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AP 8th Class Biology Notes Chapter 3 Crop Production and Management

→ Agricultural Practices : The different activities performed from ploughing upto harvesting are called agricultural parctices.

→ Animal Husbandry : The large scale rearing of animals in farms by providing them proper food, shelter and care is called animal husbandry.

→ Crop : When plants of the same kind are cultivated at one place on a large scale, it is called a crop.

→ Fertiliser : Fertilizers are chemicals which are rich in a particular nutrient.

→ Granaries : These are the huge storing places of grains, where grains are protected from the attacks of microbes, insects and rats.

→ Harvesting : The cutting of crop after it’s maturity is called harvesting.

→ Irrigation : The supply of water to crops at regular intervals is called irrigation.

→ Kharif : The crops which are grown in Kharif season (or) rainy season (June – September) (or) The time between June to September is known as Kharif.

→ Manure : The decomposed organic matter obtained from plant or animal waste.

→ Plough : The agricultural tool used for tilling or ploughing is known as plough.

→ Rabi : The crops which grow in winter season, that is from October to March (or) The time between October – March is known as Rabi.

→ Seeds : A plant’s fertilised ovules, from which a new plant may arise are called seeds.

→ Silo : The large metal containers, which can store huge quantities of grains safely are known as ‘Silo’.

→ Sowing : The method of putting seeds in the soil.

→ Storage : Grains have to be protected from pests and rats for a long time in gunny bags, silos and granaries.

→ Threshing : In the harvested crop, the grain seeds need to be seperated from the chaff. This process is called threshing.

→ Weeds : The undesirable plants that may grow naturally along with the crop, are called weeds.

→ Weedicide : The chemical which are used to control weeds are called weedicides.

→ Winnowing : The process of separation of grain and chaff by farmers with small holdings of land is called winnowing.

→ We need energy to perform our daily activities. Energy is derived from plants and animals. We get our food from plants or animals or both.

→ In order to provide food for a large population – regular production, proper management and distribution is necessary.

→ Same kind of plants cultivated at a place on a large scale is called a crop.

→ The process of yielding crops is called agriculture.

→ In our country, we have two broad cropping patterns are identified. They are

• Kharif crops; Which are sown in the rainy season (June to September)
• Rabi crops – Which are given in the winter season (October to March)

→ Before growing crops ploughing the soil properly is necessary. wooden plough is generally used for ploughing leveller is used for levelling the field. Levelling is very beneficial for sowing as well as irrigation.

→ Selection of good quality seeds is an important step in agriculture.

→ Sowing of seeds at appropriate depths and distances gives good yield. Sowing is done by seed – drills.

→ Farmers have to add manure to the fields to replenish the soil with nutrients. It is called manuring. Farmers use chemical fertilizers to improve the fertility of the soil.

→ Even though the fertilizers have helped to get better yield of crops, they become the source of water pollution.

→ The supply of water to crops at regular intervals is called irrigation. Wells, tube wells, ponds, lakes, rivers, dams and canals are the sources of irrigation in India.

→ Now a days, farmers are using modern irrigation methods like sprinklers and drip system, they save water and a boon to low water availability lands.

→ Farmers have to remove weeds from the crop fields. They use physical methods like hand picking and using Khurpi, Seed drill is also used to uproot the weeds.

→ Harvesting is the cutting of the mature crop manually or by harvester.

→ A machine called “combine” which is infact a harvester as well as thresher.

→ Proper storage of grains is necessary to protect them from pests and micro organisms.

→ Food is also obtained from animals for which animals are reared at home or farms. This large scale rearing of animals in farms is called animal husbandry.

8th Class Biology Notes

AP 8th Class Physical Science Notes Chapter 9 Some Natural Phenomena

→ Crust : The outermost layer of the earth is called the crust.

→ Discharge : Process of the flow of electric charge from one body to another.

→ Earth’s Plates : The fragments of the outermost layer of the earth are called plates.

→ Earthquake : An earthquake is a sudden shaking or trembling of the earth which lasts for a very short time, caused by a disturbance deep inside the earth’s crust.

→ Electroscope : The device that can be used to test whether an object is carrying a charge or not is known as an electroscope.

→ Lightning : Lightning is a natural phenomenon caused by the accumulation of charges in the clouds.

→ Lightning Conductor : A lightning conductor is a device used to protect buildings from the effect of lightning.

→ Negative Charge : It is a type of charge that forms on an object by gaining electrons while rubbing with another object.

→ Positive Charge : It is a type of charge that forms on an object by losing electrons while rubbing with another object.

