Class 10

MP Board Class 10th Science Solutions Chapter 8 How do Organisms Reproduce?

MP Board Class 10th Science Chapter 8 Intext Questions

Class 10th Science Chapter 8 Intext Questions Page No. 128

Question 1.
What is the importance of DNA copying in reproduction?
Answer:
The chromosomes in the nucleus of a cell contain information for inheritance of features from parents to next generation in the form of DNA (Deoxyribo Nucleic Acid) molecules. The DNA in the cell nucleus is the information source for making proteins. Hence DNA copying is important in reproduction.

Question 2.
Why is variation beneficial to the species but not necessarily for the individual?
Answer:

If a population of reproducing organisms were suited to particular niche and if the niche were drastically altered, the population could be wiped out. However, if some variations were to be present in a few individuals in these populations, there would be some chance for them to survive.

Thus, if there were a population of bacteria living in temperature waters and if the water temperature were to be increased by global warming, most of these bacteria would die, but the few variants resistant to heat would survive and grow further. Variation is thus useful for the survival of species over time. Variation is not useful for all organisms.

Class 10th Science Chapter 8 Intext Questions Page No. 133

Question 1.
How does binary fission differ from multiple fission?
Answer:
Binary fission: It is a simple kind of division which formate new individual. In binary fission, a single cell divides into two equal halves but it is possible only with very simple single cell kind. Amoeba and Bacteria divide by binary fission.

Multiple fission: Another type of simple division is multiple fission, in this, a single cell divides into many daughter, cells, e.g., Plasmodium divide by multiple fission.

Binary fission	Multiple fission
In this fission, one cell split into two equal halves during cell division. Eg: Bacteria.	Here one organism divide into many daughter cells simultaneously. Eg: yeast.

Question 2.
How will an organism be benefited if it reproduces through spores?
Answer:
The spores are covered by thick walls that protect them until they come into contact with another moist surface and can begin to grow. Thus organism be benefited if it reproduces through spores.

Question 3.
Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:
Multicellular organisms are not simply a random mass of cells but a carefully organized entity of tissues and organs are placed at definite positions in the body to form organ systems. These systems are well coordinated to perform specific functions. Hence complex organisms cannot reproduce through fragmentation.

Question 4.
Why is vegetative propagation practised for growing some types of plants?
Answer:
Advantages of vegetative propagation:

• Used in methods such as layering or grafting, to grow many plants like sugarcane, roses or grapes for agricultural purposes.
• Plants raised can bear more flowers and fruits in comparison to plants produced from seeds.
• Plants such as banana, orange, rose and jasmine which have lost the capacity to produce seeds can be propagated.
• All plants produced by vegetative propagation are genetically similar enough to the parent plant.

Question 5.
Why is DN Acopying an essential part of the process of reproduction?
Answer:
The consistency of DNA copying during reproduction is important for the maintenance of body design features that allow the organism to use that particular niche. Because of this DNA copying is an essential part of the process of reproduction.

Class 10th Science Chapter 8 Intext Questions Page No. 140

Question 1.
How is the process of pollination different from fertilization?
Answer:
Pollination is movement of pollens from one plant to another plant’s or its own plant’s stigma. It may require certain agents called pollinators such as air, water birds or some insects to perform. Fertilization, is a complex process, it involves the fusion of the male and female gametes. It occurs inside the ovule and leads to the formation of zygote.

Question 2.
What is the role of the seminal vesicles and the prostate gland?
Answer:
Along the path of the vas deferens, gland like the prostrate and the seminal vesicles add their secretions so that the sperms are now in a fluid which makes their transport easier and this fluid also provides nutrition.

Question 3.
What are the changes seen in girls at the time of puberty?
Answer:
The changes seen in girls at the time of puberty are:

1. Development of secondary sexual characteristics.
2. Growth in breast size and darkening of skin of the nipples.
3. Growth of hair in the genital area and other areas of skin like underarms, face, hands and legs.
4. Growth in the size of uterus and ovary hence, start of menstrual cycle periodically.

Question 4.
How does the embryo get nourishment inside the mother’s body?
Answer:
The embryo gets nutrition form the mother’s blood with the help of a special tissue called placenta. This is a disc which is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue on the mother’s side are blood spaces, which surround the villi. This provides a large surface area for glucose and oxygen to pass from the mother to the embryo. The developing embryo will also generate waste substances which can be removed by transferring them into the mother’s blood through the placenta.

Question 5.
If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases?
Answer:
Copper-T will helps in protecting her from sexually transmitted diseases by helping to prevent infections of diseases.

MP Board Class 10th Science Chapter 8 NCERT Textbook Exercises

Question 1.
Asexual reproduction takes place through budding in:
(a) amoeba
(b) yeast
(c) plasmodium
(d) leishmania
Answer:
(b) yeast

Question 2.
Which of the following is not system in human beings? a part of the female reproductive
(a) ovary
(b) uterus
(c) vas deferens
(d) fallopian tube
Answer:
(c) vas deferens

Question 3.
The anther contains:
(a) sepal
(b) ovules
(c) carpel
(d) pollen grains
Answer:
(d) pollen grains

Question 4.
What are the advantages of sexual reproduction over asexual reproduction?
Answer:
In case of asexual reproduction, new generations are produced by one organism. But in sexual reproduction, new generations are produced by two organisms (male and female). In case of sexual reproduction germ cells are produced in testes and these secrete a hormone testosterone. In human beings also develop special tissues for this purpose.

Question 5.
What are the functions performed by the testis in human beings?
Answer:
They are the glands where sperm and testosterone are generated and present in male body. The testes are contained in the scrotum and are composed of dense connective tissue. Functions of testes are as follows:

• It produces sperms, which contain haploid set of chromosomes of
• It produces testosterone, which initiate secondary sexual characteristics

Question 6.
Why does menstruation occur?
Answer:
Since the ovary releases one egg every month, the uterus also prepares itself every month to receive a fertilised egg. Thus its lining becomes thick and spongy. This would be required for nourishing the embryo if fertilisation had taken place. Now, however, this lining is not needed any longer. So the lining slowly breaks and comes out through the vagina as blood and mucous. This cycle takes place roughly every month and is known a menstruation. It usually lasts for about two to eight days.

Question 7.
Draw a labelled diagram of the longitudinal section of a flower.
Answer:

Longitudinal section flower.

Question 8.
What are the different methods of contraception?
Answer:

Many ways have been devised to avoid pregnancy. These contraceptive methods fall in a number of categories. One category is the creation of a mechanical barrier so that sperm does not reach the egg. Condoms on the penis or similar coverings worn in the vagina can serve this purpose.

Another category of contraceptives acts by changing the hormonal balance of the body so that eggs are not released and fertilisation cannot occur. These drugs commonly need to be taken orally as pills. However, Since they change hormonal balances, they can cause side effects too. Other contraceptive devices such as the loop or the copper-T are placed in the uterus to prevent pregnancy. Again, they can cause side effects due to irritation of the uterus. Surgery can also be used for removed of unwanted pregnancies.

Question 9.
How are the modes for reproduction different in unicellular and multicellular organisms?
Answer:
In unicellular organisms, reproduction occurs by the division of the entire cell. The modes of reproduction in unicellular organisms can be fission, budding etc. whereas in multi cellular organisms, specialised reproductive organs are present. Therefore, they can be reproduced by complex reproductive methods such as vegetative propagation, spore formation etc. In more complex multicellular organisms such as human beings and plants, the mode of reproduction is sexual reproduction.

Question 10.
How does reproduction help in providing stability to populations of species?
Answer:
Reproduction is the process of producing new individuals of the same species by existing organisms of a species. So, it helps in providing stability to population of species by giving birth to new individuals as the rate of birth must be at par with the rate of death to provide stability to population of a species.

Question 11.
What could be the reasons for adopting contraceptive methods?
Answer:
Contraceptive methods are mainly adopted because of the following reasons:

• It prevent unwanted pregnancies.
• It control rise in population and birth rate.
• It prevent sexually transmitted diseases.

MP Board Class 10th Science Chapter 8 Additional Important Questions

MP Board Class 10th Science Chapter 8 Multiple Choice Questions

Question 1.
All individuals produced by an organism are:
(a) Genetically similar
(b) Non-identical
(c) Fission
(d) Moneociuos
Answer:
(a) Genetically similar

Question 2.
Sexual reproduction is completed by _______ division:
(a) Mitotic
(b) Meiotic and mitotic both
(c) Meiosis
(d) Mitotic at some stages .
Answer:
(c) Meiosis

Question 3.
In yeast cell, division results in:
(a) Offspring
(b) Bud
(c) Clone
(d) Branch
Answer:
(b) Bud

Question 4.
Which of the following organisms undergo multiple fission?
(a) Paramecium
(b) Plasmodium
(c) Amoeba
(d) All of the above
Answer:
(b) Plasmodium

Question 5.
Hydra reproduces asexually through:
(a) Budding
(b) Binary fission
(c) Multiple fission
(d) Vegetative propagation
Answer:
(a) Budding

Question 6.
In which plant, the site of origin of new plants is node?
(a) Potato tuber
(b) Onion bulb
(c) Rhizome ginger
(d) All of the above
Answer:
(d) All of the above

Question 7.
In which of the following, asexual reproduction takes place through binary fission?
(a) Amoeba
(b) Yeast
(c) Plasmodium
(d) Leishmania
Answer:
(b) Yeast

Question 8.
Which of the following is in human beings?
(a) Ovary
(b) Uterus
(c) (a) and (b)
(d) Fallopian tube
(e) (a), (b) and (d)
Answer:
(e) (a), (b) and (d)

Question 9.
The anther, a part of male flower have:
(a) Sepals
(b) Ovules
(c) Carpel
(d) Pollen grains
Answer:
(d) Pollen grains

Question 10.
The information for making proteins is provided by:
(a) Rough endoplasmic reticulum
(b) DNA
(c) Hormones
(d) Enzymes
Answer:
(b) DNA

Question 11.
Nature of gametes are usually:
(a) Haploid
(b) Diploid
(c) Both (a) and (b)
(d) None of the above
Answer:
(a) Haploid

Question 12.
With the help of which tissues embryo gets nutrition from the mother’s blood?
(a) Zygote
(b) Uterus only
(c) Placenta
(d) None of these
Answer:
(c) Placenta

Question 13.
Which of the following is not a part of the male reproductive system in human beings?
(a) Testes
(b) Uterus
(c) Vas deferens
(d) Urethra
Answer:
(b) Uterus

Question 14.
Binary fission in some organisms occurs in definite orientation in relation to the cell structures. One such organisms is:
(a) Leishmania
(b) Plasmodium
(c) Amoeba
(d) Bacteria
Answer:
(c) Amoeba

Question 15.
Plants that have lost their capacity to produce seeds, reproduce by:
(a) Spores
(b) Vegetative propagation
(c) Fission
(d) Regeneration
Answer:
(a) Spores

Question 16.
A stamen consists of two parts namely:
(a) Anther and style
(b) Anther and filament
(c) Stigma and style
(d) Filament and style
Answer:
(b) Anther and filament

Question 17.
A bisexual flower contains:
(a) Stamens only
(b) Carpels only
(c) Either stamens or carpels
(d) Both stamens and carpels
Answer:
(d) Both stamens and carpels

Question 18.
Germinated seeds do not contains:
(a) Sepals
(b) Cotyledon
(c) Plumule
(d) Radicle
Answer:
(a) Sepals

Question 19.
A feature of reproduction that is common to amoeba, spirogyra and yeast is that:
(a) they reproduce asexually
(b) they are all unicellular
(c) they reproduce only sexually
(d) they are all multicellular
Answer:
(a) they reproduce asexually

Question 20.
Which of the part of flower ripens to form a fruit?
(a) Ovule
(b) Ovary
(c) Carpel
(d) Egg cell
Answer:
(b) Ovary

Question 21.
The testes perforin the following function/functions:
(a) Produce testosterone
(b) Produce sperms
(c) Produce male gametes and hormone
(d) Produce sperms and urine
Answer:
(b) Produce sperms

Question 22.
Where does fertilisation take place in human beings?
(a) Uterus
(b) Vagina
(c) Cervix
(d) Fallopian Tube
Answer:
(d) Fallopian Tube

Question 23.
Condom is a method of control that falls under the following category:
(a) Surgical method
(b) Hormonal method
(c) Mechanical method
(d) Chemical method
Answer:
(c) Mechanical method

Question 24.
The common passage for sperms and urine in the male reproductive system is:
(a) Ureter
(b) Seminal vesicle
(c) Urethra
(d) Vas deferens
Answer:
(c) Urethra

Question 25.
In sperm, which part dissociates after fertilization?
(a) Acrosome
(b) Tail
(c) Head
(d) Middle piece
Answer:
(b) Tail

MP Board Class 10th Science Chapter 8 Very Short Answer Type Questions

Question 1.
Which life process is not essential to maintain the life of an individual organism but important for the survival of species?
Answer:
Reproduction.

