types such as, open-range zoo, animal theme park, aquarium, etc.

Working procedure:

  1. The area of a zoological park is made comfortable for the animals to live and reproduce.
  2. In the zoological gardens, a diet chart is prepared and followed for all the animals, according to their feeding habits, health condition, and behavior.
  3. There are trained veterinary doctors who take care of the wounded and diseased animals.
  4. The scientific names, common names, and habitat of the animals, etc., are displayed in front of their enclosures. This helps the people gather information about the animals.
  5. A map is provided inside the zoo for helping the tourists.

Importance: 

  1. Zoological parks or gardens are examples of ex-situ conservation. Main scientific purpose of a zoological park is the conservation and breeding of endangered species and subsequent increase in their numbers. This is done to prevent their extinction.
  2. It plays an important role in taxonomic research, identification, and observation of different animals.  It creates awareness among the people regarding the conservation of animals.
    Names of some famous international and national zoological gardens are given in the following table.

Names of some famous zoological gardens or zoos:

The Living World Zoological Gardens Or ZoosHerbarium

Herbarium Definition: The collection of dried plant specimens mounted on a sheet of paper, so as to preserve it in such a way that it provides information about the actual plant, is called a herbarium.

Preparation of herbarium sheet: A thick sheet of paper, on which the dried samples of the plants are mounted, is called the herbarium sheet. Following are the steps involved in the preparation of a herbarium sheet.

Collection of specimen: 

  1. This is the first step involved in the preparation of a herbarium (plural: herbaria).
  2. Specimens are carefully collected at different seasons, as well as from different environments. They are also collected at different stages of their life cycle.
  3. The plant which is to be preserved should have all its parts (leaves, flowers, roots, etc.). In case of flowering plants, their different parts such as fruits, inflorescence, stem, etc., may give different information. Hence, in case of these plants, separate parts of the plant may be collected as specimens at different stages of their development.

Materials required for specimen collection:

  1. Knife,
  2. scissors or cutter (small and big),
  3. old newspapers,
  4. blotting papers and cotton,
  5. forceps,
  6. magnifying glass,
  7. polythene bags,
  8. small shovel or spade or sickle,
  9. field notebook,
  10. vasculum (a kind of case or box),
  11. empty jars and bottles with FAA solution (mixture of formaldehyde, glacial acetic acid and alcohol),
  12. pen or pencil and measuring tape,
  13. brush (small and big),
  14. pieces of cloth,
  15. plant press machine.

Drying: 

  1. Healthy shoots of a certain length, with leaves, flowers and fruits are collected from the plants.
  2. These are properly placed within two blotting papers immediately after collection. These are again kept between two sheets of old newspaper or magazine.
  3. Now these are kept in plant press machine.
  4. To make the specimen dehydrated, without any damage, the blotting papers have to be regularly changed. It is kept in this condition for 24/48/72 hours.
  5. The succulent parts are stored in bottles with FAA solution for faster dehydration.
  6. Sometimes, specimens may be dried by applying heat on moist parts.
  7. Smaller specimens are dried by placing them within folded cloth and pressing it with hot electric iron.

Mounting the sample: 

  1. After drying, the specimen is mounted on a herbarium sheet.
  2. Herbarium sheets are generally white, large, thick pieces of papers, with dimensions of about 29 x 41 cm or 30 x 45 cm.
  3. The specimens are attached to the herbarium sheet by applying glue.
  4. A single specimen is placed on a single herbarium sheet.

Labelling: After mounting, a label (identity card) is placed below the specimen, towards the lower right hand corner of the hebarium sheet. The label contains the following information

  1. Flora of
  2. Collection No
  3. Date of Collection
  4. Locality of collection
  5. Altitude
  6. Habitat
  7. Habit
  8. Distribution
  9. Abundance
  10. Collector
  11. Scientific name of the specimen
  12. Local name of the specimen
  13. Family
  14. Identifier
  15. Remarks

The Living World Specimen Conservation In A Herbarium

Preservation:

  1. Herbarium sheets are preserved in herbarium cases made of wood or steel.
  2. These are stored in cupboards to protect them from insects.
  3. The specimens are arranged according to the families. They may also be arranged according to their scientific names or the name of their species, both alphabetically.

Protection:

  1. Herbarium sheets are destroyed mostly by moulds and fungi. Hence the herbarium sheets and cases must be kept clean and proper measures should be taken to prevent the growth of fungi and moulds.
  2. 0.1% mercuric chloride, DDT spray, napthalene dust, carbon disulphide spray etc., must be used to prevent growth of harmful organisms.
  3. Sometimes warm vapours may also be used to destroy the harmful organisms.
  4. To prevent the fungal growth, the moisture of the room must be regulated.

Importance: 

  1. It helps in the comparative study of different plant species. It also provides primary information about the plant and its corresponding parts.
  2. It helps to identify similar samples of the plants and helps in taxonomic studies.
  3. It draws attention to economically important plants.
  4. It helps to conserve the newly discovered plants and their wild species.
  5. It helps to study the extinct plant species.
  6. It provides reference material for research.

The term herbarium is also used to refer to the building that houses the actual herbarium. Some of the famous herbaria are given in the following table.

Some famous herbaria:

The Living World Some Famous Herbaria

Botanical Carden

Botanical Carden Definition: Botanical gardens are the enclosed places where a wide variety of plants are scientifically cultivated and conserved.

General features:

  1. Different species of plants are cultivated and conserved in these gardens.
  2. Most of the botanical gardens are linked to several research centers, greenhouses, rock gardens, palm houses, orchid houses, herbariums, museum, etc.
  3. Botanical gardens help to educate people on conservation issues and the role of humans in environmental changes.

Do You Know:

The world’s first botanical garden was the University Gardens of Italy established in the 16th and 17th centuries. These gardens were used only for the academic study of medicinal plants.

Importance:

  1. Climate change, habitat destruction, invasive alien species, and over-exploitation are some direct threats to plant survival and the earth’s biodiversity. Botanical gardens have been set up as safe haven for plants, where they would remain safe from the above-mentioned threats. Botanical gardens are also useful for taxonomic research.
  2. Botanical gardens maintain ex-situ conservation of plants, often displayingseveral plant species. This helps to maintain a living store of genetic diversity that can support conservation and research.
  3. Botanical gardens conserve plants that might become extinct in the wild. These plants may also have commercial values.
  4. Botanical gardens store the seeds or germplasm of plants for future use, research and propagation. This is another method of ex-situ conservation of plants. It is known as seed banking.
  5. Botanical gardens make people aware about plant and biodiversity conservation. It encourages the sustainable use of plants for the benefit of all. Names of some famous international and national botanical gardens are given in the following table.

Some famous botanical gardens:

The Living World Some Famous Botanical Gardens

The international body of botanical gardens:

  • Following its establishment in 1987, the IUCN Botanic Gardens Conservation Secretariat (BGCS) began to build its membership of botanic gardens worldwide and develop a program in support of botanic gardens. In 1989, The Botanic Gardens Conservation Strategy was published.
  • In the following year, BGCS became independent from IUCN and subsequently came to be known as Botanic Gardens Conservation International (BGCI).
  • A primary concern of BGCI is to provide a means for botanic gardens in all parts of the globe to share information and news about their activities, programmes that may benefit conservation.

Acharya Jagadish Chandra Bose Indian Botanic Garden:

  • Howrah Acharya Jagadish Chandra Bose Indian Botanic Garden (formerly known as Indian Botanic Garden or Royal Botanic Garden, Calcutta) is situated at Shibpur, Howrah, near Kolkata.
  • It is the largest botanical garden in India as well as in south east Asia and one of the oldest botanic gardens in the world.
  • The garden exhibits a wide variety of rare plants and a total collection of over 12,000 specimens on a piece of over 109 hectares of land on the bank of the Hooghly river.
  • It also houses of largest herbarium of India, Central National Herbarium. It is under the administration of Botanical Survey of India (BSI).

The Living World Famous Ancient Banyan Tree

Taxonomic Key

Taxonomic Key Definition: Taxonomic key is a tool that helps scientists to identify an unknown organism on the basis of morphological similarities and dissimilarities.

Characteristics:

  1. The key is an important taxonomic aid for identification of plants and animals based on similarities and dissimilarities.
  2. It is an artificial device by which each type of taxonomic category can be identified.
  3. Keys are based on contrasting characters which are generally placed in pairs called couplets. Of the contrasting characters one is accepted while the other is rejected. Thus, the couplet represents the choice between the two options. Each statement in a key is called a Lead.
  4. Separate keys are needed for each taxonomic category (or a taxon) like family, genus, or species.
  5. Keys are more useful in identification of unknown organisms.

Importance: 

  1. Unknown organisms can be m identified quickly by using taxonomic key, since it provides a structure for sorting different species by different information given on it.
  2. During identification, it helps j to skip other species automatically that do not resemble  the species on the basis of their characteristic features.

Types of Taxonomic Key: Taxonomic keys are of two types—punched card key and dichotomous key. The latter are of two types—indented key and bracketed key.

The Living World Types Of Taxonomic Key

Punched card key:

  • A punched card key consists of a card on which holes are | punched. It may be of two types— edge-punched key | and body-punched key (polyclave). In edge-punched key, the holes are punched along the perimeter of the card.
  • While in body-punched key, holes are punched in rows on the main body of the card. In edge-punched key, attributes are represented by holes while in body-punched key, attributes are represented by the card.
  • The main advantage of using punched keys is that the attributes to be used for identification may be chosen by the user.

Importance: 

  1. Unknown organisms can be identified quickly by using taxonomic key since it provides a structure for sorting different species by different information given on it.
  2. During identification, it helps to skip other species automatically that do not resemble the species on the basis of their characteristic features.

Dichotomous key:

The term dichotomous means divided into two parts. This key gives two choices in each step. The dichotomous key consists of a pair of contrasting characteristics i.e., couplets and each statement of a couplet is called a lead (characteristic). Each lead should be numbered and easily observable. The dichotomous key is of two types—

Indented or Yoked key

Characteristics:  Indented key provides a series of two or more alternate characters arranged in successive manner. An organism may be identified by careful selection of the characters.

Example:  Some fish are identified using indented key.

1. Bony endoskeleton

  1. Body is covered with scales
  2. Barbell present ———–Labeo calbasu
  3. Barbell absent———— Labeo rohita

2. Body is not covered with scales

  1. Dorsal fin large———– Clarias gariepinus
  2. Dorsal fin small————- Heteropneustes fossilis

1. Cartilaginous endoskeleton

  1. Horizontally flat body, long tail with  spine————– Myliobatoidei
  2. Cylindrical body, heterocercal tail without  spine————- Selachimorpha

Bracketed or Parallel key:

Characteristics:  In this type, both the leads of each couplet always remain together. The characters are not further divided into sub-divisions.  Each lead of the couplets is given a number in the brackets. According to these numbers, different plants and animals can be identified.

Example: Some fish are identified using bracketed key.

  1. Bony endoskeleton————-
  2. Cartilaginous skeleton—————
  3. Body is covered with scales———-
  4. Body is not covered with scales————
  5. Barbell present Labeo calbasu—————-
  6. Barbell absent Labeo rohita————
  7. Dorsal fin large Clarias gariepinus——-
  8. Dorsal fin small Heteropneustes fossilis——-
  9. Long and narrow tail with spine Myliobatoidei———-
  10. Heterocercal tail without spine Selachimorpha———-

Advantages and disadvantages of taxonomic key: 

  1. The punched card key is useful for beginners e.g., college students, who are interested in taxonomy. But the system is costly.
  2. Dichotomous key is suitable for the taxonomists and it costs much less. The indented key gives a visual representation of the group and the users can readily obtain a clear picture about the taxon.

Concept Check:

  1. Define taxonomic key.
  2. Define museum and botanical garden.
  3. Name the tools required for the preparation of herbarium.

The Living World Notes

  • Biomolecules: Molecules involved in different metabolic processes of living organisms.
  • Blastula: An early stage of animal embryo; appears as a hollow sphere of cells.
  • Dry weight: Weight of the body excluding its water content.
  • Ecology: The branch of biology that deals with environment and its interrelationship with all the living and non-living component of it.
  • Gnstrula: This is another stage of development of animal embryo and it comes after the blastula stage.
  • Gene pool: All the genes present in a particular group of organism at a given time.
  • Germ layer; Group of cells present in the embryo which helps in the formation of all organs.
  • Germplarm: Living genetic resources such as seeds, conserved for breeding of an particular organism.
  • Macromolecules: Large biomolecules such as proteins.
  • Morula: This is an early stage of animal embryo development; appears as a sphere of cells.
  • Specimens: Organisms or objects used for scientific studies.
  • Type specimen: The specimen that serves as the reference when a new plant species is discovered and first named.

Points To Remember:

  1. The living and non-living things can be distinguished on the basis of a main feature called life. Living things have life while non-living things do not.
  2. Important characteristics of living organisms include shape and size, cellular and structural organisation, growth and development, metabolism, sensitivity, reproduction, homeostasis, coordination etc.
  3. The diversity observed in different organisms worldwide is known as biodiversity.
  4. All the cells contain their genetic material within the nucleus, mainly in the form of DNA, rarely RNA.
  5. The ability of maintaining stable internal condition, irrespective of the external environmental condition, is called homeostasis.
  6. The term ‘Taxonomy’ is composed of two Greek words ‘Taxis’ and ‘Nomos’. ‘Taxis’ means arrangement and ‘Nomos’ means law.
  7. Systematics is the branch of biology which reveals relationship among all the organisms by identification, naming, description and classification.
  8. The systematic framework of classification where taxonomic categories are arranged in specific order is known as taxonomic hierarchy.
  9. There are total seven taxonomic categories present in Linnaean hierarchy or taxonomic hierarchy and the smallest category is the species.
  10. The nomenclature that includes two categories, genus and species, is known as binomial nomenclature.
  11. Carolus Linnaeus is known as the Father of Binomial nomenclature.
  12. If one organism has more than one scientific name, then the first effectively and validly published name should be accepted. The other names are known as synonyms. This is known as the Law of Priority.
  13. Each unit of classification is known as taxon.

 

Biological Classification Important Notes

Biological Classification Introduction

Biological Classification

  • Every day we come across different kinds of organisms. We see different kinds of plants in the fields, gardens, etc. But why are the plants all so different from each other? Or why are the animals we see around us so different from each other? This is because different organisms have different characteristic features.Φ
  • Depending on these characteristics, they have been classified under different classes or groups. This is known as biological classification. In this chapter, we shall learn about the different types of classification of the organisms, that have been followed since ancient times to modern days.

Biological classification Definition: Biological classification is the scientific classification of living organisms based on their morphology, habitat, evolution similarities and differences with other organisms.

Read and Learn More: WBCHSE Notes for Class 11 Biology

  • Classification of organisms has been carried out since the dawn of human civilisation. Aristotle (384-322 BC) had classified organisms into two large groups—plant kingdom and animal kingdom.
  • But this system failed to classify several organisms, such as unicellular green algae, fungi, etc. These organisms could not be placed in any of the above-mentioned two groups.
  • It even failed to distinguish between prokaryotes and eukaryotes. Later on, several other modified classification systems have been developed.

Importance of biological classification:

Biological classification is important in following ways—

  • Identification of organisms: Using a proper classification system, organisms can be identified easily.
  • Attaining knowledge about the living world: A classification system helps to categorise organisms on the basis of their morphological features, habitat, etc.
  • Correlating different groups of organisms: Classification helps us to correlate different groups of organisms, through their similarities and dissimilarities.
  • Discovery of new species: The system of classification makes it easier to classify newly discovered species. Since the dawn of civilisation, there have been many attempts to classify living organisms. But it was done without using scientific criteria.

Biological Classification Important Notes

Cyanobacteria or Blue-Green algae (BGA)

Cyanobacteria Definition: The gram negative, unicellular prokaryotes, containing photosynthetic pigments that can carry out photosynthesis, are called cyanobacteria or cyanophyta.

  • Cyanobacteria contain photosynthetic pigments like chlorophyll and various accessory pigments. They are also called oxygenic photoautotrophs.
  • Cyanobacteria are considered to be first organisms on earth that released oxygen. With evolution, several aerobic organisms developed from the cyanobacteria.
  • They were known as | blue-green algae, myxophyceae or cyanophyceae. In 1978, ICNB (International Code of Nomenclature of | Bacteria) named them as cyanobacteria.

Distribution: Cyanobacteria are usually free-living marshy regions. They are also found in sea, under the ice, desert region, lakes, etc.

General Features

General features Structure:

  1. Shape and size: Cyanobacteria are unicellular,  They may exist as unit (Spirulina sp.), colony (Nostoc sp.) or filamentous (Oscillatoria sp.). The cells may be large, spherical or oval shaped, with size ranging from l//m to 50 pm.
  2. Protoplasm: The protoplasm is generally divided into two parts. The pigment-containing part, called chromoplasm or chromatoplasm, is present towards the periphery. While the inner region contains a colourless part called centroplasm. Thylakoids are present in the chromoplasm.
  3. Cell wall: They possess a cell wall with an inner thick peptidoglycan layer, similar to gram positive eubacteria. The outer membrane contains proteins, lipids and carotenoids. There is a mucilage layer outside the cells.
  4. Cell membrane: The lipoprotein-containing cell membrane is present just below the cell wall.
  5. Lamellasome: Similar to mesosome in bacteria, cyanobacteria have a circular extracellular appendage, that helps in respiration. These are  known as lamellasomes.
  6. Thylakoids: All cyanobacteria contain photosynthetic pigments like chlorophyll a. They also contain blue phycobilin pigments, phycocyanin and allophycocyanin. Many members also contain another pigment, phycoerythrin, making the cells appear red, or sometimes black.
  7. These phycobilins are present in some special structures, on the thylakoid membranes. These structures are known as phycobilisomes. These are highly efficient channels that transfer captured solar energy (excitation energy) to the reaction centres of photosystems.
  8. Nucleoid: Cyanobacteria lack true nuclei and other organelles like mitochondria and chloroplastids. The naked, circular DNA (DNA without histone proteins) is arranged closely within the nucleoid. Plasmids may also be present.
  9. Ribosomes: 70S ribosomes may exist and form polyribosomes.
  10. Gas vacuoles: Cyanobacterial cells contain numerous gas vacuoles.Gas vacuoles provide buoyancy to the cells, allowing the cyanobacteria to float on the surface. This allows more exposure to sunlight for photosynthesis.

Biological Classification Cyanobacterial Cell

Stored materials: Cyanobacteria also contain various types of reserve food. They include—

  1. α-granules  which are cyanophycean starch and look similar to glycogen (polyglucose) granules, which store carbon
  2. β- granules(lipid granules),
  3. Protein granules, polyphosphate or volutin granules, etc., which allow cells to accumulate energy and nutrients.

Locomotion: Cyanobacteria move by gliding along the surfaces of the substratum.

Reproduction: It reproduces by vegetative and asexual modes of reproduction. Sexual reproduction is totally absent.

  1. Vegetative reproduction: It occurs by the following : methods—
    1. Fission: Unicellular cyanobacterial cells usually divide and reproduce by fission. The cell divides into two daughter cells,
    2. Fragmentation: The filamentous members reproduce vegetatively through fragmentation. Each fragmented piece of  the filament, germinates into a new colony.
    3. Hormogonia: At the end of the growing season, filaments or trichomes break into multicellular pieces. These pieces secrete thick wall around the mass of cells, forming structures called hormogonia: These hormogonia can germinate into new filaments at the onset of favourable conditions.
  2. Asexual reproduction: Many non-motile cyanobacteria reproduce asexually by spore formation. The different types of spores formed are—
    1. Endospore: These are formed within the cell, Here the cells increase in size and their protoplasm divides to form endospores, e.g., Dermocarpa.
    2. Exospore: These are produced outside the cell.
    3. By the dissolution of the apical region of the cell surface, the protoplast gets exposed. From this exposed region of the protoplasm, round spores are formed in basipetal succession. E.g., Chaemosiphon.
    4. Akinetes: These are formed close to the heterocysts. The akinete mother cells increase in size and then develop a wall around it. Under favourable conditions, they give rise to new filaments. E.g., Cylindrospermum.
    5. Heterocyst: Some filamentous cyanobacteria form special types of thick walled cells, known as the heterocysts. The outer layer of the cell wall is made up of pectin or cellulose and the inner layer is of cellulose. They are usually yellow in colour due to the presence of carotene. They have rudimentary reproductive structures, as they can germinate into new filaments, under favourable conditions. E.g., Anobaena sp., Nostoc sp., Spirulina sp., Scytonema sp.

Biological Classification Different Types Of Cyanobacteria Nostoc Anabaena Heterocyst of Anabaena

Biological Classification Differences Between Bacteria And Cyanobacteria

Mycoplasma

Mycoplasma Definition: Mycoplasmas are simple, small, gram-negative, prokaryotic organisms that lack a cell wall and can survive without oxygen.

Nocard and Roux discovered mycoplasma in 1898. Mycoplasma is also known as PPLO (Pleuro Pneumonia Like Organism) or MLO (Mycoplasma Like Organism). They are the smallest living cells known. Many mycoplasmas are pathogenic to both animals and plants.

Example: Mycoplasma gallisepticum, Mycoplasma genitalium, Mycoplasma pneumoniae, etc.

Biological Classification Mycoplasma Cell

General Features

Cell type and structure: They are tiny, prokaryotic microorganisms (0.1-0.15/ym). The shape of mycoplasma varies from being spherical to filamentous with no cell walls. Due to absence of cell wall, they can change their structure.

This phenomenon is known as polymorphism. They contain acetylglucosamine in the cell membrane. Mesosome is absent. Ribosomes are present. Their genome consists of a double-stranded linear but coiled DNA molecule.

  • Nutrition: They can be parasitic or saprophytic.
  • Respiration: They are obligate anaerobes, usually without any electron transport system (ETS). Even if present, ETS is inactive. ‘
  • Reproduction: Mycoplasma reproduce by binary fission. Sometimes, only the nucleoid shows binary fission, without cytokinesis.
  • Effect on animal world: Several species are pathogenic in humans. These include M. pneumoniae, which causes pneumonia and other respiratory disorders. Another is M. genitalium, which causes pelvic inflammatory diseases.
  • Distribution: Actinomycetes are primarily soil dwellers. But they are also widely distributed in a diverse range of aquatic ecosystem, including sediments | obtained from seabed.

General Features

General Features Structure: Actinomycetes are gram positive, prokaryotic microorganisms that have branched hyphae. The hyphae are usually aseptate but septa may be | formed under special conditions in some forms. The hyphae form mycelium like structure. The cell wall is made up of peptidoglycan, teichoic acid, teichuronic acid and polysaccharides. The chemical composition of their cell wall is similar to that of gram-positive bacteria.

The rigid cell wall maintains the cell shape. There are tiny chromatin granules present within the cell.

  • Nutrition: They are heterotrophic in nature. Some are saprophytic aerobes growing in soils and natural habitats.
  • Respiration: They are facultatively anaerobic.
  • Temperature sensitivity: They are generally mesophilic in nature. They are active at a temperature of 35°C.
  • Effects on animal world: Some of the actinomycetes species are used to prepare antibiotics. Some actinomycetes also cause disease (Actinomycosis) in humans.

Biological Classification Different actinomycetes Slreptomyces Actinomyces

Spirochaetes

Spirochaetes Definition: The heterotrophic prokaryotes, that are j present in water and muddy regions are called j spirochaetes.

General Features

General Features Structure: Spirochaetes have long, helically coiled cells. The cells usually have diameter around 0.1 – 0.6pm and have cell wall. They are gram-negative bacteria. They are motile organisms. Spirochaetes are chemoheterotrophic in nature. They are sensitive to desiccation.

Effects on animal world: The majority of spirochaetes cause disease in man.

Examples: Leptospira sp., Borrelia sp., Treponema pallidum, etc.

Biological Classification Different Types Of Spirochaetes Leptosira Borelia

Rickettsiae

Rickettsiae Definition: Rickettsiae intracellular, gram negative coccobacillary forms that ‘ multiply within eukaryotic cells.

Rickettsiae General Features

Rickettsiae Cellular structure: Cells are 0.3 – 0.5 x 0.8 – 2.0pm in size. The cell wall is chemically similar to that of gram negative bacteria. It contains diaminopimelic acid and lacks teichoic acid.

  • Their outer membrane is composed mostly of lipopolysaccharides.
  • They lack a distinct nucleus and membrane bound organelles. Flagella, pili, mucilage capsule are absent.
  • They contain both RNA and DNA as nucleic acids. Mesosomes and ribosomes are also present in the cells.
  • Nutrition: They are parasitic in nature (obtain nutrition from host cell).
  • Growth and Reproduction: Rickettsia normally multiplies by transverse binary fission.
  • Effects on animal world: In their arthropod vectors (insects that can carry the organism), the rickettsia multiply within the epithelium of the intestinal tract. They are transmitted to man, via the arthropod saliva, through a bite. In their mammalian host (i.e., man), they are found mainly in the endothelium of the small blood vessels, particularly in those of the brain, skin and heart.
  • Rocky Mountain spotted fever is transmitted by the bite of a Dermacentor tick, which carries the rickettsia. This disease is common in warm-blooded animals, including humans.

Rickettsiae Example: Rickettsia typhi.

Kingdom Protista

  • This is the kingdom comprising unicellular, eukaryotic organisms that have features similar to those of fungi, plants or animals. The concept of kingdom Protista was given by taxonomist E. Haeckel in 1866.
  • All single-celled eukaryotes, other than green algae and red algae, have been placed under Protista. This kingdom forms a link between Kingdom Monera and the Kingdoms Fungi, Plantae and Animalia.
  • Distribution: Most of the members of this kingdom live in water or aquatic habitats. Majority of them are found as planktons within marine water, fresh water, etc.

Kingdom Protista General Features

Kingdom Protista General Features Cell type: The members of this kingdom are tiny, unicellular eukaryotes. Some of them may exist as colony or filaments.

