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AP State Syllabus AP Board 8th Class Biology Solutions Chapter 9 Production and Management of Food From Animals Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 9th Lesson Production and Management of Food From Animals

8th Class Biology 9th Lesson Production and Management of Food From Animals Textbook Questions and Answers

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Question 1.
One honey bee hive consists of different types of bees. What are they? How they differ from each other?
Answer:

1. Honey bee species are social insects, which lives in colonies.
2. A honey bee colony consists of three types of bees.
3. One queen, several thousands of workers and few hundreds of drones.
4. There is only one queen bee in a colony. The primary function of a queen is to lay eggs (800 – 1200 eggs per day). The life span of queen is two to three years.
5. A worker has 5-6 weeks and the drone has 57 days of life span.
6. There are sterile female which are called workers in the hive. These bees attend to indoor duties during first three weeks during first three weeks of their lives such as secretion of royal jelly, feeding of the brood.
7. After three weeks they attend outdoor duties like collecting nector, pollen and water.
8. Drones are the male members of the colony. They are very lazy and unable to gather food. Their main duty is participating in mating, after this they die.

Question 2.
Make a list of characters of local variety buffaloes, which gives good quantity of milk in your village.
Answer:

1. Buffaloes are most common in India and other Asian countries.
2. In general, buffaloes give more milk than cows and buffaloes milk has more fat than the cow’s milk.
3. Buffaloes are more resistant to diseases than cows.
4. There are about seven different breeds of buffaloes, Murrah, Bhadwari, Jaffrabadi, Surthi, Mehsana, Nagpri and Nili Ravi in our country.
5. Each breed of buffalo is found in specific regions of the country and they differ in their body colour, shape and length of horns and in the shape and size of their forehead.
6. Of these different breeds of buffaloes, Murrah breed of buffaloes have been recognised by Government of India, as the best native milk yielding breed.
7. It yields about 8 litres of milk/day/animal and about 1800 – 2200 litres per year.

Question 3.
Explain the process of hatching eggs under broody hen in rural areas.
Answer:

1. Hatching of eggs is an interesting process.
2. The common village hen has less care and attention.
3. So it’s productivity is low.
4. It has the capacity to lay 15 to 20 eggs an egg per day respectively.
5. After that the hen becomes a broody state (wish to hatch the egg).
6. In this process the people of the house arrange the eggs in such a way that the broody hen can sit comfortably on the eggs to hatch.
7. The hen sits on the eggs stretching its wings to provide heat to the eggs, to develop chicks inside the eggs.
8. It takes 21 days to the chicks emerges out by breaking the egg shell. This process is called incubation.
9. Almost all the eggs hatch into chicken but some eggs which do not get sufficient heat from the broody hen become rotten.

Question 4.
Write about the accessory products produced in animal husbandry.
Answer:
Milk, meat, eggs, wool are the accessory products of animal husbandry.

Question 5.
What is estuaries, how they are suitable for both marine and river fish to live?
Answer:

1. Estuary is the place where river joins the sea usually called mouth of the river. In this area fresh water (with low salt content from the river) is mixed up with sea water (with high salt content).
2. When the wave action is strong or during high tide, large amount of sea water mixes with river water. In fact sea flows deep into the river.
3. During low tide and when the wave action is weak, less amount of sea water mixes with river water.
4. Because of this, salt content of the water in the estuary changes very rapidly.
5. Animals living in the estuary should tolerate and adopt to rapid changes in the salt content of water.

Question 6.
If you have a chance to visit milk chilling centre, what doubts would you like to clarify? Please list out them.
Answer:

1. Chilling of milk means rapid cooling of raw milk to sufficiently to low temperature. So that the growth of microorganisms present in the milk is checked.
   Doubts to clarify
2. What would happen if the temperature of the milk should be reduced to less than 10°C preferably 3-4°C, in the chilling process?
3. Whether the growth of the microorganisms will be controlled as the required nutrients and the growth conditions will be favourable to them?
4. If the growth of the organisms will not be checked what would happen to the milk. Any changes would happen? Is it fit for consumption?
5. If milk has to be transported to longer distances, considerable time is involved between production and heating process.
6. During this period milk should be protected from spoilage by the action of micro-organisms the chilling process therefore considered is necessary.

Question 7.
Collect news from news papers about milk production and impurities in milk.
Prepare a note and display it on wall magazine.
Answer:
Milk Production:

1. Our government treats producing milk as an industry.
2. Generally farmers rear 1 to 5 cattle in small scale at their homes to produce milk.
3. Among cows, traditional species give 2-5 liters of milk in a day. Murrah species are reared in most of the districts in our state.
4. They give up to 8 liters of milk per day. Haryana, Jaferabad, Nagapuri are the traditional variety of cows which give good quantity of milk.
5. Jersy (England) and Holstein (Denmark) are the foreign varieties. They give 25 liters of milk per day.
6. These foreign varieties are cross breed with our traditional local varieties. They give 8-20 liters milk per day.
7. Cows play vital role in total milk production of our country.
8. Out of milk produced in our country 60% is used to prepare cheese, cova, ghee, curd, milk powder and other milk products.
9. The milk produced in dairy forms is collected from house holds and pasteurized.

Impurities in Milk

1. Addition of pure or impure water.
2. Addition of colouring matter.
3. Addition of preservatives like Sodium bicarbonate, Borax or Boric Acid, Salicylic Acid and formaldehyde.
4. Addition of substances used for thickening after dilution with water, e.g. Flour, arrowroot, chalk, carbonate of magnesium.
5. Sugar is added to raise the specific gravity of diluted skimmed milk.
6. Milk is contaminated by
   a) improper or poisonous food eaten by the animal.
   b) Poor condition of animal due to nursing.
   c) Contamination of disease germs from the cub.
   d) Absorption of bad odours.
   e) If milk has been diluted it becomes pale and blueish, so milk and cream are artificially coloured with anilines or other pigments. Annotto is the common dye used to milk cream and butter.

Question 8.
Collect information about sea weeds, sea kelp from your school library and write a note with examples.
Answer:
Sea Weeds:
Sea weeds constitute an important marine resource and are found along the Rocky intertidal and subtidal regions of the coasts of India. The Sundarbans, the Chilka lake, the deltas of Godavari and Krishna. Gulf of Mannar. Palk bay Gujarath coast and around Lakshadweep, Andaman and Nikobar Islands are areas rich in sea weeds. They are used for human consumption, as cattle and poultry feed, as manure and for industrial purposes as the source phyco colloids like Agar-agar.

Sea Kelps:
Kelps are the largest sea weeds belonging to the brown algae (phaeophyceae) in the order laminariales.
Kelps grow in under water in shallow oceans, it has a high rate of growth and it’s decay is quite efficient in yielding methane as well as sugars that can be converted into ethanol.
It has been proposed that large open ocean, kelp farms could serve as a source as renewable source of energy.
Unlike some biofuels, such as corn, ethanol, Kelp energy avoids “food verses fuel” issues and does not require irrigation.

Question 9.
Observe nearby poultry farm and findout how do they export eggs to market? What material is used for transportation?
Answer:

1. The most important component in the marketing of eggs is to handle them with care at the time of collection, transportation and in the sale counter.
2. To avoid breakage of eggs special designs are planned in the poultry forms. They are packed in tailor made egg trays which can be piled up one over the other after inserting eggs in each tray.
3. These trays protect the eggs from both vertical and horizontal friction and avoid breakage during transportation.
4. The design of the transport vehicles for the chicken are also specially designed cages to allow the requisite freedom to the birds.

Question 10.
Observe a dry honey bee hive and how the bees built it. Draw a picture. How does it look like?
Answer:

Honey Bee hive is a construction of six sided (hexagon) wax compartments made by bees to store honey and eggs.
Wax is produced by the honey bees is known as bee’s wax or honey wax. Wax is used in the production of cosmetics, shoe polish, candles and leather industry.

Question 11.
Agriculture and animal husbandry are both sides of the same coin. How can you justify this?
Answer:

1. Farmers adopt different methods of management for getting better yields in agriculture. In the same way, care is required in the management of reading animals also.
2. Since long ago, man used animals not only for obtaining food but also for agriculture, transportation.
3. People living in rural areas used to domesticate animals like cows, buffaloes, bullocks, goats, sheep, pigs, hens etc., supplying of nutritious food accommodating clear and hygienic, shelters for animals is very important issue in animal husbandry.
4. Generally villages send their cattle to rear in their fields. Where grass is easily avail-able. These animals provide him with food (milk, meat, eggs etc) and clothing. Man in turn protected, fed and took care of these domestic animals.
5. The selective relationship between man and the domestic animals has become in-dispensable – man cannot live without them and these animals cannot survive in wild.
6. Thus, the relationship is mutually dependent and beneficial.

Question 12.
How do you appreciate the uses of cattle?
Answer:

1. Cattle are not only for our food, even its excretions like dung and urine also we make them use. Cow dung is used as a cooking fuel, sanitizing cleanser, construction material, insulation and water proofing for walls and floors in rural houses, a cultural symbol in religious worship, the raw material for producing organic compost and generating electricity.
2. The urine of cows is considered an elixir of life and is used as a natural remedy for liver and heart conditions as well as for enhancing mental and physical strength and increasing longevity.
3. The utilization of cow dung and urine is a perfect example of sustainable living. An understanding of the use of cow dung and urine by the rural Indian population can illustrate the indigenous knowledge associated with these materials and alternative sources of materials for electricity generation as well as cost-effective and environmentally friendly fuel and medicines.
4. Even the ash formed from burning of dung can be used as a cleaning agent for household utensils or used as a fertilizer.
5. Cow manure contains several plant nutrients including nitrogen, potassium, calcium and magnesium. The composition of cow manure makes it ideal for several uses, including fuel, fertilizer and medicine.

Question 13.
What makes you amazing in division of work in Honey bee colony? Support your answer.
Answer:

1. Honey bees lives in colonies. It consists of three types of bees.
2. One queen bee, several thousands of worker bees and few hundreds of drones or male bees.
3. The primary function of a queen bee is to lay 800 – 1200 eggs per day.
4. The worker bees are the sterile female bees. They attend to indoor duties like secretion of royal jelly, feeding of the brood during first three weeks.
   After three weeks they attend outdoor duties like collecting nector, pollen and water.
5. Drones are the male members of the colony. They are lazy. Their main duty is participating in mating. After this they die.
6. Thus the division of work is amazing in honey bee colony.

Question 14.
Conversion of agricultural lands into fish ponds leads to food crisis and environmental pollution. Write your opinion to conduct in debate on this issue.
Answer:

1. Fish culture is not new to our country. Even today in several villages, fishes are grown in small ponds in the backyard of the houses. Besides this, growing of fishes in rice fields is also more common in villages. In fact fish production is considered as second crop for farmers, the primary crop being rice.
2. Fish culture is sometimes practised in combination with a rice crop, so that fish are grown in the water in the paddy field.
3. Growing fish in paddy field is also multi-utilitarian practice. The reason for this are the increasing use of inorganic fertilizers and insecticides in paddy fields which cause deleterious effects on fish and predation for birds, snakes etc.
4. Cultivating fish in paddy fields lower diseases like stem borers on paddy.
5. If the agricultural fields are converted to fish pond, food grains would not be sufficient for the growing population. This leads to food crisis.
6. But ours is a agricultural country. So agriculture will be done in the crop seasons, and in the unseason they can convert to fish pond.
7. The places where it is near to the water sources farmers can utilize these places for growing fish.

8th Class Biology 9th Lesson Production and Management of Food From Animals InText Questions and Answers

Question 1.
What are the food items that are obtained from animals?
Answer:
Milk, meat, eggs, honey.

Question 2.
Do we get our food only from domesticated animals? List out the food that is obtained from animals.
Answer:
We domesticated only such animals which were helpful to us. Buffalo, Cow etc. are reared for milk. Hens, goats, sheep for meat, ducks for eggs.

Question 3.
Do you know the period from which wild animals were being tamed?
Answer:

Question 4.
Why did early man domesticate only some of the animals?
Answer:
Domestic animals provide him food, (milk, meat, eggs), clothing (skin of the animals). The early man realised the capabilities of these animals, tamed and domesticated them to help him in his daily activities.

Question 5.
Why he did not domesticated animals like elephant, tiger, lion etc. or birds like eagle and owl?
Answer:
They are the wild animals and it is difficult to tame them. There is no use of these animals for his daily activities.

Question 6.
Do all the persons who won agriculture fields also rear cattle?
Answer:
All most all the persons who won agriculture fields, rear cattle also.

Question 7.
Is there any relation between agriculture and cattle rearing or animal husbandry?
Answer:
Cattle provide him food, help in agriculture and transport, man provide food for the animals from the agricultural fields.
Collect the following information from your calls.

Question 8.
Number of families in agriculture
Answer:
Less families

Question 9.
Number of families in agriculture along with animal husbandry.
Answer:
Almost all families.

Question 10.
Number of families in Animal husbandry alone.
Answer:
Less families.

LET US DO Cattle Rearing:

Question 11.
Form a group with four or five students in your class. Discuss about the reasons. Why does a farmers rear cattle?
Answer:
Farmers believe that animal husbandry is part and parcel of agriculture.

Question 12.
Where do people rear their cattle in your village?
Answer:
Generally people send their cattle to rear at the places where grass is easily available.

Question 13.
What are the cattle here?
Answer:
Cows, buffaloes, goats, sheep.

Question 14.
At which places fodder is available?
Answer:
Fodder is available in fields and open places.

Question 15.
What are places where water is available?
Answer:
Canals, ponds, wells etc.

Question 16.
Are there any differences between rearing of cows, buffaloes, goats and sheep?
Answer:
All the animals are herbivorous animals. Farmers use bullocks in agricultural practices like ploughing.

Question 17.
What are the major problems that cattle rearers generally face?
Answer:
Cattle sheds become unclean because of the remains of fodder, dung and urine. Care should be taken to prevent the growth of lice and mytes on cattle’s body.

Question 18.
Make a list of agricultural practices by using bullocks and the buffaloes.
Answer:
Ploughing, to draw the leveller, transporting agricultural goods etc.

Question 19.
Think in which way this practice is helpful to the farmer as well as field crops?
Answer:
Sheep and goats provide him meat and wool, the dung and urine becomes good manure to the field crops.

Question 20.
Where is veterinary hospital located in your area?
Answer:
It is located at the place where it is convenience to the people to bring their animals for check.

Question 21.
Who are working there and what do they do?
Answer:
The employees working in veterinary hospital are a veterinary doctor or animal husbandary assistant, a compounder and attenders. They provide treatment and health care for the cattle.

Question 22.
Meet a nearby veterinary doctor or animal husbandry assistant. Collect infor¬mation about common diseases in cattle and prepare a note on them.
Answer:

1. Gali Kuntu is a common and dangerous disease in cows and buffaloes.
2. Sheep and goats suffer from worm infections (Nattala Vyadhi)
3. Growth of lice and mytes on cattle’s body.
4. Some parasitic diseases cause damage to liver intestine.
5. Viral and bacterial diseases also effect on milk production.

Milk Production:

Question 23.
From which animals we get maximum milk production?
Answer:
Cows, buffalows

Question 24.
In which areas people use camel milk?
Answer:
Desert areas.

Question 25.
Have you ever see people taking donkeys milk? Why was it preffered?
Answer:
Yes. Donkey milk which is Rich in immunoglobulin helps human body from many viral and bacterial infections. These are mostly found in Telangana regions especially in Adilabad. The secret behind the glowing skin of Egyptian princess Cleopatra’s is donkey milk.

Question 26.
What are the types of fodder generally farmers feed the cattle with?
Answer:

1. They supply fodder from their agricultural fields.
2. They also feed the cattle with hay, green and dry grass, oil seed cakes of ground nut.

Question 27.
How farmers preserve fodder for cattle after harvesting?
Answer:
After harvesting farmers preserve fodder by arranging into heaps. This heap will be used for the cattle throughout the year.

Pasteurization :

Question 28.
Is there a milk collecting centre in your village?
Answer:
Krishna Milk Union, Vijaya Dairy Milk centres are located in our village.

Question 29.
How do they collect milk and export?
Answer:
The milk produced in dairy farm is collected from households and cattle rearers and export the milk in chilling form (cooling condition).

Question 30.
Do you know how they decide cost of milk?
Answer:
The cost of milk is decided due to fat content.

Question 31.
Do you know in which month the rate of milk production is high? Why?
Answer:
September to November the milk production is high.

Question 32.
Why the milk production is higher during those months than remaining year ? Discuss with your friends and find out the reasons.
Answer:
Milk production is slightly higher in the November and December months. Because

1. In these months food availability is rich and intaking of good food increase the milk production.
2. the climate is cool and enough hot. This sunny weather increase the milk production.
3. water resources are well in these months. Availability of food and water leads to milk production.
4. mostly animals delivered in July and August months and give milk before the summer. So the availability of milk is higher in November and December months.

Selection Procedure:

Question 33.
What care should be taken while buying cattle for milk production?
Answer:
The following points should be kept in mind.