→ Richter Scale : The power of an earthquake is expressed in terms of a magnitude on a scale called the Richter scale.

→ Seismograph : A seismograph is an instrument to record seismic waves.

→ Thunder : The loud noise produced when discharge takes place in the air between clouds is called thunder.

→ Thunderstorm : A thunderstorm is a storm with lightning and thunder.

→ Transfer of Charge : The flow of charged particles from one object to another object is called the transfer of charge.

→ Tsunami : Tsunami is a natural phenomenon caused by an earthquake in the sea.

→ Tremor : The shaking of the earth is called a fremor.

→ Lightning is caused by the accumulation of charges in the clouds.

→ When a plastic refill is rubbed with polythene, it acquires a small electric charge. These objects are called charged objects.

→ A charged balloon repelled a charged balloon.

→ A charged refill repelled a charged refill.

→ A charged balloon attracted a charged refill.

→ The electrical charges generated by rubbing are static.

→ The device that can be used to test whether an object is carrying a charge or not is known as an electroscope.

→ Electrical charge can be transferred from a charged object to another through a metal conductor.

→ The process of transferring charge from a charged object to the earth is called earthing.

→ Earthing is provided in buildings to protect us from electrical shocks due to any leakage of electrical current.

→ A lightning strike could destroy life and property.

→ During lightning and thunderstorm, no open place is safe.

→ A lightning conductor is a device used to protect buildings from the effect of lightning.

→ If a thunderstorm occurs there is always a possibility of lightning and cyclones accompanying it.

→ Earthquakes can cause damage to human life and property on a huge scale.

→ An earthquake is a sudden shaking or trembling of the earth which lasts for a very short time. It is caused by a disturbance deep inside the earth’s crust.

→ Earthquakes occur all the time, all over the earth.

→ The fragment of the outermost layer of the earth is called a plate. These plates are in continual motion.

→ The boundaries of the plates are the weak zones where earthquakes are more likely to occur. The weak zones are also known as seismic or fault zones.

→ The power of an earthquake is expressed in terms of a magnitude on a scale called the Richter scale.

→ The tremors produce waves on the surface of the earth. These are called seismic waves. The waves are recorded by an instrument called the seismograph.

→ An earthquake of magnitude 6 has thousand times more destructive energy than an earthquake of magnitude 4.

→ Earthquakes cannot be predicted.

→ An earthquake measuring 7 or more on the Richter scale can cause severe damage to life and property.

8th Class Physical Science Notes

Well-designed AP Board Solutions Class 7 Maths Chapter 1 Integers Exercise 1.3 offers step-by-step explanations to help students understand problem-solving strategies.

Integers Class 7 Exercise 1.3 Solutions – 7th Class Maths 1.3 Exercise Solutions

Question 1.
Evaluate each of the following:
a) (−30) ÷ 10.
Solution:
– 30 ÷ 10 ⇒ \(\frac{-30}{10}\) = –3

b) 50 ÷ (− 5)
Solution:
50 ÷ (−5) ⇒ \(\frac{50}{-5}\) = –10

c) (−36) ÷ (−9)
Solution:
(−36) ÷ (−9) ⇒ \(\frac{-36}{-9}\) = 4

d) (−49) ÷ (49)
Solution:
−49 ÷ (49) ⇒ \(\frac{-49}{49}\) = -1

e) 13 ÷ [(− 2) + 1]
Solution:
13 ÷ (− 2 + 1) ⇒ 13 ÷ – 1 ⇒ \(\frac{13}{-1}\) = -13

f) 0 ÷ (−12)
Solution:
0 ÷ (−12) ⇒ \(\frac{0}{-12}\) = 0

g) (−31) ÷ [(−30) + (−1)]
Solution:
(−31) ÷ [(−30 + (−1)] = − 31 ÷ (−31) = 1

h) [(−36) ÷ 12] ÷ 3
Solution:
[(−36) ÷ 12] ÷ 3 = \(\left(\frac{-36}{12}\right)\) ÷ 3 = − 3 ÷ 3 = − 1

i) [(−6) + 5)] + [(−2) + 1]
Solution:
[(−6) + 5] + [(−2) + 1] = −1 ÷ (−1) = 1

Question 2.
Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
a) a = 12, b = –4, c = 2
Solution:
Given a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
a = 12, b = –4, c = 2
LHS a ÷ (b + c) = 12 ÷ (− 4 + 2) = 12 ÷ – 2 = −6.
RHS (a ÷ b) + (a ÷ c)
= (12 ÷ −4) + (12 ÷ 2) = – 3 + 6 = 3
∴ LHS ≠ RHS

b) a = (−10), b = 1, c = 1
Solution:
LHS a ÷ (b + c) = − 10 ÷ (1 + 1) = − 10 ÷ 2 = − 5
RHS (a ÷ b) + (a ÷ c)
= (− 10 ÷ 1) + (− 10 ÷ 1)
= − 10 − 10 = − 20
∴ LHS ≠ RHS