  • Cell covering: Cell wall is usually absent; if present it may contain silica (e.g., diatoms). Cell membrane is covered by a pellicle, cuticle or shell. This forms a double-envelope system around the cells.
  • Protoplasm: The protoplasm contains a well defined nucleus and other membrane bound organelles. The cells contain 80S (60S + 40S) ribosome in the cytoplasm. On the other hand, 70S ribosome is present in mitochondria and plastids.
  • Chloroplast: The members may be photosynthetic or non-photosynthetic. Thylakoids, – containing photosynthetic pigments, are present within the j chloroplast of the photosynthetic members.
  • Locomotion: Most of them can show locomotion. The cells may contain cilia, flagella or pseudopodia. The cytoplasm, along with organelles, show a flowing j movement. This movement is called cytoplasmic j streaming or cyclosis.
  • Nutrition: Some are photoautotrophs, some heterotrophs, and some are mixotrophs (combining photosynthesis and heterotrophic nutrition). Some are holophytic, holozoic, parasitic, saphrophytic or symbiotic in nature. Some autotrophs are symbiotic in nature.
  • Respiration: Most protists are aerobic, but parasitic and ones living deep in the ocean, are anaerobic.
  • Reproduction: Protists reproduce asexually as well j as sexually.Asexual reproduction occurs by binary fission, j multiple fission, sporulation, cyst formation, etc. Sexual reproduction, on the other hand, occurs by syngamy and j conjugation. j
  • Cell division: Both mitosis and meiosis occur.
  • Life cycle patterns: The three basic life cycle patterns found in eukaryotes are represented by pr.otists. They are—Zygotic or haplontic, gametic or diplontic, sporic or haplodiplobiontic. At some point in the life cycle of many protists cysts (resistant cells) are formed.

Classification of Kingdom Protista

The members of Protista fall under the following three categories as shown in the chart below. These have been discussed under separate heads.

Biological Classification Kingdom Protista

Photosynthetic protista

These are unicellular, eukaryotic organisms, containing photosynthetic pigments. They are mainly found in aquatic ecosystems. They are mainly of three types—

Chrysophyta, Dinoflagellata and Euglenophyta.

Chrysophyta: This group includes diatoms and golden algae. They are fresh water algae as well as marine dwellers. They are pollution indicators, i.e., they cannot grow in polluted regions, hence indicate the level of pollution.

Chrysophyta General Features

  1. Cellular structure: They are unicellular, microscopic and planktonic i.e., they float passively in water currents.The cells are enclosed within a shell made up of silica, called frustule.ln diatoms, the frustule divides to form two thin halves, which fit together as the lids in a soap-case.The cell wall is made up of silica along with cellulose, that makes it hard and porous. Chromatophores are present in the cells.
    • These contain pigments like chlorophyll, /2-carotene and a special type of xanthophyll (diatomin). Fucoxanthin, which is a characteristic pigment of brown algae, is present. Due to the presence of /3-carotene and diatomin, diatoms appear golden.
  2. Locomotion: Diatoms do not have flagella. Mucilage and oil globules help them to float on water.
  3. Symmetry: Both bilateral and radial symmetry is observed among the members.
  4. Nutrition: Most diatoms are photosynthetic (autotrophic) but some have evolved to become  heterotrophic.
  5. Reproduction: Diatoms reproduce by vegetative, asexual and sexual means. Vegetative reproduction takes place by cell division.Asexual reproduction occurs by binary fission. Sexual reproduction takes place by the formation of auxospores.

Chrysophyta Examples: Melosira sp., Cyclotella sp., Striatella sp., Hydrosira sp., etc.

Biological Classification Photosynthetic Algal Protista Melosira Cyclotella

Dinoflagellata or Pyrrophyta: These are biflagellate, unicellular, marine algae.

Dinoflagellata General Features

Dinoflagellata Cellular structure: These organisms are mostly ; marine and photosynthetic. They appear yellow, green, brown, blue or red depending on the pigments present. Dinoflagellates may be covered by an outer covering outside their cells.

  • Depending on their; presence, they are of two types—unarmoured (without outer covering) and armoured (with outer covering).
  • Unarmoured dinoflagellates have a smooth pellicle or periplast. The cell wall has hard, cellulosic plates on the outer surface in case of armoured. These are known as the theca. The nucleus is large.
  • It has distinct nucleolus and nuclear envelope. Chromatin is present in chromosome form. Histone proteins are absent. Such a nucleus is known as mesokaryon.

Flagella: Most of them have two flagella. One lies longitudinally and the other transversely, in a furrow between the wall plates.

  • The flagella that is present in the longitudinal groove functions as a radar. The other flagella fits transversely in the furrow. It is used to create a rotational motion within the water.
  • Nutrition: Most of these organisms are autotrophic. They have chlorophyll a, chlorophyll c, fucoxanthin, etc., as photosynthetic pigments.
  • Stored food: Food is stored in the form of starch (polysaccharides) and oils (lipids).
  • Reproduction: Dinoflagellates show two types of reproduction— asexual and sexual. During normal condition, they reproduce asexually, by cell division.
  • But under certain stressful conditions, they undergo a different process of reproduction. Two dinoflagellates undergo fusion and form a diploid zygote.
  • This diploid zygote state remains throughout the period of stress. Once conditions become favourable, the diploid zygote breaks off into small pieces called cysts, that grow into new organisms.

Examples: Gonyaulax sp., Gymnodium sp., etc.

Biological Classification Gonyaulax

Biological Classification Noctiluca

Biological Classification Ceratium

Euglenophyta: They are unicellular, photosynthetic, fresh water protists that have features similar to those of plants and animals.

Euglenophyta General Features

  1. General Features Distribution: Majority of them are fresh water organisms, living in stagnant water.
  2. Structure: The cells of most euglenoids are spindle-shaped. There is no cell wall or other rigid structure covering the plasma membrane.
  3. Cell covering: Instead of a cell wall, they have a protein rich layer, called pellicle, which makes their body flexible.
  4. Flagella: They have two flagella, a short and a long one.The flagella are tinsel shaped and have hairs on them.Both the flagella arise from their respective basal granules.Blepharoplasts or basal bodies are present at the basal end of the flagella.
  5. Eye-spot: Eye-spot is present at the end of the flagella. It contains a red-coloured astaxanthin pigment, that acts as photoreceptor (sensitive to light). The eyespot appears to be connected to the flagellum by special strands of cytoplasm that transmit signals from one organelle to the other.
  6. Nutrition: Euglenoids are mostly autotrophic in nature. Photosynthetic euglenoids contain chloroplast just like green algae. Chlorophyll a, chlorophyll b and xanthophyll pigments are present in the chloroplast. Non-photosynthetic euglenoids are saprophytic in nature. Some of the non-photosynthetic euglenoids are holotrophic while some are mixotrophic in nature.
  7. Stored food: Euglenoids have a carbohydrate food reserve, called paramylon. It is generally present in the form of small pyrenoid granules.
  8. Locomotion: They generally swim in water, using their flagella, but may also move over wet land by creeping.
  9. Reproduction: Reproduction in euglenoids takes place by asexual means, i.e., by mitotic cell division. Example: Euglena viridis, etc

Biological Classification Euglena

Euglenoids

The euglenoids were first defined by Otto Biitschli in 1884 as the flagellate order, Euglenida. They possess flexible cell coverings, that enable them to move about freely. They also ingest their food through a structure called gullet. So earlier they were considered as animals. However, they had certain characteristics similar to plants. Botanists subsequently created the algal division Euglenophyta. Presently they are classified as both animals and plants.

Slime moulds

Slime moulds Definition: Slime moulds are saprophytic protists that do not have a specific structure (amoeboid) and can grow in humid environment.

Slime moulds Distribution: Slime moulds were formerly classified as fungi. They are found mainly in wet, marshy regions. Some of the species of slime moulds are aquatic in nature.

Slime moulds General Features

General Features Structure: During favourable conditions, they live as independent, amoeba-like cells. These structures have single nucleus and are called myxamoeba. They feed on fungi and bacteria. However, during scarcity of food and unfavourable temperature, they form an aggregated structure. This structure is called plasmodium.These structures may grow and spread over a large distance. The outer part of the body is covered by mucilage, hence the name slime mould.

  • Locomotion: They move very slowly using pesudopodia. Such a motion is called amoeboid movement.
  • Nutrition: They are saprophytic and phagotrophic in nature. Some of them are parasites.
  • Reproduction: Both asexual and sexual types of reproduction are found. Asexual reproduction takes place by binary fission, plasmotomy (type of asexual reproduction in which a multinucleate protozoan cell divides into two or more multinucleate daughter cells, without mitosis), cysts or sclerotium formation, etc.
  • Life cycle: Zygotic and gametic meiosis take place within the life cycle.

Slime moulds General Features Types: There are two main groups of slime moulds in the Kingdom Protista. They are—Plasmodial slime moulds and Cellular slime moulds.

Plasmodial slime moulds or Myxomycota or true not have a specific structure. They are large single-celled mass, without the cell wall but contain thousands of nuclei (structure called plasmodium). A slimy secretion is released by the slime mould.

Examples: Physarum sp., Fuligo sp.

Cellular slime moulds or Acrasiomycota: They spend most of their lives as haploid, single nucleus containing, amoeboid protists. They show amoeboid motion, hence called myxamoeba. Upon the release of a chemical signal, the myxamoeba aggregate into a clustered structure, known as a pseudoplasmodia. This pseudoplasmodia eventually form multicellular slugs. They contain a layer of cellulose, outside the spores.

Example: Physarum sp., Stemonitis sp., Fuligo sp., Dictyostelium sp.

Biological Classification Stemonitis Sp

Protozoa Protista

Protozoa are unicellular, heterotrophic, eukaryotic organisms. They live as predators or parasites. They are considered to be the primitive relatives of animals.

Protozoa Protista General Features

  • General Features Symmetry: The body may be asymmetrical, or radially, bilaterally or spherically symmetrical.
  • Shape: They are of different shapes—spherical or oval shaped. Some of them do not have a definite shape or structure, such as Amoeba sp.
  • Body structure: They are eukaryotic, unicellular organisms. The cell may contain one or more nucleus, They have very thin cell membrane. Sometimes the cell  membrane may be covered by pellicle (Paramoecium  sp.) or calcareous or siliciferous shells (Elphidium sp.).
  • Locomotion: Generally, members of protozoa such as Plasmodium, show contractile motion by pseudopodia, flagella or cilia. On the other hand, members like Monocystis, do not have organs for locomotion. Such organisms move about by cellular contractions, using contractile fibrils called myoneme threads.
  • Osmoregulation: Protozoa living in fresh water are subjected to a hypotonic environment. Protozoa have contractile vacuoles, water vacuoles, etc., that collect and expel excess water. Some waste products and carbon dioxide are also excreted by this vacuole, along with the water. This is how protozoa perform osmoregulation.
  • Nutrition: The mode of nutrition in most of the free-living protozoa is holozoic. Other modes of nutrition found in many protozoa are saprophytic, symbiotic or parasitic.
  • Respiration: Aerobic respiration is seen in free-living organisms, while anaerobic respiration is seen in ectoparasites (parasites living on the body surface of the host).
  • Excretion: Waste products of protozoans are nitrogenous wastes like ammonia, C02, excess water, etc. These are expelled from the body through the contractile vacuole.
  • Reproduction: Both asexual and sexual reproduction are observed in these organisms. Asexual reproduction takes place by binary fission. Sometimes, multiple fission may also occur. Sexual reproduction may occur by syngamy and conjugation. Under unfavourable conditions, they form cysts, which become active during favourable conditions.
  • Reserve food: Reserve food is generally glycogen (example,. Entamoeba sp.).

Protozoa Protista General Features Types: There are four major groups of protozoans according to the organ of locomotion. They are—

Zooflagellata, Ciliata, Sarcodina, Sporozoans

Zooflagellata (Flagellated protozoans):

Zooflagellata General Features

  1. The members of this group are either free-living or parasitic.
  2. They have one or more flagella. Some of them (parasitic forms) can cause diseases, e.g., Trypanosoma sp. causes sleeping sickness.
  3. The flagellated protozoans range from a simple oval cell with one or more flagella to collared flagellates (organisms with funnel-shaped collar of interconnected microvilli, at the base of the flagella).
  4. Most of the members have lost the ability to photosynthesise. They generally show parasitic nutrition, some may show saprophytic nutrition.
  5. Only asexual reproduction, by longitudinal binary fission, is observed. They may exist either as single cells (e.g., Paraphysomonas sp.) or as colonies (e.g., Codosiga sp.).

Biological Classification Trypanosoma

Biological Classification Leishmania

Biological Classification Amoeba

Sarcocfina (Amoeboid protozoans)

Sarcocfina General features

  1. These organisms live in fresh water, sea water or moist soil.
  2. The cells may contain more than one nucleus.
  3. The organisms move and capture their prey by stretching out their pseudopodia (false feet) as in Amoeba sp.
  4. The mode of nutrition is heterotrophic and holozoic. Some of them such as Entamoeba sp. are parasites.
  5. Marine forms have silica shells.
  6. Reproduction occurs both by asexual and sexual means. Asexual reproduction occurs by binary or multiple fission. Sexual reproduction occurs by syngamy.

Sarcocfina Example: Amoeba, etc.

Biological Classification Amoeba Proteus

Comparison among the four classes of protozoa:

Biological Classification Comparison Among The Four Classes Of Protozoa.

General Features

  1. These are aquatic, actively motile organism.
  2. Most of them are free-living,however, some may form colonies. Some may also be endoparasitic  in nature.
  3. The cilia are the main organs of locomotion.
  4. Most of these are heterotrophic and holozoic. They have modified structures for absorbing their food, such as cytostome, cytopyge or cytoproct. Digestion takes place within the food vacuoles.
  5. They have a cavity (gullet) that opens to the outside of the cell surface. The coordinated movement of rows of cilia makes any food in the water to enter the gullet.
  6. There are two nuclei in the cell, the larger one called macronucleus and smaller one called micronucleus.
  7. Both asexual and sexual reproduction are seen among the members. Asexual reproduction takes place by transverse binary fission. Sexual reproduction takes place by conjugation.

Example: Paramoecium sp.

Sporozoans:

Sporozoans General features:

  1. This group includes organisms that have an infectious spore-like stage in their life cycle.
  2. Generally, they are unicellular in nature. © They remain as endoparasites (parasites living within the body of the host) within vertebrates and invertebrates.
  3. During most of their life cycle, they are unable to move by themselves, as they lack organs for locomotion.
  4. The body is covered by a permeable cuticle or pellicle layer.
  5. The mode of nutrition is heterotrophic and saprozoic.
  6. Both asexual and sexual types of reproduction are observed in the organisms. Asexual reproduction occurs by multiple fission or schizogony. Sexual reproductiontakes place by syngamy.
  7. Life cycle is haplontic.

Sporozoans Examples: Pneumocystis carinii, Plasmodium vivax, etc.

Differences between kingdom Monera and kingdom Protista:

Biological Classification Differences Between Kingdom Monera And kingdom Protista

Kingdom Fungi

This kingdom of the biological world comprises of eukaryotic, multicellular, non-chlorophyllous, spore-forming organisms are called fungi.

Kingdom Fungi Distribution: Fungi are found in air, water, soil and, on animals and plants. They prefer to grow in warm j and humid places. Most of the fungi grow in the soil, Some of the fungi may also be aquatic. Some fungi remain as parasites, within host organisms, both intracellularly and intercellularly.

Kingdom Fungi General features

Kingdom Fungi Body Structure: Most of the fungi are simple, unicellular or multicellular and filamentous in nature, filamentous fungi consist of long, slender thread-like structures called hyphae. Due to extensive growth, the hyphae form a cottony network, known as mycelium. Some hyphae are long, slender, thread-like in nature, called hyaline hyphae. Some hyphae are filamentous in

Biological Classification Structure Of Hyphae In Mucor

Nature, filled with multinucleated cytoplasm. These are called coenocytic hyphae. Some fungi have septa or cross walls in their hyphae and are called septate. Coenocytic hyphae do not have septa (aseptate), but septa develop within them during reproduction. There may or may not be pores across these septae, that allow transport of nutrients to other parts of the mycelium.

  • Cellular structure: The cells are eukaryotic. The cell wall is composed of chitin and polysaccharides. Except chloroplast, all other organelles are present in the cell.
  • Nutrition: Most fungi are heterotrophic. They absorb soluble organic matters from dead and decayed plant or animals. Hence, they are called saprophytes. They digest the extracellular substances and derive nutrition by absorbing them. Thus, fungi show an absorptive mode of nutrition. Some fungi can grow on other living organisms and draw nutrition from them (hosts). They are called parasites. The fungi can also live as symbionts—in association with algae as lichens and with roots of higher plants as mycorrhiza.
  • Reserve food: The reserve food materials are- glycogen, starch, sucrose, maltose oil, etc.
  • Reproduction: Reproduction in fungi can be performed vegetatively through fragmentation, fission and budding. Asexual reproduction is performed by the formation of spores called conidia, sporangiospores, oidia, sclerotia, chlamydospores, aplanospores or zoospores.
  • Sexual reproduction occurs, through isogamy, anisogamy and oogamy. The gametes unite to form different kinds of spores like oospores, ascospores and basidiospores. The various spores are produced in distinct structures called fruiting bodies or fructifications. Fertilisation involves the following three steps—
  1. Protoplasmic fusion of two motile or non-motile gametes, called plasmogamy.
  2. Subsequent fusion of two nuclei within the hyphae called, karyogamy.
  3. Meiosis within the zygote, resulting in the formation of haploid spores.

Sexual reproduction in fungi

  • During sexual reproduction, two haploid hyphae of compatible mating types come together and fuse. In certain fungi, the two nuclei of opposite nature fuse immediately resulting in the formation of diploid zygotes (2n).
  • However, in others (Ascomycetes and Basidiomycetes), an intervening dikaryotic stage (n + n, i.e., two nuclei per cell) occurs; such a condition is called a dikaryon and the phase is called dikaryophase of fungus. After dikaryophase ends, the parental nuclei fuse and the cells become diploid.

Classification Of Kingdom Fungi

Different scientists have classified fungi differently. Among them, the most widely accepted classification was propunded by Gwynne-Vaughan and Barnes (1927, 1937). This classification has been discussed below.

Biological Classification kingdom Fungi

Phycomycetes

This is the class of fungi having aseptate mycelium.

Phycomycetes Distribution: Members of phycomycetes are found mostly in aquatic habitats. They may also grow as saprophytes on dead and decaying organic matters in moist and damp places. They also grow as obligate parasites on plants.

Phycomycetes Body structure:

  1. The mycelium is aseptate and coenocytic.
  2. The cell wall is made up of chitin and other polysaccharides (such as glucan, cellulose).

Phycomycetes Reproduction:

  1. Asexual reproduction takes place by zoospores (motile) or by aplanospores (non-motile). Zoospores are produced endogenously within the zoosporangium.
  2. Sexual reproduction takes place by isogamy and oogamy. The gametes are non-motile. Oospores are produced by the union of two dissimilar gametes. Zoospores are produced by the union of two similar gametangia through zygospore formation.

Phycomycetes Examples: Mucor sp., Rhizopus sp. (the bread mould) and Albugo sp. (the parasitic fungi on mustard).

Biological Classification Columella In Mucor

Ascomycetes

This is one of the three classes of fungi having septate; mycelium.

Ascomycetes Distribution: Members of ascomycetes are generally terrestrial in nature. They may also exist as saprophytes or parasites (on plants).

Ascomycetes Body structure:

  1. Vegetative body is unicellular (yeasts) or multicellular.
  2. They commonly have a J r well-developed, branched septate mycelium.
  3. The cell wall is generally composed of chitin and glucans.

Ascomycetes Reproduction:

  1. Vegetative reproduction takes place by fragmentation (in filamentous form), fission and ; budding (in unicellular form).
  2. Asexual reproduction takes place by non-motile spores, such as conida, oidia and ; chlamydospores.
  3. The sexually reproducing units are called the ascospores. Ascospores are grown inside a small if specialised sac-like structure, called ascus (plural asci).
  4. The fruiting bodies (inside which asci are developed); are called the ascocarps.
  5. There are 8 ascospores within each ascus.
  6. The ascocarps may be cleistothecium (Penicillium sp.), apothecium (Ascobolus sp.), perithecium (Daldinio sp.) or ascostroma (Elsinoe veneta).

Ascomycetes Examples: Saccharomyces cerevisiae, Penicillium notatum E Aspergillus, Claviceps, Neurospora, etc.

Types of ascocarp :

In cleistothecium, the ascocarp is small or ovoid body that is closed from all sides. In apothecium, the ascocarp is cup-shaped or disc-shaped, while perithecium, it is flask-shaped with definite apical pore. In ascostroma (also known as pseudothecium), ascii are produced in cavities containing stroma.

Biological Classification Different Types Of Fungi Aspergillus

Biological Classification Penicillium

Biological Classification Yeast

Basidiomycetes

This is another class of fungi having septate mycelium.

Basidiomycetes Body structure:

  1. Presence of well-developed, branched and septate mycelium.
  2. The mycelium is of two types—primary and secondary.
  3. The mycelial cells may contain one nucleus, called monokaryotic (or uninucleate, common in primary mycelium) or two nuclei, called dikaryotic (or binucleate, common in secondary mycelium).
  4. The secondary mycelia may organise and form fruit body.© The cell wall is mainly composed of chitin and glucans.

Basidiomycetes Nutrition: They are mostly saprophytic and parasitic.

Basidiomycetes Reproduction:

  1. Primary mycelium may reproduce asexually by oidia, conidia, chlamydospores, etc.
  2. Vegetative reproduction takes place by budding and fragmentation.
  3. Sex organs are absent. During sexual reproduction, the dikaryotic cell is formed by somatogamy, plasmogamy (dikaryotization of the cell by somatogamy of monokaryotic and dikaryotic mycelium).
  4. The dikaryotic phase persists for long period of time. Karyogamy occurs in club-shaped structures, called basidia or basidium.
  5. 4-haploid basidiospores are formed by meiosis. They are oval, cylindrical or spherical in shape. They may be unicellular and uninucleate.
  6. Basidiospores are developed exogenously on the horn-shaped or tubular outgrowths of the basidium. These outgrowths are called sterigmata.

Basidiomycetes Examples: Agaricus campestris (mushroom), Puccinia graminis (rust fungus), Ustilago (Smut) etc.

Biological Classification Agaricus

Biological Classification Microsporum

Different types of spores produced by the fungi have been depicted in the following flowchart.

Biological Classification Spore

Deuteromycetes

This is another class of fungi having septate mycelium.

Deuteromycetes Distribution: They are called the fungi imperfecti because only the asexual or vegetative phases of these j fungi are known. They are mostly terrestrial, with only a few being aquatic. They are generally decomposers and parasitic in nature, found as saprophytes in the soil.

Deuteromycetes Body structure: The vegetative body is mycelial and composed of profusely branched and septate hyphae. The sually multinucleated. In parasitic species, the hyphae grow intra or intercellularly.

Deuteromycetes Nutrition: Most of the fungi are saprophytic in nature. They act as decomposers of litter and help in mineral cycling. Some are parasites.

Deuteromycetes Reproduction:

  1. They reproduce mainly by asexual methods.
  2. The asexual reproduction takes place commonly by conidia, blastospores, chlamydospores and arthrospores.

Deuteromycetes Examples: Alternoria spv Colletotrichum sp. and Trichodermo sp.

Compare among different classes of kingdom fungi:

Biological Classification Comparison Among Different Classes Of Kingdom Fungi

Importance Of Fungi:

Fungi play a very important role in our daily life, Some fungi are useful. These are used as food, natural sources of medicine, etc. Some fungi are j directly or indirectly harmful to mankind. They spoil food and other substances and also cause diseases in; both plants and animals, Some common fungal diseases of plants are early blight of potatoes, late blight of potato, etc.

Poisonous fungi :

Some fungi like Amanita phalloides produce toxins, like a-amanitin which causes lesions of stomach cells, It also produces phalloidin which affects the liver, Fungi, like Aspergillus flavus, A. fumigatus and Penicillium islandicum can infect groundnut and mustard cakes. They produce a poisonous substance, called aflatoxin, that binds with DNA and inhibits transcription in the host plant cell. This results in the inhibition of protein synthesis, ultimately leading to cell death.

kingdom Plante

Plants are multicellular organisms composed of eukaryotic cells. The cells are organised into tissues. The cells have cell wall that contains cellulose. The cells also contain plastids and large vacuoles in the cytoplasm. They obtain nutrition by photosynthesis. They are generally autotrophic and are called producers. Stored food is starch. Growth is not distinct. They continue to grow throughout their lifetime.

Examples: mosses, ferns, conifers, and flowering plants. We shall learn about this kingdom later.

kingdom Animalia

Animals are multicellular organisms composed of eukaryotic cells. The cells are organised into tissues. The cells lack cell walls. They respond to stimuli very fast. Their growth is distinct. They grow only upto a certain period of their lifetime. They do not carry out photosynthesis and obtain nutrients from other organisms. Thus they are heterotrophic and are decomposers in nature.

Examples: sponges, worms, insects, and vertebrates. We j shall learn about this kingdom in details.

Virus Viroids And Lichens

These are also considered to be members of the living world.

Virus

Virus Definition: Viruses are microscopic, acellular organisms, that form an intermediate group between living and non-living organisms. The term ‘virus’ is derived from the same word in Latin, meaning venom or poisonous fluid. It was coined by M.W. Beijerinck in 1898. Viruses and viroids have not been placed in classification system as they do not have a cell structure and are not considered living.

Discovery of virus:

  1. D.J. Ivanowsky (1892) recognised certain microbes as causal organism of tobacco mosaic disease. These organisms were, found to be smaller than bacteria because they could pass through bacteria-proof filters.
  2. M.W. Beijerinck (1898) demonstrated that the extract of the virus infected tobacco plants could cause infection even in healthy plants. He named the fluid as contagium vivum fluidum (meaning ‘contagious living fluid’).
  3. W.M. Stanley (1935) showed that viruses could be crystallised.

However, once they infect a cell, they behave like living organisms. They take over the replication machinery of the host cell to replicate themselves, thereby killing the host cell.

Therefore, the question still remains whether the viruses are living or non-living.

Virus General features: The viruses are members of the j intermediate group of organisms, between the living and the non-living. They are inert outside their specific host cell.

As soon as they enter the host cell, they become living and start reproducing. Viruses are cannot complete their life cycle without the specific host.

Virus Shape and size:

  1. Viruses are very minute (microscopic) organisms, that have an average diameter of 20-30nm.
  2. They may be spherical (e.g., Poliovirus, Influenza virus), cylindrical (e.g., TMV), brick-shaped (e.g., Smallpox virus), or spindle-shaped (Phage virus).

