Category: Class 8

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 5th Lesson Attaining the Age of Adolescence

8th Class Biology 5th Lesson Attaining the Age of Adolescence Textbook Questions and Answers

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Question 1.
How is adolescence different from childhood?
Answer:

Adolescence	Childhood
1. It is Independent nature and very self conscious.	1. It depends upon parental assistance for basic needs.
2. Adolescents seek company of friends to share their feelings.	2. Children are learning through experimenting and communicating with other.
3. Taking decisions by critical thinking. Don’t like the supervision of elders.	3. Adults supervise and support the development process of child.
4. Lot of Stress and strain.	4. No stress, make new friends and gain new skills.
5. Rate of growth is more	5. Comparatively less.

Question 2.
Write short notes on the following.
a) Secondary sexual characters
b) Adam’s Apple.
Answer:
a) Secondary Sexual characters:

1. In adolescence age some external changes have seen in boys and girls.
2. These are called secondary sexual characters.
3. Example: in boys facial hair, moustaches and beards begin to grow. Hair starts growing on the chest of boys.
4. In girls breast begin to develop.
5. In both boys and girls hair grows in the arm pits.

b) Adam’s Apple:

1. The Adam’s apple is actually a partial growth of our voice box or larynx.
2. The larynx is made up of nine cartilages, one of which is the largest, called thyroid cartilage.
   
3. Due to the elongation of the thyroid cartilage the Adam’s apple is formed. It protrudes out in front of the neck.
4. This is caused mainly by male hormone. Testosterone during adolescence.
5. As a result muscles or chords attached to the cartilage get loosened and thickened.
6. When air passes through these chords a hoarse sound is produced.
7. This is the reason for disturbance in voice in the stage of adolescence. At the end of this stage voice get perfect.

Question 3.
List out the changes in the body that take place at the age of adolescence.
(or)
What are the changes that could be observed in adolescence phase?
Answer:

1. During this adolescence changes occur in external, internal parts of the body.
2. They show interest to spend time with peers.
3. Girls voice becomes soft, pimples also may appear on the face by the activation of oil and sweat glands.
4. There is growth and maturity in reproductive system.
5. In boys voice becomes coarse. Pimples, acne may appear on the face. Facial hair like mustache begin to grow.

Question 4.
Match the following:
1. Testes                                  ( )           a. Estrogen
2. Endocrine gland                  ( )           b. Pituitary
3. Menarche                            ( )           c. Sperm
4.Female hormone                  ( )           d. First menstruation
Answer:
1. c
2. b
3. d
4. a

Question 5.
Why acne and pimples are common in adolescents?
Answer:

1. Naturally in adolescence boys and girls feel worried of their pimples and acne.
2. The reason is the secretions of sweat and sebaceous glands are very active in adolescence.
3. Because of increased activity of these glands in the skin, boys and girls get acne and pimples.

Question 6.
What can you suggest to your classmates to keep himself/herself clean and healthy?
Answer:

1. It would better to have bath atleast twice in a day.
2. All parts of the body and inner wears should be washed and cleaned every day.
3. If cleanliness is not maintained, there are chances of having fungal, bacterial and other unwanted infections.
4. Girls should take special care of cleanliness during the time of menstrual cycle.
5. Making use of disposable napkins.

Question 7.
If you have chance to talk with a doctor, what questions you would ask about adolescent emotions and changes in the body ?
Answer:
If I have chance to talk with a doctor, I would ask about

1. How to develop positive emotions like bravery, self confidence, happiness, satisfaction, appreciation gratitude, concern and forgiveness.
2. And also how to over come the negative emotions like anger, bitterness, dissatisfaction, sadness, anxiety, fear, shame and guilt; which are needed in adolescence.

Question 8.
Some mobile phones have auditory meter to measure frequency of produced sound. By using this phone measure your friend’s voice frequency one from each class VI to X. Report your findings.
Answer:

Name	Class	His/Her Voice Frequency
Madhavi	VI	50 decibels
Kalyan	VII	52 decibels
Ravi	VIII	54 decibels
Hemanth	IX	55.5 decibels
Jalaja	VI	48 decibels
Madhu	VII	48.5 decibels
Padmaja	VIII	49 decibels
Sailaja	IX	49.5 decibels

Question 9.
Write five suggestions to improve the performance of Red Ribbon club of your school.
Answer:

1. To instill life skills.
2. To ensure that every college going youth is equipped with conceptual knowledge about various basic health aspects.
3. To increase the capacity of educational system in teaching various basic health aspects.
4. To motivate youth and build their capacity as peer educators.
5. To promote voluntary blood donations.

Question 10.
Prepare a three minute speech on behavioural changes in adolescents.
Answer:

1. Adolescence is the growing age where we may observe some changes in behaviour.
2. They are very fast in taking decisions.
3. They do not want to be forced to do any work, behave peculiarly sometimes fast and sometimes frigid.
4. Adolescents prefers to spend more times before the mirror and like to use perfumes.
5. They do not want to listen to parents suggestions and feels friends are correct but not parents.
6. They search for identify from teachers and peer groups.
7. They want more independence in taking decisions.
8. Sometimes they feel shy and sometimes feel happy.
9. They try to get romantic relationships.
10. They are more inclined towards unhealthy habits.
11. The adolescents have attraction towards opposite sex.
12. The mind of an adolescent is full of zealous acts and urge to find reasons for several things around.
13. Emotionally they are in a turbulent state all the time they get new thoughts for their life activities.
14. An adolescent feel insecure while trying to adjust to the changes in the body and the mind.
15. They seek company of friends to share their feelings even if they are of opposite sex.

Question 11.
Nature prepares human body to reproduce her generations. What do you think of it?
Answer:

1. In females, the reproductive phase of life begins usually around 10 to 12 years of age.
2. And generally it lasts till the age of 40 – 50 years.
3. The ova begin to mature with the onset of adolescence.
4. One ovum matures and is released by one of the ovaries once in about 28 to 30 days.
5. During this period the wall of the uterus becomes thick so as to receive a fertilized egg.
6. This results in pregnancy and childs’ birth.
7. If fertilization does not occur, the released egg and thickened lining of the uterus will be released with some amount of blood in woman.
8. This is called menstrual cycle.
9. Thus nature prepares the female human body to reproduce generation after generation to continue human life on the earth.
10. This is the secret of nature and is nature’s wonderful phenomena.

Question 12.
You know that early marriage is a social taboo. Prepare some slogans to prevent this. (OR)
You know about that child-marriages are social evil. Your school students are conducting a rally to educate the society. Prepare some slogans on this.
Answer:

1. Avoid child marriage – Prevent childhood.
2. Let a child be a child – stop child marriage.
3. Child marriage – a loosing game.
4. Stop child marriage – stop child abuse
5. Childhood is not for motherhood.
6. Let girls be girls but not brides.

Question 13.
13 years old Swaroop always think of his height. Can he improve his height? What do you suggest him?
Answer:
The suggestion is to take nutritious food and to do body exercise regularly to improve his height.

Question 14.
Are you angry with your parents. How do you wish your parents to be?
Answer:
When insulted or threatened unfairly by parents we get angry on them. In our opinion a parent is

1. to be a good advisor to give advice how to control stress and strain, which is needed by the adolescent.
2. to be like a guide to give guidance how to behave with opposite sex.
3. to be like a friend to give good suggestions.
4. to be like a wellwisher and always stand behind us to lead us for bright future.

Question 15.
What are your expectations about your parents and teachers?
Answer:
Parents and teachers play an important role to develop the adolescents in to healthy, productive young ones to the nation. Parents feel to develop their children in to better ones than themselves. They must have good courage, confidence, boldness, and free to solve the problems. They should not be tense and worse. Parent is to be a friend and guide towards adolescents.
Teacher is not only a master is to be a captain or a leader. Every adolescent needs mental support. Teacher is the only people who give suitable suggestions to make them free from all mental stress.

8th Class Biology 5th Lesson Attaining the Age of Adolescence InText Questions and Answers

Question 1.
Some of you also may behave like this, Why?
Answer:
At the age of 13 – 19 years, some changes like voice becomes hoarse, not caring to follow the suggestions and advises of parents, shows restlessness and growing tall.
Because at this age children will be entering into a period which is called as Adolescence, where some changes in the behaviour is seen.

Question 2.
Have you noticed that you are growing?
Answer:
At the age of adolescence growth in height takes place about 18 years of age both boys and girls reach their maximum height.

Question 3.
Have you reached the age of“Adolescence”?
a) Is mustache growing on your upper lip?
b) Did your voice change?
c) Are hairs growing under arm pit?
d) Are there pimples or acne on your face?
e) Are you taking care of your face by applying powder and combing your hairs frequently?
f) Are you feeling shy when talking with opposite sex?
g) Are you not interested to play with opposite sex which you have done earlier?
h) Are you showing restlessness while your parents suggest to do something?
Answer:
For above all the questions the answer is ‘yes’ during adolescence changes occur in external, internal parts of the body.

8th Class Biology 5th Lesson Attaining the Age of Adolescence Activities

Activity – 1

Question 1.
Observing growth rate.
a) Observe the below table and given graph, answer the following questions.

i) What have you observed from the above table?
Answer:
We have observed the height attained by boys and girls in different ages.

ii) When does growth in height nearly stop?
Answer:
In boy growth in height nearly stops at the age of 18, in girls it stops at 17 years.

iii) Which period of age according to you is the fastest growing period for girls?
Answer:
The fastest growing period for girls is between 14-17 years.

iv) Which period of age is the fastest growing period for boys?
Answer:
The fastest growing period for boys is between 16 to 18 years.

v) Who do grow faster? How can you say?
Answer:
The girls grow faster than the boys. By seeing the above graph, about 17 years of age, girls reach their maximum height.

b) Sneha is 13 years old with 125 cm tall. At the end of the growth period likely to be use the following formulae and calculate the maximum height that Sneha will reach.

Answer:

Sneha’s present height = 125 cm
At the end of the growth period Sneha is likely to be 131.5 cm.

c) Table – 1 shows that girls grow faster than boys in their adolescent period. From a group with six students in your class. Measure the height and calculate the future heights in the following table:
Answer:

Activity – 2

Question 2.
Form five groups in your class. Select at least 15 students. Collect body measurement data of the selected 15 students.
Find an average body measurements for boys and girls separately.
Answer:

Activity – 3

Question 3.
Read the following check list. Put tick (✓) mark which points reflect your behaviour.
Answer:
Check List:

Think & Discuss

I. Read the following information and answer the following questions.
Some sections of people in our society believe that during the period of menstruation women are untouchable. So, they are asked to keep a distance from others. During this time girls may be ristricted from taking bath, cooking food or going to school. In that case they may lag behind in their studies. In some sections of the society even women are also forced to stay in the huts built at the outskirts of the village.

Question 1.
In what way this kind of discrimination is harmful for girls and women?
Answer:

1. During the period of menstruation women are treated as untouchable by some sections of people in our society.
2. Generally during this period they feel weak and uncomfortable physically.
3. By this kind of discrimination, mentally also they get hurt and feel that why they have born as woman.
4. By separation, it would be known to all by this the girl may feels shy, unable to move freely with others and she becomes dull in studies.

Question 2.
Several researches have been done to prove that all these are myths and there is no scientific reason behind these. The blood and egg that is discarded would give rise to a baby if fertilization took place. This is a biological phenomena.
So how can it be impure or unclean?
Answer:

1. If fertilization took place, the uterus receives a fertilized egg and this results in pregnancy and would give rise a baby which develops in the uterus.
2. If fertilization does not occur this cause bleeding in woman, which is the unwanted and waste to the uterus. So it can be treated as impure or unclean and may create some problems in uterus.
3. During menstrual period proper care regarding health and hygiene is needed rather than following myths.

Question 3.
If young generation is trapped into such unhealthy habits, what will be the
future of our country ? What are its effects?
Answer:

1. If young generation is trapped into unhealthy habits like consuming tobacco (gutkha, cigarettes, cigar, beedi) they addicted to such social evil.
2. Todays children are tomorrows citizens so it should be avoided.
3. A famous psychiatrist Stanly Hall stated that adolescence is the age of stress and strain. By getting proper guidance from teachers, parents and elders, the adolescents be able to lead a happy meaning full life and they will save the future of the country.

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 4 Reproduction in Animals Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 4th Lesson Reproduction in Animals

8th Class Biology 4th Lesson Reproduction in Animals Textbook Questions and Answers

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Question 1.
Differentiate between:
a) Sexual Reproduction and Asexual Reproduction
b) Gametes and Zygote
c) External fertilization and Internal fertilization
d) Viviparous and Oviparous animals
Answer:
a) Sexual Reproduction and Asexual Reproduction:

Sexual Reproduction	Asexual Reproduction
1. Male and female gametes are formed.	1. No production of gametes.
2. Involves fusion of male and female gametes.	2. No fusion of gametes.
3. Involves two organisms.	3. Involves a single organism.
4. Offsprings have some characters from male parent and other from female parent. Some characters may not be present in either of the parents.	4. Produces offsprings that are identical to the parent.

b) Gametes and Zygote:

1. Millions of male gametes (sperms) are produced by the testes. These are microscopic and single celled. Sperm has a head, a middle piece and a tail.
   Ovary produces female gaffietes called ova. It is a single cell, (haploid)
2. The fusion of male and female garnet is called fertilization. The result of fertilization is the formation of a zygote. Zygote is a diploid cell. This develops mitorically and forms into an embryo, which further develops into a baby.

c) External fertilization and Internal fertilization:
The process of fertilization that occurs outside of an organism is called External fertilization.
E.g. Frog, Fish, Star fish, etc.
The process of fertilization that takes place inside the body of females is called Internal fertilization. E.g. Animals, Human beings.

d) Viviparous and Oviparous animals:
Animals which give birth to their offsprings are called Viviparous animals.
E.g. Animals, human beings.
Animals which lay eggs are called Oviparous animals.
E.g. Hen, duck, pigeon, etc.

Question 2.
Compare the reproduction in Hydra and Amoeba. Note down the differences in your notebook.
Answer:
Comparison:
Asexual Reproduction takes place in Hydra and Amoeba.

Hydra	Amoeba
1. Multicellular organism.	1. Unicellular organism.
2. Nucleus is absent.	2. Nucleus is present.
3. Buds are formed on the body surface.	3. A constriction is formed in the middle of the nucleus.
4. The bud increases in size and develops tentacles.	4. The constriction deepens divides the nucleus into two nuclei.
5. It grows in size.	5. A constriction is formed on the body wall of Amoeba in the middle.
6. This bud separates from the parent Hydra and lives independently.	6. The constriction deepens and divides the body of amoeba into two individuals (daughter amoebae).