Question 3.
Fill in the blanks

Question 4.
Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (− 2) = (− 3).
Solution:
Given a, b are integers such that a ÷ b = − 3
Given one pair (6, – 2) as 6 ÷ (− 2) = − 3
First pair:
Let a = 12, b = –4
12 ÷ –4 = – 3
(a, b) = (12, − 4)

Second pair: Let a = 36, b = – 12
36 ÷ – 12 = –3
(a, b) = (36, – 12).

Third pair:
Let a = – 30, b = 10
– 30 ÷ 10 = – 3
(a, b) = (– 30, 10)

Fourth pair:
Let a = 9, b = – 3
9 ÷ (–3) = – 3
(a, b) = (9, – 3)

Fifth Pair:
Let a = 15, b = – 5
15 ÷ (– 5) = – 3
(a, b) = (15, – 5)

Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the tempe-rature at mid-night ?
Solution:
Given the temperature at 12 noon was 10°C above zero = +10°C
Given temperature decreases at the rate of 2°C per hour until midnight.
Temperature at
1:00 PM = 10° C + (– 2° C) = 8° C
2:00 PM = 8° C + (– 2° C) =6° C
3:00 PM = 6° C + (– 2° C) = 4° C
4:00 PM = 4° C + (– 2° C) = 2° C
5:00 PM = 2° C + (– 2° C) = 0° C
6:00 PM = 0° C + (– 2° C) = – 2° C
7:00 PM = – 2° C + (– 2° C) = – 4° C
8:00 PM= – 4° C + (– 2° C) = – 6° C
9:00 PM = – 6° C + (– 2° C) = – 8° C
∴ At 9 pm the temperature will be 8° C below zero.
Temperature at
10:00 PM = – 8° C + (– 2° C) = – 10° C
11: 00 PM = – 10° C + (– 2° C) = – 12° C
12: 00 PM = – 12° C + (– 2° C) = – 14° C

Question 6.
In a class test (+3) marks are given for every correct answer and (– 2) marks are given for every incorrect answer and no marks for not attempting any question.
i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
ii) Mohini scores – 5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:
Given in a class test
Marks given for every correct answer = + 3
Marks given for every incorrect answer = – 2
Marks given for every non attempting = 0

i) Marks scored by Radhika = 20 marks
Number of correct answers by Radhika = 12
Marks for correct answers = 12 × 3 = 36
Marks for incorrect answers = 20 – 36 = – 16
Number of incorrect answers = – 16 ÷ (– 2) = 8
∴ 8 questions were attempted by her wrongly.

ii) Marks scored by Mohini = – 5
Number of correct answers = 7
Marks for correct answers = 7 × 3 = 21
Marks for incorrect answers = – 5 – 21 = – 26
Number of incorrect answers = –26 ÷ (– 2) = 13
∴ 13 questions were attempted incorrectly by Mohini.

Question 7.
An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach -350m?
Solution:
Distance descended is denoted by a negative integer
Initial height = + 10 m
Final depth = – 350 m
Total distance to be descended by the elevator = – 350 – (+ 10) = – 360 m
Time taken by the elevator to descend –6 m = 1 min
∴ Time taken by the elevator to descend
– 360 m = – 360 ÷ (– 6)
= 60 min
= 1 hour.

Well-designed AP Board Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2 offers step-by-step explanations to help students understand problem-solving strategies.

Linear Equations in One Variable Class 8 Exercise 2.2 Solutions – 8th Class Maths 2.2 Exercise Solutions

Question 1.
If you subtract \(\frac{1}{2}\) from a number and multiply the result by \(\frac{1}{2}\), you get \(\frac{1}{8}\). What is the number?
Answer:
Let the number = x
Now subtracting \(\frac{1}{2}\) from the number we get (x – \(\frac{1}{2}\)).
Now multiplying the result (x – \(\frac{1}{2}\)) by \(\frac{1}{2}\).
\(\frac{1}{2}\)(x – \(\frac{1}{2}\)) we get it \(\frac{1}{8}\)
∴ \(\frac{1}{2}\)(x – \(\frac{1}{2}\)) = \(\frac{1}{8}\)
Now multiply by ‘2’ on both sides.
We get

x – \(\frac{1}{2}\) = \(\frac{1}{4}\) (now transposing –\(\frac{1}{2}\) to RHS)
x = \(\frac{1}{4}+\frac{1}{2}=\frac{1+2}{4}=\frac{3}{4}\)
∴ The given number = \(\frac{3}{4}\)