1. Select high milk producing varieties, either traditional or hybrid.
2. Observe 2 to 3 days for average milk production.
3. Number yielding size, health, eating fodder.
4. Consult veterinary doctor, official of Director of Animal husbandry.

Some of rural people are experts in identifying high producing varieties.

Question 34.
What occasions, They decorate their cattle in your village?
Answer:
Pongal, Onam etc.

Question 35.
Do they respond when called by names ? Do you have any such experience with your pets ?
Answer:
We see the pet dogs climbing our body, sitting beside us, moves it’s tail and licks our feet when we call.

Poultry:

Question 36.
Collect information about Biogas Productions from your school library or internet and write notes on Biogas.
Answer:

1. In recent years, an alternate and better method is used to obtain energy from cattle dung and from the excreta of other animals including man and some types of organic waste material from agriculture, homes and industries.
2. This is by anaerobic fermentation of waste, to produce a gas which can be used as fuel. As this gas is produced from biological wastes, this is called BIO GAS.
3. Biogas is a mixture of gases – methane, carbondioxide and small amounts of hydro-gen, nitrogen and hydrogen sulphide. About 200 cc of biogas gives about 900 K.Cal of energy.
4. Production of biogas occurs in three stages. In the first stage, aerobic bacteria are allowed to degrade the complex molecules into simple molecules.
5. In the second stage, the simple molecules are fermented anaerobically to produce organic acids mostly to acetic acid.
6. In the final stage methane producing bacteria act on the acetic acid under anaerobic condition produce methane. The gases produced are collected into specially constructed chambers and supplied to users.
7. Use of bio gas for domestic purposes (cooking) lighting of street lights etc.
8. When bio gas is burnt, it does not pollute the environment.
9. The left over material after the production of bio gas can be used as manure in agriculture.

Question 37.
Are the hens reared in the poultry is same as our traditional varieties reared by farmers in the village?
Answer:
Farmers rear cocks and hens in villages. Most of these are local varieties (Natukollu). Poultry farms are of two types. One is for production of eggs (layers) and other for meat (Broilers).

Question 38.
Think and discuss Is genetically modified food useful or not?
Answer:
Natural wild varieties grow fully in 5 to 6 years. But broilers grow fully in just 6 to 8 weeks. This happens due to genetic modifications in the hen. So genetically modified food is useful.

Question 39.
Do you know Chicken 65? Why is this called so?
Answer:
This preparation is prepared by A.M. Buhari, in South Indian food industry in Chennai, in the year 1965. So this is called Chicken-65.

Question 40.
Have you heard about cock fight during some festival seasons. Think and discuss in your class about this type of practices which show human cruelty towards animals.
Answer:

At the time of Pongal festival, in some places like India and Tamilnadu, the birds are equipped with either metal spurs or knives tightly tied to the legs in the area where the cock’s natural spur has been partially removed. In this cock fight both the cocks would be wounded, severely bleed, this leads to the death of the birds. By watching this cock fight show, people get enjoyment. This shows human cruelty towards animals.

Question 41.
Do you know how many days a hen spends to hatch it’s eggs?
Answer:
A hen spends 21 days to hatch its eggs.

Question 42.
Prepare a detailed note on hatching eggs by observing at your village. If you need, please draw pictures also.
Answer:

Hatching of eggs into chick is called as incubation period. The full incubation period for an egg is 21 days. During this time the hen sits on her eggs and maintains a temperature of 100°F, which is needed to ensure proper embryonic development.

Question 43.
Egg is a nutritious food. Collect information about various nutrients in egg and write a note on them in your notebook.
Answer:
The nutrients present in egg are is as follows:
Choline: Healthy cell membranes
Vitamin – B: Folate and Riboflavin – converts food we eat into energy. Folate also reduces homocysteine levels and prevent birth defects.
Vitamin – A: Vision and healthy skin.
Vitamin – E and Vitamin – C also present
Luteine and Zeaxanthin: Found in yellow pigment of yolk, prevent macular degeneration.
NECC:
If you want to be healthy person eat egg everyday. This is the slogan of National Egg Co-ordination Committee. Egg is chief nutritious food which is easily available for all.

APICULTURE:

Question 44.
In what way honey bees are helpful in pollination?
Answer:
Honey bees are helpful in pollination, there after growth of food grains.

Question 45.
Generally where do you find honey bee hives in your surroundings?
Answer:
Those plants which contain nectar and pollen liked by bees are called bee flora. We find honey bee hives on the fruit trees like citrus, apple, guava, tamarind, cultivated field crops like mustered, gingelly, wheat, cotton, sunflower, vegetable plants like beans, bendi, brinjal, Trees like acacia, neem, sal and bushes, shrubs.

Question 46.
In which seasons we find honey bee hives?
Answer:
In winter and summer seasons we find honey bee hives.

Question 47.
Collection of honey from hive is a careful activity. Write a note on how people collect honey from hives. What did they do for this ?
Answer:

1. People take risk to collect honey from hives.
2. They burn some plants or wood to make smoke. They cover their body with a thick blanket.
3. They put the smoking material into a container. They take it over to the hive.
4. Working slowly and carefully, they wait for the honey bees to leave the bee hive.
5. Carefully they cut the hive with a sharp knife.
   By pressing the hive they can collect honey.

Question 48.
Ask your parents/teacher how a bear hunts bee hives for honey?
Answer:

1. Before going for hibernation (winter sleep) the bear builts fat reserves in its body and prepares itself for the winter season.
2. The bear attracted towards the bee hive and honey.
3. When it finds bee hive it drives away the bees with its forelimbs.
4. It has the advantage that its body has thick fur and avoids bee sting.
5. It chews the bee hive along with some bees in the hive swallow honeys spits out the waste.
6. This is how a bear hunts bee hives for honey.

FISHERIES:

Question 49.
Write a list of fishes that are available in your surroundings. Just write local names only.
Answer:
Marrel (Korramenu), Katla (Jella), Katrana (Bochu), Rahu (Mosu), Seer (Vanjiram) are the local varieties.

Question 50.
Do you know how to catch fish in a pond?
Answer:
Fishing rod will be helpful to catch fish.

Question 51.
How to catch fish in a large scale?
Answer:
Nylan nets will be useful to catch fish in large scale.

Question 52.
Think what will happen if mechanized fishing continuous for a long run.
Answer:
If mechanized fishing continuous at last we find no fish (extinction)

Question 53.
Ask your teacher what are the uses of Oysters?
Answer:
Oysters are the marine molluscs which produce pears. Economically pearls are important.

Question 54.
Tuna is an important fish which is available in our marine area. Collect information about Tuna and in what way it is important?
(OR)
Poultry Emu culture /Fish forms / Apiculture. Visit any one of the above industries. Get the information from farmers and prepare a note on this.
Answer:

1. Poultry are domesticated birds kept by human for the eggs and meat.
2. Generally chickens, ducks, geese, turkeys, quail are cultivated in poultry.
3. Chicken poultry’s are very common in our areas.
4. Here two types of hen’s are cultivated. Layer’s for eggs, and broiler’s for meat.
5. Poultry’s are arranged in stair manner, which save the lot of place.
6. Modern feeds for poultry consists such as soyabean oil, mineral supplements and vitamins.
7. Daily they collect the eggs and parcel in egg cases to supply.
8. Broiler chickens are domesticated and specially meant for meat production.
9. Poultry become a major economic source in rural areas.
10. It provides work and earning for rural people.

Question 55.
What is blue revolution? What are its effects? Discuss in your classroom.
Answer:
In the recent times, the Department of Agriculture and Cooperation (Blue Revolution) in the country. These schemes include the development of fresh water aquaculture through Fish Farmers Development Agencies (FFDAs) with development of brackish water aquaculture through Brackish Water Fish Farmer’s Development Agencies (BFDAs’ A shrimp and fish culture project is being implemented with World Bank assistance for the development of Shrimp culture in the states of Andhra Pradesh, Odisa and West Bengal.

8th Class Biology 9th Lesson Production and Management of Food From Animals Activities

Activity – 1

Question 1.
Form a group of 5 or 6 students. Collect different types hens and find their characters. If you want to know more details, about you need to ask hen rearers or poultry farmers in your village. Do not forget to collect information about the feed and diseases, treatment by using local technology.
Answer:

1. Production and rearing of hens on a large scale is generally called poultry. So billion hens are reared world wide for eggs and chicken.
2. Farmers rear cocks and hens in villages. Most of these are local varieties (Natukollu). Natural country varieties are good for hatching purpose, Aseel, Kadaknath, Chittagang, Longshan, Bursa are the pure local varieties.
3. Aseel (Berisa Kodi) the Indian traditional variety is meant for fighting because of its pugnacity, high stamina and majestic gait.
4. Feeding: The balanced feed must have nutrients, proteins, carbohydrates, fats, minerals, vitamins and water in sufficient quantities. Minerals like calcium, phosphorus, iodized salt, manganese and zinc are useful for better yield of eggs and meat.
5. Diseases: Poultry chicken are known to suffer from bacterial, fungal and viral diseases. Fowl cholera, salmonellasis and coryza are the common bacterial diseases. Fowl pox and Rani kher are the dreaded viral diseases.
   Fowl mite, chicken mite, fleas, licks, lice etc., are known to be present on the external surface of the poultry chicken and cause diseases and are therefore called as external parasites.
   Round worm, tape worm and coccidiosis are categorized as internal parasites.
6. Prevention and control: Prevention is always better than cure. For most of the diseases vaccines are now available. Newly born chicken should be vaccinated. A wide range of antibiotics, particularly broad spectrum sulfa drugs are widely used for treatment for poultry diseases. When a bird is identified to be suffering from a disease, it should be immediately to be isolated.

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 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

Improve Your Learning

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:

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.

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:

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:

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)

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.

SCERT AP Board 7th Class Telugu Solutions 2nd Lesson మాయాకంబళి Textbook Questions and Answers.

AP State Syllabus 7th Class Telugu 2nd Lesson Questions and Answers మాయాకంబళి

7th Class Telugu 2nd Lesson మాయాకంబళి Textbook Questions and Answers

వినడం – ఆలోచించి మాట్లాడడం

ప్రశ్న 1.
చిత్రం ద్వారా మీరేమి గమనించారో చెప్పండి.
జవాబు:
చిత్రంలోని మామ్మగారు తన వేళ్ళతో తమాషా చేసి, పిల్లలకు చూపిస్తూ కథను చెబుతున్నారు. పిల్లలు చాలా ఆశ్చర్యంగా చూస్తూ వింటున్నారు.

ప్రశ్న 2.
మీరు చదివిన కథల పుస్తకాల పేర్లు చెప్పండి.
జవాబు:
చందమామ, బాలమిత్ర, బుజ్జాయి, బాలానంద బొమ్మల పంచతంత్రం, బేతాళకథలు, అపూర్వ చింతామణి, సుజ్ఞానబోధిని, నీతికథలు, భట్టి విక్రమార్క కథలు మొదలైనవి.

ప్రశ్న 3.
కథల పుస్తకాలు చదవడం మీకు ఇష్టమా? ఎందుకు?
జవాబు:
కథల పుస్తకాలు చదవడం మాకు చాలా ఇష్టం. ఎందుకంటే కథలు ఆసక్తిని పెంచుతాయి. ఊహించని మలుపులు ఉంటాయి. కథలలోని పాత్రలు చాలా తెలివిగా, చమత్కారంగా ప్రవర్తిస్తాయి. మనకు తెలియని ఎన్నో విషయాలుంటాయి. చాలా విచిత్రమైన సమస్యలుంటాయి. వాటికి పరిష్కారాలుంటాయి. మంచి మంచి నీతులు – ఉంటాయి. అందుకే కథలంటే మాకు చాలా ఇష్టం.

Improve Your Learning (అభ్యసనాన్ని మెరుగుపరచుకుందాం)

అవగాహన – ప్రతిస్పందన

ప్రశ్న 1.
పాఠంలోని మాయాకంబళి వలే మానవ జీవితంలో సెల్‌ఫోన్, సమయం, సంపద, మాటలు వంటివి విలువైనవే. వీటిని గురించి మీ సొంత మాటలలో చెప్పండి.
జవాబు:
1) సెల్‌ఫోన్ :
ఈ రోజులలో సెల్ ఫోన్ లేనివారు లేరు. దాని వలన చాలా ప్రయోజనాలు ఉన్నాయి. నష్టాలు ఉన్నాయి. ఇంటర్నెట్లో శోధించి మనకు తెలియని ఎన్నో విషయాలు తెలుసుకోవచ్చు. ఉదాహరణకు హిమాలయాలు గురించి పాఠం విన్నాం. కాని, అవి ఎలా ఉంటాయో ఎంతలా వర్ణించి చెప్పినా మనకు వాటి స్వరూపం పూర్తిగా అర్థం కాదు. ఇంటర్నెట్ లో సెర్చ్ చేసి చూస్తే హిమాలయాలు కనబడతాయి. చక్కగా అర్థం చేసుకోవచ్చు. అలాగే శిల్పాలు, దేశాలు, కవులు, రచయితలు, నదులు, అరణ్యాలు, జంతువులు, ప్రదేశాలు ఇలా దేనినైనా మన కళ్లతో చూసి తెలుసుకోవచ్చును. సైన్సులో చెప్పినవి కూడా తెలుసుకోవచ్చు. సెల్ ఫోన్ వలన ఇలాగ అనేక ఉపయోగాలున్నాయి. అనవసర విషయాలు చూడకూడదు. సెల్ఫీలు దిగడం ప్రమాదాలలో పడడం తప్పు. అనవసరమైన ‘ఆన్లైన్ గేమ్స్’ ఆడడం కూడా చాలా తప్పు. దాని వలన ధనం, సమయం, శక్తి వృథా అవుతాయి.

2) సమయం :
సమయం చాలా విలువైనది. గడిచిపోతే తిరిగిరాదు. సమయాన్ని వృథా చేయకూడదు. అనవసరంగా కాలక్షేపం చేయకూడదు. కాలాన్ని వినియోగించుకోవాలి. సరిగా వినియోగించుకోకపోతే తర్వాత బాధపడినా ప్రయోజనం ఉండదు. చదువుకోవలసిన సమయంలో చదువుకోకపోతే జీవితమంతా బాధపడాలి. ఏ పనైనా అంతే.

3) సంపద :
మన దగ్గర సంపద అంటే డబ్బు ఉన్నపుడు చాలామంది చేరతారు. సంపద అయిపోయాక ఎవ్వరూ ‘రారు. డబ్బు సంపాదించడం కంటే దానిని కాపాడుకోవడం కష్టం. డబ్బు కోసం అబద్దాలు ఆడకూడదు. ఎవ్వరినీ మోసం చేయకూడదు. అడ్డదారులు తొక్కకూడదు. ‘డబ్బును అనవసరంగా ఖర్చు చేయకూడదు. అవసరాలలో ఉన్నవారిని ఆదుకోవాలి. ఆపదలలో ఉన్నవారికి సహాయపడాలి. డబ్బు ఉందని గర్వపడకూడదు. చెడుపనులు చేయకూడదు. అందరికీ ఉపయోగపడాలి.

4) మాటలు :
‘నోరావీపుకు దెబ్బలు తేకే’ అన్నారు. అంటే మనము చెడుమాటలు మాట్లాడితే అవమానాలు, నిందలు తప్పవు. మంచిగా మాట్లాడితే ఎవ్వరైనా స్నేహితులౌతారు. అందుకే ‘నోరు మంచిదైతే ఊరు మంచిదౌతుంది’ అన్నారు. మనకు మంచి పేరు కాని, చెడ్డపేరు కానీ తెచ్చేది మనం మాట్లాడే పద్ధతే. అందుకే ఆలోచించి మాట్లాడాలి. ఇతరులను బాధపెట్టేలా మాట్లాడకూడదు.. ఎవ్వరినీ అపహాస్యం చేయకూడదు. నలుగురితో కలుపుగోలుగా ఉండాలి. స్నేహంగా మాట్లాడాలి. మాట కత్తి కంటే పదునైనది. కఠినంగా మాట్లాడితే ఇతరుల మనసు గాయపడుతుంది. స్నేహాలు చెడిపోతాయి. బంధుత్వాలు దూరం అయిపోతాయి. ఇతరులు మనల్ని నొప్పించేలా మాట్లాడినా మౌనంగా ఉండాలి. అనవసరమైన వాద ప్రతివాదనలు పెంచుకోకూడదు. స్పష్టంగా మాట్లాడాలి. సంస్కారవంతం అయిన భాషనే ఉపయోగించాలి. మాటలలో వినయం, సంస్కారం, గౌరవం ఉట్టిపడాలి.

