Class 8

AP State Syllabus AP Board 8th Class Physical Science Important Questions Chapter 8 Combustion, Fuels and Flame

AP State Syllabus 8th Class Physical Science Important Questions 8th Lesson Combustion, Fuels and Flame

8th Class Physical Science 8th Lesson Combustion, Fuels and Flame 1 Mark Important Questions and Answers

Question 1.
What do you mean by combustion?
Answer:
A chemical process in which a material reacts with oxygen present in air to generate heat is called combustion.

Question 2.
What is ignition temperature?
Answer:
The lowest temperature at which a substance catches fire is called ignition temperature.

Question 3.
What are inflammable substances? Give examples.
Answer:
The substance which have very low ignition temperature easily catch fire are called inflammable substances, e.g.: Petrol, alcohol, LPG, etc.

Question 4.
What is explosion?
Answer:
A chemical reaction takes place with the evolution of heat, light, sound and large amount of gas is called explosion.

Question 5.
What is calorific value? What is its unit?
Answer:
Calorific value of a fuel is the amount of heat energy produced on complete combustion of 1 kg of that fuel.
It is measured in kilo joules per kg. (kJ/kg)

Question 6.
Which zone of a flame does a goldsmith use for melting gold and silver and why?
Answer:
The goldsmith would use outermost zone for melting gold and silver because it is hottest zone due to complete combustion.

Question 7.
Explain how CO₂ is able to control fires?
Answer:
When CO₂ released from the cylinder on fire, it expands enormously in volume and cools down. So, it is not only forms a blanket around fire, it also brings down the temperature of fuel. That is why it is an excellent fire extinguisher.

Question 8.
Paper by itself catches fire easily whereas a piece of paper wrapped around an aluminium pipe does not ?
Answer:
When you heat a piece of paper wrapped around an aluminium pipe the heat is taken up by aluminium pipe so the paper does not reach its ignition temperature.

Question 9.
Can the process of rusting be called combustion? Discuss.
Answer:
Rusting of Iron is a combustion reaction. Combustion is a process in which elements combine with oxygen. During formation of rust iron atoms combine oxygen in the air forming iron oxides or rust.

Question 10.
Name the products formed when a candle burns in air.
Answer:
The products formed are carbondioxide and water because wax is mixture of hydrocarbons.

8th Class Physical Science 8th Lesson Combustion, Fuels and Flame 2 Marks Important Questions and Answers

Question 1.
What are combustible and non combustible materials and give examples for them?
Answer:
Combustible materials:
The materials which burn when brought near a flame are called combustible materials, e.g. : Petrol, diesel, kerosene, etc.
Non-combustible materials:
The materials which do not burn when brought near a flame are called non-combustible materials, e.g.: Pebbles, sand, clay, iron, etc.

Question 2.
What are the different types of combustion and define them?
Answer:
Combustion is two types.

1. Spontaneous combustion:
   The type of combustion in which material suddenly bursts into flames without the application of apparent cause is called spontaneous combustion.
2. Rapid combustion:
   The type of combustion in which materials burns rapidly and produce heat and light is called rapid combustion.

Question 3.
What are the different zones present in a candles and what are the colours of those zones?
Answer:
There are three zones in a candle.

1. Outermost zone: This is the hottest part which is in blue colour.
2. Middle zone: This is moderately hot part which is in yellow colour.
3. Dark zone: This is least hot part which is in black colour.

Question 4.
In an experiment 4.5 kg of a fuel was completely burnt. The heat produced was measured to be 1,80,000 KJ. Calculate the calorific value of fuel?
Answer: The mass of fuel = 4.5 kg
The heat produced = 1,80,000 KJ
heat produced 1,80,000
The calorific value of fuel = \(\frac{\text { heat produced }}{\text { mass of the fuel }}\) = \(\frac{1,80,000}{4.5}\) = 40,000 KJ/Kg
mass of the fuel 4.5

Question 5.
LPG is better domestic fuel than wood?
Answer:

1. On burning wood it produces lot of smoke and also complete burning does not takes place whereas LPG undergo complete combustion so does not produce smoke.
2. Due to incomplete combustion wood produce harmful gas like carbon monoxide whereas LPG does not produce carbon monoxide.
3. Wood has high ignition temperature so does not burn immediately whereas LPG has low ignition temperature burns easily.

Question 6.
How do you appreciate use of fossil fuels in daily life?
Answer:
We are depend upon fossil fuels for our daily needs like cooking, transportation, running machinery and producing electricity, etc. Everywhere we go there is use of fossil fuel. Without fossil/fuels we may be in stone age. So the use of fossil fuels in daily life should be thoroughly appreciated.

Question 7.
What would happen if oxygen stops to support combustion? – Make a guess. And if it is the situation for what other fuels are useful?
Answer:
If oxygen stops to support combustion there is no other gas which will support combustion. Then fossil fuels are not useful in producing heat, energy and electricity.
So we should have to prefer alternative sources of energy like solar energy, wind energy, tidal energy, biomass energy, geothermal energy, etc. for our energy needs.

Question 8.
Let us assume that you are on the moon. If you try to focus sun light on a paper using magnifying glass, does the paper catch fire? or not? Why?
Answer:
No, moon reflects entire sunlight that falls on the surface because it acts as perfect reflector. Whereas earth is also acts as reflector but green house gases present in atmosphere absorbing the sunlight and resending on earth. So paper can be burnt on earth by using magnifying glass but it is not possible on moon.

Question 9.
Why does tap water is not used to control fire involving electrical equipment ?
Answer:

1. Tap water cannot be used to control the fire involving electric equipment because it is a good conductor of electricity.
2. It conducts electricity resulting in electrical shock to the user.

Question 10.
Explain how carbon dioxide is able to control fires.
Answer:
Carbon dioxide, being heavier than oxygen, covers the fire like blanket and also brings down the temperature of fuel. Since the contact between the fuel and oxygen is cut off the fire comes under control.

8th Class Physical Science 8th Lesson Combustion, Fuels and Flame 4 Marks Important Questions and Answers

Question 1.
How does candle works?
Answer:

1. A candle is mainly a source of light and heat.
2. A candle is made of wax in which a thick thread inserted wax in the candle melts when it is lighted by a match stick.
3. A little of wax forms vapour.
4. This vapour combines with oxygen in the air to form flame.
5. The heat of the flame melts more of the wax from the top of the candle.
6. The melted liquid wax moves upward through the thread. It also changes to vapour when it reaches the top of the wick and byrns with the flame.