Question 3.
Why do fish and frog lay more number of eggs whereas cow and human beings usually give birth to only one at a time?
Answer:

1. Fish and frog lay many eggs to increase chance of survival of the offspring and the continuation or their generation.
2. They do not take care of their young ones making them prone to predators and may even be washed away by the water force.
3. Thus the more eggs produced, the greater the chances that some will grow to maturation.
4. Female frog and fish release their eggs in the water and male animals release their sperms in the water. As fertilization takes place in the water it is external ferlitization. There is no safety for the fertilized eggs in the water so these animals lay more number of eggs.
5. Whereas cow and human beings usually give birth to only one at a time and the internal fertilization takes place in these animals. There is safety for the embryo (the offspring) in the mother’s womb until it’s birth.

Question 4.
Can animals produce offsprings even without formation of zygotes, how? Explain with suitables example.
Answer:

1. Besides Asexual and Sexual Reprodution, there is other mode of reproduction called cloning.
2. Cloning is the production of an exact copy of a cell, any other living part, or a complete organism.
3. Cloning of an animal was successfully performed for the first time by Ian Wilmut and his colleagues at the Roslin Institute in Edinburgh, Scotland.
4. They successfully cloned a sheep named Dolly. Dolly was born on 5th July 1996 and was the first mammal to be cloned.
   
5. During the process of cloning dolly, a cell was collected from the mammary gland of a female Finn Dorset Sheep.
6. Simultaneously, an egg was obtained from Scottish black face ewe.
7. The nucleus was removed from the egg. Then the nucleus of the mammary gland cell from the Finn Dorset sheep was inserted into the egg of the Scottish black face ewe whose nucleus had been removed.
8. Thus the egg produced was implanted into the Scottish black face ewe. Development of this egg followed normally and finally Dolly was born.
9. Though Dolly was given birth by the Scottish black face ewe, it was found to be absolutely identified to the Finn Dorset sheep, from which the nucleus was taken.
   
10. Since the nucleus from the egg of the Scottish black face ewe was removed, Dolly did not show any character of the Scottish black face ewe.
    Dolly was a healthy clone of the Finn Dorset sheep and produced several off-springs of her own through normal sexual means.

Question 5.
How can you identify the animal is viviparous or oviparous?
Answer:

1. Animals giving birth to young ones have epidermal hair on their skin and external ears. These animals are called viviparous animals.
   E.g. Animals, human beings, etc.
2. The animals that lay eggs do not have epidermal hair or external ears. These animals are called oviparous animals.
   E.g. Hen, Duck, Pigeon, Parrot, etc.

Question 6.
Who am I?
a) I am formed by the fusion of male and female gametes.
Answer:
Zygote: Zygote is formed by the fusion of male and female gametes. This process is
called fertilization.

b) I am a gamete that has a tail and travel to fuse with female gametes.
Answer:
Male garnets or sperm or spermatozoa:
The structure of sperm has a head, a middle piece and a tail.

c) I am a fully developed embryo inside a mother’s body.
Answer:
Offspring or baby:
The zygote divides repeatedly to give rise a ball of cells. The cells then begin to form groups that develop into tissues and organs in the body. This developing structure is termed as an Embryo.
The embryo gets embedded in the wall of the uterus for further development. It develops in the uterus. It gradually develops body parts such as hands, legs, head, eyes, ears etc. From 3 months (12 weeks) of pregnancy the embryo is called FOETUS – After the completion of this period (about 270 – 280 days) a baby (offspring) is born. This is called gestation period.

Question 7.
State the reason why most of the terrestrial animals’, fertilization takes place internally.
Answer:

1. In animals like insects, reptiles, birds and mammals, the male animals deposit the sperms inside the body of the female animals, where fertilization occurs. This is called Internal Fertilization. This is most common in terrestrial animals.
2. In majority of the animals, sexes are separate and male and female animals are distinct. This is called sexual dimorphism and animals are said to be unisexual.
3. The external features by which the males and females can be distinguished are called Secondary sexual characters.
4. There are some animals in which male and female sex organs are present in the same animal. This is called Hermaphroditism and such animals are called Hermaphrodites or Bisexual.
5. Hermaphroditism is seen in some of the members of protozoa, Coelenterata, Platy- helminthes, Nematoda, Annelida and mollusca. In these animals sperm and ova are formed in the same animal. However self fertilization is prevented by several methods.

Question 8.
Observe the following figures and write the functions of them.

Answer:
a) Testes:
Testes are the male reproductive organs and produce male gametes known as sperms or spermatozoa. Testes are egg shaped. It is connected with a pair of seminal ducts through which sperms travel and ejaculate out with the help of penis.

b) Female Reproductive system. Oviduct or fallopian tube connected with ovary. Female reproductive system contains a pair ovaries, oviducts and also called fallopian tubes and uterus.
The ovary produces female gametes called ova or eggs. In human beings, a single matured egg is released into the oviduct by one of the ovaries every month.

The ovum which is a single cell released from ovary and enters into a tube called Fallopian Tube. The end of the tube is like a funnel with several finger like structures and is also ciliated. The movement of cilia help the movement of ovum through the fallopian tube into uterus.
c) Sperm:
Human sperms are minute, microscopic and motile they have a oval head, a neck, a middle piece and a long tail.
Head consists of a large haploid nucleus. Acrosome is present in the head, which helps in fertilization.

The neck is short middle piece has several mitochondria which produce energy required for the movements of sperms. Tail piece helps in the swimming of sperm to reach the ovum during fertilization.
d) Fusion of ovum and sperm:
(Fertilization) Fertilization is of internal type. Sperm reaches the ovum in the fallopian tube. Sperm nucleus enters the ovum which is haploid.
When sperm enters the ovum the membranes of ovum becomes thicken. So that another sperm can not penetrate the ovum. This prevents double fertilization of the ovum. During fertilization, the sperm and the ovum fuse to form a zygote.

This type of internal fertilization occurs in different organisms like insects, snakes, lizards, birds and mammals etc.

Question 9.
a) By taking help of given words label the following life cycle, (eggs, adult, pupa, larva)

Answer:

b) Explain the process of metamorphosis in housefly by taking help from in the given diagram.

Answer:
Metamorphosis in the house fly: A female housefly at a time lays about 120 to 160 eggs. The eggs are laid in garbage, on dung heaps, or on a decaying animal and vegetable matter. The life history consists of 1) egg 2) larva 3) pupa and 4) adult stages. Egg: The egg is white and cylindrical on one side. It has two ribbon like longitudinal thickenings. They hatch in about 24 hours into larva.
Larva: The larva is known as a Maggot. It is white in colour. The baby of the larva has 13 segments. It has a mouth and feeds on organic matter.
Pupa: The fully grown larva moves to a dry place in the dung and changes to a pupa. The pupa is dark brown and barrel shaped. In a week, the pupa changes into an adult or imago.
Houseflies spread, germs that cause diseases like typhoid, cholera, amoebic dysentery, tuberculosis. Our food should be kept covered from houseflies. The surroundings should be clean without garbage and dung heaps. Insecticides can be used to kill the house flies.

Question 10.
Match the following.
Group – A                                                Group – B
1. Oviparous                    (  )          A) Tadpole to adult
2. Metamorphosis           (  )          B) Birds
3. Embryo                        (  )          C) Fertilisation outside the body
4. External fertilization     (  )          D) Developed Zygote
Answer:
1) B
2) A
3) D
4) C

Question 11.
What would happen if all the organisms stop the process of reproduction?
Answer:

1. Without reproduction living organisms would not survive long.
2. Different species of living organisms die due to various reasons.
   E.g. Old age, diseases, accidents, etc.
3. Imagine the death of members of a species continues and new individuals of that species are not added.
4. A stage will come when that species will disappear.
5. To ensure the continuity of the species reproduction is important.
6. Depending on the available conditions in a community, different species will reproduce continuously and increase their numbers.
7. Reproduction in a species will therefore:
   a) Replace those species that die and
   b) Allow an increase in total numbers of the species under suitable conditions.

Question 12.
Kavita found a tadpole in a pond. She collected it carefully and put it in an aquarium supposing it as a fish. After some days what did she find and why?
Answer:

1. The larva that emerges from the eggs, known as tadpole, have oval bodies and long, vertically flattened tail and are fully aquatic.
2. Tadpole lack eyelids and have cartilaginous skeleton. They take respiration through external gills, later it develops internal gills. It looks like a fish at this stage completely. The vertically flattened tails use for swimming.
3. Tadpole lack true teeth, but the jaws two elongated parallel rows of small structures called keradonts in their upper jaw and three rows of keradonts in the lower jaw as same as the fish has.
4. Tadpole are typically herbivorous, feeding mostly on algae, including diatoms, filtered the water through the gills.
5. The tadpoles may be as short as a week during metamorphosis.

Question 13.
Collect information from your library or from other sources like internet and discuss the life cycle of Honeybees in the symposium at your school.
Answer:
In the life cycle of butterfly there are four stages.

1. Egg
2. Larva
3. Pupa
4. Adult.

The cycle of changes that takes place from egg to adult is called metamorphosis.

1) Egg: The egg is the first stage in the butterfly. They are very small and round. The female butterfly lays eggs on or near the plants.
2) Larva: The larva hatches from the egg. Butterfly larva are usually called Caterpillar. Caterpillars spend most of their time eating. Butterfly do all their growing when they are caterpillars, and food gives them the energy and body building materials they need. A caterpillar’s exoskeleton can’t stretch or grow, so the caterpillar sheds its skin or molts, several times as it grows.
3) Pupa: When the caterpillar has finished growing, it forms from the outside, the pupa looks as if it’s resting. But inside, every part of the caterpillar is changing. Most of it’s organs and other body parts like head, thorax and abdomen, 3 pairs of legs, 2 pairs of wings, a pair of compound eyes, the antennae, a proboscis etc. are formed. Butterfly pupa are called chrysalises.
4) Adult: When the pupa has finished changing, it molts one last time and emerges as an adult butterfly. The adult emerges with its wings folded up against its body. The adult is the stage when butterfly mate the reproduce. Females lay their eggs on plants or other surfaces and the cycle starts all over again.

Question 14.
Sketch the diagrams of male and female reproductory systems.
Answer:

Question 15.
Draw labelled diagram of life history of frog and identify forms are herbivores.
Answer:

Parts:

1. Egg
2. Embryo before hatching
3. Hatched tadpole
4. Tadpole attached to water plant
5. Tadpole with external gills
6. Developing tadpole
7. Tadpole with fore and hind limbs
8. Tadpole changing into frog
9. Frog

Question 16.
How would you appreciate Ritwik’s work when he kept back the pigeon squab in the ventilator? If you were in Ritwik’s place what would you do?
Answer:

1. If i were in Ritwik’s place, I would like to show kindness towards the pigeon squab by keeping back the pigeon squab in the ventilator. He took great care towards it.
2. Research their needs and do what makes them happiest.
3. Check that we are not inadvertently supporting animal cruelty, which are in our surroundings.
4. Leave room for wild life habitates in the own yard by providing birds with feeders and bird bath.
5. Create a clean environment for the birds and animals.
6. Cut the usage of plastic so they can not be danger to wild life.
7. Appreciate wild life and learn more about it but do not approach them or attempt to resque them.
8. Never tolerate birds or animal cruelty. Report suspected cruelty to the authorities.
9. In still compasion in the children by demonstrating kindness and using positive training methods for the pets.
10. Keep them vaccinations’ current and visit veterinarian regularly.

Project work

Note: This project work needs patience and carefulness. Teachers should be cautious while doing this project. Care should be taken at the time of collection of eggs of frogs. From a nearby pond or slow flowing streams. If eggs are not available, you need not to worry. You can start your project after collecting Tadpoles.
To conduct this project you require:

1. Wide mouth transparent bottle / tub
2. Transparent glass
3. Dropper
4. Petridish
5. Some pebbles
6. Magnifying lens

Answer:
Step -1: Go to a nearby pond or a slow flowing stream where usually sewage stagnates during rainy season. Collect few eggs of a frog with the help of wide mouthed bottle as shown in the figure. While collecting eggs, take care that the clusters of eggs are not disturbed and isolated.

Step – 2: After collecting eggs, take a tub of 15 cm depth and a radius of 8-10 cms. Transfer the eggs along with the weeds and algae that you have collected from the pond into the tub. Carefully observe the eggs. You will find a blackish part in the middle of the eggs. That is the embryo of the frog.

Step – 3: Observe the tub daily and note down the changes in your observation book. Draw diagrams after observing for atleast once in three days.

CHANGES TAKES PLACE FROM EGG TO ADULT IN FROG

Step – 4:

Step – 5:
Try to answer these questions after your observation.
1. How many days did it take for the eggs to hatch?
Answer:
It takes 10 to 15 days for the eggs to hatch.

2. How does the tadpole look like?
Answer:
The tadpole looks like a fish.

3. When did you find gill slits in a tadpole?
Answer:
19 to 22 days.

4. On which dates did you observe?
Answer:
Heart: 28th to 30th dates.
Intestine: 31st to 3rd (31 to 33 days)
Bones: 4th to 6th (34 to 36 days)
Rectum: 31st to 3rd (31 to 33 days)
Hindlimbs: 4th to 6th (34 to 36 days)
Forelimbs: 7th to 9th (37 to 39 days)

Step – 6:

1. When did gill slits disappear?
Answer:
37 to 39 days.

2. When did the tail completely disappear?
Answer:
42 to 44 days.

3. How many days did it take for a tadpole to transform into an adult frog?
Answer:
It takes 45 to 46 days to take for a tadpole to transform into an adult frog.

8th Class Biology 4th Lesson Reproduction in Animals InText Questions and Answers

Question 1.
Do all eggs hatch into nestlings?
Answer:
Yes, all eggs hatch into nestlings.

Question 2.
Can there be pigeons if there were no eggs?
Answer:
If there were no eggs there can not be no pigeons.

Question 3.
Can there be eggs if there were no pigeons?
Answer:
If there were no pigeons, there cannot be no eggs.

Question 4.
Do all animals lay eggs?
Answer:
All animals do not lay eggs.

Question 5.
Are there any animals that give birth to young ones?
Answer:
Animals like cat, dog etc., give birth to their young ones.

Question 6.
How can we identify which animals lay eggs and which give birth to young ones?
Answer:
Animals that lay eggs do not have epidermal hair or external ears.
E.g. Crow, Pigeon, Parrot etc.
Animals giving birth to young ones have epidermal hair on their skin and external ears. E.g. Cow, Buffalo, Dog, Cat etc.

Question 7.
Are there any patterns in nature that give clues to modes of reproduction?
Answer:
There are two types of reproduction.