ప్రశ్న 2.
కథలోని మాయాకంబళి మీకు లభిస్తే దానిని ఏ విధంగా ఉపయోగిస్తారో చెప్పండి.
జవాబు:
నాకు మాయాకంబళి లభిస్తే దానిని కప్పుకొంటే నేనెవరికీ కనబడను కదా ! లంచగొండులు చేసే మోసాలను అదృశ్యరూపంలో గమనించి వారిని పోలీసులకు పట్టిస్తాను. సరుకులను కలీ చేసే వ్యాపారస్తులను అదృశ్య రూపంలో గమనించి, వారిని కూడా తూనికలు-కొలతలవారికి పట్టిస్తాను. దొంగలను కూడా అదృశ్యరూపంలో గమనించి పట్టిస్తాను. మోసాలు, దొంగతనాలు అరికట్టడానికి మాయాకంబళిని ఉపయోగిస్తాను. స్త్రీలను, పిల్లలను, బలహీనులను బాధపెట్టేవారి భండారం కూడా బయటపెడతాను. అదృశ్యరూపంలో ఎక్కడికైనా వెళ్లవచ్చుకదా ! మా స్నేహితులను ఆటపట్టిస్తాను. ‘మాయమైతే కనిపెట్టే’ ఆట ఆడుకొంటాం. చాలా రకాలుగా ‘మాయాకంబళి’తో ఆటలు ఆడుకొంటాం. ఆనందపడతాం.

ప్రశ్న 3.
మీరు చదివిన ఏదైనా కథను మీ సొంత మాటలలో చెప్పండి.
జవాబు:
అనగనగా ఒక ఊళ్లో ఒక మామ్మ, మనవడు ఉన్నారు. మనవడికి తల్లిదండ్రులిద్దరూ చనిపోయేరు. అందుచేత మామ్మకి మనవడంటే గారం. ఆ గారాబం వలన పాఠశాలకు వెళ్లేవాడు కాదు, చదువు రాలేదు. మామ్మగారు అతనిని రోజూ తిట్టేది. ఎక్కడికైనా వెళ్లి డబ్బు సంపాదించాలంటే చదువు ఉండాలని. కానీ, వినేవాడు కాదు. కనీసం బైటకి కూడా వెళ్లేవాడు కాదు. ఇంట్లోనే ఉండేవాడు. అలాగే పెద్దవాడైపోయాడు. 20 సంవత్సరాలు వచ్చేయి. అయినా. ఇల్లు వదిలేవాడు కాదు. అతని పేరు చెప్పలేదు కదూ ! అతని పేరు రాము.

“రామూ ! ఈ రోజు నువ్వు బైటకి వెళ్లి, డబ్బు సంపాదించుకురా ! వెళ్లు !” అంది మామ్మ.
“రేపు వెడతాను” అన్నాడు బద్దకంగా.
“రోజూ ఇలాగే అంటున్నావు. ఒక్క పదిరూపాయలు సంపాదించుకొనిరా ! డబ్బు విలువ తెలుస్తుంది.” అంది మామ్మ.
“పది రూపాయిలు కాదు మామ్మా ! పది. లక్షల కోట్లు సంపాదిస్తాను” అన్నాడు.
“అబ్బో ! కబుర్లకు లోటు లేదు. ఇలాగే కూర్చో ! నీకెవ్వరూ పిల్లనివ్వరు. పెళ్లి కాదు” అంది.
“కోటీశ్వరురాలు చేసుకొంటుందే మామ్మా” అన్నాడు.
“అలాగే కోతి కూడా చేసుకోదు” అంది.
“సరే ! రేపే వెడతాను. కోట్లు సంపాదించుకొని వస్తాను. కాని, నాకు మూడు మినపరొట్టెలు వేసి ఇయ్యి !” అన్నాడు.

మర్నాడు మూడు మినపరొట్టెలు, కొబ్బరి పచ్చడి, తేనె పానకం ఇచ్చింది. మనవడు ఆ రొట్టెలు పట్టుకొని బయల్దేరాడు. చాలా దూరం వెళ్లిపోయేడు. చీకటి పడే సమయానికి ఒక పెద్ద మర్రిచెట్టు దగ్గరికి చేరాడు.

“బాగా ఆకలి వేస్తోంది. పెద్దదాన్ని తినేయాలి. తేనెపానకం వేసుకొని తింటే భలే రుచిగా ఉంటుంది” అన్నాడు.

ఆ చెట్టు మీద మూడు దెయ్యాలున్నాయి. వాటికి భయపడి ఎవ్వరూ అటురారు. వీడు ధైర్యంగా వచ్చాడంటే వాడికి ఏవో మంత్రాలు వచ్చేమో ! అని దెయ్యాలు భయపడ్డాయి. పెద్ద దెయ్యం భయంతో కిందికి వచ్చింది.

“నన్ను తినకు ! నీకు దండం పెడతాను” అంది.

“దండం పెడితే కడుపునిండదమ్మా ! ఆకలిగా ఉంది తప్పదు. తినేస్తాను” అన్నాడు,

“ఒక్క క్షణం ఆగమని ఒక గిన్నె, గరిటె ఇచ్చింది.” గిన్నెలో గరిటె పెట్టి ‘వడ్డించు’ అంటే, నీకు కావలసిన ఆహార పదార్థాలు వస్తాయి. కడుపునిండా తిను ! నన్ను వదిలేయి !” అంది. తృప్తిగా తిన్నాడు. ఊళ్లోకి వెళ్లి ఒక కిరాణా వ్యాపారికి ఇచ్చి, జాగ్రత్త చేయమని, మళ్లీ చెట్టు కిందకు వచ్చాడు. కిరాణా వ్యాపారికి దాని రహస్యం కూడా చెప్పాడు. తన ఇంట్లో వారంతా తృప్తిగా తిన్నాక, ఊరందరినీ పిలిచి ఆ వ్యాపారి భోజనాలు పెట్టాడు.

అర్ధరాత్రి రాముకు మళ్లీ ఆకలి వేసింది. రెండోదాన్ని తినేస్తానన్నాడు. రెండో దెయ్యం భయపడింది. కిందకు వచ్చి బ్రతిమాలింది. ఒక సంచీ ఇచ్చింది. ఆ సంచీ దులిపితే బంగారు కాసులు పడ్డాయి. దేనిలో దులిపితే దాని నిండా పడతాయని చెప్పింది. రాము మళ్లీ ఊళ్లోకి వెళ్లి వ్యాపారికి ఇచ్చి, దీనిని ‘దులపకండి’ అన్నాడు. ‘అలాగే’ అన్నాడు. రాము వెళ్లిపోయాక ఒక గిన్నెలో దులిపాడు, గిన్నె నిండా బంగారుకాసులు పడ్డాయి. ఒక పెద్ద సంచీలో దులిపాడు. సంచి నిండా బంగారుకాసులు పడ్డాయి. . – ఊరందరికీ భోజనాలు పెట్టి, అందరికీ తలొక గుప్పెడు బంగారుకాసులిచ్చాడు. ఆ నోటా ఈ నోటా తెలిసి … కొన్ని వేలమంది జనం వచ్చేశారు. అందరికీ ఇస్తున్నాడు.

మర్నాడు ఉదయమే “చంటి దాన్ని తినేస్తాను” అన్నాడు రాము. చిన్న దెయ్యం భయపడి వచ్చింది. ఒక కర్ర – ఇచ్చింది. “వడ్డించు అంటే ఎంతమందినైనా కొడుతుంది. నువ్వు ఆగమంటేనే ఆగుతుంది” అంది. మళ్లీ పట్టుకెళ్లి వ్యాపారికి ఇచ్చాడు. “వడ్డించు అనకండి” అన్నాడు. నవ్వుతూ. “అలాగే బాబూ” అన్నాడు.

అందరికీ భోజనాలు పెట్టి, కాసులిచ్చాడు. కర్రమ ‘వడ్డించు’ అన్నాడు. అందరినీ చితక్కొడుతోంది. అందరూ గోల గోల పెట్టేస్తున్నారు. కర్ర ఆగడం లేదు. చివరికి రాము దగ్గరికి వెళ్లారు. బ్రతిమాలి తీసుకొచ్చారు. కర్రను ఆపించారు. అందరూ పారిపోయారు. మోసం చేయాలనుకొన్న వ్యాపారికి నాలుగు వడ్డించి తన గిన్నె, గరిటె, . సంచీ, కర్రతో ఇంటికి వెళ్లాడు. పేదలకు అన్నం పెట్టి, డబ్బులిస్తూ, పోషించేవాడు. ఆ డబ్బుతో ఊరిని బాగుచేశాడు. పెళ్లి చేసుకొని సుఖంగా ఉన్నాడు.

ప్రశ్న 4.
కింది గద్యాన్ని చదివి, ప్రశ్నలకు జవాబులు రాయండి.
ప్రతి ఇంట్లో ఎన్ని భాషలు. నేర్చుకున్నా మాతృభాషను విస్మరించకుండా ఉండాలి. భాష నిత్యం ప్రవహించే స్రవంతి. జీవనదిలా ప్రవహించే గోదావరి. నన్నయ కాలం నాటి గోదావరే ఇప్పుడూ ఉన్నా అందులో నీరు నిత్యం మారుతూనే ఉంటుంది. పాతనీరు పోయి కొత్తనీరు వచ్చి చేరుతూనే ఉంటుంది. భాష కూడా మార్పులకు గురవుతూనే ఉంటుంది. భాష నిలువ నీటి మడుగుగా మారితే అది సజీవ భాష కాదు అనే చెప్పాలి. అటువంటి పరిస్థితి ఏ భాషకు కలుగకూడదని గిడుగు రామ్మూర్తి గారు అన్నారు.
ప్రశ్నలు:
1) ప్రతి ఇంట్లో ఏ భాష మాట్లాడుతారు?
జవాబు:
ప్రతి ఇంట్లో తమ మాతృభాషను మాట్లాడుతారు.

2) భాష అనేది నిత్యం ఏమవుతూ వుంటుంది?
జవాబు:
భాష అనేది నిత్యం మార్పులకు లోనవుతూ ఉంటుంది.

3) భాషను ఏ నదితో పోల్చారు?
జవాబు:
భాషను గోదావరి నదితో పోల్చారు.

4) ఈ వచనంలోని మాటలు ఎవరన్నారు?
జవాబు:
ఈ వచనంలోని మాటలను గిడుగు రామ్మూర్తిగారన్నారు.

వ్యక్తీకరణ – సృజనాత్మకత

అ) కింది ప్రశ్నలకు నాలుగైదు వాక్యాలలో జవాబులు రాయండి.

ప్రశ్న 1.
భిక్షగాడు మాయా కంబళిని ఎందుకు వద్దనుకున్నాడు?
జవాబు:
బిచ్చగాడు ముసలివాడు. అది చలికాలం. అతనికి కప్పుకొందుకు కంబళి లేదు. అందుచేత శోభావతీ నగరంలో ఆ ముసలి బిచ్చగాడు చలిబాధ భరించలేక పాత కంబళిని చౌకలో కొనుక్కున్నాడు. వెంటనే దానిని కప్పుకున్నాడు. వెంటనే అదృశ్యమయ్యాడు. ఆ వార్త నగరమంతా వ్యాపించింది. చాలామంది దురాశాపరులు బిచ్చగాడి కోసం వెదకసాగారు. అది తెలిసి బిచ్చగాడు తనకు రానున్న ప్రమాదం ఊహించాడు. ఆ నగరంలో తనకు భద్రత లేదనుకున్నాడు.

ఆ కంబళి వలన తనకు అడుక్కుతినడానికి కూడా లేకపోయిందని బాధపడ్డాడు. ఊరి బయట పాడుబడిన దేవాలయం దగ్గర దొంగలు కూడా తన గురించే మాట్లాడుకోవడం గమనించాడు. ఆ రాత్రంతా ఆలోచించాడు. ఆ కంబళి కోసం హేమాహేమీలు పోటీ పడుతున్నారని గ్రహించాడు. ఆ కంబళి తనకెందుకూ పనికిరాదని నిర్ధారించుకున్నాడు. అది తనకు వద్దనుకున్నాడు. రాజు గారికి ఇచ్చేయాలని నిర్ణయించుకున్నాడు. ఇచ్చేశాడు.

ప్రశ్న 2.
ఆత్మానందుడు రాజుకు రక్షాబంధం ఎందుకు కట్టాడు?
జవాబు:
రాజుకు భిక్షగాడు మాయాకంబళిని ఇచ్చాడు. దాని మహత్తును రాజు స్వయంగా పరీక్షించి తెలుసుకున్నాడు. అంతలో ఆత్మానందుడు వచ్చాడు. రాజైన చండీదత్తుని కలుసుకున్నాడు. కంబళిని ఇమ్మన్నాడు. రాజ్యశ్రేయస్సు దృష్ట్యా అది తనవద్దే ఉండాలని రాజు అన్నాడు.

అసాధారణ శక్తుల వల్ల మంచికంటే చెడు జరగటానికే ఎక్కువ అవకాశాలున్నాయని ఆత్మానందుడు చెప్పాడు. మాయాకంబళి , వంటి మహిమాన్వితమైన వస్తువు మనిషిని పాపకార్యాలకు, నీతి బాహ్యమైన పనులకు పురిగొల్పుతుందని చెప్పాడు. ఉత్తముడైన చండీదత్తుడు భ్రష్టుడుకావడం తనకిష్టం లేదని ఆత్మానందుడు చెప్పాడు. ఎన్ని విధాల చెప్పినా రాజు వినలేదు. స్వానుభవంతో తప్ప రాజుకు. విషయం బోధపడదని చెప్పాడు. రాజు క్షేమం కోసం రాజు చేతికి రక్షాబంధం కట్టాడు. ఆ రక్షాబంధనం వలన మాయాకంబళి కప్పుకొని మాయమైనవారు రాజుకు కనబడతారు. దాని వలననే చంచల, విక్రముల మోసాన్ని కనిపెట్టి, రాజు తనను, రాజ్యాన్ని రక్షించుకొన్నాడు.

ప్రశ్న 3.
కలువకొలను సదానంద గురించి మీరు తెలుసుకున్న అంశాలను రాయండి.
జవాబు:
కలువకొలను సదానంద

జననం : చిత్తూరు జిల్లా పాకాలలో 22.2. 1939లో జన్మించారు.
తల్లిదండ్రులు : నాగమ్మ, కృష్ణపిళ్లే దంపతులు.
వృత్తి : ఉపాధ్యాయుడు – 1992లో జాతీయ ఉత్తమ ఉపాధ్యాయుడుగా ఎంపికయ్యారు.
రచనలు (పిల్లలకథలు): శివానందలహరి, విందుభోజనం, చల్లని తల్లి, నీతికథా మంజరి, తుస్సన్న మహిమలు, పరాగభూమి మొదలైనవి 200 కథలు, 2 నవలలు, 100 కి పైగా గేయాలు, కొన్ని గేయ కథలు రచించారు. చందమామ కథలు, వార్తా పత్రికలలో కథానికలు వ్రాశారు.

బహుమతులు : ‘బంగారు నడచిన బాట’ నవలకు 1966లో కేంద్రప్రభుత్వ విద్యాశాఖ బహుమతి వచ్చింది.
‘నవ్వే పెదవులు-ఏడ్చే కళ్లు’ కథాసంపుటికి 1976లో ఆంధ్రప్రదేశ్ సాహిత్య అకాడమీ అవార్డు లభించింది. ‘అడవితల్లి’ పిల్లల నవలకు 2010లో కేంద్రసాహిత్య, బాలసాహితీ అవార్డు లభించింది.

ఆ) కింది ప్రశ్నలకు 8 నుండి 10 వాక్యాలలో జవాబులు రాయండి.

ప్రశ్న 1.
మాయాకంబళిని రాజు మొదట ఎందుకు కావాలనుకున్నాడు? తరువాత ఎందుకు తిరిగి ఇచ్చేశాడు?
జవాబు:
రాజ్యశ్రేయస్సు కోసం బిచ్చగాని వద్ద మాయాకంబళిని రాజు తీసుకొన్నాడు. అది తన దగ్గర ఉంటే దొంగలను, మోసగాళ్లను పట్టుకోవచ్చుననుకొన్నాడు. అందుచేతనే ఆత్మానందుడు ఆ కంబళిని అడిగినా రాజు ఇవ్వలేదు. ఆ కంబళి వలన పాపకార్యాలు, నీతి బాహ్యమైన పనులు పెరిగిపోతాయని చెప్పినా వినలేదు. ఆ కంబళి వలన రాజుకే ప్రమాదమన్నా పట్టించుకోలేదు.

ఆ కంబళి సహాయంతో నేరాలు, అక్రమాలు అరికట్టాడు. రాజనర్తకి చంచల కోరికపై ఆమెకు ఆ కంబళి ఇచ్చాడు. ఆత్మానందుడు కట్టిన రక్షాబంధం వలన కంబళి కప్పుకొని మాయమైన చంచలను చూశాడు. ఆమె కోశాధికారి విక్రముడి వద్దకు వెళ్లింది.

విక్రముడు ఆ కంబళి కప్పుకొని, అదృశ్యరూపంలో రాజుగారి పడకగదిలోకి వచ్చాడు. రాజును చంపాలను కున్నాడు. అది గ్రహించిన చండీదత్తుడు కత్తి దూశాడు. విక్రముని చంపాడు. చంచలను బంధించాడు, దేశ బహిష్కార శిక్షను విధించాడు.