8th Class Physical Science 8th Lesson Combustion, Fuels and Flame Important Questions and Answers

Question 1.
Spirit burns quickly like petrol but sodium metal and white phosphorous burns without any ignition.
Complete the following table and rewrite it in the table.

Answer:

Question 2.

Answer the following questions.
i) Name the fuel having highest calorific value.
ii) How much heat energy is released when one kg of petrol burnt?
iii) Name two fuels which causes less pollution.
iv) Mention any one of the alternate energy source which is not mentioned in the above table.
Answer:
i) Hydrogen
ii) 45,000 Kilo Joules
iii) Hydrogen, L.P.G
iv) Solar power, Gobar gas, Wind power, Bio-diesel.

AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion InText Questions and Answers.

8th Class Maths 5th Lesson Comparing Quantities Using Proportion InText Questions and Answers

Do this

Question 1.
How much compound interest is earned by investing Rs. 20000 for 6 years at 5% per annum compounded annually? (Page No. 114)
Answer:
P = Rs. 20,000; R = 5%; n = 6 years

∴ Compound Interest = Amount – Principal = 26802 – 20,000
∴ C.I. = Rs. 6802 /-

Question 2.
Find compound interest on Rs. 12600 for 2 years at 10% per annum compounded annually.    (Page No. 114)
Answer:
P = Rs. 12,600; R = 10%; n = 2 years

∴ Compound Interest = Amount – Principal = 15,246 – 12,600
∴ C.I. = Rs. 2646 /-

Question 3.
Find the number of conversion times the interest is compounded and rate for each.
i) A sum taken for 1\(\frac{1}{2}\) years at 8% per annum is compounded half yearly.
ii) A sum taken for 2 years at 4% per annum is compounded half yearly.     (Page No. 115)
Answer:
Compound interest will be calculated for every 6 months.
There will be 3 periods in 1\(\frac{1}{2}\) year.
∴ n = 3
∴ Rate of interest for half yearly = \(\frac{1}{2}\) × 8% = 4%
∴ R = 4%; n = 3
ii) C.I. should be calculated for every 6 months.
There will be 4 time periods in 2 years.
∴ n = 4
∴ Rate of interest for half yearly = \(\frac{1}{2}\) × 4% = 2%
∴ n = 4 ; R = 2%

Try These

Question 1.
Find the ratio of gear of your bicycle.       (Page No. 96)

Count the number of teeth on the chain wheel and the number of teeth for the sprocket wheel.
{number of teeth on the chain wheel} : {number of teeth of sprocket wheel}
This is called gear ratio. Write how many times sprocket wheel turns for every time the chain wheel rotates once.
Answer:
The ratio between the rotations of chain wheel and sprocket wheel is 4 : 1.

Question 2.
Collect newspaper cuttings related to percentages of any five different situations.  (Page No. 96)
Answer:
1) Bharti to sell 5% stake for $ 1.2b:
New Delhi, May 3: The country’s largest telecom operator Bharti Airtel said on Friday that it will sell 5 per cent stake to Doha – based Qatar Foundation Endowment (QFE) for $1.26 billion (Rs. 6,796 crores) to fund its future growth plans.
The deal will bring cash for the company at a time when its balance sheet is stretched and there is threat of Bharti Airtel having to pay hefty fees to regulatory authorities as government is re-looking at past policies.

2) Indian Firms Mop – Up Down By 36% In FY13:
New Delhi: Indian companies raised nearly Rs. 31,000 crore from the public issuance of equity and debt in 2012 – 13, a slump of 36 per cent from the preceding year.
According to latest data available with market regulator Sebi (Securities and Exchange Board of India), a total of Rs. 30,859 crore worth of fresh capital were mopped – up from equity and debt market during 2012 – 13, which was way below than Rs. 48,468 crore garnered in 2011 -12. Going by the statistics, it was mostly debt market that was leveraged to meet the funding requirements of businesses in the past fiscal as compared to capital raised through sale of shares through instruments like initial public offering (IPO) and rights issue. A total of Rs. 15,386 crore were raised from the debt market via 11 issues in 2012 – 13, much lower than Rs. 35,611 crore garnered through 20 issues in the preceding fiscal.

3) IT – ITeS sector employs 2.97m people in FY13:
New Delhi: The total number of professionals working in India’s $100 billion IT – information technology enabled services (IT – ITeS) sector grew by 7 per cent to 2.97 million in the last fiscal, Parliament was informed on Friday.
The IT – ITeS sector, which contributes about 8 per cent to the country’s economy, provided employment to 2.77 million professionals in 2011 -12 fiscal, minister of state for communications and IT Milind Deora said. “The Indian IT – ITeS industry has been progressively growing and is able to secure new projects from various foreign coun¬tries,” Mr Deora said. During the 2012 -13 fiscal, 6,40,000 professionals were employed in the domestic market.

4) For RBI, it’s not all is well yet:
Slashes repo rate by 0.25%; rules out any more cuts; raises red flag on CAD DC Correspondent Mumbai, May 3:
The RBI cut the repo rate (rate at which it lends to banks) by a quarter per cent on Friday to 7.25 per cent from 7.75 per cent, but this will not be passed on to the consumers by way of lower personal loans for housing etc., immediately according to bankers.
It also raised the growth rate from 5.2 per cent projected in January to 5.7 per cent for 2013 -14 and lowered the inflation rate to 5.5 per cent for the year.
RBI governor Dr D. Subbarao said based on the current and prospective assessment of various economic factors and the dismal 4.5 per cent lowest growth rate in the last quarter, it was decided to cut the policy rate by 25 basis points.

5) Markets sink on RBI’s Bearish outlook on rate:
DC Correspondent Mumbai, May 3:
In a highly volatile trading session, the markets retreated from their three month high led by interest rate sensitive banking, auto and real estate sector stocks after the Reserve Bank of India (RBI) cautioned that the room for further monetary policy easing is limited.
The Sensex closed 19,575.64, sliding 160.13 points or 0.81 per cent while the Nifty dropped 55.35 points or 0.92 per cent to end the week at 5,944.

Question 3.
Find the compound ratios of the following. (Page No. 99)
a) 3 : 4 and 2 : 3
b) 4 : 5 and 4 : 5
c) 5 : 7 and 2 : 9
Answer:
Compound ratio of a : b and c : d is ac : bd.

