AP Inter 2nd Year Maths 2A Formulas Chapter 5 ప్రస్తారాలు-సంయోగాలు

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 5 ప్రస్తారాలు-సంయోగాలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A ప్రస్తారాలు-సంయోగాలు Formulas

→ ప్రాథమిక సూత్రం: ఒక పని W1 ను ‘m’ విభిన్న విధాలుగానూ, మరొక పని W2 ను ‘n’ విభిన్న విధాలుగానూ చేయగలిగితే, ఈ రెండు పనులునూ ఒకేసారి ‘mn’ విభిన్న విధాలుగా చేయవచ్చు.

→ n అనేది ఋణేతర పూర్ణాంకం అయిన (i) 0! = 1 (ii) n! = n(n – 1)!

→ ఇచ్చిన వస్తువుల నుంచి (ఒకే విధంగా లేక విభిన్లు) కొన్ని లేదా అన్నీ ఎంచుకొని ఒక వరసలో (సరళరేఖలో) అమర్చడాన్ని ఒక ‘రేఖీయ ప్రస్తారం’ లేదా ‘ప్రస్తారం’ అంటాం.

→ n, r లు ధన పూర్ణాంకాలు, r ≤ n అయినప్పుడు ‘n’ విభిన్న వస్తువుల నుంచి ఒక్కొక్కసారి ‘r’ వస్తువుల చొప్పున తీసుకొంటే వచ్చే ప్రస్తారాల సంఖ్యను nPr తో సూచిస్తాం.
nPr = \(\frac{n !}{(n-r) !}\), 0 ≤ r ≤ n

AP Inter 2nd Year Maths 2A Formulas Chapter 5 ప్రస్తారాలు-సంయోగాలు

→ n, r లు ధన పూర్ణాంకాలు, r ≤ n అయిన

  • nPr = n . (n-1)P(r-1), r ≥ 1
  • nPr = (n) . (n – 1) (n-2)P(r-1), r ≥ 2
  • nPr = (n-1)Pr + r . (n-1)P(r-1)

→ n అసరూప వస్తువుల నుంచి ‘r’ వస్తువుల చొప్పున తీసుకొంటే వచ్చే ప్రసారాలలో

  • నిర్దేశించిన ఒక వస్తువు ఉండే ప్రస్తారాల సంఖ్య (r) (n-1)P(r-1)
  • నిర్దేశించిన ఒక వస్తువు లేని ప్రస్తారాల సంఖ్య (n-1)Pr
  • నిర్దేశించిన ఒక వస్తువు నిర్ధేశించిన స్థానంలో ఉండే ప్రస్తారాల సంఖ్య = (n-1)P(r-1)

→ A లో వున్న విభిన్న మూలకాల సంఖ్య n(A); B లో వున్న విభిన్న మూలకాల సంఖ్య n(B) అయితే n(A) ≤ n(B) అయినపుడు A నుండి B కు గల విభిన్న అన్వేక ప్రమేయాల సంఖ్య n(B)Pn(A)

→ n(A) = n(B) అయినపుడు A నుండి B కు గల ద్విగుణప్రమేయాల సంఖ్య n(A)!

→ A నుండి B కు గల ప్రమేయాల సంఖ్య [n(B)]n(A)

→ పునరావృతాన్ని అనుమతించినపుడు n విభిన్న వస్తువులతో, r వస్తువులుండేటట్లు ఏర్పరచగల ప్రస్తారాల సంఖ్య nr.

→ n విభిన్న శూన్యేతర 1, 2, 3, ……, 9 అంకెలను ఉపయోగించి పునరావృతం లేకుండా ఏర్పరచగల r స్థానాలు గల సంఖ్యల మొత్తం (n-1)P(r-1) × (దత్త అంకెల మొత్తం) × (111…. r సార్లు).

→ పైన చెప్పిన అంశంలోని n విభిన్న పూర్ణాంకాలలో ‘0’ కూడా ఉన్నప్పుడు ఏర్పరచగల 7 స్థానాలున్న సంఖ్యల మొత్తం = (n-1)P(r-1) × దత్త అంకెల మొత్తం × (111……. r సార్లు) (n-2)P(r-2) × (దత్త అంకెలమొత్తం × (111 – (r – 1) సార్లు)

→ n విభిన్న వస్తువుల నుంచి వచ్చే వృత్తాకార ప్రస్తారాల సంఖ్య = (n – 1)!

→ n విభిన్న వస్తువులతో ఏర్పడే పువ్వుల దండలు, పూసల గొలుసులు వంటి వేలాడే రకం వృత్తాకార ప్రస్తారాల సంఖ్య \(\frac{1}{2}\) (n – 1)!

→ ఇచ్చిన n వస్తువులలో p వస్తువులు ఒక రకంగానూ, q వస్తువులు మరొక రకంగానూ, r వస్తువులు వేరొక రకంగానూ ఉంటూ మిగిలిన వస్తువులు విభిన్నంగా ఉంటే, ఈ n వస్తువులను అమర్చడం ద్వారా వచ్చే ప్రస్తారాల సంఖ్య \(\frac{n !}{p ! q ! r !}\)

AP Inter 2nd Year Maths 2A Formulas Chapter 5 ప్రస్తారాలు-సంయోగాలు

→ n విభిన్న వస్తువుల నుండి ‘r’ వస్తువుల వంతున తీసుకొంటే వచ్చే సంయోగాల సంఖ్యను nCr తో సూచిస్తాం. మరియు nCr = \(\frac{n !}{(n-r) ! r !}\), 0 ≤ r ≤ n

→ n, r లు ధన పూర్ణాంకాలు, 0 ≤ r ≤ n అయితే nCr = nC(n-r)

→ n, r, s ధన పూర్ణాంకాలు 0 ≤ r ≤ n, 0 ≤ s ≤ n మరియు nCr = nCs అయితే r = s లేదా r + s = n అవుతుంది.

→ (m ≠ n) అయినపుడు (m + n) విభిన్న వస్తువుల నుండి m, n వస్తువులు ఉన్న రెండు భాగాలుగా విభజించే విధానాల సంఖ్య (m + n)Cm = (m + n)Cn = \(\frac{(m+n) !}{m ! n !}\)

→ ఇట్లే m, n, p లు విభిన్న ధన పూర్ణాంకాలయినప్పుడు (m + n + p) వస్తువులను m, n, p వస్తువులున్న 3 భాగాలుగా విభజించే విధానాల సంఖ్య \(\frac{(m+n+p) !}{m ! n ! p !}\)

→ ‘mn’ విభిన్న వస్తువులను ‘m’ సమభాగాలుగా (ఒక్కొక్క భాగంలో n వస్తువులుండే విధంగా) విభజించే విధాల సంఖ్య \(\frac{(m n) !}{(n !)^m m !}\)

→ (mn) విభిన్న వస్తువులను m వ్యక్తులకు సమానంగా పంచే విధాల సంఖ్య \(\frac{(m n) !}{(n !)^n}\)

→ ఒక రకం సరూప వస్తువులు p, మరొక రకం సరూప వస్తువులు q, వేరొక రకం సరూప వస్తువులు r ఇచ్చినపుడు వాటి నుంచి ఒకటి లేదా అంత కంటే ఎక్కువ వస్తువులను ఎంచుకొనే విధానాల సంఖ్య (p + 1) (q + 1) (r + 1) – 1.

→ m ధన పూర్ణాంకాం మరియు \(p_1^{\alpha_i} \cdot p_2^{\alpha_2} \ldots \ldots p_k^{\alpha_k}\) అవుతూ, p1, p2, ……., pk లు విభిన్న ప్రధాన సంఖ్యలు α1, α2, ….., αk ధన పూర్ణాంకాలు అయినపుడు m కు గల ధన భాజకాల సంఖ్య (α1 + 1) (α2 + 1) …… (αk + 1), (1, m లతో కలిపి)

AP Inter 2nd Year Maths 2A Formulas Chapter 5 ప్రస్తారాలు-సంయోగాలు

→ n విభిన్న వస్తువుల నుండి 0, 1, 2,…… లేదా n వస్తువులతో ఏర్పడే సంయోగాల సంఖ్య nC0 + nC1 + nC2 + …… + nCr = 2n

→ n విభిన్న వస్తువుల నుండి ఒకటి లేదా అంతకంటే ఎక్కువ వస్తువులతో ఏర్పడే సంయోగాల సంఖ్య = 2n – 1

→ n భుజాలున్న బహుభుజిలో కర్ణాల సంఖ్య = \(\frac{n(n-3)}{2}\)

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment.

AP State Syllabus SSC 10th Class Biology Important Questions 9th Lesson Our Environment

10th Class Biology 9th Lesson Our Environment 1 Mark Important Questions and Answers

Question 1.
Suggest one alternative method in place of pesticides to protect crops?
Answer:
Alternative methods for using pesticides to save the crops from pests:

  1. Rotation of crops
  2. Studying the life histories of pests
  3. Biological Control
  4. Sterility
  5. Genetic Strains

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 2.
By taking two plants of your surroundings as examples, explain how they protect themselves against the animals which eat them.
Answer:

  1. Neem Tree: Neem leaves contain an alkaloid Nimbin to protect themselves from the animals which eat them.
  2. Cactus: They have thorns to protect themselves.
  3. Datura: Datura leaves gives bad odour.

Question 3.
Identify one food chain from your surroundings. Name the producers and different levels of consumers in that food Chain.
Answer:
Grass → Insects → Frog → Snake.
Producers – Grass.
Primary Consumers – Insects.
Secondary Consumers – Frog
Tertiary consumers – Snake.

Aquatic Plants → Insects → Fish → Crane.
Producers – Aquatic Plant.
Primary Consumers – Insects
Secondary Consumers – Fish
Tertiary consumers – Crane

Question 4.
Write the names of producers and consumers in the food chain, you have observed.
Answer:
Producers – Plants, Green Algae
Consumers – All Animals.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 5.
Write any two slogans to promote awareness among the people about Ecofriendly programs.
Answer:
a) Lets go green to get global clean.
b) If you disturb the nature, the nature will disturb you.
c) The best solution to arrest pollution is plantation.
d) Reduce the usage of plastic and reduce the pollution.

Question 6.
What happens if decomposers are removed from the food web?
Answer:

  1. If decomposers are removed from the food web then the biological cycles are not completed.
  2. If the decomposers are not present in an ecosystem the remains of the other organisms accumulate.

Question 7.
Observe the following given below. Draw the pyramid of numbers.
Grass → Goat → Man
Answer:
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 1

Question 8.
We can’t expect the world without sparrows. So how should be our concern towards their conservation?
Answer:

  1. Sparrows are useful to control harmful insects like locust which damage food grains.
  2. Chemical pesticides are the cause for destruction of sparrows and useful insects.
  3. By using biological methods we can conserve the sparrow population.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 9.
Human being is modifying agriculture lands and lakes into residential areas. What is its effect on Bio-diversity?
Answer:

  1. The shelter may not be provided for migratory birds.
  2. Food chain get disturbed.
  3. Decrease in the ground water level.

Question 10.
How do you protect the plants, which were planted in “Haritha Haaram” programme in your school?
Answer:
We protect the plants:

  1. Watering of plants at regular intervals.
  2. Fencing or gaurding of plants.
  3. Adoption of plants.
  4. Providing organic manure.

Question 11.
The figure given below represents a food pyramid. Study it and answer the following questions.
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 2
i) Which trophic level has maximum energy?
Answer:
T1 (or) Primary producers (green plants)
ii) Give one example for T4 trophic level.
Answer:
Lion, tiger, hawk, etc.

Question 12.
“We can’t imagine the world without insects and birds, conserve them.
Answer:
Methods to conserve insects and birds:

  1. Avoid indiscriminate usage of pesticides
  2. Protect the natural habitats of insects and birds.
  3. Development of bird sanctuaries
  4. Everyone should follow environmental ethics.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 13.
What is environment?
Answer:
The sum of physical and biological factors along with their chemical interactions that affect an organism is called environment.

Question 14.
What is biosphere?
Answer:
The world of living things is called biosphere.
(OR)
The life supporting zone on the earth is called biosphere.

Question 15.
What are the physical or abiotic factors in a biosphere?
Answer:
Land, air, water, sunlight, humidity etc. are the physical or abiotic factors in a biosphere.

Question 16.
What does a food chain show?
Answer:
Food chain shows that how the energy is passed from one organism to another.

Question 17.
How the terrestrial ecosystems are determined?
Answer:
The terrestrial ecosystems are determined largely by the variations in climatic conditions between the poles and equator.

Question 18.
Where did the Kilimanjaro mountain located?
Answer:
The Kilimanjaro mountain is located in equatorial Africa (present in Tanzania, Africa).

Question 19.
What is the main source of energy for all the organisms in an ecosystem?
Answer:
Sunlight is the main source of energy for all the organisms in an ecosystem.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 20.
What is food web?
Answer:
The network of a large number of food chains existing in an ecosystem is called a food web.

Question 21.
What does a food web indicate?
Answer:
A food web indicates that the number of possible links for food in an ecosystem and reflects the fact that the whole community is a complex inter-connected unit.

Question 22.
What is ‘niche’?
Answer:
The position of organisms in a food web.

Question 23.
What does the word ‘niche’ denotes?
Answer:
‘Niche’ is the term used to describe not only the animals position in the food web and what it eats but also its mode of life.

Question 24.
What is an ecological pyramid?
Answer:
The graphic representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called “Ecological pyramid”.

Question 25.
Who was the first one to introduce “Ecological pyramid”?
Answer:
Ecological pyramid was first introduced by a British Ecologist Charles Elton in 1927.

Question 26.
In ecological pyramids the producers are represented at?
Answer:
The producers are represented at the base of the ecological pyramids.

Question 27.
What is pyramid of number?
Answer:
A graphical representation designed to show the number of organisms at each tropic level in a given ecosystem is called “pyramid of number”.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 28.
What does the pyramid of number show?
Answer:
Pyramid of number shows the population of organisms at each tropic level in a food chain.

Question 29.
What does each bar represent in a pyramid of number?
Answer:
In a pyramid of number, each bar represents the number of individuals at each tropic level in a food chain.

Question 30.
When does the pyramid of number not look like a pyramid at all?
Answer:
If the producer is a large plant such as tree or if one of the organisms at any tropic level is very small, then the pyramid of number does not look like a pyramid.

Question 31.
What is biomass?
Answer:
Any type of plant or animal material that can be converted into energy is called biomass.

Question 32.
What is biofuels?
Answer:
The materials which are used for energy production are known as biofuels.