మాయాకంబళి వలన మంచి చేయవచ్చుననుకున్నాడు కాని, దాని వలన మానవులలో స్వార్థం, నీచత్వం, . చెడు ఆలోచనలు పెరిగి ప్రాణాలకే ముప్పు కలుగుతుందని రాజు గ్రహించాడు. ఆ మాటలే ఆత్మానందుడికి చెప్పాడు. కంబళిని తిరిగి ఆత్మానందుడికి ఇచ్చేశాడు.

ప్రశ్న 2.
మాయాకంబళి కథా సారాంశాన్ని మీ సొంతమాటలలో రాయండి.
జవాబు:
పూర్వం హిమాలయ పర్వతాలలో ఒక వృద్ధ యోగి ఉండేవాడు. అతని పేరు ఆత్మానందుడు. ఆయన చాలా మహిమలు కలవాడు. ఆయనకు భక్తుల తాకిడి ఎక్కువయింది. వారి నుండి తప్పించుకొనేందుకు తన పాత కంబళి మీద మంత్రజలం చల్లాడు. దాని మహిమతో ఆ కంబళి కప్పుకుంటే అదృశ్యమైపోతాడు.

ఒకరోజు ఆత్మానందుడు ఒక లోయలోకి పడిపోయాడు. ఆ కంబళి పడిపోయింది. అది ఒక వ్యాపారికి దొరికింది. ఆ వ్యాపారి శోభావతీ నగరంలో సంతకు వచ్చాడు. సంతలో దానిని ఒక ముసలి బిచ్చగాడు చౌక ధరకు కొన్నాడు. చలికి తట్టుకోలేక అది కప్పుకొని మాయమయ్యాడు. ఈ వార్త అందరికీ తెలిసింది.

దొంగలు, మోసగాళ్లే కాకుండా హేమాహేమీలు కూడా ఆ బిచ్చగాని కోసం వెతుకుతున్నారు. అది బిచ్చగాడు గమనించాడు. తనకు భద్రత లేదని తెలుసుకున్నాడు. కనీసం అడుక్కొని తినే అవకాశం కూడా లేనందుకు బాధ పడ్డాడు. అదృశ్యరూపంలో, చండీదత్త మహారాజు వద్దకు వెళ్లాడు. సభలో ప్రత్యక్షమయ్యాడు. మాయాకంబళిని రాజుకు ఇచ్చేశాడు. రాజు అతనికి జీవించడానికి సదుపాయాలు కల్పించాడు.

అంతలో ఆత్మానందుడు వచ్చాడు. రాజును కంబళి ఇమ్మన్నాడు. రాజు ఇవ్వలేదు. ఆ మాయాకంబళితో అన్యాయాలు, అక్రమాలను రాజు నివారించాడు.

రాజ్యంలో శాంతి ఏర్పడింది. రాజనర్తకి చంచల కోరగా మాయాకంబళిని ఆమెకు ఇచ్చాడు. ఆత్మానందుడు కట్టిన రక్షాబంధం మహిమతో చంచలను గమనించాడు. ఆమె మాయాకంబళిని కప్పుకొని మాయమై, కోశాధికారి విక్రముని దగ్గరకు వెళ్లడం చూశాడు. ఆ రాత్రి విక్రముడు మాయాకంబళిని కప్పుకొని, రాజును చంపడానికి వచ్చాడు. అది గమనించిన రాజు కత్తిదూసి, అతనిని చంపాడు. చంచలను రాజ్యం నుండి బహిష్కరించాడు.

మాయాకంబళి వలన కలిగే ప్రమాదం తెలుసుకున్నాడు. వెంటనే హిమాలయాలకు వెళ్లి, ఆత్మానందుడికి ఆ మాయాకంబళిని ఇచ్చేశాడు.

ప్రశ్న 3.
విద్యార్థులారా ! మీరు మీ ఊహాశక్తితో ఒక కథను రాయండి.
జవాబు:
ఒకరోజు రెండు రాళ్లు మాట్లాడుకుంటున్నాయి.
“అడుగో ! శిల్పి వస్తున్నాడేవ్” అంది ఒక రాయి.
“అతనంటే నాకు భయం. ఒళ్లంతా చెక్కేస్తాడు” అంది మరొక రాయి.
“ఆ దెబ్బలు భరిస్తేనే కదా ! మనకు గౌరవం దక్కుతుంది.” అంది మొదటి రాయి.
“గౌరవం లేదు. గాడిద గుడ్డూ లేదు. నేను భరించలేను, పారిపోతున్నాను.” అని క్రిందికి దొర్లిపోయింది.

పది సంవత్సరాల తర్వాత, గుడిలో వినాయక విగ్రహం ఎదురుగా మెట్ల క్రింద రాయి ఉంది. ఆ రాయి వినాయకుని కాపాడమని ప్రార్థించింది. అందరూ తనను తొక్కుతూ వెడుతున్నారని, ఆ బాధ భరించలేకపోతున్నానని, విముక్తి కల్గించమని ప్రార్థించింది. వినాయక విగ్రహం పకపకా నవ్వింది. పది సంవత్సరాల క్రితం మనిద్దరం ప్రక్క ప్రక్కనే ఉండేవాళ్లం.

“నువ్వు ఉలిదెబ్బలకు భయపడి పారిపోయావు. నేను భరించాను. అందుకే ఈ రోజు పూజలందుకొంటున్నాను. కష్టపడితే ఫలితం వస్తుంది. చిన్నప్పుడు కష్టపడి చదువుకొంటే, జీవితమంతా సుఖపడవచ్చు. చిన్నప్పుడు చదువుకు భయపడితే, నీలాగే జీవితమంతా బాధపడాలి. అందుకే మొదట కష్టపడు. తర్వాత సుఖపడు అన్నారు. ఎవరూ ఎవరినీ రక్షించలేరు. నీ జీవితానికి నువ్వే కర్తవు. “జాలి పడడం’ తప్ప నేనేం చేయలేను” అంది వినాయక శిల్పంగా మారిన రాయి.

భాషాంశాలు

అ) కింద గీత గీసిన పదాలకు అర్ధాలు రాయండి. వాటితో సొంతవాక్యాలు రాయండి.

1. పక్షులు యథేచ్ఛగా ఆకాశంలో తిరుగుతున్నాయి.
ఉదా : యథేచ్ఛగా = స్వేచ్ఛగా
సొంతవాక్యం : మనం స్వేచ్ఛగా మనలోని భావాలు చెప్పాలి.

2. బియ్యం చౌక ధరకు అమ్ముతున్నారు.
చౌక = తక్కువ ధర
సొంతవాక్యం : ప్రభుత్వం తక్కువ ధరకు నాణ్యమైన సరుకులు ఇస్తుంది.

3. చెడ్డ పనులు ఎవరైనా చేస్తే అభ్యంతరం చెప్పాలి.
అభ్యంతరం : ఆటంకం
సొంతవాక్యం : మంచిపనికి ఆటంకం కల్గించకూడదు.

4. చలి వేసినప్పుడు గొంగళి కప్పుకుంటాను.
గొంగళి = రగ్గు
సొంతవాక్యం : శీతాకాలంలో రగ్గు చాలా అవసరం.

ఆ) కింది ప్రకృతికి వికృతిని జతపరచండి.

జవాబు:

ఇ) కింది వాక్యాలలో సమానార్థక పదాలు గుర్తించి రాయండి.

1. పర్వతాలు ఎత్తుగా ఉన్నాయి. ఆ శైలము మీద చెట్లు ఉన్నాయి. ఆ గిరులు మంచుతో కప్పబడి ఉన్నాయి.
జవాబు:
పర్వతము, శైలము, గిరి

2. ఆ పట్టణంలో రాజు ఎంతో ఉత్తముడు. అందుకే ఆ భూపాలుడ్ని అందరూ నరేంద్రుడు అని పిలుస్తారు.
జవాబు:
రాజు, భూపాలుడు, నరేంద్రుడు

3. రాజు కరవాలంతో యుద్ధం చేస్తాడు. ఆ ఖడ్గం శత్రువుల తలను ఖండిస్తుంది.
జవాబు:
కరవాలం, ఖడ్గం

ఈ) కింది పదాలను సొంతవాక్యాలలో ప్రయోగించండి.
ఉదా : ఏకాగ్రత = అవధానము
సొంతవాక్యం : ఏ విషయం మీదైనా ఏకాగ్రత పెట్టినట్లయితే విజయం సాధిస్తారు.

1. ప్రశాంతత = శాంతం
సొంతవాక్యం : ప్రశాంతత లేకపోతే ఎంత సంపద ఉన్నా ప్రయోజనం లేదు.

2. తిరుగుముఖం = వెనుకకు ప్రయాణం కావడం
సొంతవాక్యం : కరోనా తిరుగుముఖం పట్టిందనుకొంటే, మళ్లీ చెలరేగిపోతోంది.

3. ప్రలోభ పెట్టడం = లంచం ఇచ్చి వశపరచుకోవడం
సొంతవాక్యం : తప్పుడు పని కోసం ఇతరులను ప్రలోభపెట్టడం తప్పు.

4. శ్రేయస్సు = మేలు
సొంతవాక్యం : గురువులు ఎప్పుడూ శిష్యుల శ్రేయస్సునే కోరతారు.

5. దురాశాపరులు = చెడ్డదైన ఆశ కలవారు
సొంతవాక్యం : దురాశాపరులు తమ అవసరం కోసం ఎంత తప్పుగానైనా ప్రవర్తిస్తారు.

6. కంటపడకుండా = ఇతరులు గమనించకుండా
సొంతవాక్యం : పాఠశాలకు ఆలస్యంగా వచ్చిన చందు ప్రధానోపాధ్యాయురాలి కంటపడకుండా జాగ్రత్త పడ్డాడు.

ఉ) ఎటునుంచి చదివినా ఒకే పదం వచ్చే ‘కచిక’ పదాలకు (పద భ్రమణం) వాక్యాల ఆధారంగా జవాబులు రాయండి.
ఉదా : నటులు చేసేది – నటన
1. చింతకాయ రుచి
జవాబు:
పులుపు

2. శరీరాన్ని కప్పే వస్త్రానికి మరొక పేరు
జవాబు:
వలువ

3. గాలి, వెలుతురు కోసం ఇంటికి పెట్టేది.
జవాబు:
కిటికి

4. ఇది తీయడమంటే నిద్రపోతున్నాడని అర్థం
జవాబు:
కునుకు

5. సముద్రపు ఆల్చిప్పలో వుండేది
జవాబు:
ముత్యము

ఇలాంటి మరికొన్ని పదాలను సేకరించి రాయండి.
1. గరగ
2. విరివి
3. కలక
4. జలజ
5. కచిక
6. మహిమ

ఇటువంటి వాక్యం : సినిమాకురా పరాకు మానిసి.

వ్యాకరణాంశాలు

అ) కింది వాక్యాలను గమనించండి.

1. పోతన భాగవతం రచించాడు.
2. ఆవు పాలు ఇచ్చింది.
3. రాజు వేటకు వెళ్ళాడు.
4. రాము అన్నం తిన్నాడు.
5. గీత పుస్తకం తెచ్చింది.

క్రియకు ముందు భాగంలో ఎవరు? ఏది? అని ప్రశ్నిస్తే వచ్చే జవాబును గమనించండి.
ఉదా : భాగవతాన్ని ఎవరు రచించారు? – పోతన

1. పాలు ఇచ్చే జంతువు ఏది?
జవాబు:
ఆవు

2. వేటకు ఎవరు వెళ్ళారు?
జవాబు:
రాజు

3. అన్నం ఎవరు తిన్నారు?
జవాబు:
రాము

4. పుస్తకం ఎవరు తెచ్చారు?
జవాబు:
గీత

క్రియను ఎవరు? ఏది? అని ప్రశ్నించినప్పుడు వస్తున్న సమాధానాన్ని కర్తగా చెప్పవచ్చు.

కింది వాక్యాలను చదవండి. కర్తను గుర్తించి గీతగీసి, పక్కన రాయండి.
ఉదా : ఆత్మానందుడు కంబళిని భుజాన వేసుకొని వెళ్ళాడు. (ఆత్మానందుడు)
1. భిక్షగాడు మాయాకంబళితో అదృశ్యమయ్యాడు. (భిక్షగాడు)
2. విక్రముడు కత్తి పట్టుకొని రాజు మందిరానికి వెళ్ళాడు. (విక్రముడు)
3. చండీదత్తుడు విక్రముడిని సంహరించాడు. (చండీదత్తుడు)
4. భటులు చంచలను బంధించారు. (భటులు)

ఉత్వసంధి

ఆ) కింది వాక్యాలను చదవండి. గీత గీసిన పదాలను పరిశీలించి, విడదీసి రాయండి.
1. భద్రాచల రాముడతడు.
2. వినాయక చవితికి సెలవిచ్చారు.
3. బడి గంటలు మ్రోగుచున్నవి.
4. మనిషికి కొంచెమైనా ధర్మగుణం ఉండాలి.
5. గంగి గోవు పాలు పద్య భావమేమి.
6. ప్రజలందరూ కరోనాకు భయపడుతున్నారు.

సంధిరూపం – విడదీసిన రూపం
ఉదా : రాముడతడు = రాముడు + అతడు
1. మ్రోగుచున్నవి = మ్రోగుచు + ఉన్నవి (ఉకార వికల్ప సంధి)
2. కొంచెమైనా = కొంచము + ఐనా (ఉకార వికల్ప సంధి)
3. భావమేమి = భావము + ఏమి (ఉకార వికల్ప సంధి)
4. ప్రజలందరూ = ప్రజలు + అందరూ (ఉకార వికల్ప సంధి)

పై ఉదాహరణలు గమనించినప్పుడు పూర్వ స్వరంగా ‘ఉ’ ఉంది. కనుక దీనిని ఉత్వసంధి అని చెప్పవచ్చు. ఈ పాఠంలో ఉన్న ఉకారసంధి పదాలను గుర్తించి విడదీసి రాయండి.
ఉదా : నగరమంతా = నగరము + అంతా (ఉకార వికల్ప సంధి)
1. కలవాడని = కలవాడు + అని (ఉకార వికల్ప సంధి)
2. ఎవరైనా = ఎవరు + ఐనా (ఉకార వికల్ప సంధి)
3. రాజయిన = రాజు + అయిన (ఉకార వికల్ప సంధి)
4. పనికిరాదని = పనికిరాదు + అని (ఉకార వికల్ప సంధి)
5. ప్రత్యక్షమయ్యాడు = ప్రత్యక్షము + అయ్యాడు (ఉకార వికల్ప సంధి)

ఇ) కింది గద్యాన్ని చదివి, భాషాభాగాలను గుర్తించండి.

సంతోష్, శ్యామల జంతు ప్రదర్శనశాలకు వెళ్లారు. వారు అక్కడ క్రూర మృగాలైన సింహం, పులి, పొడవైన దంతాలు కలిగిన ఏనుగును చూశారు. శ్యామల సంతోషాన్ని ఆపుకోలేక అబ్బో ! ఎంత పెద్ద జంతువులో ! అని ఆశ్చర్యపోయింది. పచ్చని రామచిలుకలు, అందమైన నెమళ్ళను చూశారు. తరువాత అక్కడ ఉన్న దుకాణంలోకి వెళ్ళి తినుబండారాలు తిన్నారు. “ఓ…… సంతోష్ ఇటువైపు చూడు అవి ఎంత బాగున్నాయో !” అని కౌజుపిట్టలను శ్యామల, సంతోష్ కి చూపించింది. ప్రదర్శనశాలలో ఉన్న ఒక లేడి కాలికి గాయం కాగా దానిని చూసి సంతోష్ అయ్యయ్యో ! లేడి కాలి నుండి రక్తం కారుతోందని సానుభూతిని వ్యక్తం చేశాడు.
1. నామవాచకం :
సంతోష్, శ్యామల, జంతుప్రదర్శనశాల, మృగాలు, సింహం, పులి, దంతాలు, ఏనుగు, జంతువు, రామచిలుక, నెమళ్లు, దుకాణం, తినుబండారాలు, కౌజుపిట్టలు, లేడి, కాలు, గాయం, రక్తం.

2. సర్వనామం :
వారు, అక్కడ, అవి, దానిని, ఒక

3. క్రియ :
వెళ్లారు, చూశారు, పోయింది, వెళ్లి, తిన్నారు, చూపించింది, చూసి, చేశాడు.

4. విశేషణం :
క్రూర, పొడవైన, పెద్ద, పచ్చని, అందమైన, బాగు

5. అవ్యయం :
అబ్బో, ఓ, అయ్యయ్యో

మాయాకంబళి పాఠంలోని భాషాభాగాలను ఐదింటిని గుర్తించి రాయండి.