1. Asexual reproduction and
2. Sexual reproduction.

Question 8.
Names of some animals are listed below. Observe carefully and fill the table. Deer, Leopard, Pig, Fish, Buffalo, Giraffe, Frog, Sparrow, Lizard, Crow, Snake, Elephant, Cat.
Answer:

Animals that have external ears	Animals that do not have external ears
Deer	Fish
Leopard	Frog
Pig	Sparrow
Buffalo	Lizard
Giraffe	Crow
Elephant	Snake
Cat

You can also add some more names of animals you know to this table.

Question 9.
Think how animals could hear without external ears?
Answer:
Though the animals do not have ears to hear, they can sense the surrounding by its body.

Question 10.
Read the names of animals given below and try to fill the table given below.
Cow, Rat, Crow, Pig, Fox, Hen, Camel, Duck, Frog, Elephant, Buffalo, Pigeon, Cat, Peacock, Lizard. You can also add a few more animals to this list.
Answer:

Name of Animals	Presence of external ears (Yes/No)	Presence of epidermal hairs on the skin/Feathers on their wings
Cow	Yes	Epidermal hair
Rat	Yes	Epidermal hair
Crow	No	Feathers on their wings
Pig	Yes	Epidermal hair
Fox	Yes	Epidermal hair
Hen	No	Feathers on their wings
Camel	Yes	Epidermal hair
Duck	No	Feathers on their wings
Frog	No	No hair, no feathers
Elephant	Yes	No hair
Buffalo	Yes	Epidermal hair
Pigeon	No	Feathers on their wings
Cat	Yes	Epidermal hair
Peacock	No	Feathers on their wings
Lizard	No	No feathers, no wings.

8th Class Biology 4th Lesson Reproduction in Animals Activities

Activity – 1

Question 1.
Draw the diagram of Hydra. Compare it with the figure below recall what you have observed in the first slide ?

Compare slide 1 & 2 to observe which part of it’s body develops a swelling?
Answer:
The body wall develops swelling. Observe all the remaining slides.
a) What have you observed in slide/picture 1, 2 and 3?
Answer:
Picture 1, 2 the body surface of hydra has smooth surface.
Picture 3, a swelling is formed on it’s body surface.

b) What is the main difference between slide 1 and 2 as well as 3 and 4?
Answer:
Slide 1 and 2 Hydra body is smooth.
Slide. 3, a swelling is formed. Swelling increases in size, tentacles are formed which is called bud.
Slide 4, the bud is cut off and separated from parent Hydra and can live individually.

c) What does swelling (bulge) develop into?
Answer:
The swelling develops into a bud.

Activity – 2

Question 2.
Observe the given diagram carefully and fill the following table:

i)

Changes in the Nucleus/Body structure
1st diagram	Nucleus is round.
2nd diagram	Constriction in the centre of nucleus.
3rd diagram	Nucleus divides into two daughter nuclei. On the centre of the body surface a constriction is formed.
4th diagram	The constriction deepens.
5th diagram	The constriction still deepens and ready to cut into two parts.
6th diagram	Two daughter amoebae are formed.

ii) How many amoebae are formed at the end ?
Answer:
Two amoebae are formed at the end.
Male flower – its parts:

1. Calyx (sepals)
2. Corolla (Petals)
3. Androecium (stamen)(male part)
4. Pollen grain (male gametes)

Female Flower – Its parts.

1. Calyx (Sepals)
2. Corolla (Petals)
3. Gynoecium (Ovary) female part
4. Ovules (Future seeds)

1) What would happen if fusion of sperm and ova doesn’t takes place?
Answer:
If fusion of sperm and ova doesn’t takes place fertilization would not happen.

2) Why animals give birth to their babies?
Answer:
To continue their species on the earth.

3) What happens if each couple give birth to more than two babies?
Answer:
The population increases.

4) Is it necessary to control population?
Answer:
The rapidly increasing population is posing a number of problems as our resources in nature do not increase proportionately. On the other hand they diminish. So it is necessary to control population.

Activity – 3

Question 3.
Observation of resemblance in Parents & Children.
Table given below will help you to note the similar and dissimilar. Fill in the table.

You can ask your teacher and know why sometimes no characters match with your father or mother.
Answer:

1. The ability of an organism to produce a new generation of individuals of the same species is called reproduction.
2. That means the characteristics of parental organisms are being transferred to their next generation in the process of reproduction.
3. It involves the transmission of genetic material (chromosomes) from the parental generation to the next generation.
4. In some methods of reproduction the genetic material of the parent and the offspring next generation will be exactly same.
5. Whereas in some methods the characters from two parents (male and female) recombine to form a new individual.
6. In this process some characters of one parent and remaining characters from the other parent are seen in the offsprings. Some characters will be new which are not seen in either of the parents.
7. This happens because of chromosome recombination. The process of reproduction ensures continuation of race and the perpetuation of characteristics of the species and particularly the parental organisms.

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 3 Story of Microorganisms 1 Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 3rd Lesson Story of Microorganisms 1

8th Class Biology 3rd Lesson Story of Microorganisms 1 Textbook Questions and Answers

Improve Your Learning

Question 1.
Which organisms are interlinked between living and non-living organisms? Why do you think so?
Answer:
Viruses are an interesting type of microorganisms. They usually made up of crystalized proteins. They behave like nonliving things when they are outside of a living cell. But they behave like living organisms when they are inside host living cells and reproduce just like bacteria. Hence viruses can also call as connecting links between living and nonliving things.

Question 2.
What are microorganisms? Where do you find them?
Answer:
We can see several organisms in our surroundings but we cannot see many of them with our unaided eyes. They can be seen only with the help of microscope only. They are called microorganisms. They can found in air, water, soil and all living organisms.

Question 3.
What type of microorganisms we can observe in pond water?
Answer:
Usually pond water contains bacteria, phytoplanktons, algal members, fungi, rotifers, hydra etc. Collect some pond water with greenish scrapings on a slide and we can observe different algal members like Spirogyra, Chara and Chlamydomonas through the microscope.

Question 4.
Whether microorganisms are useful or harmful. How? Explain.
Answer:

1. Microorganisms are useful and some are harmful.
2. Some microorganisms are useful in formation process, medicine preparation and increase soil fertility.
3. Some microorganisms are harmful by causing diseases and spoling food items.

Question 5.
How are the human actions causing the death of useful bacteria and fungi? What will happen if it continuous?
Answer:

1. Soil is highly rich in microorganisms such a bacteria, fungi, protozoa, micro arthropods.
2. The top eight inches of soil of one acre many contain as much as five and half tons of fungi and bacteria.
3. This is very much useful for growing crops.
4. Excess use of pesticides kills these bacteria.
5. Thus human actions causing death of useful bacteria and fungi.
6. If it happens continue, then it causes to ecological imbalance.

Question 6.
Why the cooked food spoil soon but not uncooked food ? Give your reasons.
Answer:

1. Cooking of food items makes the proteins in the food materials coagulate.
2. It also degrades the protective surface of the food.
3. Thus the cooked foods can be easily inhabited by microoganisms.
4. So they can be spoiled in less time than the uncooked food.

Question 7.
What questions would you like to ask your teacher to know about different shapes of bacteria ?
Answer:

1. Where can we find bacteria?
2. How can we see bacteria?
3. What do we call the round shaped bacteria?
4. What do we call the spring shaped bacteria?
5. What is the name of coma shape bacteria?
6. How many types of bacteria do we find in nature?
7. What is the shape of Lactobacillus bacteria? How is it useful?
8. How is septicemia bacteria harmful?
9. Which type of bacteria is responsible for food poisoning?
10. Which bacteria is present in root nodules of leguminous plants? How do they useful?
11. Name the bacteria that causes leprosy.
12. Which type of bacteria is responsible for tuberculosis?
13. What is the shape of Bacillus thuringiensis bacteria how is it useful to plants?
14. What is the shape of staphylococci bacteria? In what way it affect the health of people?

Question 8.
What will happen if you add buttermilk to chilled milk?
Answer:

1. Lacto bacillus bacteria is responsible for the formation of curd.
2. When we add buttermilk to luke warm milk it takes 2 or 3 hours time to form curd.
3. But if we add buttermilk to chilled milk it takes more time or curd would not form.
4. Curdling indicates that the increase in number of bacteria in milk.
5. In chilled condition the number of bacteria do not increase in number there by curd would not be formed.

Question 9.
How do you observe Lactobacillus bacterium?
Answer:
Take one or two drops of buttermilk on a slide and spread it. Heat the slide slightly on a lamp (3-4 sec¬onds). Add a few drops of crystal violet. Leave it for 30-60 seconds and wash the slide gently with water.
Observe the slide under the Compound Microscope to see the Lactobacillus bacterium.

Question 10.
Visit any bakery or milk chilling center near your school with the help of your teacher or parents. Learn about some techniques to culture and usage of some Microorganisms and prepare a note on them.
Answer:
The Milk Collection Station is a specially designed, integrated unit, which combines the several functions of a milk collection centre. It measures the weight, fat content and gives the price of the milk brought in by the each producer. The equipment is particularly useful for the milk cooperatives / milk collection centres as it can also maintain a summary of milk supplied. This state of the art equipment operates both on battery and mains and is able to process and record 120-150 milk collection per hour. An Electronic Milk weighing Unit, the Electronic Milk Tester and Data Processor Unit are main components of the system. The membership code of individual mem¬bers is entered automatically by member identity card / manually by an electronic key-board.

Question 11.
Observe some permanent slides of microorganisms in your school lab with the help of microscope. Draw its picture.
Answer:

Question 12.
Prepare a model of any microorganism. And write a note on them.
Answer:

Question 13.
Why should we clean our hands with soap before eating ?
Answer:

1. We touch the objects.
2. Microbes are present on them.
3. When we touch them, they will inhabit our hands.
4. Washing our hands with soap kills all the microbes.
5. And makes our hands clean and hygenic.
6. So we should clean our hands with soap before eating.

8th Class Biology 3rd Lesson Story of Microorganisms 1 Activities

Activity – 2

Question 1.
Identify the fungi present, in rotten vegetables.
Answer:
Take some rotten part of vegetable or black spoiled part of bread or coconut with help of a needle on a slide, Put a drop of water. Place a cover slip on it and we can see the following microorganisms through microscope.

Activity – 6

Question 2.
Observe different soil microorganisms through microscope and draw rough sketches.
Answer:
Collect some soil from the field in a beaker or in a glass. Add some water to it and stir it. Wait for some time to allow the soil particles to settle down. Take a drop of water on a slide and we can observe the following microorganisms.

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 2 Cell: The Basic Unit of Life Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 2nd Lesson Cell: The Basic Unit of Life

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life Textbook Questions and Answers

Improve Your Learning

Question 1.
Who discovered the cell for the first time?
Answer:
It was the year 1665 Robert Hooke, a British scientist observed thin slices of cork under a simple magnifying device which he had made himself. He observed that the cork resembled the structure of a honey comb consisting of many empty spaces or empty box like structures. He thought it was made up of very small cavities, Robert Hooke called these cavities “Cell”.

Question 2.
Name the factors on which shape of the cells depend.
Answer:
The shape and size of the cells vary considerably but all of these cells ultimately determined by the specific function of the cells.
e.g.: Amoeba is changing its shape for specific functions like collection of food and locomotion.
The shape of the cell may vary for giving definite structure to the organism, e.g,: Epidermal cells.

Question 3.
Distinguish between unicellular and multi cellular organisms.
Answer:

Unicellular	Multicellular
1) An organism composed of just one cell.	1) An organism composed of more than cell.
2) Many of nature’s simplest creatures called “unicellular organisms”.	2) More biologically advanced creatures, called “multicellular organisms”.
3) Cell is an individual form no gathering to perform tasks, but they live together.	3) Different kinds of cells are joined together to perform specialized tasks.
4) The single cell a unicellular organism possesses, the smaller its body, e.g.: Amoeba, Chlamydomonas.	4) The more cells a multicellular organism possesses, the larger its body, e.g.: Fish, Neem tree.

Question 4.
How will you prepare slide without drying quickly?
Answer:
Preparation of slide is a technique to observe microscopic structures. Microscopic slide is prepared on a 2 mm thick. Thin flat plant material directly placed in a drop of water on the glass slide. A drop of glycerin is added to the water to keep the material for longer time. Glycerine saves the material from drying quickly.

Question 5.
Deekshith said that, “we can’t see cells with unaided eye.” Is the statement true or false? Explain.
Answer:
We can’t see cell with naked eye is true. All living things formed by microscopic cells which are visible through microscope only but the egg of birds is visible without microscope. The size of the cells in living organism may be as small as the millionth of a meter.
Most of the cells either in unicellular and multicellular are small in size to perform all the life processes perfectly. The smallest cell 0.1 to 0.5 micrometers found in bacteria. Some of the cells can be seen with naked eye. The largest cell 17 cm x 18 cm egg of Ostrich.

Question 6.
Correct the statement and if necessary rewrite. (OR)
What is cell wall and what are its functions?
a) Cell wall is essential in plant cells.
b) Nucleus controls cell activity.
c) Unicellular organisms perform all life processes like respiration, excretion, growth and reproduction.
d) To observe nucleus and organelles clearly, staining is not necessary.
Answer:
a) Cell wall is essential in plant cells.
Plant cell walls are essential for plant life and also have numerous industrial applications, ranging from wood to nutraceuticals.
The cell wall is the tough, usually flexible but sometimes fairly rigid layer that surrounds some types of cells. It is located outside the cell membrane and provides these cells with structural support and protection, in addition to acting as a filtering mechanism. A major function of the cell wall is to act as a pressure vessel, preventing over-expansion when water enters the cell. Cell wall is found in plants, bacteria, fungi, algae, archaea. Animals and protozoa do not have cell walls. “Plant cell wall is an essential component of biotic stress response mechanisms.”

b) Nucleus controls cell activity.
By containing the instructions for protein products in the DNA of the nucleus. All “control” work in the cell is carried out by proteins, such as enzymes, though DNA codes for other structural material, only protein has metabolic and behavioural control in the organism’s cells. Thus, the nucleus is the cell’s control center.

c) Unicellular organism perform all life processes like respiration, excretion, growth and reproduction.
All living organisms perform some basic life processes like respiration, excretion, etc., to sustain its life and improve its race. Unicellular organisms also perform all life processes.

d) To observe nucleus and organelles clearly, staining is not necessary.
To observe nucleus and organelles clearly, staining is necessary. Staining is a technique to get attached color to different parts of a cell. This helps to highlight particular areas in the cell.

Question 7.
Describe the structure of Nucleus.
Answer:
The nucleus is the largest cellular organelle in animals. In mammalian cells, the average diameter of the nucleus is approximately 6 micrometers (pm), which occupies about 10% of the total cell volume.

Question 8.
Explain the functions of Nucleus.
Answer:
Functions:

1. The main function of the cell nucleus is to control gene expression and mediate the replication of DNA during the cell cycle.
2. The nucleus provides a site for genetic transcription that is segregated from the location of translation in the cytoplasm, allowing levels of gene regulation that are not available to prokaryotes.