Question 2.
How a species can get a danger of being extinct?
Answer:
If individuals of any species stops reproducing, then that species can get a danger of being extinct.

Question 3.
How an individual is able to make a copy of itself?
Answer:
DNA copying is a process at cellular level which enables an individual to make copy of it self.

Question 4.
Write the name of process by which Hydra reproduces.
Answer:
Budding only.

Question 5.
Generally, how many individuals are involved in asexual reproduction?
Answer:
One.

Question 6.
Write the name of some common method of asexual reproduction.
Answer:
Vegetative propagation, budding, fragmentation and spore formation.

Question 7.
Which type of flower is called unisexual flowers?
Answer:
A flower which have either male or female reproductive parts is called unisexual flowers.

Question 8.
What is pollination?
Answer:
The transfer of pollen grains from the anther to the stigma of the same or of another flower of the same kind is known as pollination.

Question 9.
What do you understand by term fertilisation?
Answer:
The fusion of male and female gametes is termed as fertilisation.

Question 10.
How seed is dispersed?
Answer:
Seed dispersal takes place by means of wind, water and animals.

MP Board Class 10th Science Chapter 8 Short Answer Type Questions

Question 1.
How does plasmodium undergo fission?
Ans.
Plasmodium divides into many daughter cells through multiple fission.

Question 2.
How spirogyra reproduces by fragmentation?
Answer:
An individual spirogyra breaks up into many smaller pieces, each fragment grows into new individual.

Question 3.
Which cells are responsible for budding in hydra?
Answer:
Regenerative cells.

Question 4.
Name the structure into which following develops: the plumule and radicle?
Answer:
Plumule develops to shoot while radicle form root of a plant.

Question 5.
On which plant can you find buds on its leaves?
Answer:
Bryophyllum.

Question 6.
Write the scientific name of the bread mould.
Answer:
Rhizopus.

Question 7.
Where are the testes located in human beings?
Answer:
In abdominal cavity, in scrotum.

Question 8.
For what specific reason have the testes specific location?
Answer:
As testes, requires lesser temperature, to produce sperm than of abdominal cavity.

Question 9.
Correlate the rate of general body growth and maturation of reproductive tissue during puberty.
Answer:
When reproductive tissues (organs) begin to mature, body growth rate slows down.

Question 10.
Where does the zygote get implanted in human beings?
Answer:
In the wall of uterus.

Question 11.
Which two important substances are delivered to developing embryo through placenta?
Answer:
Glucose and oxygen.

Question 12.
How change in hormonal balance prevents pregnancy?
Answer:
It prevents the release of eggs.

Question 13.
Name the tissue in mother’s body that provides nutrition to developing embryo?
Answer:
Placenta provides nutrition to developing embryo.

Question 14.
Write one side effect of loop placed in uterus.
Answer:
It may cause permanent irritation and excessive and prolong bleeding in uterus.

Question 15.
Which structures need to be blocked in males and females respectively to prevent pregnancy?
Answer:
Vas deferens in male (vasectomy), fallopian tube in female (tubectomy).

Question 16.
Why is children sex ratio alarmingly declining in our country.
Answer:
Abortions based on sex selections.

Question 17.
Name the chemical methods of preventing pregnancy.
Answer:
Morning over oral pills.

Question 18.
Name some of the devices used as mechanical method for preventing pregnancy.
Answer:
Loop, copper T, condoms.

Question 19.
Name the only mammal(s) which lays eggs.
Answer:
Echidna and duck-billed platypus.

Question 20.
What is parthenogenesis?
Answer:
Parthenogenesis is a type of asexual reproduction. In this case, embryo development takes places without fertilisation. A few species of insects, bees, wasps, birds and lizards (e.gKomodo dragon lizard) reproduce this way.

Question 21.
Give an example of an organism which reproduces by:
(a) Fragmentation
(b) Spore formation
(c) Stems
Answer:
(i) Spirogyra.
(ii) Bacteria, fungi (rhizopus), moss, algae.
(iii) Plants like potato (tuber), onion (bulb) reproduce by vegetative propagation of stems.

Question 22.
Discuss various artificial vegetative propagation techniques.
Answer:
Various artificial vegetative propagation techniques are:

1. Cutting
2. Layering
3. Grafting
4. Tissue culture

Question 23.
What is grafting? What are different types of grafting techniques?
Answer:
In grafting, one part of a plant is inserted into another plant in a way that both of them will unite and grow together as a single plant. Different methods of grafting are:

• Approach grafting
• Cleft grafting
• Bud grafting
• Tongue grafting

Question 24.
Name some:

1. Plants which are reproduced by vegetative propagation.
2. Plants which have unisexual flowers.
3. Plants which have bisexual flowers.
4. Plants with self-pollination.
5. Plants that do cross-pollination.

Answer:

1. Rose, sweet potatoes, bryophyllum.
2. Coconut, papaya, watermelon.
3. Lily, rose, sunflower.
4. Beans, peas, tomatoes.
5. Grasses, catkins, maple trees.

Question 25.
What is germination?
Answer:
The seed contains the future plant or embryo which develops into a seedling under appropriate conditions. This process is known as germination.

Question 26.
What is cross-pollination?
Answer:
Cross-pollination is the process of transfer of pollen from the anther of a flower to stigma of a flower of another plant of the same species or closely related species.

Question 27.
Explain hormonal pills of contraception.
Answer:
Oral contraceptives: In this method, tablets or drugs are taken orally by females to check pregnancy These contain small doses of hormones in forms of pills that prevent the release of eggs and thus, fortilisation cannot occur.

MP Board Class 10th Science Chapter 8 Long Answer Type Questions

Question 1.
Why simply copying of DNA in a dividing cells not enough to maintain continuity of life?
Answer:
Copying of DNA preserve and pass specific characters of a generation to next generation offsprings. In reproduction, it is very important to create DNA copy. It determines the body design of an individual. But variation in genotype is also important,, because sometimes existing genotype don’t find its potential to survive in changing surroundings. So, genotype must have some alterations which are caused by variations only. Hence, simply copying of DNA in a dividing cells is not enough to maintain continuity of life.

Question 2.
Describe in brief the fragmentation mode of asexual reproduction.
Answer:
Fragmentation: Many lower organisms, use fragmentation mode of asexual reproduction for its growth e.g., algae. When water and nutrients are available in sufficient amount algae grow and multiply rapidly by fragmentation. An algae breaks up to multiple fragments. These fragments or pieces grow into new individuals.

Question 3.
Explain budding in yeast.
Answer:
The yeast is a single-celled organism. The small bulb-like projection come out from the yeast cell in favourable time and is called a bud. The bud gradually grows and gets dettached from the parent cell and forms a new yeast cell. The new yeast cell grows, matures and produces more yeast cells.

Question 4.
Describe the process of implantation.
Answer:
A week after the sperm fertilizes the egg, the fertilized egg (zygote) undergo development and become a multicelled blastocyst. The blastocyst fix itself into the lining of the uterus, called the endometrium. The hormone estrogen causes the endometrium to become thick and rich with blood. Progesterone and other hormone released by the ovaries, keeps the endometrium thick with blood so that the blastocyst can absorb nutrients from uterus. This process is called implantation.

Question 5.
Explain the following.

1. Hermaphrodites
2. Unisexual
3. Syngamy

Answer:

1. Hermaphrodites are bisexual organisms which possess both male and female reproductive organs. Examples: earthworm, leech, starfish.
2. Animals which have different male and female individuals as birds, mammals etc.
3. The process of fusion of male gamete with female gamete is called syngamy.

Question 6.
What is contraception? Discuss natural and barrier method of contraception.
Answer:
Contraception or birth control methods include: condoms, the diaphragm, the contraceptive pill, implants, IUDs (intrauterine devices), sterilization and the morning after pill and many more some of best methods are given below:

• Natural method: It involves avoiding the chances of meeting of sperms and ovum. In this method, the sexual intercourse is avoided by the couple from day 10th to 17th of the menstrual cycle of female as in this period, ovulation is expected and therefore, the chances of fertilisation are very high.
• Barrier method: In this method, the fertilisation of ovum and sperm is checked out with the help of artificially developed barriers. Barriers are developed for both males and females. Most common barrier available in market are condoms.

Question 7.
Describe implants and surgical methods of contraception
Answer:
Contraceptive devices are also developed as the loop or copper-T to prevent pregnancy. Surgical methods are also used to block the gamete transfer. It includes the blocking of vas deferens to prevent the transfer of Sperms known as vasectomy. Similarly, tubectomy in the fallopian tubes of the female can be blocked so that the egg will not reach the uterus.

Question 8.
Discuss fertilization in flowering plants.
Answer:
There are two main procedures of completing fertilization in flowering plants, which are:
(i) Pollination
(ii) Fertilisation

(i) Pollination: Pollination is a very important part of the life cycle of a flowering plant which results in seeds that grow into new plants. It is part of the sexual reproduction process of flowering plants. Flowers are the structures of flowering plants that contain all the specialized parts needed for sexual reproduction. Plants have gametes, which contain half the normal number of chromosomes for that plant species. Male gametes are found inside tiny pollen grains on the anthers of flowers. Female gametes are found in the ovules of a flower. Pollination is the process that brings these male and female gametes together. The wind or animals, especially insects and birds, pick up pollen from the male anthers and carry it to the female stigma. Flowers have to encourage animals to pollinate them.

(ii) Fertilisation: After pollination, when pollen has landed on the stigma of a suitable flower of the same species, various process occurs in the making of seeds. A pollen grain on the stigma grows a tiny tube, all the way down the style to the ovary. This pollen tube carries a male gamete to meet a female gamete in an ovule. In a process called fertilization, the two gametes join. The fertilised ovule form a seed, which contains a food store and an embryo that grow into a new plant. The ovary develops into a fruit to protect the seed.

Question 9.
Inside womb, how does a child receive food, oxygen and water? Discuss.
Answer:
As a mother eats something the nutrient like glucose, proteins, fats, vitamins, etc. are absorbed into the mother’s blood by the small intestine. The nutrients flow to the placenta, and then transferred to the baby’s bloodstream through the umbilical cord. The baby’s waste products (like CO2) are disposed of in the mother’s blood stream as well. In the placenta, the mothers blood flows into a network of blood Vessels and capillaries. Molecules in the mother’s blood like glucose, proteins, fats, oxygen etc. flow out of the mother’s blood supply and are absorbed into another network of blood vessels and capillaries containing the baby’s blood supply. The baby’s blood then flows through the umbilical cord back to the baby. It is the complete process of baby’s nutrition inside womb.

Question 10.
Discuss the advantages and disadvantage of autogamy or self¬pollination.
Answer:
Advantages of autogamy:

It is a sure method of seed formation. Scent and Nectar are not needed by the flower to attract insects. Parent characteristics are preserved in off spring’s. Small quantity of pollen is required for pollination. Flowers need not be large or attractive. Disadvantages of autogamy plants lose their vigor in their future generations due to repeated self-pollination. Since, there is no variation, no genetic improvement occurs in offsprings. Weak characteristics of the plant are inherited by the next generations.