Biological Classification Shape Of Virus

Virus Structure: Virus does not contain protoplasm or cell membrane. Virus consists of two parts—

  1. Nucleic acid (centrally placed) and
  2. Protein coat, sometimes with an additional envelope, made of lipoprotein. The protein coat, called capsid, is made up of small subunits, called capsomeres. These capsomeres are arranged in helical or polyhedral shapes. The capsid protects nucleic acid.
  3. Viruses contain either RNA or DNA as nucleic acid. A virus is a nucleoprotein (containing nucleic acid and protein) particle and the genetic material is infectious. In general, viruses that infect plants have single-stranded RNA.
  4. On the other hand, viruses that infect animals have either single or double-stranded RNA or double-stranded DNA. Bacteriophages (viruses that invade the bacteria) are usually double-stranded DNA viruses. Structural details is discussed under separate heads below.

Virus Living features: Presence of—

  1. Nucleic acid,
  2. Reproduction and
  3. Mutability.

Virus Non-living features:

  1. Lack of protoplast. The cell structure is not complete.
  2. Absence of respiration,
  3. High specific gravity like non-living things.

Virus General features Types: According to the presence of the nucleic acids and the host cell, the virus are classified into the following groups—

Biological Classification Virus

Structure of a virus

Biological Classification Structure Of A Virus

Outer covering

Outer covering Capsid: The protein coat surrounding the genome in a virus is called capsid. The capsid together, along with the enclosed nucleic acid, is called nucleocapsid. The capsid is made up of a number of protein subunits, called capsomeres.

Biological Classification Structure Of Virus

Outer covering Envelope: Many mammalian viruses have a protein, carbohydrate or lipid bilayer surrounding the j nucleocapsid. This loose covering is called envelope, The envelope is a part of the host cell membrane. It: is made up of several units called peplomers. The envelope may possess several outgrowths called spikes.

  • The virus which have lipid envelope outside the capsid are called lipoviruses. The viruses which do not have envelope are called naked virus (Example—TMV).

Internal or inner core

Nucleoid: The nucleic acids enclosed within the capsid region make up the nucleoid. Viruses contain only one type of nucleic acid, either DNA or RNA. DNA containing viruses are called deoxyviruses, whereas viruses having RNA are called riboviruses.

  • They vary in the structure of their nucleic acid. The nucleoid forms viral genome. The nucleoid consists of about 1000-250,000 nucleotide pairs. The nucleoid remains inactive as long as it does not come in contact with the host cell, The amount of nucleic acid of a virus usually depends on its size.

Genetic material of plant virus:

Most of the plant viruses have RNA either single (TMV) or double stranded (Rice ragged stunt viruses), except a few have DNA, either single (Gemini viruses) or double stranded (Dahlia mosaic virus). Animal viruses have mostly double stranded DNA or either single (Polio virus) or double stranded RNA (Reo virus) and bacteriophages contain mostly double stranded DNA, but they also have single stranded RNA or single stranded DNA.

Enzymes: Hemagglutinin esterase, integrase, reverse transcriptase, viral neuraminidase, polymerase, lysozyme, etc., are some of the enzymes found in viruses.

Differences between plant and animal virus:

Biological Classification Differences Between Plant And Animal Virus

Viral Reproduction: Reproduction in virus takes place when its nucleic acid replicates within the host cell. The process of reproduction is slightly different in bacteriophage, from that of the other virus.

Reproduction in virus, other than bacteriophage: Principal events involved in viral replication, other than that in bacteriophage, are given below—

  1. Infection phase: It consists of two stages—
    1. Attachment: The first step in viral infection of a cell is attachment of the virus to the cell surface. Attachment is via temperature-independent ionic interactions. There are specific receptor molecules (generally protein, carbohydrate or lipid in nature), on the cell surface. Virus have a protein called viral attachment protein. This protein recognises the specific receptor on the cell and attaches the virus. Cells without the specific receptors are not susceptible to the virus,
    2. Penetration: After attachment, the virus has to enter or penetrate the host cell. The virus enters the cell in a variety of ways according to the nature of the virus. Plant viruses enter the host cell either via vectors (aphid, mites, etc.) or through damaged cell membrane. Animal viruses enter the host cell through processes like phagocytosis, endocytosis, etc.
  2. Eclipse phase: It occurs in two stages—
    1. Uncoating: The nucleic acid must be uncoated so that viral replication can begin. As soon as the nucleic acid becomes uncoated, the eclipse phase begins. This phase continues until new infectious virions are made.
    2. Synthesis of viral nucleic acid and protein: In this phase, the viral nucleic acid is replicated. Some 0 Assembly/maturation: New virus particles are assembled using the synthesised nucleic acids and coat proteins. Thus, the virus particles attain a maturation phase that follows the initial assembly process.
  3. Lysis (Release): Virus may be released due to cell lysis, or, if enveloped, may form bud on the cell.

 

Biological Classification Viral Replication

Reproduction in bacteriophage: Reproduction in j bacteriophage takes place by two cycles

  1. Lytic cycle
  2. Lysogenic cycle. The former type is seen in virulent | and the latter is found in avirulent or temperate phage. These cycles have been discussed under separate heads.

Lytic Cycle:

The cyclic reproduction mechanism, by which virulent | bacteriophage produces progeny virus by replication is called lytic cycle. It is common in T-even phages (T2, T4, etc.) which attack Escherichia coli. It takes place for about 20-30 minutes.

Lysogenic Cycle:

The cyclic reproduction mechanism by which phage virus introduces its nucleic acid into the host cell, divides to form progeny viruses is called lysogenic cycle. Lwoff (1953) discovered this cycle in Lambda (A) phages attacking £ coli. The phage involved in this cycle is called temperate phage, while the bacterium is of lysogenic strain.

Differences between lytic and lysogenic cycle:

Biological Classification Differences Between Lytic And Lysogenic Cycle

Biological Classification Reproduction Of A Typical Virus Bacteriophage

Characteristics of some important viruses

Characteristics of some of the viruses have been discussed below in details—

Bacteriophage

Bacteriophage Definition: Bacteriophages are viruses that infect bacteria, reproduce within them and later on lyse them to release the progeny phage.

Bacteriophage Example: T2 phage, lambda phage, phage MS2, ΦX 174 phage, T4 phage.

Bacteriophage Structure: T4 is one of the largest bacteriophages; it j is approximately ranging from 25nm-200nm long and 80-100nm wide. They are made up of four important parts—head, neck, tail and end plate or basal plate.

  • Head: All phages contain a structure called head, which varies in size and shape.
  1. Some are icosahedral (with 20 sides) and others are filamentous. The head is covered by a bilayered protein membrane called capsid.
  2. Capsid is made up of several units called capsomeres. It encloses nucleic acid and acts as the protective covering.
  3. The nucleic acid is about 0.05nm long double stranded DNA, containing hydroxymethylated cytosine.

Biological Classification Structure Of Bacteriophage

  • Neck: It is a small part that extends just after the head. It is made up of neck tube and collar.
  • Tail: Some phages have another structure called tail attached to the phage head. The tail has a hollow tube, called core. The nucleic acid is passed to the host cell during infection.
  • T4 tail is surrounded by a Y contractile sheath, called tail sheath, which contracts during infection of the bacterium. At the end of tail, phages like T4 have another structure called base plate. One or more tail fibres remain attached to it.
  • End plate or base plate: The tail has a plate-like structure at its end. This structure is called end plate or base plate. Six spikes are arranged, along with a long tail fibre, on this plate. The base plate and tail fibres are involved in the binding of the phage to the bacterial cell. All phages do not have base plates and tail fibres.

Biological Classification Bacteriophage

Tobacco mosaic virus

Tobacco mosaic virus Definition: The rod-shaped plant virus, that contains coiled RNA and causes mosaic diseases in tobacco plants are called tobacco mosaic virus (TMV).

Shape and size: TMV has a cylindrical hollow tube-like structure, with a length of 300nm and diameter 15-18 nm.

Tobacco mosaic virus Structure:

  1. The virus is composed of single strand of RNA, wrapped inside a capsid.
  2. The virus particle that the virus can attain the cylindrical structure.
  3. The capsid is composed of about 2130 capsomeres (70 j angstrom x 20 angstrom).
  4. The capsomeres are arranged like bricks in a cylindrical chimney-like structure. This gives the capsid a shape of spiral staircase.
  5. Each capsomere contains 158 amino acids.
  6. The single-stranded RNA is arranged spirally (as a spring), having 6400-7300 ; nucleotide units.
  7. The RNA strand encodes four : proteins. These include two proteins that replicate the ; viral RNA, a protein that transports the RNA from cell to cell, spreading the infection, and the capsid protein.

Biological Classification Structure Of Tobacco Mosaic Virus

HIV, AIDS and AIDS wasting syndrome

HIV or Human Immunodeficiency Virus is a type of j retrovirus, that causes the disease AIDS (Acquired j Immunodeficiency Syndrome) in humans. AIDS wasting syndrome is a condition in which a person suffering; from AIDS loses at least 10% of his body weight, especially that of the muscles.

Tobacco mosaic virus Importance of virus

Viruses can be both useful and harmful and hence, have both advantages as well as disadvantages. Some of these are as follows—

Tobacco mosaic virus Importance of virus Advantages:

  • Use as bactericide: Bacteriophages are sometimes j used in ‘polluted water as bactericides. They act as scavengers that eradicate the bacterial population j present in the water. Similarly, they can also kill the disease-causing bacteria.
  • Use in scientific experiments: In space research,  lysogenic phage cultures are used as radiation detector, Avirulent or temperate phages are useful in studying genetic recombination (transduction) and are used ; widely in biotechnological research.
  • Treatment and prevention of diseases: Sometimes bacteriophages are used in therapy and prophylaxis of some bacterial diseases. Viruses are utilised in the production of vaccines used to develop immunity against viral infection.

Importance of virus Disadvantages:

Causal agents of various diseases: Viruses are responsible for causing various diseases of both plants and animals. Some of the viral diseases in plants are tobacco mosaic, yellow vein mosaic of lady’s finger, leaf roll of potato, leaf curl of papaya, etc. The plant viruses cause damage to different parts like root, leaf, fruit, seed, etc. These incur economic losses by hampering the quality and quantity of the plant products. Some viral diseases in animals include small pox, meningitis, pneumonia, mumps, bronchitis, etc. In case of animals, the diseases may even prove fatal.

Oncovirus: Some of the viruses are responsible for causing cancer. These viruses are called oncoviruses.

Viroids

Viroids Definition: The tiny, non-coiled, infectious, ssRNA containing particles that can cause diseases in plants are called viroids.

Viroids General features

Viroids Structure: The viroids do not have capsid (i.e., protein coat) around the RNA molecule.Their genome contains a circular RNA strand extensively base paired within itself. So, they can resist RNase attack.

Viroids Genome:

  1. Viroids have small, low-molecular weight, circular, single stranded RNA molecule. The ssRNA is made up of about 250-390 bases. This number of bases is insufficient to code even for one enzyme, required for replication. Some of the nucleotides even lack AUG(initiation codon).
  2. The molecular weight is about 75,000 to 1,20,000 Da.
  3. There are five domains in the viroid genome—TL (Terminal Left), P domain (Pathogenicity Domain), CCC
  4. Viruses can be both useful and harmful and hence, have both advantages as well as disadvantages. Some of these are as follows— leaf curl of papaya, etc. The plant viruses (Central Conserved Region), TR (Terminal Rigid), V Domain.

Viroids Nature:

  1. The viroids are smaller and simpler than virus particles.
  2. The viroids lack any protein coat and remain as free RNA.
  3. The size of RNA is 25-370 bases in viroids as compared to 3.2-20 kb in viruses.
  4. The ability to mutate is more in case of viroids, than in viruses.
  5. They generally reproduce under high temperature. They reproduce after attaching to the host cell DNA.
  6. Its RNA cannot be degraded by RNase enzyme.

Viroids that cause diseases in plants: Nearly twenty plant diseases are known to have caused by viroids. Some of such viroids are —Potato Spindle Tuber Viroid (PSTVd), Coconut Cadang-Cadang Viroid (CCVd), Avocado Sunblotch Viroid, etc.

Only two diseases such as Scrapie disease of sheep and Kuru disease of human are confirmed to be caused by viroids.

Biological Classification Differences Between Virus And Viroids

Lichens

The term ‘Lichen’ was given by Theophrastus. Its symbiotic characteristic was however, stated by, Schwander in 1867.

Lichens Definition: Lichens are symbiotic associations of fungi and algae.

Lichens Distribution: Lichens can grow over a range of habitats, such as stones, bark of trees, soil, even in water.

Lichens General features

  1. General features Structure: Lichens are composite thalloid, generally leaf-like or cylindrical in shape. The thalloid may be of different colours—grey, green, yellow, pink or brown. There are three layers in the structure of the thalloid— upper and lower layers are made up of fungal hyphae, with the middle cortex layer made up of algae. Algal layer is also known as gonidial layer.
  2. Nutrition: They depend on atmospheric source of nutrition. The lichens that grow on soil or rock, absorb nutrients from their substrate.
  3. Reproduction: Lichen reproduces by all the three means—vegetative, asexual, and sexual.Vegetative reproduction takes place by fragmentation, decaying of older parts, by soredia and isidia. Asexual reproduction takes place by the formation of oidia. Sexual reproduction takes place by the formation of ascospores or basidiospores. Only fungal component is involved in sexual reproduction.
  4. Components: The algal partner (known as photobiont or phycobiont) may be cyanobacteria (Nostoc sp.) or green algae (Trebouxia sp.).
    • Generally the fungal partner (known as mycobiont) occupies most of the thallus and produces its own reproductive structures. It absorbs water and mineral salts and transports them to the algae.
    • In about 98% lichens, fungal component is a member of ascomycetes, while the rest are those from basidiomycetes and deuteromycetes.
    • The algal component, on the other hand, manufactures the food ; through photosynthesis. This food probably diffuses out from the algal body and is absorbed by the fungal component.
  5. Classification of lichens: Lichens are classified on the basis of mutualism and morphological structure.

Types of lichen according to mutualism:

Biological Classification Types Of Lichen According To Mutualism

Types of lichen according to morphology:

Biological Classification Types Of Lichen According To Morphology

Importance of lichens: The lichens are beneficial as well as harmful to mankind. They are used as food and fodder, and medicines. They also have various kinds of industrial uses.

  • As food: Some species of Parmelia sp. are used as curry powder in India, while various others are used as food in other countries. Lichens like Lecanora saxicola and Aspicilia calcarea, etc., are used as food by several other organisms.
  • As fodder: Ramalina fraxinea, R. fastigiato, Evernia prunastri, etc., are used as fodder for animals.
  • As medicine: Lichens are medicinally important due to the presence of lichenin and some bitter alkaloids or astringent substances. They are used in the treatment of diseases like jaundice, diarrhoea, epilepsy, etc.
  • Industrial uses: Lichens are also used in different kinds of industries—
  1. Tanning industry: Some lichens like Lobaria pulmonaria and Cetraria islandica are used in tanning leather.
  2. Brewery and distillation: Lichens like Lobaria \ pulmonaria are used in brewing of beer. In Russia and Sweden, Usnea florida, Cladonia rangiferina and Ramalina fraxinea are used in production of alcohol due to their rich lichenin content.
  3. Preparation of dye: Dyes obtained from some lichens have been used since ancient times for colouring fabrics, etc. Litmus, an acid-base indicator dye (substance that shows colour change in the presence of an acid or base), is extracted from Roccella tinctoria, R. montagnei and also from Lasallia pustulata.
  4. Cosmetics and perfumery: Certain aromatic compounds (organic compounds with characteristic odour) are available in lichen and are extracted.
    These are used in the preparation of cosmetics and perfumes. Essential oils (oils present in cosmetics) extracted from Ramalina sp. and Evernia sp. are used in the manufacture of cosmetic soaps.
  5. Harmful activities:
    1. Lichens like Amphiloma sp. and Cladonia sp. grow on mosses and cause total destruction of moss colonies.
    2. Different crustose lichens cause severe damage to window glasses and marble stones of old buildings,
    3. Lichens like Letharia vulpina (wolf moss) are highly poisonous due to the presence of vulpinic acid.

Ecological importance of lichens

  1. Pioneer of rock vegetation: Due to their ability to grow using minimum nutrients and water, crustose lichens can colonise fast.
  2. Accumulation of radioactive substances: Lichens can absorb different radioactive substances from the environment through bioremediation.
  3. Sensitivity to air pollutants: Lichens are very sensitive to air pollutants like S02, CO, C02, etc. Thus lichens are important bioindicators of these gaseous pollutants.

Notes

  • Amitosis: A type of cell division, which involves simple
  • Bioremediation: Using ‘ plants or microorganisms cleavage of the nucleus, followed by division of the j (naturally growing or artificially introduced) to consume cytoplasm. j or break down pollutants in the environment.
  • Biota: Total collection of organisms of a particular region, at a specific time period.
  • Chemosynthetic: Organisms which obtain energy by oxidising inorganic molecules into organic ones.
  • Equatorial plane: Plane present between the poles of a cell, that remains perpendicular to the spindle apparatus in the dividing cell.
  • Febrile: Feverish
  • Heterotrophic: Organisms which cannot produce their own food.
  • Histone: A type of protein that is wrapped around by DNA molecule to form chromatin fibre.
  • Holozoic: A type of nutrition involving ingestion, digestion, absorption and assimilation of the food. Osmolability: Unstable towards changes in osmotic pressure.
  • Peptidoglycan: A polymer made of sugar and amino ! acids, that forms the basic component of the cell wall.
  • Pseudopodia: A type of transient organ of locomotion j produced by protoplasmic extension, present in some organisms.
  • Ruminant animals: Animals that acquire nutrients from ! plant-based food, by fermenting it in a special stomach before digestion. This fermented ingesta is regurgitated I and chewed by these animals.
  • Syngamy: Fusion of two cells, or their nuclei, during ; reproduction.

Points To Remember:

  1. Carolus Linnaeus (1753) classified living organisms into two kingdoms—Plant kingdom and Animal kingdom.
  2. R. H. Whittaker (1969) classified living organisms into five kingdoms. These kingdoms are Monera, Protista, Fungi, Plantae and Animalia.
  3. Kingdom Monera consists of only prokaryotes. Kingdom Protista, . Fungi, Plantae and Animalia consist of eukaryotes.
  4. Bacteria under the kingdom Monera was first discovered by Leeuwenhoek (1676).
  5. Cell wall of bacteria is composed of peptidoglycan or \ murein.
  6. The coiled structure formed by the invagination of j plasma membrane of Gram positive and Gram \ negative bacteria is called mesosome.
  7. The circular DNA molecule present in bacterial cell as I a substitute of nucleus is called nucleoid or genophore.
  8. Some bacteria (like Chlorobium) possess the j photosynthetic pigment bacteriochlorophyll. These bacteria are capable of photosynthesis.
  9. The protist which has characteristics of both animal j and fungi is called slime mould (Myxomycetes).
  10. Ascomycetes and basidiomycetes of kingdom Fungi i are called sac fungi and club fungi respectively.
  11. Edible mushroom is Agaricus bisporus.
  12. Each basidiocarp (mushrooms) has two parts—stalk like structure is called stipe and umbrella like structure is called pileus.
  13. The causative fungus of Black stem rust disease of wheat is Puccinia graminis tritici.
  14. Late blight of potato is caused by the fungus Phytopthora infestans.
  15. Lichen is a symbiotic organism formed by the association of algae and fungi. Its fungal part is mycobiont and algal part is phycobiont.
  16. Mycoplasma are prokaryotes that lack cell wall.
  17. Silica can be found in the cell wall of diatoms.
  18. The protozoa Plasmodium has four species. Each of them is a carrier of malaria. Such as
    1. Plasmodium vivax— Benign malaria
    2. Plasmodium falciparum—Malignant malaria
    3. Plasmodium malariae—Quartan malaria
    4. Plasmodium ovale—Mild tertian malaria
  19. The term virus was coined by the Dutch microbiologist Martinus Beijerinck. In the early 20th century. Frederick Twort discovered that bacteria could be attacked by viruses.
  20. Viruses are intermediate between living and non-living organisms.
  21. The viruses which invade the body of harmful bacteria and kill them are called bacteriophages.

 

WBBSE Solutions For Class 8 Maths Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts

Dividing A Line Segment Into Three Or Five Equal Parts

Question 1. I with the help of pencil, scale and compass, trisect a straight line 9 cm long and will measure the length of each part by scale.
Solution:

WBBSE Solutions For Class 8 Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts Angle BAC Is Drawn

A 9 cm long straight line AB is drawn. A On point A of any measure ∠BAC is drawn. At AB on point B equal to ∠BAC, ∠ABX is drawn.

From straight line AC taking same radius two equal parts AD and DE are cut.

From straight line BD taking same radius two equal parts BF and FG cut. E, F and D, G are joined. EF and DG both straight lines cut AB straight line respectively points P and Q points.

AB straight line is divided into three equal parts on points P and Q.

i.e., AP = PQ = QB = \(\frac{1}{3}\) AB = \(\frac{1}{3}\) × 9 cm = 3 cm

Length of each part is 3 cm.

Read and Learn More WBBSE Solutions For Class 8 Maths

Dividing A Line Segment Exercise 23.1

Question 1. Rihana draw a straight line PQ of length 10 cm. I divided the segment PQ into five equal parts by scale and compass. Let’s verify by scale that each part of the segment is 2 cm or not.
Solution:

WBBSE Solutions For Class 8 Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts Line Segment PQ Into Five Equal Parts By Scale And Compass

Given

Rihana has drawn in her copy a 10 cm long straight line PQ. I divided the segment PQ into five equal parts by scale and compass.

At PQ on point P an acute angle ∠RPQ is drawn and on PQ straight line on the opposite side of ∠RPQ equal to that angle ∠PQS is drawn.

Now from straight line PR taking the same radius four equal parts PA, AB, BC and CD are cut.

Similarly the straight line QS, taking same radius, is cut into four equal parts QE, EF, FG and GH. A,H; B,G; C,H and D,E are joined which cut PQ straight line at points I, J, K and L respectively.

PQ straight line is divided by I, J, K and L points into five equal parts and length of each part 2 is cm,

i.e., PI = IJ=JK=KL=LQ= \(\frac{1}{5}\) PQ = \(\frac{1}{5}\)×10 cm

= 2 cm.

Each part of the segment is 2 cm.

WBBSE Class 8 Maths Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts

Question 2. Ajij divided a line segment XY of length 12 cm in some parts in such a way that each part is of length 2.4 cm. First, decide how many parts are there and then divide the line XY into that number of equal parts.
Solution:

Given

Ajij divided a line segment XY of length 12 cm in some parts in such a way that each part is of length 2.4 cm.

To divide 12 long straight line into parts of length 2.4, it should be cut into 5 equal parts.

∴ Ajij will cut XY straight line into 5 equal parts.

WBBSE Solutions For Class 8 Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts Ajij Will Cut XY Straight Line Into 5 Equal Parts

At point X on XY an acute angle ∠ZXY is drawn and on XY straight line on the opposite side of ∠ZXY, equal to that angle ∠XYP is drawn.

Now from straight line XY taking same radius, it is cut into 4 equal parts XA, AB, BC and CD.

Similarly taking the same radius YP straight line is cut into 4 equal parts YE, EF, FG and GH. A, H; B,G; C,H and D,E are joined by which XY straight line is divided into five equal parts at points l, J, K and L; and length of each part is 2.4 cm, i.e.,

XI = IJ = JK= KL= LY = \(\frac{1}{5}\) XY = \(\frac{1}{5}\)×12 cm

=2.4 cm.

Question 3. Anoara drew a triangle ABC in her exercise book. We bisect side BC by compass and then draw the median AD. I trisect AD in AE, EF and FD. Now we join B and F by scale and extend it to X such that it intersects AC at X.
Solution:

Given

Anoara drew a triangle ABC in her exercise book. We bisect side BC by compass and then draw the median AD.

I trisect AD in AE, EF and FD. Now we join B and F by scale and extend it to X such that it intersects AC at X.

Measuring by scale we see,

AX =  CX [Let’s put the number]

WBBSE Solutions For Class 8 Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts Anoara Drew A Triangle ABC In Her Exercise Book

In ΔABC side BC is bisected with the help of a pencil and compass and the AD median is drawn. Median AD is divided into three parts AE, EF and FD with help of scale and compass.

Now, with help of scale, I join two points B and F, and extend them which cut a straight line AC at point X.

Measuring by scale we see,

AX = CX, i.e., AX= 1 CX

Question 4. Let’s divide a line segment of length 12.6 cm into 7 equal parts, by scale and compass. Using the above construction, let’s draw an equilateral triangle of length 7.2 cm.
Solution:

WBBSE Solutions For Class 8 Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts Equilateral Triangle

A 12.5 cm long straight line AB is drawn. At point A on AB an acute angle ∠XAB is drawn and on AB on the opposite side of ∠XAB, equal to that angle, ∠ABY is drawn.

Now from AX straight line taking same radius six equal parts AC, CD, DE, EF, FG and GH are cut off.

Similarly by taking same radius BY straight line is cut into six equal parts Bl, IJ, JK, KL, LM and MO. C,0 ; D,M; E,L; F,K; G,J and H,l are joined by which AB which straight line is divided into 7 equal parts respectively P, Q, R, S, T and U points and length of each part is 1.8 cm. i.e.,

AP = PQ = QR = RS = ST = TU = UB = 1.8 cm

AS = 7.2 cm. Now taking radius equal to AS and taking centre A and S two arcs are drawn which cut each other at W point. A, W and S, W are joined.

ASW equilateral triangle is formed whose length of each side is 7.2 cm.

Question 5. Rama pradhan drew a parallelogram ABCD where AB = 6 cm, BC = 9 cm and ABC = 60°. Let’s select two points P and Q on the diagonal BD such that BP = PQ = QD. Joining the points A, P; P, C; C, Q and Q, A, let’s write the nature of quadrilateral APCQ.
Solution:

WBBSE Solutions For Class 8 Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts Rama Pradhan Drew A Parallelogram ABCD

Given

Rama pradhan drew a parallelogram ABCD where AB = 6 cm, BC = 9 cm and ABC = 60°.