Question 33.
What is Pyramid of biomass?
Answer:
A graphical representation designed to show the quantity of living matter at each tropic level in a given ecosystem is called “Pyramid of biomass”.

Question 34.
Why Pyramid of biomass inverted in case of aquatic ecosystem?
Answer:

  1. In an aquatic ecosystem, the biomass of phytoplankton is quite negligible as compared to that of the crustaceans and small herbivorous fish that feed on these producers.
  2. The biomass of large carnivorous fish living on small fishes is still greater. This makes the pyramid of biomass inverted.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 35.
How much percentage of the biomass is transferred from one tropic level to the next in a food chain?
Answer:
The percentage of the biomass transferred from one tropic level to the next level in food chain is nearly 10 – 20%.

Question 36.
When does the species at the top of the pyramid get more energy?
Answer:
The species at the top of the pyramid get, more energy when the steps in a food chain are fewer.

Question 37.
What are bio-geochemical cycles?
Answer:
Flow of materials between organisms and their environment is called cycling of materials or bio-geochemical cycles.

Question 38.
What is ecological efficiency?
Answer:
The ratio between energy flows at different tropic levels among the food chain expressed as percentage is called ecological efficiency.

Question 39.
What is ten per cent law?
Answer:
During the transfer of energy from one tropic level to the next, only about ten per cent of the energy from organic matter is stored as flesh. This is called “Ten per cent law”.

Question 40.
Where do Kolleru one of the largest fresh water lakes in India exists?
Answer:
Kolleru is one of the largest fresh water lakes in India exists between West Godavari and Krishna districts of Andhra Pradesh.

Question 41.
What is the catchment area of the lake Kolleru?
Answer:
A catchment area of the Kolleru lake extends up to 6121 Km2.

Question 42.
Through which The lake Kolleru discharges its excess water into Bay of Bengal?
Answer:
The lake Kolleru discharges its excess water into Bay of Bengal through the twisty channel called Upputeru which is about 65 km long.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 43.
When did Government of Andhra Pradesh had declared the lake as Bird Sanctuary?
Answer:
In November 1999, Government of Andhra Pradesh had declared the lake as Bird Sanctuary.

Question 44.
What is the number of species of birds being hosted by the Kolleru?
Answer:
The Kolleru lake is hosting 193 species of birds.

Question 45.
What are the major sources of pollution in Kolleru lake?
Answer:
The major sources of pollution are agricultural runoff containing residues of several agrochemicals, fertilizers, fish tank discharges, industrial effluents containing chemical residues and different types of organic substances, municipal and domestic sewage.

Question 46.
What is the objective of “Operation Kolleru” by the ministry of environment and forest, Government of India?
Answer:
The objective of operation Kolleru by the ministry of environment and forest, Government of India is to bring back the ecological balance of Kolleru lake which is a gift of nature.

Question 47.
What is Bioaccumulation?
Answer:
The process of entering of pollutants in a food chain is known as Bioaccumulation.

Question 48.
What is Biomagnification?
Answer:
The tendency of pollutants to concentrate as they move from one tropic level to the next is known as Biomagnification.

Question 49.
What are pesticides?
Answer:
The chemical materials used to control pests that attack crop plants or live as parasites on the body of farm animals are called pesticides.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 50.
What is a perfect pesticide?
Answer:
The perfect pesticide is one which destroys a particular pest and is completely harmless to every other form of life.

Question 51.
Why did the egg breaking among the peregrines increase?
Answer:
Egg breaking among the peregrines increased due to their disturbed behaviour caused by the nerve poisons that entered into their tissues through food chain.

Question 52.
Why did the aquatic biota is being contaminated?
Answer:
The aquatic biota is being contaminated with heavy metals due to industrialization and anthropogenic activities.

Question 53.
Why fish are considered to be the bioindicators of metal contamination?
Answer:
Fish are considered to be the bioindicators of metal contamination in environmental monitoring because fish species are strongly respond to stress conditions.

Question 54.
Where did Edulabad water reservoir located?
Answer:
Edulabad water reservoir is located in urban areas of Ranga Reddy district of Telangana.

Question 55.
Which fish species is grown in Edulabad water reservoir?
Answer:
Cyprinus carpio (common scale carp) is the fish species grown in Edulabad water reservoir.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 56.
What are the effects of bioaccumulation of metals in human beings that eat cyprinus carpio?
Answer:
The bioaccumulation of various metals in cyprinus carpio cause disorders. Such as hypertensions, sporadic fever, renal damage, nausea, etc.

Question 57.
In which country sparrows were hunted extensively in 1958?
Answer:
In China sparrows were hunted extensively in 1958.

Question 58.
In your opinion what are the effective methods to control pests?
Answer:
Rotation of crops, biological control, developing genetically modified plants are the effective methods to control pests in my opinon.

Question 59.
Why the temperatures are very high during the day and cold during the nights in deserts?
Answer:
In deserts, the rainfall and humidity are very low, so the sun’s rays easily penetrate the atmosphere making ground temperatures very high during the day. But the nights are often cold as the earth loses heat rapidly.

Question 60.
How can we draw a food chain?
Answer:
We can draw a food chain by connecting the pictures or names of organisms by putting arrows between them. These arrows should always point from food to the feeder.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 61.
How many types of ecological pyramids are there in practice? Name them.
Answer:
There are three types of ecological pyramids. They are:

  1. Pyramid of number
  2. Pyramid of biomass and
  3. Pyramid of energy.

Question 62.
Which process helps to convert the solar energy into suitable form of energy for animals to consume?
Answer:
Photosynthesis helps to convert the solar energy into suitable form of energy (food) for animals to consume.

10th Class Biology 9th Lesson Our Environment 2 Marks Important Questions and Answers

Question 1.
Grass → Grasshopper → Frog → Snake → Hawk
What will happen if we remove Frog from the above food chain? Explain.
Answer:

  1. Frog is secondary consumer in this food chain.
  2. If we remove frog from the food chain, the number of grasshopper will increase on other hand the number of snakes which depend on frogs will decrease.
  3. Hence, ecological balance may be damaged.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 2.
Observe the diagram and answer the following.
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 4i) Write any two food chains from the diagram.
ii) What are the secondary consumers in the food chain that are written by you?
Answer:
i) a) Plants → Goat → Tiger
b) Plants → Rabbit → Wolf / Fox
ii) Tiger, Wolf / Fox

Question 3.
Study the given paragraph and answer the questions.

Solar energy from sun enters into the producers of an ecosystem. No organisms except green plants and photosynthetic bacteria can absorb solar energy and convert it into chemical energy.

A) What are the producers mentioned in the given paragraph?
B) What form is energy converted into photosynthesis? In Photosynthesis, which form is energy converted into?
Answer:
A) Green plants and photosynthetic bacteria.
B) In photosynthesis, the light (or) solar energy is converted into chemical energy.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 4.
Explain the flow chart given below.
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 5Answer:
It is the pyramid of biomass.

  1. In this pyramid 10% of the food will reach to the next trophical level and so on at each level.
  2. It would take 1000 kg of phytoplankton to provide 100 kg of zooplankton and to form 1 kg of human tissue, 10 kg of frog is needed.
  3. The fewer the steps in the food chain, the more energy will be for the species at the top.

Question 5.
Observe the pyramid of number which is given below and answer the questions.
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 6i) As per the number of organisms in the tropic level, which group of organisms
are more in number and which are less in number?
ii) What happens if Secondary consumers disappear?
Answer:
i) If producers are more in number, then tertiary consumers are less in number,
ii) If secondary consumers disappear the primary consumers increase in number and the tertiary consumers found no food to live. It leads to death.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 6.
Explain in brief about the alternate methods to be followed to prevent the harmful effects of over usage of pesticides.
(OR)
Mention any four effective methods of controlling pests, which are less harmful on environment based on biological principles.
Answer:
Some of alternative pest control methods are

  1. Rotation of Crop: Growing different crops on a particular piece of land in successive years.
  2. Studying the life histories of the pests: When this is done it is sometimes possible to sow the crops at a time when least damage will be caused.
  3. Biological Control: Introducing Natural predator or parasite of the pest.
  4. Sterility: Rendering the males of a pest species sterile.
  5. Genetic Strains: The development of genetic strains (genetically modified plants) which are resistant to certain pest.
  6. Environmental ethics: People need to know besides laws regarding environment there are some basic ethics what is right and what is wrong in view of environment.

Question 7.
Write any 4 slogans on the necessity of forests and on their conservation.
Answer:

  1. Save the trees, save the earth. We are the guardians of nature’s birth.
  2. Don’t destroy the greenary and don’t spoil the scenery.
  3. Don’t make trees rare, we should keep them with care.
  4. To live for future in rest, saving forest is the best.

Question 8.
How does the given below concepts differs?
(a) Bioaccumulation b) Biomagnification
Answer:
a) Bioaccumulation: The process of entry of pollutants into a food chain is known as bioaccumulation.
b) Biomagnification: It is the tendency of pollutants to concentrate as they move from one tropic level to the next is known as biomagnification.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 9.
The biomass of a producer in an ecosystem is calculated as 3500 kgs. Calculate the biomass of primary, secondary, tertiary consumers.
Answer:
In a food chain roughly 90% of the food is lost at each step. So if the biomass of a producer in an ecosystem is calculated as 3500 kgs. the biomass of primary consumer as will be 350 kgs. and of secondary consumer is 35 kgs and biomass of tertiary consumer is 3.5 kgs.
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 7

Question 10.
Write a short note on food chain and food web.
Answer:

  1. Food chain is a pathway along which food is transferred from one tropic level to another tropic level beginning with producers.
  2. It shows who eats what in a particular habitat.
  3. The arrows between each item in the chain always point from the food to the feeder.
  4. For example
    Grass → Rabbit → Snake → Hawk
  5. The elaborate interconnected feeding relationships in an ecosystem is said to be food web.
  6. Many of the food chains in an ecosystem are crosslinked to form food web.
  7. For example,
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 8
  8. Food chain and food web help us to understand the food relations among living things.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 11.
Write a short notes on ecological pyramids.
Answer:

  1. The graphical representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called “Ecological pyramid”.
  2. It was first introduced by a British Ecologist Charles Elton in 1927.
  3. In the ecological pyramid, the producers (First tropic level) are represented at the base, and the successive tropic levels (primary, secondary and tertiary consumers) are represented one above the other with top carnivores at the tip.
  4. There are three types of pyramids.
    i) Pyramid of number ii) Pyramid of biomass and iii) Pyramid of energy.
  5. Pyramid of number shows the population of organisms at each tropic level in a food chain.
  6. Pyramid of biomass represents the available food as a source of energy at each tropic level in the food chain.
  7. Pyramid of energy represents the available energy at each tropic level in food chain.

Question 12.
Write a short notes on pyramid of number.
Answer:

  1. Pyramid of number is a graphical representation designed to show the number of organisms at each tropic level in a given ecosystem.
  2. The shape of this pyramid varies from eco-system to ecosystem.
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 9
  3. In aquatic and grassland ecosystems, numerous small autotrophs support lesser herbivores which support further small number of carnivores and hence the pyramid structure is upright.
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 10
  4. In forest ecosystem, less number of producers support greater number of herbivores who in turn support a fewer number of carnivores. Hence the pyramid structure is partly upright.
  5. In parasitic food chain, one primary producer supports numerous parasites which support still more hyperparasites. Hence the pyramid structure is inverted.
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 11

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 13.
Write a short notes on pyramid of biomass.
Answer:

  1. Pyramid of biomass is a graphical representation designed to show the quantity of living matter (biomass) at each tropic level in a given ecosystem.
  2. In terrestrial ecosystems, the biomass progressively decreases from producers to top carnivores. Hence the pyramid structure is upright.
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 12
  3. In an aquatic ecosystem, the biomass of phytoplankton (producers) is quite negligible as compared to that of crustaceans and small herbivorous fish that feed on these producers. The biomass of large carnivorous fish living on small fishes is still greater. This makes the pyramid of biomass inverted.
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 13

Question 14.
How do pesticides cause Bioaccumulation and Biomagnification?
(OR)
What are the effects of pesticides on environment?
Answer:

  1. Pesticides are the toxic chemicals used to destroy pest and insects which damage our crops and stored foods.
  2. These pesticides vary in their length of life as toxic materials.
  3. Some of the pesticides are degradable that can be broken down into harmless substances in a comparatively short time and others are non-degradable.
  4. Non-degradable pesticides accumulate in the bodies of animal and pass right through food web.
  5. Thus the pesticides cause bioaccumulation.
  6. These accumulated pesticides concentrate as they move from one tropic level to the next, thus leads to biomagnification.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 15.
List out some human activities which altered the communities of plants and animals in their natural ecosystem.
Answer:

  1. Industrialization
  2. Damming rivers
  3. Draining marshes
  4. Re-claiming land from the sea
  5. Cutting down forests
  6. Using chemical fertilisers and pesticides
  7. Building towns, cities, canals and motor ways.

Question 16.
What kind of changes may come in 2m ecosystem due to development of a large town?
Answer:
The following changes are expected due to development of a large town.

  1. Some plants and animal species will die out.
  2. Some will adapt to the new conditions sufficiently to survive in reduced numbers.
  3. Some will benefit by the new conditions and will increase in numbers.

Question 17.
Write a comparative note on pyramid of number and pyramid of biomass.
(OR)
Write the differences between pyramid of number and pyramid of biomass.
Answer:

Pyramid of numberPyramid of biomass
1. Pyramid of number is a graphical representation designed to show the number of organisms at each tropic level in a given ecosystem.1. Pyramid of biomass is a graphical representation designed to show the quantity of living matter at each tropic level in a given ecosystem.
2. It shows the population of organisms at each tropic level in a food chain.2. It represents the available food as a source of energy at each tropic level in the food chain.
3. This pyramid sometimes does not look like a pyramid at all. It may be upright like in grassland ecosystem, partly upright like in forest ecosystem or inverted like in parasitic ecosystem.3. This pyramid may be upright like in terrestrial ecosystem or inverted like in aquatic ecosystem.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 18.
Write a comparative note on pyramid of biomass and pyramid of energy.
(OR)
What are the differences between pyramid of biomass and pyramid of energy?
Answer:

Pyramid of biomassPyramid of energy
1. Pyramid of biomass is a graphical representation designed to show the quantity of living matter at each tropic level in a given ecosystem.1. Pyramid of energy is a graphical representation designed to show the quantity of energy present at each tropic level in a given ecosystem.
2. It represents the available food as a source of energy at each tropic level in the food chain.2. It represents the available energy at each tropic level in the food chain.
3. This pyramid may be upright like in terrestrial ecosystem or inverted like in aquatic ecosystem.3. This pyramid is always upright as only 10% of energy from one level is transfering to the next level.