Question 9.
What is difference between cells in onion peel and cells in Spinach?
Answer:
Cells in onion peel arranged systematically with prominent nucleus. Cells in spinach are in different sizes and shapes without nucleus to perform nutrition.

Question 10.
Label parts of diagrammes given below. And identify which one is plant cell and which one is animal cell.

Answer:
A. Nucleus
B. Cytoplasm
C. Cell membrane
D. Vacuole
E. Nucleus
F. Cell wall
G. Cell membrane
H. Vocuole
I. Vacuole

Question 11.
What questions will you pose to know diversity in cells?
Answer:

1. Are all the cells similar in shape and size?
2. Do you find nuclei in all the cells?
3. How many different types of cells could you see?
4. What are the different shapes of the cells?
5. Do all the cells one of the same in length?

Question 12.
If you want to know about unicellular and multicellular organisms, what questions will you pose?
Answer:

1. What do you mean by unicelluar organism?
2. What do you mean by multicellular organism?
3. Give examples for unicellular and multicellular organisms.
4. What are the differences between unicellular and multicellular organism?

Question 13.
Get some floating slime from a puddle, pick a very small amount of slime and put it on a slide. Separate out one fiber and look at it through the microscope. Draw the diagram what you have observed.
Answer:

Question 14.
Collect different kinds of leaves from your surroundings and observe the shapes of the epidermal cells under microscope. Make a table which contains serial number, name of the leaf, shape of the leaf, shape of the epidermal cells. Do not forget to write specific findings below the table.
Answer:

Question 15.
Make sketches of animal and plant cells which you observed under microscope.
Answer:

Question 16.
Ameer said “Bigger onion has larger cells when compared to the cells of smaller onions”! Do you agree with his statement or not? Explain why.
Answer:
The sizes of the cells in living organisms are too small to be seen with naked eye. The size of the cell is related to its function. The cell has to perform similar function in all living organisms.
The size of the onion depends upon the number of cells and not the size of the cell. Cells are of different shape, size and number.
Hence, I don’t agree with Ameer. “Bigger onions have larger cells when compared to the cells of smaller onions”.

Question 17.
How do you appreciate the fact that a huge elephant, man and trees are made of cells, which are very small and we can look at them through microscope?
Answer:
The size of the cells in living organisms may be as small as the millionth of a meter (micron) or may be as large as a few centimeters. Majority of the cells are too small to be seen with naked eye.
The size of the cell is related to its function. For example, nerve cell both in man and animals are long and branched. They perform the same functions that of transferring message.
The size of the organism is depending upon the number of cells and not the size of the cell. Cells are of different shape, size and number.

Question 18.
Deepak said, “A plant can’t stand erect without cell wall”. Support this statement.
Answer:
Deepak said, “A plant can’t stand erect without cell wall.” We support this statement with the following reasons.
i) Plant cells differ from those of animals in having an additional layer around the cell membrane.
ii) We called if as ‘cell wall’.
iii) Cell wall gives strength and rigidity to planks.
So a plant can’t stand erect without cell wall.

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life InText Questions and Answers

Question 1.
Make different questions to know cells and cell organelles.
Answer:

1. What are the structures present in the cells?
2. Why cells are considered to be structural and functional unit of life?
3. What is the need of cell wall in plant cells?
4. Did we see the cells with naked eye?

Question 2.
Prepare different questions to know the discovery of cell.
Answer:

1. In which year cell discovered?
2. Name the scientist who observed cells.
3. What type techniques used to observe the cell?
4. Is there any special devices used to study the cell?
5. Name the different devices used for discovery of cell.
6. Can we see living cells under the microscope?

Question 3.
Prepare permanent slides of Onion cell, cheek cell and compare practically.
Answer:
Take inner layer of onion and cheek cells. Stain them with saffranin or methylene blue and keep coverslip.
Observe both the slides under microscope. Cell membrane and cell wall is present in both the cells. Dark stained nucleus is presented in the centre of the cell.
Comparison between onion and cheek cells:

Onion cells	Cheek cells
Cells arranged compactly	Cells arranged loosely
Boundary of onion is cell membrane	Boundary of cheek cell is cell wall
Cytoplasm and nucleus is present	Cytoplasm surrounds the nucleus
Cell organelles present in cytoplasm	Cell organelles present in cytoplasm.

Question 4.
Explain diversity in leaf cells practically.
Answer:
Take a section of Neem leaf on the slide and put a drop of water, cover it with a coverslip and observe it under the microscope. We can see different types of tissues present in the leaf. The section of leaf shows the following features.
Three groups of tissues arranged in leaf.
The first group epidermis where the cells are barrel shaped and covered by waxy material for protection.
The second group mesophyll tissue where the cells contain chloroplasts for nutrition. Cells arranged loosely with air spaces and stomata for exchange of gases.
The third group vascular tissue where the cells are thick walled to transport water and food.

Question 5.
Make a sketch of blood cells.
Answer:

Question 6.
Draw a neat diagram of Clilamydomonas.
Answer:

Question 7.
Make a sketch of Amoeba. (OR)
a) Draw a labelled diagram of Amoeba.
b) What are pseudopodia?
Answer:
a)

b) The projections on the body of amoeba which help in locomotion and collecting food.

Question 8.
Have you listened to the words of the cell? Guess how big a cell is? Is the number and sizes of cells in both man and elephant the same? Are the cells of an elephant bigger than that of a man?
Answer:
The size of the cells in living organisms may be as small as the millionth of a meter (micron) or may be as large as a few centimeters. Majority of the cells are too small to be seen with naked eye.
The size of the cell is related to its function. For example, nerve cell in both man and elephant are long and branched. They perform the same functions that of transferring message.
The size of the organism is depending upon the number of cells and not the size of the cell. Cells are of different shape, size and number. Hence size of cells in both elephant and man are same. The number of cells are more in elephant than man.

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life Activities

Activity – 2

Question 1.
Prepare a slide of an onion peel and find out the special characters.
Answer:
Peel an onion and cut out a small fleshy portion from the bulb. Break this into two small parts and try to separate them. We notice a thin film like material holding the pieces together. Take out small portion and spread it evenly on a slide. Cover it with a cover slip and observe it under microscope.
Draw the figure what you have observed.
Cells are arranged side by side without any gaps.

Activity – 4

Question 2.
Observation of the Nucleus in onion peel cells. (OR)
Sahitya is trying to observe the nucleus in the cells of onion peel. Explain the procedure to be followed for the experiment.
Answer:
Peel a membrane an onion now keep this membrane on a slide and add 1-2 drops of the stain (saffranin, methylene blue or red ink). Cover this with a cover slip and leave it for about five minutes. Then add water drop-wise from one side of the cover slip while soaking the extra water with a filter paper from the other side. This will help in washing away the extra stain. Now observe this slide under a microscope. The blue spot observed within the cell is the nucleus.

Activity – 5

Question 3.
Observe the following pictures and answer the questions given below.

i) What are the structures present in the cells ?
Answer:
Cell wall in onion cell and cell membrane in cheek cell.
Cytoplasm and nucleus are common in both the cells.

ii) Did you see a tiny dark stained thing in all the cells ?
Answer:
Yes there is a tiny dark stained nucleus is common in both the cells.

iii) Are they located in the center of the cell in both the cells ?
Answer:
Nucleus located in center of both the cells.

iv) What is the difference between boundary of onion cell and cheek cell ?
Answer:
Cell wall is the boundary of onion and cell membrane is the boundary of cheek cell.

Activity – 6

Question 4.
Collect leaves stems and roots of different plants from the field and take sections to study different types of cells and tissues present in leaf and stems practically.
Answer:
Take a section of grass leaf on the slide and put a drop of water, cover it with a cover slip and observe it under the microscope. We can see different types of tissues present in the leaf.
The section of root or stem shows the following features.
Four groups of cells can be observed.
First group is outermost layer called epidermis.
It protects stem or root externally.
Major portion of stem or root has second group of cells.
This group synthesizes food and preserve food.
Third group of cells transport water and food.
Fourth group of cells placed centrally.

Question 5.
a) Observe the following cells and collect permanent slides.

b) Fill the following table with help of your teacher.

Name of the cell	Shape of the cell	Name of the parts observed in it
RBC	Biconcave	Blood tissue
Smooth Muscle Cell	Rod	Muscles
Nerve Cell	Tree	Brain, spinal cord and nerves
Bone Cell	Star	All bones
White Blood cell	Amoeboid	Blood tissue

i) Are there any similarities in shape of the cells?
Answer:
No similarity is found in the shape and size of the above cells.

ii) Do you find nuclei in all the cells?
Answer:
In human RBC nucleus is absent. Muscle, Nerve, Bone and White Blood Cells consist nucleus.

iii) Do you know, which cell is the longest in all animals?
Answer:
Nerve cell is the longest in all animals.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers InText Questions and Answers.

8th Class Maths 15th Lesson Playing with Numbers InText Questions and Answers

Do this

Question 1.
Write the place value of numbers underlined.     (Pg. No: 312)
i) 29879   ii) 10344   iii) 98725
Answer:
i) 29879
Place value of 8 = 8 × 100 = 800
Place value of 2 = 2 × 10,000 = 20,000
ii) 10344
Place value of 4 = 4 × 1 = 4
Place value of 3 = 3 × 100 = 300
iii) 98725
Place value of 5 = 5 × 1 = 5
Place value of 8 = 8 × 1000 = 8,000

Question 2.
Write the following numbers in expanded form,        (Pg. No: 313)
i) 65    ii)    74    iii) 153    iv) 612
Answer:
Number Expanded form
i) 65 = 60 + 5 = (6 × 10¹) + (5 × 10⁰)
ii) 74 = 70 + 4 = (7 × 10¹) + (4 × 10⁰)
iii) 153 = 100 + 50 + 3 = (1 × 10²) + (5 × 10¹) + (3 × 10⁰)
iv) 612 = 600 + 10 + 2 = (6 × 10²) + (1 × 10¹) + (2 × 10⁰)

Question 3.
Write the following in standard notation.       (Pg. No: 313)
i) 10 × 9 + 4     ii) 100 × 7 + 10 × 4 + 3
Answer:
Expanded form General form
i) 10 × 9 + 4 = 90 + 4 = 94
ii) 100 × 7 + 10 × 4 + 3 = 700 + 40 + 3 = 743

Question 4.
Fill in the blanks.       (Pg. No: 313)
Answer:
i) 100 × 3 + 10 × ——— + 7 = 357 (5)
ii) 100 × 4 + 10 × 5 + 1 = ——— (451)
iii) 100 × ——— + 10 × 3 + 7 = 737 (7)
iv) 100 × ——— + 10 × q + r = \(\overline{\mathrm{pqr}}\) (p)
v) 100 × x + 10 × y + z = ——— (\(\overline{\mathrm{xyz}}\))
Do you know?

Question 5.
The number 8281807978777675747372717069686766656463626160595857565554535251504948474645444342414039383736353433323130292827262524232221201918 1716151413121110987654321 is written by starting at 82 and writing backwards to 1 and see that it is a prime number.        (Pg. No: 313)
Answer:
No.of digits in the given number are 155.

Question 6.
Write all the factors of the following numbers.       (Pg. No: 314)
a) 24    b) 15   c) 21   d) 27   e) 12   f) 20   g) 18   h) 23   i) 36
Answer:
a) Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
b) Factors of 15 = 1, 3, 5, 15
c) Factors of 21 = 1, 3, 7, 21
d) Factors of 27 = 1, 3, 9, 27
e) Factors of 12 = 1, 2, 3, 4, 6, 12
f) Factors of 20 = 1, 2, 4, 5, 10, 20
g) Factors of 18 = 1, 2, 3, 6, 9, 18
h) Factors of 23 = 1, 23
i) Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36

Question 7.
Write first five multiples of given numbers     (Pg. No: 314)
a) 5   b) 8   c) 9
Answer:
a) Multiples of 5 = 5, 10, 15, 20, 25
b) Multiples of 8 = 8, 16, 24, 32, 40
c) Multiples of 9 = 9, 18, 27, 36, 45

Question 8.
Factorize the following numbers into prime factors.    (Pg. No: 314)
a) 72    b) 158   c) 243
Answer:
a) 72 = 2 × 2 × 2 × 3 × 3
b) 158 = 2 × 7 × 9
c) 243 = 7 × 7 × 7

Question 9.
Check whether the following given numbers are divisible by 10 or not.   (Pg. No: 315)
a) 3860   b) 234   c) 1200   d) 10³   e) 10 + 280 + 20
Answer:
a) 3860, c) 1200, d) 103, e) 10 + 280 + 20 are divisible by ’10’.
[∵ the units digit of above numbers is ‘0’]
b) 234, is not divisible by 10.
[∵ its unit digit is 4]

Question 10.
Check whether the given numbers are divisible by 10 or not.    (Pg. No: 315)
a) 10¹⁰   b) 2¹⁰   c) 10³ + 10¹
Answer:
a) 10¹⁰ = 10000000000
b) 2¹⁰ = 1024
c) 10³ + 10¹ = 1000 + 10 = 1010
∴ a) 10¹⁰, c) 10³ + 10¹ are divisible by ’10’.
[∵ Their units digits are ‘0’.]
b) 1024 is not divisible by 10.
[∵ Its units digit is 4.]

Question 11.
Check whether the given numbers are divisible by 5 or not      (Pg. No: 315)
a) 205   b) 4560    c) 402    d) 105    e) 235785
Answer:
a) 205, b) 4560, d) 105, e) 235785 are divisible by 5.
[∵ The units digit of the above numbers are either 0 (or) 5.]
c) 402 is not divisible by 5.
[∵ Its units digit is 2.]

Question 12.
Check whether the given numbers which are divisible by 3 or 9 or by both,      (Pg. No: 318)
a) 3663    b) 186    c) 342    d) 18871    e) 120    f) 3789    g) 4542    h) 5779782
Answer:

Question 13.
Check whether the given numbers are divisible by 6 or not.      (Pg. No: 320)
a) 1632    b) 456     c) 1008     d) 789     e) 369    f) 258
Answer:

Question 14.
Check whether the given numbers are divisible by 6 or not.     (Pg. No: 320)
a) 458 + 676    b) 6³    c) 6² + 6³    d) 2² × 3²
Answer:

Question 15.
Can you arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in an order so that the number formed by first two digits is divisible by 2, the number formed by first three digits is divisible by 3, the number formed by first four digits is divisible by 4 and so on upto nine digits?
Solution: The order 123654987 looks promising check and verify.     (Pg. No : 320)
Answer:

∴ This number can’t continue upto ‘9’.
→ 123654987

∴ The given number 123654987 is not divisible by all the numbers like 2, 3, 4, 5,……… 9.