1. నామవాచకం :
పర్వతాలు, ఆత్మానందుడు, కంబళి, జలం, భుజం, కాలు, శిల, లోయ, కనుమ, వ్యాపారి, సంత, మనిషి, మూట, శోభావతి, నగరం, బిచ్చగాడు, వార్త, రాజు, చండీదత్తుడు, ప్రత్యర్థి, పొద, ముష్టివాడు, దేవాలయం, రాత్రి, సభ, కొండ, రాజ్యం , వస్తువు, కార్యం , ఉత్తముడు, భ్రష్టుడు, మందిరం, నర్తకి, చంచల, విక్రముడు, అధికారి, కత్తి, రాణి, దేశం

2. సర్వనామం :
ఆయన, తమకు, తన, దానికి, ఈ, ఒక, అతను, ఆ, అది, దానిని, వాటిని, వారు, అందుకు, ఇలాంటి, నువ్వు, నేను, ఆమె, కాబట్టి, వాడు.

3. క్రియ :
వచ్చి, పెట్టి, చేశాడు, వేసుకొని, జారి, పడిపోయి, కోల్పోయాడు, పడిపోయింది, పోతూ, చూశాడు, వెళ్లాడు, పడుతూ, కప్పుకొన్నాడు, గమనించి, ఆశ్చర్యపోయారు, పాకింది, ఉన్నాయి, సాగారు, విని, గ్రహించి, మసలసాగాడు, చేరుకున్నాడు, వచ్చారు, మాట్లాడుకున్నారు, ఆలోచించి, కప్పుకుని, వెళ్లి, తీసి, అయ్యాడు, ఇచ్చాడు, పరీక్షించి, తెలుసుకుని, చేసి, తీసుకున్నాడు, కోలుకొని, తెలుసుకుని, కలుసుకుని, అన్నాడు, చెబుతున్నాను, ఉన్నాయి, చెప్పినా, నిట్టూర్చాడు, ఒప్పుకొని, చెప్పి, వెళ్లిపోయాడు, పోయాయి, చేకూరింది, అడిగాడు, అని, కోరింది, వచ్చి, వెళ్లు, పట్టింది, చూసి, తోచింది, ఆలోచిస్తూ, వెళ్లక, పడుకున్నాడు, గ్రహించి, లేచి, దూసి, పడ్డాడు, పోయాు , బంధించి, తెచ్చారు, బహిష్కరించాడు, ఇచ్చేశాడు, కట్టాడు, అని, నవ్వుతూ, తీసుకున్నాడు.

4. విశేషణం :
పూర్వం, వృద్ధ, గొప్పు, తరచు, చెడు, పాత, ఏకాంతం, మంచి, కఠినం, తగ్గి, కొంత, స్పష్టం, శయన, చివాలున, తేతిక, సిద్ధం, ముప్పు.

5. అవ్యయం :
యథా, ప్రతి

గ్రంథాలయంలో మీరు చదివిన 5 కథలను అందులోని నీతిని కింది పట్టికలో రాయండి.

నీతిపద్యం

తే॥గీ॥ | సర్వ తీర్థాభిగమనంబు సర్వవేద
సమధిగమము సత్యంబుతో సరియుగావు
ఎఱగు మెల్ల ధర్మంబుల కెందు పెద్ద
యండ్రు సత్యంబు ధర్మజ్ఞులైన మునులు.

భావం :
పుణ్య నదులలో స్నానం చేయడం, సమస్త వేదాలు అధ్యయనం చేయడం ఒక్క సత్యానికి సాటిరావు. అన్ని ధర్మాల కంటే నిజం పలకడమే గొప్ప ధర్మం అని మునులు చెపుతారు. నన్నయ ఈ చిన్న పద్యంలో సత్యాన్ని దాని ప్రాధాన్యతను గురించి ఎంతో చక్కగా చెప్పాడు.

అశ్వమేధం వంటి యాగాలు చేయడం రాజులకే సాధ్యం. పుణ్యక్షేత్రాలు దర్శించడానికి ధనం ఉండాలి. నిజం చెప్పడానికి మాత్రం ఏమీ ఖర్చు లేదు. అయితే దాని ఫలితం ఎంతో విలువైనది. ఒక్క సత్యవాక్కు ఎంతటి పుణ్యాన్ని కలిగిస్తుందో, ఒక అబద్దం అంతకంటే ఎక్కువ పాపాన్ని, నష్టాన్ని కలిగిస్తుందని గ్రహించాలి.

మీకు తెలుసా?

శతవిధాల :
శతవిధాలుగా అంటే వంద విధాలుగా చెప్పడం. ఏదైనా ఒక విషయాన్ని అనేక రకాలుగా తెలియజేస్తున్నారని తెలిపే పదబంధం.

స్వానుభవం :
తనకు తాను ఏదైనా చేయడం ద్వారా అలవడే నేర్పు.

ఉపాధ్యాయులకు సూచనలు

1. విద్యార్థుల చేత పాఠాన్ని చక్కగా చదివించాలి.
   విద్యార్థులకు కథల పట్ల ఆసక్తి కల్గించేలా మరిన్ని కథలు సేకరించి చదివించండి.
   రచయిత కలువకొలను సదానంద కథలను సేకరించండి.

కవి పరిచయం

కవి పేరు : కలువకొలను సదానంద
జననం : చిత్తూరు జిల్లా పాకాలలో 22.2.1939లో జన్మించారు.
తల్లిదండ్రులు : నాగమ్మ, కృష్ణపిశె దంపతులు.
వృత్తి : ఉపాధ్యాయుడు – 1992లో జాతీయ ఉత్తమ ఉపాధ్యాయుడుగా ఎంపికయ్యారు.

రచనలు (పిల్లలకథలు) : శివానందలహరి, విందుభోజనం, చల్లని తల్లి, నీతికథా మంజరి, తుస్సన్న మహిమలు, పరాగభూమి మొదలైనవి 200 కథలు, 2 నవలలు, 100 కి పైగా గేయాలు, కొన్ని గేయ కథలు రచించారు. చందమామ కథలు, వార్తా పత్రికలలో కథానికలు వ్రాశారు.

బహుమతులు : ‘బంగారు నడచిన బాట’ నవలకు 1966లో కేంద్రప్రభుత్వ విద్యాశాఖ బహుమతి వచ్చింది. ‘నవ్వే పెదవులు-ఏడ్చే కళ్లు’ కథాసంపుటికి 1976లో ఆంధ్రప్రదేశ్ సాహిత్య అకాడమీ అవార్డు లభించింది. ‘అడవితల్లి’ పిల్లల నవలకు 2010లో కేంద్ర సాహిత్య, బాలసాహితీ అవార్డు లభించింది.
(ఈ పాఠ్యాంశం ‘మాయాకంబళి’ సంపుటి నుండి గ్రహించబడింది.)

పదాలు – అర్థాలు

1. పూర్వం…… దెబ్బతీయవచ్చని వారి ఆలోచన.
అర్థాలు :
పర్వతాలు = కొండలు
వృద్ధుడు – ముసలివాడు
ఏకరువు పెట్టు = గట్టిగా చెప్పు
మార్గము = దారి
జలం = నీరు
రహస్యం = గుట్టు
యథేచ్ఛ = ఇష్టం వచ్చినట్లు
అవినీతి = నీతిలేకపోవడం
ప్రత్యర్థి = వ్యతిరేకి
కనుమ = లోయ
ధర = వెల
మడిచి = మడత పెట్టి
చౌక = తక్కువ ధర
అదృశ్యం = కనబడకుండా పోవడం
వార్త = విషయం
స్పర్థ = విరోధం
పెచ్చుమీరు = బాగా పెరిగిపోవు
మహత్తు = మహిమ
అదృశ్యకరణి = అదృశ్యం చేసేది

2. కంబళి సహాయంతో …… ప్రత్యక్షమయ్యాడు.
అర్థాలు :
మసలడం = తిరగడం
భద్రత = భయంలేని స్థితి
పాడుబడిన = పాడైపోయిన
దేవాలయం గుడి – శిల రాయి
హేమాహేమీలు = గొప్పవారు
తహతహలాడడం = ఆత్రుత పడడం
రూఢీ = నిర్ధారణ
ప్రత్యక్షం = కనబడడం

3. మహాప్రభూ …… తీసుకొని వెళ్లు అన్నాడు.
అర్థాలు :
ప్రాణహాని = ప్రాణానికి ప్రమాదం
స్వయంగా = తనంతట తానుగా
ఏకాంతం = ఒక్కరే ఉన్న స్థితి
అభ్యంతరం = ఆటంకం
శ్రేయస్సు = మేలు, అభివృద్ధి
శక్తి = బలము
మహిమాన్వితం = మహిమతో కూడిన
కార్యము = పని
నీతిబాహ్యము = నీతిలేనిది
పురికొల్పడం = ప్రోత్సహించడం
భ్రష్టుడు = సర్వనాశనమైనవాడు
ఏకీభవించడం = కలియడం
స్వానుభవం = తన అనుభవం
నిట్టూర్పు = నిరాశతో గాలిని విడవడం
రక్షబంధం = రక్ష కొరకు కట్టే దారం
అవినీతిపరులు = నీతిలేనివారు
అక్రమం = చట్టబద్ధం కానిది, సరైనది కానిది
మందిరం = గది
నర్తకి = నాట్యం చేసే స్త్రీ
ఆనందభరితుడు = ఆనందంలో మునిగినవాడు
ముచ్చట = కోరిక
నిరాకరించడం = ఒప్పుకోకపోవడం
శయన మందిరం = పడకగది

4. రాజు తన మందిరానికి … దానిని తీసుకొన్నాడు.
అర్థాలు :
తిరుగుముఖం = వెనుకకు ప్రయాణం
కోశాధికారి = భాండాగార రక్షకుడు (ధనాన్ని రక్షించే అధికారి)
చివాలున = వెంటనే
ప్రలోభపెట్టడం = దేనినో ఆశ చూపి లొంగ దీసుకోవడం
పరిణామం = మార్పు
బహిష్కరించడం = వెళ్లకొట్టడం, వెలివేయడం
స్వార్థం = తన గురించిన ఆలోచన

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 2nd Lesson Linear Equations in One Variable InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable InText Questions and Answers.

8th Class Maths 2nd Lesson Linear Equations in One Variable InText Questions and Answers

Do this

Question 1.
Which of the following are linear equations:        [Page No. 35]
i) 4x + 6 = 8
ii) 4x – 5y = 9
iii) 5x² + 6xy – 4y² = 16
iv) xy + yz + zx = 11
v) 3x + 2y – 6 = 0
vi) 3 = 2x + y
vii) 7p + 6q + 13s = 11
Answer:
(i), (ii), (v), (vi), (vii) are the linear equations.

Question 2.
Which of the following are simple equations?        [Page No. 36]
i) 3x + 5 = 14
ii) 3x – 6 = x + 2
iii) 3 = 2x + y
iv) \(\frac{x}{3}\) + 5 = 0
v) x² + 5x + 3 = 0
vi) 5m – 6n = 0
vii) 7p + 6q + 13s = 11
viii) 13t – 26 = 39
Answer:
(i), (ii), (iv), (viii) are the simple equations.
Since these are all in the form of ax + b = 0.

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 8 Production and Management of Food From Plants Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 8th Lesson Production and Management of Food From Plants

8th Class Biology 8th Lesson Production and Management of Food From Plants Textbook Questions and Answers

Improve Your Learning

Question 1.
State reasons why wheat is cultivated in Kharif.
Answer:

1. The crops grown in the rainy season are termed as Kharif in the months of June to October.
2. If we cultivate wheat crop in the month of July it takes 8 – 10 weeks for growing.
3. After that flowering will take place. By that time, it would be October.
4. Then the night duration extends more than 12\(\frac{1}{2}\) hours. Wheat plants flowering takes place only in long night durations.
5. Crop production is based on the flowering of plant. If flowering of plant is more, the crop production also is more.
6. Wheat is important cereal crop gained a lot through Green Revolution by developing high yielding hybrid strains.

Question 2.
Ramaiah’s field is flattened. Somaiah’s field has many up and downs. Who will get more crop ?
Answer:

1. Generally the fields have a lot of ups and downs even after ploughing. So a leveller is used for levelling the soil.
2. By levelling the soil, it becomes flattened, water and nutrients can be reached to every part of the land. It also helps in sowing seeds and planting.
3. Because Ramaiah’s field is flattened, he will get more produce than Somaiah.

Question 3.
What are the advantages of ploughing?
Answer:
Before growing crops ploughing the soil properly is necessary.
Advantages:

1. Ploughing loosens the soil and it helps in easier transportation of air and water.
2. Water is stored deeply for a long time as the soil is soft.
3. Roots penetrate in the deep and can respire well as the air enters easily into the soil.
4. Friendly microorganisms and earthworms can grow well when the soil is soft.
5. Some foe microorganisms die due to the sun rays.

Question 4.
Treating with fungicides before sowing the seed is necessary. Why?
Answer:

1. Sometimes farmers wash seeds with chemicals to protect from pests.
2. Medication is done to keep away the seeds from the harmful microorganisms like bacteria, fungi etc.
3. So, generally farmers treat the seeds with fungicides before sowing the seeds before making them to germinate.

Question 5.
Why do farmers dry the paddy crop after cutting them from fields?
Answer:

1. Naturally food produce can be damaged by fungi, pests, rats and bacteria.
2. If moisture is also there in the grains, it helps to develop moulds (fungi).
3. Such grains neither germinates nor suitable to eat.
4. To overcome this problem farmers dry the grains for 2-3 days in sun.
5. After drying they keep the grains in jute bags and preserve them in godowns.

Question 6.
Give some examples of plants that grow after replanting.
Answer:
Seeds that are broadcast in a plot, grow into seedlings. When these plants grow to certain height, farmers pick out the plants (seedlings) from the plots, make bundles and are sown in proper distances. This is called transplantation (replanted)
E.g.: Rice (Sri Vari), Tobacco, Onions, Chillies etc.

Question 7.
Rahim removed weeds in his crop field, but David did not. Guess who get more yield. Why?
Answer:

1. Rahim get more yield than David. Because
2. Weeds are the unwanted or undesirable plants which grow in the fields and compete with crop plants for water, nutrients, light and space.
3. Because of these plants the crop plants may not grow properly. So they should be removed.
4. Otherwise the yield of the crop will be reduced.
5. Weeds give shelter for insects, pests and microorganisms and serve as a host for them.
6. Weeds are capable of germinating and growing fastern than crop plants. They flower and form seeds much earlier than the crop plants.
7. Some weeds disperse pollen grain to air which in turn causes respiratory diseases.

Question 8.
What is natural manure? How to prepare it? Give two examples.
Answer:
Natural manure: A manure made by the decomposition of plants and animal (organic) wastes is called natural manure or natural fertilizer or Bio Fertilizer.
Preparation:

1. These fertilizers are formed by decomposing plant and animal wastes.
2. In rural areas farmers keep these plant and animal wastes outside the village in open space.
3. Some bacteria like Azatobacter, Nitrobacter decompose and it becomes manure which contain nutrients.
4. Wherever the manure is added to the soil, there it provides nutrients to the plants. Examples : Vermi compost, Dung fertilizer.

Question 9.
Why do farmers plough their field during summer?
Answer:

1. Ploughing loosens the soil and it helps in easier transportation of air and water.
2. In summer temperature is very high. So the soil becomes dry. Then the soil becomes very loosly.
3. Some foe microorganisms die due to the sunrays.

Question 10.
Rajendar cultivated cotton crop in his field. He did not get sufficient yield. Can you guess the reasons? (OR)
A farmer cultivated cotton crop in his field. He did not get sufficient yield. Guess any four causes for it.
Answer:

1. Farmers in our state generally purchase seeds in the nearby market. The grains that are available in the packets play vital role in agriculture.
2. Sometimes the rate of germination of the seeds is not up to the mark, which was labelled on the packet.
3. Sometimes never germinate too.
4. At times, seeds grow into plants but they be sterile.
5. And sometimes the crop may be attacked by the larva of spotted brown boll-worm and pinkish boll-worms.
6. The larva of brown catter pillar sucks the juice from the leaves. The buds and the fruits of cotton plants drop off from the infected plants.
7. The larva of pinkish boll worm of cotton make hole in the stem, flower buds, flowers and fruits. As a result flower buds, cotton bolls drop off.
8. May be these reasons Rajendar did not get sufficient yield.

Question 11.
I am a plant. I grow in crop fields. Farmers pluck me. I do not know the reason. Can you tell who am I ?
Answer:

1. We observe some other plants growing along with the crop plants These are undesirable plants called weeds. They should be removed immediately.
2. The weeds, compete with the crop plants for nutrients, water and light, so he crop plants may not grow properly. This is the reason why they (weeds) should be removed.
3. Besides competition for food, light and water they also work as a carrier for different diseases. They also serve as host for different pests.
4. Some weeds disperse pollen grains to air which in turn causes respiratory diseases.
5. So the farmers pluck the weeds from crop plants by using different methods.