Question 19.
Write a comparative note on pyramid of number and pyramid of energy?
(OR)
What are the differences between pyramid of number and pyramid of energy?
Answer:

Pyramid of numberPyramid of energy
1. Pyramid of number is a graphical representation designed to show the number of organisms at each tropic level in a given ecosystem.1. Pyramid of energy is a graphical representation designed to show the quantity of energy present at each tropic level in a given ecosystem.
2. It shows the population of organisms at each tropic level in a food chain.2. It shows the available energy at each tropic level in a food chain.
3. This pyramid sometimes does not look like a pyramid at all. It may be upright like in a grassland ecosystem partly upright like in forest ecosystem or inverted like in parasitic ecosystem.3. This pyramid is always upright, as only 10% of energy is transfering from one tropic level to other.

Question 20.
What is ecological efficiency? Write a short notes on Ten per cent law?
Answer:
Ecological efficiency: The ratio between energy flows at different tropic levels along the food chain expressed as percentage is called “ecological efficiency”.
Ten per cent law:

  1. The amount of energy transferred decreases with successive tropic levels.
  2. Slobodkin (1959) suggested that the transfer of energy from one tropic level to the next is of the order of 10% and this is called “Gross ecological efficiency”.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 21.
Geetha said “a given species may occupy more than one tropic level in the same ecosystem at the same time”. Do you support her or not? Explain your answer with example.
Answer:
I support her for the following reason.

  1. A snake eating a mouse in a field or lawn occupies the third tropic level.
    Plant → Mouse → Snake
  2. When the snake eats a frog in the same field, it occupies the fourth tropic level in a food chain. It is because the frog feed on some of the insects that depend on the plants.
    Plant → Insect → Frog → Snake
  3. Thus, a given species may occupy more than one tropic level in the same ecosystem at the same time.
  4. This is to satisfy its food needs, as it cannot do so by occupying one tropic level.

Question 22.
If we introduce a man into a forest ecosystem, at which level of food chain we will place him? Explain your answer.
Answer:

  1. If we introduce a man into a forest ecosystem, he can fit for any level of consumers of food chain.
  2. He may feed on plant parts such as fruits. Then we can place him at primary consumer level.
    Plant → Man
  3. He may feed on some of the herbivorous organisms such as rabbit, then we can place him at secondary consumers level.
    Plant → Rabbit → Man
  4. He may also feed on some of the carnivorous organisms such as insectivorous birds then we can place him at tertiary consumers level.
    Plant → Insect → Bird → Man
  5. This is possible to place him at any level of consumers, as he is an omnivore, who feed on both plant originated and animal originated foods.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 23.
Draw the ecological pyramids for the given food chain.
Banyan Tree → Herbivorous birds → Carnivorous birds.
Answer:
1) Pyramid of number
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 14
2) Pyramid of biomass
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 15
3) Pyramid of energy
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 16

Question 24.
Write briefly about Minamata disease.
Answer:

  1. Minamata disease was first discovered in Minamata city in Kumamoto prefecture, Japan, in 1956.
  2. It was caused by the release of methyl mercury in the industrial waste water from the Chisso corporation’s chemical factory, which continued from 1932 to 1968.
  3. This highly toxic chemical bioaccumulated in shellfish and fish in Minimata Bay and the Shiranui Sea, which, when eaten by the local populace, resulted in mercury poisoning.
  4. While cat, dog, pig and humans death continued for 36 years.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 25.
What are trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:

  1. Trophic levels is the feeding position in a food chain.
  2. It is the functional level occupied by an organism in a food chain.
  3. Examples of trophic levels include ‘herbivores’ and ‘decomposers’
  4. An example of food chain depicting various trophic levels is as follows:
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 17

Question 26.
What is the role of decomposers in the ecosystem? (OR)
How decomposers help in cleaning the environment?
Answer:

  1. If the decomposers are not present in an ecosystem the remains of the other organisms accumulate.
  2. Eventually the world would run out of carbon dioxide or nitrate or phosphate or other inorganisms material essential for life.
  3. The decomposers breakdown the organic waste products and dead remains of organisms into the inorganic substances needed by the producers.
  4. Most decomposition is carried out by saprophytic fungi, by bacteria and by invertebrates.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

10th Class Biology 9th Lesson Our Environment 4 Marks Important Questions and Answers

Question 1.
Prepare some slogans about ‘Vanam – Manam’ programme to display in your school rally?
Answer:

  1. Save paper – Save trees
  2. Plant a tree – Plant a life
  3. Saving trees is our duty
  4. Think green – Go green
  5. If we protect plants – they protects us
  6. Conserve plants – Conserve life
  7. Plant a tree – get the air free
  8. Plant a tree – Reduce the pollution.
  9. Tree on – Global warming gone.
  10. If cut a tree – It kills a life.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 2.
Read the information about Kolleru lake in the given table and answer the following questions.

ClassesArea in 1967 (Km2)Area In 2004 (Km2)
Lake-water spread area70.7062.65
Lake with sparse weed047.45
Lake with dense weed015.20
Lake-liable to flood in rainy season100.970
Aquaculture ponds099.74
Rice fields8.4016.62
Encroachment0.311.37
Total180.38180.38

a) In which year, lake water spread area is more?
b) Why do you think weeds are more in the lake?
c) Guess the reasons for decrease in the lake area.
d) What measures are to be taken to control pollution in the lake?
Answer:
a) In the year 1967
b) Addition of excessive nutrients from aquaculture ponds and rice fields.
c) Aquaculture ponds, Ricefields and Encroachment are the reasons for decrease in lake area.
d) Anthropogenic activities are to be controlled in the lake catchment area. (Or)
Fish ponds are to be removed in the lake catchment area. (Or)
Agricultural practices in the lake area should be minimised as per the norms of government.

Question 3.
Observe the following pyramid of biomass and answer the following questions.
a) This pyramid shows a decrease in the biomass as we move up, why the biomass is decreasing?
Answer:
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 18The pyramid of biomass for the given food chain, at each step 90% of the food is lost. That means 1000 kg of phytoplankton to produce 100 kg of Zooplankton to form 10 kg of fish to produce 1kg of human tissues. The fewer the steps in the food chain the more energy will be for the species at the top.

b) Give some examples of producers and primary consumers.
Answer:
Examples for producers: Plants, Grass, Diatoms.
Examples for primary consumers: Grasshopper, Rabbit, Deer,

c) Where do producers get the energy from?
Answer:
From the sun.

d) How much biomass is lost at each step?
Answer:
90%

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 4.
Observe the following diagram and answer the following questions.
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 4i) Name the primary producers in the given food web.
ii) Prepare any one food – chain from the diagram.
iii) What are the tertiary consumers?
iv) Write names of any two herbivores.
Answer:
i) Plants, Grass, Trees Phytoplanktons etc.
ii) Grass → Rabbit → Fox → Tiger
iii) Tiger, Vulture, Crane, Owl, Peacock etc.
(OR)
The animals which are at 4th trophic level in a food chain are called as Tertiary Consumers.
iv) Rabbit, Deer, Goat, Cow

Question 5.
What is number pyramid? What does it indicate?
Answer:

  1. The number of organisms in a food chain can be represented graphically in a pyramid of number.
  2. Each bar represents the number of individuals at each tropic level in a food chain.
  3. At each link in a food chain, from the first order consumers to the large carnivores, there is normally an increase in size but decrease in number.
  4. For example in a wood, the aphids are very small and occur in astronomical numbers.
  5. The lady birds which feed on them are distinctly larger and not so numerous.
  6. The insectivorous birds which feed on the lady birds are larger still and are only present in a small number and there may only be a single pair of hawks of much larger size than the insectivorous birds on which they prey.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 6.
Draw the diagram of number pyramid keeping foxes as third consumers. What are the consequences if their number increases?
Answer:

  1. If the number of foxes increases, then the competition for food will be very severe and less amount of food will be available for them.
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 19
  2. As a result some of the foxes may not get enough food and die due to starvation.
  3. This reduces the population of foxes and very few foxes will be left in the forest.
  4. This increases the chances of survival of secondary consumers birds, hence their number increases.
  5. This increases the availability of food for foxes. Very soon a balance will be established between the number of secondary consumers and foxes.

Question 7.
What reasons are responsible for decrease in number of top carnivores and biomass starting production in a food chain?
(OR)
Why the number of organisms get decreased as we move from producers to consumer levels?
Answer:

  1. In a food chain as we move from producers to different levels of consumers the energy available will decrease gradually.
  2. Only ten percent of the energy present in one tropic level transfer to another tropic level.
  3. Biomass also decreases gradually as only 10 – 20% of the biomass is transferred from one tropic level to the next in a food chain.
  4. As there is less energy of less biomass available at top levels, number of organisms also less generally.
  5. So the number of organisms get decreased as we move from producers to different level of consumer.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 8.
Show food chain of different organisms, number of pyramid of your school.
Answer:
Food chain of different organisms in our school:
Plant → aphids → spiders thirds.
AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 20

  1. The pyramid of organisms in a food chain can be represented graphically in a pyramid of number.
  2. Each bar represents the number of individuals at each tropic level in a food chain.
  3. At each link in a food chain, from the first order consumers to the large carnivores, there is normally an increase in size, but decrease in number.

Question 9.
What determines the terrestrial ecosystems on the earth?
Answer:

  1. The terrestrial ecosystems on the earth are being determined largely by the variations in climatic conditions between the poles and equator.
  2. The main climatic influences which determine these ecosystems are rainfall, temperature and availability of light from the sun.
  3. For instance, forests are usually associated with high rainfall, but the type Is influenced by temperature and light.
  4. The same applies to deserts which occur in regions where rainfall is extremely low.
  5. Thus, the climatic conditions along the horizontal climatic regions determined the terrestrial ecosystems on the earth.
  6. If we move from equatorial region to the polar region, we can come across tropical rain forests, savannah, deciduous forest, coniferous forests and then tundras respectively.
  7. Similarly altitude of the place is also a determining factor.
  8. If we climb a mountain such as Kilimanjaro in equatorial Africa, we can go through a comparable system of ecosystems, starting with tropical rain forest at the base and ending with perpetual snow and ice at the summit.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 10.
“All the energy in the ecosystem is ultimately derived from sunlight.” Justify.
Answer:

  1. All the organisms in an ecosystem derive energy from food.
  2. The food by its nature is the chemical energy and by in its stored form, it is the potential energy.
  3. In an ecosystem, all the consumers at any level depend upon producers for their food either directly or indirectly.
  4. The producers in any ecosystem are nothing but photosynthetic organisms such as plants, phytoplanktons and photosynthetic bacteria.
  5. Energy enters the producers in the ecosystem from the sun in the form of solar energy during photosynthesis.
  6. From the producers, the chemical energy passes to the consumers from one tropic level to the next through food.
  7. For example, in a grassland ecosystem, grass traps the solar energy and stores in its body.
  8. When this grass is eaten and assimilated by insects this stored energy enters into the body of insects.
  9. From the insects it will pass to frog, from them to snake and so on to eagle.
  10. Thus, all the energy in the ecosystem is ultimately derived from sunlight.

Question 11.
What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Answer:

  1. The tendency of pollutants to concentrate as they move from one trophic level to the next trophic level is known as Biomagnification.
  2. Plants absorb pesticides, heavy metals from the soil.
  3. The primary consumers when eat these plants the remaining of pesticides and heavy metals enter their bodies.
  4. As these chemicals are not degradable, they accumulate in the bodies of organisms of all trophic levels in the food chain.
  5. Most of the plants products which we eat are grown in fields in which pesticides and fertilisers have been used.
  6. These are absorbed by the plants and cannot be removed by washing or other means.
  7. Human beings are at the top level of the food chain these chemicals get accumulated in our bodies and cause various disorders.
  8. Levels of biological magnification would increase as the trophic level increases.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 12.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels ? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:

  1. If we remove producers from ecosystem, herbivores will not survive and the entire ecosystem collapse.
  2. Removing herbivores result in increase number of producers and carnivores would not get food.
  3. Removing carnivores result in increase of herbivores to unsustainable levels.
  4. If we remove decomposers from ecosystem waste material and animal dead remains would pile up and nutrients would not be available to the producers.
  5. Some or the other damage would be caused to the ecosystem if the organisms of any trophic level is removed.
  6. However impact of removing producers or decomposers would be serve as the whole ecosystem would collapse.
  7. Without plants sun’s energy cannot be converted to chemical energy which is the basis of life on earth.
  8. Without decomposers the nutrients cannot be recycled and made available to producers.

Question 13.
Every organism has got the right to live on this planet. Write slogans to motivate the people on preservation of biodiversity.
Answer:

  1. Live and let live.
  2. Conserve nature – conserve life.
  3. Clean the environment, live happily.
  4. Think eco-friendly and live eco-friendly.
  5. If we protect the environment, it protect us.
  6. Reduce pollution – conserve the biodiversity.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 14.
Write some friendly ecosystem activities you will conduct in your school.
Answer:

  1. Forming eco-clubs: These clubs consists of student representatives from each class. They will take up the eco-friendly activities and encourage the people of that village to follow environment friendly activities.
  2. Setting up garden at school: This ensures the school and its premises green through planting of flowering plants, vegetables and fruit trees. It is a symbol of biodiversity because various plants and animals inhabit the garden.
  3. Electricity conservation programme: To save energy the school implements certain hours to be switched off habit. This switching off programme for one hour from 3.30 p.m. to 4.30 p.m. help conserve electricity in every classroom.
  4. Pollution prevention programme: A ‘no burning of trash policy should be implemented in the school. Waste materials are recycled and properly disposed to ensure a clean, waste-free environment.
  5. Making compost by organic wastes: By digging a pit at one corner of the school and throwing the organic waste particularly of mid day meal waste into pit and covering with soil layers prepares compost which can be used as manure for plants. This creates a clean environment in the school.
  6. Using cloth bags instead of polythene bags by pupil.
  7. Collection of solid waste materials and proper management of its helps in reducing soil pollution.
  8. Children should be encouraged to follow ‘3R’ system i.e. Reduce, Re use and Recycle different substances.