Let’s select two points P and Q on the diagonal BD such that BP = PQ = QD. Joining the points A, P; P, C; C, Q and Q, A,

ABCD is a parallelogram whose sides AB = CD = 6 cm and BC = AD = 9 cm and ABC = 60°. I choose with help of pencil, scale and compass, on BD diagonal of the parallelogram ABCD two points at P and Q such that BP = PQ = QD.

Now joining A,P; P,C; C,Q and Q, A, APCQ quadrilateral is formed. In it ∠AQC = 130°, ∠PAQ = ∠PCQ = 50° and AP = CQ and AQ = CP.

∴ It is a parallelogram.

Question 6. Sujata drew three line segments of length 4 cm, 6 cm and 10 cm. By scale and compass, Rahul bisects the 1st line segment, trisects the 2nd line segment and divides the 3rd line segment into five equal parts. Sabnarh drew a triangle PQR taking \(\frac{1}{3}\) th of 1 1 the first line segment, \(\frac{1}{5}\) th of the 2nd line segment and \(\frac{1}{2}\) th of the 3rd line segment. Let’s classify the triangles on the basis of their sides.
Solution:

WBBSE Solutions For Class 8 Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts Sujata Drew Three Line Segments 1

WBBSE Solutions For Class 8 Chapter 23 Dividing A Line Segment Into Three Or Five Equal Parts Shabnam Has Draw Euilateral Triangle

Given

Sujata drew three line segments of length 4 cm, 6 cm and 10 cm.

By scale and compass, Rahul bisects the 1st line segment, trisects the 2nd line segment and divides the 3rd line segment into five equal parts.

Sabnarh drew a triangle PQR taking \(\frac{1}{3}\) th of 1 1 the first line segment, \(\frac{1}{5}\) th of the 2nd line segment and \(\frac{1}{2}\) th of the 3rd line segment.

Half of the length of the first straight line = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) × 4 cm = 2 cm

= \(\frac{1}{3}\) rd of the length of the 2nd straight line = \(\frac{1}{3}\) PQ = \(\frac{1}{3}\) × 6 cm = 2 cm

= \(\frac{1}{5}\) th of length of the third straight line = \(\frac{1}{5}\) XY = \(\frac{1}{5}\) × 10 cm = 2 cm

∵ Half of the length of the first straight line, \(\frac{1}{3}\) rd of that of the second and \(\frac{1}{5}\) th of that of the third, are equal.

∴ PQR is an equilateral triangle whose length of each side is 2 cm.

∴ Shabnam has drawn equilateral triangles.

WBBSE Solutions For Class 8 Maths Chapter 24 A Few Interesting Problems

A Few Interesting Problems

Question 1. Interesting problems with matchsticks.

1. I constructed an equilateral triangle with three matchsticks.
Solution:

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Equilateral Triangle With Three Matchsticks

2. My brother constructed 6 equilateral triangles with 12 matchsticks.
Solution:

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems 6 Equilateral Triangle

Replacing only 4 sticks from the 12 sticks I will construct 3 equilateral triangles where their areas are not equal.

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems 3 Equilateral Triangle

The first triangle is A, the Second triangle is (A + B) and 3rd triangle is (A + B + C). All three are equilateral triangles whose measurements are not equal.

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. Megha arranged 26 matchsticks as shown in the figure beside.
Solution:

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Arrangement With 26 Matchsticks

Replacing only 14 matchsticks in this arrangement I constructed 3 squares where their areas are not same.

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems 14 Matchsticks In This Arrangement

A, B, and C are three squares whose areas are not the same.

WBBSE Solutions For Class 8 Maths Chapter 24 A Few Interesting Problems

Question 3. Rokeya had 20 matchsticks. She made a square-type house with 4 matchsticks as the figure beside and she also made a square-type fence around the garden with the remaining 16 sticks.
Solution:

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Rokeya Had 20 Matchsticks

Given

Rokeya had 20 matchsticks. She made a square-type house with 4 matchsticks as the figure beside and she also made a square-type fence around the garden with the remaining 16 sticks.

Let’s divide the garden into five equal shapes and sizes with 10 other matchsticks.

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Divide The Garden Into Five Equal Shapes And Sizes

I inserted Rokeya’s arrangement with 10 new sticks 1, 2, 3, 4, 5, 6, 7,8, 9, and 10; and divided this garden into five equal parts A, B, C, D, and E.

Question 4. Let’s put numbers from 1 to 19 in the circles of the following wheel in such a way that the sum of the numbers in 3 circles along a straight line segment is 30.
Solution:

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Straight Line Segment Is 30

Question 5. 

1. 

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Missing Number Question 5.1

Solution:

= \(\frac{3}{2}-1=\frac{1}{2}\)

= \(\frac{8}{3}-2=\frac{2}{3}\)

= \(\frac{19}{3}-3=\frac{10}{3}\)

2. 

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Missing Number Question 5.2

Solution:

(4+9) × 2 = 26

(9 + 16) × 2 = 50

(16 + 4) × 2 = 40

Question 6. 

1.

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Missing Number Question 6.1

Solution:

\((2)^2+(4)^2=20\),

\((1)^2+(5)^2=26\),

\((3)^2+(9)^2=90\)

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Missing Number Question 6.2

\((1)^3+(2)^2=9\),

\((4)^3+(5)^3=189\),

\((2)^3+(3)^3=35\)

Question 7.

1. 

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Missing Number Question 7.1

Solution:

1. 7×3 + 8 = 29

2. 4×3 + 7 = 19

3. 5×? + 6 = 31

∴ ? = (31 – 6)÷5 = 5

2.

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Missing Number Question 7.2

Solution:

1. 4×2 – 1 =7

2. 5×3-3 = 12

3. 6×7 – ? = 39

∴ ? = 42 – 39 = 3

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Missing Number Question 7.3

Solution:

1. (10-9)+(15-12)

Solution:

(10-9)+(15-12) = 1+3

(10-9)+(15-12) = 4

(10-9)+(15-12) = 4

2. (16-20)+(28-12)

Solution:

(16-20)+(28-12)  =-4+16

(16-20)+(28-12)  =12

(16-20)+(28-12) =12

3. (15-16)+(23-11)

Solution:

(15-16)+(23-11) =-1+12

(15-16)+(23-11) =11

(15-16)+(23-11) =11

Let’s see the rules of the game and find out the correct numbers.

Question 1. If the sign ‘÷’ stands for the sign ‘×’ and the sign ‘+’ stands for the sign ‘÷’ and the sign ‘#’ stands for the sign ‘+’, then let’s write which of the following numbers will be the value of 2÷5+5#100.

  1. 100
  2. 102
  3. 108
  4. 105

Solution:

2 × 5 ÷ 5 + 100

= 2 × 1 + 100

= 102

Question 2. If 7 * 1= 64 and 3*9=144, then let’s write which one of the following numbers will be the value of 5*6 :

  1. 2
  2. 45
  3. 101
  4. 121

Solution:

7*1 = 64

= \((7+1)^2\) = 64 :

= \((3+9)^2\) = 144

∴  5 * 6

= \((5+6)^2\)

= 121

Question: 3. If 84 ‘+’ 72 = 45 and 73 ‘×’ 41=43, then find the value of 94 ‘×’ 72 from the following:

  1. 55
  2. 59
  3. 56
  4. 66

1. 84 = 8-4 = 4

Solution:

72 = 7-2 = 5

∴ Number = 45

2. 73 = 7-3= 4

Solution:

41=4-1=3

∴ Number = 43

3. 94 = 9-4 = 5

Solution:

72 = 7-2 = 5

∴ Number = 55

Question 4. If the sign ‘÷’ and the sign ‘+’ and the numbers ‘6’ and ‘3’ interchange their positions, then find which one of the following relations is true-

1. 3+6-2=5

Solution:

=6÷3 + 2

= 2 + 2

= 4

2. 6÷3+2=8

Solution:

= 3 + 6 ÷ 2

= 3 + 3

= 6

3. 3+6÷5=7

Solution:

6÷3 + 5

=2+5

=7

4. 3÷6+1 =6

Solution:

= 6 + 3÷1

= 6 + 3

= 9

∴ 3. Relation is true

Question 5. If the sign ‘+’ and the sign ‘-‘ and the numbers ‘4’ and ‘8’ interchange their positions, then find which of the following relations is true

1. 4+8-12=16

Solution:

8-4+12

=20-4

= 16

2. 4-8+12=6

Solution:

8 + 4-12

= 12-12

= 0

3. 8+4-12=24

Solution:

4 – 8 + 12

= -4+12

= 8.

4. 8-4+12=8

Solution:

4 + 8-12

= 12-12

= 0

∴ 1. Relation is true

Question 6. Let’s find out a few more interesting numbers –

Solution:

Let’s see why 142857 is a (Revolving Number) –

142857×1 = 142857

142857×2 = 285714

142857×3 = 428571

142857×4 = 571428

142857×5= 714285

142857×6= 857142

Question 7. Let’s find 24 using a one-digit number thrice: 33-3 = 24 Let’s find 24 using another one-digit number thrice except 3.

Solution:

8 + 8 + 8 = 24

8. Let’s find 30 using a one-digit number thrice: 33+3 = 30. Let’s find 30 using another one-digit number thrice except 3.

Solution:

6 x 6 – 6 = 30

5 x 5 + 5 = 30

9. Imran arranged 8 pieces of papers in two columns writing the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9.

Solution:

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Columns Writing The Numbers

WBBSE Solutions Class 8 Chapter 24 A Few Interesting Problems Columns Replacing Only 2 Pieces Of Paper

Let’s try to get the same sum from the columns replacing only 2 pieces of paper.

Question 10. Maria’s father kept an envelope with some money on the table for 10.buy books and go to the office. Maria returned home and saw that 98 was written on the envelope. So she went to the bookshop and bought a book priced at Rs. 92. However, at the time of paying the money, she saw that there were less than 92 rupees in the envelope. How did this happen? Let’s think and write.
Solution:

Given

Maria’s father kept an envelope with some money on the table for 10.buy books and go to the office.

Maria returned home and saw that 98 is written on the envelope. So she went to the bookshop and bought a book priced at Rs. 92.

However, at the time of paying the money, she saw that there were less than 92 rupees in the envelope.

Maria’s father had written 86 on the envelope but Maria had seen the envelope in othe pposite direction.

WBBSE Solutions For Class 8 Maths Chapter 22 Construction Of Parallel Lines

Construction Of Parallel Lines

Question 1. With a scale and pencil let’s join two points C and Q and produce the line obtained on both sides. Thus, I got a straight-line PR.
Solution:

WBBSE Solutions For Class 8 Chapter 22 Construction Of Parallel Lines Criteria Of Congruency

Now I prove logically step-by-step that PR||AB.

I joined D,Q and O,Q.

In Δ s CDQ and DOQ.

DC = OQ, CQ= DO and DQ is their common side.

∴ ΔCDQ = ΔDOQ (by s-s-s criteria of congruency)

∴ ∠CQD = ∠QDO, but they are alternate angles.

∴ CQ || DO

Therefore, PR || AB

Read and Learn More WBBSE Solutions For Class 8 Maths

∴ I got straight line PR through the point C, which is parallel to the straight line AB, i.e., I draw a line PR through the point C which was not on the given line AB and parallel to the line PR AB.

Parallel Lines Exercise  22.1

Question 1. Let’s see how many line segment parallel to the line XY are possible to draw through the point Z which is not on XY.
Solution:

WBBSE Solutions For Class 8 Chapter 22 Construction Of Parallel Lines Line Segment Parallel To The Line XY

On straight line XY a point Q is taken. Points Z and Q are joined, Consequently ∠ZQY is formed. Now on point Z on the opposite side of ∠ZQY ∠QZP is drawn equal to angle ∠ZQY.

Joining P and Z with scale and extending on both sides AB straight line is got.

∵ ∠PZQ = ∠ZQY, are alternate angles, AB // XY

Now outside the line XY from a point Z, no other parallel line can be drawn because the alternate angle at Z will not be equal in measurement.

∴ From point Z parallel to XY only one straight line can be drawn.

Question 2. Habib has drawn a line segment PQ in his exercise book and he has also considered a point R outside the line segment PQ. Let’s draw a line parallel to the line segment PQ that passes through R by a scale and a compass.
Solution:

Given

Habib has drawn a line segment PQ in his exercise book and he has also considered a point R outside the line segment PQ.

On straight line PQ a point S is taken Both points. R and S are joined with scale, consequently, ∠RSQ is drawn.

Now on point R on opposite side of ∠RSQ ∠SRT is drawn equal to ∠RSQ.

∵ ∠SRT = ∠RSQ, alternate angles

∴ PQ and TU straight lines are parallel, i.e., PQ // TU

WBBSE Solutions For Class 8 Maths Chapter 22 Construction Of Parallel Lines

Question 3. By a scale and a compass, Megha draws an angle ∠ABC = 60° between rays BA and BC respectively. Let’s take two points P and Q. Let’s draw a straight line through point P parallel to the ray BC and also draw a straight line through the point Q parallel to the ray BA. Let D be the intersecting point. Let’s write the type, of which quadrilateral PBQD is.
Solution:

WBBSE Solutions For Class 8 Chapter 22 Construction Of Parallel Corresponding Angles

Given

By a scale and a compass, Megha draws an angle ∠ABC = 60° between rays BA and BC respectively. Let’s take two points P and Q.

Let’s draw a straight line through point P parallel to the ray BC and also draw a straight line through the point Q parallel to the ray BA. Let D be the intersecting point.

∠ABC = 60°

At AB on point P ∠APE is drawn equal to ∠ABC. At BC on point Q ∠FQC is drawn equal to ∠ABC. PE and QF are joined which cut each other at D.

∠ABC = ∠APE (Corresponding angles)

∴ PE//BC

∠ABC = ∠FQC (Corresponding angles)

∴ AB // FQ

Now in PBQD,

BP // QD and BQ // PD

∴ PBQD is a parallelogram.

WBBSE Solutions For Class 8 Maths Chapter 19 Formation Of An Equation And Its Solution

Formation Of An Equation And Its Solution

Today we will play an interesting game. Shibani has collected some marbles in a bag made of cloth. Dhruba, Mahua, Ashoke, Murad and I will play with marbles in the room on the roof of Sibani’s house.

The game is like this : first Shibani will distribute some marbles to Dhruba and Mahua in a certain way and then I will tell the number of marbles each of them has got.

To Mahua, Shibani gives 18 marbles more than twice the number of marbles she gives to Dhruba. Let’s find the number of marbles Sibani might give to Mahua.

Let Shibani gives × marbles to Dhruba.

∴ Shibani gives – (2 × x + 18) = 12x + 18I marbles to Mahua.

Number of marbles 1 2 3 4 5 6 7 ……………… x
given to Dhruba

Number of marbles 20 22 24 26 28 30 32 12x + 18 marbles to Mahua.
given to Mahua.

∴ Shibani can give 20, 22, marbles to Mahua.

I come to know that Shibani has given 108 marbles to Mahua.

2x + 18 = 108

or, 2x = 108 – 18

or, 2x = 90

or, x= 90/2

∴ x = 45

So, Shibani has given 45 marbles to Dhruba.

∴ x = 45 is the root of the equation : 2x + 18 = 108.

Read and Learn More WBBSE Solutions For Class 8 Maths

Now, Shibani gives Mahua 4 less marbles than half of marbles she gives to Murad. Let’s find out the number of marbles which Shibani can give to Mahua.

Let’s think that Shibani gives x marbles to Murad.

∴ Shibani gives \(\frac{x}{4}\)-4  marbles to Mahua.

Number of marbles given 8 10 12 20 28 ……… n to Ashoke

Number of marbles given 0 1 2 6 10 \(\frac{n}{2}\)-4 to Mahua

After counting, we see that Shibani has given 86 marbles to Mahua.

= \(\frac{x}{2}-4=86\)

∴ Shibani has given 180 to Murad.

∴ x = 180

We get that the root of the equation.

= \(\frac{x}{2}-4=86\) is 180

Again, Shibani gives to Mahua 3 less marbles than \(\frac{5}{2}\) part of marbles she gives to Ashoke. Let’s find the number of marbles which Shibani can give to Mahua.

Let’s think that Shibani gives x marbles to Ashoke.

∴ Shibani gives \(\frac{5 x}{2}-3\) marbles to Mahua.

Number of marbles 2 4 8 10 20 ………………. n given to Ashoke

Number of marbles 2 7 17 22 47 ……………… \(\frac{5 x}{2}-3\) given to Mahua

If Shibani gives 127 marbles to Mahua, let’s find the number of marbles Shibani has given to Ashoke.

Let’s find the root of the equation

= \(\frac{5 x}{2}-3=127\)

WBBSE Class 8 Maths Chapter 19 Formation Of An Equation And Its Solution

Formation Of An Equation Exercise 19.1

Question 1. Shibani has given some marbles to Murad and me. If the number of marbles of Murad is 2 less than \(\frac{x}{2}\) part of the marbles mine, then let’s calculate the possible number of marbles of Murad.
Solution:

Given

Shibani has given some marbles to Murad and me. If the number of marbles of Murad is 2 less than \(\frac{x}{2}\) part of the marbles of mine

Let Shibani has given me x marbles.

∴ Murad has been given \(\frac{7 x}{3}-2\) marbles.

Number of marbles 3 6 9 15 18 21

Number of Murad’s marbles 5 12 19 33 40 47

∴ Murad can be given 5,12,19,33,40,47……………marbles.

B.T.P.,

= \(\frac{7 x}{3}-2=40\)

or, \(\frac{7 x}{3}=40+2\)

or, x = \(\frac{42 \times 3}{7}\)

or, x = 18

If Murad has 40 marbles, then Shibani has given me 18 marbles.

WBBSE Solutions Class 8 Chapter 19 Formation Of An Equation And Its Solution 4n Number Of Sticks

This type of n number of designs contain 4n number of sticks.

If we use 80 sticks then the number of squares is 20

Therefore, 4n = 80. The root of the equation is 20

Again, Rokeya made another design –

WBBSE Solutions Class 8 Chapter 19 Formation Of An Equation And Its Solution H Number Of Sticks

To make such n number of ‘H’ the number of sticks will be needed is (3 x n + 2).

Let’s form an equation and find the root of the equation to find the number of ‘IT if we form H,s by 35 sticks. Rokeya made another design

WBBSE Solutions Class 8 Chapter 19 Formation Of An Equation And Its Solution K Number Of Sticks

This type of n number of designs contain 4n number of sticks. Now I will make another design with matchsticks and find the number of sticks required to from n number of such designs.

WBBSE Solutions Class 8 Chapter 19 Formation Of An Equation And Its Solution Equation

Make equation and write in language.

x = -19

2x + 38 = 0

t = 21

5t-105 = 0

All of us will contribute some money for organising a feast. I have 2, 100 rupee notes. I want to change these two notes. I want to change these two notes in the form of 5 and 10 rupee notes.

Shibani’s father changed my money and gave me 32 notes in total.

Let’s form an equation and solve it to get how many notes of each type I have.

Let, I have x number of 5 rupee notes.

∴ I have (32 – x) number of 10 rupee notes.

Value of 1 number of 5 rupee notes is Rs. 5.

∴ Value of 1 number of 10 rupee notes is Rs. 10.

Value of x number of 5 rupee notes is Rs 5x.

∴ Value of (32 – x) number of 10 rupee notes is Rs. 10(32 – x).

So I have Rs. 5x and Rs. 10(32 – x) and my total money is Rs. 200.

’ We have, 5x + 10 (32-x) = 200

or, 5x + 320 – 10x = 200

or, 5x – 10x = 200-320

or, -5x = -120 -120

or, x = \(\frac{-120}{-5}\)

∴  x = 24

So I have 24 number of 5 rupee notes and 32-24 = 8 number of 10 rupee notes.

Question 2. If I change 3 hundred rupee notes in the form of 5 and 10 rupee notes then I get 48 notes in total. Let’s find out how many notes of each type I would have.
Solution:

Given

If I change 3 hundred rupee notes in the form of 5 and 10 rupee notes then I get 48 notes in total.

Let I have x notes of Rs. 5.

∴ I have 10 number of (48-x) notes.

Value of 5 notes of Rs. x is Rs. 5x

Value of 10 notes of Rs. (48-x) is Rs. 10 (48-x)

B.T.P.,

5x + 10 (48 – x) = 300

or, 5x + 480 – 10x = 300

or, -5x = 300 – 480

or, -5x = -180 .

or, x = \(\frac{-180}{-5}\)

or, x = 36

∴ I have 36 notes of Rs. 5 and (48-36) = 12 notes of 10.

Question 3. There are 35 students in Dhruba’s class. Their average age is 14 years. New 7 more students got admission and then the average age becomes is 13.9 years. Let’s find the equation and calculate the average age of the seven new students of Dhruba’s class.
Solution:

Given

There are 35 students in Dhruba’s class. Their average age is 14 years. New 7 more students got admission and then the average age becomes is 13.9 years.

Let the average age of the 7 new students is x years.

∴ Total age of the 7 new students is (7xx) years = 7x years.

Total age of the previous 35 students was 35×14 years = 490 years.

Now the total number of students is Dhruba’s class is (35+7)

= 42

∴ Total age of these 42 students is = (7x + 490) years.

Average age of these 42 students is = 13.9 years.

∴ Total age of these 42 students is = (13.9 x 42) years.

B.T.P., 7x +490 = 13.9×42

or, 7x = 13.9 × 42 – 490

or, x = \(\frac{13.9 \times 42-490}{7}\)

∴ x = 13.4

∴The average age of the 7 new students is h 3.4l years.

Question 4. Manash has written a fraction whose denominator is 1 more than twice of the numerator. If we add 4 to the numerator and the denominator, then the fraction will be \(\frac{7}{11}\) Let’s form an equation and find out the fraction written by Manash.

Solution: 

Manash has written a fraction whose denominator is 1 more than twice of the numerator. If we add 4 to the numerator and the denominator, then the fraction will be \(\frac{7}{11}\)

Let the numerator of the fraction = x

Hence the denominator = 2x + 1

∴ The fraction = \(\frac{x}{2 x+1}\)

Let’s add 4 to both the numerator and the denominator of the fraction and let’s see what we get,

= \(\frac{x+4}{2 x+1+4}=\frac{x+4}{2 x+1+4}=\frac{x+4}{2 x+5}\)

B.T.P.,

= \(\frac{x+4}{2 x+5}=\frac{7}{11}\)

or, 14x+35 = 11x+44

or, 14x-11x = 44-35

or, 3x = 9

∴ x = 3

The required fraction  = \(\frac{x}{2 x+1}=\frac{3}{2 x 3+1}=\frac{3}{7}\)

Question 5. Ashoke has written a fraction whose numerator is 2 less than the denominator. If we add 1 to the numerator and to the denominator then the fraction will be \(\frac{4}{5}\); let’s form an equation and hence find the fraction written by Ashoke.

Solution:

Given

Ashoke has written a fraction whose numerator is 2 less than the denominator. If we add 1 to the numerator and to the denominator then the fraction will be \(\frac{4}{5}\);

Let denominator is x.

∴ Number of fraction = x-2

∴ Fractions = \(\frac{x-2}{x}\)

B.T.P.,

= \(\frac{x-2+1}{x+1}=\frac{4}{5}\)

or,  \(\frac{x-1}{x+1}=\frac{4}{5}\)

or, 5x – 5 = 4x + 4

or, 5x – 4x = 4 + 5

or, x = 9

∴ Denominator = 9

∴ Numerator = 9-2 = 7

∴ Fractions = \(\frac{7}{9}\)

Question 6. Murad has written a two digit number where the sum of the two digits is 11; if we add 63 to the number then the digits change their positions. Let’s form an equation and hence try to find the two digit number written by Murad.

Given

Murad has written a two digit number where the sum of the two digits is 11; if we add 63 to the number then the digits change their positions.

Let the digit in the unit’s position of the two digit number is x.

∴ Digit in tenth place = (11 — x)

∴ Two digit number = 10x number in tenth place + number in unit place

= 10 × (11-x) + x

= 110 – 10x + x

Two digit number = 110- 9x

If we interchange the digits of the two digit number, then the number becomes

= 10 × x+11-x

= 10x +11-x

= 9x + 11

B.T.P.,

= 110-9x+63 = 9x+11

or, -9x-9x = 11-110-63

or, -18x = -162

or, x = \(\frac{-162}{-18}\) = 9

∴ x = 9

∴ Digit in unit place = 9

∴ Digit in tenth place = 11 -9 = 2

∴ Two digit number = 10×2 + 9

= 20 + 9

Two digit number = 29

Question 7. Half of a number is 6 more than \(\frac{1}{5}\) part of that number. Let’s form an equation and hence find the number.
Solution:

Given

Half of a number is 6 more than \(\frac{1}{5}\) part of that number.

Let the number be x.

∴ Half of the number = \(\frac{x}{2}\)

Half of the number is 6 more than of that \(\frac{1}{5}\) number.

Therefore, \(\frac{x}{2}-\frac{x}{5}\) = 6

or, \(\frac{5 x-2 x}{10}c\) = 6

or, \(\frac{3 x}{10}\) = 6

∴ x = 20

The required number is 20

Formation Of An Equation Exercise 19.1

Let’s form equation and work out.

Question 1. 2 more than twice the number that Sima has written is equal to 5 less than thrice the number she has written. Find the number she has written.

Solution:

Given

2 more than twice the number that Sima has written is equal to 5 less than thrice the number she has written.

Let Sima has written number x.

∴ Twice of the numbers = 2x

∴ Thrice of the number = 3x

B.T.P., required equation –

2x + 2 = 3x – 5

or, 2x – 3x = – 5 – 2

or, -x = -7

or, x = 7

∴ Number written by Sima = 7

Equation = 2x + 2 = 3x – 5, Number = 7

Question 2. Let’s write down three consecutive integers such that 5 less than the sum of the numbers is equal to 11 more than twice the second number. Let’s find the three consecutive integers.

Solution:

Let three consecutive numbers be x, x + 1 and x + 2.

Twice of the second number = 2 (x + 1).

B.T.P., required equation –

x + (x + 1) + (x + 2)-5 = 2 (x + 1)+11

or, x + x+1+x + 2- 5 = 2x + 2+11

or, 3x – 2x = 2 + 11 + 5 – 1-2

or, x = 15

∴ First number = 15

Second number = 15 + 1 = 16

Third number = 15 + 2 = 17

Required equation : x + x+1 + x + 2- 5 = 2 (x+1) +11

Three consecutive numbers are 15, 16 and 17.