Question 16.
Check whether the given numbers are divisible by 4 or 8 or by both 4 and 8.
a) 464    b) 782     c) 3688    d) 100     e) 1000    f) 387856    g)4⁴     h) 8³ (Pg. No: 321)
Answer:
If a number is divisible by 4 then the last two digits of the number must be divisible by 4.
If the last 3 digits of a number is divisible by 8 then it is divisible by 8.

Question 17.
Check whether the given numbers are divisible by 7. (Pg. No: 322)
a) 322     b) 588     c) 952     d) 553    e) 448
Answer:

All the given numbers are divisible by ‘7’.

Question 18.
Check whether the given numbers are divisible by 11.    (Pg. No: 323)
i) 4867216      ii) 12221     iii) 100001
Answer:
If the difference between the sum of digits of odd places and even places is divisible by 11, then entire number is divisible by 11.

Question 19.
Take different pairs of numbers and check the above four rules.      (Pg. No: 325)
Answer:
a) Consider a factor of 36, say 9.
Factors of 9 are 1,3,9.
∴ 36 is divisible by 1, 3, 9.
∴ 36 is also divisible by all the factors of 9.
b) Let us consider a number 60. It is divisible by 5 and 6. It is also divisible by 5 x 6 = 30 Where 5, 6 are co-primes.
c) Take two numbers 25, 30. These numbers are both divisible by 5.
The number 25 + 30 = 55 is also divisible by 5.
d) Take two numbers 36, 54. These numbers are both divisible by 9.
Their difference i.e., 54 – 36 = 18 is also divisible by 9.

Question 20.
144 is divisible by 12. Is it divisible by the factors of 12? Verify.     (Pg. No : 325)
Answer:
Factors of 12 = 1, 2, 3, 4, 6, 12.
If 12 is a factor of 144 then 144 is divisible by all the factors of 12.

Question 21.
Check whether 2³ + 2⁴ + 2⁵ is divisible by 2. Explain.       (Pg. No : 325)
Answer:
2³ + 2⁴ + 2⁵ = 8 + 16 + 32 = 56 is an even.
∴ 56 is divisible by 2.

Question 22.
Check whether 33 – 32 is divisible by 3. Explain    (Pg. No : 325)
Answer:
33 – 32 = 27 – 9 = 18 → 1 + 8 = 9 ⇒ \(\frac{9}{3}\) (R = 0)
∴ It is divisible by ‘3’.

Question 23.
Check the result if the numbers chosen were       (Pg. No : 328)
i) 37    ii) 60    iii) 18   iv) 89
Answer:
i) If the digits are interchanged in 37 then it becomes as 73.
∴ 37 + 73 = 110 → \(\frac{110}{11}\) (R = 0)
It is divisible by ’11’.

Question 24.
In a cricket team there are 11 players. The selection board purchased 10x + y T-shirts to players. They again purchased ‘10y + x’ T-shirts and total T-shirts were distributed to players equally. How many T-shirts will be left over after they distributed equally to 11 players ? How many each one will get?     (Pg. No : 328)
Answer:
No.of players in the team = 11
No.of T- shirts are purchased at first = 10x + y
No. of T – shirts are purchased for the 2nd time = 10y + x
Sum of the T – shirts = (10x + y) + (10y + x)
= 11x + 11y = ll(x + y)
∴ 11(x + y) T – shirts are distributed among 11 players then each will get ll(x + y)
\(\frac{11(x + y)}{11}\) = x + y
Remaining T – shirts = Purchased T – shirts – 11 (No.of T-shirts got by each)
= 11(x + y) – 11(x + y)
= 0

Question 25.
In a basket there are ‘10a + b’ fruits (a ≠ 0 and a > b). Among them ‘10b + a’ fruits are rotten. The remaining fruits distributed to 9 persons equally. How many fruits are left over after equal distribution? How many fruits would each child get?      (Pg. No: 328)
Answer:
No. of fruits in a basket = 10a + b
No. of fruits are rotten = 10b + a
Remaining fruits to be distributed = (10a + b) – (10b + a)
= 10a + b – 10b – a
= 9a – 9b = 9(a – b)
∴ 9(a – b) fruits are distributed among ‘9’ Children
then each will get = 9(a – b) ÷ 9 = \(\frac{9(a – b)}{9}\) = (a – b)
No. of fruits left over after distribution
= Total no. of fruits distributed – No.of fruits got by each
= 9(a – b) – 9(a – b) = 0

Question 26.
Check in the above activity with the following numbers.      (Pg. No: 329)
i) 657    ii) 473     iii) 167    iv) 135
Answer:

Question 27.
If 21358AB is divisible by 99, find die values of A and B.     (Pg. No: 331)
Answer:
If 21358AB is divisible by 99, then it is divisible by 9 and 11.
If 21358AB is divisible by 9 then the sum of the digits is divisible by 9.
2 + 1 + 3 + 5 + 8 + A + B = 9 × 3 say
⇒ 19 + A + B = 27
⇒ A + B = 27-19 = 8
A + B = 8 …… (1)
If 21358AB is divisible by ‘ll’ then the difference of sum of even and odd digits will be divisible by’ll’.
2 1 3 5 8 A B
∴ (2 + 3 + 8 + B) – (1 + 5 + A) = 11 × 1 say
⇒ 13 + B – 6 – A = 11
⇒ B – A = 11 – 7 = 4 ……. (2)
From (1) & (2) A = 2, B = 6
∴ The required number is 21358AB = 2135826 which is divisible by 99.

Question 28.
Find the values of A and B of file number 4AB8 (A, B are digits) which is divisible by 2, 3, 4, 6, 8 and 9.       (Pg. No: 331)
Answer:
Given number is 4AB8.
4AB8 → \(\frac{8}{2}\) (R = 0) so, it is divisible by ‘2’.
4AB8 → If it is divisible by ‘3’, sum of all the digits should be a multiple of ‘3’. .
∴ 4 + A + B + 8 = 3 or 6 or 9 or 12 or 15 …….

4AB8 → If it is divisible by 9, sum of all the digits should be a multiple of ‘9’.
∴ 4 + A + B + 8 = 9 or 18 or 27 or 36
⇒ A + B + 12 = 9 ∣18∣ 27∣ 36 ……. (3)
From (1) & (3)
A + B + 12 = 9 or 18 say
If A + B + 12 = 9
A + B = 9 – 12 = -3
It is impossible
If A + B + 12 = 18
A + B = 18 – 12 = 6
∴ A + B = 6
If A = 4 & B = 2
4AB8 = 4428
4AB8 → 4428 → \(\frac{428}{8}\) (R ≠ 0)
∴ A = 4 & B = 2 are not possible.
If A = 2& B = 4
4AB8 → 4248 → \(\frac{248}{8}\) (R = 0)
∴ A = 2 and B = 4

Question 29.
By using the above method check whether 7810364 is divisible by 4 or not.        (Pg. No: 333)
Answer:
Given number = 7810364

Sum of product of place values and remainders of place values = 0 + 0 + 0 + 0 + 0 + 12 + 4
→ \(\frac{16}{4}\) (R = 0)
∴ 7810364 is divisible by ‘4’.

Question 30.
By using the above method check whether 963451 is divisible by 6 or not.     (Pg. No: 333)
Answer:
The given number = 963451

Sum of product of place values and remainders of place values
= 36 + 24 + 12 + 16 + 20 + 1 → \(\frac{109}{6}\) (R ≠ 0)
∴ 963451 is not divisible by ‘6’.

Try these

Question 1.
In the division 56 Z ÷ 10 leaves remainder 6, what might be the value of Z.     (Pg. No: 315)
Answer:
Let 56Z, Z = 0, 1,2, 3, 4, ….. , 9 say.
To obtain remainder ‘6’ when divided by 10, Z = 6
\(\frac{566}{10}\) = \(\frac{560+6}{10}\)
Remainder is 6.

∴ Z = 6

Question 2.
If 4B ÷ 5 leaves remainder 1, what might be the value of B?    (Pg. No : 316)
Answer:
If 4B is divided by 5 then remainder should be ‘1’,
∴ B = {0, 1, 2, ….. , 9}
i.e., 40, 41, 42, …… , 49
From the above numbers we have to take 41 and 46.
41 and 46 are divided by 5 and leaves the remainder 1.
∴ B = {1, 6}

Question 3.
If 76C ÷ 5 leaves remainder 2, what might be the value of C?     (Pg. No: 316)
Answer:
To get remainder 2, when 76C is divided by 5 take C = {0, 1,……, 9}.
If C = 2, 7 then
76C = 762 or 767 are divided by 5 leaves the remainder 2.

Question 4.
“If a number is divisible by 10, it is also divisible by 5.” Is the statement true? Give reasons.     (Pg. No : 316)
Answer:
The given statement is true.
∵ When a number is divisible by ’10’, then its units digit should be ‘0’.
Similarly the units digit of a number is 5 or 0, then it is divisible by 5.
∴ The number which is divisible by 10 is also divisible by 5.

Question 5.
“If a number is divisible by 5, it is also divisible by 10.” Is the statement is true or false? Give reasons.     (Pg. No : 316)
Answer:
The given statement is false.
∵ If a number is divisible by 5, then its units digit must be ‘5’ or ‘0’. But in case of 10, it . must be ‘0’ only.
∴ The number which is divisible by ‘5’ is need not be divisible by ’10’.

Question 6.
Check whether the given numbers are divisible by 4 or 8 or by both 4 and 8.     (Pg. No : 321)
a) 4² × 8² b) 10³ c) 10⁵ + 10⁴ + 10³ d) 4³ + 4² + 4¹ – 2²
Answer:

Question 7.
Take a four digit general number, make the divisibility rule for ‘7’.     (Pg. No : 322)
Answer:
Let the 4 – digited number be ‘abcd’ say.
The remainders when divided by ‘7’,

∴ If (6a + 2b + 3c + d) is divisible by 7 then the 4 – digited number be divisible by ‘7’.

Question 8.
Check your rule with the number 3192 which is a multiple of 7.      (Pg. No : 322)
Answer:
The given number is 3192
⇒ a = 3, b = 1, c = 9, d = 2
6a + 2b + 3c + d = 6 × 3 + 2 × l + 3 × 9 + 2.
= 18 + 2 + 27 + 2
= 49 → \(\frac{49}{7}\) (R = 0)
∴ 3192 is divisible by 7’according to my law.

Question 9.
1) Verify whether 789789 is divisible by 11 or not.     (Pg. No: 323)
2) Verify whether 348348348348 is divisible by 11 or not.
3) Take an even palindrome i.e. 135531 check whether this number is divisible by 11 or not.
4) Verify whether 1234321 is divisible by 11 or not.
Answer:

Question 10.
Check whether 1576 × 1577 × 1578 is divisible by 3 or not.     (Pg. No : 325)
Answer:
The given number is 1576 × 1577 × 1578.
The product of any 3 consecutive numbers is divisible by ‘3’.
Ex : 4 × 5 × 6 = 120 → \(\frac{120}{3}\) (R = 0)
∴ 1576 × 1577 × 1578 is divisible by ’3’.

Question 11.
Check the above method applicable for the divisibility of 11 by taking 10-digit number.     (Pg. No : 326)
Answer:
The largest 10 – digited number = 9,99,99,99,999
D C B A
∴ 9/999/999/999
⇒ B + D = 9 + 999 = 1008
A + C = 999 + 999 = 1998
∴ (A + C) – (B + D) = 990 → \(\frac{990}{11}\) (R = 0)
∴ The largest 10 – digited number should be divisible by 11 according to this method.

Question 12.
Take a three digit number and make the new numbers by replacing its digits as (ABC, BCA, CAB). Now add these three numbers. For what numbers the sum of these three numbers is divisible?      (Pg. No : 329)
Answer:

Question 13.
If YE × ME = TTT find the numerical value of Y + E + M + T.
[Hint: TTT = 100T + 10T + T = T(111) = T(37 × 3)]      (Pg. No: 332)
Answer:
TTT = 100T + 10T + T
= T(111)
= T(37 × 3)
∴ YE × ME = T(37 × 3)
∴ T ={1, 2, 3, ….., 9}
But T = {3, 6, 9} are multiples of 3.
T(37 × 3) = 3(111), 6(111), 9(111) are divisible by 3.
∴ YE × ME = 333∣666∣999
YE × ME = 999 = 27 × 37
∴ Y = 2, M = 3, E = 7, T = 3
∴ Y + E + M + T = 2 + 3 + 7 + 3 = 15

Question 14.
If cost of 88 articles is A733B, find the values of A and B.       (Pg. No: 334)
Answer:
If A733B is divisible by 88 then it is divisible by 8 × 11.
Divisibility of 11:
⇒ A733B → (A + 3 + B) – (7 + 3) = 0
⇒ A + B = 7 ……. (1)
Divisibility of 8:
⇒ A733B ⇒ \(\frac{33B}{8}\)
∴ \(\frac{336}{8}\) (R = 0) (If B = 6 then it is divisible by 8)
∴ B = 6 ……. (2)
From (1), (2)
∴ A = 1, B = 6

Question 15.
Check whether 456456456456 is divisible by 7, 11 and 13.     (Pg. No: 334)
Answer:
∴ The given number = 456456456456
456456456456 = 456 (1001001001)
= 456 × (7 × 11 × 13) × (1000001)
∴ 456456456456 is divisible by 7, 11 and 13.

Think, Discuss and Write

Question 1.
Find the digit in the units place of a number if it is divided by 5 and 2 leaves the remainders 3 and 1 respectively.   (Pg. No: 316)
Answer:
If a number is divided by 5 and 2 leaves the remainders 3 and 1 respectively, then its units digit be 3.
Ex: \(\frac{13}{5}\) ⇒ (R = 3), \(\frac{13}{2}\) ⇒ (R = 1)
\(\frac{23}{5}\) ⇒ (R = 3), \(\frac{23}{2}\) ⇒ (R = 1)
∴ The unit’s digit of a required number be 3.

Question 2.
Take a two digit number reverse the digits and get another number. Subtract smaller number from bigger number. Is the difference of those two numbers is always divisible by 9?      (Pg. No : 328)
Answer:

Question 3.
1) Can we conclude 10²ⁿ – 1 is divisible by both 9 and 11? Explain.     (Pg. No: 333)
2) Is 10²ⁿ⁺¹ – 1 is divisible by 11 or not? Explain.
Answer:

Question 4.
Verify a⁵ + b⁵ is divisible by (a + b) by taking different natural numbers for ‘a’ and ‘b’.    (Pg. No : 334)
Answer:

∴ a⁵ + b⁵ is divisible by (a + b).
∴ (a⁵ + b⁵) is divisible by (a + b) for all the values of a, b.

Question 5.
Can we conclude (a²ⁿ⁺¹ + b²ⁿ⁺¹) is divisible by (a + b)?      (Pg. No : 334)
Answer:

a²ⁿ⁺¹ + b²ⁿ⁺¹ is divisible by (a + b) for all the values of ‘n’.

AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers.

8th Class Maths 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers

Do this

Question 1.
Find the total surface area of the following cuboid.      [Page No. 298]

Answer:
i) l = 4 cm, b = 4 cm, h = 10 cm.
The total surface area of a cuboid = 2 (lb + bh + lh)
= 2 (4 × 4 + 4 × 10 + 4 × 10) = 2(16 + 40 + 40)
= 2 × 96
= 192 Sq. cms.

ii) l = 6 cm, b = 4 cm, h = 2 cm.
The total surface area of a cuboid = 2 (lb + bh + lh)
= 2(6 × 4 + 4 × 2 + 6 × 2)
= 2 (24 + 8 + 12)
= 2 × 44
= 88 sq. cms.

Question 2.
Let us find the volume of a cuboid whose length, breadth and height are 6 cm, 4 cm and 5 cm respectively.      [Page No. 287]

Let place 1 cubic centimeter blocks along the length of the cuboid. How many blocks can we place along the length? 6 blocks, as the length of the cuboid is 6 cm.
How many blocks can we place along its breadth? 4 blocks, as the breadth of the cuboid is 4 cm. So there are 6 × 4 blocks can be placed in a layer.
How many layers of blocks can be placed in the cuboid? 5 layers, as the height of the cuboid is 5 cm. Each layer has 6 × 4 blocks. So, all the 5 layers will have 6 × 4 × 5 blocks i.e. length × breadth × height.
This discussion leads us to the formula for the volume of a cuboid.
Volume of a cuboid = length × breadth × height      [Page No. 305]
Answer:
The dimensions of a cuboid are 6 cm, 4 cm, 5 cm respectively.
∴ Volume (V) = lbh
= 6 × 4 × 5.
= 120 cm³

Question 3.
Arrange 64 unit cubes in as many ways as you can to form a cuboid. Find the surface area of each arrangement. Can solid cuboid of same volume have same surface area? [Page No. 306]
Answer:
No. of cuboids are formed using 64 unit cubes
64 = 1 × 64 ……. (1)
= 2 × 32 …….. (2)
= 4 × 16 …….. (3)
1) l = 64 cm, b = 1 cm, h = 1 cm.
The total surface area of a cuboid, A = 2 (lb + bh + lh)
= 2 (64 × 1 + 1 × 1 + 1 × 64)
= 2 (64 + 1 + 64)
= 2 × 129
= 258 Sq. cm.

2) l = 32 cm, b = 2 cm, h = 1 cm.
A = 2 (lb + bh + lh)
= 2 (32 × 2 + 2 × 1 + 32 × 1)
= 2 (64 + 2 + 32)
= 2 × 98 = 196 Sq. cm.

3) l = 16 cm, b = 4 cm, h = 1 cm.
A = 2 (lb + bh + lh)
= 2 (16 × 4 + 4 × 1 + 16 × 1)
= 2 (64 + 4 + 16)
= 2 × 84 = 168 Sq. cm.

No, the volume of a cuboid is not same as the surface area of a cuboid.

Try These

Question 1.
Find the surface area of cube ‘A’ and lateral surface area of cube ‘B’.      [Page No. 300]

Answer:
a = 10 cm.
The total surface area of a figure ‘A’ = 6a²
= 6 × (10)²
= 6 × 100 = 600 Sq. cm.
Lateral surface area of a figure ‘B’ – 4a²
= 4 × (8)² [∵ a = 8 cm.]
= 4 × 64 = 256 Sq. cm.

Question 2.
Two cubes each with side ‘b’ are joined to form a cuboid as shown in the given fig. What is the total surface area of this cuboid?      [Page No. 300]

Answer:

Total surface area of a cuboid = 2 (lb + bh + lh)
= 2 (2b × b + b × b + 2b × b)
= 2 (2b² + b² + 2b²)
= 2(5b²) = 10b² Sq. cm.

Question 3.
How will you arrange 12 cubes of equal lengths to form a cuboid of smallest surface area?      [Page No. 300]

Answer:
We can’t obtain the least total surface area by arranging 12 cubes by side by side.

∴ A = 2 (lb + bh + lh)
= 2 (12 × 1 + 1 × 1 + 12 × 1)
= 2 (12 + 1 + 12)
= 2 × 25 = 50 Sq. cm.
We can obtain the least total surface area by arranging 3 cubes by 4 cubes.
∴ A = 2 (lb + bh + lh)
= 2 (3 × 1 + 1 × 4 + 3 × 4) (∵ l = 3; b = 1; h = 4)
= 2 (3 + 4 + 12)
= 2 × 19
= 38 Sq. cm.

Question 4.
The surface area of a cube of 4 × 4 × 4 dimensions is painted. The cube is cut into 64 equal cubes. How many cubes have
(a) 1 face painted? (b) 2 faces painted? (c) 3 faces painted? (d) no face painted?       [Page No. 300]
Answer:
If the 4 × 4 × 4 cube is divided into 64 equal cubes then the length of its each side = 1 unit.

[∵ \(\frac{4 \times 4 \times 4}{64}\) = 1]
a) No.of cubes (a = 4) have painted 1 face = 6(a – 2)² = 6(4 – 2)² = 6 × 4 = 24
b) No.of cubes have painted 2 faces = 12(a – 2) = 12(4 – 2) = 24
c) No.of cubes have painted 3 faces = 4 × a = 4 × 2 = 8
d) No.of cubes have painted no faces = (a – 2)³ = (4 – 2)³ = (2)³ = 8

Think, Discuss and Write

Question 1.
Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base.      [Page No. 299]
Answer:
Total surface area of a cuboid = L.S.A + 2 × Area of base
= 2h (l + b) + 2 × lb
= 2lh + 2bh + 2lb
= 2 (lb + bh + lh)
We can conclude that total surface area of a cuboid = L.S.A + 2 × Area of base

Question 2.
If we change the position of cuboid from Fig. (i) to Fig. (ii) do the lateral surface areas become equal?     [Page No. 299]

Answer:
There will be no change in the L.S.A of a cuboid if its positions are changed.

Question 3.
Draw a figure of cuboid whose dimensions are l, b, h are equal. Derive the formula for LSA and TSA.       [Page No. 299]
Answer:

Lateral surface area of a cuboid
= 4 × (areas of 4 faces)
= 2 (l × h) + 2 × (b × h) (1 + 2 + 3 + 4 faces)
= 2h(l + b) sq.units (1 = 3, 4 = 2)
∴ Total surface area of a cuboid
= 4 × (Area of 4 faces) + (Areas of upper & lower faces)
= 2h (l + b) + 2 (lb)
= 2lh + 2bh + 27b
= 2 (lb + bh + lh) sq.units.

AP State Syllabus 8th Class Maths Solutions 13th Lesson Visualizing 3-D in 2-D InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions and Answers.

8th Class Maths 13th Lesson Visualizing 3-D in 2-D InText Questions and Answers

Do This

Question 1.
Name some 3 – Dimensional objects.   [Page No. 282]
Answer:

1. Cube
2. Cylinder
3. Sphere
4. Cuboid
5. Cone

Question 2.
Give some examples of 2 – Dimensional objects.     [Page No. 282]
Answer:

1. Square
2. Rectangle
3. Line segment
4. Circle
5. Triangle

Question 3.
Draw a kite in your notebook. Is it 2 – D or 3 – D object?      [Page No. 282]
Answer:

Kite is a 2 – D object.

Question 4.
Identify some objects which are in cube or cuboid shape.      [Page No. 282]
Answer:
Shapes of Cube                           Shapes of Cuboid
a) Chalk piece box                       a) Duster
b) Dice                                         b) Cell phone (Cuboidal in shape)
c) Cube shaped cake                   c) Plasma T.V.

Question 5.
How many dimensions that a circle and sphere have?      [Page No. 282]
Answer:
Circle has 2 dimensions.
Sphere has 3 dimensions.

Question 6.
Identify the faces, edges and vertices of given figures.      [Page No. 288]

Answer:

Question 7.
Write the names of the prisms given below:     [Page No. 290]

Answer:
(i) Cube
(ii) Triangular prism
(iii) Pentagonal prism
(iv) Hexagonal prism
(v) Rectangular prism

Question 8.
Write the names of the pyramids given below:       [Page No. 290]

Answer:
(i) Square pyramid
(ii) Pentagonal pyramid
(iii) Hexagonal pyramid

Question 9.
Fill the table:      [Page No. 290]

Answer:

Question 10.
Explain the difference between prism and pyramid.      [Page No. 290]
Answer:
Upper and lower sides of a prism are equal in number. But, in a pyramid the base is a plane and all the edges are coincide in a single point on the top.

Try These

Question 1.
Name three things which are the examples of polyhedron. [Page No. 287]
Answer:

Question 2.
Name three things which are the examples of non-polyhedron. [Page No. 287]
Answer:

Think, Discuss and Write

Question 1.
How to find area and perimeter of top view and bottom view of the given figure.       [Page No. 283]

Answer:
Let the side of each face be ‘1’ unit say.
Shapes of different positions (I)                      Their areas (II)
1. Front view                                                    A = (1 × 1) + (1 × 1) + (1 × 1) = 3 Sq. Units
2. Top view                                                      A = (1 + 1 + 1) × (1 + 1) = 3 × 2 = 6 Sq. Units
3. Bottom view                                                A = (1 + 1 + 1) × (1 + 1) = 3 × 2 = 6 Sq. Units
Perimeters (III)
1. —————>                                             1 + 1 + 1 = 3 Units
2. —————>                                             2(l + b) = 2 (3 + 2) = 2 × 5 = 10 Units
3. —————>                                             2(l + b) = 2 (3 + 2) = 2 × 5 = 10 Units

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions and Answers.

8th Class Maths 12th Lesson Factorisation InText Questions and Answers

Do this

Question 1.
Express the given numbers in the form of product of primes. [Page No. 267]
(i) 48      (ii) 72      (iii) 96
(i) 48
Answer:
48 = 2 × 2 × 2 × 2 × 3

ii) 72
Answer:
72 = 2 × 2 × 2 × 3 × 3

iii) 96
Answer:
96 = 2 × 2 × 2 × 2 × 2 × 3

Question 2.
Find the factors of following:      [Page No. 268]
(i) 8x²yz     (ii) 2xy (x + y)       (iii) 3x + y³z
Answer:
i) 8x²yz = 2 × 2 × 2 × x × x × y × z
ii) 2xy (x + y) = 2 × x × y × (x + y)
iii) 3x + y³z = (3 × x) + (y × y × y × z)

Question 3.
Factorise:      [Page No. 270]
(i) 9a² – 6a
(ii) 15a³b – 35ab³
(iii) 7lm – 21lmn
Answer:
(i) 9a² – 6a = 3 × 3 × a × a – 2 × 3 × a
= 3 × a (3a – 2)
∴ 9a² – 6a = 3a (3a – 2)

ii) 15a³b – 35ab³
= 3 × 5 × a × a × a × b – 7 × 5 × a × b × b × b
= 5 × a × b [3 × a × a – 7 × b × b]
= 5ab [3a² – 7b²]

iii) 7lm – 21lmn
= 7 × l × m – 7 × 3 × m × n × l
= 7 × l × m [1 – 3n]
= 7lm [1 – 3n]

Question 4.
Factorise:
i) 5xy + 5x + 4y + 4
ii) 3ab + 3a + 2b + 2 [Pg. No. 271]
Answer:
i) 5xy + 5x + 4y + 4
= (5xy + 5x) + (4y + 4)
= 5x(y + 1) + 4(y + 1)
= (y + 1) (5x + 4)

ii) 3ab + 3a + 2b + 2
= [3 × a × b + 3 × a] + [2 × b + 2]
= 3 × a [b + 1] + 2 [b + 1]
= (b + 1) (3a + 2)

Think, Discuss and Write

While solving some problems containing algebraic expressions in different operations, some students solved as given below. Gan you identify the errors made by them? Write correct answers.     [Page No. 279]
Question 1.
Srilekha solved the given equation as shown below.
3x + 4x + x + 2x = 90
9x = 90 Therefore x = 10 What could say about the correctness of the solution?
Can you identify where Srilekha has gone wrong?
Answer:
Srilekha’s solution is wrong,
∵ 3x + 4x + x + 2x = 90
10x = 90
x = \(\frac{90}{10}\)
∴ x = 9

Question 2.
Abraham did the following.      [Page No. 280]
For x = -4, 7x = 7 – 4 = -3.
Answer:
Abraham’s solution is wrong.
∴ If x = -4
⇒ 7x = 7(-4) = -28

Question 3.
John and Reshma have done the multiplication of an algebraic expression by the following methods: verify whose multiplication is correct.      [Page No. 280]

Answer:

∴ John’s solutions are wrong and Reshma’s solutions are correct.

Question 4.
Harmeet does the division as (a + 5) ÷ 5 = a + 1 His friend Srikar done the same (a + 5) ÷ 5 = a/5 + 1 and his friend Rosy did it this way (a + 5) ÷ 5 = a Can you guess who has done it correctly? Justify!     [Page No. 280]
Answer:
The solutions of Harmeet, Rosy are wrong.
(a + 5) ÷ 5 = \(\frac{a+5}{5}\)
= \(\frac{a}{5}\) + \(\frac{5}{5}\)
= \(\left(\frac{a}{5}+1\right)\)
∴ Srikar had done it correctly.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions InText Questions and Answers.

8th Class Maths 11th Lesson Algebraic Expressions InText Questions and Answers

Do this

Question 1.
Find the number of terms in following algebraic expressions.
5xy², 5xy³ – 9x, 3xy + 4y – 8, 9x² + 2x + pq + q.    [Page No. 248]
Answer:

Question 2.
Take different values for x and find values of 3x + 5.     [Page No. 248]
Answer:
If x = 1 then 3x + 5 = 3(1) + 5 = 3 + 5 = 8
If x = 2 then 3x + 5 = 3(2) + 5 = 6 + 5 = 11
If x = 3 then 3x + 5 = 3(3) + 5 = 9 + 5 = 14

Question 3.
Find the like terms in the following: ax²y, 2x, 5y², -9x², -6x, 7xy, 18y².    [Pg. No. 249]
Answer:
Like terms are (2x, – 6x) (5y², 18y²).