Question 15.
What is Ecological pyramid? Describe different types of Ecological pyramids.
Answer:

  1. The graphical representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called ecological pyramid.
    AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment 9
  2. There are three types of ecological pyramids. They are
    1) Pyramid of number, 2) Pyramid of biomass and 3) Pyramid of energy.
  3. Pyramid of number is a graphical representation designed to show the number of organisms at each tropic level in a given ecosystem.
  4. The shape of this pyramid varies from ecosystem to ecosystem.
  5. In forest ecosystem the pyramid structure is partly upright and in parasitic food chain is inverted.
  6. Pyramid of biomass is a graphical representation designed to show the quantity of living matter (bio mass) at each trophic level in a given ecosystem.
  7. In terrestrial ecosystems, the biomass progressively decreases from producers to top carnivores hence the pyramid structure is upright whereas in aquatic ecosystem it is inverted.
  8. Pyramid of energy is a graphical representation designed to show the quantity of energy present at each tropic level in a given ecosystem. The pyramid of energy is always upright.

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

Question 16.
Collect information regarding pesticides commonly used in your area and prepare a chart showing pesticide and common name and on which crop and pest it is commonly used.
Answer:

S.No.PesticideCrop
1.ImidaclopriolCotton, ChilliesAphid, White fly, Jassids, Thrips
PaddyBPH, WBPH, GWT
2.TriazophosCottonBollworm
PaddyLeaf folder, Green leaf hopper, Hispa
3.ChlorpyriphosPaddyLeaf roller, Hispa gall midge, Stem borer, Whorl maggot
CottonAphid, Bollworm, White fly
4.MonocrotophosPaddyBrown plant hopper, Green leaf hopper, Leaf roller, Yellow stem borer
Maize Bengal gram Green gram Black gramShoot fly Pod borer
SugarcaneEarly shoot borer, Mealy bug
Cotton Oil seeds VegetablesInsects
5.AcephateCottonJassid, Bollworms
SunflowerAphids
6.DichlorvosPaddyLeaf roller
SugarcanePyrilla
Oil seeds, VegetablesInsect pests
7.AcetamipridCottonAphids, Jassids, White flies

AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

SCERT AP Board 8th Class Social Solutions 2nd Lesson Energy from the Sun Textbook Questions and Answers.

AP State Syllabus 8th Class Social Studies Solutions 2nd Lesson Energy from the Sun

8th Class Social Studies 2nd Lesson Energy from the Sun Textbook Questions and Answers

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 1.
Correct the false statements.
a) If a place is closer to the sea, irrespective of its distance from the equator, it will always be cooler.
Answer:
True

b) As you go up higher from the earth, it becomes warmer because the sun is closer to you.
Answer:
False.
Correction: As you go up higher from the earth, it becomes cooler, because the temperature decreases at the rate of 6°C for every ascent of 1000 metres.

c) Sun heats the air first and then the earth.
Answer:
False.
Correction: Sun heats the earth first and then the air.

d) Global warming is related to oxygen.
Answer:
False.
Correction: Global warming is related to carbon-di-oxide.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 2.
What’s the difference between the highest temperatue in Table 2 and the lowest temperature in Table 1?

Table 1

MonthMaximum°CMinimum °C
Jan3017
Feb3319
Mar3722
Apr3926
May3926
Jun3525
Jul3324
Aug3324
Sep3323
Oct3223
Nov3020
Dec2918

Table – 2

MonthMaximum°CMinimum °C
Jan3219
Feb3221
Mar3223
Apr3325
May3326
Jun3024
Jul2924
Aug2824
Sep2924
Oct3224
Nov3322
Dec3221

Answer:
The highest temperature in Table 2 = 33°C
The lowest temperature in Table 1 = 17°C
The difference = 16°C

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 3.
Suppose, the temperature in Moscow was – 8°C at 10 AM on 6 December. Twenty-four hours later it was 12°C higher. What was the temperature at 10 AM on 7 December?
Answer:
The temperature was 4°C.

Question 4.
Delhi and Mumbai are both situated on plains and their height above sea level is less than 300 metres. Why is there so much difference in their monthly average temperatures? In which months are the average temperatures in these two cities most similar? Can you explain?
Answer:
Mumbai experiences moderate climate while Delhi experiences extreme climate. This is because Mumbai’s coastal nature and tropical location ensures moderate temperature throughout the year. Delhi’s distance from the sea gives it an extreme type of climate.
The average temperatures in these two cities are more similar in the months of August and September.

Question 5.
Given below are the average monthly minimum and maximum temperatures of Jodhpur. Make a line graph of them. Which are the hottest and coldest months of the year?
Table: Average Monthly Maximum Temperatures in Jodhpur, Rajasthan (°C).

MonthJanFebMarAprMayJunJulAugSepOctNovDec
Minimum091217222729272524201411
Maximum252833384240363335363127

Answer:
a) April, May and June are the hottest months in the year.
b) December, January and February are the coldest months in the year.
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 1

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 6.
Given here are the average maximum temperatures of three places: A, B, and C. Make graphs of them. What can you guess about each place by looking at the Table and graphs?

PlaceJanFebMarAprMayJunJulAugSepOctNovDec
A232633384139343333332925
B-3161217212524211482
C313233323229292930303031

Answer:
A and C are in hot regions and B is in cold region.
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 2

Question 7.
Give three possible explanations for the differences between the average temperatures in Thiruvananthapuram and Shimla in January.
Answer:

  1. Tiruvananthapuram is situated on sea coast.
  2. Shimla is on high altitude.
  3. Tiruvananthapuram is near to the equator and Shimla is far away from the equator.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 8.
Between Bhopal, Delhi, Mumbai and Shimla, which two places show a similar temperature pattern? How can you explain the similarity between these two places?
Answer:
Bhopal and Delhi show a similar temperature pattern. This is because these two are far away from the sea.

Question 9.
Look at the graph of Minimum – Maximum temperature given below and answer the questions below.
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 3a) What is the average highest temperature in July?
Answer:
28°C

b) How warm does it usually get in December?
Answer:
26°C

c) How cool does it usually get in June?
Answer:
20°C

d) Is there a bigger difference between night and day temperature in May or in August?
Answer:
In the month of May

e) When is summer?
Answer:
March, April and May months.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 10.
Nithin says thermal power is better to use, but Padmaja says that Solar Energy is better. Which of them do you support? Why?
(OR)
How is solar energy better than thermal power?
Answer:
Solar energy is better than thermal power because solar energy is clean, breath-takingly abundant and is a responsible renewable resource to meet much of the world’s energy needs, as well as a fundamental need of our body for vitamin D.

Question 11.
Read the para under the title “Height and Temperature” and comment on it.

At the peak of summer some people go from the plains to hilly places such as Ooty or Shimla to avoid the heat. Even in the summer months, the temperatures are low on high hills. The highest parts of a mountain generally have the lowest temperatures. Temperatures decreases with elevation (height).

Answer:
The temperature decreases with altitude. On an average, the temperature decreases by 3.5°F for every 1000 ft of altitude (6.5°C for every 1000 metres).
Additional information: The lapse rate depends on, if the air is dry or moist – in dry air, the temperature decreases more quickly, in moist air more slowly.

8th Class Social Studies 2nd Lesson Energy from the Sun InText Questions and Answers

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 1.
Can you point out the difference between insolation and radiation? (Textbook Page No. 19)
Answer:
Insolation: The radiation received on the surface of a body is called insolation.
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 4Radiation: When a body gives out energy it is called radiation.

Question 2.
What will happen if the atmosphere gets more polluted with smoke and duct? (Textbook Page No. 19)
Answer:
Some amount of solar rays are absorbed or reflected away by smoke or dust in the atmosphere. If the atmosphere gets more polluted with smoke and dust, it may absorb more heat. As a result the temperatures on the earth may raise. If it reflects more, there will not be heat on the earth. Both are dangerous to ‘the life’ on ‘the earth’.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 3.
Can you tell why there is difference in heating of land and water? (Textbook Page No. 21)
Answer:
The water is transparent and lets energy pass through. On the other land, the sun heats up only the top layer of the land. This is because the land is solid. A solid is opaque and does not let energy through. This is the difference in heating of land and water.

Question 4.
a) Where will the rays fall more slanting – in Japan or North Pole? (Textbook Page No. 20)
Answer:
In North pole.

b) Where will Sun’s rays fall more intensively, in Andhra Pradesh or Rajasthan?
Answer:
In Andhra Pradesh.

c) If the Earth is flat and not curved, then which will get more heat – Japan or the Equator, or both equally?
Answer:
Both equally,

d) Look at the globe and say which countries wifi be hotter and which will be cooler?
Answer:

  1. Australia, Indonesia, Philippines, Cambodia, India, Chile, Brazil, Sudan, Uganda, etc. -hotter.
  2. North America, Iceland, Scandinavia and Russia, etc. – cooler.

Question 5.
To get an idea of other temperatures, measure and note the temperature of the following things. Before you start measuring, guess the temperature of each one. (Textbook Page No. 23)

ThingTemperature, °C
GuessMeasurement
Water in bucket
Ice
Giass of cold water
Warm bath water

Answer:

ThingTemperature, °C
GuessMeasurement
Water in bucket25°C35°C
Ice0°C0°C
Glass of cold water15°C10°C
Warm bath water70°C76°C

Question 6.
It is safer and advisable to use thermometers that have a scale of -10°C to 110°C. Using such a thermometer, also measure and note the temperature of boiling water and hot tea. (Textbook Page No. 23)
Answer:

  1. Temperature of boiling water: 100°C
  2. Temperature of hot tea: 95°C,

Question 7.
Look at the number line below, you can see how + and – numbers are marked. (Textbook Page No. 24)
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 13
a) Which temperature is greater: 5°C or -5°C?
Answer:
5°C

b) At which of these two temperatures will we feel colder?
Answer:
-5°C

c) How many degrees difference is there between -5° and 5°?
Answer:
10°C

d) Write in short form each of the following temperatures:
i) 88 degrees below zero, Celsius
Answer:
-88°C
ii) 38 degrees above freezing, Celsius
Answer:
38°C
iii) 32 degrees below freezing, Celsius
Answer:
-32°C

e) Did you note the temperature in your classroom today? 88 degrees below zero Celsius is how many degrees lower than the temperature you measured?
Answer:
Classroom temperature = 28°C It is 116°C low.

f) The temperature of a normal human body is 37°C. How much hotter than normal body temperature is 50°C?
Answer:
13°C

g) How much colder than the normal body temperature is -5°C?
Answer:
42°C

h) Arrange the following temperatures from the highest to the lowest:
12°C, -16°C, 29°C, 0°C, – 4°C.
Answer:
29°C, 12°C, 0°C, -4°C – 16°C.

i) At which of the above temperatures will we feel hottest?
Answer:
At 29°C

j) At which of the above temperatures will we feel coldest?
Answer:
At -16°C.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 8.
Have you been to places that have different climate than where you live? Describe it in the class. (Textbook Page No. 18)
Answer:
I live in Vijayawada It is very hot place. I have been to Ooty during last vacation. It is a hill station in Tamilnadu. It is situated on Nilgiri hills. I went from Vijayawada to Coimbatore and there to Ooty. It is ‘queen of hill stations’. It is too cool as it is situated on high altitude. There I stayed in a hotel opposite to ‘Horse Race Course’. There I visited Dodabetta peak, Love Dale, Wildlife Sanctuary, Botanical Gardens, Emerald lake etc., in Ooty. There I enjoyed the cold in summer. I took a resolution on the 1st January of this year that I would visit Ooty often in my life. I love the memories of my trip.

Question 9.
Table 1: (Average monthly temperature of Ananthapuram)

MonthMaximum,°CMinimum, °C
Jan3017
Feb3319
Mar3722
Apr3926
May3926
Jun3525
Jul3324
Aug3324
Sep3323
Oct3223
Nov3020
Dec2918

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 5
Use the data in the Table 1 and plot the average minimum monthly temperatures for Ananthapuram on the same graph paper in which the graph of maximum temperatures is drawn for your understanding. The first two months have already been done for you in the Graph 1.
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 6
Look at the data and the graph and answer the following questions about Ananthapuram:
a) How cold does it usually get in November in Ananthapuram?
Answer:
20°C

b) Which month has the highest maximum temperature in Ananthapuram?
Answer:
April and May, 39°C

c) What is the difference between the highest maximum temperature and the lowest maximum temperature in the year?
Answer:
22°C.

d) Which three months are the hottest in Ananthapuram?
Answer:
March, April and May.

e) Which three months are the coldest?
Answer:
December, January and February

f) What is the average maximum temperature in January in Ananthapuram?
Answer:
30°C

g) From June through December, the average minimum monthly temperature keeps falling in Ananthapuram. Does the average maximum monthly temperature also keep falling?
Answer:
Yes, It is also falling.

h) What is the difference between the maximum and minimum temperature in May?
Answer:
13°C

i) What is the difference between the maximum and minimum temperature in August?
Answer:
9°C

j) Based on your answers to the above two questions, is there a larger difference between the maximum and minimum temperatures in the summer or in rainy season in Ananthapuram?
Answer:
There is not much variation.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 10.
Table 2: (Average monthly temperature)

MonthMaximum,°CMinimum, °C
Jan3219
Feb3221
Mar3223
Apr3325
May3326
Jun3024
Jul2924
Aug2824
Sep2924
Oct3224
Nov3322
Dec3221

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 7
The average of maximum and minimum monthly temperatures are plotted on the Graph 2.
a) Which month has the lowest minimum temperature in Visakhapatnam? How much is it?
Answer:
In the month of January; It is 19°C.

b) Which is the hottest month in Visakhapatnam? How much was the average maximum temperature for that month?
Answer:
April, May and November months are too hot; 33°C.

c) Compare the temperatures of Ananthapuram and Visakhapatnam to answer the following:
i) In January, which place is colder?
Answer:
Ananthapuram
ii) In June, which place is hotter?
Answer:
Ananthapuram
iii) In which place, Ananthapuram and Visakhapatnam does the temperature remain more or less the same throughout the year?
Answer:
Visakhapatnam

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 11.
Observe the following graphs.
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 8AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 12
Now answer the following questions.
a) Can you think of any other reason for this state of temperature?
Answer:
Due to altitude, mean sea level, etc. the state of temperature differs.

b) What will happen if the inversion occurs?
Answer:
When inversion occurs cold air underlies warmer air.

c) How many metres higher than Delhi is Shimla?
Answer:
Shimla is 1900 metres higher than Delhi.

d) Based on the difference in elevation, calculate the likely difference in temperature between the two places.
Answer:
Approximately 12°C.

e) Which month has the highest maximum temperature in Shimla? How much is the temperature?
Answer:
May month, it is 22°C.

f) Which month has the highest maximum temperature in Deihi? How much is it?
Answer:
May month. It is 40°C.

g) In September, the average maximum temperature in Shimla is ———–°C while in Delhi, it is ———–°C.
Answer:
17°C, 34°C.

h) Which is colder: Delhi in January or Shimla in July?
Answer:
Delhi in January is colder.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 12.
Sometimes, especially in mornings of winters, coal temperatures are found near the ground. You can see dew drops on the grass due to condensation. The cooler temperatures near the ground level are due to less amount of insolation received due to the shorter days and excessive radiation due to the longer nights. This is known as inversion. (Textbook Page No. 28)
a) Can you think of any other reason for this state of temperature?
Answer:
I find no other reason.

b) What will happen if the inversion occurs?
Answer:
When inversion occurs cold air underlies warmer air.