Question 2.
The perimeter of a rectangular swimm- ingpool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Answer:
Let the breadth of rectangle = x m
then twice of it = 2x m
then its length = 2 more than its twice = 2x + 2

Then Perimeter = (2x + 2) + x + (2x + 2) + x = 154
(∵ Perimeter = l + b + l + b)

Now grouping variables and constants
(2x + x + 2x + x) + (2 + 2) = 154
⇒ 6x + 4 = 154 transposing + 4 to RHS
⇒ 6x= 154 – 4= 150
⇒ 6x = 150 (dividing both sides by 6)

⇒ x = 25 m
∴ Breadth of rectangle = x = 25 m
Length of rectangle = 2x + 2
= 2(25) + 2 = 50 +2 = 52 m
∴ Length = 52 m; Breadth = 25 m

Question 3.
The base of an isosceles triangle is \(\frac{4}{3}\) cm. The perimeter of the triangle is 4\(\frac{2}{15}\)cm. What is the length of either of the remaining equal sides?
Answer:
Let △ABC is an Isosceles triangle. In which AB = AC.

Let AB = AC = x cm and base BC = \(\frac{4}{3}\) cm
Then Perimeter of △ABC = 4\(\frac{2}{15}\)
= \(\frac{15(4)+2}{15}=\frac{60+2}{15}=\frac{62}{15}\)
= \(\frac{62}{15}\)

Perimeter of △ABC
= AB + BC + AC = \(\frac{62}{15}\)

∴ Length of remaining equal sides = \(\frac{7}{5}\) cm.

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Answer:
Let the smaller number = x
Then the greater number = x + 15( ∵ exceeds by 15)
Sum of both numbers = x + (x + 15) = 95
⇒ 2x + 15,= 95
⇒ 2x = 95 – 15 = 80 (transposing + 15 to RHS)
⇒ \(\frac{2 \mathrm{x}}{2}=\frac{80}{2}\) (now dividing by 2)
⇒ x = 40
So the smaller number = 40 and
greater number = x + 15 = 40 +15 = 55

Question 5.
Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Answer:
Ratio of two numbers = 5:3
Let the numbers are 5x, 3x
Then the difference = 5x – 3x = 2x = 18
∴ 2x = 18 (dividing by 2)
⇒ \(\frac{2 x}{2}=\frac{18}{2}\)
⇒ x = 9
Then the number 6x = 5 (9) = 45
Other number = 3x = 3 (9) = 27
So, the given numbers are 45, 27.

Question 6.
Three consecutive integers add up to 51. What are these integers?
Answer:
Let the three consecutive integers are
x, x + 1, x + 2.
Sum of them = 51
∴ x + x + 1+ x + 2 = 51
⇒ (x + x + x) + (1+ 2) = 51
⇒ 3x + 3 = 51 (transposing +3 to RHS)
⇒ 3x = 51 – 3 = 48
⇒ 3x = 48
Dividing both sides by ‘3’,
we get = \(\frac{3 x}{3}=\frac{48}{3}\)
⇒ x = 16
then x + 1 = 16 +1 = 17 and x + 2 = 16 + 2 = 18
∴ Three consecutive integers are 16, 17, 18

Checking:
Sum of 16,17 and 18 are 16+17 + 18 = 51 (verified.)

Question 7.
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Answer:
Sum of three consecutive multiples of 8 = 888.
Let one multiple of 8 = 8x (say)
Then the other two consecutive multiples would be 8(x + 1) and 8(x + 2).
Sum of the above three are
8x + 8(x + 1) + 8(x + 2) = 888
⇒ 8x + 8x + 8 + 8x + 16 = 888 (multiplicative distribution law)
⇒ (8x + 8x + 8x) + (8 + 16) = 888
⇒ 24x + 24 = 888 (transposing + 24 to RHS)
⇒ 24x = 888 – 24 = 864
∴ \(\frac{24 x}{24}=\frac{864}{24}\) (dividing by 24 on both sides)
⇒ x = 36
Then the multiples of 8,
that are 8x = 8 (36) = 288
and 8 (x + 1) = 8x + 8 = 288 + 8 = 296
and 8 (x + 2) = 8x + 16 = 288 + 16 = 304

Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Answer:
Let the 3 consecutive integers are
x, (x + 1) and (x + 2). then multiplying ‘x’ by 2
we get (x)(2) = 2x then multiplying (x + 1) by (3)
we get (x + 1) 3 = 3x + 3 then multiplying (x + 2) by ‘4’ ,
we get (x + 2) 4 = 4x + 8

Sum of the above
= (2x) + (3x + 3) + (4x + 8)
which is equal to 74.
2x + 3x + 3 + 4x + 8 = 74
⇒ (2x + 3x + 4x) + (3 + 8) = 74
⇒ 9x + 11 = 74 (transposing + 11 to RHS)
⇒ 9x = 74 – 11 = 63
⇒ 9x = 63 (dividing both sides by ‘9’ we get)
\(\frac{9 x}{9}=\frac{63}{9}\)
⇒ x = 7
∴ The given 3 consecutive integers are 7, 8, 9

Checking: 2(7) + 3(8) + 4(9)
= 14 + 24 + 36
= 74 (verified).

Question 9.
The ages of Rahul and Haroon are in the ratio 5:7. Fpur years later the sum of their ages will be 56 years. What are their present ages?
Answer:
Ratio of ages of Rahul and Haroon = 5: 7
then let age of Rahul = 5x years
and age of Haroon = 7x years.
Four years later age of Rahul
= (5x + 4) years
Four years later age of Haroon
= (7x + 4) years
Sum of their ages after four years = 5x + 4 + 7x + 4 = 56
⇒ (5x + 7x) + (4 + 4) = 56
⇒ 12x + 8 = 56 (transposing + 8 to RHS)
we get 12x = 56 – 8 = 48
⇒ 12x = 48

⇒ x = 4 years
then present age of Rahul
= 5x = 5(4) = 20 years
present age of Haroon
= 7x = 7(4) = 28 years

Checking:
12(4) + 8 = 56 ⇒ 56 = 56 (verified).

Question 10.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of giris. What is the total class strength?
Answer:
Ratio of number of Boys and Girls = 7:5
then let the number of boys = 7x (say)
and the number of girls = 5x
From the data given
No. of boys = 8 more than no. of girls
∴ 7x = 5x + 8
⇒ 7x – 5x = 8 (transposing + 5x to LHS)
⇒ 2x = 8 (dividing both sides by 2)
⇒ \(\frac{2 x}{2}=\frac{8}{2} = 4\) = 4
∴ x = 4 then total students = Boys + Girls
= 7x + 5x = 12x = 12 (4) = 48
∴ Total class strength = 48

Question 11.
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Answer:
Let Baichung’s age = ‘P’ years then Baichung’s Father age = P + 29
(∵ Father is 29 years older) then Baichung’s grand Father age
= P + 29 + 26(v Baichung’s Father is 26 younger than grand father)
Sum of all three ages =135 years
∴ p + p + 29 + P + 29 + 26 = 135
⇒ P + P + P + 29 + 29 + 26 = 135
⇒ 3P + 84 = 135
⇒ 3P = 135 – 84 (Transposing + 84 to RHS)
⇒ 3P = 51
⇒ P = \(\frac{51}{3}\)
⇒ P = 17
So Baichung’s age =17 years
Baichung’s father age = 17 + 29 = 46 years
Baichung’s grand father age = 46 + 26
= 72 years

Checking:
17 + 46 + 72 = 135 years (verified).

Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Answer:
Let Ravi’s present age = x years
After 15 years Ravi’s age will be = (x + 15) years.
From the data given After 15 years Ravi’s age
Four times of his present age
∴ (x + 15) = 4(x)
⇒ x + 15 = 4x
⇒ 15 = 4x – x (Transposing + x to RHS)
⇒ 15 = 3x (Re – writing the equation in j standard form)
⇒ 3x = 15 (dividing by 3 on both sides)
⇒ x = \(\frac{15}{3}\),
So Ravi present age = 5 year

Checking: After 15 years Ravi’s age
= 5 + 15 = 20 = 4(5) (verified)

Question 13.
A rational number is such that when you multiply it by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, you get \(\frac{-7}{12}\). What is the number?
Answer:
Let the number = ‘N’
When you multiply it by \(\frac{5}{2}\);
you get = (N)\(\frac{5}{2}=\frac{5 \mathrm{~N}}{2}\)
By adding \(\frac{2}{3}\) to the product,
we get \(\frac{5 N}{2}+\frac{2}{3}\)
then the result we get = \(\frac{-7}{12}\)