Question 3. Let’s find a number such that one-fourth of the number is 1 less than one-third of the number.

Solution:

Let the number is = x

= \(\frac{1}{3}\) rd of the number = \(\frac{x}{3}\)

= \(\frac{1}{4}\) th of the number =\(\frac{x}{4}\)

B.T.P.,

= \(\frac{x}{3}-\frac{x}{4}=1\)

or, \(\frac{4 x-3 x}{12}=1\)

or, x = 12

Required equation : \(\frac{x}{3}-\frac{x}{4}=1\)

Required number = 12

Question 4. Let’s find a fraction whose denominator is 2 more than the numerator; and when 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes equal to \(\frac{7}{3}\)

Solution.

Let the numerator of the fraction is x.

∴ The denominator of fraction = x + 2.

∴ Fraction = \(\frac{x}{x+2}\)

B.T.P., required equation-

= \(\frac{x+3}{x+2-3}=\frac{7}{3}\)

= \(\frac{x+3}{x-1}=\frac{7}{3}\)

or, 7x-7 = 3x+9

or, 7x-3x = 9+7

or, 4x = 16

or, x = \(\frac{16}{4}\) = 4

Required equation = \(\frac{x+3}{x+2-3}=\frac{7}{3}\) and fraction = \(\frac{4}{6}\)

Question 5. Sucheta wrote a fraction whose denominator is 3 more than the numerator. Again, she added 2 to the numerator and subtracted 1 from the denominator and also subtracted 1 from the numerator and added 2 to the denominator to get 2 fractions whose product is \(\frac{2}{5}\). Let’s write the fraction written by Sucheta.

Solution:

Given

Sucheta wrote a fraction whose denominator is 3 more than the numerator. Again, she added 2 to the numerator and subtracted 1 from the denominator and also subtracted 1 from the numerator and added 2 to the denominator to get 2 fractions whose product is \(\frac{2}{5}\).

Let the numerator of fraction is x.

∴ Denominator = x + 3

∴ Fraction = \(\)

B.T.P., required equation

= \(\frac{x+2}{x+3-1} \times \frac{x-1}{x+3+2}=\frac{2}{5}\)

= \(\frac{x+2}{x+2} \times \frac{x-1}{x+5}=\frac{2}{5}\)

= \(\frac{x-1}{x+5}=\frac{2}{5}\)

or, 5x – 5 = 2x + 10

or, 5x – 2x = 10 + 5

or, 3x = 15

or X= \(\frac{15}{3}\) = 5

∴ Numerator of the fraction = 5 and denominator =5 + 3 = 8

∴ Required fraction = \(\frac{5}{8}\)

Required equation :  \(\frac{x+2}{x+3-1} \times \frac{x-1}{x+3+2}=\frac{2}{5}\)

Fraction written by Sucheta = \(\frac{5}{8}\)

Question 6. Raju wrote a number having two digits where the digit in the tenth position is thrice the digit in the unit position and the new number obtained by reversing the positions of the digits is 36 less than the initial number. Let’s find the number written by Raju.

Solution:

Given

Raju wrote a number having two digits where the digit in the tenth position is thrice the digit in the unit position and the new number obtained by reversing the positions of the digits is 36 less than the initial number.

Let the digit in the unit’s place of the two digit number is x.

∴ Digit in the tenth’s place = 3x

∴ Two digit number = 10x digit in the tenth’s place + digit in the unit’s place.

= 10 × 3x + x

= 30x + x

Number got by interchanging the digits of the two digit number-

= 10 × x + 3x

= 10x + 3x

B.T.P., required equation

30x + x – 36 = 10x + 3x

or, 31 x – 13x = 36

or, 18x = 36

or, x = \(\frac{36}{18}\) = 2

∴ Digit in the units place of the two digit number = 2 and digit in the tenths place = 3×2 = 6

Required number = 10×6 + 2 = 62

Required equation : 30x + x – 36 = 10x +3x

Required number = 62

The number written by Raju = 62

Question 7. If the sum of two numbers is 89 and their difference is 15, let’s find the two numbers.

Solution:

Given

If the sum of two numbers is 89 and their difference is 15,

Let the greater number is x.

∴ The smaller number = 89 – x

B.T.P., required equation –

= x – (89 – x) = 15

or, x – 89 + x = 15

or, 2x = 15 + 89

or, 2x = 104

or, x  = \(\frac{104}{2}\)= 52

∴ Greater number = 52, smaller number = 89 – 52 = 37

Required equation : x – (89-x) = 15

Required number : 52 and 37.

Question 8. Divide 830 into two parts in such a way that 30% of one part will be 4 more than 40% of the other.

Solution:

Let one part is x.

∴ Second part = (830 – x)

30% of one part = 30% of x

= \(\frac{x \times 30}{100}=\frac{3 x}{10}\)

40% of second part = 40% of (830 – x)

= \(\frac{(830-x) \times 40}{100}=\frac{2(830-x)}{5}\)

B.T.P., required equation-

= \(\frac{x \times 30}{100}=\frac{(830-x) \times 40}{100}+4\)

= \(\frac{3 x}{10}=\frac{2(830-x)}{5}+4\)

= \( \frac{3 x}{10}=\frac{1680-2 x+20}{5}\)

= \(\frac{3 x}{2}=\frac{1660-2 x}{1}\)

= \(\frac{3 x}{2}=\frac{1660-2 x}{1}\)

or, 3x = 3360-4x

or, 3x+4x = 3360

or, x = \(\frac{3360}{7}\) = 480

∴ One part = 480

Second part = 830 – 480 = 350

Required equation: = \(\frac{x \times 30}{100}=\frac{(830-x) \times 40}{100}+4\)

Two parts are respectively 480 and 350.

Question 9. Divide 56 into two parts so that thrice of the first part becomes 48 more than one third of the second part.

Solution:

Let the first part is x.

∴ Second part = 56 – x

Thrice of the first part = 3x

= \(\frac{1}{3}\) rd of the second part = (56 – x)

∴ B.T.P., required equation –

3x = \(\frac{56-x}{3}\) + 48

or, 9x = 56 – x + 48 x 3

or, 9x + x = 56 + 144

or, 10x = 200

or x= \(\frac{200}{20}\) =20

First part is 20 and the second part is 56-20 = 36 ,

Required equation : 3x = \(\frac{56-x}{3}\)+ 48

Required numbers are 20 and 36.

Question 10. \(\frac{1}{5}\) part of a pole lies in mud, \(\frac{3}{5}\) part lies in water and the remaining 5 metres lies above the water. Let’s find the length of the pole and write it.

Solution:

Given

\(\frac{1}{5}\) part of a pole lies in mud, \(\frac{3}{5}\) part lies in water and the remaining 5 metres lies above the water.

Let the length of the pole is x meters.

∴ Part of pole in mud = \(\frac{1}{5}\) part x m of = \(\frac{x}{5}\)

∴ Part of pole in water = \(\frac{3}{5}\) part x m of = \(\frac{3x}{5}\)m

B.T.P., required equation-

= \(x-\left(\frac{x}{5}+\frac{3 x}{5}\right)=5\) = 5

= \(x-\frac{4 x}{5}\) = 5

= \(\frac{5 x-4 x}{5}\) = 5

or, x = 25

∴ length of the pole = 25 meters

B.T.P., required equation : \(x-\frac{4 x}{5}\) = 5

Length of the pole = 25 m

Question 11. At present, my father’s age is 7 times of my age. After 10 years my father’s age will be 3 times of my age. Let’s find and write the present age of my father and me.

Solution:

Given

At present, my father’s age is 7 times of my age. After 10 years my father’s age will be 3 times of my age.

Let my present age is x years.

∴ Present age of my father = 7x years.

10 years later my age – (x + 10) years.

10 years later father’s age = (7x + 10) years.

B.T.P., required equation-

7x + 10 = 3 (x + 10)

or, 7x + 10 = 3x + 30

or, 7x – 3x = 30 – 10

or, 4x = 20

or, x = \(\frac{20}{4}\) = 5

My present age = 5 years

Present age of my father = 7 × 5 years = 35 years

Required equation : 7x + 10 = 3 (x + 10)

My present age is 5 years and the present age of my father is 35 years.

Question 12. My uncle changed a 1000 rupee cheque from a bank in the form of 5 rupee and 10 rupee notes, if he received 137 notes in total, then find the number of 5 rupee notes that he got.

Solution:

Given

My uncle changed a 1000 rupee cheque from a bank in the form of 5 rupee and 10 rupee notes, if he received 137 notes in total

Let uncle got 10 number of Rs. x notes.

∴ (137 – x) number of Rs. 5 notes got.

B.T.P., required equation –

10x + 5 (137-x) = 1000

or, 10x + 685 – 5x = 1000

or, 5x = 1000 – 685

or, 5x = 315

or, x = \(\frac{315}{5}\) = 63

∴ Number of 5 rupee notes = (137-x) = 7 (137 – 63) = 74

Required equation : 10x + 5 (137 – x) = 100

Number of Rs. 5 notes = 74

Question 13. Salem uncle of our village used half of his savings to buy a house after his retirement from a government job. One day after falling into trouble, he sold his house and got 5% more than its cost price. If he would have taken 3450 rupees more than he would get 8% more than his cost price. Let’s find the amount of money Salem uncle used to buy his house, and his entire savings.

Solution:

Given

Salem uncle of our village used half of his savings to buy a house after his retirement from a government job. One day after falling into trouble, he sold his house and got 5% more than its cost price. If he would have taken 3450 rupees more than he would get 8% more than his cost price.

Let the total savings of Salem uncle be Rs. x

Expenditure on buying house = \(\frac{1}{2}\) part of Rs. x Rs. = Rs. \(\frac{x}{2}\)

5% of the cost price of house = 5% of Rs. \(\frac{x}{2}\) = Rs. \(\frac{x}{2} \times \frac{5}{100}\) x

8% of the cost price of house = 8% of Rs. \(\frac{x}{2}\) = Rs. \(\frac{x}{2} \times \frac{8}{100}\)

B.T.P., required equation –

= \(\frac{x}{2} \times \frac{5}{100}+3450=\frac{x}{2} \times \frac{8}{100}\)

or, \(\frac{x}{40}+3450=\frac{x}{25}\)

or, \(\frac{x}{40}-\frac{x}{25}=-3450\)

or, \(\frac{5 x-8 x}{200}=-3450\)

or, \(\frac{-3 x}{200}=-3450\)

or, \(x=\frac{3450 \times 200}{2}\)

or, x = 230000

Cost price of the house = = \(\frac{230000}{2}\) = Rs.115000

Amount of the savings of Salem Uncle = Rs. 230000

Required equation : – = \(\frac{x}{2} \times \frac{5}{100}+3450=\frac{x}{2} \times \frac{8}{100}\)

Salem uncle bought house of Rs. 115000 and his total savings was Rs. 230000.

Question 14. There was food provision for twenty days in a refugee camp of the village Gopalpur. After 7 days, 100 more refugees joined the refugee camp and they consumed the food grains within 11 days. Let’s write the number of refugees that were present initially.

Solution.

Given

There was food provision for twenty days in a refugee camp of the village Gopalpur. After 7 days, 100 more refugees joined the refugee camp and they consumed the food grains within 11 days.

Let there were x number of people in the refugee camp.

∴ For x people 20 days’ food was there. After 7 days for x people (20-7) or 13 days’ food was there.

7 days later number of people = (x + 100)

∴ (x + 100) people will eat x people’s food for 13 days.

B.T.P., required equation

= \(\frac{(20-7) x}{x+100}=11\)

or, \(\frac{13 x}{x+100}=11\)

or, 13x = 11x+1100

or, 13x-11x=1100

or, 2x = 1100

or, x = \(\frac{1100}{2}\) = 550

∴ Earlier there were 550 number of people in the refugee camp.

Requied equation \(\frac{(20-7) x}{x+100}=11\) and 550 number of people.

Question 15. Let’s find the roots of the equations :

1. \(\frac{5}{3 x+4}=\frac{4}{5(x-3)}\)

Solution:

or, 25x-75 = 12x+16

or, 25x-12x = 16+75

or, 13x = 91

or, x = \(\frac{91}{13}\) = 7

or, x = 7

2. 14(x-2)+3(x+5) = (x+8)+5

Solution:

or, 25x-75 = 12x+16

or, 25x-12 = 16+75

or, 13x = 91

or, x = \(\frac{42}{14}\) = 3

∴ x = 3

3. \(\frac{x}{2}+5=\frac{x}{3}+7\)

Solution:

= \(\frac{x}{2}+5=\frac{x}{3}+7\)

= \(\frac{x}{2}-\frac{x}{3}=7-5\)

or, \(\frac{3 x-2 x}{6}=2\)

or, x = 12

4. \(\frac{x+1}{8}+\frac{x-2}{5}=\frac{x+3}{10}+\frac{3 x-1}{20}\)

Solution:

or, \(\frac{5(x+1)+8(x-2)}{40}=\frac{2(x+3)+3 x-1}{-1^{20}}\)

or, \(\frac{5 x+5+8 x-16}{2}=\frac{2 x+6+3 x-1}{1}\)

or, \(\frac{13 x-11}{2}=\frac{5 x+5}{1}\)

or, 13x-11 = 10x+10

or, 13x-10x = 10+11

or, 3x = 21

or, x = \(\frac{21}{3}\) = 7

∴ x = 7

5. \(\frac{x+1}{4}+3=\frac{2 x+4}{5}+2\)

Solution:

or, \(\frac{x+1+12}{4}=\frac{2 x+4+10}{5}\)

or, \(\frac{x+13}{4}=\frac{2 x+14}{5}\)

or, 16x+2 = 3x+80

or, 16x-3x = 80-2

or, 13x = 78

or, x = \(\frac{78}{13}\) = 6

∴ x = 6

6. \(\frac{3}{5}(x-4)-\frac{1}{3}(2 x-9)=\frac{1}{4}(x-1)-2\)

Solution:

or, \(\frac{3(x-4)}{5}-\frac{(2 x-9)}{3}=\frac{(x-1)}{4}-2\)

or, \(\frac{9 x-36-10 x+45}{15}=\frac{x-1-8}{4}\)

or, \(\frac{9-x}{15}=\frac{x-9}{4}\)

or, 15x-135 = 36-4x

or, 15x+4x = 36+135

or, 19x = 171

or, x = \(\frac{171}{19}\) = 9

∴ x = 9

7. \(\frac{x+5}{3}+\frac{2 x-1}{7}=4\)

Solution:

or, \(\frac{7 x+35+6 x-3}{21}=4\)

or, 13x+32 = 84

or, 13x = 84-32

or, 13x = 52

or, x = \(\frac{52}{13}\) = 4

∴ x = 4

8. 25 + 3(4x-5) +8(x+2) = x+3

Solution:

or, 25 + 12x-15 + 8x+16 = x + 3

or, 12x + 8x-x = 3 + 15-25-16

or, 20x =18-41

or, 19x = -23

or, \(x=\frac{-23}{19}=-1 \frac{4}{19}\)

∴  x = \(-1 \frac{4}{19}\)

9. \(\frac{x-8}{3}+\frac{2 x+2}{12}+\frac{2 x-1}{18}=3\)

Solution:

or, \(\frac{12(x-8)+3(2 x+2)+2(2 x-1)}{36}=3\)

or, 12x – 96 + 6x + 6 + 4x – 2 = 108

or, 22x = 108+ 96+ 2-6

or, 22x = 206 – 6

or, 22x = 200

or, \(x=\frac{200}{22}=\frac{100}{11}\)

∴ \(x=9 \frac{1}{11}\)

10. \(\frac{t+12}{6}-t=6 \frac{1}{2}-\frac{1}{12}\)

Solution:

or, \(\frac{t+12-6 t}{6}=\frac{13}{2}-\frac{1}{12}\)

or, \(\frac{12-5 t}{6_1}=\frac{78-1}{72_2}\)

or, \(\frac{12-5 t}{1}=\frac{77}{2}\)

or, 24 -10t = 77

or, -10t = 77 – 24

or, -10t = 53

or, t = \(-\frac{53}{10}\)

or, t = \(-5\frac{3}{10}\)

11. \(\frac{x+1}{2}-\frac{5 x+9}{28}=\frac{x+6}{21}+5-\frac{x-12}{3}\)

Solution:

or, \(\frac{14 x+14-5 x-9}{28}=\frac{x+6+105-7 x+84}{21}\)

or, \(\frac{14 x+14-5 x-9}{-28}=\frac{x+6+105-7 x+84}{-243}\)

or, \(\frac{9 x+5}{4}=\frac{195-6 x}{3}\)

or, 27x + 15 = 780 – 24x

or, 27x + 24x = 780 – 15

or, 51x = 765

or, x = \(\frac{765}{51}\) = 15

∴ x = 15

12. \(\frac{9 x+5}{14}+\frac{8 x-7}{7}=\frac{18 x+11}{28}+\frac{5}{4}\)

Solution:

or, \(\frac{9 x+5+16 x-14}{14}=\frac{18 x+11+35}{28}\)

or, \(\frac{25 x-9}{1}=\frac{18 x+46}{2}\)

or, 50x-18 = 18x+46

or, 50x-18x = 46+18

or, 32x = 64

or, x = \(\frac{64}{32}\) = 2

or, x = 2

13. \(\frac{3 y+1}{16}+\frac{2 x-3}{7}=\frac{y+3}{8}+\frac{3 y-1}{14}\)

Solution:

or, \(\frac{3 y+1}{16}=\frac{y+3}{8}=\frac{y+3}{8}+\frac{3 y-1}{14}\)

or, \(\frac{3 y+1-2 y-6}{16}=\frac{3 y-1-4 y+6}{14}\)

or, \(\frac{y-5}{8}=\frac{5-y}{7}\)

or, 7y – 35 = 40 + 8y

or, 7y + 8y = 40 + 35

or, 15y = 75

or, y = \(\frac{75}{15}\) = 5

∴ y = 5

14. 5x -(4x-7)(3x-5) = 6-3(4x-9)(x-1)

Solution:

or, 5x-12 x2+ 20x + 21 x – 35 = 6-3 (4x2-4x – 9x + 9)

or, -12x2 + 46x – 35 = 6 – 12x2 + 12x + 27x – 27

or, -12x2 + 12x2 + 46x – 12x – 27x = 6 + 35-27

or, 46x – 39x = 41-27

or, 7x = 14

or, x =\(\frac{14}{7}\)= 2

∴ x = 2

15. 3(x-4)2 + 5(x-3)2=(2x-5) (4x-1)-40

Solution:

or, 3 (x2 – 8x + 16) +5 (x2 – 6x + 9) = 8x2 – 2x – 20x + 5-40

or, 3x2 – 24x + 48 + 5x2 – 30x + 45 = 8x2 – 22x – 35

or, 8x2 – 8x2 – 24x – 30x + 22x =-35 – 48 – 45

or, -32x = – 128

or x = \(\frac{-128}{-32}\) =4

∴ x = 4

16. 3(y-5)2+5y = (2y-3)2-(y+1)2 + 1

Solution:

or, 3 (y-5)2 + 5y = (2y-3)2– (y+1 )2+1

or, 3(y2-10y+25) + 5y = 4y2– 12y + 9 – (y2 + 2y+ 1) + 1

or, 3y2 – 30y + 75 + 5y = 4y2 – 12y+ 9- y2– 2y – 1 + 1

or, 3y2+y2-4y2-30y + 12y + 2y = 9 – 75

or, -11 y = -66

or, y = \(\frac{-66}{-11}\) = 6

Question 17. Let’s write mathematical stories and form equations.

Solution:

x = 5 → 2x-10 = 0

y = -11 → 5y+55 = 0

t = \(\frac{7}{8}\) → 8t-7 = 0

x = 24 → 3x-72 = 0

x = 7 → 7x-49 = 0

WBBSE Solutions For Class 8 Maths Chapter 17 Time And Work

Time And Work

There are 18 looms in Mansur’s factory in Shantipur. But last week 3 looms were closed. So Last week only 165 dhotis and sarees were weaved. This week all the looms are running.

Question 1. Let’s calculate in a proportional method how many dhotis and sarees will be weaved this week in Mansur’s factory.

Solution:

Given

There are 18 looms in Mansur’s factory in Shantipur. But last week 3 looms were closed. So Last week only 165 dhotis and sarees were weaved. This week all the looms are running.

The problem in mathematical language is

WBBSE Solutions For Class 8 Chapter 17 Time And Work Time Number Of Looms
If the time is constant and the number of looms increases or decrease, then the number of dhotis and sarees will Increase or Decrease (increase/decrease).

∴ There is a direct proportion between the number of looms and the number of dhotis and sarees.

∴ The direct proportion is –

15 : 18:: 165:? (Required number of dhotis and sarees)

∴ The required number of dhotis and sarees is = \(\frac{165 \times 18}{15}\) = 198.

So in this week, all looms being run, 198 dhotis and sarees will be made.

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. If all the looms run, calculate in how many days 594 dhotis and sarees will be made.

Solution:

In mathematical language, the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Number Of Looms And Number Of Dhotis

The number of looms being constant, as the number of dhoti-sarees increases/decreases, the required time will Increase or Decrease (increase / decrease)

∴ The direct proportion is –

198 : 594:: 7:?

∴ Required time = \(\frac{594 \times 18}{198}\) days = 21 days

Question 3. If all the looms, i.e., 18 looms would run, 594 dhotis and sarees can be woven in 21 days. Let’s calculate how many looms will run to weave 594 dhotis and sarees in 14 days.

Solution:

Given

If all the looms, i.e., 18 looms would run, 594 dhotis and sarees can be woven in 21 days.

In mathematical language the problem is

WBBSE Solutions For Class 8 Chapter 17 Time And Work Number Of Dhotis And Sarees And Time

If the number of dhotis and sarees is constant then the number of looms will Increase (increase/decrease) with the increase of time and the number of looms will Decrease (increase / decrease) with the decrease of time.

The number of looms and time are inversely proportional.

The inverse proportion is –

21 : 14:: ? (required number of looms) : 18 ∴ 14 : 21:: 18 : ? (required number of looms)

∴ Required number of looms = \(\frac{21\times18}{14}\) = 27

∴ In 21 days, to weave 594 dhotis and sarees 27 looms, i.e., (27 – 18) = 9 more looms will run.

So we see that there is a relation between time and work in Monsur’s loom. Let’s write the relation between the required time, the amount of work, and the number of workers is the chart given below.

WBBSE Class 8 Maths Chapter 17 Time And Work

Question 4. In a village of Niamatpur 15 men can cultivate 10 bighas of land in a week. Let’s calculate how many men can cultivate 18 bighas of land in a week.

Solution:

Given

In a village of Niamatpur 15 men can cultivate 10 bighas of land in a week.

In mathematical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Time And Cultivation Land

As work increases or decreases, the number of workers will, time being constant, Increase or Decrease (increase/decrease).

The amount of work and the number of workers are in Direct (direct / inverse) relation.

Let’s calculate in the proportional method.

So, 10 : 18 :: 15 : ? (required number of workers)

∴ Required number of people = \(\frac{15 \times 18}{10}\) = 27 people

∴ 27 workers can cultivate 18 bighas of land in a week.

Let’s do the above problem by unitary method.

Question 5. 250 volunteers of Bakultala gram panchayat have constructed half of a dam in 24 days. Let’s calculate how many men will be needed to construct the rest half part in 20 days.

Solution:

Given

250 volunteers of Bakultala gram panchayat have constructed half of a dam in 24 days.

The problem in mathematical language is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Amount Of Work And Time

Keeping work constant, with an increase or decrease of time, the number of workers will respectively d] or CH (increase/decrease).

∴ The time and the number of workers are in (direct/inverse) relation.

∴ The inverse proportion is

24 : 20 :: ? (the required number of workers) : 250

∴  20 : 24 :: 250 : ? (the required number of workers) .

∴ The required number of workers = \(\frac{250 \times 24}{20}\)

So, additional workers 300-250 = 50 need to be absorbed in the team of workers. Let’s calculate this problem by unitary method.

Time And Work Exercise 17.1

Question 1. In Aman’s factory 216 parts of machine are made in 3 days. Let’s calculate how many parts of machine will be made in 7 days.

Solution:

Given

In Aman’s factory 216 parts of machine are made in 3 days.

In mathematical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Days Number Of Machine Parts

If number of days increases or decreases, number of machine parts will increase or decrease.

∴ Days and number of machine parts are in direct proportion.

∴ 3 : 7 :: 216 : ?

or, \(\frac{3}{7}=\frac{216}{?}\)

or, Required number of parts = \(\frac{7 \times 216}{3}\) = 504

Question 2. In loom factory at Atpur 12 looms can weave 380 sarees. During the puja season for doing more work 3 looms have been established. Let’s make a proportion and calculate how many sarees can be woven in this month.

Solution:

Given

In loom factory at Atpur 12 looms can weave 380 sarees. During the puja season for doing more work 3 looms have been established.

In mathematical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Number Of Looms And Number Of Sarees

If number of looms increases, number of sarees will increase.

∴ Number of looms and number of sarees are in direct proportion.

∴ 12 : 15 :: 380 : ?

or, Required number of sarees = \(\frac{15 \times 380}{12}\)

Question 3. Let’s make the story of mathematics and find out the relation seeing the above chart.

Solution:

WBBSE Solutions For Class 8 Chapter 17 Time And Work Time And Amount Of Work

In a factory a given number of looms can weave 45 m cloth in 25 days, in 15 days the same number of looms will weave how much cloth?

Relation : If number of days increases, length of cloth will increase.

∴ Days and length of cloth are in direct proportion.

∴ 25 : 15 :: 45 : ?

or, Length of cloth woven in 15 days =\(\frac{15 \times 45}{25}\) = 27 m

Question 4. We can see after 15 days of digging a canal of 1200 metre length, 3/4 parts of the canal have been dug. Let’s calculate how many days are needed to dig the rest of the part.

Solution:

Given

We can see after 15 days of digging a canal of 1200 metre length, 3/4 parts of the canal have been dug.

Length of the canal dug in 15 days

Length of remaining part of canal = (1200 – 900) = 300 m

In mathematical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Length Of Canal And Days

Relation : If length of canal increases, days will increase.

∴ Length of canal and days are in direct relation.