Question 13.
Observe the following graph.
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 10
a) Which of the three places shown in the graph is located near to the Equator?
Answer:
Singapore

b) What is the average yearly temperature in that place?
Answer:
27.8°C

c) Does it usually get much warmer in the summer than in the winter there?
Answer:
No, there is slight difference.

d) Is summer in Vladivostok warmer than the winter in Singapore?
Answer:
No, there is much difference.

e) Does it usually get warmer in July in Singapore or in Shanghai?
Answer:
Singapore

f) Which of the three places on the graph has the most extreme climate?
Answer:
Vladivostok

g) What is the warmest month in Shanghai?
Answer:
July and August

h) What is the average yearly temperature there?
Answer:
15.3°C

i) Which month has the lowest average maximum temperature in this place?
Answer:
January and February.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 14.
Use the maps in your Atlas to find out the latitude and the average temperature in January of each of the following places: The first one has already been done for you.

PlaceLat.Temp, in January
Vijayawada, A.P.17° Nbetween 20 and 22.5° C
Agra, U.P.
Madurai, T.N
Nagpur, M.H

a) According to this map, there is no place in India that has an average January temperature higher than 30°C. (Remember, this is the average. There may be some January days in some places that do get hotter than 30°C.)
Look at the map and find out which parts of India usually have the highest average temperature (in January).
b) If you look north from this place on the map, is the average January temperature higher or lower?
Answer:

PlaceLat.Temp, in January
Vijayawada, A.P.17°Nbetween 20 and 22.5° C
Agra, U.P.27°N22.3°C- 8°C
Madurai, T.N9.93°N30°C – 20°C
Nagpur, MH21 °N28°C – 12°C

a) Southern parts of India have the highest average temperature.
b) It is lower.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 15.
Why is the North cooler in winter? (Textbook Page No. 30)
Look at the following table, it shows sunrise and sunset in different cities in India on 10th January.

PlaceSunriseSunset
Visakhapatnam, A.P.6:295:38
Agra, U.P.7:095:42
Madurai, T.N.6:376:12
Nagpur, MH6:535:48
Hyderabad, T.S.6:495:58
Kohima, Nagaland6:024:40

Answer the questions below:
a) In which of these six cities does the sun rise first?
Answer:
Kohima (Nagaland)

b) In which of these cities does the sun set last?
Answer:
Madurai.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 16.
Find out the locations of Singapore, Shanghai and Vladivostok on the following picture of the globe.
Answer:
AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun 11

Project work
Question 1.
You know that the Sun is the source of heat on the Earth. But why do you think the heat varies from morning to night or from season to season and from place to place? We are listing some variations here. Try to think a reason for it and discuss in the class before you proceed to read this chapter. (Textbook Page No. 18)

  1. Cool in the early morning and warmer in the afternoon.
  2. Warm in summer and cool in winter.
  3. Cool on hill tops and warm in the plains.
  4. Warm in Equatorial region and cold in Polar region.

Answer:
The heat on the earth varies from morning to night or from season to season and from place to place. The reasons are latitude, altitude, distance from the sea, ocean currents, mountain barriers, air masses and prevailing wind system.
Reason 1: In the morning the Sun’s rays are slanting on a particular place. In the afternoon they are straight on the same place. This is due to rotation.

Reason 2: During the summer the Sun’s rays hit the earth at a steep angle. The light does not spread out as much, thus increasing the amount of energy hitting any given spot. Also the long day light hours allow the earth plenty of time to reach warm temperatures. This is due to revolution.

Reason 3: The average rate of decrease of the temperature of normal air with the increase in height. It is equal to 6°C/km. (Normal Lapse Rate)

Reason 4: The Sun’s rays fall straight on the equatorial region and starting on the poles.
This is due to curvature of the earth.

AP Board 8th Class Social Studies Solutions Chapter 2 Energy from the Sun

Question 2.
Measure the temperature for a week in different months throughout the year. You will be able to see the temperature differences that occur between the summer, winter, monsoon, and other seasons. (Textbook Page No. 23)
a) For the next week, measure the air temperature each day at the same time and place (Remember to choose a place that Is in the shade). Each day before you measure, write down your guess. Keep your record in a separate notebook.
Place: ———–
Time: ———–
Month: ———–

DateAir Temperature, °C
GuessMeasurement

b) Record the temperature every day for one week for few months.
c) Calculate the weekly average temperatures.
d) Discuss the variations between different weeks.
Answer:
Place: Bengaluru
Time: 12 Noon
Month: January

Air Temperature, °C
DateGuessMeasurement
18.1.201628°C29°C
19.1.201627°C30°C
20.1.201629°C30°C
21.1.201629°C30 C
22.1.201628°C30°C
23.1.201627°C30°C
24.1.201628°C30°C

AP Inter 2nd Year Maths 2A Formulas Chapter 4 సమీకరణ వాదం

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 4 సమీకరణ వాదం to solve questions creatively.

AP Intermediate 2nd Year Maths 2A సమీకరణ వాదం Formulas

→ n రుణేతర పూర్ణసంఖ్య; a0, a1, a2, ……, an లు వాస్తవ లేదా సంకీర్ణ సంఖ్యలు, an ≠ 0 అయితే, అప్పుడు f(x) = a0 xn + a1 xn-1 + a2 xn-2 + ….. + an సమాసాన్ని x లో nవ తరగతి బహుపది అంటాం.

→ f(x) = a0 xn + a1 xn-1 + a2 xn-2 + ….. + an = 0 సమీకరణాన్ని nవ తరగతి బీజీయ సమీకరణం లేదా బహుపది సమీకరణం అంటాం. ఇచ్చట a0 ≠ 0

→ f(α) = 0 అయితే, సంకీర్ణ సంఖ్య α ను బహుపది f(x) = 0 సమీకరణానికి మూలం అని అంటాం.

→ f(α) = 0 అయిన f(x) సమాసానికి (x – α) ఒక కారణాంకం అవుతుంది.

AP Inter 2nd Year Maths 2A Formulas Chapter 4 సమీకరణ వాదం

→ సమీకరణ మూలాలు, గుణకాల మధ్య సంబంధం:
(i) x3 + p1x2 + p2x + p3 = 0 సమీకరణ మూలాలు α1, α2, α3, అయితే, అప్పుడు

  • s1 = α1 + α2 + α3 = -p1
  • s2 = α1α2 + α2α3 + α3α1 = p2
  • s3 = α1α2α3 = -p3

(ii) x4 + p1x3 + p2x2 + p3x + p4 = 0 సమీకరణ మూలాలు α1, α2, α3, α4 అయిన, అప్పుడు

  • s1 = α1 + α2 + α3 + α4 = -p1
  • s2 = α1α2 + α2α3 + α3α4 + α1α3 + α1α4 + α2α4 = p2
  • s3 = α1α2α3 + α2α3α4 + α3α4α1 + α1α2α4 = -p3
  • s4 = α1α2α3α4 = p4

→ ఒక ఘన సమీకరణ మూలాలు

  • అంకశ్రేఢిలో ఉంటే వాటిని a – d, a, a + d గా తీసుకొంటాం.
  • గుణశ్రేఢిలో ఉంటే \(\frac{a}{r}\), a, ar గా తీసుకుంటాం.
  • హరాత్మక శ్రేఢిలో ఉంటే వాటిని \(\frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}\) గా తీసుకొంటాం.

→ ఒక ద్వివర్గ సమీకరణ మూలాలు

  • అంకశ్రేఢిలో ఉంటే వాటిని a – 3d, a – d, a + d, a + 3d గా తీసుకొంటాం.
  • గుణశ్రేఢిలో ఉంటే వాటిని \(\frac{a}{r^3}, \frac{a}{r}\), ar, ar3 గా తీసుకొంటాం.
  • హరాత్మక శ్రేఢిలో ఉంటే వాటిని \(\frac{1}{a-3 d}, \frac{1}{a-d}, \frac{1}{a+d}, \frac{1}{a+3 d}\) గా తీసుకొంటాం.

→ వాస్తవ సంఖ్యలు గుణకాలుగా గల బహుపదీయ సమీకరణానికి సంకీర్ణ మూలాలు సంయుగ్మంగా ఉంటాయి.

AP Inter 2nd Year Maths 2A Formulas Chapter 4 సమీకరణ వాదం

→ అకరణీయ సంఖ్యలు గుణకాలుగా గల బహుపదీయ సమీకరణానికి కరణీయ మూలాలు సంయుగ్మాలు.
ఉదా: 2 + √3 ఒక మూలమైతే, 2 – √3 కూడా మూలం అవుతుంది.

→ α1, α2, ….., αn లు f(x) = 0 కు మూలాలైన,

  • 1, -α2, ……, -αn లు మూలాలుగా గల బహుపది సమీకరణం f(-x) = 0 అవుతుంది.
  • 1, kα2, ….. kαn లు మూలాలుగా గల సమీకరణం f(\(\frac{x}{k}\)) = 0, (k ≠ 0)
  • \(\frac{1}{\alpha_1}, \frac{1}{\alpha_2}, \ldots \frac{1}{\alpha_n}\) లు మూలాలుగా గల సమీకరణం f(\(\frac{1}{x}\)) = 0
  • α1 + h, α2 + h, ……., αn + h లు మూలాలుగా గల సమీకరణం f(x – h) = 0
  • α1 – h, α2 – h, ……, αn – h లు మూలాలుగా గల సమీకరణం f(x + h) = 0
  • \(\alpha_1^2, \alpha_2^2, \ldots \alpha_n^2\) లు మూలాలుగా గల సమీకరణం f(√x) = 0

→ f(x) = p0 xn + p1 xn-1 + p2 xn-2 + …… + pn = 0 సమీకరణంలో రెండవ పదం తొలగింపు చేయటానికి f(x) = 0 ను f(x + h) = 0 గా రూపాంతరం చెందించాలి. ఇచ్చట h = \(\frac{-p_1}{\text { (n) } p_0}\) అవుతుంది.

→ f(x) = 0 బహుపది సమీకరణంలో x బదులు \(\frac{1}{x}\) ప్రతిక్షేపించినా, ఆ సమీకరణంలో మార్పు లేనట్లయితే f(x) = 0 ను వ్యుత్కృమ సమీకరణం (Reciprocal equation) అంటారు.

→ f(x) = 0 వ్యుత్కృమ సమీకరణంలోని గుణకాలన్నీ pi = pn-i (i = 0, 1, 2, …., n) పాటిస్తే మొదటి కోవకు (Class one) చెందిన వ్యుత్త మ సమీకరణమనీ; pi = -pn-i; పాటిస్తే రెండో కోవకు (Class two) చెందిన వృత్తమ సమీకరణమనీ అంటాం.

→ మొదటి కోవకు చెందిన బేసి వ్యుత్ప్రమ సమీకరణానికి -1 ఒక మూలం అవుతుంది. ఈ సందర్భంలో రెండో కోవకు చెందిన వ్యుతమ సమీకరణానికి 1 మూలం అవుతుంది.

AP Inter 2nd Year Maths 2A Formulas Chapter 4 సమీకరణ వాదం

→ రెండో కోవకు చెందిన సరిపరిమాణ వ్యుతమ సమీకరణానికి 1, -1 లు మూలాలు అవుతాయి.

→ f(x) = 0 సమీకరణం n వ తరగతికి చెందినదై, దానిలో ‘r’ వ పదాన్ని తొలగించటానికి దానిని f(x + h) = 0 కు రూపాంతరం చెందించిన, (h స్థిరాంకం) f(n-r+1) (h) = 0 కావలయును (i.e.,.) f(x) యొక్క (n – r + 1) వ అవకలనం x = h వద్ద సున్నా కావలయును.

→ f(x) = 0 మూలాన్ని కనుక్కోవటానికి f(x) = 0 ని తృప్తిపరిచే x విలువను కనుక్కోవాలి. కొన్ని సందర్భాలలో పరిశీలన ద్వారా ఈ పని చేయవచ్చు. ఈ పద్ధతిని యత్న-దోష పద్ధతి అంటాం.

AP Inter 2nd Year Maths 2A Formulas Chapter 3 వర్గసమాసాలు

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 3 వర్గసమాసాలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A వర్గసమాసాలు Formulas

→ a, b, c లు వాస్తవ సంఖ్యలు లేదా సంకీర్ణ సంఖ్యలు అయి, a ≠ 0 అయినపుడు ax2 + bx + c రూపంలోని బహుపదిని, చలరాశి x లో వర్గ సమాసం అంటాం.
ఉదా: 4x2 – 2x + 3

→ aα2 + bα + c = 0 అయితే, సంకీర్ణ సంఖ్య ‘α’ ను ax2 + bx + c వర్గ సమాసానికి సున్న అంటాం.

→ a, b, c లు వాస్తవ సంఖ్యలు లేదా సంకీర్ణ సంఖ్యలు అయి, a ≠ 0 అయినపుడు ax2 + bx + c = 0 రూపంలో ఉన్న సమీకరణాన్ని చలరాశి x లో వర్గ సమీకరణం అంటాం. a, b, c లను ఈ సమీకరణ గుణకాలు అంటాం.
ఉదా: 2x2 – 5x + 6 = 0

→ aα2 + bα + c = 0 అయితే, సంకీర్ణ సంఖ్య α ను ax2 + bx + c = 0 సమీకరణానికి మూలం అనిగానీ, సాధన అనిగానీ అంటాం.

→ ax2 + bx + c = 0 వర్గ సమీకరణం మూలాలు \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

AP Inter 2nd Year Maths 2A Formulas Chapter 3 వర్గసమాసాలు

→ α, β లు ax2 + bx + c = 0 లు వర్గ సమీకరణ మూలాలు అయితే α + β = \(\frac{-b}{a}\); αβ = \(\frac{c}{a}\)

→ α, β లు మూలాలు గల వర్గ సమీకరణం x2 – (α + β)x + αβ = 0

→ వర్గ సమీకరణం మూలాల స్వభావం: ∆ = b2 – 4ac ని వర్గ సమాసం ax2 + bx + c, వర్గ సమీకరణం ax2 + bx + c = 0 ల “విచక్షణి” అంటాం.