MP Board Class 10th Science Chapter 8 Textbook Activities

Class 10 Science Activity 8.1 Page No. 129

• Dissolve about 10 gin of sugar in 100 mL of water.
• lake 20 mL of this solution in a test tube and add a pinch of yeasl granules to it.
• Put a cotton plug on the mouth of the test tube and keep it in a warm place.
• Alter 1 or 2 hours, put a small drop of yeast culture from the test tube on a slide and cover it with a coverslip.
• Observe the slide under a microscope.

Observations:

• Formation of yeast cells can be seen. Some of them, shows chain budding.

Class 10 Science Activity 8.2 Page No. 129

• Wet a slice of bread, and keep it in a cool, moist and dark place.
• Observe the surface of the slice with a magnifying glass.
• Record your observation for a week.

Observations:

• A layer of while cottony mass is seen over the surface of slice. These inercase in size and number and after a week, the layer turns black show ing formation of spores or sporangia.

Class 10 Science Activity 8.3 Page No. 129

• Observ e a permanent slide of Amoeba under a microscope.
• Similarly observe another permanent slide of Amoeba show-ing binary fission.
• Now, compare the observation of both the slides.

Observations:

• The permanent slide of amoeba shows normal cytoplasm and nucleus. Nucleus can be seen dividing and construction in cytoplasm can also be seen. The binary fission with two daughter cells is observed in the other slide.

Class 10 Science Activity 8.4 Page No. 129

• Collect water from a lake or pond that appears dark green and contains filamentous structures.
• Put one or two filaments on a slide.
• Put a drop of glycerin on these filaments and cover it with a coverslip
• Observe the slide under a microscope.
• Can you identify different tissues in the Spimgyra filaments.

Observations:

• Spirogyra filament consists of many cells which are attached linearly to form a filament.

Class 10 Science Activity 8.5 Page No. 132

• Take a potato and observe its surface. Can notches be seen?
• Cut the potato into small pieces such that some pieces contain a notch or bud and some do not.
• Spread some cotton on a tray and wet it. Place the potato pieces on this cotton. Note where the pieces with the buds are placed.
• Observe changes taking place in these potato pieces over the next few days. Make sure that the cotton is, kept moistened.
• Which arc the potato pieces that give rise to fresh green shoots and roots.

Observations:

• The,potato undergoes various changes in few days. The buds in notches show growth of young shoots and roots. The pieces which do not have eye buds do not show any growth.

Class 10 Science Activity 8.6 Page No. 132

• Select a money-plant.
• Cut some pieces such that they contain at least one leaf.
• Cut out some other portions between two leaves.
• Dip one end of all the pieces in water and observe over the next few days.
• Which ones grow’ and give rise to fresh leaves?
• What can you conclude from your observations ?

Observations:

• The leaves at the nodes show formation of fresh leaves. The formation of branch from axillary buds axil of leaf is also observ ed.
• The leaves that undergo photosynthesis show tendency to grow into a new plant through vegetative propagation.

Class 10 Science Activity 8.7 Page No. 135

• Soak a few seeds of Bengal gram (chana) and keep them overnight.
• Drain the excess water and cover the seeds with a wet cloth and leave them for a day. Make sure that the seeds do not become dry.
• Cut open the seeds carefully and observe the different parts.
• Compare your observations with the Fig. 8.2 and sec if you can identify all the parts.

Observations:

• The parts identified includes- cotyledon which stores food, plumule which is a future shoot radicle that is a future root.

Germination.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.
Solution:
Since, diameter of the cylinder = 10 cm 10
∴ Radius of the cylinder (r) = \(\frac{10}{2}\) cm = 5cm
⇒ Length of wire in one round = 2πr
= 2 × 3.14 × 5 cm = 31.4 cm
∵ Diameter of wire = 3 mm = \(\frac{3}{10}\) cm
∴ The thickness of cylinder covered in one round = \(\frac{3}{10}\) cm
⇒ Number of rounds (turns) of the wire to cover 12 cm = \(\frac{12}{3 / 10}=12 \times \frac{10}{3}\) = 40
∴ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds
= l = 40 × 31.4 cm = 1256 cm
Now, radius of the wire = \(\frac{3}{2}\) mm = \(\frac{3}{20}\) cm
∴ Volume of wire = πr²l
= 3.14 × \(\frac{3}{20} \times \frac{3}{20}\) × 1256 cm³
∵ Density of wire = 8.88 g/cm³
∴ Mass of the wire = [Volume of the wire] × density

Question 2.
A right triangle, whose sides are 3cm and 4cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).
Solution:
Let us consider the right ABAC, right angled at A such that AB = 3 cm, AC = 4 cm
∴ Hypotenuse BC = \(\sqrt{3^{2}+4^{2}}\) = 5cm
Obviously, we have obtained two cones on the same base AA’ such that radius = DA or DA’.
Now,

Question 3.
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Solution:
Dimensions of the cistern are 150 cm, 120 cm and 100 cm.
∴ Volume of the cistern = 150 × 120 × 110 cm³ = 1980000 cm³
Volume of water contained in the cistern = 129600 cm³
∴ Free space (volume) which is not filled with water = (1980000 – 129600) cm³
= 1850400 cm³
Now, volume of one brick
= (22.5 × 7.5 × 6.5) cm³ = 1096.875 cm³
∴ Volume of water absorbed by one brick
= \(\frac{1}{17}\) × 1096.875 cm³
Let n bricks can be put in the cistern.
∴ Volume of water absorbed by n bricks
= \(\frac{n}{17}\) × 1096.875 cm³
∴ Volume occupied by n bricks = [free space in the cistern + volume of water absorbed by n bricks]
⇒ [n × 1096.875] = [1850400 + \(\frac{n}{17}\)(1096.875)]
⇒ 1096.875 n – \(\frac{n}{17}\)(1096.875) = 1850400
⇒ (n – \(\frac{n}{17}\)) × 1096.875 = 1850400
⇒ \(\frac{16}{17} n=\frac{1850400}{1096.875} \Rightarrow n=\frac{1850400}{1096.875} \times \frac{17}{16}\)
= 1792.4102 ≈ 1792
Thus, 1792 bricks can be put in the cistern.

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the norma water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Volume of three rivers = 3 {(Surface area of a river) × Depth}

Since 0.7236 km³ ≠ 9.728 km³
∴ The additional water in the three rivers is not equivalent to the rainfall.

Question 5.
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see figure).

Solution:
We have,
For the cylindrical part:
Diameter = 8 cm ⇒ Radius (r) = 4 cm
Height = 10 cm

Question 6.
Derive the formula for the curved surface area and total surface area of the frustum of a cone.
Solution:

Since, ∆OC₁Q ~ ∆OC₂S [By AA similarity]

Now, the total surface area of the frustum = (curved surface area) + (base surface area) + (top surface area)
= πl(r₁ + r₂) + πr₂² + πr₁² = π [(r₁ + r₂)l + r₁² + r₂²]

Question 7.
Derive the formula for the volume of the frustum of a cone.
Solution:
We have,
[Volume of the frustum RPQS] = [Volume of right circular cone OPQ] – [Volume of right circular cone ORS]

Since, ∆OC₁Q ~ ∆OC₂S [By AA similarity]

From (1) and (2), we have
Volume of the frustum RPQS

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos48° – sin42°
(iv) cosec31°- sec59°
Solution:

(iii) cos 48° – sin 42°
cos 48° = cos (90° – 42°) = sin 42°
[∵ cos (90° – A) = sin A]
∴ cos 48° – sin 42° = sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
cosec 31° = cosec (90° – 59°) = sec 59° [ ∵ cosec (90° – A) = sec A]
∴ cosec 31° – sec 59° = sec 59° – sec 59° = 0

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution;
(i) L.H.S. = tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan 23° tan 42° tan (90° – 23°)
= cot 42° tan 23° tan 42° cot 23° [ ∵ tan (90° – A) = cot A]

(ii) L.H.S. = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin38° sin(90° – 38°)
= cos 38° sin 38° – sin 38° cos 38° [ ∵ sin(90° – A) = cosA and cos(90° – A)= sinA]
= 0 = R.H.S.
⇒ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot(A – 18°)
cot(90 – 2A) = cot (A – 18)
90 – 2A = A – 18
90 + 18 = A + 2A
3A= 108
A = \(\frac {108}{3}\)
A = 36°.

Question 4.
If tan A = cotB, prove that A + B = 90°.
Solution:
tan A = cot B and cot B = tan (90° – B) [∵ tan (90° – θ) = cot θ]
∴ A = 90° – B ⇒ A + B = 90°

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A= cosec (A – 20)
cosec (90 – 4A) = cosec (A – 20)
90 – 4A = A – 20
90 + 20 =A + 4A
110 = 5A
A = \(\frac{110}{5}\)
A = 22°.

Question 6.
If A, 6 and C are interior angles of a triangle ABC, then show that \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)
Solution:
Since, sum of the angles of ∆ABC is 180° i.e.,
A + B + C = 180°
∴ B + C = 180° – A
Dividing both sides by 2, we get

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75° = sin (90 – 23) + cos (90 – 15)
[sin(90 – θ) = cosθ
cos(90 – θ) = sinθ]
= cos 23 + sin 15

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Solution:
(i) Let the vertices of the triangle be A(2, 3), B(-1, 0) and C(2, – 4)
Here, x₁ = 2, y₁ = 3
x₂ = -1, y₂ = 0
x₃ = 2, y₃ = -4
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [2{0 – (-4)} + (-1){-4 – (3)} + 2{3 – 0}]
= \(\frac{1}{2}\) [2(0 + 4) + (-1)(-4 – 3) + 2(3)]
= \(\frac{1}{2}\) [8 + 7 + 6] = \(\frac{1}{2}\) [21] = \(\frac{21}{2}\) sq. units

(ii) Let the vertices of the triangle be A(-5, -1), B(3, -5) and C(5, 2)
Here, x₁ = -5, y₁ = -1
x₂ = 3, y₂ = -5
x₃ = 5, y₃ = 2
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [-5{-5 – 2} + 3{2 – (-1)} + 5{-1 – (-5)}]
= \(\frac{1}{2}\) [-5{-7} + 3{2 + 1} + 5{-1 + 5}]
= \(\frac{1}{2}\) [35 + 3(3) + 5(4)]
= \(\frac{1}{2}\) [35 + 9 + 20] = \(\frac{1}{2}\) × 64 = 32 sq. units

Question 2.
In each of the following find the value of’k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5)
Solution:
The given three points will be collinear if the triangle formed by them has zero area.
(i) Let A(7, -2), B(5, 1) and C(3, k) be the vertices of a triangle.
∴ The given points will be collinear, if ar (∆ABC) = 0
or \(\frac{1}{2}\) [7(1 – k) + 5(k + 2) + 3(-2 – 1)] = 0
⇒ 7 – 7k + 5k + 10 + (-6) – 3 = 0
⇒ 17 – 9 + 5k – 7k = 0
⇒ 8 – 2k = 0 ⇒ 2k = 8 ⇒ k = \(\frac{8}{2}\) = 4
The required value of k = 4.
(ii) \(\frac{1}{2}\) [8(- 4 + 5) + k(- 5 -1) + 2(1 + 4)] = 0
⇒ 8 – 6k + 10 = 0 ⇒ 6k = 18 ⇒ k = 3.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let the vertices of the triangle be A(0, -1), B(2, 1) and C(0, 3).
Let D, E and F be the mid-points of the sides BC, CA and AB respectively.
∴ Coordinates of D are


ar(∆DEF) : ar(∆ABC) = 1 : 4.

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution:
Let A(- 4, – 2), B(- 3, – 5), C(3, – 2) and D(2, 3) be the vertices of the quadrilateral.
Let us join diagonal BD.