∴ 900 : 300 :: 15 : ?

or, \(\frac{900}{300}=\frac{15}{?}\)

∴ Required days = \(\frac{300 \times 15}{900}\)

Question 5. 3 tractors can plough 18 bighas in a day. By rule of three method let’s calculate how many bighas can be ploughed in a day.

Solution:

Given

3 tractors can plough 18 bighas in a day.

In mathemetical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Number Of Tractors Quantity Of Land In Bighas

Relation : If number of tractors increases, quantity of land will increase.

∴ Number of tractors and quantity of land are in direct proportion.

or, Required quantity of land = g Bighas = 42 Bighas

Question 6. In Kusum’s factory 35 men can cast 10 ton iron parts in a week. The owner has got an order of casting 14 ton iron parts in a week. Let’s find out the proportion and calculate how many new men will be needed.

Solution.

Given

In Kusum’s factory 35 men can cast 10 ton iron parts in a week. The owner has got an order of casting 14 ton iron parts in a week.

In mathematical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Number Of People Quantity Of Iron

Relation : If quantity of iron increases, number of people will increase.

∴ Quantity of land and number of people are in direct proportion.

∴ \(\frac{10}{14}=\frac{35}{?}\)

or, Required number of people = \(\frac{14 \times 35}{10}\) = 49

Number of extra people = (49 – 35) = 14

So 14 more men will be appointed.

Question 7. Let’s make the story in mathematical language and find out the relation seeing the above chart.

Solution:

WBBSE Solutions For Class 8 Chapter 17 Time And Work Number Of Men And Amount Of Work

If 9 people make 6 cycles every day then 72 people will make how many cycles?

Relation : If the number of people increases, amount of work will increase.

∴ Number of people and number of cycles are in direct proportion.

∴ 9 : 72 :: 6 : ?

or, \(\frac{9}{72}=\frac{6}{?}\)

or, Required number of cycles = \(\frac{72 \times 6}{9}\) = 48

Question 8. A pond will be dug in our locality. 24 men need 12 days to cut that pond. Let’s find out the proportion and find relation how many men will be needed to dig that pond in 8 days.

Solution:

Given

A pond will be dug in our locality. 24 men need 12 days to cut that pond.

In mathematical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Days Number Of People

Relation : If days decrease, number of people will increase.

∴ Days and number of people are in inverse proportion.

∴ \(\frac{12}{8}=\frac{?}{24}\)

∴ Required number of men = \(\frac{12 \times 24}{8}\) = 36

To dig that pond in 8 days, more men will be needed.

Question 9. 5 members in a bulb manufacturing co-operative factory can make 10000 bulbs in 12 days. Based on a sudden order, 10000 bulbs have to be made in 9 days. Let’s calculate how many extra members will be appointed to supply the bulbs according to the agreement.

Solution:

Given

5 members in a bulb manufacturing co-operative factory can make 10000 bulbs in 12 days. Based on a sudden order, 10000 bulbs have to be made in 9 days.

In mathematical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Days And Number Of People 2

Relation : If days decrease, number of people will increase.

∴ Days and number of people are in inverse proportion.

∴ \(\frac{12}{9}=\frac{?}{45}\)

∴ Required number of people = \(\frac{12 \times 45}{9}\)   = 60

More number of members = (60-45) = 15. To complete the work in time, 15 more members will be appointed.

Question 10. 250 men need 18 days to dig a pond of 50 metre length and 35 metre breath. Let’s calculate how many days will be needed by 300 men to dig a pond of 70 metre length and 40 metre breadth of the same depth.

Solution:

Given

250 men need 18 days to dig a pond of 50 metre length and 35 metre breath.

Relation : If days decrease, number of people will increase.

WBBSE Solutions For Class 8 Chapter 17 Time And Work Number Of Men And Area Of Pond

Relation : If number of men increases, days will decrease and if area of pond increases, days will increase. So number of people and days are in inverse proportion and area of pond and days are in direct proportion.

∴ Required number of days To dig this kind of pond of length 70 m and breadth 40 m it will take 300 people days.

Question 11. Anawarabibi and Mihir uncle started to plaster three rooms of same size. But Harun. chacha, Anawarabibi and Mihir uncle finished the work in 10 days, 12 days and 15 days respectively. If the three members worked together in one room, the work would finish in less time. How will we calculate the number of days for plastering 1 room where three men work together?

Solution:

Given

Anawarabibi and Mihir uncle started to plaster three rooms of same size. But Harun. chacha, Anawarabibi and Mihir uncle finished the work in 10 days, 12 days and 15 days respectively. If the three members worked together in one room, the work would finish in less time.

Let’s see how many parts of the work was done by every one in a day.

Haran chacha plasters 1 room in 10 days.

Haran chacha plasters \(\frac{1}{10}\) part of the total work in 1 day.

Anawarabibi plasters 1 room in 12 days.

Anawarabibi plasters \(\frac{1}{12}\) part of the total work in 1 day.

Mihir uncle plasters 1 room in 15 days.

Mihir uncle plasters \(\frac{1}{15}\) part of the total work in* 1 day.

∴ Three men together plaster part of a room in 1 day

= \(\frac{1}{10}\) part + \(\frac{1}{10}\) part + \(\frac{1}{10}\) part

= \(\frac{6+5+4}{60}\) part

= \(\frac{15}{60}\) part

= \(\frac{1}{4}\) part

∴ Three of them together do \(\frac{1}{4}\) part of the total work in a day.

∴ The total work, i.e., 1 part is done in 1 ÷ \(\frac{1}{4}\) = 4 days.

∴ If three of them work, together, then they can complete the work in 4 days.

Question 12. Let’s calculate the number of days Mihlr uncle takes to complete the remaining \(\frac{1}{5}\) part of the work.

Solution:

Given

\(\frac{1}{5}\)

Mihir uncle does 1 part in 15 days.

Mihir uncle does \(\frac{1}{5}\) part in = \(15 \times \frac{1}{5}\) =3 days.

∴ Mihir uncle .alone completes the rest of the work in 3 days.

∴ If the work goes on such way, the required time to complete the work is = 2+2+4=7 days.

Question 13. Let’s calculate the part of the total work done by each of the if the work would go on such way.

Solution:

Harunchacha did the \(\left(\frac{1}{10} \times 2\right)\) part = \(\frac{1}{5}\) part of the total work.

Anawarabibi did the \(\left(\frac{1}{12} \times 4\right)\) part = \(\frac{1}{3}\) part of the total work.

Mihir uncle did the \(\left(\frac{1}{15} \times 7\right)\) part = \(\frac{7}{15}\) part of the total work. 15

Question 14. Bulu and Tathagata can do a work separately in 20 days and 30 days respectively. After working 7 days together both of them left away. Then Tatai came and completed the rest of the work alone in 10 days. Let’s make proportion and calculate how many days Tatai will take to complete the work alone.

Solution:

Given

Bulu and Tathagata can do a work separately in 20 days and 30 days respectively. After working 7 days together both of them left away. Then Tatai came and completed the rest of the work alone in 10 days.

Let’s calculate the remaining part of the total work after 7 days when Bulu and Tathagata left.

Bulu alone does \(\frac{1}{20}\) part in 1 day.

Tathagata alone does \(\frac{1}{30}\) part in 1 day.

∴ Together they do \(\left(\frac{1}{20}+\frac{1}{30}\right)\) part in 1 day.

= \(\frac{3+2}{60} \text { part }=\frac{5}{60} \text { part }=\frac{1}{12} \text { part }\)

∴ Bulu and Tathagata do =\(\frac{1}{12}\) part × 7 = \(\frac{7}{12}\) part in 7 days.

∴ After 7 days when Bulu and Tathagata left the remaining work is

= \(\left(1-\frac{7}{12}\right) \text { part }=\frac{5}{12} \text { part }\)

Tatai does the remaining \(\frac{5}{12}\) part in 10 days.

∴ The problem in mathematical language is

WBBSE Solutions For Class 8 Chapter 17 Time And Work Amount Of Work And Time

If the amount of work increases or decreases, the required time will (increase/decrease) Increase or Decrease.

∴ The amount of work and time are in Direct (direct/inverse)proportion.

The direct proportion is

= \(\frac{5}{12}\) : 1 : 10 : ?

∴ Required time days = \(\frac{10 \times 1}{\frac{5}{12}}\) days=  \(10 \times \frac{12}{5}\) = 24 days

∴ Tatai alone will do the work in 24 days.

Question 15. An empty tank is filled up with two pipes separately in 12 minutes and 15 minutes respctively. Let’s calculate the time required to fill up the half full tank while two pipes are opened together.

Solution:

Given

An empty tank is filled up with two pipes separately in 12 minutes and 15 minutes respctively.

It takes 1 minute to fill up \(\frac{1}{12}\) part of the empty tank by the first pipe.

By the second pipe it takes 1 minute to fill up \(\frac{5}{12}\) part of the empty tank.

∴ If the two pipes are opened together, it takes 1 minute to fill up the \(\left(\frac{1}{12}+\frac{1}{15}\right)\) part

= \(\frac{5+4}{60} \text { part }\)

= \(\frac{9}{60} \text { part }=\frac{3}{20} \text { part }\)

Half of the tank is full.

∴ Two pipes together will fill  \(\left(1-\frac{1}{2}\right)\) part = \(\frac{1}{2}\) part of the tank.

Two pipes together fill  \(\frac{3}{20}\) part of the tank in 1 minute.

Two pipes together fill 1 part in 1× \(\frac{20}{3}\) minute.

Two pipes together fill \(\frac{1}{2}\) part in \(1 \times \frac{20}{3} \times \frac{1}{2}\) minute

= \(\frac{10}{3}\) minute = \(3 \frac{1}{3}\) minute

= 3 minute 20 seconds.

Let’s calculate using proportion. ,

∴ The problem in mathematical language is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Amount Of Water And Time

The amount of water and time are in Direct (direct/inverse) proportion.

∴ Using direct proportion, we get –

= \(\frac{3}{20}: \frac{1}{2}:: 1: ? \)(Required time) –

∴ Required time = \(1 \times \frac{1}{2} \times \frac{20}{3}\)

= \(\frac{10}{3}\) minutes = 3 minutes 20 seconds.

The time required to fill up the half full tank while two pipes are opened together = 3 minutes 20 seconds.

Time And Work Exercise 17.2

Question 1. Priya and Debu individually complete a work in 10 hours and 12 hours respectively. If they do the work together, let’s calculate in how many hours they will complete the work.

Solution:

Given

Priya and Debu individually complete a work in 10 hours and 12 hours respectively.

Priya can do \(\frac{1}{10}\) part of the in 1 hour. Debu cah do \(\frac{1}{12}\) part in 1 hour.

∴ Both can do \(\frac{11}{60}\) part of the work in 1 hour.

Both can do 1 part work in \(1 \div \frac{11}{60}\) hours.

= \(\frac{60}{11}\)

= \(5 \frac{5}{11}\)

If they work for \(5 \frac{5}{11}\) hours, the work will be completed.

Question 2. My elder brother, my elder sister and me together will paint our windows. My elder brother, my elder sister and me can complete the work separately in 12, 4 & 6 days respectively. If we do the work together then let’s calculate’ and write how many days we will take to complete the work.

Solution:

Given

My elder brother, my elder sister and me together will paint our windows. My elder brother, my elder sister and me can complete the work separately in 12, 4 & 6 days respectively.

My sister can do \(\frac{1}{12}\) part in 1 day.

My brother can do \(\frac{1}{4}\) part in 1 day. I can do \(\frac{1}{6}\) part in 1 day.

Three of us can do \(\left(\frac{1}{12}+\frac{1}{4}+\frac{1}{6}\right)\) part = \(\frac{1+3+2}{12}\) part in 1 day.

= \(\frac{6}{12} \text { part }\)

= \(\frac{1}{2} \text { part }\)

∵ Three of us can do \(\frac{1}{2}\) part in 1 day.

∴ Three of us can do 1 part in \(1 \div \frac{1}{2}\) days = 2 days.

If the three of us work together, then in 2 days we will complete the work.

Question 3. Abani and Anawar complete a work separately in 20 and 25 days respectively. After 10 days of their working together, they both left; then Sukhen came and completed the remaining work in 3 days. If Sukhen alone would do the work, let’s calculate how many days he would take to complete the work.

Solution:

Given

Abani and Anawar complete a work separately in 20 and 25 days respectively. After 10 days of their working together, they both left; then Sukhen came and completed the remaining work in 3 days.

Abani works \(\frac{1}{20}\) part in 1 day. Anawar works \(\frac{1}{25}\) part in 1 day.

Both of them together can do \(\left(\frac{1}{20}+\frac{1}{25}\right)\) part in 1 day

= \(\frac{5+4}{100} \text { part }\)

= \(\frac{9}{100} \text { part }\)

∴ Both of them together can do in 10 days = \(10 \times \frac{9}{100} \text { part }\)

= \(\frac{9}{10} \text { part }\)

Remaining work = \(\left(1-\frac{9}{10}\right) \text { part }\)

Remaining work = \(\frac{10-9}{10} \text { part }=\frac{1}{10} \text { part }\)

Remaining work =  Sukhen does \(\frac{1}{10}\)part in 3 days.

∴  Sukhen does 1 part in 3 x 10 days.

∴ Sukhen does 1 part in 30 days. If Sukhen works alone, then he will complete the work in 30 days.

Question 4. There are two pipes for taking water from the municipality water tank. The tank can be made empty in 4 hours by the two pipes separately. If both the pipes remain opened, let’s calculate when the full tank will be empty.

Solution:

Given

There are two pipes for taking water from the municipality water tank. The tank can be made empty in 4 hours by the two pipes separately. If both the pipes remain opened

Both pipes will together empty

Both pipes can empty part in 1 hour.

= \(\left(\frac{1}{4}+\frac{1}{4}\right) \text { part }\)

= \(\frac{1+1}{4} \text { part }\)

= \(\frac{2}{4} \text { part }\)

= \(\frac{1}{2} \text { part }\)

∵ Both pipes can empty \(\frac{1}{2}\)part in 1 hour.

∴ Both pipes can empty 1 part in 2 hours. If both pipes are opened then in 2 hours the full tank will be empty.

Question 5. In our tank there are 3 pipes. With these three pipes separately the tank can be filled up in 18, 21 and 24 hours respectively

1. If the 3 pipes remain open together, let’s make proportion and calculate when the tank will be filled up with water.

Solution:

Given

In our tank there are 3 pipes. With these three pipes separately the tank can be filled up in 18, 21 and 24 hours respectively

If the three pipes remain open then in 1 hour the tank will fill up

= \(\left(\frac{1}{18}+\frac{1}{21}+\frac{1}{24}\right)\) part

= \(\frac{28+24+21}{504}\) part

= \(\frac{73}{504}\) part

In mathematical language the problem is

WBBSE Solutions For Class 8 Chapter 17 Time And Work Part Of Tank And Time

If part of the tank increases, time will increase.

∴ Here is direct proportion.

∴ \(\frac{\frac{73}{504}}{1}=\frac{1}{?}\)

or, \(\frac{73}{504}=\frac{1}{?}\) hrs.

∴ Required time = \(\frac{504}{73}\) hrs.

= \(6 \frac{66}{73}\) hrs.

If the three pipes remain open then in \(6 \frac{66}{73}\) hours the tank will be filled up.

2. If the first two pipes would remain open, let’s calculate the time required to fill up the tank with water.

Solution:

First two pipes will fill in 1 hour \(\left(\frac{1}{18}+\frac{1}{21}\right)\)

= \(\frac{7+6}{126}\)

= \(\frac{13}{126}\)

First two pipes fill up \(\frac{13}{126}\) part in 1 hour.

∴ First two pipes will fill 1 part in \(1 \div \frac{13}{126}\) hours

= \(\frac{126}{13}\) hours

= \(9 \frac{9}{13}\) hours

If first two pipes remain open, tank will require \(9 \frac{9}{13}\) hours to fill up.

Question 6. Rehana’s tank can be filled up in 30 minutes for by the municipality water supply pipe, opening all the taps, three hours. The can work for 4 hours with the water of the full tank. In one day the water supply pipe remains open for 25 minute. Let’s calculate how long they will work with that water.

Solution:

Given

Rehana’s tank can be filled up in 30 minutes for by the municipality water supply pipe, opening all the taps, three hours. The can work for 4 hours with the water of the full tank. In one day the water supply pipe remains open for 25 minute.

∵ Municipality pipe fills up 1 part in 30 minutes.

∴ Municipality pipe fills up \(\frac{1}{30}\) part in 1 minutes.

∴ Municipality pipe fills up \(\frac{25}{30}\) part in 25 minutes.

∴ Municipality pipe fills up \(\frac{5}{6}\) part in 25 minutes.

In mathematical language the problem is –

WBBSE Solutions For Class 8 Chapter 17 Time And Work Part Of Tank And Time To Work With Water

If part of tank decreases, time will decrease.

∴ Here is direct proportion.

∴ \(\frac{1}{\frac{5}{6}}=\frac{4}{?}\)

or, \(\frac{6}{5}=\frac{4}{?}\)

∴ Required time = \(\frac{4 \times 5}{6}\) hours

= \(\frac{10}{3}\) hours

= \(3 \frac{1}{3}\) hours

With that water they can work for \(3 \frac{1}{3}\) hours.

Question 7. Ruma and Rohit can complete a work in 20 days. Rohit and Sobha can complete that work in 15 days. Let’s calculate in how many days they will complete the work together.

Solution:

Given

Ruma and Rohit can do \(\frac{1}{20}\) part in 1 day. Rohit and Sobha can do in 1 day. Ruma and Sobha can do \(\frac{1}{15}\) part in 1 day. Ruma and Sobha can do \(\frac{1}{20}\) part in 1 day.

∴ 2 × (Ruma + Rohit + Sobha) can do \(\left(\frac{1}{20}+\frac{1}{15}+\frac{1}{20}\right)\)  part in 1 day part

= \(\frac{3+4+3}{60}\) part

= \(\frac{10}{60}\) part

= \(\frac{1}{6}\) part

∴ Ruma + Rohit + Sobha can do \(\frac{1}{2 \times 6}\) part in 1 day

= \(\frac{1}{12}\)

∵ Three of then can do \(\frac{1}{12}\) part in 1 day.

∴ All three together can complete the work in 12 day.

∴ Ruma can do \(\left(\frac{1}{12}-\frac{1}{15}\right)\) part in 1 day

= \(\frac{5-4}{60} \text { part }\)

= \(\frac{1}{60} \text { part }\)

Ruma can do \(\frac{1}{60} \text { part }\) part in 1 day.

∴ Ruma can complete the work alone in 60 days.

Ruma can do \(\left(\frac{1}{12}-\frac{1}{20}\right)\) part in 1 day

= \(\frac{5-3}{60}\) part

= \(\frac{2}{6}\) part

= \(\frac{1}{30}\) part

Rohit can do \(\frac{1}{30}\) part in 1 day.

∴ Rohit can complete the work alone in 30 days.

Sobha can do in 1 day \(\left(\frac{1}{12}-\frac{1}{20}\right)\) part

= \(\frac{5-3}{60}\)

= \(\frac{2}{6} \)

=\(\frac{1}{30} \)

Sobha can complete \(\frac{1}{30}\) part in 1 day.

∴ Sobha can do the work alone in 30 days.

Question 8. Alok, Kalam and Joseph individually can complete a work in 10, 12 and 15 days respectively. They started doing the work together. After 3 days Kalam had to go. Let’s make proportion and calculate in how many days Alok and Joseph will complete the remaining work.

Solution:

Given

Alok, Kalam and Joseph individually can complete a work in 10, 12 and 15 days respectively. They started doing the work together. After 3 days Kalam had to go.

Alok can do \(\frac{1}{10}\) part in 1 day.

Kalam can do \(\frac{1}{12}\) part in 1 day.

Joseph can do \(\frac{1}{15}\) part in 1 day.

All three together can do \(\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\right)\)

= \(\frac{6+5+4}{60}\)

= \(\frac{15}{60}\) part

= \(\frac{1}{4}\) part

All three together can do 3 × \(\frac{1}{4}\) part =\(\frac{3}{4}\) part in 3 days.

Remaining work \(\left(1-\frac{3}{4}\right)\) part

= \(\frac{4-3}{4}\) part = \(\frac{1}{4}\)

Alok and Joseph can do together \(\left(\frac{1}{10}+\frac{1}{15}\right)\) part in 1 day

= \(\frac{3+2}{30}\) part

= \(\frac{5}{30}\) part

= \(\frac{1}{6}\) part

Alok and Joseph can together work \(\frac{1}{6}\) part in 1 day.

∴ Alok and Joseph can together work 1 part in 6 days.

∴ Alok and Joseph can together work \(\frac{1}{4}\)  part in \(\frac{1}{6}\)×6 days

= \(\frac{3}{2}\) days

= \(1 \frac{1}{2}\) days

Alok and Joseph will complete the remaining work in \(1 \frac{1}{2}\)days.

Question 9. Mary and David can do a work individually in 10 days and 15 days respectively. At first Mary alone worked for 4 days, then David alone worked for 5 days and left. Maria came and completed the remaining work in 4 days. If Mary, David and Maria would work together, let’s calculate in how many days they woild complete the work.

Solution:

Given

Mary and David can do a work individually in 10 days and 15 days respectively. At first Mary alone worked for 4 days, then David alone worked for 5 days and left. Maria came and completed the remaining work in 4 days.

Mary can do \(\frac{1}{10}\) part in 1 day.

∴ Mary can do \(\frac{4}{10}\) part = \(\frac{2}{5}\) part in 4 days.

David alone can do \(\frac{1}{15}\) part in 1 day.

∴ David alone can do \(\frac{5}{15}\) part = \(\frac{1}{3}\) part in 5 days.

Both can do \(\left(\frac{2}{5}+\frac{1}{3}\right)\) part of the total work

= \(\frac{6+5}{15}\) part

= \(\frac{11}{15}\) part

Remaining work = \(\left(1-\frac{11}{15}\right)\)

= \(\frac{15-11}{15}\) part

= \(\frac{4}{15}\) part

Maria can do \(\frac{4}{15}\) part in 4 days.

∴ Maria can do  \(\frac{4}{4 \times 15}\) part = \(\frac{1}{15}\) part in 1 day.

Mary, David and Maria together can do \(\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{15}\right)\) part in 1 day.

= \(\frac{3+2+2}{30}\) part

= \(\frac{7}{30}\) part

All three can do \(\frac{7}{30}\) part of the work in 1 day.

∴ All three can do 1part of  the work in \(\frac{30}{7}\) days.

∴ All three can complete the work in \(4 \frac{2}{7}\) days.

If Mary, David and Maria do the work together, then in \(4 \frac{2}{7}\) days it will be completed.

Question 10. A municipality made a tank for water preservation and added pumps with it. The pumps separately can fill the empty tank in 16, 20 & 30 hours respectively. Today when the 3 pumps started together at 7 am, \(\frac{1}{3}\) part of the tank was filled with water. After 1 hour 36 minutes the first pump, and after 2 more hours, the third pump were stopped.

Solution:

Given

A municipality made a tank for water preservation and added pumps with it. The pumps separately can fill the empty tank in 16, 20 & 30 hours respectively. Today when the 3 pumps started together at 7 am, \(\frac{1}{3}\) part of the tank was filled with water. After 1 hour 36 minutes the first pump, and after 2 more hours, the third pump were stopped.

First pump can fill \(\frac{1}{16}\) part in 1 hour.

Second pump can fill \(\frac{17}{20}\) part in 1 hour.

Third pump can fill \(\frac{1}{30}\) part in 1 hour.

All three pumps can fill

= \(\left(\frac{1}{16}+\frac{1}{20}+\frac{1}{30}\right)\) part in 1 hour

= \(\frac{15+12+8}{240}\) part

= \(\frac{35}{240}\) part

= \(\frac{7}{48}\) part

1 hour 36 minutes = 96 minutes

∵ All three pumps fill = \(\frac{7}{48}\) part in 60 minutes.

∴ All three pumps fill \(\frac{7 \times 96}{48 \times 60}\) part in 96 minutes.

= \(\frac{7}{30}\)

Second and third pump fill = \(\left(\frac{1}{20}+\frac{1}{30}\right)\) part in 1 hour.

= \(\frac{3+2}{60}\) part

= \(\frac{5}{60}\) part

= \(\frac{1}{12}\) part

∴ Second and third pump fill = \(2 \times \frac{1}{12}\) part part in 2 hours.

= \(\frac{1}{6}\) part

Now, the filled part of the tank = \(\left(\frac{1}{3}+\frac{1}{30}+\frac{1}{6}\right)\) part

= \(\frac{10+7+5}{30}\) part

= \(\frac{22}{30}\) part

= \(\frac{11}{15}\) part

Remaining to be filled = \(\left(1-\frac{11}{15}\right)\) part

= \(\frac{15-11}{15}\) part

= \(\frac{4}{15}\) part

First pump fills 1 part in 20 hours.

∴ Second pump will fill

= \(\frac{4}{15}\) part in

= \(\frac{4}{15} \times 20\) hours

= \(\frac{16}{3}\) hours

= 5 hours 20 minutes

1. Let’s calculate when the tank was totally filled up.

Solution:

Time required to fill the tank fully

= 1 hour 36 minutes + 2 hours + 5 hours 20 minutes

= 8 hours 56 minutes

Morning 7 am + 8 hours 56 minutes = Evening 3 hours 56 minutes

The tank will be filled completely at 3 o’clock 56 minutes.

2. Let’s calculate how much of the tank the second pump filled up.

Solution:

Time of second pump

= 1 hours 36 minutes + 2 hours + 5 hours 20 minutes

= 8 hours 56 minutes

= (8 x 60 + 56) minutes (480 + 56) minutes

= 536 minutes

Second pump can fill up = \(\frac{1}{20}\) part in 60 minutes.

∴ Second pump can fill up = \(\frac{1}{20 \times 60}\) part in 1 minutes.

∴  Second pump can fill up = \(\frac{1 \times 536}{20 \times 60}\) part in 536 minutes.

∴ Second pump can fill up = \(\frac{67}{150}\) part in 536 minutes.

Second pump filled up = \(\frac{67}{150}\) part of the tank.

3. When the third pump is stopped, let’s calculate how much of the tank was filled with water.

Solution:

When third pump was was stopped, \(\frac{4}{15}\) part was empty.