Question 4.
Write 3 like terms for 5pq².     [Pg. No. 249]
Answer:
Like terms of 5pq² are – 3pq², pq², \(\frac{\mathrm{pq}^{2}}{2}\)etc.,

Question 5.
If A = 2y² + 3x – x², B = 3x² – y² and C = 5x² – 3xy then find          [Pg. No. 250]
(i) A + B (ii) A – B (iii) B + C (iv) B – C (v) A + B + C (vi) A + B – C
Answer:
A = 2y² + 3x – x², B = 3x² – y², C = 5x² – 3xy
i) A + B = (2y² + 3x – x²) + (3x² – y²)
= (2y² – y²) + 3x + (3x² – x²)
∴ A + B = y² + 3x + 2x² = 2x² + 3x + y²

ii) A – B = (2y² + 3x – x²) – (3x² – y²)
= 2y² + 3x – x² – 3x² + y²
∴ A – B = 3y² + 3x – 4x²

iii) B + C = (3x² – y²) + (5x² – 3xy)
= 3x² + 5x² – y² – 3xy
∴ B + C = 8x² – y² – 3xy

iv) B – C = (3x² – y²) – (5x² – 3xy)
= 3x² – y² – 5x² + 3xy
∴ B – C = – 2x² – y² + 3xy

v) A + B + C = A + (B + C)
= (2y² + 3x – x²) + (8x² – y² – 3xy)
= (8x² – x²) + (2y² – y²) + 3x – 3xy
∴ A + B + C = 7x² + y² + 3x – 3xy

vi) A + B – C = A + (B – C)
= (2y² + 3x – x²) + (-2x² – y² + 3xy)
= (2y² – y²) + (-x² – 2x²) + 3x + 3xy :
∴ A + B – C = y² – 3x² + 3x + 3xy

Question 6.
Complete the table:       [Page No. 253]

Answer:

Question 7.
Check whether you always get a monomial when two monomials are multiplied.        [Page No. 253]
Answer:
Yes, the product of two monomials is always a monomial.
Ex: 2xy × 5y = 10xy is a monomial.

Question 8.
Product of two monomials is a monomial? Check.     [Pg. No. 253]
Answer:
Yes, the product of two monomials is a monomial.
∵ 2x × y = 2xy

Question 9.
Find the product: (i) 3x(4ax + 8by) (ii) 4a²b(a – 3b) (iii) (p + Sq²) pq (iv) (m³ + n³) 5mn²       [Pg. No. 255]
Answer:
i) 3x (4ax + 8by) = 3x × 4ax + 3x × 8by
= 12ax² + 24bxy

ii) 4a²b (a – 3b) = 4a²b × a – 4a²b × 3b
= 4a³b – 12a²b²

iii) (p + 3q²) pq = p × pq + 3q² × pq
= p²q + 3pq³

iv) (m³ + n³) 5mn² = m³ × 5mn² + n³ × 5mn²
= 5m⁴n² + 5mn⁵

Question 10.
Find the number of maximum terms in the product of a monomial and a binomial?       [Pg. No. 255]
Answer:
The no.of terms in the product of a monomial and a binomial are two (2).

Question 11.
Find the product:       [Pg. No. 257]
(i) (a – b) (2a + 4b)
(ii) (3x + 2y) (3y – 4x)
(iii) (2m – l)(2l – m)
(iv) (k + 3m)(3m – k)
Answer:
i) (a – b) (2a + 4b) = a(2a + 4b) – b(2a + 4b)
= (a × 2a + a × 4b) – (b × 2a + b × 4b)
= 2a² + 4ab – (2ab + 4b²)
= 2a² + 4ab – 2ab – 4b²
= 2a² + 2ab – 4b²

ii) (3x + 2y) (3y – 4x) = 3x(3y – 4x) + 2y(3y – 4x)
= 9xy – 12x² + 6y² – 8xy
= xy – 12x² + 6y²

iii) (2m – l) (2l – m) = 2m(2l – m) – l(2l – m)
= 2m × 2l – 2m × m – l × 2l + l × m
= 4lm – 2m² – 2l² + lm
= 5lm – 2m² – 2l²

iv) (k + 3m) (3m – k) = k(3m – k) + 3m(3m – k)
= k × 3m – k × k + 3m × 3m – 3m × k
= 3km – k² + 9m² – 3km
= 9m² – k²

Question 12.
How many number of terms will be there in the product df two binomials?        [Page No. 257]
Answer:
No. of terms in the product of two binomials are 4.
Ex: (a + b) (c + d) = ac + ad + be + bd

Question 13.
Verify the following are identities by taking a, b, c as positive integers.    [Pg. No. 260]
(i) (a – b)² = a² – 2ab + b²
(ii) (a + b) (a – b) = a² – b²
(iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Answer:
i) (a – b)² = a² – 2ab + b²
a = 3, b = 1
⇒ (3 – 1)² = (3)² – 2 × 3 × 1 + 1²
⇒ (2)² = 9 – 6 + 1
⇒ 4 = 4
∴ (i) is an identity,

ii) (a + b) (a – b) = a² – b²
a = 2, b = 1
⇒ (2 + 1) (2 – 1) = (2)² – (1)²
⇒ 3 × 1 = 4 – 1
⇒ 3 = 3
∴ (ii) is an identity.

iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
a = 1, b = 2, c = 0
⇒ (1 + 2 + 0)² = 1² + 2² + 0² + 2 × 1 × 2 + 2 × 2 × 0 + 2 × 0 × 1
⇒ (3)² = 1 + 4 + 0 + 4 + 0 + 0
⇒ 9 = 1 + 4 + 4
⇒ 9 = 9
∴ (iii) is an identity.

Question 14.
Now take x = 2, a = 1 and b = 3, verify the identity (x + a) (x + b) s x + (a + b)x + ab.        [Pg. No. 260]
i) What do you observe? Is LHS = RHS?
ii) Take different values for x, a and b for verification of the above identity.
iii) Is it always LHS = RHS for all values of a and b?
Answer:
i) (x + a) (x + b) = x² + (a + b)x + ab
x = 2, a = 1, b = 3 then
⇒ (2 + 1) (2 + 3) = 2² + (1 + 3)² + 1 × 3
⇒ 3 × 5 = 4 + 4x² + 3
⇒ 15 = 4 + 8 + 3 ⇒ 15 = 15
∴ LHS = RHS

ii) x = 0, a = 1, b = 2 then
⇒ (0 + 1) (0 + 2) = 0² + (1 + 2) 0 + 1 × 2
⇒ 1 × 2 = 0 + 0 + 2
⇒ 2 = 2
∴ LHS = RHS for different values of x, a, b.

iii) LHS = RHS for all the values of a, b.

Question 15.
Consider (x + p) (x + q) = x + (p + q)x + pq.
(i) Put q instead of ‘p’ what do you observe?
(ii) Put p instead of ‘q’ what do you observe?
(iii) What identities you observed in your results?            [Pg. No. 261]
Answer:
i) (x + p) (x + q) = x² + (p + q)x + pq …… (1)
Substitute q instead of p in (1).
⇒ (x + q) (x + q) = x² + (q + q)x + q × q
⇒ (x + q)² = x² + 2qx + q²

ii) Substitute ‘p’ instead of q in (1).
⇒ (x + p) (x + p) = x² + (p + p)x + p × p
⇒ (x + p) = x² + 2px + p²

iii) ∴ I observe the following identities.
(x + q)² = x² + 2qx + q²
(x + p)² = x² + 2px + p²

Question 16.
Find: (i) (5m + 7n)²
(ii) (6kl + 7mn)²
(iii) (5a² + 6b²)²
(iv) 302²
(v) 807²
(vi) 704²
(vii) Verify the identity: (a – b)² = a² – 2ab + b², where a = 3m and b = 5n.         [Pg. No. 261]
Answer:
i) (5m + 7n)² is in the form of (a + b)².
(a + b)² = a² + 2ab + b² [a = 5m, b = 7n]
(5m + 7n)² = (5m)² + 2 × 5m × 7n + (7n)²
= (5m × 5m) + 70 mn + 7n × 7n
= 25m² + 70mn + 49n²

ii) (6kl + 7mn)²
We know that (a + b)² = a² + 2ab + b²
∴ (6kl + 7mn)² = (6kl)² + 2 × 6kl × 7mn + (7mn)²
= 36 k²l² + 84 klmn + 49 m²n²

iii) (5a² + 6b²)²
a = 5a², b = 6b²
(5a² + 6b²)² = (5a²)² + 2 × 5a² × 6b² + (6b²)²
= (5a² × 5a²) + 60a²b² + (6b² × 6b²)
= 25a⁴ + 60a²b² + 36b⁴

iv) (302)² = (300 + 2)²
a = 300, b = 2
∴ (300 + 2)² = (300)² + 2 × 300 × 2 + (2)²
= (300 × 300) + 1200 + (2 × 2)
= 90,000 + 1200 + 4
= 91,204

v) (807)² = (800 + 7)²
a = 800, b = 7
∴ (800 + 7)² = (800)² + 2 × 800 × 7 + (7)²
= (800 × 800) + 11,200 + (7 × 7)
= 6,40,000 + 11,200 + 49
= 6,51,249

vi) (704)² = (700 + 4)²
a = 700, b = 4
∴ (700 + 4)² = (700)² + 2 × 700 × 4 + 4²
= (700 × 700) + 5600 +(4 × 4)
= 4,90,000 + 5600 + 16
= 4,95,616

vii) (a – b)² = a² – 2ab + b² …… (1)
Substitute a = 3m, b = 5n in (1).
LHS = (3m – 5n)² = (3m)² – 2 × 3m × 5n + (5n)²
= 9m² – 30mn + 25n²
RHS = (3m)² – 2 × 3m × 5n + (5n)²
= 9m² – 30mn + 25n²
∴ LHS = RHS

Question 17.
Find:
(i)(9m – 2n)²
(ii) (6pq – 7rs)²
(iii) (5x² – 6y²)²
(iv) 292²
(v) 897²
(vi) 794²        [Pg. No. 262]
Answer:
i) (9m – 2n)² is in the form of (a – b)².
(a – b)² = a² – 2ab + b²
(9m – 2n)² = (9m)² – 2 × 9m × 2n + (2n)²
= (9m × 9m) – 36mn + (2n × 2n)
= 81m² – 36mn + 4n²

ii) (6pq – 7rs)²
a = 6pq, b = 7rs
(6pq – 7rs)² = (6pq)² – 2 × 6pq × 7rs + (7rs)²
= (6pq × 6pq) – 84pqrs + (7rs × 7rs)
= 36p²q² – 84pqrs + 49r²s²

iii) (5x² – 6y²)² = (5x²)² – 2 × 5x² × 6y² + (6y²)²
= (5x² × 5x²) – 60x²y² + (6y² × 6y²)
= 25x⁴ – 60x²y² + 36y⁴

iv) (292)² = (300 – 8)²
a = 300, b = 8
∴ (300 – 8)² = (300)² – 2 × 300 × 8 + (8)² = (300 × 300) – 4800 + (8 × 8)
= 90,000 – 4800 + 64
= 90,064 – 4800
= 85,264

v) (897)² = (900 – 3)²
= (900)² – 2 × 900 × 3 + (3)²
= 8,10,000 – 5400 + 9
= 8,10,009 – 5400
= 8,04,609

vi) (794)² = (800 – 6)²
= (800)² – 2 × 800 × 6 + (6)²
= 6,40,000 – 9600 + 36
= 6,40,036 – 9600
= 6,30,436

Question 18.
Find:
(i) (6m + 7n) (6m – 7n)
(ii) (5a + 10b) (5a – 10b)
(iii) (3x² + 4y²) (3x² – 4y²)
(iv) 106 × 94
(v) 592 × 608
(vi) 92² – 8²
(vi) 984² – 16²      [Pg. No. 262]
Answer:
i) (6m + 7n) (6m -,7n) is in the form of (a + b) (a – b). (a + b) (a – b) = a² – b²,
here a = 6m, b = 7n
(6m + 7n) (6m – 7n) = (6m)² – (7n)²
= 6m × 6m – 7n × 7n
= 36m² – 49n²

ii) (5a + 10b) (5a – 10b) = (5a)² – (10b)² [∵ (a + b) (a – b) = a² – b²]
= 5a × 5a – 10b × 10b
= 25a² – 100b²

iii) (3x² + 4y²) (3x² – 4y²)
= (3x²)² – (4y²)²
= 3x² × 3x² – 4y² × 4y²
= 9x⁴-16y⁴ [∵ (a + b) (a – b) = a² – b²]

iv) 106 × 94 = (100 + 6) (100 – 6)
= 100² – 6² = 100 × 100 – 6 × 6 [∵ (a + b) (a- b) = a²– b²]
= 10,000 – 36
= 9,964

v) 592 × 608 = (600 – 8) (600 + 8)
= (600)² – (8)²
= 600 × 600 – 8 × 8
= 3,60,000 – 64
= 3,59,936

vi) 92² – 8² is in the form of a² – b² = (a + b) (a – b).
92² – 8² = (92 + 8)(92 – 8)
= 100 × 84
= 8400

vii) 9842 – 162 = (984 + 16) (984 – 16)
= (1000) (968) [∵ (a + b)(a – b) = a² – b²]
= 9,68,000

Try These

Question 1.
Write an algebraic expression using speed and time; simple interest to be paid, using principal and the rate of simple interest.    [Pg. No. 251]
Answer:
Distance = speed × time
d = s × t

Question 2.
Can you think of two more such situations, where we can express in algebraic expressions?     [Pg. No. 251]
Answer:
Algebraic expressions are used in the following situations:
i) Area of a triangle = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) bh
ii) Perimeter of a rectangle = 2(length + breadth) = 2(l + b)

Think, Discuss and Write

Question 1.
Sheela says the sum of 2pq and 4pq is 8p²q² is she right? Give your explanation.      [Pg. No. 249]
Answer:
The sum of 2pq and 4pq = 2pq + 4pq = 6pq
According to Sheela’s solution it is 8p²q².
6pq ≠ 8p²q²
Sheela’s solution is wrong.

Question 2.
Rehman added 4x and 7y and got 1 lxy. Do you agree with Rehman?     [Pg. No. 249]
Answer:
The sum of 4x and 7y
= (4x) + (7y)
= 4x + 7y ≠ 11xy
I do not agree with Rehman’s solution.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions InText Questions and Answers.

8th Class Maths 10th Lesson Direct and Inverse Proportions InText Questions and Answers

Do this

Question 1.
Write five more such situations where change in one quantity leads to change in another quantity.     [Page No. 231]
Answer:
The change in one quantity leads to change in another quantity will see in the following situations.

1. If speed increases then time decreases.
2. In a family, the number of persons are increased then their consumption will also increases.
3. If water consumption increases then water levels decreases.
4. If the capacity of worker’s increases then time decreases.
5. If thickness of a wire increases then its resistance decreases.

Question 2.
Write three situations where you see direct proportion.      [Page No. 233]
Answer:

1. The relation between number of students to number of teachers.
2. Number of buffaloes to their consumption of grass.
3. Number of workers to length of wall.

Question 3.
Let us consider different squares of sides 2, 3, 4 and 5 cm. Find the areas of the squares and fill the table.