Now, ar(∆ABD)

Question 5.
You have studied in class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(A, -6), B(3, -2) and C(5, 2).
Solution:
Here, the vertices of the triangle are A(4, -6), B(3, -2) and C(5, 2).
Let D be the midpoint of BC.
∴ The coordinates of the point D are
\(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\) or ( 4, 0)
AD divides the triangle ABC into two parts i.e., ∆ABD and ∆ACD.

ar(∆ADC) = \(\frac{1}{2}\) [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)]
= \(\frac{1}{2}\) [-8 + 32 – 30] = \(\frac{1}{2}\) [-6] = -3
= 3 sq. units (numerically) ………… (2)
From (1) and (2),
ar(∆ABD) = ar(∆ADC)
Thus, a median divides the triangle into two triangles of equal areas.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 15 Probability Ex 15.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1

Question 1.
Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = _________
(ii) The probability of an event that cannot happen is ______.Such an event is called _________.
(iii) The probability of an event that is certain to happen is ______ Such an event is called ______
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
(v) The probability of an event is greater than or equal to _______ and less than or equal to ______.
Solution:
(i) 1 : Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) 0, impossible: The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) 1, certain: The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
(iv) 1: The sum of the probabilities of all the elementary events of an experiment is 1.
(v) 0, 1: The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) It depends on various factors such as whether the car will start or not. So, the probability of car will start does not equal to the probability of car will not start.
∴ The outcomes are not equally likely.
(ii) It depends on the player’s ability. So, probability that the player shot the ball is not the same as the probability that the player misses the shot.
(iii) The outcomes are equally likely as the probability of answer either right or wrong is \(\frac{1}{2}\)
(iv) The outcomes are equally likely as the probability of ‘newly born baby to be either bay or girls’ is \(\frac{1}{2}\) .

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
Since on tossing a coin, the outcomes ‘head’ and ‘tail’ are equally likely, the result of tossing a coin is completely unpredictable and so it is a fairway.

Question 4.
Which of the following cannot be the probability of an event?
(A) \(\frac{2}{3}\)
(B) -1.5
(C) 15%
(D) 0.7
Solution:
Since, the probability of an event cannot be negative.
∴ -1.5 cannot be the probability of an event.

Question 5.
If P(E) = 0.05, what is the probability of ‘not E’?
Solution:
∵ P(E) + P(not E) = 1
∴ 0.05 + P(not E) = 1 ⇒ P(not E) = 0.95
Thus, probability of ‘not E’ = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Since there are only lemon flavoured candies in the bag.
∴ Taking out orange flavoured candy is not possible.
⇒ Probability of taking out an orange flavoured candy = 0.

(ii) Probability of taking out a lemon flavoured candy = 1.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let the probability of 2 students having same birthday = P(SB)
And the probability of 2 students not having the same birthday = P(NSB)
∴ P(SB) + P(NSB) = 1
⇒ P(SB) + 0.992 = 1 ⇒ P(SB) = 1 – 0.992 = 0.008

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution:
Total number of balls = 3 + 5 = 8
∴ umber of possible outcomes = 8
(i) ∵ There are 3 red balls.
∴ Number of favourable outcomes = 3
∴ P (red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}\)
= \(\frac{3}{8}\)
(ii) Probability of the ball drawn which is not red = 1 – P(red) = \(1-\frac{3}{8}=\frac{8-3}{8}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total number of marbles = 5 + 8 + 4 = 17
∴ Number of all possible outcomes = 17
(i) ∵ Number of red marbles = 5
∴ Number of favourable outcomes = 5
∴ Probability of red marbles, P(red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{5}{17}\)

(ii) Number of white marbles = 8
∴ Probability of white marbles, P(white) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{8}{17}\)
_ Number of favourable outcomes _ 8 Number of all possible outcomes 17

(iii) Number of green marbles = 4 Number of marbles which are not green
= 17-4 = 13
i.e., Favourable outcomes = 13
∴ Probability of marbles ‘not green’, P(not greeen)
\(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50p coin?
(ii) will not be a ₹ 5 coin?
Solution:
Number of coins 50 p = 100, ₹ 1 = 50 ₹ 2 = 20, ₹ 5 = 10
Total number of coins = 100 + 50 + 20 +10 = 180
∴ Total possible outcomes = 180

(i) For a 50 p coin:
Favourable outcomes = 100

(ii) For not a ₹ 5 coin:
Y Number of ₹ 5 coins = 10
∴ Number of ‘not ₹ 5’ coins = 180 – 10 = 170
⇒ Favourable outcomes = 170

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Number of male fishes = 5
Number of female fishes = 8
∴ Total number of fishes = 5 + 8 = 13
⇒ Total number of outcomes = 13
For a male fish:
Number of favourable outcomes = 5

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution:
Total number marked = 8
∴ Total number of possible outcomes = 8
(i) When pointer points at 8:
Number of favourable outcomes = 1

(ii) When pointer points at an odd number:
∵ Odd numbers are 1, 3, 5 and 7
∴ Total odd numbers from 1 to 8 = 4
⇒ Number of favourable outcomes = 4

(iii) When pointer points at a number greater than 2:
∵ The numbers 3, 4, 5, 6, 7 and 8 are greater than 2
∴ Total numbers greater than 2 = 6
⇒ Number of favourable outcomes = 6

(iv) When pointer points at a number less than 9:
∵ The numbers 1, 2, 3, 4, 5, 6, 7 and 8 are less than 9.
∴ Total numbers less than 9 = 8
∴ Number of favourable outcomes = 8

Question 13.
A die is thrown once. Find the probability of getting:
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Since, numbers on a die are 1, 2, 3, 4, 5 and 6.
∴ Total number of possible outcomes = 6
(i) Since 2, 3 and 5 are prime number.
∴ Favourable outcomes = 3

(ii) Since the numbers between 2 and 6 are 3, 4 and 5
∴ Favourable outcomes = 3

(iii) Since 1, 3 and 5 are odd numbers.
⇒ Favourable outcomes = 3

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Solution:
Number of cards in deck = 52
∴ Total number of possible outcomes = 52
(i) ∵ Number of red colour kings = 2
[∵ King of diamond and heart is red]
Number of favourable outcomes = 2

(ii) For a face card:
∵ 4 kings, 4 queens and 4 jacks are face cards
∴ Number of face cards = 12
⇒ Number of favourable outcomes = 12

(iii) Since, cards of diamond and heart are red
∴ There are 2 kings, 2 queens, 2 jacks i.e., 6 cards are red face cards.
∴ Number of favorable outcomes = 6

(iv) Since, there is only 1 jack of hearts.
∴ Number of favourable outcomes = 1

(v) There are 13 spades in a pack of 52 cards.
∴ Number of favourable outcomes = 13

(vi) ∵ There is only one queen of diamond.
∴ Number of favourable outcomes = 1

Question 15.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
We have five cards.
∴ Total number of possible outcomes = 5
(i) ∵ Number of queen = 1
∴ Number of favourable outcomes = 1

(ii) The queen is drawn and put aside.
∴ Only 5 – 1 = 4 cards are left.
∴ Total number of possible outcomes = 4
(a) ∵ There is only one ace.
∴ Number of favourable outcomes = 1

(b) Since, the only queen has been put aside already.
∴ Number of favourable outcomes = 0

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
We have number of good pens = 132 and number of defective pens = 12
∴ Total number of pens = 132 + 12 = 144 = Total possible outcomes
There are 132 good pens.
∴ Number of favourable outcomes = 132

Question 17.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
Since, there are 20 bulbs in the lot.
Total number of possible outcomes = 20
(i) ∵ Number of defective bulbs = 4
∴ Favourable outcomes = 4

(ii) ∵ The bulb drawn above is not included in the lot.
∴ Number of remaining bulbs = 20 – 1 = 19.
⇒ Total number of possible outcomes = 19.
∵ Number of bulbs which are not defective = 19 – 4 = 15
⇒ Number of favourable outcomes = 15

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
We have total number of discs = 90
∴ Total number of possible outcomes = 90
(i) Since the two-digit numbers are 10, 11, 12, ………, 90.
∴ Number of two-digit numbers = 90 – 9 = 81
∴ Number of favourable outcomes = 81

(ii) Perfect square from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
∴ Number of perfect squares = 9
∴ Number of favourable outcomes = 9

(iii) Numbers divisible by 5 from 1 to 90 are 5, 10,15, 20, 25, 30, 35,40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
i. e., There are 18 numbers from (1 to 90) which are divisible by 5.
∴ Numbers of favourable outcomes = 18

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Since there are six faces of the given die and these faces are marked with letters

∴ Total number of letters = 6
∴ Total number of possible outcomes = 6
(i) ∵ Number of faces having the letter A = 2
∴ Number of favourable outcomes = 2

(ii) ∵ Number of faces having the letter D = 1
∴ Number of favourable outcomes = 1

Question 20.
20. Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution:
Here, area of the rectangle = 3m × 2m = 6 m²
And, the area of the circle = πr²

Question 21.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it? (ii) She will not buy it?
Solution:
Total number of ball pens = 144
⇒ Total number of possible outcomes = 144
(i) Since there are 20 defective pens.
∴ Number of good pens = 144 – 20 = 124
⇒ Number of favourable outcomes = 124

(ii) Probability that Nuri will not buy it = 1 – [Probability that she will buy it]
= \(1-\frac{31}{36}=\frac{36-31}{36}=\frac{5}{36}\)

Question 22.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. An event is defined as the sum of the two numbers appearing on the top of the dice.
(i) Complete the following table

(ii) A student argues that’there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\) Do you agree with this argument? Justify your answer.
Solution:
∵ The two dice are thrown together.
∴ Following are the possible outcomes :
(1, 1) ; (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1) ; (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1) ; (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6,.2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ Total number of possible outcomes is 6 × 6 = 36
(i) (a) The sum on two dice is 3 for (1, 2) and (2, 1)
∴ Number of favourable outcomes = 2
⇒ P(3) = \(\frac{2}{36}\)

(b) The sum on two dice is 4 for (1, 3), (2, 2) and (3, 1).
∴ Number of favourable outcomes = 3
⇒ P(4) = \(\frac{3}{36}\)

(c) The sum on two dice is 5 for (1, 4), (2, 3), (3, 2) and (4,1)
∴ Number of favourable outcomes = 4
⇒ P(5) = \(\frac{5}{36}\)

(d) The sum on two dice is 6 for (1, 5), (2, 4), (3, 3), (4, 2) and (5,1)
∴ Number of favourable outcomes = 5
⇒ P(6) = \(\frac{5}{36}\)

(e) The sum on two dice is 7 for (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1)
∴ Number of favourable outcomes = 6
⇒ P(7) = \(\frac{62}{36}\)

(f) The sum on two dice is 9 for (3, 6), (4, 5), (5, 4) and (6, 3)
∴ Number of favourable outcomes = 4
⇒ P(9) = \(\frac{4}{36}\)

(g) The sum on two dice is 10 for (4, 6), (5, 5), (6,4)
∴ Number of favourable outcomes = 3
⇒ P(10) = \(\frac{3}{36}\)

(h) The sum on two dice is 11 for (5, 6) and (6,5)
∴ Number of favourable outcomes = 2
⇒ P(11) = \(\frac{2}{36}\)

Thus, the complete table is as follows:

(ii) No. The number of all possible outcomes is 36 not 11.
∴ The argument is not correct.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Let T denotes the tail and H denotes the head.
∴ All the possible outcomes are:
{H H H, H H T, H T T, T T T, T T H, T H T, T H H, H T H)
∴ Number of all possible outcomes = 8
Let the event that Hanif will lose the game denoted by E.
∴ Favourable events are: {HHT, HTH, THH, THT, TTH, HTT}
⇒ Number of favourable outcomes = 6
∴ P(E) = \(\frac{6}{8}=\frac{3}{4}\)

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Solution:
Since, throwing a die twice or throwing two dice simultaneously is the same.
∴ All possible outcomes are:
(1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1) ; (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6, 2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ All possible outcomes = 36
(i) Let E be the event that 5 does not come up either time.
∴ Numebr of favourable outcomes = [36 – (5 + 6)] = 25
∴ P(E) = \(\frac{25}{36}\)
(ii) Let N be the event that 5 will come up at least once, then number of favourable outcomes = 5 + 6 = 11
∴ P(N) = \(\frac{11}{36}\)

Question 25.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
(ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{3}\)
Solution:
(i) Given argument is not correct. Because, if two coins are tossed simultaneously then four outcomes are possible (HH, HT, TH, TT). So total outcomes is 4.
∴ The required probability = \(\frac{1}{4}\).
(ii) Given argument is correct.
Since, total numebr of possible outcomes = 6
Odd numbers = 3 and even numbers = 3
So, favourable outcomes = 3 (in both the cases even or odd).
∴ Probability = \(\frac{3}{6}=\frac{1}{2}\)

MP Board Class 10th Science Solutions Chapter 9 Heredity and Evolution

MP Board Class 10th Science Chapter 9 Intext Questions

Class 10th Science Chapter 9 Intext Questions Page No.143

Question 1.
If a trait Aexists in 10% of a population of an asexually reproducing species and trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait ‘B’ is likely to have arisen earlier. Because in asexually reproducing species, small differences are seen due to DNA replication.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
Depending on the nature of variations, different individuals would have different kinds of advantages. Bacteria that can withstand heat will better in a heat wave, as selection of variants by environmental factors forms the basis for evolutionary process.