∴ The given number = \(\frac{-1}{2}\)

Checking: Number = \(\frac{-1}{2}\)
Multiplying \(\frac{5}{2}=\frac{-1}{2} \times \frac{5}{2}=\frac{-5}{4}\)
Adding \(\frac{2}{3}\frac{3(-5)+4(2)}{12}=\frac{-15+8}{12}\) we get \(\frac{-5}{4}+\frac{2}{3}\)
= \(\frac{3(-5)+4(2)}{12}=\frac{-15+8}{12}\)
= \(\frac{-7}{12}\)(verified)

Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakslimi is ₹ 4,00,000. How many notes of each denomination does she have?
Answer:
Ratio of currency note of denomination ₹ 100, ₹ 50, ₹ 10 = 2: 3: 5
So let the number of 100/- notes = 2x
Then the total value of 100 rupee notes in 2x number = (2x) (100)
= 200 x rupees …………(1)
Now let the number of ₹ 50/- notes = 3x (∵ 2 : 3 : 5 ratio)
Then the total value of 50 rupee notes in 3x number = (3x) (50)
= 150x rupees …….(2)
Now let the number of 10/- notes = 5x (∵ 2 : 3 : 5 ratio)
Then the total value of 10 rupee notes in 5x number = (5x) (10)
= 50x rupees …….(3)
∴ Total cash with Lakshmi = 4,00,000/-
From the above 200x + 150x + 50x = 4,00,000/-
⇒ 400x = 4,00,000/-
(dividing by 400 on both sides)
⇒ x = \(\frac{4,00,000}{400}\) = 1000
∴ Denomination of notes = 100 rupees notes = 2x = 2(1000) = 200
50 rupee notes = 3x = 3(1000) = 3000
10 rupee notes = 5x = 5 (1000) = 5000

Checking:
Value of 100 rupees denomination = (100) (2000) = 2,00,000
Value of 50 rupees denomination
= 50(3000)= 1,50,000
Value of 10 rupees denomination
= 10 (5000) = 50,000
Total = 4,00,000
Herice verified.

Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Answer:
Among the coins
let the number of 5 rupee coins = x (Say)
Then the number of 2 rupee coins = 3x (∵ 3 times to 5 rupee coins)
Rest of coins are 1 rupee coins.
Given that total number of coins = 160
∴ 1 rupee coins would be 160 – no. of (5 rupee coins + 2 rupee coins)
= 160 – (x + 3x) = 160 – 4x
Now,
the value of 5 rupee coins = 5 × its number = 5(x) = 5x /- rupees
The value of 2 rupee coins = 2 × its number = 2(3x) = 6x /- rupees
The value of 1 rupee coins = 1 × its number = 1(160 – 4x)= 160-4x
Value of total ₹ 1, ₹ 2, ₹ 5. coins
= 300 rupees
So from above 5x + 6x + 160 – 4x = 300
⇒ (5x + 6x – 4x) + 160 = 300
⇒ 7x = 300 – 160 (Transposing + 160 to RHS)
⇒ 7x = 140 (dividing by 7 on both sides)
⇒ \(\frac{7 x}{7}=\frac{140}{7}\) = 20
∴ x = 20
Then number of ₹ 5 coins = x = 20 …….. (1)
Then number of ₹ 2 coins = 3x = 3(20) = 60 ……….(2)
Then number of ₹ 1 coins = 160 – 4 (20) = 80…… (3)

Checking:
Value of 5 rupee coins = 20 × 5 = 100
Value of 2 rupee coins = 60 × 2 = 120
Value of 1 rupee coins = 80 × 1 = 80
Total Value = 100 + 120 + 80 = 300
Hence verified.

Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.
Answer:
The total participants = 63
Let the number of winners = x (say)
Then the number of loosers = 63 – x
So cost of prize to each winner = 100/-
Then cost of total winners
= (100) (x) = 100x rupees ……….(1)
Cost of prize to each looser = 25 rupees
The cost of prizes to the loosers
= (63 – x) (25)
= (63) (25) – x (25)
= (1575 – 25x) rupees ……..(2)
Total prize money distributed = ₹ 3000/-
∴ (1) – (2) = ₹ 3000/-
100x + 1575 – 25x = ₹ 3000
(100x – 25x) + 1575 = ₹ 3000 (Transposing + 1575 to RHS)
75x = 3000 – 1575 = 1425
∴ x = \(\frac{1425}{75}\) = 19
∴ So total number of winners =19 and number of looser = 63 – 19 = 44

Checking:
Prize for winner @100/- each = 19 × 100 = 1900
Prize for looser @ 25/- each = 44 × 25= 1100
Total Prize Amount = 1900 + 1100 = 3000
Hence verified.