Question 4.
Give examples for compound ratio from daily life.     (Page No. 99)
Answer:
Examples for compound ratio from daily life:
i) To compare the ratio of tickets of 8th class students (Boys & Girls) is 3:4 and the ratio of tickets of 7th class students is 4 : 5.
ii) The comparision between two situations is 4 men can do a piece of work in 12 days, the same work 6 men can do in 8 days.
iii) Time – distance – speed.
iv) Men – days – their capacities etc.

Question 5.
Fill the selling price for each.     (Page No. 104)

Answer:

Question 6.
i) Estimate 20% of Rs. 357.30 ii) Estimate 15% of Rs. 375.50      (Page No. 105)
Answer:

ii) 15% of 375.50 = \(\frac{15}{100}\) × 375.50 = 15 × 3.7550 = Rs. 56.325

Question 7.
Complete the table.     (Page No. 105)

Answer:

Think, discuss and write

Question 1.
Two times a number is 100% increase in the number. If we take half the number what would be the decrease in percent?    (Page No. 101)
Answer:
Increase percent of 2 times of a number = \(\frac{(2-1)}{1}\) × 100 = 1 × 100 = 100%
Half of the number = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Decrease in percent = \(\frac{\frac{1}{2}}{1}\) × 100 = \(\frac{1}{2}\) × 100 = 50%

Question 2.
By what percent is Rs. 2000 less than Rs. 2400? Is it the same as the percent by which Rs. 2400 is more than Rs. 2000? (Page No. 101)
Answer:
Decrease in percent of Rs. 2000 less than Rs. 2400

Increase in percent of Rs. 2400 more than Rs. 2000

Question 3.
Preethi went to a shop to buy a dress. Its marked price is Rs. 2500. Shop owner gave 5% discount on it. On further insistence, he gave 3% more discount. What will be the final discount she obtained? Will it be equal to a single discount of 8%? Think, discuss with your friends and write it in your notebook. (Page No. 105)
Answer:
Marked price of a dress selected by Preethi = Rs. 2500
After allowing 5% of discount then S.P = M.P. – Discount%
= 2500 – \(\frac{5}{100}\) × 2500 = 2500 – 125 = Rs. 2375
Again 3% discount is allowed on Rs. 2375 then
S.P = 2375 – 3% of 2375
= 2375 – \(\frac{3}{100}\) × 2375 = 2375 – 71.25 = Rs. 2303.75
If 8% discount is allowed then S.P =

The S.P’s of both cases are not equal.
Discount on 5% + Discount on 3% = 125 + 71.25 = Rs. 196.25
Discount on 8% = Rs. 200
∴ Discounts are not equal which are obtained by Preethi.

Question 4.
What happens if cost price = selling price. Do we get any such situations in our daily life?
It is easy to find profit % or loss% in the above situations. But it will be more meaningful if we express them in percentages. Profit % is an example of increase percent of cost price and loss % is an example of decrease percent of cost price. (Page No. 106)
Answer:
If selling price is equal to cost price then either profit or loss will not be occurred.
In our daily life S.P. will not be equal to C.P. Then profit or loss will be occurred.
∴ Profit % = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100;
Loss % = \(\frac{\text { Loss }}{\text { C.P. }}\) × 100.

Question 5.
A shop keeper sold two TV sets at Rs. 9,900 each. He sold one at a profit of 10% and the other at a loss of 10%. Oh the whole whether he gets profit or loss? If so what is its percentage? (Page No. 108)
Answer:
S.P of each T.V = Rs. 9,900
S.P of both T.Vs = 2 × 9,900 = Rs. 19,800
10% profit is allowed on first then C.P. =

10% loss is allowed on second then C.P.

C.P. of both T.V.’s = 9000 + 11000 = Rs. 20,000
Here C.P > S.P then loss will be occurred.
∴ Loss = C.P – S.P = 20000 – 19,800 = 200

Question 6.
What will happen if interest is compounded quarterly? How many conversion periods will be there? What about the quarter year rate – how much will it be of the annual rate? Discuss with your friends. (Page No. 115)
Answer:
Here C.I will be calculated for every 3 months. So, 4 time periods will be occurred in 1 year.
Rate of Interest (R) = \(\frac{R}{4}\) [∵ \(\frac{12}{3}\) = 4]

A = P\(\left[1+\frac{R}{400}\right]^{4}\)

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 3 Story of Microorganisms 2 Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 3rd Lesson Story of Microorganisms 2

8th Class Biology 3rd Lesson Story of Microorganisms 2 Textbook Questions and Answers

Improve Your Learning

Question 1.
How do vaccines works in our body?
Answer:
When a vaccine is given which contains weak microorganisms our body learn to fight them by producing antibodies whenever danger comes. When the disease causing microorganisms enter in our body, the already present antibodies fight and eliminate them. And the disease is controlled. The weakened disease causing microorganisms which are injected into our bodies are called as vaccine.

Question 2.
What are the differences between Antibiotic and Vaccine?
Answer:

Question 3.
Invention of pencillin protected the world from deaths during first world war, Explain.
Answer:

1. Pencillin was invented by Dr. Alexander Flemming.
2. He was an army doctor in First World War.
3. In the First World War many injured soliders died because of bacterial infection of wounds.
4. Pencillin killed many disease causing bacteria.
5. So the invention of pencillin protect the world from deaths during the first World War.

Question 4.
Take three bowls and mark as A,B,C. Pour lukewarm milk in bowl A, hot milk in bowl B, cold milk in bowl C. Add one tea spoon of curd or butter milk in three bowls and stir them slightly. Cover the bowls with lids. Keep the bowls undisturbed for five to six hours. In which bowl milk turned into curd ? Give your reasons.
Answer:
After six hours the milk in the bowl A converted into curd.
This conversion is happened by fermentation. Fermentaion is one kind of enzy-matic action in anaerobic respiration. Fermentation needs the help of microbes like yeast or some other bacteria. This process occurs at a particular temperature. Be-cause of enzymes are inactive at low and high temperatures. Hence the milk in bowl A converted into curd by the bacterium Lactobacillus.

Question 5.
Collect more information about scientists who invented and discovered other facts related to Microorganisms. How these discoveries helped mankind ? Make a chart presentation and paste it on your classroom wall magazine.
Answer:

Question 6.
Make an album of scientists and their discoveries related to Microorganisms.
Answer:

Question 7.
Visit the veterinary hospital and prepare a list of cattle diseases by asking questions, to the doctor.
Answer:

Question 8.
What is pasteurisation? What is the use of it?
Answer:

1. Pasteurisation is a process which helps in the preservation of the milk.
2. It is discovered by “Louis Pasteur.” Hence it is called as pasteurisation.
3. In this process milk is heated up to 70° for 15 to 30 seconds and suddenly chilled and stored.
4. This prevents the growth of the microorganisms.