Class 10th Science Chapter 9 Intext Questions Page No. 147

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
Mendel used a number 0 contrasting visible characters of garden peas-round/wrinkled seeds, tall/short plants, white/violet flowers and so on. He took pea plants with different characteristics a tall plant and a shot plant, produced progeny by crossing then, and calculated the percentages of tall or short progeny.

In the first place, there were no halfway characteristics in this first generation, or F₁ progeny no medium-height plants. All plants were tall. This means that only one of the parental traits was seen, not some mixture of the two, so the next question was were the tall plants in the F₁ generation exactly the same as the tall plants of the parent generation? Mendelian experiments test this by getting both the parental plans and these F₁ tall plants to reproduce by self-pollination.

The progeny of the parental plants are of course, all tall. However the second generation, or F₂ progeny of the F₁ tall plants are not all tall. Instead, one quarter of them are short. This indicates that both the tallness and shortness traits were inherited in the F₁ plants, but only the tallness trait was expressed. This led Mendel to propose that two copies of factor (now called genes) controlling traits are present in sexually reproducing organism. These two may be identified or may be different, depending on the percentage. A pattern of inheritance can be worked out with this assumption as shown in figure.

In this explanation, both TT and Tt are tall plants, while only it is a short plant. Traits like T are called dominant traits, while those that behave like ‘t’ are called recessive traits.

Question 2.

How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel crossed

1. pure breeding tall plants having round seeds and
2. pure breeding short plants having wrinkled seeds.

The plants of F1 generation were all tall with round seeds indicating that the traits of tallness and round seeds were dominant.

While self breeding, of F1 yielded plants with characters of 9 tall round seeded, 3 tall wrinkled seeded, 3 short round seeded and one short wrinkled seeded. Tall wrinkled and short round seeded plants are new combinations which can develop only when the traits are inherited independently.

Question 3.
A man with blood group A marries a woman with blood group C and their daughter has blood group O. Is this information enougl to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
From this information, it is not possible to tell which of the traits-blood group A or O is dominant. AA group becomes AO. So this information is complete.

Question 4.
How is the sex of the child determined in human beings?
Answer:
Pair of sex chromosomes determines the particular sex of a child. In human, the males have one X and one Y chromosome and the females have two X chromosomes therefore, the females are XX and the males are XY. The gametes receive half of the chromosomes. The child gametes have 22 autosomes and either X or Y sex chromosome in males while X in females.
Type of male gametes: 22 + X Or 22 + Y.
Type of female gamete: 22 + X.
This is the basis of sex determination in human beings.

Class 10th Science Chapter 9 Intext questions Page No. 150

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:
A big reason could be accurate copying of DNA and limited variations. Variation causes in generation of new traits and non preserving of existing parental traits. Individuals with a particular trait may also increase due to natural selection that is trait offers some survival advantage. Genetic drift which is caused by genes governing that trait become common in a population.

Question 2.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Variation is not hereditary from generation to generation. In case of Asexual reproduction DNA will not transfer to germ cells. So experiences of an individual during its lifetime cannot be passed on to its progeny and cannot direct evolution.

Question 3.
Why are the small numbers of survivingjigers a cause of worry from the point of view of genetics?
Answer:
Tigers are adopted to their environment as per genes. If tigers number is decreasing number of genes also decrease. So its generation is becoming less.

Class 10th Science Chapter 9 Intext Questions Page No. 151

Question 1.
What factors could lead to the rise of a new species?
Answer:
Variations, non copying of DNA, natural selection, genetic drift and acquisition of traits during the life time of an individual can give rise to new species.

Question 2.
Will geographical isolation be a major factor in the speciation of a self-pollinating plant species? Why or why not?
Answer:
Geographical isolation will not a major factor in the speciation of a self pollinating plant species, because self pollination is taking place in one plant. In cross pollination it is a major factor.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
Geographical isolation will not be a major factor in the speciation of an organism that reproduces asexually because it is a major factor in organisms that reproduces sexually.

Class 10th Science Chapter 9 Intext Questions Page No. 156

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
The characteristics in different organisms would be similar because they are inherited from a common ancestor. As an example, consider the fact that mammals have four limbs, as do birds, reptiles and amphibians. The basic structure of the limbs is similar through it has been modified to perform different functions in various vertebrates. Such a homologous characteristic helps to identify an evolutionary relationship between apparently different species.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
The wing of a butterfly and the wing of a bat be is not considered homologous organs because the designs of the two wings, their structure and components are very different.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
Fossils are the remains of organisms that once existed on earth.

We can gather information about the development of the structures from simple structured to complex structured organisms. They tell us about the phases of evolutions through which they must have undergone in order to sustain themselves in the competitive environment.

Class 10th Science Chapter 9 Intext Questions Page No. 158

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Ans.
A species is a group of organisms that are capable of interbreeding to produce a fertile offspring. Skin colour, looks and size are all variety of features present in human beings. These features are genetic but also environmentally controlled. Various human races are formed based on these features. All human races have more than enough similarities to be classified as same species. Therefore, all human beings are a single species as humans of different colour, size and looks are capable of reproduction and can produce a fertile off spring.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:
Bacteria is a better body design, because even though it a simple organism, it can survive in hot springs, at the bottom of sea and even in coldest ice covered place such as Antarctica.

MP Board Class 10th Science Chapter 9 NCERT Textbook Exercises

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as:
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW. Genetic make up of tall plant.

Question 2.
An example of homologous organs is:
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer:
(d) Both organs in all option have same basic structural design but have different functions and appearance.

Question 3.
In evolutionary terms, we have more in common with:
(a) a Chinese school-boy
(b) a chimpanzee
(c) a spider
(d) a bacterium
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
No, we cannot say anything about whether the light eye colour trait is dominant or recessive. As this information is not sufficient. For considering a trait as dominant or recessive, we need data of at least last three generations.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:
An example will help with this A brother and a sister are closely related. They have common ancestors in the first generation before them, namely their parents. A girl and her first cousin are also related, but less than the girl and her brother. This is because cousins have common ancestors, their grandparents, in the second generation before them, not in the first one. We can now appreciate that classification of species is in fact a reflection of their evolutionary relationship.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:

Consider the fact that mammals have four limbs, as do birds, reptiles and amphibians. The basic structure of the limbs is similar though it has been modified to perform different functions in various vertebrates. This is an example of homologous characteristic.

We find that the wings of bats are skin folds stretched mainly between elongated fingers. But the wings of birds are feathery covering all along the arm. The design of the two wings, their structure and components, are thus very different. They look similar because they have a common use for flying, but their origins are not common. This makes them analogous characteristics, rather than homologous characteristics.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
There are variety of genes that govern coat colour of a dog. At least eleven identified gene series (A, B, C, D, E, F, G, M, P, S, T) that influence coat colour in dog.

A dog inherits one gene from each of its parents. The dominant gene gets expressed in the phenotype. For example, in the B series, a dog can be genetically black or brown.

Let us assume that one parent is homozygous black (BB), while the other parent is homozygous brown (bb)

In this case, all the off springs will be heterozygous (Bb).
Since black (B) is dominant, all the offsprings will be black. However, they will have both B and b alleles.
If such heterozygous pups are crossed, they will produce 25% homozygous black (BB), 50% heterozygous black (Bb), and 25% homozygous brown (bb) offsprings.

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
Analysis of the organ structure in fossils allows us to make estimates of how far back evolutionary relationships go. The wild cabbage plant is a good example. Humans have over more two thousand years, cultivated wild cabbage as a food plant, and generated different vegetables from it by selection. This is of course, artificial selection rather than natural selection. Kale, cauliflower. Broccoli, cabbage, Red cabbage and Kohl rabi all these have same ancestor.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
As we all know, that there occurs a time when our planet was lifeless, at that time intial matter to develop life was water, sand and some atmospheric gases as CO₂, CH4 and nitrogen. The evidence for the origin of life from inanimate matter, was provided through an experiment, conducted in 1953, by Stanley L. Miller and Harold C. Urey. In experiment, they assembled an atmosphere containing molecules like ammonia, methane and hydrogen sulphide, but not oxygen.

This was similar to atmosphere that existed on early earth . This was maintained at a temperature just below 100°C and sparks were passed through the mixture of gases to simulate lightning. At the end of a week, 15% of the carbon from methane, had been converted to simple compounds of carbon including amino acids which make up protein molecules and support the life in basic form.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:

Change in on-reproductive tissues cannot be passed on to the DNA of the germ cell. Therefore the experiences of an individual during its lifetime cannot be passed on to its progeny and cannot direct evolution.

Ex: If we breed a group of mice, all their progeny will have tails, as expected. Now, if the tails of these mice are removed by surgery in each generation, do these tailless mice have tailless progeny? The answer is no and it makes sense because removal of the tail cannot change the genes of the germ cells of the mice. Hence sexual reproduction gives rise to more viable variations than asexual reproduction.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Genetic inheritance begins at the time of conception, progeny inherited 23 chromosomes from female parent and 23 from male parent. Together it form 22 pairs of autosomal chromosomes and a pair of sex chromosomes (either XX in case of female, or XY in male). Homologous chromosomes have the same genes in the same positions, but may have different alleles (varieties) of those genes. An individual has two copies of alleles, and that can be homozygous (both copies the same) or heterozygous (the two copies are different) for given gene.

Hence, in human beings, equal genetic contribution of male and female parents is ensured in the progeny through inheritance of equal number of chromosomes from both parents. Females have a equal pair of two X sex chromosomes and males have a pair of one X and one Y sex chromosome. As fertilisation takes place, the male gamete (haploid) fuses with the female gamete (haploid) resulting in formation of the diploid zygote. The zygote in the progeny receive an equal contribution of genetic traits from the parental generations.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Only variations that confer an advantage to an individual organism will survive in a population we agree to this statement variation is convenient for survival. This provides diversity for organisms.

MP Board Class 10th Science Chapter 9 Additional Important Questions

MP Board Class 10th Science Chapter 9 Multiple Choice Questions

Question 1.
Fossil archaeopteryx exhibits connection between:
(a) Amphibian and fish
(b) Reptiles and fish
(c) Reptile and birds
(d) Birds and mammals
Answer:
(c) Reptile and birds

Question 2.
The sex of the human child depends on the sex chromosome present in the:
(a) Egg
(b) Spenn
(c) Both (a) and (b)
(d) None of these
Answer:
(b) Spenn

Question 3.
Genetic information is carried out by long chain of molecules made up of:
(a) Enzymes
(b) DNA
(c) Amino acids
(d) Proteins
Answer:
(b) DNA

Question 4.
Which one of the following represents a ratio of monohybrid cross?
(a) 9 : 7
(b) 3 : 1
(c) 1 : 1 : 1 : 1
(d) 9 : 3 : 3 : 1
Answer:
(b) 3 : 1

Question 5.
On which plant Mendel carried his experiments of inheritance?
(a) Cow pea
(b) Wild pea
(c) Garden pea
(d) Pigeon pea
Answer:
(c) Garden pea

Question 6.
A gamete certains which of the following?
(a) Both alleles of a gene
(b) Only one allele of a gene
(c) All alleles of a gene
(d) No allele of a gene
Answer:
(b) Only one allele of a gene

Question 7.
Chromosomes are made up of
(a) Proteins
(b) DNA
(c) RNA
(d) All of these.
Answer:
(d) All of these.