Question 12.
What do you observe in the experiment of dropping a fist of Bengal gram seeds in water?
a) What are the differences you observed in both the seeds?
Answer:
We can observe some seeds floated on water, the remaining sank under the water. The seeds which are floated are wrinkled and rough shaped but the sank seeds are round and smooth.
The floated seeds are light in weight but the sank seeds are more in weight.

b) Do you know why the floated seeds are lighter in weight?
Answer:
The floated seeds are not healthy, so they are lighter in weight.

c) Which seeds germinate well? Why?
Answer:
The seeds which sank in water germinate well because they are healthy.

d) Which seeds would not germinate properly? Why?
Answer:
The seeds which are wrinkled and rough would not germinate properly. Because the cotyledon inside the seed would not develop healthy.

Question 13.
Go to your nearest fertilizer shop and collect the information about chemical fertilizers and fill the table. Copy the following table in your notebook.
Answer:

Question 14.
Prepare a flow chart from ploughing to yielding in paddy.
Answer:

Question 15.
How do you appreciate the irrigation systems used in the drought prone areas?
Answer:

1. This method is employed when the availability of water is poor.
2. As the water reaches the plants drop by drop this is called Drip irrigation.
3. A long tube followed by small tubes attached to a motor. The tubes are made holes. So the water comes out from the tube.
4. The holes are arranged in such a way that it provide water exactly at the place where plant roots could receive water.
5. The man’s best technical method of utilizing the water in farming where the conditions are pravailing and in the areas where the availability of water is scanty.

Question 16.
Narendra sprayed over dose of pesticides on his cotton crop. Ramesh says it is a hazard to biodiversity and crop yield. Can you support Ramesh? How?
Answer:

1. In agriculture pests damage the crops. Almost all crops are generally effected by pests.
2. Wheat, Paddy and suagarcane are generally affected by fungal diseases. Groundnut is affected by Tikka disease. The catterpillars of spotted brown boll worm and Pinkish boll worm affect the crop.
3. A wide variety of agricultural and garden pesticides are available. A few derived from neem tobacco and chrysanthemum (Chamanthi) are less dangerous to other living organisms.
4. A wide variety of inorganic and organic pesticides are commonly used D.D.T. (Dichloro diphenoxy Trichloro ethane) BHC (Benzene Hexa Chloride), Chlordane, Endrin, Aldrin, Endosulfan and Diazinon pesticides are usually dusted or sprayed on crops or put in the soil.
5. But pesticides should not use unwisely. They get into the bodies of plants and animals in the soil and water. When these plants are eaten by animals like fish the pesticide get into their bodies.
6. A bird that eat the fish might get a concentrated lethal dose.
7. D.D.T. also accumulate in the egg shells, weakening them and making the shells break before hatching. It is observed D.D.T. is present in the milk of buffaloes and cows.
8. In this way pesticides are passed down the food chain and accumulate in the bodies of higher animals including human beings causing sickness and sometimes death.

Question 17.
Venkatesh observed the irrigation method for paddy field. He wanted to follow the same practice for his Maize crop. What suggestions do you give him?
Answer:

1. Paddy is grown as a Kharif or a Rabi crop. It requires high temperature of 22°C to 32°C and heavy rain fall. It is cultivated heavily in Kharif season.
2. Maize is cultivated in both Kharif and Rabi seasons but heavily in Kharif season. This requires high temeprature (35°C) with moderate rainfall.
3. So Venkatesh can follow the ame practice for his maize crop as the requirements of both the crops are almost same.

8th Class Biology 8th Lesson Production and Management of Food From Plants InText Questions and Answers

Question 1.
Look at the picture given below and write the constituents in it.

Answer:
Nitrogen (20%), Phosphorus (5%), Potash (10%)

Question 2.
Which manure is beneficial?
Answer:
Natural manure is beneficial.

Question 3.
Let us compare both, which manure is beneficial.

Observe the table carefully, discuss with your teacher and conclude which fertilizer is best to the farmers and why?
Answer:
Natural fertilizer is the best because this is made by the decomposition of plants and animal (organic) waste. Deposits of humus layer is found in the soil with less amount of Nitrogen, Phosphorus and potash deposits in the soil.
a) When do farmers irrigate the land?
Answer:
After applying manure farmers irrigate the land.

b) List out the water resources of your village.
Answer:
Wells, ponds, canals, tanks are the water resources.

c) Are they useful to your farmers?
Answer:
Yes. They are useful.

d) In what way the farmers of your village get water to the fields ?
Answer:
Farmers irrigate their fields either manually using bullocks or by using pumps.

Question 4.
What are the reasons for high production in Japan?
Answer:
Japan has cooler temperature.

Question 5.
What are the reasons for low production in India?
Answer:
High temperature and uncertainty of rainfall.

8th Class Biology 8th Lesson Production and Management of Food From Plants Activities

Activity – 1

Question 1.
CROPS IN INDIA:
Observe the following India map.

a) Are there many crops that are grow in most of the parts of our country? What are they?
Answer:
India is a unique position to grow almost every possible crop. It is the land of producing a variety of cereals like paddy, wheat, jowar, bajra, maize and ragi, pulses, spices, fruits, vegetables, oil seeds, fibre crops etc.

b) Why such crops are grown all over the country?
Answer:
India is an agricultural country and people derive their livelihood from agriculture. Agriculture is the back bone of Indian economy.

c) From the above map, which of them are grown in your village?
Answer:
Rice, pulses, banana, vegetables and leafy vegetables
a) Country: India
b) State: Andhra Pradesh
c) Your village: Khajipalem

d) How many days are required for getting the crop?
Answer:
Nearly 120 days.

e) Is time period for all crops are same?
Answer:
The time peirod for all the crops are not same. They are different from one crop to another.

f) Which crop needs more duration?
Answer:
Rice and Wheat.

Activity – 2

Question 2.
DURATION OF CROP:
a) Write the information in the table.

1. Example for long term crops:
Answer:
Jowar, red gram.

2. Example for short term crops:
Answer:
Pulses like green gram, black gram etc., and onions.

Activity – 3

Question 3.
WHERE DO CROPS ARE GROWN.
Discuss in groups and make a list of these things for the following table.

1. In which season do you find more varieties of vegetables in the market?
Answer:
Rainy season.

2. Generally farmers grow varieties of vegetables during rainy season. Can you guess the reason?
Answer:
In rainy season ponds, wells, rivers, ditches are pooled with water.

3. If we cultivate wheat in the month of the November what will happen?
Answer:
We get hot climate from February onwards. It is suitable for maturing the grains. That is the reason wheat is cultivated in the Rabi season only.

Activity – 4

Question 4.
PRODUCTION OF PADDY:
a) Go and collect the information through your nearest farmer and fill the following table.

a) In which season farmers get more benefits?
Answer:
Kharif season

b) Are there any other crops which are growing both Kharif and Rabi seasons?
Answer:
Paddy, wheat and maize

c) In which seasons farmers generally get good quality of seeds.
Answer:
Kharif and Rabi

d) The quantity of grains is higher in Kharif season than Rabi season. Do you agree this? Give your reasons.
Answer:
The climate, (the temperature, humidity with abundance of water supply) will be suitable in Kharif season

e) Do you know about third crop?
Answer:
The third crop season known as Zayad, grown in the months of April, May and June.

Activity – 5

Question 5.
SELECTION OF SEEDS:
Take some water in glass. Drop a fist of seeds in it. You cam observe some seeds will float on water. Collect those seeds and observe with hand lens and comparing with seeds sink under the water. Write your observations in the table.

a) What are the differences you observed in both seeds?
Answer:
Some seeds sank in water, some seeds floated.

b) Do you know why the floated seeds are light in weight?
Answer:
They are unhealthy seeds.

Activity – 6

Question 6.
GERMINATION AND SELECTION:
Show both the seeds in different pots and provide water uniformly, observe the growth of the plants in two pots and make a report.
1. Which seeds germinate well? Why?
Answer:
The seeds which are smooth and round germinate well because they are healthy seeds. Biology

2. Which seeds do not germinate properly? Why?
Answer:
The seeds which are wrinkled and rough do not germinate properly because they are unhealthy.

3. Were all the seeds were tested like this?
Answer:
The crop plant like Rice, wheat etc.

4. Do you know how the paddy seeds germinate?
Answer:
There are different stages in sprouting of the soaked rice seeds before it is planted.

5. Observe a sprout of paddy. Cam you say which part become root? Which part become shoot im the picture?
Answer:
Coleoptile become shoot and the part beneath the ground is root.

Activity – 7

Question 7.
SOWING METHODS
Collect information from the nearby farmers and fill in the table.

1. Why the seedlings are replanted at proper distance?
Answer:
They get water, mineral and sunlight equally when they are replanted at proper distance.

2. Do all the crops grow when replanted? Why not?
Answer:
Mostly all plants will grow.

Activity – 8

Question 8.
CROPS AND DISEASES:
Form a group with 4 to 5 of your classmates, visit nearby field, discuss with farmers about diseases effected by, and how to control them. If you do not know the name of the disease, write its local name or its characters.

1. Do all the farmers use the same pesticides for the same crop?
Answer:
For different crops different pesticides are used.

2. Is there any disease that you find in all fields?
Answer:
No

3. Where do they buy pesticides?
Answer:
From government and private agencies.

4. What are the appliances that they use to spray pesticides?
Answer:
Sprayer or dusters.

5. Did you find any other living organisms dying along with pests due to pesticides ? What are they ?
Answer:
Yes. Caterpillars, sparrows etc.

Activity – 9

Question 9.
IDENTIFICATION OF PESTS:
Observe the plants in a nearby field or in your school garden. Closely observe the leaves and stems to collect the following information. If the character is present put a ‘S’ mark and if there is no character put ‘X’ mark.
Name of the plant/crop : Rice (Blast of Rice)
Place: Prakasam

a. Do all the leaves of plant have spots?
Answer:
Yes, all the leaves of plant have spots.

b. Draw the leaf with these spots.
Answer:

c. What is your reason for the leaves which have cutting edges?
Answer:
Eaten by grasshoppers.

d. Do you find any twilted leaves with insects? How are they?
Answer:
Infection occurs on leaf sheaths.

e. Are the scars on the stems is same as spots on leaves?
Answer:
Yes, the scars on the stems is same as spots on leaves.

f. Collect powdery substance of the spots on leaves and observe under microscope. Write down your observations.
Answer:
Fungus produce small spores known as conidia.

Activity – 10

Question 10.
PEST CONTROLING PRACTICES:
In your village farmers control pests by using different pesticides and insecticides for different crops. For this they use different practices. Ask your elders the names of pesticides that they use in the following pest controlling practices.
Answer:

1. Spraying: Endrin, Diethane, M-45, Eldrine.
2. Dusting: Aldrin, D.D.T.
3. Put in the soil: Zinc, Sulphur, Phosphorous, fluorine
4. Burning and picking are also the practices where they use these: Sugarcane, citrus
5. Bio pesticides: Neem water

Observe the following pest controlling practices
a. Which of the above practices is good?
Answer:
A farmer remove the affected leaves from the plant and burnt them.

b. Why do you think so?
Answer:
By burning the affected leaves the pests will be controlled.

c. Why did the farmer use two pesticides at a time?
Answer:
At the first time the pests will be controlled by spraying pesticide, but if we use unwisely, pests become resistant to the pesticides.

d. What will we do to solve the problem?
Answer:
Pesticides will be used as per the requirement for the disease.

e. Farmers add manure to the soil.
What they used to add?
Answer:
They used to add nutrients to the soil.

f. Have you a compost pit in your school / house?
Answer:
Yes, there is a compost pit in our school. All the waste materials like dry leaves, fruits peel etc. will dump into the pit.

Activity – 11

Question 11.
WHEN SHOULD FARMERS IRRIGATE THE HELD?
Consult to the farmers and fill the table with the information to how and when they provide water to various crops.

a. Are all the crops provided with equal amount of water?
Answer:
No. Irrigation should be done according to nature of the soil and the type of crop to be grown.

b. Why do farmers provide more water to the summer crops?
Answer:
Summer season is the hottest climate so the crops which grow in summer requires high quantity of water.

Weedling:
a. Why should they (weeds) be removed?
Answer:
The weeds compet with the primary crops for nutrients, water and light because of these plants crop plants may not grow properly. So they should be removed.

Question 12.
How sprinklers and drip system are used and write down their merits and demerits.
Answer:
When the availability of water is poor, drip irrigation system is used. In this the water reaches the plant drop by drop through sprinklers. So that water comes out from the sprinklers wetting exactly the place of the roots of the plant.
Advantages:

1. Maximum use of available water.
2. No water being available to weeds.
3. Maximum crop yield.
4. Efficiency use of fertilizers.
5. Less weed growth.
6. Low labour and low operation cost.
7. No soil erosion.
8. Improved infiltration in soil with low in take.

Disadvantages:

1. Sensitivity to clogging.
2. Moisture distribution problem.
3. Salinity hazards
4. High cost compared to furrow.
5. High skills is required for design, install and operation.

Activity – 12

Question 13.
Ask your nearby nursery and know the weeds that grow in different crops. Make a table in your notebook.

Harvesting of paddy:
a. If the paddy is not dried well enough. What will happen?
Answer:
If moisture is there in paddy grains it helps to develop moulds (fungi). Such grains neither germinate nor suitable to eat.

b. Where do farmers harvest the crops in your village ?
Answer:
Farmers generally used to harvest by using traditional methods.

c. Is harvesting same for all crops?
Answer:
Yes.

Activity – 13

Question 14.
Find out the methods of harvesting in and around our village and fill the table.
Answer:

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions and Answers.

8th Class Maths 1st Lesson Rational Numbers InText Questions and Answers

Do this

Question 1.
Consider the following collection of numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125. Write these numbers under the appropriate category. [A number can be written in more than one group]  (Page No. 2)
Answer:
i) Natural numbers 1, 22
ii) Whole numbers 0, 1, 22
iii) Integers 0, 1, 22, -5, -2
iv) Rational numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125 etc.
Would you leave out any of the given numbers from rational numbers? No
Is every natural number, whole number and integer is a rational number? Yes

Question 2.
Fill the blanks in the table.     (Page No. 6)
Answer:

Question 3.
Complete the following table.     (Page No. 9)
Answer:

Question 4.
Complete the following table.      (Page No. 13)
Answer:

Question 5.
Complete the following table.      (Page No. 16)
Answer:

Question 6.
Complete the following table.      (Page No. 17)
Answer:

Question 7.
Represent – \(\frac{13}{5}\) on the number line.     (Page No. 22)
Answer:
Representing – \(\frac{13}{5}\) on the number line.

Try These

Question 1.
Hamid says \(\frac{5}{3}\) is a rational number and 5 is only a natural number. Shikha says both are rational numbers. With whom do you agree?       (Page No. 3)
Answer:
I would not agree with Hamid’s argument. Since \(\frac{5}{3}\) is a rational number. But ‘5’ is not only
a natural number, it is also a rational number.
Since every natural number is a rational number,
According to Shikha’s opinion \(\frac{5}{3}\), 5 are rational numbers.
∴ I agree with Shikha’s opinion.

Question 2.
Give an example to satisfy the following statements.        (Page No.3)
i) All natural numbers are whole numbers but all whole numbers need not be natural numbers.
ii) All whole numbers are integers but all integers are not whole numbers.
iii) All integers are rational numbers but all rational numbers need not be integers.
Answer:
i) ‘0’ is not a natural number.
∴ Every whole number is not a natural number. (∵ N ⊂ W)
ii) -2, -3, -4 are not whole numbers.
∴ All integers are not whole numbers. (∵ W ⊂ Z)
iii) \(\frac{2}{3}\), \(\frac{7}{4}\) are not integers.
∴ Every rational number is not an integer. (∵ Z ⊂ Q)

Question 3.
If we exclude zero from the set of integers is it closed under division? Check the same for natural numbers.    (Page No. 6)
Answer:
If ‘0’ is subtracted from the set of integers then it becomes Z – {0}.
Closure property under division on integers.
Ex: -4 ÷ 2 = -2 is an integer.
3 ÷ 5 = \(\frac{3}{5}\) is not an integer.
∴ Set of integers doesn’t satisfy closure property under division.
Closure property under division on natural numbers.
Ex: 2 ÷ 4 = \(\frac{1}{2}\) is not a natural number.
∴ Set of natural numbers doesn’t satisfy closure property under division.

Question 4.
Find using distributivity.     (Page No. 16)
A) \(\left\{\frac{7}{5} \times\left(\frac{-3}{10}\right)\right\}+\left\{\frac{7}{5} \times\left(\frac{9}{10}\right)\right\}\)
B) \(\left\{\frac{9}{16} \times 3\right\}+\left\{\frac{9}{16} \times-19\right\}\)
Answer:
Distributive law: a × (b + c) = ab + ac
A)

B)

Question 5.
Write the rational number for the points labelled with letters, on the number line.       (Page No. 22)
i)

ii)

Answer:
i) A = \(\frac{1}{5}\), B = \(\frac{4}{5}\), C = \(\frac{5}{5}\) = 1, D = \(\frac{7}{5}\), E = \(\frac{8}{5}\), F = \(\frac{10}{5}\) = 2.
ii) S = \(\frac{-6}{4}\), R = \(\frac{-6}{4}\), Q = \(\frac{-3}{4}\), P = \(\frac{-1}{4}\)

Think, discuss and write

Question 1.
If a property holds good with respect to addition for rational numbers, whether it holds good for integers? And for whole numbers? Which one holds good and which doesn’t hold good?     (Page No. 15)
Answer:
Under addition the properties which are followed by set of rational numbers are also followed by integers.