→ α, β లు వర్గ సమీకరణం ax2 + bx + c = 0 కి మూలాలు అనుకోండి.
సందర్భం 1: a, b, c లు వాస్తవ సంఖ్యలు అయితే అప్పుడు

  • ∆ = 0 ⇔ α = β = \(\frac{-b}{2 a}\) (ax2 + bx + c = 0) కు ద్విరుక్త మూలం
  • ∆ > 0 ⇔ α, β లు విభిన్న వాస్తవ సంఖ్యలు.
  • ∆ < 0 ⇔ α, β లు వాస్తవేతర పరస్పర సంయుగ్మ సంకీర్ణ సంఖ్యలు.

సందర్భం 2: a, b, c లు అకరణీయ సంఖ్యలు అయితే అప్పుడు

  • ∆ = 0 ⇔ α, β లు సరిసమానమైన అకరణీయ సంఖ్యలు (= \(\frac{-b}{2 a}\), ద్విరుక్త మూలం).
  • ∆ > 0 శూన్యేతర అకరణీయ సంఖ్య యొక్క వర్గం ⇔ α, β లు అకరణీయ సంఖ్యలు.
  • ∆ ధనాత్మకం, కానీ అకరణీయ సంఖ్య యొక్క వర్గం కాదు ⇔ α, β లు సంయుగ్మ కరణులు.
  • ∆ < 0 ⇔ α, β లు వాస్తవేతర పరస్పర సంయుగ్మ సంకీర్ణ సంఖ్యలు.

→ \(a_1 x^2+b_1 x+c_1=0, a_2 x^2+b_2 x+c_2=0\) వర్గ సమీకరణాల మూలాలు, ఏకీభవించడానికి ఆవశ్యక, పర్యాప్త నియమం \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

→ \(a_1 x^2+b_1 x+c_1=0, a_2 x^2+b_2 x+c_2=0\) వర్గ సమీకరణాలకు ఉమ్మడి మూలం ఉండటానికి ఆవశ్యక, పర్యాప్త నియమం (c1a2 – c2a1)2 = (a1b2 – a2b1) (b1c2 – b2c1) మరియు ఉమ్మడి మూలం = \(\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\)

→ f(x) = ax2 + bx + c = 0 వర్గ సమీకరణంకు మూలాలు α, β లు అనుకుందాం.

  • c ≠ 0 అయితే, అప్పుడు αβ ≠ 0, \(\frac{1}{\alpha}, \frac{1}{\beta}\) లు మూలాలుగా గల సమీకరణం f(\(\frac{1}{x}\)) = 0;
  • α + k, β + k లు మూలాలుగా గల సమీకరణం f(x – k) = 0.
  • α – k, β – k లు మూలాలుగా గల సమీకరణం f(x + k) = 0.
  • -α, -β లు మూలాలు గల సమీకరణం f(-x) = 0.
  • kα, kβ లు మూలాలు గల సమీకరణం f(\(\frac{x}{k}\)) = 0

→ a, b, c ∈ R, a ≠ 0 అనుకొందాం. అప్పుడు R లోని అన్ని x లకు ax2 + bx + c = 0, ‘a’ లకు ఒకే గుర్తు ఉంటేనే ax2 + bx + c = 0 మూలాలు వాస్తవేతర సంకీర్ణ సంఖ్యలవుతాయి.

AP Inter 2nd Year Maths 2A Formulas Chapter 3 వర్గసమాసాలు

→ α, β లు ax2 + bx + c = 0 వర్గ సమీకరణం మూలాలై, α < β అయితే, అప్పుడు

  • α < x < β అయినప్పుడు, ax2 + bx + c, ‘a’ లకు వ్యతిరేక గుర్తులు ఉంటాయి.
  • x < α లేదా x > β అయినప్పుడు, ax2 + bx + c, ‘a’ లకు ఒకే గుర్తులు ఉంటాయి.

→ a < 0 అయితే, ax2 + bx + c వర్గ సమాసానికి, x = \(\frac{-b}{2a}\) వద్ద పరమ గరిష్ఠ విలువ ఉంటుంది.
గరిష్ట విలువ = \(\frac{4 a c-b^2}{4 a}\)

→ a > 0 అయితే, ax2 + bx + c వర్గ సమాసానికి x = \(\frac{-b}{2a}\) వద్ద పరమ కనిష్ఠ విలువ ఉంటుంది.
కనిష్ట విలువ = \(\frac{4 a c-b^2}{4 a}\)

AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Exercise 4.4

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles Ex 4.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.4

Question 1.
In the given figure, two lines p ∥ q and r are transversal, If ∠3 = 135°, then find the remaining angles.
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 1
Answer:
Given ∠3 = 135
∠1 = ∠3 = 135° (vertically opposite angles)
∠1 = ∠5 = 135° (corresponding angles)

∠3 = ∠8 = 135° (corresponding angles)
∠3 = ∠5 = 135° (Alternate interior angles)

∠1 + ∠7 = 180° (co-exterior angles are supplementary)
135° + ∠7 = 180°
∠7 = 180°- 135° = 45°
∠6 = ∠7 = 45° (vertically opposite angles)
∠6 = ∠2 = 45° (corresponding angles)
∠4 = ∠7 = 45° (corresponding angles)
∴ ∠1 = 135°, ∠2 = 45°, ∠3 = 135°, ∠4 = 45°, ∠5 = 135°, ∠6 = 45°, ∠7 = 45°, ∠8 = 135°

Question 2.
In the given figure, \(\overleftrightarrow{\mathbf{AB}}\) || \(\overleftrightarrow{\mathbf{CD}}\) and \(\overleftrightarrow{\mathbf{DE}}\) is a transversal. Find x.
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 2
Answer:
Given \(\overleftrightarrow{\mathbf{AB}}\) || \(\overleftrightarrow{\mathbf{CD}}\) and \(\overleftrightarrow{\mathbf{DE}}\) is a transversal line.
∠ABE = ∠CDB (corresponding angles)
∠ABF + ∠FBE = ∠CDB (we know ∠ABE = ∠ABF + ∠FBE)
From the figure ∠ABF = 35°,
∠FBE = x° and ∠CDB =100°
35° + x = 100°
⇒ 35° + x – 35° = 100° – 35°
∴ x = 65°

AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Exercise 4.4

Question 3.
In the given figure, m || n and p is transversal. Find x and y.
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 3
Answer:
Given m || n and p is transversal.
So, ∠x = 120° (vertically opposite angles)
45 + y = x (corresponding angles)
⇒ 45 + y = 120° (we know ∠x = 120°)
⇒ 45 + y – 45 = 120-45°
⇒ y = 75°
x = 120° and y = 75°

Question 4.
In the given figure, \(\overrightarrow{\mathbf{A B}}\|\overrightarrow{\mathbf{C D}}\| \overrightarrow{\mathbf{F E}}\). Find x, y and ∠AEC.
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 4
Answer:
Given \(\overrightarrow{\mathbf{A B}}\|\overrightarrow{\mathbf{C D}}\| \overrightarrow{\mathbf{F E}}\)
∠x = 20° (Alternate interior angles)
∠y = 33° (Alternate interior angles)
∠AEC = ∠AEF + ∠FEC
∠AEC = ∠x + ∠y
∠AEC = 20° + 33° = 53°
∴ x = 20°, y = 33° and ∠AEC = 53°

Question 5.
In the given figure, a transversal t intersects two lines p and q. Check whether p ∥q or not.
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 5
Answer:
If co-interior angles are supplementary, then the lines are parallel.
100° + 80° = 180° (co-interior angles supplementary)
So, p and q are parallel to each other.

Question 6.
In the given figure if l ∥m, then find x, y and z.
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 6
Answer:
Given l ∥m and AC is transversal.
z = 40° (Alternate interior angles)
l ∥m and AB is transversal. , and x + (y + z) = 180° (co-interior angles are supplementary)
x + y + z = 180°
(x + y) + 40° = 180° (co-interior angles are supplementary) (∵ z = 40°)
x + y + 40° = 180°
x + y + 40°-40° = 180° – 40°
∴ x + y = 140°
But x = y (given)
2x = 140°
∴ x = \(\frac{140^{\circ}}{2}\) = 70° = y
So, x = 70°, y = 70° and z = 40°

AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Exercise 4.4

Question 7.
In the given figure p, q, r and s are parallel lines and t is a transversal. Find x, y and z.
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 7
Answer:
p ∥ q
So, 80° + x = 180° (co-interior angles)
80° + x-80° = 180°-80°
x = 100°

q ∥ r
x + y = 180° (co-interior angles)
100° + y = 180° (we know x – 100°)
100° + y – 100° = 180°- 100°
y = 80°

r∥s
y = z (Alternate exterior angles)
y = z = 80° (we know y = 80°)
x = 100°, y = 80° and z = 80°

Question 8.
In the given figure \(\overrightarrow{\mathbf{A B}} \| \overrightarrow{\mathbf{C D}}\) and E is a point in between them. Find x + y + z. (Hint : Draw a parallel line to \(\overrightarrow{\mathbf{A B}}\) through E)
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 8
Answer:
Given \(\overrightarrow{\mathbf{A B}} \| \overrightarrow{\mathbf{C D}}\)
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 9
Draw \(\overrightarrow{\mathrm{EF}} \| \overrightarrow{\mathrm{AB}}\)
∠BAE = x,
∠AEF = p
∠FEC = q,
∠ECD = z
∠AEC = ∠AEF + ∠FEC
y = p + q

AB ∥ EF and AE is a transversal.
∠BAE + ∠AEF = 180° (co-interior angles are supplementary)
x + p = 180° ……………….(1)
CD ∥ EF and EC is a transversal.
∠FEC + ∠ECD = 180° (co-interior angles are supplementary’)
q + z = 180° …………….(2)

By adding (1) and (2)
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 10
x + y. + z = 360° (we know y = p + q)

Question 9.
Identify the pair of parallel lines in the given figure and write them.
AP Board 7th Class Maths Solutions Chapter Chapter 4 Lines and Angles Ex 4.4 11
Answer:
∠A = ∠E = 60° So, AL ∥ EH
∠B = ∠D = 90° So, BK ∥ DL
∠C = ∠F = 50° So, CJ ∥ FG

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.2

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers

Form a pair of linear equations for each of the following problems and find their solution.
Question 1.
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly income.
Answer:
Given ratio of incomes of two persons = 9 : 7
So let the incomes of each = Rs. 9x and Rs. 7x
and ratio of expenditures = 4 : 3
So let the expenditures of each = 4y and 3y
then earnings of each = (income – expenditure) of each
⇒ 9x – 4y = Rs. 2000 and 7x – 3y = 200
∴ 9x – 4y = 7x – 3y = 2000
⇒ 9x – 7x = 4y – 3y
⇒ y = 2x
now putting y = 2x in 9x – 4y = 2000 we get
9x – 4(2x) = 2000 ⇒ x = 2000
∴ Income of each = 9x = 9(2000) = 18000
and 7x = 7(2000) = 14,000

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2

Question 2.
The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Answer:
Let the digit in units place be x
and the digit in tens place be y
then the value of the number = 10y + x
Number obtained by reversing the digits = 10x + y
By problem,
(10y + x) + (10x + y) = 66
and x – y = 2
⇒ 11x – 11y = 66 and x – y = 2
⇒ x + y = 6 and x – y = 2
Solving these two equations
x + y = 6
x – y = 2
(+) 2x = 8
x = \(\frac{8}{2}\) = 4
Substituting x = 4 in x + y = 6
we get 4 + y = 6 ⇒ y = 2
Substituting x, y values in equations (10y + x) & (10x + y),
We get 10y + x
= 10(2) + 4 = 20 + 4 = 24
and 10x + y = 10(4) + 2
= 40 + 2 = 42
∴ The number is 42 or 24
Thus we have two such numbers.

Question 3.
The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Answer:
Let the pair of supplementary angles be x and y [and x > y]
then we have x + y = 180° …… (1)
By problem, x = y + 18°
⇒ x – y = 18° …… (2)
Solving the equations (1) and (2) we get
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 1
and x = \(\frac{198}{2}\) = 99°
Substituting x = 99° in equation (2) we get
99° – y° = 18°
⇒ y° = 99° – 18 = 81°
∴ The angles are 99° and 81°.

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2

Question 4.
The taxi charges in Hyderabad are fixed, along with the charge for the distance covered. For a distance of 10 km., the charge paid is Rs. 220. For a journey of 15 km. the charge paid is Rs. 310.
i) What are the fixed charges and charge per km?
ii) How much does a person have to pay for travelling a distance of 25km?
Answer:
Let the fixed charge be = Rs. x.
and the charge per one km = Rs. y.
By problem, x + 10y = 220 x + 15y = 310
Solving (1) and (2) we get
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 2
∴ y = \(\frac{-90}{-5}\) = 18
i.e., charge per one km = Rs. 18
Substituting y = 18 in equation (1) we get
x + 10 × 18 = 220
⇒ x = 220 – 180
⇒ x = Rs. 40
∴ Fixed charge = Rs. 40;
Charge per km = Rs. 18.

ii) Now, the charge for travelling a distance of 25 km = 25 × 18
= Rs. 450 + 40
= Rs. 490

Question 5.
A fraction becomes equal to \(\frac{4}{5}\) if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes equal to \(\frac{1}{2}\). What is the fraction?
Answer:
Let the numerator of the fraction = x
and the denominator of the fraction = y
By problem,
\(\frac{x+1}{y+1}\) = \(\frac{4}{5}\) and \(\frac{x-5}{y-5}\) = \(\frac{1}{2}\)
⇒ 5(x + 1) = 4(y + 1) and 2(x – 5) = 1(y – 5)
5x + 5 = 4y + 4 and 2x – 10 = y – 5
⇒ 5x – 4y = 4 – 5 and 2x – y = – 5 + 10
⇒ 5x – 4y = – 1 …… (1)
and 2x – y = 5 …… (2)
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 3
∴ y = \(\frac{-27}{-3}\) = 9
Substituting y = 9 in equation (2) we get
2x – 9 = 5
⇒ 2x = 5 + 9
⇒ 2x = 14 and
x = \(\frac{14}{2}\) = 7
Thus the fraction is \(\frac{x}{y}\) = \(\frac{7}{9}\)

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2

Question 6.
Places A and B are 100 km apart on a highway One car starts from A and another from B at the same time at different speeds. If the cars travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Answer:
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 5
Let the speed of the car which started from the place A = x kmph
and B = y kmph
Distance travelled by first car in 5h = 5x and in 1h = x
The distance covered by second car in 5h = 5y and in 1h = y
By problem when travelled in same direction,
5x – 5y = 100 ⇒ x – y = 20 …… (1)
and when travelled towards each other
x + y = 100 ……. (2)
Solving (1) and (2),
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 4
∴ x = \(\frac{120}{2}\) = 60
Substituting x = 60 in equation (1) we get
60 – y = 20
⇒ y = 60 – 20 = 40 kmph
Thus the speed of the cars are 60 kmph and 40 kmph.