Question 9.
How do you appreciate Edward Jenner’s experiment?
Answer:

Edward Jenner was a man of the Enlightenment and believed in the intellectual and moral improvement of mankind.
Possessing that spirit, he predicted in 1801 that “the annihilation of small pox-the most dreadful scourge of the human race-will be the final result of this practice” of vaccination. And he was right. Since the Enlightenment, the core branches of science and medicine have generally developed rationally. As a result, the World Health Organization was able to declare in 1980 that smallpox had been eradicated worldwide. Only two laboratory samples emained, one in Russia and one in the United States.

Question 10.
Visit a nearby milk chilling centre. Observe the process and make a report on it.
Answer:
The Milk Collection Station is a specially designed, integrated unit, which combines the several functions of a milk collection centre. It measures the weight, fat content and gives the price of the milk brought in by the each producer. The equipment is particularly useful for the milk cooperatives / milk collection centres as it can also maintain a summary of milk supplied. This state of the art equipment operates both on battery and mains and is able to process and record 120-150 milk collection per hour. An Electronic Milk Weighing Unit, the Electronic Milk Tester and Data Processor Unit are main components of the system. The membership code of individual mem¬bers is entered automatically by member identity card / manually by an electronic key-board.

Question 11.
“Prevention is better than cure” comment.
Answer:
We heard about “prevention is better than cure” since our childhood – brushing of teeth twice to prevent tooth decay, washing of hands before eating to prevent stomach upset, do not eat uncovered road side food items, get vaccine against polio to free from disability, get rid of stagnate water to prevent malaria like that follow the life skills to avoid dangerous AIDS. It’s better to take care that a problem does not happen than to have to solve the problem afterwards. It’s easier to stop something bad from happening in the first place than to fix the damage after it has happened.

Question 12.
Raheem tells to his neighbours, “stagnation of’ sewage in our surroundings is harmful to our health. ” Do you support this? Why?
Answer:
Stagnation of sewage in our surroundings is harmful to our health in different ways. Sewage consist different wastes of organic matters. Microbes decompose all organic wastes. Disease transmitting carriers prefer to grow on decomposing organic matters. Hence sewage provides good shelter to mosquitoes, houseflies and other insects which transmit life threatening diseases like encephalitis, malaria, cholera, typhoid and other epidemics. Therefore sewage stagnation is more harmful to our health.

Question 13.
Jeevan said that “If there are no microorganisms earth will remain with wastes. ” Will you agree this statement? Why?
Answer:
Yes, I will agree this statement. Because of:

1. Microorganisms present in the soil, air and water act upon wastes around us.
2. They decomposes them.
3. They are converted into simple substances.
4. Thus microorganisms help us in cleaning the environment.
5. So if there are no microorganisms earth will remain with wastes.

Question 14.
Pranavi is suffering from serious illness. Doctor prescribed antibiotics for five days. After three days of usage she stopped taking antibiotics. Is it right or not, discuss.
Answer:
No, it is not right. Doctor prescribed anitbiotics for five days. Because the doctor knows that the correct days of the recovering of her illness. After three days of usage she stopped taking antibiotics. But her illness is not completely recovered. It is breaking out after some time. So it is not right.

Question 15.
What are the precautions taken to eradicate malaria?
Answer:

1. Malaria is caused by the pathogen piasmodium, is transmitted by female Anopheles mosquito.
   
2. Methods used to prevent malaria include medications, mosquito elimination and the prevention of bites.
3. Using of indoor mosquito repellents, residual sprays, taking care not to store water.
4. Providing awareness how to control mosquito population.

Question 16.
One medical store owner is giving antibiotics to his customer who is suffering from fever without doctor’s prescription? But the customer’s daughter Malathi is telling her father not to take antibiotics without doctor’s prescription. Whom do you support and why?
Answer:
We must use antibiotics prescribed by qualified doctor only. If you use antibiotics without consulting a qualified doctor it may harm us. Unnecessary use of antibiotics affects blood cells which fight infections. Sometimes they may kill useful bacteria too in our intestine and this increases the resistance towards antibiotics.

8th Class Biology 3rd Lesson Story of Microorganisms 2 InText Questions and Answers

Question 1.
What diseases can be prevented if we control mosquitoes?
Answer:
Malaria, Dengue, Chickungunya, Japanese Encephalitis.

Question 2.
Which diseases can we prevented by vaccination?
Answer:
Tuberculosis, Chickenpox, Measles, Polio, Swineflu.

Question 3.
Name the diseases which are transmitted by contaminated water?
Answer:
Cholera, Typhoid

Question 4.
Can you name some diseases which are transmited by Air?
Answer:
Tuberculosis, Chickenpox, Measles, Polio, Swineflu.

8th Class Biology 3rd Lesson Story of Microorganisms 2 Activities

Activity – 1

Question 1.
Take some lukewarm milk in two small bowls. Add a few drops of butter milk or little curd in one of the bowls. In the second bowl do not add anything. Keep the two bowls in a warm place and observe the milk in the bowls after 5 to 6 hours.
a) What changes did you observe?
Answer:
In the first bowl milk changed into curd. There is no change in the second bowl.
b) What is the reason for this?
Answer:
We add curd or butter milk in the first bowl. Curd or buttermilk contain bacterium named lactobacillus. Lactobacillus converts the milk into curd.

Activity – 2

Question 2.
Take 100 grams of maida in a bowl, add one or two spoons of yeast powder, add some water and knead it to make dough. Keep the dough in a warm place. Observe the dough after 3 – 4 hours.
a) What changes do you observe in the dough ?
Answer:
We observe that the dough rises and increase its volume.
b) What might be the reason ? Discuss with your friends and write about it.
Answer:
When yeast is added to the maida dough, the dough rises. This is due to the produc¬tion of carbondioxide gas during the process of fermentation. Bubbles of the gas fill the dough, increase its volume and make it spongy in nature.

Activity – 3

Question 3.
Commercial use of microorganisms.
Take some water in two separate beakers. Add 5 to 10 spoons of sugar to each beaker, then add 2 to 3 spoons of yeast to one of the bowls only. Close both of the bowls with lids and keep them in a warmplace. After 3 to 4 hours remove the lids and smell the contents.
a) What differences did you observe between the two bowls?
Answer:
We feel a new variety of smell from one of the bowls. This is the characteristic smell of alcohol. There is no smell from another bowl.
b) What will be the reason for the odour in yeast mixed bowl?
Answer:
Sugars are converted into alcohol by yeast. This process of conversion of sugars into alcohol is known as fermentation.