Thoroughly analyzing AP Inter 2nd Year Maths 2A Model Papers and TS Inter 2nd Year Maths 2A Question Paper March 2019 in Telugu Medium helps students identify their strengths and weaknesses.

TS Inter 2nd Year Maths 2A Question Paper March 2019 in Telugu

Time: 3 Hours
Maximum Marks: 75

గమనిక : ఈ ప్రశ్నాపత్రంలో A, B, C అనే మూడు విభాగాలు ఉన్నాయి.

Section – A 10 × 2 = 20

I. అతిస్వల్ప సమాధాన తరహా ప్రశ్నలు.

1. అన్ని ప్రశ్నలకు సమాధానములు వ్రాయండి.
2. ప్రతి ప్రశ్నకు రెండు మార్కులు.

ప్రశ్న 1.
Z = (cos θ, sin θ) అయితే (Z – \(\frac{1}{Z}\) ని కనుక్కోండి.
సాధన:
Z = (cos θ, sin θ)
⇒ Z = cos θ + i sin θ
\(\frac{1}{Z}\) = \(\frac{1}{(\cos \theta+i \sin \theta)}\) = cos θ – i sinθ
Z – \(\frac{1}{Z}\) = (cos + i sin θ) (cos – i sin θ)
= cos θ + i sin – cos + i sin θ
= 2i sin θ.
∴ Z – \(\frac{1}{Z}\) = 2i sin θ.

ప్రశ్న 2.
Z = x + iy, |Z| = 1, అయితే Z యొక్క బిందు పథాన్ని కనుక్కోండి.
సాధన:
Z = x + iy
|Z| = 1
⇒ \(\sqrt{x^2+y^2}\) = 1
⇒ x² + y² = 1
Z యొక్క బిందు పథము
∴ x² + y² = 1.

ప్రశ్న 3.
ఏకకపు (ఒకటి) ఘనమూలాలు I, W, W² అయితే (1 – W + W²)³ విలువ కనుక్కోండి.
సాధన:
1 (ఏకకపు) ఘనమూలాలు 1, W, W² కావున
∴ 1 + W + W² = మరియు W³ = 1
(1 – W + W²)³ = (- W – W)³ = (-2W)³ = -8W³ = -8(1) = -8
∴ (1 – W + W²)³ = – 8.

ప్రశ్న 4.
x² + x – 12 ≤ 0 యొక్క సాధన సమితిని బీజీయ పద్ధతి ద్వారా కనుక్కోండి.
సాధన:
x² + x – 12 ≤ 0
⇒ x² + 4x – 3x – 12 ≤ 0
⇒ x(x + 4) – 3(x + 4) ≤ 0
⇒ (x + 4) (x – 3) ≤ 0
⇒ x ∈ [-4, 3]
∴ x² + x – 12 ≤ 0 కు సాధన సమితి [-4, 3].

ప్రశ్న 5.
4x³ – 6x² + 7x + 3 = 0 మూలాలు α, β, γ అయితే αβ + βγ + γα విలువను కనుక్కోండి.
సాధన:
4x³ – 6x² + 7x + 3 = 0 …….. (1)
(1) యొక్క మూలాలు α, β, γ కావున
∴ αβ + βy + yα = s₂ = \(\frac{7}{4}\).

ప్రశ్న 6.
6 వేర్వేరు రంగుల పూసలతో, ఎన్ని రకాలుగా గొలుసులు తయారుచేయవచ్చు ?
సాధన:
The number of ways of preparing a chain with 6 different colomed beads = \(\frac{1}{2}\) (n – 1)!
= \(\frac{1}{2}\) (6 – 1)!
= \(\frac{1}{2}\) (5) !
= \(\frac{1}{2}\) (120)
= 60.

ప్రశ్న 7.
ⁿC₄ = 210 అయితే n విలువను కనుక్కోండి.
సాధన:
ⁿC₄ = 210
⇒ \(\frac{n(n-1)(n-2)(n-3)}{4!}\) = 5.6.7
⇒ n(n – 1) (n-2) (n – 3) = 24.5.6.7
= 7.8.9.10
= 10.9.8.7.
∴ n = 10.