Question 7.
Two angles are complementary. The larger angle is 3° less than twice the measure of the smaller angle. Find the measure of each angle.
Answer:
Let the pair of complementary angles be x° and y° with x° > y°
then x° + y° = 90° and
By problem
x = 2y – 3° ⇒ x – 2y = – 3°
Solving these two equations we get,
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 6
∴ y = \(\frac{93}{3}\) = 31°
Substituting y = 31°in x + y = 90° we get
x + 31° = 90°
⇒ x = 90° – 31° = 59°
The angles are 59° and 31°.

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2

Question 8.
An algebra textbook has a total of 1382 pages. It is broken up into two parts. The second part of the book has 64 pages more than the first part. How many pages are in each part of the book?
Answer:
Let the first part of the book contains x pages
and the second part of the book contains y pages By problem,
x + y = 1382 ….. (1)
y = x + 64 ⇒ x – y = -64 …… (2)
Solving equations (1) and (2) we get
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 7
∴ x = \(\frac{1318}{2}\) = 659
Substituting x = 659 in equation (1) we get
659 + y = 1382
⇒ y = 1382 – 659 = 723
∴ The number of pages in the first part = 659
Second part = 723

Question 9.
A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the other is 80% solution. How much of each should be used to obtain 100 ml of a 68% solution?
Answer:
Let the first solution contains 50% acid.
Second solution contains 80% acid.
Let x ml of 1st solution and y ml of second solution are added.
Then x + y = 100
Acid content in the ‘mix’ is 50% of x + 80% of y = 68%
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 8
∴ y = \(\frac{180}{3}\) = 60
Substituting y = 60 in equation (1) we get
x + 60 = 100
⇒ x = 100 – 60 = 40
∴ Quantity of first solution = 40 ml
Quantity of second solution = 60 ml

AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2

Question 10.
Suppose you have Rs. 12000 to invest. You have to invest some amount at 10% and the rest at 15%. How much should be invested at each rate to yield 12% on the total amount invested ?
Answer:
Let the amount to be invested @ 10% be Rs. x
and the amount to be invested @ 15% be Rs. y
By problem x + y = 12000 ……. (1)
Also 10% of x + 15% of y = 12% of 12000
AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2 9
⇒ y = \(\frac{-24000}{-5}\) = Rs. 4800
Substituting y = 4800 in equation (1) we get
x + 4800 = 12000
⇒ x = 12000 – 4800 = 7200
The invested @ 10% = Rs. 7200
@ 15% = Rs. 4800

AP Inter 2nd Year Maths 2A Formulas Chapter 2 డిమోయర్ సిద్ధాంతం

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 2 డిమోయర్ సిద్ధాంతం to solve questions creatively.

AP Intermediate 2nd Year Maths 2A డిమోయర్ సిద్ధాంతం Formulas

→ ‘n’ పూర్ణాంకము అయితే, (cos θ + i sin θ)n = cos nθ + i sin nθ

→ n అకరణీయ సంఖ్య అయితే, (cis θ)n కు ఒక విలువ cis(nθ)

→ 1 యొక్క n -వ మూలాలు: cis(\(\frac{2 k \pi}{n}\)), k = 0, 1, 2, 3, … (n – 1)

AP Inter 2nd Year Maths 2A Formulas Chapter 2 డిమోయర్ సిద్ధాంతం

→ 1 యొక్క ఘనమూలాలు: ఏకకపు ఘనమూలాలు 1, ω, ω2

  • ω3 = 1
  • 1 + ω + ω2 = 0
  • ω = \(\frac{-1+i \sqrt{3}}{2}\), ω2 = \(\frac{-1-i \sqrt{3}}{2}\)
  • 1, ω, ω2 లు గుణశ్రేఢిలో ఉన్నవి.

→ z0 = r0 cis θ0 ≠ 0 అయితే Z0 యొక్క n-వ మూలాలు \(\alpha_k=r_0^{1 / n} {cis}\left(\frac{2 k \pi+\theta_0}{n}\right)\), k = 0, 1, 2,… (n – 1).

AP Inter 2nd Year Maths 2A Formulas Chapter 1 సంకీర్ణ సంఖ్యలు

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 1 సంకీర్ణ సంఖ్యలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A సంకీర్ణ సంఖ్యలు Formulas

→ సంకీర్ణ సంఖ్య నిర్వచనం: x, y ∈ R కు z = x + iy రూపంలో \(\sqrt{-1}\) అనగా i2 = -1 రూపంలో సంఖ్యలను సంకీర్ణ సంఖ్యలు అందురు. ఇందులో x ను z యొక్క వాస్తవ భాగమని మరియు y ను z కు కల్పిత భాగమని అందురు. వీటిని Re(z), Im(z) తో సూచిస్తాం.
x = 0 మరియు y ≠ 0 ⇔ Re(z) = 0 మరియు Im(z) ≠ 0 అయిన Z = x + iy ను శుద్ధ కల్పిత సంఖ్య (i.e) y ≠ 0 ⇔ Im(z) ≠ 0 అయిన z= x + iy వాస్తవము కాని సంఖ్య అగును.

→ వాస్తవసంఖ్యల క్రమయుగ్మాన్ని సంకీర్ణ సంఖ్య అంటాం. సంకీర్ణ సంఖ్యా సమితిని C తో సూచిస్తాం. అంటే C = {(a, b)/ a ∈ R, b ∈ R} = R × R

→ a = c, b = d అయితే రెండు సంకీర్ణ సంఖ్యలు z1 = (a, b), z2 = (c, d) లకు సమానం అంటాం.

→ z1 = (a, b), z2 = (c, d) అయితే

  • z1 + z2 = (a + c, b + d)
  • z1, z2 = (a – c, b – d)
  • z1z2 = (ac – bd, ad + bc)
  • \(\frac{z_1}{z_2}=\left(\frac{a c+b d}{c^2+d^2}, \frac{b c-a d}{c^2+d^2}\right)\)

AP Inter 2nd Year Maths 2A Formulas Chapter 1 సంకీర్ణ సంఖ్యలు

→ మాపము – ఆయామం: సంకీర్ణ సంఖ్య z = x + iy ను P(x, y) బిందువు రూపంలో XOY-తలంలో సూచించిన ఆ తలమును ఆర్లాండ్ తలము అందురు.
|OP| పొడవును z యొక్క మాపము అని, దీనిని |z| చే సూచించెదరు మరియు ∠XOP = θ ను z కు ఆయామం లేదా Arg Z తో సూచిస్తాం.
|z| = r = \(\sqrt{x^2+y^2}\)
Re(z) = r cos θ ⇔ x = r cos θ …..(1)
మరియు Im(z) = r sin θ ⇔ y = r sin θ ……(2)
(1), (2) ల నుండి ‘θ’ ను కనుగొందుము. θ ∈ (-π, π) ను θ కు ప్రధాన ఆయామము అంటారు.
0 = 0 + i0 యొక్క ఆయామము నిర్వచించలేము. Z యొక్క ఆయామము ఏకైకము కాదు.
2nπ + θ, n ∈ z కూడ z కు ఆయామమే.
మరియు -π < θ ≤ π విలువను z యొక్క ప్రధాన ఆయామం అందురు. దీనిని amp(z) లేదా Arg(z) గా సూచించెదరు.

→ సంకీర్ణ సంఖ్యకు సంయుగ్మం: x, y ∈ R మరియు ఏదేని సంకీర్ణ సంఖ్య z = x + iy కు సంయుగ్మం x + i(-y) ⇒ x – iy అని నిర్వచించి, \(\bar{z}\) తో సూచిస్తాం. ఏదేని సంకీర్ణ సంఖ్య మరియు వీని సంయుగ్మముల లబ్దము మరియు మొత్తములు ఖచ్చితముగా వాస్తవములు.

→ మాపము, ఆయామం మరియు సంయుగ్మ సంకీర్ణ సంఖ్యలు కొన్ని ధర్మములు:

  • │\(\bar{z}\)│= |z|
  • z + \(\bar{z}\) = 2 Re(z) మరియు z – \(\bar{z}\) = 2 Im(z)
  • \(\overline{z_1 z_2}=\overline{z_1} \overline{z_2}\)
  • \(\left(\frac{\overline{z_1}}{z_2}\right)=\frac{\bar{z}_1}{\bar{z}_2^{\prime}}\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\), (z2 ≠ 0), |z1z2| = |z1| |z2|
  • \(z \bar{z}=|z|^2\) మరియు z ≠ 0, z-1 = \(\frac{\bar{z}}{|z|^2}\)
  • z = |\(\bar{z}\)|; amp(\(\bar{z}\)) = 2π – amp(z)
  • |z1 + z2| ≤ |z1| + |z2|, |z1 – z2| ≤ |z1| + |z2|
  • |z1 – z2| ≥ ||z1| – |z2||
  • amp (z1) – amp (z2) = 2π యొక్క ధన పూర్ణాంకముల లబ్ధము అయిన (vii) మరియు (viii) లు వాస్తవములు.
  • amp (z1z2) = amp (z1) + amp (z2) + nπ, n ∈ {-1, 0, 1}
  • amp(\(\frac{z_1}{z_2}\)) = amp (z1) – amp (z2) + nπ, n ∈ {-1, 0, 1}
  • \(\frac{1}{{cis} \alpha}\) = cis (-α)
  • cis α . cis β = cis(α + β)
  • \(\frac{{cis} \alpha}{{cis} \beta}\) = cis(α – β)

AP Inter 2nd Year Maths 2A Formulas Chapter 1 సంకీర్ణ సంఖ్యలు

→ డిమోయర్ సిద్ధాంతము:

  • ‘n’ ఏదేని పూర్ణాంకం అయితే (cos θ + i sin θ)n = cos nθ + i sin nθ
  • n అకరణీయ సంఖ్య అయితే (cos θ + i sin θ)n కు ఒక విలువ cos nθ + i sin nθ
  • z0 = r0 cis θ0 ≠ 0 అయితే \(z_0^{1 / n}=r_0^{1 / n} {cis}\left(\frac{2 k \pi+\theta_0}{n}\right)\), k = 0, 1, 2, ……., (n – 1)

→ ఒకటి యొక్క ఘనమూలాలు:

  • ఏకకపు ఘనమూలాలు 1, ω = \(\frac{-1+\sqrt{3 i}}{2}\) మరియు ω2 = \(\frac{-1-\sqrt{3 i}}{2}\)
  • 1 + ω + ω2 = 0 మరియు ω3 = 1; 1 + ω = -ω2, 1 + ω2 = -1
  • ఒకటి వాస్తవము గాని విలువలు ఒకటి మరియొక దాని వర్గమునకు సమానమగును.
  • ఒకటి యొక్క వాస్తవముగాని విలువలు α, β అయిన α + β = -1, αβ = 1, α2 = -β, β2 = α మరియు α3 = β3 = 1
  • \((-1)^{1 / 3}\) మూలాలు -1, -ω, -ω2

సూత్రాలు:
→ z యొక్క మాపము = \(\sqrt{x^2+y^2}\)

→ \(\sqrt{a+i b}\) = (x + iy) అయిన

  • \(\sqrt{a+i b}+\sqrt{a-i b}=\sqrt{2 a+2 \sqrt{a^2+b^2}}\)
  • \(\sqrt{a+i b}-\sqrt{a-i b}=i \sqrt{2 \sqrt{a^2+b^2-2 a}}\)

→ a + ib యొక్క సంయుగ్మము = a – ib

→ a – ib యొక్క సంయుగ్మము = a + ib

AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations

SCERT AP Board 6th Class Social Solutions 6th Lesson Early Civilisations Textbook Questions and Answers.

AP State Syllabus 6th Class Social Studies Solutions 6th Lesson Early Civilisations

6th Class Social Studies 6th Lesson Early Civilisations Textbook Questions and Answers

Improve Your Learning

Question 1.
Write the similarities between Indus and Vedic civilisation.
Answer:
Religious features: Both Harappan people and Vedic people worshipped nature and Hindu gods such as Shiva and Sakti.
Baked bricks, pottery, and wheeled carts were used both by Harappan and Vedic people.

Question 2.
Who participated in the excavations of the Indus valley civilisation?
Answer:
In 1850, the British engineers while laying a railway line between Karachi and Lahore excavated the Indus valley civilization. In 1920 the archeologists realised that it was a very did civilisation.

AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations

Question 3.
Describe the economic life of the Indus people.
Answer:

  1. The economy of the Indus valley civilization was based upon agriculture and trade.
  2. Weaving cotton and woolen clothes were their main occupation. Indus people were the earliest to produce cotton.
  3. Production of baked bricks was another occupation of Indus people.
  4. Crops such as wheat, barley, peas were raised by the Indus people.
  5. They carried trade with other countries like Egypt, Mesopotamia, Afghanistan, and Iran also.
  6. They carried their international trade through Lothal port in the Arabian Sea.

Question 4.
How did the Indus people build their houses?
Answer:
The buildings of Indus people were well planned. It is clear that they possessed a flourishing urban culture. There were wide roads, public wells. Houses were made of bricks which were found in so many other areas across this civilization also. There were two storied buildings also. Every house was connected directly to an excellent main drainage system. A great tank was built for a public bath which now was called as “the Great Bath”. Large granaries and quarters for labor were specially built in the Indus civilization.

Question 5.
The drainage system of Indus was exemplary. How?
Answer:

  1. The Indus civilization had an elaborate sanitary and drainage system. They maintained a highly efficient drainage system.
  2. Importance was given to the cleanliness and public health.
  3. They maintained an underground drainage system for the streets. They were connected to the main drainage system.