Activity – 4

Question 4.
Visit nearby PHC and collect information about vaccination given to 0-15 years chil-dren. Meet a doctor or a health worker and ask what types of vaccines are there ? Which disease can be prevented ? When it should be taken? List them out.
Answer:

Activity – 5

Question 5.
Take two pots or dig two pits in the corner of the garden at home or at your school ground. Fill them up to half with loose soil. Put some biological wastes like fallen leaves, vegetable wastes, waste papers, etc., in one of them. Fill the second one with plastic wastes, polythene bags and with some empty glass bottles.
Answer:
Now cover both the pits with soil and sprinkle water twice a day up to three weeks and observe the changes in both the pits and record.
After three weeks the pit which consists organic wastes decomposed into inorganic matter. But the pit which is filled with polythene bags, plastics and glass bottles is remain same no change is observed. Because of plastics and polythene materials are non biodegradables.

Activity – 6

Question 6.
Meet a doctor of your locality and ask him about the different types of diseases, caused by different microorganisms. Note them down. (OR)
Read the following table and answer the following questions.


Questions:
a) Name the viral diseases that can be prevented by controlling mosquito population.
Answer:
Dengue, Chikungunya, Japanese encephalitis.

b) Which bacterial disease / diseases cannot be controlled by vaccination (acc.to above table) ?
Answer:
Cholera, Typhoid.

c) What would you do to control mosquito population ?
Answer:
Use mosquito repellents, control breeding of mosquitoes by not allowing water stag¬nation in our surroundings.

d) Name the viral disease that spreads through water.
Answer:
Polio.

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

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

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

Do this

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

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

ii)

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

iii)

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

iv)

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

v)

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

vi)

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

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

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

Answer:
Area of a trapezium

fig (ii)

Answer:
Area of a trapezium

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

Graph Sheet:

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

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

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

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

Answer:

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

Area of trapezium PBCQ:

Area of △QCD:

Area of △DER:

Area of △ERA:

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

Answer:

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

Area of △QBC :

Area of △DRC :

Area of trapezium EPRD:

Area of △EPA :

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

Try these

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

Answer:
Area of a parallelogram ABCD

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

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

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

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

ii)

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

iii)

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

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

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

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

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

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

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

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

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

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

Think, discuss and write

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

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

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

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

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

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

= 25 × 26 – 5 × 6
= 650 – 30
= 620

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

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

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

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

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

Question 7.

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

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

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

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

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

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

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

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

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.2

Question 1.
Observe the following tables and fmd which pair of variables (x and y) are in inverse proportion

Solution:
i) From the given table if the value of x is decreases then the value of ‘y’ is increases.
∴ x, y are in inverse proportion.
ii) From the given table if the value of x is increases then the value of y is decreases.
∴ x, y are in inverse proportion.
iii) From the given table the value of x is decreases then the value of y is increases.
∴ x, y are in inverse proportion.

Question 2.
A school wants to spend ₹6000 to purchase books. Using this data, fill the following table.

Solution:

Question 3.
Take a squared paper and arrange 48 squares in different number of rows as shown below.

What do you observe? As R increases, C decreases
(i) Is R₁:R₂ = C₂:C₁?
(ii) Is R₃:R₄ = C₄:C₃?
(iii) Is R and C inversely proportional to each other?
(iv) Do this activity with 36 squares.
Solution:
(i) Is R₁:R₂ = C₂:C₁
⇒ 2 : 3 = 16 : 24

(ii) Is R₃:R₄ = C₄:C₃

R₁:R₂ = C₂:C₁

(iii) R₃:R₄ = C₄:C₃
⇒ 4 : 6 = 8 : 12
\(\frac{4}{6}=\frac{8}{12}=\frac{4 \times 2}{6 \times 2}=\frac{4}{6} \Rightarrow \frac{4}{6}=\frac{4}{6}\) =
∴ R₃:R₄ = C₄:C₃

(iv) Do this activity with 36 squares.

From the above table we can conclude that if number of rows are increases then number of columns are decreases.

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 7 Different Ecosystems Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 7th Lesson Different Ecosystems

8th Class Biology 7th Lesson Different Ecosystems Textbook Questions and Answers

Improve Your Learning

Question 1.
Define an ecosystem. Explain it with a suitable example.
Answer:

1. An ecosystem can be defined as a functional unit of nature, where living organisms interact among themselves and also with the surrounding physical environment.
2. For example, Mangroves are one of the most productive ecosystem on earth, deriving nourishment from terrestrial fresh water and tidal salt water.
3. Coringa mangrove is situated south of Kakinada Bay and is about 150 km south of Visakhapatnam.
4. It is named after the river coringa. Coringa mangroves receives fresh water from coringa and Gaderu rivers and salt waters from Kakinada Bay.
5. Biotic components in coringa:
   Producers: Mangrove, Spirogyra, Euglena, Oscilatoria, Blue Green Algae, Ulothrix etc.
   Consumers: Shrimp, crab, hydra, protozoans, mussel, snails, turtle, daphnia, brittle word, tube worm etc.
   Decomposers: Detritus feeding bacteria etc.
6. Abiotic components: Salt and fresh water, air, sunlight, soil, etc.

Question 2.
Explain how diversity of living organisms helps in enriching any ecosystem.
Answer:

1. The existence of the ecosystem depends on the continued survival of the organisms in the ecosystem.
2. All organisms require energy for growth, reproduction and survival.
3. This energy is obtained by the organisms from the food they consume.
4. Plants are the producers producing food in any ecosystem. The animals present in the ecosystem are consumers as they consume food from plants.
5. Some of the organisms in the ecosystem such as bacteria and fungi, obtained their nutritional requirements by decomposing the dead bodies of both producers and consumers.
6. They retain nutrients to the soil for the plants to use. As the cycle begins again.
7. Like this diversity of living organisms helps in enriching any ecosystem.

Question 3.
What happens when two animals having similar habits share one ecosystem?
Answer:
When two animals having similar habits, sharing one ecosystem , only the stronger and better equipped animal can survive, while the weaker one die or eliminated from the ecosystem. This is called ‘Survival of the fittest’.