∴ Filled up part = \(\left(1-\frac{4}{15}\right)\)

= \(\frac{15-4}{15}\)

= \(\frac{11}{15}\)

When the third pump was stopped, \(\frac{11}{15}\) part of the tank was filled with water.

Question 11. My friend Rina can do garden work alone in 4 hours. I can do that work alone in 9 hours. But if we together do that work, let’s calculate how much time will be needed with number.

Solution: 

Given

My friend Rina can do garden work alone in 4 hours. I can do that work alone in 9 hours.

Rina can do \(\frac{1}{4}\) part of the work in 1 hour.

I can do\(\frac{1}{3}\) part of the work in 1 hour.

Both can do together

= \(\left(\frac{1}{4}+\frac{4}{15}\right)\) part of the work in 1 hour

= \(\frac{3+4}{12}\) part

= \(\frac{7}{12}\) part

Both can do \(\frac{7}{12}\) part of the work in 1 hour.

∴ Both can do the work in \(\frac{12}{7}\) hours

= \(1 \frac{5}{7}\) hours.

Both can do the work in = \(1 \frac{5}{7}\) hours.

WBBSE Solutions For Class 8 Maths Chapter 21 Construction Of Triangles

Construction Of Triangles

If I draw a triangle, lengths of whose two sides are 5 cm and 4 cm and the measurement of the angle opposite to the side of length 4 cm is 45°, then let’s see what type of triangle I shall get.

It is seen that it is not possible to draw any triangle with these conditions.

WBBSE Solutions Class 8 Chapter 21 Construction Of Triangles Measurement Of The Angles

  1. Firstly, 30° arid 45° are drawn and a straight line 6 cm long is drawn.
  2. Now straight line AX is drawn and from it 6 cm long is cut off.
  3. On points, A and B of straight line AB, two angles equal to 45° ∠YAB and ∠ZBX are drawn respectively.
  4. Now on point B equal to 30° is drawn AY straight line on the same side of BZ ZPBZ. PB and AY cut each other at C point

In ΔABC = 6 cm

Read and Learn More WBBSE Solutions For Class 8 Maths

∠CAB = 45 and ∠ACB = 30°

Proof: ∠XBZ = ∠XAY (By construction)

∴ ∠BZ//AY (∵ Corresponding angles)

∴ ∠ZBP = Alternate angle ∠BCA

∵ ∠ZBP = 30°

∴∠BCA = 30°

∴ In ΔABC ∠CAB = 45°, ∠ACB = 30°

and the opposite side of 30° is AB = 6 cm

Construction Of Triangles Exercise 21.1

Question 1. If I draw a triangle whose 2 sides are 5 cm and 4 cm and the measurement of the angle opposite to the side of length 4 cm is 45°, then let’s see what type of triangle I shall get. It is seen that it isn’t possible to draw any triangle with these conditions.

Solution:

Given

If I draw a triangle whose 2 sides are 5 cm and 4 cm and the measurement of the angle opposite to the side of length 4 cm is 45°

  1. With the help of scale two 5 cm long straight lines are drawn.
  2. With the help of a scale, a ray AX is drawn. On the ray off. Taking AX at point A an angle of 45° ∠XAY is drawn.
  3. From the ray A, X a straight line AB equal to 5 cm is cut off. Taking B as the centre, taking a radius equal to the straight line 4 cm, an arc is drawn which cuts ray AX at point C. B, and C are joined. An ABC is drawn whose AB = 5 cm, ∠BAC = 45° and BC = 4 cm and the angle opposite to the side BC ∠BAC = 45°.

WBBSE Solutions Class 8 Chapter 21 Construction Of Triangles Measurement Of The Angle Opposite To The Side Of Length

Question 2. But why is it so? Sometimes we are getting one triangle, sometimes two triangles and sometimes no triangle. It is seen that the perpendicular distance from B on the AX line segment is BM = cm = I say.

Solution:

WBBSE Solutions Class 8 Chapter 21 Construction Of Triangles Perpendicular Distance From B On AX Line Segment

We see that perpendicular distance AX from point B to the straight line

BM = 3.6 cm l = let

a = 5cm

b = 4cm

We see that if b > a then |a  1 | triangle can be drawn.

If l < b < a then 2 triangles can be drawn.

If b = a then 1 triangle can be drawn.

If b < l then no triangle can be drawn.

If b, = l then 1  triangle can be drawn.

If a = b, i.e., try to form a triangle whose two sides a = 5 cm, b = 5 cm and the opposite angle to the 5 cm long side is ∠x= 100°

WBBSE Solutions For Class 8 Maths Chapter 21 Construction Of Triangles

Question 3. Let’s see whether such a triangle can be drawn.

Solution:

WBBSE Solutions Class 8 Chapter 21 Construction Of Triangles Perpendicular Distance From B On AX Line Segment 2

straight line AC is drawn and on point A equal to angle 100°, ∠BAC is drawn.

From the AD side, a 5 cm long AB is cut off. Now taking B as the centre 5 cm straight line equal to taking as radius, on a point on AC an arc is drawn then it is seen that cuts at only on one point of AC, A. So it is not possible to draw a triangle.

Question 4. If a < b, i.e., try to form a triangle whose two sides a = 5 cm, b = 4 cm and the opposite angle of 4the  cm long side is x = 100°. Let’s see if such a triangle can be drawn.

Solution:

WBBSE Solutions Class 8 Chapter 21 Construction Of Triangles Opposite Angle Of 4 cm Long Side

Given

if a < b, i.e., try to form a triangle whose two sides a = 5 cm, b = 4 cm and the opposite angle of 4 cm long side is x = 100°.

A straight line AC is drawn and on point A equal to angle 100°, ∠BAC is drawn. From the AD side, a 5 cm long AB is cut off.

Now taking point B as a centre and taking a radius length equal to a 4 cm straight line AC cut an arc on any point then it is seen that the AC arc doesn’t cut at any point.

So it is not possible to construct a triangle.

Construction Of Triangles Exercise 21.2

Question 1. Let’s draw a triangle whose two sides are 6 cm and 7 cm and the measurement of the angle opposite to the side of length 7 cm is 60°. Let’s write what will be the measurement of the sides to form a triangle.

Solution:

WBBSE Solutions Class 8 Chapter 21 Construction Of Triangles Measurement Of The Sides To Form A Triangle

Measurement of the sides to form the triangle

A straight line AX is drawn and on point A equal to angle 60°, ∠DAX is drawn. From side AD, AB is cut equal to 6 cm long. Taking B as the centre and equal to 7 cm long taking as radius point AX is cut at C. B, and C are joined. ABC is the required triangle.

From point B to AC, if the length of the side is more than the length of the perpendicular drawn then two triangles will be formed.

Question 2. Let’s construct a triangle whose lengths of two sides are 6 cm and 9 cm and the measurement of the angle opposite to the side of length 9 cm is 105°. Let’s write for what length of sides we will not be able to construct two triangles.

Solution:

WBBSE Solutions Class 8 Chapter 21 Construction Of Triangles Measurement Of The Angle Opposite

A straight line AX is drawn and at AX on point A equal to angle 105°, ∠DAX is drawn. From side AD AB is cut equal to 6 cm long. Taking B as centre equal to 9 cm length taking as radius AX is cut at point C. B, and C are joined. ABC is the required triangle.

WBBSE Solutions For Class 8 Maths Chapter 20 Geometrical Proofs

Geometrical Proofs

Mitali gave me two sticks of length 3 cm and 5 cm respectively.

I took a stick of length 2 cm and attached it with sticks given by Mitali and we got a figure like the given figure beside.

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Two Sticks Of Length 3cm And 5cm Respectively

But it is seen that we are not able to form a triangle. Here it is seen – that 3 cm + 2 cm [I] 5 cm (put =/</> ). Jasmin replaced the stick of length 2 cm by a stick of length 3 cm and she tried to form a figure like a triangle and she is able to make the figure.

Here it is seen that 3 cm + 3 cm > 5 cm ( put =/</>)

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Triangle With Three Sticks Respectively

Sneha made a figure Triangle with three sticks 3 cm, 5 cm and 4 cm respectively.

Here it is seen that 4 cm + 3 cm g] 5 cm ( put =/</>).

I am able to form a shape like a triangle with two sides of lengths 3 cm, 5 cm and a third stick of length 6 cm or 7 cm, i.e., it is seen that the figures like triangles are formed with the sticks of length (3 cm,6 cm & 5 cm) and (3 cm, 7 cm & 5 cm).

Read and Learn More WBBSE Solutions For Class 8 Maths

Sneha, by placing the shapes like this triangle on the exercise book, she got some triangles, of which the sum of length of third sides Big (big / small).

Now Sucheta tried to make a shape like a triangle with the sticks of lengths 9 cm, 3 cm, and 5 cm and she got –

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Sneha By Placing The Shapes Like This Triangle On The Excerise Book

Here the sum of lengths of any two sides is not greater than the length of the third side.

We got,

1. The length of the sticks which makes a figure like a triangle with the sticks of length 5 cm and 3 cm is Big (less / more) than 5 cm – 3 cm = 2 cm and then small (less/more) (5 + 3) cm = 8 cm

2. The sum of the length of any two sides of a triangle is then the length of the third side Big (less/more)

Now Anik gave two sticks of length 2 cm and 4 cm.

Now let’s try to make a figure like a triangle with sticks of different lengths and let’s see what I get.

It is seen that

1. It is possible to construct a figure like a triangle with the (4-2) cm = 2 cm Big (less/more) long stick and (4 + 2) cm = 6 cm small (less/more) long stick 2 cm and 4 cm long sticks.

2. Again it is also seen that the sum of the lengths of any two sides is Big (less/more) than the length of the third side.

WBBSE Solutions For Class 8 Maths Chapter 20 Geometrical Proofs

Geometrical Proofs Exercise

Question 1. Let’s see what the length of the sticks can make a figure like a triangle with the sticks of length 4 cm and 6 cm.

Solution:

It is possible to make a triangle with sticks of length greater than (6-4) cm = 2 cm and less than less than (6+4) cm = 10, i.e., from 3 to 9 cm triangle can be made with sticks of any length.

Question 2. If the length of two sides of triangle are 3 cm and 6 cm then let’s find the length of the third side that will lie between which two numbers. Let’s write the relation between the sum of the lengths of any two sticks with the length of the third stick.

Solution:

Given

If the length of two sides of triangle are 3 cm and 6 cm then let’s find the length of the third side that will lie between which two numbers.

(6-3) cm = 3 cm greater than

∴ The length of the third side can be 4 cm or 5 cm.

Aminur has drawn some triangles on the exercise books. He said let’s try to find the relation between the length of each side of these triangles.

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Lengths Of Two Sides Is Big

AB = 2.6 cmBC = 4.6 cm, CA = 3.4 cm

AB+BC > CA [Put >/<]

BC+CA > BC [Put >/<]

CA+AB > [put >/>]

Similarly measuring the length of the sides of  ΔPQR & ΔXYZ, it is seen that the sum of the lengths of two sides is Big than the length of third side.

Question 3. Megha has drawn PQR triangle. I will prove with the help of facts that PQ + QR > PR ; OR + RP > PQ and RP + PQ > QR.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Megha Has Drawn PQR Triangle

Given : Let the greatest side of PQR triangle is QR.

We have to prove that RP + PQ > QR.

Construction : From P, the highest point of ΔPQR, PS a perpendicular is drawn on OR which cuts, QR at S i.e., PS ⊥ QR.

Proof : In ΔPSR ∠PSR=1 right angle (Constructed)

∴ ∠PSR is a right angle and ∠SPR is an acute angle.

∴ ∠PSR> ∠SPR

So, PR > SR …………. (1)

In ΔPSQ < PSQ = 1 right angle (∵ PS⊥QR)

∴ ∠PSQ right angle and ∠QPS acute angle ∠

∴ PSQ>∠QPS

So : PQ > QS …………… (2)

by adding equations (1) and (2)

PR + PQ > SR + QS

or, PR + PQ > QR Proved

Geometrical Proofs Exercise

Question 1. Let’s prove logically step by step that the difference of the lengths of any two sides of triangle is less than the length of the third side.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs AD Perpendicular BC

Let the greatest side of ΔABC is BC.

We have to prove that AB-AC<BC.

Construction : From A, the highest point of ΔABC, AD a perpendicular is drawn on BC which cuts BC at D, i.e., AD ⊥ BC

Proof : In ΔABD ∠ADB=1 right angle (∵ AD ⊥ BC)

∴ ∠ADB > ∠BAD

∴ AB > BD ………………..(1)

In ΔADC ∠ADC=1 right angle

∴ ∠ADC is a right angle and ∠DAC is an acute angle.

∴ ∠ADC > ∠DAC

∴ AC > DC ………………(2)

(1) — (2) AB – AC < BD + DC

or, AB – AC < BC Proved

Question 2. Let’s observe the lengths of the following sides and let’s write in which cases it is possible to draw a triangle –

1. (3 cm, 6 cm and 8 cm)

Solution:

Given

3 cm + 6 cm > 8 cm

∴ It is possible to construct a triangle.

2. (8 cm, 6 cm and 15 cm)

Solution:

Given

8 cm + 6 cm < 15 cm

∴ It is not possible to construct a triangle.

3. (2.7 cm, 6.1 cm and 8.8 cm) .

Solution:

Given

2.7 cm + 6.1 cm – 8.8 cm

∴ It is possible to construct a triangle.

4. (2.5 cm, 8 cm and 6 cm) .

Solution:

Given

2.5 cm + 6 cm > 8 cm

∴ It is possible to construct a triangle.

Geometrical Proofs Exercise

Question 1. D is any, point on the side BC of A ABC. Let’s prove that AB + BC + CA > 2AD.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Delta ABC And D Is A Point On BC

Given

In A ABC, D is a point on BC.

Prove that AB + BC + CA > 2AD

Construction : A straight line AD is drawn.

Proof : In ΔABD,

AB + BD > AD …………(1)

(∴ The sum of two sides of a triangle is greater than the third side.)

Again, In ΔADC,

DC + AC > AD …………….(2)

by . adding equations (1) and (2) .

AB + BD + DC + AC > AD + AD

or, AB + BC + AC > 2AD Proved

Question 2. O is any point inside ΔABC. Let’s prove that:

  1. AB + AC > OB + OC
  2. AB + BC + AC > OA + OB + OC

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs AO Is Extended Which Cuts BC At Q

Construction : AO is extended which cuts BC at Q. Similarly BO and CO are extended which cut AC and AB at points R and P respectively.

Proof : In ΔABQ, AB + AQ > BQ >

or, AB + AQ > OB + OQ ………….(1)

In A COQ, OQ + CQ > OC ………..(2)

By adding equations (1) and (2)

AB + AQ + OQ + CQ > OB > OB + OQ + OC

or, AB + (AQ + CQ) > OB + OC

or, AB + AC > OB + OC ……………(3)

∵ BC > OA ………………………(4)

By adding equations (3) and (4)

AB + AC + BC > OB + OC + OA Proved

Question 3. Let’s prove that the perimeter of a triangle is more than the sum of lengths of the medians.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Perimeter Of A Triangle

Let the three medians of A ABC are AD, BE and CF.

We have to prove that AB + BC + CA > AD + BE + CF.

Proof :  ∵ BA + AC > 2AD …………….(1)

BC + AC > 2CF …………………..(2)

and AB + BC > 2BE ………………….(3)

By adding (1), (2) and (3)

AB + AC + BC + AC + AB +BC > 2AD + 2CF + 2BE

or, 2(AB + BC + AC) > 2 (AD + CF + BE)

or, AB + BC + AC > AD + BE + CF Proved

Question 4. P is any point inside A ABC. Let’s prove that,

  1. AB + BP > AB
  2. AB + BC + AC < 2 (AP + BP + CP)

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs In Delta ABC And P Is Any Point

Given

In ΔABC, P is any point. Prove that (1) AP + BP > AB (2) AB + BC + AC < 2 (AP + BC + CP)

Construction : A, P ; B, P and C, P are joined.

Proof: In A ABP

(1) AP + BP > AB (∵ The sum of any two sides of a triangle is greater than the third side.)

In A ABP, AB<AP + BP……………………….. (1)

In A BPC, BC < BP + PC………………. (2)

and in A APC, AC < AP + PC …………….. (3)

by adding (1), (2) and (3)

AB + BC + AC < 2AP + 2BP + 2PC

or, AB + BC + AC < 2 (AP + BP + PC) (2) Proved

Question 6. Let’s prove that the sum of lengths of two diagonals is greater than the sum of the lengths of any two opposite sides of a quadrilateral.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Lengths Of Two Diagonals

Let ABCD is a quadrilateral whose two diagonals are AC and BD. Let’s prove that AC + BD > AD + BC.

Proof : In AOD, OA + OD > AD …………..(1)

In BOC, OB + OC > BC ………….(2)

by adding (1) and (2)

OA + OD + OB + OC > AD + BC or, (OA + OC) + (OB + OD) > AD + BC

or, AC + BD > AD + BC Proved

Question 7. Let’s prove that the sum of lengths of two diagonals of a quadrilateral is more tham the semi-perimeter of the quadrilateral.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Lengths Of Two Diagonals Of A Quadrilateral

Let ABCD is a quadrilateral whose two diagonals AC and BD cut each other at 0.

Proof: In AOD, AO + OD > AD ………………… (1)

In AOB, AO + OB > AB ………………… (2)

In BOC, OB + OC > BC ………………… (3)

In COD, OC + OD > CD ………………… (4)

by adding (1), (2), (3) and (4)

AO + OD + AO + OB + OB + OC + OC + OD > AD + AB + BC + CD

or, 2A0 + 20C + 20D + 20B > AB + BC + CD + AD

or, 2 (A0 + OC) + 2(0D + OB) > AB + BC + CD + AD

or, 2AC + 2BD > AB + BC + CD + AD

or, 2 (AC + BD) > AB + BC + CD + AD

or, AC + BD > (AB + BC + CD + AD) Proved

Question 8. From a point inside a quadrilateral (not on any diagonal) we join the vertices of the quadrilateral. Let’s prove that the sum of lengths of these line segments is greater than the sum of lengths of the diagonals. Now let’s see for which position of the point inside the quadrilateral the sum of lengths of the line segment obtained by the vertices of the quadrilateral with that point will be the smallest.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Inside ABCD Quadrilateral

Given

From a point inside a quadrilateral (not on any diagonal) we join the vertices of the quadrilateral.

Let inside ABCD quadrilateral 0 is any point and AC and BD are the two diagonals.

Let’s prove that: OA + OB + OC + OB > AC + BD

Construction : O, A; O, B; C; and O, D

Proof: From figure (1) In ΔAOC, OA + OC > AC ……………….(1)

and in ΔBOD, OB + OD > BD ……………….(2),

From (1) + (2), OA + OC + OB + OD > AC + BD (Proved)

From figure (2) In ΔAOB, AO + OB > AB ………………….(1)

(The sum of lengths of any two sides of a triangle is greater than length of the third side.)

In A AOC, BO + OC > BC ……………….. (2) (For the above reason)

In A GOD, OC + OD > CD ………………. (3) (For the above reason)

In A ODA, OD + OA > AD ……………………. (4) (For the above reason)

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Inside ABCD Quadrilateral Point

Now, from A + B + C + D

AO+OB+BO+OC+OC+OD+OD+OA>AB+BC+AD

or, 2A0+20C+20B+20D>AB+BC+CD+CA

or, AO+OG+OB+OD> \(\frac{1}{2}\) (AB+BC+GD+GA)………….(A)

From figure (3) In ΔAOD, AO + OD > AD ……………(1)

In ΔAOB, AO + OB > AB ………..(2)

In ΔBOC, OB + OC > BC ………………(3)

In ΔCOD, OC + OD > CD ……………. (4)

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Obtained By Vertices Of Quadrilateral

by adding (1+2+3+4)

AO+OD+AO+OB+OB+OC+OC+OD>AD+AB+BC+CD

or, 2AC + 20C + 20D+20B>AD+AB+BC+CD

or, 2(AO+OC)+2(OD+OB)>AD+AB+BC+CD

or, 2AC =2BD>AD+AB+BC+CD

or, AC+BD> (AB+BC+CD+AD)

AC+BD> (AB+BC+CD+AD) ………………….(B)

From (A) and (B)

AC + BD < OA + OB + OC + OD

So we see that if the point inside the quadrilateral is the point of intersection of two diagonals then the sum of lengths of line segments obtained by vertices of quadrilateral with that point will be the smallest.

Salema and Bibhas have made-

I drew these figures on a thick art paper and named them. I measured the interior angles b>Ca protractor and got :

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Salema And Bibhas

Measuring by protractor, we see, In ΔABC ∠ABC = 60 degree ∠ABC = 60 degree and ∠ACB= 60 degree

By measuring angles of quadilateral HAND we see ∠AHD = 90 degree ∠HAN = 90 degree ∠AND = 90 degree ∠HDN = 90 degree

By measuring angles of pentagon CAMEL we see ∠ACL = 90 degree ∠CLE = 110 degree ∠LEM =100 degree ∠EMA = 90 degree and ∠MAC = 140 degree

By measuring angles of quadrilateral REST we see

∠ERT = 100 degree ∠RES = 80 degree ∠EST = 110 degree and ∠RTS = 70 degree

Now I add the measurement of the interior angles of each polygon and then let’s try to find a general formula.

Sum of the measurement of the interior angles of ΔABC = ∠BAC + ∠ABC +∠ACB = 180 degree = 2 right angles

Sum of the measurement of the interior angles of HAND = ∠AHD + ∠HAN + ∠AND + ∠BAC = 360 degree = 4 right angles

Sum of the interior angles of CAMEL = ∠ACL + ∠CAM + ∠AME + ∠MEL + ∠ELC = 540 degree = 6 right angles

Sum of the interior angles of REST = ∠ERS + ∠RES + ∠EST + ∠RTS = 360° degree= 4 right angles

Let’s try to find out the relation between the interior angles and the number of sides of a polygon.

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Polygon Number Of Sides

Question 9. The number of sides of a regular polygon is 18. Let’s find the measurement of each interior angle and each exterior angle.

Solution:

Given

The number of sides of a regular polygon is 18.

Sum of measurements of 18 interior angles of the polygon is = 2 (18-2) × 90°

= 2 × 16 × 90°= 2880°

∴ Measure of each interior angle is = 2880° -18 = 160°

Sum of measurements of all exterior angles is = 360°

∴ Measurement of each exterior angle is = 360, 18=20°

Alternatively, measurement of 1 interior angle + 1 measurement of exterior angle = 180°

∴ Measurement of 1 exterior angle = 360° 18 = 20°

Measurement of 1 interior angle = 180° – 20° = 160°

Question 10. Measurement of each exterior angle of a regular polygon is 144°. Let’s find the number of sides of the polygon.

Solution:

Given

Measurement of each exterior angle of a regular polygon is 144°.

Measurement of 1 interior angle + measurement of 1 exterior angle = 180

∴ Measurement of 1 exterior angle = 180°

-144° = 36°

∴ Number of sides of the regular polygon = 360° ÷ 360° = 10

Geometrical Proofs Exercise

Question 1. Let’s prove logically the sum of four interior angles of a quadrilateral is 360°.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs Interior Angles

Let ABCD is a quadrilateral.

Let’s prove that ∠A + ∠B + ∠C + ∠D = 360°

Construction : Diagonal BD is drawn.

Proof : In ΔABD, ∠A + ∠ABD + ∠ADB = 180°

(∵ Sum of all angles of a triangle is 180°)

Again, In ΔBCD ZBDC + ZCBD + ZC = 180°

ΔABD+ ABCD= ∠A+ ∠ABD+ ∠ADB+ ∠BDC+ ∠CBD+ ∠C= ∠A+ ∠B+ ∠C+ ∠D

∴ ΔABD + ABCD = ∠A+∠B+∠C+∠D = 180°+180°

∴ ABCD =∠A+ ∠B+ ∠C+ ∠D= 360° Proved

Question 2. Let’s write by calculation the sum of the measurements of eight interior angles of an octagon.

Solution:

Sum of measurements of the interior angles of an octagon = 2 (8-2) × 90°

Sum = 2 (8-2) × 90°

= 2 × 6 × 90°

= 10802°

The sum of the measurements of eight interior angles of an octagon = 10802°

Question 3. Let’s write the measurement of each interior and enterior angle of a regular polygon with 10 sides.

Solution:

Sum of measurements of 10 interior angles of a decagon = 2 (10-2) × 90° .

= 2 × 8 × 90° = 1440°

∴ Measurement of each interior angle = 1440° ÷ 10 = 144°

Measurement of each exterior angle = \(\frac{360^{\circ}}{10}\)= 36°

Question 4. Let’s write by calculation the number of sides of a regular polygon where measurement of each angle is 120°.

Solution:

Each interior angle of the polygon = 120°

∴ Measurement of each exterior angle of the polygon = 180° – 120°

Number of sides of the regular polygon = \(\frac{360^{\circ}}{10}\) = 6

Geometrical Proofs Exercise

Question 1. Let’s write the sum of the interior angles of the following polygons:

  1. Pentagon
  2. Hexagon
  3. Heptagon
  4. Octagon
  5. Decagon
  6. A polygon with 12 sides.

1. Pentagon

Solution:

Sum of measurements of the interior angles of a pentagon

= 2 (5-2) × 90°

= 2 × 3 × 90° = 540°

pentagon = 540°

2. Hexagon

Solution:

Sum of measurements of the interior angles of a hexagon

= 2(6-2) × 90°

= 2 × 4 × 90°

= 720°

Hexagon = 720°

3. Heptagon

Solution:

Sum of measurements of the interior angles of a heptagon = 2

(7-2) × 90°

= 900°

Heptagon = 900°

4. Octagon

Solution:

Sum of measurements of the interior angles of an octagon

= 2 (8-2) × 90°

= 2 × 6 × 90° = 1080°

Octagon = 1080°

5. Decagon

Solution:

Sum of measurements of the interior angles of a decagon = 2

(10-2) × 90°

= 2×8×90°

= 1440°

Decagon = 1440°

6. Polygon with 12 sides

Solution:

Sum of measurements of the interior angles of a polygon with 12 sides

= 2 (12-2) × 90°

= 1800°

Polygon with 12 sides = 1800°

Question 2. Three angles of a quadrilateral are 104.5°, 65° and 72.5°; find the measurement of the fourth angle.