Question 8.
Pea plants were more suitable than cats for Mendel’s experiments because:
(a) Cats have many genetic traits
(b) No pedigree record of cats
(c) Pea plants can be self-pollinated or fertilised
(d) Pea plants favour cross pollination.
Answer:
(c) Pea plants can be self-pollinated or fertilised

Question 9.
In a cross Tt × Tt, the percentage of offsprings produced having same phenotype as the parents would be:
(a) 50%
(b) 100%
(c) 25%
(d) 0%
Answer:
(a) 50%

Question 10.
Who proposed the laws of heredity?
(a) Darwin
(b) Mendel
(c) Morgan
(d) Dalton
Answer:
(b) Mendel

MP Board Class 10th Science Chapter 9 Very Short Answer Type Questions

Question 1.
Define heredity.
Answer:
The process by which traits and characteristics are reliably inherited or passed from the parents to the offspring is called heredity.

Question 2.
What is a gene?
Answer:
Gene is a functional segment of DNA on a chromosome occupying specific position, which carries out a specific biological function.

Question 3.
Name the plant on which Mendel performed his experiments.
Answer:
Garden pea (Pisum sativum).

Question 4.
Define the term variation.
Answer:
Variation: There are differences found in structure, function, behaviour and genetic make up of different individuals of the same parentage, variety, race and species. These differences refer to variation.

Question 5.
Write the expanded form of DNA.
Answer:
Deoxyribonucleic acid (DNA).

Question 6.
Define genetics.
Answer:
The branch of biology which deals with heredity and variations, is known as genetics.

Question 7.
Define the term offspring.
Answer:
Offspring is an individual formed as a result of sexual reproduction involving the formation and fusion of two gametes. The genotype of an offspring is different from either of the parents due to shuffling of chromosomes and their genes.

Question 8.
What are reciprocal crosses?
Answer:
They are two types of crosses involving two groups of individuals where the male of one group is crossed with the female of the other and vice versa.

Question 9.
Where are the genes located? What is the chemical nature of gene?
Answer:
Genes are located at a specific position on a chromosome. Chemical Nature of Gene: Chemically, gene is a segment of deoxyribonucleic acid (DNA) consisting of specific sequence of the nucleotides. The sequence of the constituent nucleotides determines the functional property of a gene.

MP Board Class 10th Science Chapter 9 Short Answer Type Questions

Question 1.
Define genetics. What is the contribution of Mendel in this branch of Biology?
Answer:
Genetics is the branch of science of heredity and variations which deals with the study of the transmission of traits from parents to the offsprings and the occurrence of differences among the individuals.

Contribution of Mendel: Mendel did his experiments on garden pea (Pisum sativum) and discovered the scientific principles, which govern patterns of inheritance i.e., the principle of inheritance. He explained that contrasting characters are controlled by units which he called ‘Factors Today, these factors are called genes.

Question 2.
Differentiate between inherited and acquired traits.
Answer:
Inherited traits:

1. The traits which are inherited from the parents (Father and Mother) to the offsprings (progeny) are called inherited traits.
2. These traits are due to genetic make up of the progeny.

Acquired characters

1. These traits cannot be passed on to their future generations.
2. These traits develop in response to the environment.

Question 3.
What are Mendel’s laws of inheritance?
Answer:
Law of dominance: When two homozygous individuals with one or more sets of contrasting characters are crossed the characters that appear in the F, hybrids are dominant characters.

Law of segregation: Contrasting characters brought together in hybrid remain together without being contaminated and when gametes are formed from the hybrid, the two separate out from each other and only one enters each gamete.

Law of independent assortment: In inheritance of more than one pair of contrasting characters simultaneously, the factors for each pair of characters assort independently of other pair’s.

Question 4.
How did life originate on earth?
Answer:
Life originated on earth from inorganic elements and compounds under extreme atmospheric conditions (such as very high temperature, electric discharges, reducing atmosphere etc.) by formation of complex organic compounds such as amino acids.

Question 5.
Why did Mendel choose garden pea for his experiments?
Answer:
Due to the following reasons, Mendel selected garden pea for his experiment:

1. Garden pea flowers are normally self-pollinated but can be easily cross-pollinated.
2. Many varieties with distinguished contrasting characters e.g., smooth seed coat, wrinkled seed coat are available.
3. A large number of progeny can be produced in a short duration.
4. Its flowers can be easily handled for experimentation.

Question 6.
What are the factors which help in speciation?

1. Genetic drift: Due to genetic drift, there will be accumulation of # different changes in each sub-populations. The levels of gene flow ’ between them will decrease if they are further isolated, it will be more on a small sub-population.
2. Over generations, genetic drift will accumulate, causing different changes in the populations.
3. Natural selection may also operate differently in the different geographical location.
4. Together, genetic and natural selection will make the population more and more different from each other. As a result, members will be incapable of reproducing with each other. Changes may be due to change in DNA or number of chromosomes.

Question 7.
Does geographical isolation of individuals of a species lead to formation of a new species? Provide a suitable explanation.
Answer:
Yes, geographical isolation of sub-populations of a population of a species leads to genetic drift. This may impose limitations to.sexual reproduction of the separated population. Slowly, the separated individuals will reproduce among themselves and generate new variations. Continuous accumulation of those variations through a few generations may ultimately lead to the formation of a new species.

Question 8.
What tools have been used to study human evolution?
Answer:
The tools used for tracing evolutionary line are:

1. Excavating time – dating and study of fossils.
2. Determining DNA sequences.

MP Board Class 10th Science Chapter 9 Long Answer Type Questions

Question 1.
A husband have 46 chromosomes, his wife has 46 chromosomes. Then, why don’t their offspring have 46 pairs of chromosomes, which is obtained by the fusion of male and female gametes? Support your answer with a neat illustration.
Answer:
At the stage of gamete formation, meiosis division (reduction division) occurs. As a result, each gamete receives half number of chromosomes of the parent. So, when male gamete (sperm) fuses with egg, original number of chromosomes of the parent is received by the zygote.

Fertilisation

Question 2.
Name the characters studied by Mendel in garden pea.
Answer:

Question 3.
Explain the terms:
Monohybrid cross, dihybrid cross, monohybrid ratio and dihybrid ratio.
Answer:
Monohybrid cross: Monohybrid cross is that cross which is made to study the inheritance of a single pair of genes or factors of a character.

Dihybrid cross: It is a cross which is made to study the inheritance of two pairs of genes or two characters.

Monohybrid ratio: It is the ratio which is obtained in the F2 generation when a monohybrid cross is made. It is usually 3 : 1 (Phenotypic ratio) or 1 : 2 : 1 (genotypic ratio).

Dihybrid ratio: It is the ratio, which is obtained in the F2 generation when a dihybrid cross is studied. It is usually 9 : 3 : 3 : 1 (phenotypic ratio).

Question 4.
Describe any three methods of tracing evolutionary relationships among organisms.
Answer:
The following methods help us in tracing evolutionary relationships:

(i) Study of homologous organs: Organs which have similar structure and origin are called homologous organs. For example: limbs of birds, frog, human may look different but they have similar structure and origin. Such homologous organs help to identify an evolutionary relationship between apparently diffejent species.

(ii) Study of analogous organs: Analogous organs are similar in function but differ in structure and origin. For example’. forelimbs of birds and bats are used for flying but their origins and components are not common. Thus, study of analogous organs reveals difference in their ancestry and their evolutionary relationship.

(iii) Study of fossils: All impressions, casting of body or hard remains of ancient life in the sedimentary rocks are called fossils. Study of fossils helps in finding out:
(a) Interrelationship of ancient life.
(b) Correlation of forms of life existing today and their line of evolution from ancient life.

Question 5.
A tall pea plant bearing violet flowers is crossed with short pea plant bearing white flowers. Work out the F1 and F2 generations. Give F2 ratio.
Answer:
Parents: Tall pea plant with violet flower × Short pea plant with white flower

Question 6.
Given below is the experiment carried out by Mendel to study inheritance of two traits in garden pea:
(а) What do A, B, C, D, E, F and G represent in these boxes?
(b) State the objective for which Mendel performed this experiment.

Independent inheritance of two separate traits, shape, and colour of seeds.
Answer:
(a) A = gamete (Ry) of round green plant.
B = gamete (rY) of wrinkled yellow plant.
C = (RrYy).
D = 9, E = 3, F = 3, G = 1.
(b) To show independent inheritance of traits or to prove law of independent assortment.

MP Board Class 10th Science Chapter 9 NCERT Textbook Activities

Class 10 Science Activity 9.1 Page No. 143

Observe the ears of all the students in the class. Prepare a list of students having free or attached earlobes and calculate the percentage of students having each (Fig. 9.1). Find out about the earlobes of the parents of each student in the class. Correlate the earlobe type of each student with that of their parents. Based on this evidence, suggest a possible rule for the inheritance of earlobe types.

(a) Free and (b) attached earlobes. The lowest part of the ear, called the earlobe, is closely attached to the side of the head in some of us, and not in others. Free and attached earlobes are two variants found in human populations.

Observations:

• Earlobes can be free or attached. The genes for earlobe inheritance consists of two alleles. Both the alleles for attached and free earlobes can be present in a single human being, the one which is dominant shows and the recessive one do not express itself,

Class 10 Science Activity 9.2 Page No. 144

• In Fig. 9.2, what experiment would we do to confirm that the F2 generation did in fact have a 1 : 2 : 1 ratio of TT, Tt and tt trait combinations?

Inheritance of traits over two generations.

Observations:
The cross fertilisation of pea plants showing different traits can be done at F2 stage and the number of plants for particular trait (height here) can be studied and used to confirm 1 : 2 : 1 ratio of TT, Tt and it

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 10 Circles Ex 10.2

In questions 1 to 3, choose the correct option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
(A): ∵ QT is a tangent to the circle at T and OT is radius

∴ OT⊥QT
Also, OQ = 25 cm and QT = 24 cm
∴ Using Pythagoras theorem, we get
OQ² = QT² + OT²
⇒ OT² = OQ² – QT² = 25² – 24² = 49
⇒ OT = 7
Thus, the required radius is 7 cm.

Question 2.
In figure, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ =110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
(B): TQ and TP are tangents to a circle with centre O and ∠POQ = 110°
∴ OP⊥PT and OQ⊥QT
⇒ ∠OPT = 90° and ∠OQT = 90°
Now, in the quadrilateral TPOQ, we get
∠PTQ + 90° + 110° + 90° = 360° [Angle sum property of a quadrilateral]
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 360° – 290° = 70°

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
(A) : Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.
∴ OA⊥AP and OB⊥BP
⇒ ∠OAP = ∠OBP = 90°

Now, in quadrilateral PAOB, we have
∠BPA + ∠PAO + ∠AOB + ∠OBP = 360°
⇒ 80° + 90° + ∠AOB + 90° = 360°
⇒ 260° + ∠AOB = 360°
⇒ ∠AOB = 360° – 260° ⇒ ∠AOB = 100°
In right ∆OAP and right ∆OBP, we have
OP = OP [Common]
∠OAP = ∠OBP [Each 90°]
OA = OB [Radii of the same circle]
∴ ∆OAP ≅ ∆OBP [By RHS congruency]
⇒ ∠POA = ∠POB [By CPCT]
∴ ∠POA = \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 100° = 50°

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
In the figure, PQ is diameter of the given circle and O is its centre.
Let tangents AB and CD be drawn at the end points of the diameter PQ.
Since, the tangents at a point to a circle is perpendicular to the radius through the point.