Question 4.
What is the difference between habitat and ecosystem?
Answer:
Habitat is the natural living place of an organism or a group of organisms. Land and water are the major habitates.
An ecosystem is a Natural unit and has both Abiotic and biotic components, which interact and influence each other.

Question 5.
Who am I?
1. I am the base of food chain.
Answer:
Green plants.
2. I depend on plants for food.
Answer:
Consumers (Herbivorous Animals)
3. I break down the remains of dead plants and animals.
Answer:
Decomposers.
Ex: Bacteria, Fungi.

Question 6.
Which of the following is producer and why?
а) fox b) fungus c) chicken d) grass.
Answer:

1. Grass is the producer.
2. All green plants produce food materials with the help of carbondioxide and water in the presence of sunlight. So plants are called producers.
3. As grass is the green plant and produce food for other animals, it is called producer.

Question 7.
What do you understand by food web? Describe your own food web with the help of diagrammatic representation.
Answer:

1. A food web consists of several interlinked food chains and each organism in the food web will be a member of more than one food chain.
2. For example rats and insects eat seeds and other plant parts. As their food.
3. Insects are eaten by frogs and lizards.
4. Rats and frogs are eaten by snakes.
5. Lizards and snakes are eaten by birds.
6. Birds are eaten by fox, wolves. These are eaten by tigers and lions.
7. Thus a single plant or an animal may become food for more than one animal.
8. Similiarly an animal may consume more than one type of food depending on its taste and availability in the ecosystem.
9. Thus each organism in an ecosystem may be a member of more than one food chain.
10. When we looked at these relationships between various organisms for food in the ecosystem, it appears that several food chains are interlinked with each other forming a food web.

FOOD WEB

Question 8.
An ecosystem that had mice. What happens if more cats were added to it?
Answer:

1. In an ecosystem if the mice and the cats were existing equally, smooth balance would be maintained. When reproduction took place in these animals, generations would be continued then the ecosystem would be maintained healthy.
2. When more cats were added in that ecosystem all the mice would be eaten away by the cats it leads to the end of mice population.
3. Because of lack of food, the cats, either have to leave the ecosystem or they die.
4. If there is no continuity of the animals, the ecosystem would be destroyed.

Question 9.
List out producers (plants, bushes, trees), consumers (herbivores and carnivores) and decomposers that you observed in your agriculture field or school garden.
Answer:
Producers:
Plants – Grass plants, creepers like pumpkin, bottlegourd, etc.
Bushes – Rose, Jasmine, chrysanthemum, marigold.
Trees – Palm, coconut, mango, drumstick, lemon, sweet lemon, guava.
Consumers:
Herbivores – Goat, sheep, buffalo, ox, rats, butterflies, etc.
Carnivores – Crow, dogs, snakes, frogs, lizards.
Decomposers – Fungi (mushroom), Bacteria.

Question 10.
In grassland ecosystem, rabbit eats only plants. They eat plants faster than the plants can grow back. What must happen to bring the ecosystem into balance?
Answer:
The animals like fox, wolves, tigers, etc. which feed on rabbits will be introduced in that grass land ecosystem, then the rabbits will be controlled by them. Thus the ecosystem will comes into balance.

Question 11.
Plant, Tiger, Rabbit, Fox, Hawk.
Did you find any connection among the above list of things. If we remove Rabbit from the list what will happen?
Answer:

1. Plant, tiger, rabbit, fox, hawk these are the animals living in an ecosystem and are interdepending one on the other, and maintaining a food web – Plants → Rabbit → Hawk → Fox → Tiger.
2. A delicate balance is seen in nature between members of different species.
3. Any disturbance in this balance affects the organisms in a food web.
4. If we remove rabbit from the list the other animals like tiger, fox, hawk which are depending one on the other for food will die because of no food.
5. All the organisms, big or small, have right to live on this planet as man. We should respect this and allow other organisms to live and share the wealth of this planet.
6. “LIVE AND LET LIVE” should become our motto.

Question 12.
What do you understand by interdependency of animals and plants ? How do you appreciate ?
Answer:

1. An ecosystem is made up of groups of living things and their environments.
   
2. Living things like plants, animals and microorganisms are known as biotic components and others like soil, water, sunlight etc are called as abiotic components of the ecosystem.
3. All these organisms live together and interact with one another in many ways.
4. There is a feeding relationship between plants and animals. Along with this an interdependence between plants and animals for space, reproduction, shelter, etc.
5. All organisms in an ecosystem derive energy from food to live.
6. The sun is the main source of energy for all living things.
7. Plants being autotrophic, trap this energy through a process called photosynthesis and produce food to all living organisms. They are known as Primary producers.
8. Animals as they can not prepare food, they consume plants directly or indirectly and called consumers.
9. Living organisms like fungi and bacteria which are called decomposers, decay and decompose the dead animals of producers and consumers, and valuable nutrients to the soil for plants to use, as the cycle begins again.
10. A delicate balance is seen in nature between plants and animals by interdependence one to the other for thousands of years, which is unreachable to the human brain.

8th Class Biology 7th Lesson Different Ecosystems InText Questions and Answers

Question 1.
What is a Habitat?
Answer:

1. The dwelling place for plants and animals is called habitat.
2. One habitat shared by different types of plants and animals.
3. Try to add more such points to your list.
4. The natural home for plants and animals is called a habitat.
5. Habitat is the environment of an animal or plant.
6. Habitat is a suitable place for plants and animals to live.
7. Habitat is the origin for plants and animals.

Question 2.
Draw the diagram of Interdependence between the biotic components and answer the following questions.
Answer:
Interdependence between the biotic components:

1. What do the arrows in the figure indicate?
Answer:
The animals are depending one on the other for food.

2. Trace the path from grass to tiger. You may trace out other paths as well.
Answer:
Grass → grass hopper → frog → snake.
Carrot and grass → rabbit → fox → tiger
Plants → deer → bear → tiger
Seeds → squirrel → eagle → tiger

3. On how many organisms is rabbit dependent? Write their names.
Answer:
Carrot, grass.

4. How many organisms depend on rabbit? Write their names.
Answer:
Snake, fox, eagle, tiger.

5. Where do plants get their food from?
Answer:
Plants are autotrophs they can prepare their food from carbondioxide and water, in the presence of sunlight.

6. What other things do animals need for their survival?
Answer:
Abiotic components like soil, water, sunlight, etc.

8th Class Biology 7th Lesson Different Ecosystems Activities

Activity – 1

Lab Activity
Answer:
Aim: Study an ecosystem at your school/ home garden to understand it’s structure. Material Required : Measuring tape string, small sticks, hand lens, hand towel.
Procedure: To know about structure of the ecosystem we have to follow the following procedure.