Question 2.
Write the numbers whose multiplicative inverses are the numbers themselves.      (Page No. 15)
Answer:
The number T is multiplicative inverse of itself.
∵ 1 × \(\frac{1}{1}\) = 1 ⇒ 1 × 1 = 1
∴ The multiplicative inverse of 1 is 1.

Question 3.
Can you find the reciprocal of ‘0’ (zero)? Is there any rational number such that when it is multiplied by ‘0’ gives ‘1’?
          (Page No. 15)
Answer:
The reciprocal of ‘0’ is \(\frac{1}{0}\).
But the value of \(\frac{1}{0}\) is not defined.
∴ There is no number is found when it is multiplied ‘0’ gives 1.
∵ 0 × (Any number) = 0

∴ No, there is no number is found in place of ‘A’.

Question 4.
Express the following in decimal form.     (Page No. 28)
i) \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\), \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\)
ii) Which of the above are terminating and which are non-terminating decimals?
iii) Write the denominators of above rational numbers as the product of primes.
iv) If the denominators of the above simplest rational numbers has no prime divisors other than 2 and 5 what do you observe?
Answer:
i) \(\frac{7}{5}\) = 0.4,
\(\frac{3}{4}\) = 0.75,
\(\frac{23}{10}\) = 2.3,
\(\frac{5}{3}\) = 1.66… = \(1 . \overline{6}\),
\(\frac{17}{6}\) = 2.833… = \(2.8 \overline{3}\),
\(\frac{22}{7}\) = 3.142
ii) From the above decimals \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\) are terminating decimals.
While \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\) are non-terminating decimals
iii) By writing the denominators of above decimals as a product of primes is

iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals.

Question 5.
Convert the decimals \(0 . \overline{9}\), \(14 . \overline{5}\) and \(1.2 \overline{4}\) to rational form. Can you find any easy method other than formal method?     (Page No. 31)
Answer:
Let x = \(0 . \overline{9}\)
⇒ x = 0.999 ……. (1)
The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 0.999
10x = 9.999 …….. (2)
From (1) & (2)

∴ x = 1 or \(0 . \overline{9}\) = 1
Second Method:
\(0 . \overline{9}\) = 0 + \(0 . \overline{9}\)
= 0 + \(\frac{9}{9}\)
= 0 + 1
= 1

Let x = \(14 . \overline{5}\)
⇒ x = 14.55 …….. (1)
The periodicity of the equation (1) is 1.
So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 14.55
10x = 145.55 …….. (2)

Second Method:

Let x = \(1.2 \overline{4}\)
⇒ x= 1.244 …….. (1)
Here periodicity of equation (1) is 1. So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 1.244
10 x = 12.44 …….. (2)

Second Method:

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots InText Questions and Answers.

8th Class Maths 6th Lesson Square Roots and Cube Roots InText Questions and Answers

Do this

Question 1.
Find the perfect squares between (i) 100 and 150 (ii) 150 and 200      [Page No. 124]
Answer:
i) The perfect squares between 100 and 150 are = 121, 144
ii) Perfect squares between 150 and 200 = 169, 196

Question 2.
Is 56 a perfect square? Give reasons.      [Page No. 124]
Answer:
Product of primes of 56 = 8 × 7 = (2 × 2) × 2 × 7
56 is not a perfect square. Since it can’t be written as product of two same numbers.

Question 3.
How many non perfect square numbers are there between 9² and 10²?      [Page No. 128]
Answer:
No. of non perfect square numbers between 9² and 10² are
= 2 × base of first number = 2 × 9 = 18
They are 82, 83, ……. 99.

Question 4.
How many non perfect square numbers are there between 15² and 16²?     [Page No. 128]
Answer:
No. of non perfect square numbers between 15 and 16 are = 2 × base of first number = 2 × 15 = 30
They are 226, 227, ……. 255,

Question 5.
Check whether the following numbers form pythagorean triplet.     [Page No. 129]
(i) 2, 3, 4
(ii) 6, 8, 10
(iii) 9, 10, 11
(iv) 8,15, 17
Answer:

Question 6.
Take a pythagorean triplet. Write their multiples. Check whether these multiples form a pythagorean triplet.      [Page No. 129]
Answer:
3, 4, 5 are pythagorean triplets.

From 6,8,10
⇒ 10² = 8² + 6²
⇒ 100 = 64 + 36
⇒ 100 = 100 (T)
From 9, 12, 5
⇒ 15² = 9² + 12²
⇒ 225 = 81 + 144
⇒ 225 = 225 (T)
∴ The multiples of pythagorean triplets are also pythagorean triplets.

Question 7.
By subtraction of successive odd numbers And whether the following numbers are perfect squares or not.        [Page No. 131]
(i) 55 (ii) 90 (iii) 121
Answer:
(i) √55
Step 1 → 55 – 1 = 54 (1st odd number be subtracted)
Step 2 → 54 – 3 = 51 (2nd odd number be subtracted)
Step 3 → 51 – 5 = 46 (3rd odd number be subtracted)
Step 4 → 46 – 7 = 39 (4th odd number be subtracted)
Step 5 → 39 – 9 = 30 (5th odd number be subtracted)
Step 6 → 30 – 11 = 19 (6th odd number be subtracted)
Step 7 → 19 – 13 = 6 (7th odd number be subtracted)
∴ 55 is not a perfect square number.
(∵ difference of consecutive odd numbers is not equal to ‘0’)

ii) √90
Step 1 → 90 – 1 =89 (1st odd number be subtracted)
Step 2 → 89 – 3 = 86 (2nd odd number be subtracted)
Step 3 → 86 – 5 = 81 (3rd odd number be subtracted)
Step 4 → 81 – 7 = 74 (4th odd number be subtracted)
Step 5 → 74 – 9 = 65 (5th odd number be subtracted)
Step 6 → 65 – 11 = 54 (6th odd number be subtracted)
Step 7 → 54 – 13 = 41 (7th odd number be subtracted)
Step 8 → 41 – 15 = 26 (8th odd number be subtracted)
Step 9 → 26 – 17 = 9 (9th odd number be subtracted)
∴ 90 is not a perfect square number.
(∵ difference of consecutive odd numbers is not equal to ‘0’)

iii) √121
Step 1 → 121 – 1 = 120 (1st odd number is subtracted)
Step 2 → 120 – 3 = 117 (2nd odd number is subtracted)
Step 3 → 117 – 5 = 112 (3rd odd number is subtracted)
Step 4 → 112 – 7 = 105 (4th odd number is subtracted)
Step 5 → 105 – 9 = 96 (5th odd number is subtracted)
Step 6 → 96 – 11 = 85 (6th odd number is subtracted)
Step 7 → 85 – 13 = 72 (7th odd number is subtracted)
Step 8 → 72 – 15 = 57 (8th odd number is subtracted)
Step 9 → 57 – 17 = 40 (9th odd number is subtracted)
Step 10 → 40 – 19 = 21 (10th odd number is subtracted)
Step 11 → 21 – 21 = 0 (11th odd number is subtracted)
∴ At the 11th step, the difference of consecutive odd numbers is ‘0’
121 is a perfect square number.
∴ √121 = \(\sqrt{11 \times 11}\) = 11 (∵ It ends at 11th step)

Question 8.
Which of the following are perfect cubes?     [Page No. 143]
(i) 243    (ii) 400    (iii) 500   (iv) 512     (v) 729
Answer:

∴ 512 and 729 are perfect cubes.

Try These

Question 1.
Guess and give reason which of the following numbers are perfect squares. Verify from the above table. (Refer table in Text Page no: 124)         [Page No. 124]
(i) 84   (ii) 108   (iii) 271   (iv) 240    (v) 529
Answer:
(i), (ii), (iii), (iv) are not perfect squares.
(v) 529 = 23 × 23
∴ 529 is a perfect square number.

Question 2.
Which of the following have one in its units place?     [Page No. 125]
(i) 126²    (ii) 179²    (iii) 281²     (iv) 363²
Answer:

Question 3.
Which of the following have 6 in the units place?
(i) 116²    (ii) 228²    (iii) 324²    (iv) 363²        [Page No. 125]
Answer:
i) 1162 ⇒ (6)² = 36 units digit = 6
ii) 2282 ⇒ (8)² = 64 units digit = 4
iii) 3242 ⇒ (4)² = 16 units digit = 6
iv) 3632 ⇒ (3)² = 9 units digit = 9
∴ Numbers which are having ‘6’ in its unit’s digit are: (i) 116² (iii) 324²

Question 4.
Guess, how many digits are there in the squares of i) 72   ii) 103    iii) 1000        [Page No. 125]
Answer:

Question 5.

27 lies between 20 and 30
27² lies between 20² and 30²
Now find what would be 27² from the following perfect squares.      [Page No. 125]
(i)329      (ii) 525     (iii) 529    (iv) 729
Answer:
The value of (27)² = 27 × 27 = 729

Question 6.
Rehan says there are 37 non square numbers between 9² and 11². Is he right? Give your reason.       [Page No. 128 ]
Answer:
No. of (integers) non perfect square numbers between 9² and 11²
= 82, 83, ……. 100 …… 120 = 39
But 100 is a perfect square number.
∴ Required non perfect square numbers are = 39 – 1 = 38
∴ No, his assumption is wrong.

Question 7.
Is 81 a perfect cube?      [Page No. 140]
Answer:
81 = 3 × 3 × 3 × 3 = 3⁴
No, 81 is not a perfect cube.
[∵ 81 can’t be written as product of 3 same numbers.]

Question 8.
Is 125 a perfect cube?       [Page No.140]
Answer:
125 = 5 × 5 × 5 = (5)³
Yes, 125 is a perfect cube.
[∵ It can be written as product of 3 same numbers]

Question 9.
Find the digit in units place of each of the following numbers.      [Page No. 141]
(i) 75³   (ii) 123³    (iii) 157³    (iv) 198³    (v) 206³
Answer:

Think, Discuss and Write

Question 1.
Vaishnavi claims that the square of even numbers are even and that of odd are odd. Do you agree with her? Justify.  [Page No. 125]
Answer:
The square of an even number is an even
∵ The product of two even numbers is always an even.
Ex: (4)² = 4 × 4 = 16 is ah even.
The square of an odd number is an odd.
∵ The product of two odd numbers is an odd number.
Ex: 11² = 11 × 11 = 121 is an odd.

Question 2.
Observe and complete the table:      [Page No. 125]

Answer:

Question 3.
How many perfect cube numbers are present between 1 and 100,1 and 500,1 and 1000?     [Page No. 140]
Answer:
Perfect cube numbers between 1 and 100 = 8, 27, 64
Perfect cube numbers between 1 and 500 = 8, 27, 64, 125, 216, 343
Perfect cube numbers between 1 and 1000 = 8, 27, 64, 125, 216, 343, 512, 729

Question 4.
How many perfect cubes are there between 500 and 1000?      [Page No. 140]
Answer:
Perfect cubes between 500 and 1000 = 512 and 729

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 SSC 10th Class Maths Solutions 6th Lesson Progressions InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions InText Questions and Answers.

10th Class Maths 6th Lesson Progressions InText Questions and Answers

Question 1.
Write three examples for finite A.P and three for infinite A.P. (Page No. 130)
Answer:
Examples for finite A.P
i) 3, 5, 7, 9, 11 where a = 3; d = 2.
ii) x, x + a, x + 2a, x + 3a, x + 4a, x + 5a where a = x; d = a.
iii) \(\frac{1}{9}\), \(\frac{2}{9}\), \(\frac{3}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{6}{9}\), \(\frac{7}{9}\), ….
where a = \(\frac{1}{9}\); d = \(\frac{1}{9}\).

Examples for infinite A.P
i) 10, 20, 30, 40, ……
where a = 10, d = 10.
ii) 5.5, 6.6, 7.7, 8.8, 9.9,……
where a = 5.5; d = 1.1.
iii) -100, -95, -90, -85,…..
where a = – 100, d = 5.

Do these

Question 1.
Write three examples for finite A.P and three for infinite A.P. (Page No. 130)
Answer:
Examples for finite A.P
i) 3, 5, 7, 9, 11 where a = 3; d = 2.
ii) x, x + a, x + 2a, x + 3a, x + 4a, x + 5a where a = x; d = a.
iii) \(\frac{1}{9}\), \(\frac{2}{9}\), \(\frac{3}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{6}{9}\), \(\frac{7}{9}\), ….
where a = \(\frac{1}{9}\); d = \(\frac{1}{9}\).
Examples for infinite A.P
i) 10, 20, 30, 40, ……
where a = 10, d = 10.
ii) 5.5, 6.6, 7.7, 8.8, 9.9,……
where a = 5.5; d = 1.1.
iii) -100, -95, -90, -85,…..
where a = – 100, d = 5.

(Page Nos. 131, 132)

Question 2.
Take any Arithmetic Progression.
Answer:
4, 7, 10, 13, 16, ……

Question 3.
Add a fixed number to each and every tetm of A.P. Write the resulting numbers as a list.
Answer:
4, 7, 10, 13, 16, …….
Adding ‘5’ to each term of the above A.P. we get
4 + 5, 7 + 5, 10 + 5, 13 + 5, 16 + 5,…
9, 12, 15, 18, 21, ……
In the list obtained the first term
a₁ = 9; a₂ = 12, a₃ = 15, a₄ = 18,
Also a₂ – a₁ = 12 – 9 = 3
a₃ – a₂ = 15 – 12 = 3
a₄ – a₃ = 18 – 15 = 3
……………………………………
i.e.,
d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = …. = 3
∴ The resulting list forms an A.P.

Question 4.
Similarly subtract’a fixed number from each and every term of A.P. Write the resulting numbers as a list.
Answer:
4, 7, 10, 13, 16, ……
Subtracting ‘2’ from the each term of A.P in given series, we get
4 – 2, 7 – 2, 10- 2, 13 – 2, 16 – 2, ……
2, 5, 8, 11, 14, ……
In the list obtained, the first term
a₁ = 2 , a₂ = 5, a₃ = 8, a₄ = 11,
Also a₂ – a₁ = 5 – 2 = 3
a₃ – a₂ = 8 – 5 = 3
a₄ – a₃ = 11 – 8 = 3
……………………………………
i.e., d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = 3
∴ The resulting list forms an A.P.

Question 5.
Multiply and divide each term of A.P by a fixed number and write the resulting numbers as a list.
Answer:
4, 7, 10, 13, 16, ……
Multiplying each term by 3, we get
4 × 3, 7 × 3, 10 × 3, 13 × 3, 16 × 3, ……
12, 21, 30, 39, 48, …….
In the list obtained the first term a₁ = 12 and a₂ = 21, a₃ = 30, ….
Also a₂ – a₁ = a₃ – a₂ = …… = 9
∴ The resulting list also forms an A.P.
Now divide every term by 7, we get
\(\frac{4}{7}\), \(\frac{7}{7}\), \(\frac{10}{7}\), \(\frac{13}{7}\), \(\frac{16}{7}\), …… is the resulting list.

Question 6.
Check whether the resulting lists are AP in each case.
Answer:
The first term
∴ d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = ….. = \(\frac{3}{7}\)
and the above list forms an A.P.

Question 7.
What is your conclusion?
Answer:
If a₁, a₂, a₃, …… are in A.P, then

i.e., “If each term of an A.P is added/ multiplied / divided by a fixed number, the resulting terms also form an A.P” and fixed term is subtracted from each term of an A.P, then the resulting terms also form an A.P.

Try these

Question 1.
i) Which of these are arithmetic progressions and why?  (Page No. 128)
a) 2, 3, 5, 7, 8, 10, 15, ……
Answer:
2, 3, 5, 7, 8, 10, 15, …… is not an A.P.
∵ a₂ – a₁ = 3 – 2 = 1
a₃ – a₂ = 5 – 3 = 2
a₄ – a₃ = 7 – 5 = 2
i.e., The difference between any two successive terms is not same throughout the series.
(or)
Every number is not formed by adding a fixed number to its preceding term.

b) 2, 5, 7, 10, 12, 15 ……
Answer: The given list does not form an A.P, since each term is not obtained by adding a fixed number to its preceding term.

c) -1,-3,-5,-7, ……
Answer: -1,-3,-5,-7,….. is an A.P.
a₂ – a₁ = – 3 – (- 1) = -3 + 1 = -2
a₃ – a₂ = – 5 – (-3) = -5 + 3 = -2
a₄ – a₃ = – 7 – (- 5) = -7 + 5 = -2
Every number is formed by adding a fixed number to its preceding term,

ii) Write 3 more Arithmetic Progressions.
Answer:
a) a = -7;d = -3 and
A.P. is-7, – 10, – 13, – 16, …….
b) a = 15; d = 4 and
A.P. is 15, 19, 23, 27, 31, …….
c) a = 100; d = 50 and
A.P. is 100, 150, 200, 250, ……..