Question 6.
“Devotion of God is belief”. Comment on Indus people’s deities.
Answer:

  1. Indus people believed in God. They believed in image worshipping.
  2. They worshipped Pasupathi (Shiva) and the mother of the Goddess.
  3. They accepted the pipal trees as sacred.
  4. They worshipped neem trees, sun, fire, earth, water, and snakes also.
  5. Fire pits and yagna vatikas are also found.
  6. Swastik symbols were found in the archeological remains of the Indus valley civilization.

AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations

Question 7.
How many Vedas are there? What are they?
Answer:
There are four Vedas in the Vedic culture. They are:

  1. Rig Veda,
  2. Yajur Veda,
  3. Sama Veda and
  4. Atharvana Veda.

Question 8.
“Veda means superior knowledge”. Comment on it.
Answer:

  1. The term ‘Veda’ means “Superior Knowledge” in Sanskrit.
  2. They are the most ancient religious texts which define truth for Hindus.
  3. They are the most important sources of knowledge about the Aryans and the Vedic Period.
    They are called shrutis which means hearing.
  4. Vedas are the universal truths that were discovered or realized by learned people.
  5. They formed the bedrock for Indian philosophy and Yoga.
  6. Vedas contain deeper spiritual and scientific knowledge.
  7. The Vedic traditions are continued till today and are still unbroken.
  8. Research on vedas is carried out in many Indian and foreign Universities also. Swami Dayananda Saraswati gave a call “Go back to the Vedas”.

Thus we can say the Vedas mean superior knowledge.

Question 9.
Write about the social life of the early Vedic period.
Answer:
The Vedic literature is the major source to understand the Vedic period. The period of Rig Veda is called the early Vedic period.
The social life of early Vedic period:

  1. The family was regarded as the basic unit of society.
  2. The father was the head of the family.
  3. The joint family system was in practice at that time.
  4. The prisoners of war were called dasas and they were like slaves.
  5. Monogamy was the usual practice of the Vedic period.

AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations

Question 10.
What do you know about the political life of the early Vedic period?
Answer:

  1. The polity of the Early Vedic period was basically a tribal polity with the tribal chief in the center. The tribe was called ‘Jana’, and the chief of that Jana was called ‘Rajan’.
  2. Rajan looked after the affairs of the tribe with the help of other tribal members. There are two tribal assemblies 1) Sabha and 2) Samiti.
  3. Rajan was not a dictator, Kingship was not hereditary.

Question 11.
What is the importance of epics?
Answer:

  1. The epics tell us the right way to life.
  2. They tell us what the importance of speaking the truth is.
  3. They tell us that why we should have the basic manners in our lives in an interesting manner.
  4. They tell us the importance of becoming a good human being.
  5. We should read these epics because they fill our minds with knowledge and wisdom.
  6. Epics tell us that how should we behave in society.

In Ramayana Rama is an ideal ruler, ideal brother, ideal son. Sita is ideal for Indian womanhood.
Mahabharata is a victory of Dharma over Adharma.

Question 12.
Locate the following rivers on the outline map of India
a) River Indus b) River Ganges c) River Yamuna
Answer:
AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations 1

AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations

Question 13.
What were the reasons for the decline of the Indus valley civilization?
Answer:

  1. Some are of the opinion that the invasion of the Aryans was the cause for the decline of Indus valley civilization.
  2. Change of course of the river Indus changed the area into the sand and affected the fertility and this led to the decline of the Indus valley civilization.
  3. Due to changes in climatic conditions, the area lost fertility and people were compelled to move to another area.
  4. Constant use of firewood for brick manufacture caused deforestation and low rainfall.

Project Work

Make a project on the similarities of Indus civilization and Vedic civilization.
Answer:
The Indus Valley civilization and the Vedic civilization had the following similarities.

  • Both worshipped Hindu Gods such as Sakthi and Siva.
  • Both believed in some sort of life after death.

They also played pre-Indian games such as gambling, dicing, and horse cart riding. The remains of a stone sculpture of a priest say that their society of most similar to that of Vedic culture.

6th Class Social Studies 6th Lesson Early Civilisations InText Questions and Answers

Let’s Do

Question 1.
Compare the economic activity of Indus people with the present. (Textbook Page No. 66)
AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations 2
Answer:

Economic activityIndus timesPresent
ExportsCotton goods, pottery agricultural products.Gems, jewellary, ready-made garments metals, cotton, yarn, and fabrics.
ImportsJade, agate gold, silver, copper, tin.Petrol, paper, precious stones, plastic wool, medicines, machinery, edible oil.
CropsWheat, barley, dates cotton, mustard.Foodgrains, cotton, jute, sugarcane tobacco, oilseeds, coffee, tea, coconut fruits, vegetables.
Domestication of animalsSheep, goat, buffalo, dog, cat, pig, elephant.Cow, buffalo, oxen, sheep, goat, horse, camel, elephant, pig etc.
CraftsJewellery, Art, necklaces, bangles, combs, idol-making.Shawls, woodwork, pottery, leather, jute, shell, brass handicrafts, bamboo handicrafts.

Question 2.
Compare the metals used in the Indus civilization with the present day. (Textbook Page No. 67)
AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations 3
Answer:

Metals used in the Indus civilizationMetals used now
Gold, copper, silver, bronze metals were the metals known to Indus Valley Civilization people.At present we are using gold, copper, silver, bronze, platinum, brass and Iron metals.

AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations

Question 3.
List out the names of famous books and its writers from your school library. (Textbook Page No. 72)
Answer:

  1. The Story of My Experiments with Truth – Mahatma Gandhi
  2. The Discovery of India. – Jawaharlal Nehru
  3. ………………………………………………………………….
  4. ………………………………………………………………….(Student Activity)

Think and Respond

(Textbook Page No. 65)

Question 1.
Do you find any difference between the planning of Indus cities and the present cities? How?
Answer:

  1. The Indus valley cities were established on a good plan. The modern Indian cities are not based on any such plans.
  2. Both the Indus valley and the modern Indian cities have a drainage system but the modern drainage system does not work properly when compared to the Indus valley cities.
  3. There were citadels in the Indus valley civilization while there is no citadels in the modern Indian cities, (maybe because of population increase)

Question 2.
Was the Indus drainage system more advanced? Why?
Answer:
The Indus drainage system is well planned underground drainage system. The drainage system which was followed by the Indus people was a great achievement. Every house is connected with pipes which carry waste into the main drains.

AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations

(Textbook Page No. 69)

Question 3.
Do you find any similarities among these ancient (Mesopotamia, Egypt, Indus valley, and China) civilizations?
Answer:
Yes, we can find similarities among these ancient civilizations.

  1. All the civilisations grew along mighty rivers.
  2. All the civilisations have similarities in their settlements.
  3. All the civilisation developed religions to explain the forces of nature and their roles in the world.
  4. All the civilizations made efforts in the fields of science and technology.

Question 4.
In what ways is the Indus valley civilisation more advanced among them?
Answer:
In the Indus valley civilisation a strong and centralized government was established. The people in the Indus valley civilisation built their houses with dried or baked bricks. They built two-storeyed buildings also. Every house had a well for water and bathrooms with pipes that carried waste into the main drains. The cities of the Indus valley civilisation were well planned. They were protected by forts. They had wide roads, public wells. They had a well-planned underground drainage system which shows the importance given to cleanliness and public health. We don’t find this type of development in other civilizations of that time.

Question 5.
Write the names of religions in our country. (Textbook Page No. 71)
Answer:
Hinduism, Buddhism, Jainism, Sikhism, Islam, Christianity, Zoroastrianism, Judaism, Animism, Dony polo, and Nirankari.

AP Board 6th Class Social Studies Solutions Chapter 6 Early Civilisations

Question 6.
How are our public representatives elected? (Textbook Page No. 72)
Answer:
Our public representatives are elected in both direct and indirect elections.

AP Board 7th Class Maths Solutions Chapter 6 Data Handling Ex 6.1

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.1

Question 1.
Find the Arithmetic Mean of the following.
(i) 4, 5, 11, 8
Answer:
Given data: 4,5, 11, 8.
Sum of observations = 4 + 5 + 11 + 8 = 28
Number of observations = 4
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{28}{4}\) = 7.

AP Board 7th Class Maths Solutions Chapter 6 Data Handling Ex 6.1

(ii) 10, 15, 21, 12, 17
Answer:
Given data: 10, 15, 21, 12, 17
Sum of observations 10 + 15 + 21 + 12 + 17 = 75
Number of observations = 5
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{75}{5}\) = 15

(iii) \(\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \frac{3}{2}, \frac{5}{4}\)
Answer:
Given data: \(\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \frac{3}{2}, \frac{5}{4}\)
Sum of observations = \(\frac{1}{4}+\frac{1}{2}+\frac{3}{4}+\frac{3}{2}+\frac{5}{4}\) = \(\frac{1+2+3+6+5}{4}\) = \(\frac{17}{4}\)
Number of observations = 5
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{\frac{17}{4}}{5}=\frac{17}{4} \div \frac{5}{1}=\frac{17}{4} \times \frac{1}{5}=\frac{17}{20}\)
∴ Arithmetic Mean = \(\frac{17}{20}\)

Question 2.
Amounts donated by eight students to ‘NIVAR’ cyclone effected people are ₹300, ₹450, ₹700, ₹650, ₹400, ₹750, ₹900 and ₹850. Find the Arithmetic Mean of amounts donated.
Answer:
Given data, ₹300, ₹450, ₹700, ₹650, ₹400, ₹750, ₹900 and ₹850.
Sum of observations = 300 + 450 + 700 +650 +400 + 750 + 900 + 850 = 5000
Number of observations = 8
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{5000}{8}\) = ₹625
∴ Arithmetic Mean of amount donated = ₹625

Question 3.
The number of passengers who travelled in APSRTC bus from Eluru to Rangapuram in 5 trips in a day are 35, 42, 28, 41 and 44. What is the average of number of passengers travelled per trip ?
AP Board 7th Class Maths Solutions Chapter 6 Data Handling Ex 6.1 1
Answer:
Given data are 35, 42, 28, 41, 44.
Sum of observations = 35 + 42 + 28 + 41 + 44 = 190
Number of observations = 5
∴ Arithmetic Mean (or) Average = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{190}{5}\) = 38
∴Average of number of passengers travelled per trip = 38.

AP Board 7th Class Maths Solutions Chapter 6 Data Handling Ex 6.1

Question 4.
Find Arithmetic mean of factors of 24.
Answer:
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
Sum of (factors) observations = 1 + 2 + 3 + 4 + 6 + 8+ 12 + 24 = 60
Number of observations = 8
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{60}{8}\) = 7.5
∴ Arithmetic Mean of factors of 24 is 7.5

Question 5.
Find the Arithmetic Mean of x, x + 1 and x + 2.
Answer:
Given data x, x + 1, x + 2.
Sum of observations = x + x + 1 + x + 2 = 3x + 3
Number of observations = 3
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{3 x+3}{3}\) = \(\frac{3(x+1)}{3}\)
∴ Arithmetic mean = x + 1

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.1

10th Class Maths 6th Lesson Progressions Ex 6.1 Textbook Questions and Answers

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
i) The taxi fare after each km when the fare is Rs. 20 for the first km and rises by Rs. 8 for each additional km.
Answer:
Fare for the first km = Rs. 20 = a
Fare for each km after the first = Rs. 8 = d
∴ The fares would be 20, 28, 36, 44, …….
The above list forms an A.P.
Since each term in the list, starting from the second can be obtained by adding ‘8’ to its preceding term.

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1

ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\)th of the air remaining in the cylinder at a time.
Answer:
Let the amount of air initially present in the cylinder be 1024 lit.
First it removes \(\frac{1}{4}\)th of the volume
i.e., \(\frac{1}{4}\) × 1024 = 256
∴ Remaining air present in the cylinder = 768
At second time it removes \(\frac{1}{4}\)th of 768
i.e., \(\frac{1}{4}\) × 768 = 192
∴ Remaining air in the cylinder = 768 – 192 = 576
Again at third time it removes \(\frac{1}{4}\)th of 576
i.e., \(\frac{1}{4}\) × 576 = 144
Remaining air in the cylinder = 576 – 144 = 432
i.e., the volume of the air present in the cylinder after 1st, 2nd, 3rd,… times is 1024, 768, 576, 432, …..
Here, a2 – a1 = 768 – 1024 = – 256
a3 – a2 = 576 – 768 = – 192
a4 – a3 = 432 – 576 = – 144 .
Thus the difference between any two successive terms is not equal to a fixed number.
∴ The given situation doesn’t show an A.P.

iii) The cost of digging a well, after, every metre of digging, when it costs ? 150 for the first metre and rises by ? 50 for each subsequent metre.
Answer:
Cost for digging the first metre = Rs. 150
Cost for digging subsequent metres = Rs. 50 each.
i.e.,
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 1
The list is 150, 200, 250, 300, 350, ……..
Here d = a2 – a1 = a3 – a2 = a4 – a3 = ……. = 50
∴ The given situation represents an A.P.

iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8 % per annum.
Answer:
Amount deposited initially = P = Rs. 10,000
Rate of interest = R = 8% p.a [at C.I.]
∴ \(A=P\left(1+\frac{R}{100}\right)^{n}\)
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 2
The terms 10800, 11664, 12597.12, ……. a2 – a1 = 800
Here, a = 10,000                                     a3 – a2 = 864
But, a2 – a1 ≠ a3 – a2 ≠ a4 – a3                a4 – a3 = 953.12
∴ The given situation doesn’t represent an A.P.

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
i) a = 10, d = 10
ii) a = -2, d = 0
iii) a = 4, d = – 3
iv) a = – 1, d = 1/2
v) a = – 1.25, d = – 0.25
Answer:
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 3

Question 3.
For the following A.Ps, write the first term and the common difference:
i) 3, 1, – 1, – 3,….
ii) – 5, – 1, 3, 7,….
iii) \(\frac{1}{3}\), \(\frac{5}{3}\), \(\frac{9}{3}\), \(\frac{13}{3}\), ……..
iv) 0.6, 1.7, 2.8, 3.9,…
Answer:
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 4

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1

Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
i) 2, 4, 8, 16, …….
ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …….
iii) – 1.2, – 3.2, – 5.2, – 7.2,……
iv) -10,-6, -2, 2, …….
v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …….
vi) 0.2, 0.22, 0.222, 0.2222, ……
vii) 0, -4, -8, -12, …….
viii) –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\)
ix) 1, 3, 9, 27,…..
x) a, 2a, 3a, 4a,….
xi) a, a2, a3, a4, …..
xii) √2, √8, √18, √32, …….
xiii) √3, √6, √9, √12, …….
Answer:
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 5AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.1 6