1. Use the tape to measure a square area that is one meter long and one meter wide. It can be on grass, bare dirt or side wall.
2. Mark the edges of the square with the help of string/chalk.
3. Observe the study area (that has been marked). Look for the plants and animals that live there. Use the hand lens.
4. Record all the living organisms you see. You can even dig to go deeper to find out other living organisms that may be present there.

Observation / Findings: We find –

1. Plants like grass, herbs, shrubs, guava, neem and creepers.
2. Insects like ants, grass hoppers, butterflies, mosquitoes, houseflies, locusts, etc.
3. Animals like cat, dog, buffalo, frog, lizards, garden lizards, snake.
4. Mushroom, algae.
5. Deeper layers of soil we find earth worm, leech, rats, bandicoots, rabbits, etc.
6. Birds like crow, parrot, mynah, etc.

Discussion:
1. What living things did you find in your study area? Try to count them if possible.
Answer:
Grass, creepers, shrubs, herbs, trees, herbivores, carnivores, fungi.
2. Which kind of living thing was most common in your study area?
Answer:
Plants.
3. How was your study area different from those of other student groups?
Answer:
Living conditions, food, animals and plants are different.
4. Other than the living organisms what other things can you record from your study area?
Answer:
Soil, water, sunlight (temperature) are recorded.

Activity – 2

Question 2.
Observe the food web given below figure.

Answer:
The diagram showing food web.
Now answer the following Questions.
1. Which are the producers in the food web?
Answer:
Grass, rice plants, maize, bushes.
2. Which are consumers?
Answer:
Fish, frog, birds, rats, rabbit, deer, tadpole, larva ,sheep, cat, fox, tiger, crane, eagle, snake, owl, peacock, insects, lion.
3. Where does the food web start from?
Answer:
Food web starts from green plants.
4. Name the organism where the food web ends.
Answer:
Crane, eagle, owl, peacock, lion.
5. What happens when plants and animals die in a food web?
Answer:
When plants and animals die, they are decayed and decomposed by Decomposers like bacteria and fungi. They return nutrients to the soil for plants to use, as the cycle begin again. This is the reason ‘Decomposers are also called as recyclers.

Activity – 3

Question 3.
Collect the information forests of Andhra Pradesh and write the flora and fauna and fill up the following table:
Answer:
Forests of Andhra Pradesh

Name of the Forest – Kondapalli Reserve Forest

Investigations:

1. Do all the forest have same type of vegetation?
Answer:
No, there are mainly trees that show much species diversity and greater degrees stratification.
2. Are producers afforest ecosystem higher than its consumers?
Answer:
Trees are higher than consumers, besides trees shrubs and ground vegetation also there.
3. Do all the forests have same type of animals?
Answer:
No, the availability of food and environment different type of animals are present in different forest.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.1

Question 1.
The cost of 5 meters of a particular quality of cloth is ₹ 210. Find the cost of(i) 2 (ii) 4
(iii) 10 (iv) 13 meters of cloth of the same quality.
Solution:
The cost of 5 m of a cloth = ₹ 210
The length of a cloth and its price are in direct proportion.
i) \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)
Here x₁ = 5, y₁ = 210
x₂ = 2, y₂ = ?

Question 2.
Fill the table.

Solution:

Question 3.
48 bags of paddy costs ₹ 16, 800 then find the cost of 36 bags of paddy.
Solution:
Number of bags of paddy and their cost are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 16,800
x₂ = 36 y₂ = ?

= 3 × 4200
y₂ = ₹ 12600
∴ The cost of 36 bags of paddy = ₹ 12600

Question 4.
The monthly average expenditure of a family with 4 members is 2,800. Find the
monthly average expenditure ofa family with only 3 members.
Solution:
Number of family members and their expenditure are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 4
y₁ = 2,800
x₂ = 3 y₂ = ?

= 3 × 700 = 2100
y₂ = ₹ 2100
The expenditure for 3 members = ₹ 2100

Question 5.
In a ship of length 28 m, height of its mast is 12 m. If the height of the mast in its model is
9 cm what is the length of the model ship?
Solution:
The length of ship and the height of its mast are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 28
y₁ = 12
x₂ = ? y₂ = 9

x₂ = 7 × 3 = 21
∴ The length of model ship = 21 m

Question 6.
A vertical pole of 5.6 m height casts a shadow 3.2 m long. At the same time find (j) the
length of the shadow cast by another pole 10.5 m high (ii) the height of a pole which casts
a shadow 5m long.
Solution:
length of a vertical pole and length of its shadow are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

i) x₁ = 5.6
y₁ = 3.2
x₂ = 10.5 y₂ = ?

∴The length of the shadow = 6 cm

ii) x₁ = 5.6 m x₂ = ?
y₁ = 3.2 m y₂ = 5
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

∴ x₂ = 8.75

Question 7.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Time and distance are in direct proportion
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 14 km , x₂ = ?
y₁ = 25min = \(\frac{25}{60} \mathrm{hr}=\frac{5}{12} \mathrm{hr}\) = y₂ = 5hrs
⇒ \(x_{2}=\frac{x_{1} \times y_{2}}{y_{1}}=\frac{14 \times 5}{5}=\frac{14 \times \not 5 \times 12}{\not 5}\)
= 168 km
∴ Lorry travelled in 5 hrs = 168km

Question 8.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 16 \(\frac { 2 }{ 3 }\) kilograms?
Solution:
Number of pages and their weight are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 12 km , x₂ = ?
y₁ = 40 gm
y₂ = 16 \(\frac { 2 }{ 3 }\) gm = \(\frac { 50 }{ 3 }\) x 1000 gm
= \(\frac { 50000 }{ 3 }\) gm

From (1)

∴ Number of pages = 5000

Question 9.
A train moves at a constant speed of 75 km/hr.
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of 250 km.
Solution:
Speed of the train = 75 km/hr
i) The distance travelled in 20 min.
d = s x t = 75 x 20 min
= 75 x = 25 km
= \(75 \times \frac{20}{60}=\frac{75}{3}\) = 25 km

ii) Time taken to travel 250 km
t = \(\frac{d}{s}=\frac{250}{75}\)
t = \(\frac{10}{3}\) hrs