Solution:

Given

Three angles of a quadrilateral are 104.5°, 65° and 72.5°

Sum of measurements of three angles of a quadrilateral.

= 104.5° + 65° + 72.5°

= 242°

Sum of measurements of all four angles of a quadrilateral = 360°

∴ Measurements of the 4th angle of the quadrilateral = 360° – 242°

= 118°

Question 3. Four angles of a pentagon are respectively 65°, 89°, 132° and 116°. Let’s write the measurement of the fifth angle.

Solution:

Given

Four angles of a pentagon are respectively 65°, 89°, 132° and 116°.

Sum of 4 angles of a pentagon = 65° + 89° + 132° + 116° = 402°

Sum of all the interior angles of a pentagon = 2 (5- 2) × 90° .

= 2×3×90° =540°

Measurement of the 5th angle of the pentagon = 540°-402° = 138°

Question 4. Let’s write whether the measurement of 3 angles of a convex v quadrilateral being 68°, 70° and 75° is possible or not.

Solution:

Sum of 3 angles of a quadrilateral = 68° + 70° + 75° = 213°

∵ Sum of 4 angles of a quadrilateral = 360°

∵ 213° < 360°

∴ The three angles of a convex quadrilateral can be respectively 68°, 70° and 75°

Question 5. Let’s write whether the five angles of a convex hexagon 120°, 70°, 95°, 78° and 160° is possible or not.

Solution:

Sum of 5 angles of a hexagon = 120°+70°+95°+78°+160° = 523°

Sum of six angles of a hexagon = 2 (6-2) × 90°

= 2 × 4 × 90°

= 720°

∵ Measure of the 6th angle of a hexagon = 720° – 523° = 197°

∵ 197° > 180°, which is impossible.

∴ The five angles of a convex hexagon can’t be respectively 120°,

70°, 95°, 78° and 160°. *

Question 6. Let’s write the measurement of each interior and exterior angle of the following polygons :

1. Pentagon

Solution:

Sum of measurements of the interior angles of a pentagon

= 2 (5-2) × 90°

= 2 × 3 × 90°

= 540°

∴ Measurement of each interior angle of a regular pentagon

= 54° ÷ 5

= 108°

Interior angle + Exterior angle = 180°

Measurement of each interior angle of a reguiar pentagon= 180°

108° =72°

2. Hexagon

Solution:

Sum of measurements of the interior angles of a hexagon = 2 (6-2) × 90°

= 2 × 4 × 90°

= 720°

∴ Measurement of each interior angle of an regular hexagon = 720°÷6 = 120°

∴ Measurement of each exterior angle of a regular hexagon = 180° -120° = 60°

3. Octagon

Solution:

Sum of measurements of the interior angles of an Octagon

= 2 (8-2) x 90° = 2 × 6 × 90° = 1080°

∴ Measurement of each interior angle = 1080° + 8 = 135°

∴ Measurement of each exterior angle of the octagon = 180° – 135° = 45°

4. Polygon with 9 sides

Solution:

Sum of the interior angles of a polygon with 9 sides = 2 (9-2) × 90° = 1260° .

∴ Measurement of each interior angle of a polygon with 9 sides = 1260°-9= 140° –

∴ Measurement of each exterior angle of a polygon with 9 sides = 180°-120° = 40°

5. Decagon

Solution:

Sum of measurements of the interior angles of a Decagon

= 2 (10-2) x 90° = 2 × 8 × 90° = 1440°

∴ Measurement of each interior angle of a polygon with 10 sides = 1440° – 10 = 144°

∴ Measurement of each exterior angle of a regular polygon with 10 sides

= 180°-144° = 36°

6. Polygon with 18 sides

Solution:

Sum of measurements of the interior angles of a regular polygon with 18 sides.

= 2 (18-2) × 90°

= 2 × 16 × 90°

= 2880°

∴ Measurement of each interior angle of a polygon with 18 sides = 2880° ÷ 18 = 160°

∴ Measurement of each exterior angle of a polygon with 18 sides = 180°-160°=20°

Question 7. Let’s find whether the following measurements are possible or not for an exterior angle of a regular polygon – (write yes /no):

  1. 10°
  2. 13°
  3. 18°
  4. 35°

1. 6°

Solution:

The measurement of each exterior angle of a regular polygon can be 6°

2. 10°

Solution:

The measurement of each exterior angle of a regular polygon can be 10°.

3. 13°

Solution:

The measurement of each exterior angle of a regular polygon can be .13°.

4. 18°

Solution:

The measurement of each exterior angle of a regular polygon can be 18°.

5. 35°

Solution:

The measurement of each exterior angle of a regular polygon can be 35°.

Question 8. Let’s find whether the following measurement of each interior angle is possible or not for a regular polygon

Solution:

  1. 80°
  2. 100°
  3. 120°
  4. 144°
  5. 155°
  6. 160°

1. 80°

Solution:

The measurement of each interior angle of a regular polygon can be 80° .

2. 100°

Solution:

The measurement of each interior angle of a regular polygon can be 100°.

3. 120°

Solution:

The measurement of each interior angle of a regular polygon can be 120°.

4. 144°

Solution:

The measurement of each interior angle of a regular polygon can be 144°.

5. 155°

Solution:

The measurement of each interior angle of a regular polygon can be 155°.

6. 160°

Solution:

The measurement of each interior angle of a regular polygon can be 160°.

Question 9. Each exterior angle of a regular polygon is 60°. What is the number of sides of it?

Solution:

Given

Each exterior angle of a regular polygon is 60°.

The measurement of each exterior angle of a regular polygon can be = 60°

∴ Number of sides of the regular polygon = \(\frac{360^{\circ}}{60^{\circ}}\)= 6

Question 10. Each interior angle of a regular polygon is 135°. What is the number of sides of that polygon?

Solution:

Given

Each interior angle of a regular polygon is 135°.

The measurement of each interior angle of a regular polygon can be = 135°

∴ The measurement of each exterior angle of a regular polygon can be = 180° – 135° = 45°

∴ Number of sides of the regular polygon = \(\frac{360^{\circ}}{45^{\circ}}\)= 8

Question 11. The ratio of measurements of an interior angle and an exterior angle of a polygon is 3:2. Let’s write the number of sides of the polygon.

Solution:

Given

Ratio of measurements of each interior and exterior angles of a regular polygon = 3:2

Sum of ratio = 3 + 2 = 5

∴ Measurement of exterior angle = \(\frac{2}{5}\) × 180° = 72°

∴ Number of sides of the regular polygon = \(\frac{360^{\circ}}{72^{\circ}}\) = 5

Question 12. The sum of measurements of all exterior angles of a polygon is 18000°. Let’s write the number of sides of the polygon.

Solution:

Given

The sum of measurements of all exterior angles of a polygon is 18000°.

Let the number of sides’ of the polygon = n

Sum of measurements of all the interior angles of the polygon = 2 (n-2) × 9

B.T.P.,

2 (n-2) × 90° = 1800°

or’ \(n-2=\frac{1800^{\circ}}{2 \times 90^{\circ}}\)

or, n – 2 = 10

or, n -2 = 10 + 2 = 12

∴ Number of sides of the polygon = 12

Question 13. Measurement of each of 5 interior angles is 172° and the measurement of each of the other interior angles of a polygon is 160°. Let’s write number of sides of the polygon.

Solution:

Given

Measurement of each of 5 interior angles is 172° and the measurement of each of the other interior angles of a polygon is 160°.

Let the number of sides of the polygon be n.

∴ Sum of measurements of the interior anqles of the polyqon = 2 (n-2) × 90°

Total of five interior angles ofthe polygon = 5 × 172° = 860°

∴ Measurements of the remaining interior angles = (n- 5) × 160°

2(n-2)×90° = 860°-(n-5)×160°

or, 2(n-2)×90°-(n-5)×160°= 860°

or, 20°{9(n-2)-8(n-5)} = 860°

or, 9x-18-8n+40 = 43

or, n = 43+18-40

or, n = 21

∴ n = 21

∴ Number of sides of polygon = 21

Question 14. ABODE is a regular pentagon. Let’s prove the AABC is an isosceles triangle, BE and CD are two parallel line segments.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs ABODE Is A Regular Pentagon

Given : ABODE is a regular pentagon.

We have to prove that :

1. ABC is an isoceles triangle and

2. BE//CD.

Proof : In ΔABC

AB = BC (∵ABCDE is a regular pentagon)

∴ ΔABC is an isoceles triangle.

Question 15. ABCDEF is a regular hexagon. The bisector of ∠BAF interects DE at X. Let’s write the measurement of ∠AXD.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs ABCDEF Is A Regular Hexagon

Given

∴ ABCDEF is a regular hexagon.

∴ Measurement of each of its interior angles

= \(\frac{2(6-2) \times 90^{\circ}}{6}\) = \(\frac{2 \times 4 \times 90^{\circ}}{6}=120^{\circ}\)

∴ ∠B= ∠C= ∠D = 120°

∴ ∠B + ∠C + ∠D = 3 x 120° = 360°

∠BAX = \(\frac{120^{\circ}}{2^{\circ}}\) = 60° (∵ ∠A is the bisector of AX.)

∵ ABCDX is a pentagon

∴ Sum of measurements of the interior angles of ABCDX = 2(5-2) × 906 = 2 × 3 × 90° = 540°

Sum of measurements of the 4 angles of ABCDX = 360° + 60° = 420°

∴ Pentagon AXD = 540° – 420° = 120°

Manasi constructs some triangles with some sticks like the figure beside. The bases of these triangles are in a same straight line and have same vertex.

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs These Triangles Are In A Same Straight Line And Have Same Vertex

Question 17. Let’s try to measure the lengths of the sides of the triangle, i.e., the lengths of the sides only from the measurement of the angles of the triangle.

Solution:

It is seen that in ΔABC, ∠ACB is an obtuse angle.

∴ ABC is an acute angle (obtuse angle/acute angle)

We know that the opposite side of greater angle is than the opposite side of smaller angle Big (less/more), i.e, AB > AC [ Put >/<]

Again, measuring by protractor we see, side AD is perpendicular to FC, i.e, AD⊥FC

∴ ∠ADB = 90°

∴ In ΔADC,∠ACD acute angle (is an obtuse/acute angle), i.e, AC>AD [Put >/< ]

∴ It is seen the relation between sides, AC, AD and AB is AD < AC < AB.

Geometrical Proofs Exercise

Question 1. Two persons, one of them is coming along south direction to reach to east-west road and the other person is coming along the south-east direction, starting from the same place. Let’s calculate which person reaches first to the road.
Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs BC Is An East West Straight Line

Given

Two persons, one of them is coming along south direction to reach to east-west road and the other person is coming along the south-east direction, starting from the same place.

Let BC is an east-west straight line.

Both persons started journey together from point A to C. First person traversed in the southern direction distance AB, while the second traversed BC distance in the eastern direction.

Second person traversed AC distance in south-east.

Distance covered by the first person = AB + BC

Distance covered by the second person = AC

∵ AB + BC > AC (∵ The sum of measurements of two sides of a triangle is greater than that of its third side.)

∴ The second person reached first.

Question 2. In a quadrilateral ABCD, AB=AD and BC = DC; and DP is the smallest distance drawn from D on AC. Let’s prove that B5 P and D are collinear.

Solution:

Given : In quadrilateral ABCD, AB = AD and BC = DC, the shortest distance from point D to side AC is DP.

Let’s prove that three points B, P, D are collinear.

Construction : DB is extended B to point forwards.

Proof: ∵ DP is at the shortest distance from AC.

∴ DP ⊥ AC

∴ ∠APD = 90° and ∠DPC = 90°

∴ APB = 90° (∵ the Vertically Opposite Angle of ∠DPC is ∠APB)

∴ APD + APB = 90° = 180°

∴ BPD = 180°

Question 3. AD is a median of the triangle ABC. The smallest distance of AD from B and C are BP and CQ respectively. Let’s prove that BP = CQ.

Solution:

WBBSE Solutions Class 8 Chapter 20 Geometrical Proofs ABC Is The Median Of AD

Given: ABC is the median of AD. The shortest distance of AD From points B and C is BP and CQ.

∴ BP ⊥ AD and CQ ⊥ AD

∴ ∠BPD = ∠CQD = 90°

In equiangular ΔBPD and equiangular ΔCQD, angle BD = angle DC (∵ AD is the median ∴  BD = DC)

And PD is a common side.

∴  ΔBPD = ΔCQD (R-H-S congruency)

∴  BP = CQ Proved

WBBSE Solutions Class 8 Chapter 18 Graphs

Graphs

Let’s find out the points in the graph paper.

We can see many points in the graph paper. Let’s try to write the co-ordinates of these points.

We see, A point is 4 units away from the origin on the x-axis.

WBBSE Solutions Class 8 Chapter 18 Graphs Coordinate Of These Points

Question 1. What are the coordinates of point A ?
Solution:

The coordinates of point A

The distance of point A from the y-axis is 4 units and from the x-axis is 0 unit.

∴ The coordinates of point A are (4, 0).

Understand, the coordinates of the point B are (9,0); the coordinates of point C are (14,0)

The coordinates of the point (D) are (18, 0).

So we got, the y- coordinate of any point on x-axis is 0 (zero).

We see that the point E is situated at 3 units away from the origin and it is on the y-axis.

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. What are the coordinates of point E ?
Solution:

The coordinates of point E

Point E lies on y-axis and 3 units away from x-axis but it is at 0 unit away from y-axis.

∴ The coordinates of point E are (0, 3).

Now we understand, the coordinates, of point F are (0.7).

The coordinates of point G are (0,10).

The coordinates of point H are (0, 14).

So we got, the x-coordinate of any point on the y-axis is 0

Let’s write the coordinates of the points I and J from the graph paper.

We see, the point I is 5 units away from the y-axis and 3 units away from the x-axis.

∴ The coordinates of point I are (5, 3).

The point J is 10 units away from the y-axis and 3 units away from the x-axis.

∴ The coordinates of J are (10,3).

The coordinates of K are (13,3)

The coordinates of L are (17,3)

Similarly let’s write the coordinates of the points M, N, P, Q, R, S and T.

The coordinates of M are (3,4)

The coordinates of N are (3,6)

The coordinates of P are (3,9)

The coordinates of Q are (3,13)

The coordinates of R are (8,9)

The coordinates of S are (10,11)

The coordinates of T are (14,12)

Now let’s join the points of the graph paper with a pencil and see which three or more than three points are collinear.

We see that the points A, B, C, and D lie on the X axis and they are collinear (collinear/noncollinear).

WBBSE Solutions Class 8 Chapter 18 Graphs Collinear Or Noncollinear

The Points E,F,G and H are on the y-axis and they are collinear.

The points I, J, K, and L are collinear and M, N,P and Q are collinear. However, the points R, S and T are noncollinear.

WBBSE Solutions Class 8 Chapter 18 Graphs

Graphs Exercise 18.1

Question 1. Let’s plot the points A (4, 0), B (0, 6), C (2, 5), D (7, 1), E(6, 5) and F (10, 5) on a graph sheet.

Given

A (4, 0), B (0, 6), C (2, 5), D (7, 1), E(6, 5) and F (10, 5)

Coordinates of point A are (4, 0)

Coordinates of point B are (0, 6)

Coordinates of point C are (2, 5)

Coordinates of point D are (7, 1)

Coordinates of point E are (6, 5)

Coordinates of point F are (10, 5)

WBBSE Solutions Class 8 Chapter 18 Graphs Coordinates Of Point

Question 2. Let’s plot the points (1, 1), (3, 7), (9, 1) and (12, 1) on a graph sheet and verify whether they lie on a line.
Solution:

Given

(1, 1), (3, 7), (9, 1) and (12, 1)

WBBSE Solutions Class 8 Chapter 18 Graphs Llie On A Line

Question 3. On a graph sheet let’s write 4 collinear points and write the coordinates of the points. Similarly, plotting the points on the graph sheet and joining them we get some pictures. What will we call that?
Solution:

I took on the graph paper 4 collinear points A, B, C and D.

WBBSE Solutions Class 8 Chapter 18 Graphs Collinear Points

Plotting the points on the graph sheet from the coordinates of the points and joining the points we get a picture which is called a (Graph).

If joining the points we get a straight line segment, then this graph is called a linear graph.

1. Let’s plot the points A(2,0), B(6, 0) and C(4, .3) on the graph paper and join the points A, B ; B, C and C, A. Let’s find out what we have got.

Joining the points we have got a [triangle (triangle/quadrilateral).

2. Joining the points P(2,2) Q(4,4), R(6,6) we got a (collinear/non collinear) graph.

WBBSE Solutions Class 8 Chapter 18 Graphs Joining The Points In Collinear Or Noncollinear

3. Let’s join the points P(1,2), A (2,3), T(3,4), and H(4,5) and see whether we have got a linear graph.
Solution:

Given

P(1,2), A (2,3), T(3,4), and H(4,5)

Path is a linear graph

WBBSE Solutions Class 8 Chapter 18 Graphs Path Is A Linear Graph

Coordinates of P are (1, 2)

Coordinates of A are (2, 3)

Coordinates of T are (3, 4)

Coordinates of H are (4, 5)

Graphs Exercise 18.2

Question 1. I took 4 copies for Rs. 20. Let’s write the number and price of exercise books on the following table and draw a graph from the data.

WBBSE Solutions Class 8 Chapter 18 Graphs Data In Exercise Books

Let’s find out from the graph the price of 6 exercise books and the number of exercise books for Rs. 45.
Solution:

Given

I took 4 copies for Rs. 20

WBBSE Solutions Class 8 Chapter 18 Graphs Price Of 6 Exercise Books And The Number Of Exercise Books

1. First draw x- and y-axes.

2. Let us consider the length of one smallest square along x-axis = copy and the length of one smallest square along the y-axis = 1 copy and the length of one smallest square along the y-axis = Re. 1

3. From the data of graph plot points. (4, 20), (8, 40), (10, 50) and (12, 60)

4. Joining the points got AB straight line. So it is a linear graph.

5. On x-axis took the 6 exercise books and drew a ray parallel to the y-axis which cuts straight line at P point. Coordinates of P are (6, 30).

∴ Coordinates of point P are (6,30).

6. On the y-axis from the point of Rs. 45 drew a ray parallel to x- axis which cuts AB at Q. Coordinates of Q are (9, 45)

∴ Got from the graph that 9 exercise books are got in 45 rupees.

Graphs Exercise 18.3

Question 1. Let’s draw a graph of time and distance on the graph paper based on the information given below and find the distance covered in 4 hours and find how long it will take to cover 150 km.

WBBSE Solutions Class 8 Chapter 18 Graphs Time And Distance
Solution:

1. First of all, draw on the graph paper x- and y-axes.

2. Let distance is measured on x-axis and time along y-axis. Let the length of one smallest square along x-axis = 5 km and the length of one smallest square along y-axis = 1 hour

3. Plot the points on the graph paper (50, 2), (75, 3) and (125, 5) from the data given above.

4. Joining the above points got AB straight line. So it is a linear graph.

5. On the y-axis from the point of 4 hours drew a ray parallel to the x-axis which cuts AB at P.

Coordinates of point P are (100, 4)

∴ Got from the graph that in 4 hours, 100 km distance is covered.

6. On x-axis from the point of 150 km drew a ray parallel to y-axis which cuts AB at point Q. Coordinates of point Q are (150, 6).

∴ To cover 150 distance 6 hours are required.

WBBSE Solutions Class 8 Chapter 18 Graphs Coordinates Of Point P And Q

Graphs Exercise 18.3

Question 1. For the graph paper beside we consider: 1 unit = length of 2 sides of the smallest squares along both the axes. Let’s write the coordinates of the points.
Solution:

Given

For the graph paper beside we consider: 1 unit = length of 2 sides of the smallest squares along both the axes.

Coordinates of A are (2, 0)

Coordinates of B are (5, 0)

Coordinates of C are (0, 4)

Coordinates of D are (0, 6)

Coordinates of E are (0, 8)

Coordinates of F are (4, 2)

Coordinates of G are (7, 2)

Coordinates of-H are (10, 2)

Coordinates of I are (2, 3)

Coordinates of J are (3, 4)

Coordinates of K are (4, 5)

WBBSE Solutions Class 8 Chapter 18 Graphs Smallest Squares

2. Let’s find which three points are collinear.
Solution:

CDE, UK, FGH, CIF, and JFB are collinear.

3. Let’s find which three points are non-collinear.
Solution:

A, B and C are non-collinear.

Question 2. Let’s plot the points (1,0), (0, 5), (2, 1), (3,3), (1,3), (2,5) and (0,0) on the graph paper.
Solution:

Given

(1,0), (0, 5), (2, 1), (3,3), (1,3), (2,5) and (0,0)

O = (0, 0)

A = (1,0)

B = (0; 5)

C = (2, 1)

D = (3, 3)

E = (1, 3)

F = (2, 5)

WBBSE Solutions Class 8 Chapter 18 Graphs Plot The Point And Graph Paper

Question 3.1. Let’s put the points (1,1), (2,2) and (6,6) on the graph paper and see whether they are collinear.

2. Let’s put three non-collinear points on the graph paper.Let’s put three more

3. collinear points other than the above three collinear points on the graph paper and write their coordinates.

A=(1,1)

B=(2,2)

C=(6,6)

A,B and C are collinear.

WBBSE Solutions Class 8 Chapter 18 Graphs ABC Are Collinear

2. Lets plot any three non-collinear are points on the graph paper.
Solution:

D = (13, 5)

E = (15, 10)

F = (20, 10)

D, E and F are three non-collinear points.

3. After the three points above, we plot three collinear points and write their coordinates.
Solution:

P = (5, 10)

Q = (10, 15)

R = (15, 20)

P,Q and R are three collinear points.

Let 1 guava = length of 2 sides of the smallest square along x- axis and 1 rupee = length of a side of one smallest square along y-axis.

1. Let’s write the relation between the number of guava and the price of guava from graph.
Solution:
There is direct relation between the number of guavas and price.

2. Let’s write the price of 4 guavas.
Solution:

4 guavas cost Rs. 12.

WBBSE Solutions Class 8 Chapter 18 Graphs Number Of Guavas And Prise Of Guavas

3. Let’s write the number of guavas in Rs. 30 from the graph.
Solution:

10 guavas are bought in Rs. 30.

4. Let’s write the number of guavas in Rs. 9.
Solution:

3 guavas are bought in Rs.9.

Question 5. Let’s see the graph beside and find the answers to the questions given below

Let 1 hdur = length of the 2 sides of the smallest squares along x- axis and 5 km = length of a side of the smallest square along y-axis.

1. Let’s write the relation between time and distance.
Solution:

Distance is in direct relation with time.

2. Let’s write the distance covered in 3. hours.
Solution:

90 km distance is covered in 3 hours.

3. Let’s write the time required to cover 120 km.
Solution:

To cover 120 km distance 4 hours are required.

4. Let’s find the velocity in km per hour.
Solution:

Velocity’is 30 kmph.

5. Let’s find and write the distance covered in 2 hours 30 minutes from the graph.
Solution:

In 2 hours 30 minutes, 75 km distance is covered.

6. Let’s find and write the time required to cover 45 km.
Solution:

To cover 45 km distance 1 hour 30 minutes are-required.

WBBSE Solutions Class 8 Chapter 18 Graphs Coordinates Of Point

Question 6. Let’s draw the graph on the graph paper of the following data and see whether it is a linear graph.

WBBSE Solutions Class 8 Chapter 18 Graphs Number Of Pencil And Prise Of Pencile
Solution:

  1. Let on the x-axis the length of one smallest square = 1. Pencil. Let along the y-axis the length of one smallest square = Rs. 1
  2. Plot the points (3, 6), (5, 10), (7, 14) and (8, 16) on graph from the above data.
  3. Joining the points got AB straight line. So it is a linear graph.

WBBSE Solutions Class 8 Chapter 18 Graphs Points Got AB Straight Line And Linear Q6

Question 7. Let’s draw the graph on the graph paper of the foilwing data and see whether it is a linear graph.

WBBSE Solutions Class 8 Chapter 18 Graphs Time And Distance In Data
Solution:

  1. Let the length of the smallest square on the axis = 1 hour; on the y-axis the length of one smallest square = 5 km.
  2. Let on the plot the points (2, 40), (4, 80), (6, 120) and (8, 160)on the graph from the above data.
  3. Joining the points got AB straight line. So it is a linear graph.

WBBSE Solutions Class 8 Chapter 18 Graphs Points Got AB Straight Line And Linear Q7

Question 8. Let’s draw the graph of the following data on the graph paper and see whether it is a linear graph.

WBBSE Solutions Class 8 Chapter 18 Graphs Number Of Bags And Prise Of Bags
Solution:

  1. Let on the the x-axis the length of 5 sides of the smallest square =1 bag. Let on the y-axis the length of one smallest square = Rs. 5
  2. Plot the points (1, 50), (2,100), (3, 150) and (4, 200) on the graph paper from the above data.
  3. Joining the points got AB straight line. So the graph is linear.

WBBSE Solutions Class 8 Chapter 18 Graphs Points Got AB Straight Line And Linear Q8

Question 9. Let’s, draw the graph of the following data on the graph paper and see whether it is a linear graph.

WBBSE Solutions Class 8 Chapter 18 Graphs Number Of Books And Prise Of Books
Solution:

  1. Let on the x-axis the length of 5 sides of the smallest square = 1 book and on the y-axis the length of one side of the smallest square = Rs. 5.
  2. Plot the points (2, 50), (3, 75), (5, 125) and (8, 200) on the
    graph paper from the above data.
  3. Joining the points got AB straight line. So the graph is linear.

WBBSE Solutions Class 8 Chapter 18 Graphs Points Got AB Straight Line And Linear Q9

Question 10. Let’s draw the graph on the graph paper of the following data and see whether it is a linear graph.

WBBSE Solutions Class 8 Chapter 18 Graphs Number Of Overs And Score At End Of Overs
Solution:

  1. Let on the x-axis the length of 5 sides of the smallest square = 1 over. On the y-axis the length of one side of the smallest square = 1 run.
  2. Plot the points (1, 4), (3, 12), (5, 20) and (7, 24) on the graph paper from the above data.
  3. Joining the points got AB graph. So the graph is not linear.

WBBSE Solutions Class 8 Chapter 18 Graphs Points Got AB Graph And Non Linear