Think & Discuss

(Page No. 129)

Question 1.
Think how each of the list given above form an A.P. Discuss with your friends.
a) Heights (in cm) of some students of a school standing in a queue in the morning assembly are 147,148, 149,…, 157.
Answer:
The given list forms an A.P, since each term starting from the second is obtained by adding a fixed number + 1 to its preceding term.

b) Minimum temperatures (in degree ‘ Celsius) recorded for a week, in the month of January in a city, arranged in ascending order are -3.1, -3.0, -2.9, -2.8, -2.7, -2.6, -2.5, …….
Answer:
The given list forms an A.P, since every term starting from the second is obtained by adding a fixed number +0.1 to its preceding term.

c) The balance money (in Rs.) after paying 5% of the total loan of Rs. 1000 every month is 950, 900, 850, 800, …, 50.
Answer:
The given list forms an A.P, since each term starting from the second is obtained by adding a fixed number (-50) to its preceding term.

d) Cash prizes (in Rs.) given by a school to the toppers of Classes I to XII are 200, 250, 300, 350, ….., 750 respectively.
Answer:
The given list forms an A.P, since each term starting from the second is obtained by adding a fixed number 50 to its preceding term.

e) Total savings (in Rs.) after every month for 10 months when Rs. 50 are saved each mouth are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
Answer:
The given list forms an A.P, since every term starting from the second term is obtained by adding a fixed number 50 to its preceding term.

Question 2.
Find the common difference of each of the above lists. Think when is it positive?
Answer:
Common difference d = a₂ – a₁
a) 148 – 147 = 1
b) -3.0 – (-3.1) = 0.1
c) 900 – 950 = -50
d) 250 – 200 = 50
e) 100 – 50 = 50
Common difference is positive when a₂ > a₁

Question 3.
Make a positive Arithmetic Progression in which the common difference is a small positive quantity.
Answer:
a = 50 ; d = 0.5 then A.P is 50, 50.5, 51, 51.5, 52, ……

Question 4.
Make an A.P in which the common difference is big (large) positive quantity.
Answer:
a = 100; d = 1000 then A.P. is 100, 1100, 2100, 3100, 4100, ……

Question 5.
Make an A.P in which the common difference is negative.
Answer:
a = 80, d = -7
then A.P. is 80, 73, 66, 59, 52, ……

Do these

(Page No. 143)

Find the sum of indicated number of terms in each of the following A.Ps.
i) 16, 11, 6, …..; 23 terms.
Answer:
Given: 16, 11, 6, …..; S₂₃
t₁ = a = 16; t₂ = 11; t₃ = 6,
d = t₂ – t₁ = 11 – 16 = -5

= -23 × 39 = -897

ii) -0.5, -1.0, -1.5,…..; 10 terms.
Answer:
Given : -0.5, -1.0, -1.5, …. S₁₀
a = – 0.5
d = t₂ – t₁ = (-1.0) – (-0.5)

iii) -1, \(\frac{1}{4}\), \(\frac{3}{2}\), …… ;10 terms.
Answer:
Given: -1, \(\frac{1}{4}\), \(\frac{3}{2}\), …… ;S₁₀.
a = – 1
d = t₂ – t₁ = \(\frac{1}{4}\) – (-1) = 1 + \(\frac{1}{4}\) = \(\frac{5}{4}\)

Do these

(Page No. 149)

Find which of the following are not G.P.

Question 1.
6, 12, 24, 48, ……
Answer:
Given: 6, 12, 24, 48, ……
a₁ = a = 6; a₂ = 12; a₃ = 24, a₄ = 48,…

The given list is of the form
a, ar, ar², ar³,
∴ The given numbers are in G.P.

Question 2.
1, 4, 9, 16, ……
Answer:
Given: 1, 4, 9, 16, …..
a₁ = a = 1
a₂ = 4; a₃ = 9, a₄ = 16

∴ The given numbers do not form a G.P.

Question 3.
1, -1, 1, -1, …..
Answer:
Given: 1, -1, 1, -1, …….
a₁ = a = 1
a₂ = -1; a₃ = 1, a₄ = -1, …..

∴ The given list forms a G.P.

Question 4.
-4, -20, -100, -500, ……
Answer:
Given: -4, -20, – 100, -500, ……
a₁ = a = -4, a₂ = -20, a₃ = -100, a₄ = -500,

∴ The given list forms a G.P.

Think & Discuss

(Page No. 149)

Question 1.
Explain why each of the lists above is a G.P.
i) 1, 4, 16, 64, 256, …….
Answer:
Here
a = 1 = a₁; a₂ = 4; a₃ = 16; a₄ = 64,….

i.e., Common ratio r = 4.

ii) 550, 605, 665.5, ……..
Answer:
The given series is in G.P. Since every term can be obtained by multiplying its preceding term by a fixed number ‘1.1’.

iii) 256, 128, 64, 32,…….
Answer:
The given series forms a G.P.
Since every term, starting from the second can be obtained by multiplying its preceding term by a fixed number \(\frac{1}{2}\).

iv) 18, 16.2, 14.58, 13.122, …….
Answer:
The given list forms a G.P.
Since each term, starting from the second can be obtained by multiplying its preceding term by a fixed number 0.9.
here

Question 2.
To know about a G.P. what is minimum information that we need?
Answer:
To know whether a number pattern forms a G.P or not, we should check that the ratio between the successive terms is equal or not.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles InText Questions and Answers.

10th Class Maths 8th Lesson Similar Triangles InText Questions and Answers

Do these

(Page No. 194)

Question 1.
Fill in the blanks with similar/not similar.
i) All squares are ………. (similar)
ii) All equilateral triangles are ………. (similar)
iii) All isosceles triangles are ………. (similar)
iv) Two polygons with same number of sides are ………, if their corresponding angles are equal and corresponding sides are equal. (similar)
v) Reduced and enlarged photographs of an object are ………. (similar)
vi) Rhombus and squares are ……… to each other. (not similar)

Question 2.
Write True / False for the following statements.
i) Any two similar figures are congruent.
Answer:
False
ii) Any two congruent figures are similar.
Answer:
True
iii) Two polygons are similar if their corresponding angles are equal.
Answer:
False

Question 3.
Give two different examples of pair of
i) Similar figures
ii) Non-similar figures
Answer:
i) Similar figures:
a) Any two circles
b) Any two squares
c) Any two equilateral triangles
ii) Non-similar figures:
a) A square and a rhombus
b) A square and a rectangle

Question 4.
What value(s) of x will make DE || AB, in the given figure?  (Page No. 200)
AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x.

Answer:
Given : In AABC, DE // AB AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x
By Basic proportionality theorem,
If DE // AB then we should have
\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = \(\frac{\mathrm{CE}}{\mathrm{EB}}\)
\(\frac{x+3}{8x+9}\) = \(\frac{x}{3x+4}\)
⇒ (x + 3) (3x + 4) = x (8x + 9)
⇒ x (3x + 4) + 3 (3x + 4) – 8x² + 9x
⇒ 3x² + 4x + 9x + 12 = 8x² + 9x
⇒ 8x² + 9x – 3x² – 4x – 9x -12 = 0
⇒ 5x² – 4x – 12 = 0
⇒ 5x² – 10x + 6x – 12 = 0
⇒ 5x (x – 2) + 6 (x – 2) = 0
⇒ (5x + 6) (x – 2) = 0
⇒ 5x + 6 = 0 or x – 2 = 0
⇒ x = \(\frac{-6}{5}\) or x = 2;
x cannot be negative.
∴ The value x = 2 will make DE // AB.

Question 5.
In △ABC, DE || BC. AD = x, DB = x – 2, AE = x + 2 and EC = x – 1. Find the value of x.    (Page No. 200)

Answer:
Given: In △ABC, DE // BC
∴ By Basic proportionality theorem, we have
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
\(\frac{\mathrm{x}}{\mathrm{x}-2}\) = \(\frac{x+2}{x-1}\)
⇒ x (x – 1) = (x + 2) (x – 2)
⇒ x² – x = x² – 4
⇒ -x = -4
∴ x = 4

Try This

Question 1.
E and F are points on the sides PQ and PR respectively of △PQR. For each of the following, state whether EF || QR or not?      (Page No. 197)
i) PE = 3.9 cm, EQ = 3 cm,
PF = 3.6 cm and FR = 2.4 cm.
Answer:

Here
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{3.9}{3}\) = \(\frac{1.3}{1}\)
\(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{3.6}{2.4}\) = \(\frac{3}{2}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) ≠ \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
Hence, EF is not parallel to QR.

ii) PE = 4 cm, QE = 4.5 cm,
Answer:
Here
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{4}{4.5}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PF}}{\mathrm{RF}}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{RF}}\)
∴ EF // QR
Hence, EF is parallel to QR.

iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 1.8 cm and PF = 3.6 cm.
Answer:

Given: PQ = 1.28 cm, PE = 1.8 cm
⇒ EQ = PE – PQ = 1.8 – 1.28
⇒ EQ = 0.52 cm
Also, PR = 2.56 cm, PE = 3.6 cm, FR = PF – PR = 3.6 cm – 2.56 cm
FR = 1.04 cm

∴ EF // QR (By converse of Basic proportionality theorem)
Hence, EF is parallel to QR.

Question 2.
In the following figures DE || BC.    (Page No. 198)
i) Find EC.

Answer:
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{AE}{EC}\)
⇒ \(\frac{1.5}{3}\) = \(\frac{1}{EC}\)
∴ EC = \(\frac{3}{1.5}\) = 2 cm

ii) Find AD.

Answer:

Think & Discuss

Question 1.
Can you give some more examples from your daily life where scale factor is used?    (Page No. 192)
Answer:
Scale factor is used in drawing maps, designing machines and in sculpture, etc.

Question 2.
Can you say that a square and a rhombus are similar? Discuss with your friends.Write why the conditions are not sufficient.       (Page No. 193)
Answer:
A square □ ABCD and a rhombus ▱ PQRS are not similar.
Though the ratio of their corresponding sides are equal, the corresponding angles are not equal.

but ∠A ≠ ∠P; ∠B ≠ ∠Q
∠C ≠ ∠R; ∠D ≠ ∠S

Try This

(Page No. 207)

Question 1.
Are the triangles similar ? If so, name the criterion of similarity. Write the similarity relation in symbolic form.
i)

!! ∠G = ∠I alt. int. angles for the ∠F = ∠K parallel lines GF // KI
Answer:
∠FHG = ∠IHK (Vertically opp. angle)
∴ ∠GHF and ∠IKH are similar by AAA similarity rule.
△GHF ~ △IKH.

ii)

Answer:
\(\frac{PQ}{LM}\) = \(\frac{6}{3}\) = 2;
\(\frac{QR}{MN}\) = \(\frac{10}{4}\) = 2.5;
\(\frac{PQ}{LM}\) ≠ \(\frac{QR}{MN}\)
△PQR and △LMN are not similar.
△PQR ~ △LMN

iii)

Answer:
∠A = ∠A (Common)
\(\frac{AB}{AX}\) = \(\frac{5}{3}\); \(\frac{AC}{AY}\) = \(\frac{5}{3}\)
∴ △ABC and △AXY are similar by SAS similarity condition.
△ABC ~ △AXY.

iv)

Answer:

∴ △ABC and △APJ are not similar.

v)

Answer:
∠A = ∠B = 90°
∠AOQ = ∠POB (∵ Vertically opposite angles)
∠Q = ∠P (alternate interior angles)
∴ △AOQ and △BOP are similar by AAA criterion.
△AOQ ~ △BOP.

vi)

Answer:
△ABC and △QPR are similar by AAA similarity condition.
△ABC ~ △QPR.

vii)

Answer:
∠A = ∠P
\(\frac{AB}{PQ}\) = \(\frac{2}{5}\); \(\frac{AC}{PR}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ \(\frac{AB}{PQ}\) ≠ \(\frac{AC}{PR}\)
Hence not similar.

viii)

Answer:
\(\frac{AB}{PQ}\) = \(\frac{6}{2.5}\); \(\frac{AC}{PR}\) = \(\frac{10}{5}\)
∴ △ABC and △PQR are not similar.

Question 2.
Explain why the triangles are similar and then find the value of x.
i)

Answer:
Given: In △PQR and △LTS
∠Q = ∠T; ∠R = ∠S = 90°
∴ ∠P = ∠T
(by angle sum property of triangles)
Hence, △PQR ~ △LTS [∵ AAA]

ii)

Answer:
Given: In △ABC and △PQC
∠B = ∠Q
[∵ ∠PQC = 180°- 110° = 70° – linear pair of angles]
∠C = ∠C [∵ Common]
∠A = ∠P [∵ Angle Sum property of triangles]
△ABC ~ △PQC by AAA similarity condition.
Then the ratio of their corresponding sides are equal.

iii)

Answer:
Given: In △ABC and △ECD
∠A = ∠E
∠ACB = ∠ECD [∵ Vertically opposite]
∴ ∠B = ∠D [∵ angle sum property]
∴ △ABC ~ △EDC

iv)

Answer:
Given: In △RAB and △RST
∠R = ∠R (common)
∠A = ∠S and ∠B = ∠T
[∵ Pair of corresponding angles for AB // ST]
∴ △RAB ~ △RST [∵ AAA similarity]

v)

Answer:
Given: In △PQR and △PMN
∠P = ∠P [∵ Common]
∠Q = ∠M [∵ Pair of corresponding angles for MN // QR]
∠R = ∠N
△PQR ~ △PMN [∵ AAA similarity]

[From the figure, PR = 4 + x]
⇒ 3 × 4 = 4 + x
⇒ x = 12 – 4 = 8

vi)

Answer:
Given: In △XYZ and △XBA,
∠X = ∠X [∵ Common]
∠B = ∠Y [∵ Pair of corresponding ∠A = ∠Z angles for AB // ZY]
∴ △XYZ ~ △XBA [∵ AAA similarity]

[From the figure, XZ = 7.5 + x]
\(\frac{3}{2}\) = \(\frac{7.5+\mathrm{x}}{\mathrm{x}}\)
3x = 15 + 2x;
3x – 2x = 15
x = 15

vii)

Answer:
Given: With the given conditions, we can’t find the value of x.
Note: If it is given that ∠A = ∠E then
we can say that △ABC ~ △EDC by AAA rule

viii)

Answer:
In △ABC and △BEC
∠ABC = ∠CEB (given)
∠C = ∠C (Common angle)
∴ △ABC ~ △BEC
(A.A. Criterion similarity)

Think & Discuss

(Page No. 203)

Question 1.
Discuss with your friends that in what way similarity of triangles is dif¬ferent from similarity of other polygons?
Answer:
In two triangles if the corresponding angles are equal then they are similar, whereas in two polygons if the corre-sponding angles are equal, they may not be similar, i.e., In triangles,
(Pairs of corresponding angles are equal) ⇔ (Ratio of corresponding sides are equal).
But this is not so with respect to polygons.

Do This

Question 1.
In △ACB, ∠C = 90° and CD ⊥ AB. Prove that \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}\) = \(\frac{BD}{AD}\).    (Page No. 218)

Answer:
Proof: △ADC and △CDB are similar.

[Ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides.]
From (1) and (2),
\(\frac{BD}{AD}\) = \(\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}\) (Q.E.D.)

Question 2.
A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12 m high. Find the width of the street.    (Page No. 218)
Answer:

Let A and D be the windows on the either sides of the street.
From Pythagoras theorem,
AC² = AB² + BC²
15² = 9² + BC²
BC² = 225 – 81
BC = √144 = 12 ….. (1)
Also, CD² = DE² + CE²
15² = 12² + CE²
CE² = 225 – 144
CE = √81 = 9
∴ BE = BC + CE = 12 + 9 = 21
Width of the street = 21 m.

Question 3.
In the given figure if AD ⊥ BC, prove that AB² + CD² = BD² + AC².    (Page No. 219)

Answer:
Given: In △ABC, AD ⊥ BC.
R.T.P: AB² + CD² = BD² + AC²
Proof: △ABD is a right angled triangle
AB² – BD² = AD² ……. (1)
△ACD is a right angle triangle
AC² – CD² = AD²
From (1) and (2)
AB² – BD² = AC² – CD²
AB² + CD² = BD² + AC²

Think & Discuss

Question 1.
For a right angled triangle with integer sides atleast one of its measurements must be an even number. Why? Discuss this with your friends and teachers.    (Page No. 215)
Answer:

Let l, m, n are integer sides of a right
angled triangle.
then l² – m² + n²
⇒ n = l² – m² = (l + m) (l – m)
Now
Case I: Both l, m are even the (l + m) is even then (l + m) (l – m) is also even. So ‘n’ is even. Here all are even.
Case II: Both l, m are odd then (l + m) and (l – m) become even. Then the product of even numbers is even so ‘n’ is even.
Here only ‘n’ is even.
Case III: If we consider l is even, m is’ odd then ‘n’ will be odd. So here T is even. We observe in all above three cases at least one of l, m, n is even; Hence proved.