∴ PQ⊥AB
⇒ ∠APQ = 90°
And PQ⊥CD
⇒ ∠PQD = 90° ⇒ ∠APQ = ∠PQD
But they form a pair of alternate angles.
∴ AB || CD
Hence, the two tangents are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
In the figure, the centre of the circle is O and tangent AB touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.
Join OP.
Since, tangent at a point to a circle is perpendicular to the radius through that point.

∴ OP⊥AB
⇒ ∠OPB = 90° ……….. (1)
But by construction, PQ⊥AB
⇒ ∠QPB = 90° ………….. (2)
From (1) and (2),
∠QPB = ∠OPB
which is possible only when O and Q coincide. Thus, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
∵ The tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OTA = 90°
Now, in the right ∆OTA, we have
OA² = OT² + AT² [Pythagoras theorem]

⇒ OT² = 5² – 4²
⇒ OT² = (5 – 4)(5 + 4)
⇒ OT² = 1 × 9 = 9 = 3²
⇒ OT = 3
Thus, the radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
In the figure, O is the common centre, of the given concentric circles.
AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.
Since, OP is the radius of the smaller circle.
∴ OP⊥AB ⇒ ∠APO = 90°
Also, radius perpendicular to a chord bisects the chord.

∴ OP bisects AB
⇒ AP = \(\frac{1}{2}\) AB
Now, in right ∆APO,
OA² = AP² + OP²
⇒ 5² = AP² + 3² ⇒ AP² = 5² – 3²
⇒ AP² = 4² ⇒ AP = 4 cm
⇒ \(\frac{1}{2}\) AB = 4 ⇒ AB = 2 × 4 = 8 cm
Hence, the required length of the chord AB is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure).

Prove that AB + CD = AD + BC
Solution:
Since, the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touches the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.
∴ AP = AS, BP = BQ,
DR = DS and CR = CQ
Adding them, we get
(AP + BP) + (CR + RD) = (BQ + QQ) + (DS + SA)
⇒ AB + CD = BC + DA

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.

Prove that ∠AOB = 90°.
Solution:
∵ The tangents drawn to a circle from an external point are equal.
∴ AP = AC ……… (1)
Join OC.
In ∆PAO and ∆CAO, we have
AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC [From (1)]
⇒ ∆PAO ≅ ∆CAO [SSS congruency]
∴ ∠PAO = ∠CAO
⇒ ∠PAC = 2∠CAO …………. (2)
Similarly, ∠CBQ = 2∠CBO ……………… (3)

Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
2∠CAO + 2∠CBO = 180° [From (2) and (3)]
⇒ ∠CAO + ∠CBO = \(\frac{180^{\circ}}{2}\) = 90° …………… (4)
Also, in ∆AOB,
∠BAO + ∠OBA + ∠AOB = 180° [Sum of angles of a triangle]
⇒ ∠CAO + ∠CBO + ∠AOB = 180°
⇒ 90° + ∠AOB = 180° [From (4)]
⇒ ∠AOB = 180° – 90°
⇒ ∠AOB = 90°

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution:
Let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right ∆OAP and right ∆OBP, we have
PA = PB [Tangents to circle from an external point]
OA = OB [Radii of the same circle]
OP = OP [Common]
⇒ ∆OAP ≅ ∆OBP [By SSS congruency]
∴ ∠OPA = ∠OPB [By CPCT]
and ∠AOP = ∠BOP
⇒ ∠APB = 2∠OPA and ∠AOB = 2∠AOP
In right ∆OAP,
∠AOP + ∠OPA + ∠PAO = 180°
⇒ ∠AOP = 180° – 90° – ∠OPA
⇒ ∠AOP = 90° – ∠OPA
⇒ 2∠AOP = 180° – 2∠OPA
⇒ ∠AOB = 180° – ∠APB
⇒ ∠AOB + ∠APB = 180°

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre O.
Since, tangents to a circle from an external point are equal in length

∴ AP = AS
BP = BQ
CR = CQ
DR = DS
On adding, we get
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
But AB = CD [Opposite sides of parallelogram]
and BC = AD
∴ AB + CD = AD + BC ⇒ 2AB = 2BC
⇒ AB = BC
Similarly, AB = DA and DA = CD
Thus, AB = BC = CD = DA
Hence, ABCD is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Solution:
Here ∆ABC circumscribes the circle with centre O. Also, radius = 4 cm
Let AC and AB touches the circle at E and F, respectively and join OE and OF.
∵ The sides BC, CA and AB touches the circle at D, E and F respectively.
∴ BF = BD = 8 cm
[ ∵ Tangents to a circle from an external point are equal]
CD = CE = 6 cm
AF = AE = x cm (say)

∴ The sides of the ∆ABC are 14 cm, (x + 6) cm and (x + 8) cm
Perimeter of ∆ABC
= [14 + (x + 6) + (x + 8)] cm
= [14 + 6 + 8 + 2x] cm
= (28 + 2x) cm
⇒ Semi perimeter of ∆ABC,
s = \(\frac{1}{2}\) [28 + 2x] cm = (14 + x) cm
∴ s – a = (14 + x) – (8 + x) = 6
s – b = (14 + x) – (14) = x
s – c = (14 + x) – (6 + x) = 8
where, a = AB, b = BC, c = AC

Squaring both sides, we get
(14 + x)² = (14 + x)3x
⇒ 196 + x² + 28x = 42x + 3x²
⇒ 2x² + 14x – 196 = 0
⇒ x² + 7x – 98 = 0
⇒ (x – 7)(x + 14) = 0
⇒ x – 7 = 0 or x + 14 = 0
⇒ x = 7 or x = -14
But x = -14 is rejected.
∴ x = 7
Thus, AB = 8 + 7 = 15 cm, BC = 8 + 6 = 14 cm and CA = 6 + 7 = 13 cm

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
We have a circle with centre O. A quadrilateral ABCD is such that the sides AB, BC, CD and DA touches the circle at P, Q, R and S respectively.
Join OP, OQ, OR and OS.
We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2
∠3 = ∠4
∠5 = ∠6 and ∠7 = ∠8
Also, the sum of all the angles around a point is 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
∴ 2(∠1 + ∠8 + ∠5 + ∠4) = 360°
⇒ (∠1 + ∠8 + ∠5 + ∠4) = 180° …………. (1)
and 2(∠2 + ∠3 + ∠6 + ∠7) = 360°
⇒ (∠2 + ∠3 + ∠6 + ∠7) = 180° ……………. (2)
Since, ∠2 + ∠3 = ∠AOB, ∠6 + ∠7 = ∠COD, ∠1 + ∠8 = ∠AOD and ∠4 + ∠5 = ∠BOC
∴ From (1) and (2), we have
∠AOD + ∠BOC = 180°
and ∠AOB + ∠COD = 180°

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and 8(3, 7).
Solution:
Let the required ratio be k : 1 and the point C divide them in the above ratio.
∴ Coordinates of C are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\)
Since, the point C lies on the given line 2x + y – 4 = 0.
∴ We have \(2\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)-4\) = 0
⇒ 2(3k + 2) + (7k – 2) = 4 × (k + 1)
⇒ 6k + 4 + 7k – 4k – 4 – 2 = 0
⇒ (6 + 7 – 4)k + (-2) = 0 ⇒ 9k – 2 = 0
⇒ k = \(\frac{2}{9}\)
The required ratio = k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9

Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
Let the given points be A(x, y), B( 1, 2) and C(7, 0) are collinear.
The points A, B and C will be collinear if area of ∆ABC = 0
⇒ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0 or x + 3y – 7 = 0, which is the required relation between x and y.

Question 3.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
Let the points are A (6, -6), B(3, -7) and C(3, 3)
Let P(x, y) be the centre of the circle. Since, the circle is passing through A, B and C.
∴ AP = BP = CP
Taking AP = BP, we have AP² = BP²
⇒ (x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
⇒ x² – 12x + 36 + y² + 12y + 36 = x² – 6x + 9 + y² + 14y + 49
⇒ – 12x + 6x + 12y – 14y + 72 – 58 = 0
⇒ – 6x – 2y + 14 = 0
⇒ 3x + y – 7 = 0 ……………….. (1)
Taking BP = CP, we have BP² = CP²
⇒ (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
⇒ x² – 6x + 9 + y² + 14y + 49 = x² – 6x + 9 + y² – 6y + 9
⇒ – 6x + 6x + 14y + 6y + 58 -18 = 0
⇒ 20y + 40 = 0
⇒ y = \(\frac{-40}{20}\) = -2
From (1) and (2), 3x – 2 – 7 = 0
⇒ 3x = 9 ⇒ x = 3
i.e., x = 3 and y = -2
∴ The required centre is (3, -2).

Question 4.
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
Let we have a square ABCD such that A(-1, 2) and C(3, 2) are the opposite vertices. Let B(x, y) be an unknown vertex.

Since, all sides of a square are equal.
∴ AB = BC ⇒ AB² = BC2²
⇒ (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²
⇒ x² + 2x + 1 + y² – 4y + 4
⇒ x² – 6x + 9 + y² – 4y + 4
⇒ 2x + 1 = -6x + 9
⇒ 8x = 8 ⇒ x = 1  …………………. (1)
Since, each angle of a square = 90°.
∴ ∆ABC is a right angled triangle.
∴ Using Pythagoras theorem, we have AB² + BC² = AC²
⇒ (x + 1)² + (y – 2)²] + [(x – 3)² + (y – 2)²]
= [(3 + 1)² + (2 – 2)²]
⇒ [x² + 2x + 1 + y² – 4y + 4] + [x² – 6x + 9 + y² – 4y + 4]
= [4² + 0²]
⇒ 2x² + 2y² + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16
⇒ 2x² + 2y² – 4x – 8y + 2 = 0
⇒ x² + y² – 2x – 4y + 1 = 0 …………….. (2)
Substituting the value of x from (1) into (2), we have
1 + y² – 2 – 4y + 1 = 0
⇒ y² – 4y + 2 – 2 = 0
⇒ y² – y = 0
⇒ y(y – 4) = 0
⇒ y = 0 or y = 4
Hence, the required other two vertices are (1, 0) and (1, 4).

Question 5.
The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

Solution:
(i) By taking A as the origin and AD and AB as the coordinate axes. We have P(4, 6), Q( 3, 2) and R( 6, 5) as the vertices of ∆PQR.
(ii) By taking C as the origin and CB and CD as the coordinate axes, then the vertices of ∆PQR are P(-12, – 2), Q(-13, – 6) and R(- 10, – 3)
Case I: When P(4, 6), Q(3, 2) and R(6, 5) are the vertices.
∴ ar(∆PQR) = \(\frac{1}{2}\) [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]
= \(\frac{1}{2}\) [-12 – 3 + 24] = \(\frac{9}{2}\) sq. units
Case II: When P(-12, -2), Q(-13, -6) and R(-10, -3) are the vertices.
∴ ar(∆PQR)
= \(\frac{1}{2}\) [-12(- 6 + 3) + (-13)(- 3 + 2) + (-10)(-2 + 6)]
= \(\frac{1}{2}\) [-12(-3) + (-13)(-1) + (-10) × (4)]
= \(\frac{1}{2}\) [36 + 13 – 40] = \(\frac{9}{2}\) sq. units
Thus, in both cases, the area of ∆PQR is the same.

Question 6.
The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. [Recall “The converse of basis proportionality theorem”, and “theorem of similar triangles taking their areas and corresponding sides”]
Solution:

Question 7.
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC.
(i) The median from A meets BCat D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1
(iv) What do you observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]
(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.
Solution:
We have the vertices of ∆ABC as A (4, 2), B(6, 5) and C(1, 4).
(i) Since AD is a median

∴ Coordinates of D are


Also, CR : RF = 2 : 1 i.e., the point R divides CF in the ratio 2 : 1
∴ Coordinates of R are

(iv) We observe that P, Q and R represent the same point.
(v) Here, we have A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC. Also AD, BE and CF are its medians.
∴ D, E and F are the mid points of BC, CA and AB respectively.
We know, the centroid is a point on a median, dividing it in the ratio 2 : 1.
Considering the median AD, coordinates of

Question 8.
ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Solution:
We have a rectangle whose vertices are A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1).
∵ P is mid-point of AB
∴ Coordinates of P are


We see that PQ = QR = RS = SP i.e., all sides of PQRS are equal.
∴ It can be a square or a rhombus.
But PR ≠ QS i.e., its diagonals are not equal.
∴ PQRS is a rhombus.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) The sides are : 7 an, 24 cm, 25 cm
Here, (7 cm)² = 49 cm²
(24 cm)² = 576 cm²
(25 cm)² = 625 cm²
∵ (49 + 576)cm² = 625 cm²
∴ It is a right triangle.
Hypotenuse = 25 cm.