Question 10.
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?
Solution:
The scale of the design of a microchip
= 40 : 1
The length of the design = 18 cm
The actual length of microchip = ?
The length of the design and actual length of the microchip are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 40 km , x₂ = 18
y₁ = 1
y₂ = ?
⇒ \(\frac{40}{1}=\frac{18}{y_{2}}\)
⇒ \(\frac{18}{40}=\frac{9}{20}\) cm
∴ The original (actual) length of the microchip = [latexs]\frac{9}{20}[/latex]cm

Question 11.
The average age of consisting doctors and lawyers is 40. If the doctors average age is 35 and the lawyers average age is 50, fmd the ratio of the number of doctors to the number of lawyers.
Solution:
Let the number of doctors = x
Number of lawyers = y
The average age of doctors = 35
The total age of doctors = 35 × x
= 35 x years
The average age of lawyers = 50
∴ The total age of lawyers = 50 x y
= 50y
According to the sum
\(\frac{35 x+50 y}{x+y}\) = 40
⇒ 35x + 50y = 40x + 40y
⇒ 40x – 35x = 50y – 40y
⇒ 5x = lOy
⇒ \(\frac{x}{y}=\frac{10}{5}\) (or)
x : y = 2 : 1
∴ The ratio of number of doctors to lawyers = 2:1

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.4

Question 1.
Check whether 25110 is divisible by 45.
Solution:
The given number = 25110
If 25110 is divisible by 45 then it should be divisible by 5 and 9.

∴ The number 25110 is divisible by 45

Question 2.
Check whether 61479 is divisible by 81.
Solution:
If 61479 is divisible by 81 then it is divisible by 9.
If the sum of the digits of a number is dívisible by 9 then the entire number is divisible by 9.
∴ 61479 → 6 + 1 + 4 + 7 + 9 → \(\frac { 27 }{ 9 }\) (R = 0)
∴ 61479 is divisible by 81. [∵ 9 is factor of 81]

Question 3.
Check whether 864 is divisible by 36? Verif,’ whether 864 is divisible by all the factors of 36 ?
Solution:
864 is divisible by 2 and 3.
∴ 864 is divisible by 6.
∴ 864 is divisible by 36 [ ∵ 6 is the factor of 36]
∴ Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18. 36.


∴ 864 is divisible by all the factor of 36.

Question 4.
Check whether 756 is divisible by 42? Verify whether 756 is divisible by all the factors of 42?
Solution:
756 is divisible by 2 and 3.
∴ 756 is divisible by 6.
2a + 3b + c = 2 x 7 + 3 x 5 + 6 = 14 + 15 + 6 → \(\frac { 35 }{ 7 }\) (R = 0)
∴ 756 is divisible by 7.
∴ 756 is divisible by 42. [ ∵ 6, 7 are the factors of 42]
Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42.


∴ 756 is divisible by all the factor of 42.

Question 5.
Check whether 2156 is divisible by 11 and 7? Verify whether 2156 is divisible by product of 11 and 7?
Solution:

Question 6.
Check whether 1435 is divisible by 5 and 7? Verify if 1435 is divisible by the product of 5 and 7?
Solution:

Question 7.
Check whether 456 and 618 are divisible by 6’? Also check whether 6 divides the sum of 456 and 618 ‘?
Solution:

Question 8.
Check whether 876 and 345 are divisible by 3. Also check whether 3 divides the difference of 876 and 345?
Solution:

Question 9.
Check whether 2² + 2³+2⁴ is divisible by 2 or 4 or by both 2 and 4’?
Solution:

∴ 2² + 2³+2⁴ is divisible by both 2 and 4.

Question 10.
Check whether 32² is divisible by 4 or 8 or by both 4 and 8’?
Solution:

32² is divisible by 4 and 8

Question 11.
If A679B is a 5-dit number is divisible by 72 find ‘A’ and ‘B”?
Solution:
If A679B is divisible by 72 then it should be divisible by 8 and 9.
[ ∵ 8, 9 are the factors of 72]
A679B is divisible by 9 then
A + 6 + 7 + 9 + B = A + B + 22 = 27 (= 9 x 3)
=A + B = 5 ……………. (1)
A679B → \(\frac{79 \mathrm{~B}}{8}\) [From B (2,4,6,8) we take B = 2]
= \(\frac{792}{8}\) (R = 0)
∴ B = 2
From (1) ⇒ A + 2 = 5
∴ A = 3, B = 2

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

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

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

Improve Your Learning

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Activity – 2

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

Activity – 6

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

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

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

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

Do this

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

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

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

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

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

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

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

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

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

Try These

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

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

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

Answer:

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

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

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

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

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

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

Think, Discuss and Write

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

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

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

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

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

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.3

Question 1.
Check whether the given numbers are divisible by ‘6’ or not?
(a) 273432
(b) 100533
(c) 784076
(d) 24684
Solution:
if a number is divisible by ‘6’, it has to be divisible by 2 and 3.

Question 2.
Check whether the given numbers are divisible by ‘4’ or not?
(a) 3024
(b) 1000
(c) 412
(d) 56240
Solution:

Question 3.
Check whether the given numbers are divisible by ‘8’ or not?
(a) 4808
(b) 1324
(c) 1000
(d) 76728
Solution:

Question 4.
Check whether the given numbers are divisible by ‘7’ or not?
(a) 427
(b) 3514
(e) 861
(d) 4676
Solution:

Question 5.
Check whether the given numbers are divisible by ‘11’ or not?
(a) 786764
(b) 536393
(c) 110011
(d) 1210121
(e) 758043
(f) 8338472
(g) 54678
(h) 13431
(i) 423423
(j) 168861
Solution:

Question 6.
If a number is divisible by ‘8’, then it also divisible by ‘4’. also Explain?
Solution:
If a number is divisible by 8 it ¡s also divisible by 4.
∴ If a number is divisible by 8, then it ¡s also divisible by the factors of 8.
Factors of 8 = 1, 2, 4, 8.
∴ The number which is divisible 8, is also divisible by 4.

Question 7.
A 3-digit number 4A3 is added to another 3-digit number 984 to give four digit number 13B7, which is divisible by 11. Find (A + B).
Solution:
The given 3 – digited numbers are = 4A3, 984
∴ 4A3 + 984 = 13B7. If It is divisible by 11 then,
⇒ 1 3 B 7
(1 + B) – (3 + 7)
⇒ (B+1) – 10 = 0 ⇒ B – 9 = 0
∴ B = 9

⇒ A + 8 = 9 ⇒ A = 9 – 8 = 1
∴ A = 1
A + B= 1+9
∴ A + B = 10