What do you observe? Do you find any change in the area of the square with a change in its side? Further, find the ratio between the area of a square to the length of its side. Is the ratio same? Obviously not.
∴ This variation is not a direct proportion.     [Page No. 233]
Answer:

From the above table the ratios are not equal.
∴ So the change is not in direct proportion.
If the measure of side of a square will be change then its area also be changed.

Question 4.
The following are rectangles of equal breadth on a graph paper. Find the area for each rectangle and fill in the table.


Is the area directly proportional to length?      [Page No. 233]

Answer:
Yes, the area is directly proportional to its length.

Question 5.
Take a graph paper make same rectangles of same length and different width. Find the area for each. What can you conclude about the breadth and area?        [Page No. 233]
Answer:

Area of first rectangle (A₁) = 3 × 1 = 3 sq. cm.
Area of second rectangle (A₂) = 3 × 2 = 6 sq. cm.
∴ The relation between the areas of rectangle and breadths is in direct proportion.
[∵ \(\frac{1}{3}\) = \(\frac{2}{6}\)]

Question 6.
Measure the distance in the given map and using that calculate actual distance between (i) Vijayawada and Visakhapatnam, (ii) Tirupati and Warangal. (Scale is given)        [Page No. 235]

Answer:
i) The distance between Vijayawada and Visakhapatnam = 2 cm
According to the sum
1 cm = 300 km then 2 cm = ?
1 …… 300
2 …… ? (x)
⇒ x = 2 × 300 = 600 km
The distance between the above two cities is 600 km.
ii) The distance between Tirupathi and Warangal = 3 cm
But given that 1 cm = 300 km
3 cm = ? (x)
x = 3 × 300 = 900 km
∴ The distance between Tirupathi and Warangal = 900 km.

Question 7.
Write three situations where you see inverse proportion.      [Page No. 238]
Answer:
i) Time – work capacity
ii) Speed – distance
iii) Time – speed

Question 8.
To make rectangles of different dimensions on a squared paper using 12 adjacent squares. Calculate length and breadth of each of the rectangles so formed. Note down the values in the following table.

What do you observe? As length increases, breadth decreases and vice-versa (for constant area).
Are length and breadth inversely proportional to each other?      [Page No. 238]
Answer:
In a rectangle if length is increases then breadth is decreases and vise-versa.
∴ Length and breadth of a rectangle are in inverse proportion.

Think, discuss and write

Question 1.
Can we say that every variation is a proportion.
A book consists of 100 pages. How do the number of pages read and the number of pages left over in the book vary?

What happened to the number of left over pages, when completed pages are gradually increasing? Are they vary inversely? Explain.         [Page No. 239]
Answer:
In every situation number of pages read and number of pages left over in the book are in inverse proportion.
If number of pages read are increases then number of pages left are decreases.

AP State Syllabus 8th Class Maths Solutions 9th Lesson Area of Plane Figures InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures InText Questions and Answers.

8th Class Maths 9th Lesson Area of Plane Figures InText Questions and Answers

Do this

Question 1.
Find the area of the following figures:     [Page No. 200]
i)

Answer:
Area of a parallelogram = b × h = 7 × 4 = 28 sq.cm.

ii)

Answer:
Area of a triangle = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 7 × 4
= 14 sq.cm.

iii)

Answer:
Area of a triangle = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 5 × 4
= 10 sq.cm.

iv)

Answer:
Area of rhombus = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × (4+4) × (3+3)
[∴ d₁ = 4 + 4 = 8, d₂ = 3 + 3 = 6]
= \(\frac{1}{2}\) × 8 × 6
= 24 cm²

v)

Answer:
Area of a rectangle = l × b
= 20 × 14 = 280 sq.cm

vi)

Answer:
Area of a square = s²
= s × s
= 5 × 5 = 25 cm²

Question 2.
The measurements of some plane figures are given in the table below. However, they are incomplete. Find the missing information.     [Page No. 200]
Answer:

Question 3.
Find the area of the following trapezium.      [Page No. 204]
fig (i)

Answer:
Area of a trapezium

fig (ii)

Answer:
Area of a trapezium

Question 4.
Area of a trapezium is 16 cm². Length of one parallel side is 5 cm and distance between two parallel sides is 4 cm. Find the length of the other parallel side. Try to draw this trapezium on a graph paper and check the area.
[Page No. 204]
Answer:
Given that
Area of a trapezium = 16 sq.cm
Length of one of the parallel sides is a = 5 cm; h = 4 cm
Length of 2nd parallel side (b) = ?
A = \(\frac{1}{2}\)h(a + b)

Graph Sheet:

Area of parallelogram ABCD = 12 sq.cm + (S + P) + (Q + R) + (W + T) + (V + U)
= 12 + 1 + 1 + 1 + 1
= 12 + 4
= 16 sq.cm

Question 5.
ABCD is a parallelogram whose area is 100 sq.cm. P is any point insile the parallelogram (see fig.) find tie area of △APB + △CPD.       [Page No. 204]

Answer:
Area of parallelogram ABCD = 100 sq.cm
From the given figure,
ar (△APB) + ar (△CPD) = ar (△PD) + ar (△BPC)

Question 6.
The following details are noted in meters in the field book of a surveyor. Find the area of the fields.     [Page No. 213]
i)

Answer:

From the above figure
i) A, B, C, D, E are the vertices of pentagonal field,
ii) AD is the diagonal.
iii) Now the area of the field = Areas of 4 triangles and a trapezium.
PQ = AQ – AP = 50 – 30 = 20
QD = AD – AQ = 140 – 50 = 90
RD = AD – AR = 140 – 80 = 60
Area of △APB:

Area of trapezium PBCQ:

Area of △QCD:

Area of △DER:

Area of △ERA:

∴ Area of the field = ar △APB + ar trapezium PBCQ + ar △QCD + ar △DER + ar △ERA
= 450 + 800 + 2250 + 1500 + 2000 = 7000 sq. units
ii)

Answer:

From the above figure
i) A, B, C, D, E are the vertices of a pentagonal field.
ii) AC is the diagonal.
iii) The area of a field is equal to areas of 4 triangles and a trapezium.
QC = AC – AQ = 160 – 90 = 70
RC = AC – AR = 160 – 130 = 30
PR = AR – AP = 130 – 60 = 70
Area of △AQB:

Area of △QBC :

Area of △DRC :

Area of trapezium EPRD:

Area of △EPA :

∴ Area of the field = ar △AQB + ar △QBC + ar △DRC + ar trapezium EPRD + ar △EPA
= 2700 + 2100 + 450 + 2450 + 1200 = 8900 sq. units

Try these

Question 1.
We know that parallelogram is also a quadrilateral. Let us split such a quadrilateral into two triangles. Find their areas and subsequently that of the parallelogram. Does this process in turn with the formula that you already know?   [Page No. 209]

Answer:
Area of a parallelogram ABCD

Area of parallelogram ABCD
= base x height
= bh sq. units
(OR)

Area of parallelogram ABCD
= ar △ABC + ar △ACD
= \(\frac{1}{2}\) BC × h₁ + \(\frac{1}{2}\) AD × h₂
= \(\frac{1}{2}\) bh + \(\frac{1}{2}\) bh [∵ h₁ = h₂]
= bh sq. units.
∴ This process in turn with already known formula.

Question 2.
Find the area of following quadrilaterals.      [Page No. 213]
i)

Answer:
d = 6 cm, h₁ = 3 cm, h₂ = 5 cm
Area of a quadrilateral
= \(\frac{1}{2}\)d(h₁ + h₂)
= \(\frac{1}{2}\) × 6 (3 + 5) = 3(8) = 24 cm²

ii)

Answer:
d₁ = 7 cm; d₂ = 6 cm
Area of a rhombus A = \(\frac{1}{2}\) d₁d₂
= \(\frac{1}{2}\) × 7 × 6
= 7 × 3 = 21 cm²

iii)

Answer:
Area of a parallelogram (A) = bh
(∵ The given fig. is a parallelogram in which two opposite sides are parallel)
Area of a parallelogram = 2 ar AADC
= 2 × \(\frac{1}{2}\) × 8 × 2 = 16 Sq. cm.
[∵ Area of a parallelogram = ar △ADC + ar △ABC. But ar △ABC = ar △ADC]

Question 3.
i) Divide the following polygon into parts (triangles and trapezium) to find out its area.     [Page No. 214]

Answer:
FI is a diagonal of polygon EFGHI.
If perpendiculars GA, HB are drawn on the diagonal FI, then the given figure pentagon is divided into 4 parts.
∴ Area of a pentagon EFGHI = ar △AFG + ar AGHB + ar △BHI + ar △EFI.

NQ is a diagonal of polygon MNOPQR. Here the polygon is divided into two parts.
∴ Area of a hexagon MNOPQR = ar NOPQ + ar MNQR.

ii) Polygon ABCDE is divided into parts as shown in the figure. Find the area.     [Page No. 215].

If AD = 8 cm, AH = 6 cm, AF = 3 cm and perpendiculars BF = 2 cm, GH = 3 cm and EG = 2.5 cm.
Answer:
Area of polygon ABCDE = ar △AFB + ar FBCH + ar △HCD + ar △AED

So, the area of polygon ABCDE = 3 + 7.5 + 3 + 10 = 23.5 sq.cm

iii) Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm.   [Page No. 215].

NA, OD, QC and RB are perpendiculars to diagonal MP.
Answer:
Area of MNOPQR
= ar △MAN + ar ADON + ar △DOP + ar △CQP + ar BCQR + ar △MBR
Hence CP = MP – MC = 9 – 6 = 3 cm
BC = MC – MB = 6 – 4 = 2 cm
AB = MB – MA = 4 – 2 = 2 cm
DP = MP – MD = 9 – 7 = 2 cm
AD = MD – MA = 7 – 2 = 5 cm

= 2.5 + (2.5 × 5.5) + 3 + 3 + 4.5 + (2 × 2.5)
= 2.5 + 13.75 + 3 + 3 + 4.5 + 5
= 31.75 sq.cms

Think, discuss and write

Question 1.
A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?    [Page No. 213]
Answer:

No, we cannot divide a trapezium into two congruent triangles.
∵ From the adjacent figure,
△ABC ≆ △ADC

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.6 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.6

Question 1.
Find the sum of integers which are divisible by 5 from 1 to 100.
Solution:
Numbers which are divisible by 5 from 1 to 100 are 5, 10, 15, …………………95, 100.
∴ Sum of the above numbers = 5+10 + ……………..+ 95 + 100
= 5[1 + 2 + ………………….+ 20]
= 5 [ \(\frac{20 \times(20+1)}{2}\) ]
= \(\frac{5 \times 20 \times 21}{2}\) [∵ Sum of ‘n’ natural numbers = \(\frac{n(n+1)}{2}\) & n = 20 ]
= 1050

Question 2.
Find the sum of integers which are divisible by 2 from 11 to 50.
Solution:
. Numbers which are divisible by 2 from 11 to 50 are 12, 14,48, 50.
Sum of the numbers = 12 + 14 + ……….. + 48 + 50 ‘
= (2 + 4 + ……….. + 50) – (2 + 4 + ……….. + 10)
= 2(1 + 2 +……….. + 25) – 2 (1 + 2 + ……….. + 5)

= 25 × 26 – 5 × 6
= 650 – 30
= 620

Question 3.
Find the sum of integers which are divisible by 2 and 3 from 1 to 50.
Solution:
Numbers which are divisible by 2 and 3 i.-e., which are divisible by 6 from 1 to 50 are 6,12 …………….48.
Sum of the numbers = 6 + 12 + ……..+ 48
= 6(1 + 2 +……… + 8)
= 6 \(\left[\frac{8(8+1)}{2}\right]\)
= 3 × 8 × 9 = 216

Question 4.
(n³ – n) is divisible by 3. Explain the reason.
Solution:

∴ If n = 4, (n³ – n) is divisible by 3.
∴ (n³ – n) is divisible by all the values of n.
Method 2:
n³ – n = n(n² – 1)
= n(n + 1)(n – 1)
∴ (n³ – n) is divisible by ‘3’ for all the values of n.
[∵ (n – 1), n, (n + 1) are three consecutive odd numbers]

Question 5.
Sum of ‘n’ odd number of consecutive numbers is divisible by ‘n’. Explain the reason.
Solution:
Sum of n’ consecutive odd numbers = n²
Since n is a factor of n², It Is divisible by ‘n’.

Question 6.
Is 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ divisible by 5? Explain.
Solution:
Sum of units digit of number 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹
= 1 + 8 + 7 + 4
= 20 → \(\frac{20}{5}\)(R = 0)
∴ 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ is divisible by 5.

Question 7.

Find the number of rectangles of the given figure?
Solution:

∴ No.of rectangles in the given figure = 1 + 2 + 3 + 4 + 5 + 6 = 21

Question 8.
Rahul’s father wants to deposit sorne amount of money every year on the day of Rahul’s birthday. On his 1st birth day Rs.100, on his 2nd birth day Rs.300, on his 3 birth day Rs.600, on his 4th birthday Rs. 1000 and so on. What is the amount deposited by his father on Rahul’s 15th birthday.
Solution:

Rahul’s father deposits on every year 200, 300, 400 more than before year.
Then he deposits ₹ 10,500 on 14th birthday.
∴ The amount deposits on 15th birthday
= 10,500 + 1,500
= ₹ 12,000/-

Question 9.
Find the sum of integers from 1 to 100 which are divisible by 2 or 5.
Solution:
Sum of the numbers which are divisible by 2 from 1 to 100
= 2 + 4 + ……….. + 100
= 2(1 + 2 + ………… +50)
= 2 × \(\frac{50 \times(50+1)}{2} \)
= 50 × 51 = 2550
Sum of the numbers which are dMsible by 5froin I to 100
= 5 + 10 + ……….. + 100
= 5(1 + 2 +……….. +20)
= 5 × \(\frac{20 \times(20+1)}{2}\)
=5 × 10 × 21
=1050

Sum of the numbers which are.divisible by both 2 and 5 = 2550 + 1050 =3600
∴ Sum ol the numbers which are divisible by 2 or 5 from 1 to 100
= 10 + 20 + ………..+ 100 ( L.C.M of 2, 5 is 10)
=10(1 + 2 + ………..+ 10)
= 10 × \(\frac{10 \times(10+1)}{2}\)
= 5 × 10 × 11 .
= 550
∴ The sum of required numbers 3600—550 3050

Question 10.
Find the sum of integers from 11 to 1000 which are divisible by 3.
Solution:
Sum ol the numbers which are divisible by 3 from lito 1000
= 12 + 15+ ……….. +099
= 3(4 + 5 + ……….. +333)
= 3(1 + 2 + ……….. + 333) – 3(1 + 2+3)

= 999 × 167 – 9 × 2
= 166833 – 18
= 166815