(ii) The sides are: 3 cm, 8 cm, 6 cm
Here, (3 cm)² = 9 cm²
(8 cm)² = 64 cm²
(6 cm)² = 36 cm²
∵ (9 + 36) ≠ 64 cm²
∴ It is not a right triangle.

(iii) The sides are : 50 cm, 80 cm, 100 cm
Here, (50 cm)² = 2500 cm²
(80 cm)² = 6400 cm²
(100 cm)² = 10000 cm²
∵ (2500 + 6400) cm² ≠ 10000 cm²
∴ It is not a right triangle.

(iv) The sides are : 13 cm, 12 cm, 5 cm
Here, (13 cm)² = 169 cm²
(12 cm)² = 144 cm²
(5 cm)² = 25 cm²
∵ (144 + 25)cm² = 169 cm²
∴ It is a right triangle.
Hypotenuse = 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR.
Solution:
In ∆QMP and ∆QPR,
∠QMP = ∠QPR [Each 90°]
∠Q = ∠Q [Common]
⇒ ∆QMP ~ ∆QPR …. (1) [AA similarity]
Again, in ∆PMR and ∆QPR,
∠PMR = ∠QPR [Each = 90°]
∠R = ∠R [Common]
⇒ ∆PMR ~ ∆QPR …… (2) [AA similarity]
From (1) and (2), we have

Question 3.
In the figure, ABD is a triangle, right angled at A and AC ⊥ BD.

Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
Solution:

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
We have, right ∆ABC such that ∠C = 90° and AC = BC.

∴ By Pythagoras theorem, we have AB² = AC² + BC² = AC² + AC² = 2AC²
[∵ BC = AC (given)]
Thus, AB² = 2AC²

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.
Solution:
We have, an isosceles AABC such that BC = AC.

Also, AB² = 2AC²
∴ AB² = AC² + AC²
But AC =BC
∴ AB² = AC² + BC²
∴ Using the converse of Pythagoras theorem, ∠ACB = 90°
f.e., ∆ABC is a right angled triangle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
In equilateral triangle, altitude bisects the base.
⇒ AD = DB

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Let us have a rhombus ABCD.

∵ Diagonal of a rhombus bisect each other at right angles.
∴ OA = OC and OB = OD
Also, ∠AOB = ∠BOC [Each = 90°]
And ∠COD = ∠DOA [Each = 90°]
In right ∆AOB, we have,
AB² = OA² + OB² …… (1)
[Using Pythagoras theorem]
Similarly, in right ∆BOC,
BC² = OB² + OC² …… (2)
In right ∆COD,
CD² = OC² + OD² …… (3)
In right ∆AOD,
DA² = OD² + OA² ……. (4)
Adding (1), (2), (3) and (4)
AB² + BC² + CD² + DA²
= [OA² + OB²] + [OB² + OC²] + [OC² + OD²] + [OD² + OA²]
= 2OA² + 2OB² + 2 OC² + 2OD² = 2[OA² + OB² + OC² + OD²]
= 2[OA² + OB² + OA² + OB²]

Thus, sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Question 8.
In the figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF²

Solution:
We have a point in the interior of a ∆ABC such that
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
(i) Let us join OA, OB and OC.

In right ∆OAF, by Pythagoras theorem
OA² = OF² + AF² …(1)
Similarly, from right triangle ODB and OEC, we have
OB² = BD² + OD², …(2)
and OC² = CE² + OE² …(3)
Adding (1), (2) and (3), we get OA² + OB² + OC²
= (AF² + OF²) + (BD² + OD²) + (CE² + OE²)
⇒ OA² + OB² + OC²
= AF² + BD² + CE² + (OF² + OD² + OE²)
⇒ OA² + OB² + OC² – (OD² + OE² + OF²)
= AF² + BD² +CE²
⇒ OA² + OB² + OC² – OD² – OE² – OF²
= AF² + BD² + CE²

(ii) In right triangle OBD and triangle OCD, by Pythagoras theorem:
OB² = OD² + BD² and OC² = OD² + CD²
⇒ OB² – OC² = OD² + BD² – OD² – CD²
⇒ OB² – OC² = BD² – CD² ….. (1)
Similarly, we have
OC² – OA² = CE² – AE² …… (2)
and OA² – OB² = AF² – BF² ….. (3)
Adding (1), (2) and (3), we get (OB² – OC²) + (OC² – OA²) + (OA² – OB²) = (BD² – CD²) + (CE² – AE²) + (AF² – BF²)
⇒ 0 = BD² + CE² + AF² – (CD² + AE² + BF²)
⇒ BD² + CE² + AF² = CD² + AE² + BF²
or AF² + BD² + CE² = AE² + BF² + CD²

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Let PQ be the ladder and PR be the wall

⇒ PQ = 10 m, PR = 8 m
Now, in the right ∆PQR, PQ² = PR² + QR²
⇒ 10² = 8² + QR²
[using Pythagoras theorem]
⇒ QR² = 10² – 8² = (10 + 8)(10 – 8)
= 18 × 2 = 36
QR = \(\sqrt{36}\) = 6m
Thus, the distance of the foot of the ladder from the base to the wall is 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB is the wire and BC is the vertical pole. The point A is the stake.

Now, in the right AABC, using Pythagoras Theorem, we have

Thus, the stake is required to be taken at \(6 \sqrt{7}\)m from the base of the pole to make the wire taut.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after \(1 \frac{1}{2}\) hours?
Solution:
Let the point A represent the airport. Plane-I fly towards North,
∴ Distance of the plane-I from the airport after \(1 \frac{1}{2}\) hours = speed × time

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops?
Solution:
Let the two poles AB and CD are such that the distance between their feet AC = 12m.
∵ Height of pole-1, AB = 11 m

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
Solution:
We have a right ∆ABC such that ∠C = 90°.
Also, D and E are points on CA and CB respectively.
rain We have a right ∆ABC such that ∠C = 90°.

Let us join AE and BD.
In right ∆ACB, using Pythagoras theorem
AB² = AC² + BC² …… (1)
In right ∆DCE, using Pythagoras theorem,
DE² = CD² + CE² …… (2)
Adding (1) and (2), we get
AB² + DE² = [AC² + BC²] + [CD² + CE²]
= AC² + BC² + CD² + CE² = [AC² + CE²] + [BC² + CD²] …. (3)
In right ∆ACE,
AC² + CE² = AE² …… (4)
In right ∆BCD,
BC² + CD² = BD² …….. (5)
From (3), (4) and (5), we have AB² + DE² = AE² + BD² or AE² + BD² = AB² + DE²

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB² = 2AC² + BC².

Solution:

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD² = 7AB².
Solution:
We have an equilateral ∆ABC; in which D is a point on BC such that BD = \(\frac{1}{3}\) BC.

Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
We have an equilateral ∆ABC, in which AD ⊥ BC.
Since, an altitude in an equilateral A, bisects the corresponding side.

∴ D is the mid-point

Question 17.
Tick the correct answer and justify : In ∆ABC, AB = \(6 \sqrt{3}\) cm, AC = 12 cm and BC = 6 cm. The angle B is
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution:
(C): We have, AB = \(6 \sqrt{3}\) cm, AC = 12 cm, and BC = 6 cm

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0; 3x – 9y – 2 = 0
(ii) 2x + y = 5; 3x + 2y = 8
(iii) 3x – 5y = 20; 6x -10y = 40
(iv) x – 3y – 7 = 0; 3x – 3y – 15 = 0
Solution:
(i) For x – 3y – 3 = 0, 3x – 9y – 2 = 0
∴ a₁ = 1, b₁ = – 3, C₁ = – 3, a₂ = 3, b₂ = – 9, C₂ = -2

∴ The given system has no solution.

(ii) 2x + y – 5 = 0, 3x + 2y – 8 = 0
∴ a₁ = 2, b₁ = 1, c₁ = -5, a₂ = 3, b₂ = 2, c₂ = -8

∴ The given system has a unique solution.
Now, using cross multiplication method, we have

(iii) For 3x – 5y – 20 = 0, 6x – 10y – 40 = 0

∴ The given system of linear equation has infinitely many solutions

(iv) For x – 3y – 7 = 0, 3x – 3y – 15 = 0

∴ The given statement has unique solution. Now, using cross multiplication method, we have

Question 2.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?
2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1
Solution:
(i) We have, 2x + 3y = 7 and
(a – b) x + (a + b) y = (3a + b – 2)
⇒ 2x + 3y – 7 = 0
and (a – b) x + (a + b) y -(3a + b – 2) = 0
a₁ = 2, b₁ = 3, c₁ = – 7, a₂ = (a- b) , b₂ = (a + b) , c₂ = -(3a + b – 2)
For infinite number of solutions,

From the first two equations, we get

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9; 3x + 2y = 4
Solution:
Method-1 [Substitution method]:
8x + 5y = 9 … (1)
3x + 2y = 4 … (2)

Question 4.
Form the pair of linear equations in the following problems and find their solutions (if they exist)by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
(i) Let the fixed charges = ₹ x
and charges of food per day = ₹ y
For student A : Number of days = 20
∴ Cost of food for 20 days = ₹ 20y
According to the problem,
x + 20y = 1000
⇒ x + 20y – 1000 = 0 …. (1)
For student B : Number of days = 26
Cost of food for 26 days = ₹ 26y
According to the problem,
x + 26y = 1180
⇒ x + 26y – 1180 = 0 … (2)
Solving these by cross multiplication, we get

Thus x = 400 and y = 30
∴ Fixed charges = ₹ 400
and cost of food per day = ₹ 30

(ii) Let the numerator = x
and the denominator = y

From equations (1) and (2) , we have

(iii) Let the number of correct answers = ₹ x
and the number of wrong answers = ₹ y
Case-I: Marks for all correct answers
= (3 × x) = 3x
Mark for all wrong answers = (1 × y) = y
∴ According to the condition :
3x – y = 40 ⇒ 3x – y – 40 = 0
Case-II: Mark for all correct answers
= (4 × x)=
Marks for all wrong answers = (2 × y)
= 2y
∴ According to the condition :
4x – 2y = 50
⇒ 2x – y = 25 ⇒ 2x – y – 25 = 0 … (2)
From (1)and (2) , we have a₁ = 3, b₁ = -1, c₁ = -40, a₂ = 2, b₂ = -1, c₂ = -25

Now, total number of questions = [Number of correct answers] + [Number of wrong answers]
= 15 + 5 = 20
Thus, required number of questions = 20.

(iv) Let the speed of car-I be x km/hr.
and the speed of car-II be y km/hr.
Case-I:

Distance travelled by car-I = AC
∵ AC = time × speed = 5 × x km, AC = 5x
Distance travelled by car-II, BC = 5y
Since AB = AC – BC,
100 = 5x – 5y
⇒ 5x – 5y – 100 = 0
⇒ x – y – 20 = 0 …. (1)


Thus, speed of car-I = 60 km/hr
Speed of car-II = 40 km/hr

(v) Let the length of the rectangle = x units
and the breadth of the rectangle = y units
∴ Area of rectangle = x × y = xy

Condition-II:
(Length + 3) × (Breadth + 2) = Area + 67
⇒ (x + 3) (y + 2)= xy + 67 ⇒ 2x + 3y + 6 = 67
⇒ 2x + 3y – 61 = 0….(2)
Now, using cross multiplication method in (1)and (2) , where

Thus, length of the rectangle = 17 units and breadth of the rectangle = 9 units.