ప్రశ్న 8.
(2x + 3y + z)⁷ విస్తరణలోని పదాల సంఖ్యను కనుక్కోండి.
సాధన:
(2x + 3y + z)⁷ విస్తరణలోని పదాల సంఖ్య
= \(\frac{(7 + 1)(7 + 2)}{2}\)
= \(\frac{8.9}{2}\) = 36.

ప్రశ్న 9.
5, 12, 3, 18, 6, 8, 2, 10 దత్తాంశానికి విస్తృతి కనుక్కోండి.
సాధన:
అవర్గీకృత దత్తాంశం 5, 12, 3, 18, 6, 8, 2, 10
మధ్యమము \(\bar{z}\) = \(\frac{5+12+3+18+6+8+2+10}{8}\)
= \(\frac{64}{8}\) = 8
విస్తృతి a² = \(\frac{1}{8}\) [(5 – 8)² + (12 – 8)² + (3 – 8)² + (18 – 8)² + (6 – 8)² + (8 – 8)² + (2 – 8)² + (10 – 8)]²
= \(\frac{1}{2}\) [9 + 16 + 25 + 100 + 4 + 0 + 36 + 4]
= \(\frac{1}{8}\) (194)
= 24.25.

ప్రశ్న 10.
ఒక ద్విపద విభాజనం అంకమధ్యమం, విస్తృతి వరుసగా 4, 3. ఆ విభాజనాన్ని సందానించి, P(x ≥ 1) ని కనుక్కోండి.
సాధన:
n, p లు పరామితులుగా x ద్విపద విభాజనాన్ని పాటిస్తుంది.
Let x follows a binomial distribution with parameters n and p.
మధ్యమము = np = 4 ……. (1)
విస్తృతి = npq = 3
\(\frac{n p q}{n p}\) = \(\frac{3}{4}\) ⇒ q = \(\frac{3}{4}\).
p = 1 – q = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\).
(1) నుండి
n (\(\frac{1}{4}\)) = 4 ⇒ n = 16.
p(x ≥ 1) = 1 – p(x = 0)
= 1 – 16_Co(\(\frac{1}{4}\))⁰ (\(\frac{3}{4}\))^{16 – 0}
= 1 – (\(\frac{3}{4}\))¹⁶.

Section – B 5 × 4 = 20

II. స్వల్ప సమాధాన తరహా ప్రశ్నలు.

1. ఏవైనా అయిదు ప్రశ్నలకు సమాధానాలు వ్రాయండి.
2. ప్రతి ప్రశ్నకు నాలుగు మార్కులు.

ప్రశ్న 11.
x + iy = \(\frac{1}{1+\cos θ+i \sin θ}\) అయితే 4x² – 1 = 0 అని చూపండి.
సాధన:
x + iy = \(\frac{1}{1+\cos θ+i \sin θ}\)
= \(\frac{1}{2 \cos ^2 \frac{\theta}{2}+i .2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\)
= \(\frac{1}{2 \cos ^2 \frac{\theta}{2}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)}\)
= \(\frac{1}{2 \cos \frac{\theta}{2}}\left(\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}\right)\)
= \(\frac{1}{2}\) – i\(\frac{1}{2}\) tan \(\frac{\theta}{2}\)
ఇరువైపులా వాస్తవ భాగాలను పోల్చగా
x = \(\frac{1}{2}\)
⇒ x² = \(\frac{1}{4}\)
⇒ 4x² = 1
⇒ 4x² – 1 = 0.

ప్రశ్న 12.
c² ≠ ab అయితే (c² – ab)x² – 2(a² – bc)x + (b² – ac) = 0 సమీకరణం మూలాలు సమానమైతే అప్పుడు a³ + b³ + c³ = 3 abc లేదా a = 0 అని చూపండి.
సాధన:
(c² – ab) x² – 2 (a² – bc) x + (b² – ac) = 0 ……… (1)
(1) కు సమాన మూలాలు కావున
∴ 4(a² – bc)² – 4(c² – ab) (b² – ac) = 0
⇒ (a² – bc)² – (c² – ab) (b² – ac) = 0
⇒ a⁴ – 2a²bc + b²c² – (b²c² – ac³ + ab³ – a²bc = 0
⇒ a⁴ – 2a²bc + b²c² – b²c² + ac³ + ab³ – a²bc = 0
⇒ a⁴ + ab³ + ac³ – 3a²bc = 0.
⇒ a(a³ + b³ + c³ – 3abc) = 0
⇒ a = 0 (లేదా) a³ + b³ + c³ – 3abc = 0
⇒ a = 0 (లేదా) a³ + b³ + c³ = 3abc.

ప్రశ్న 13.
MASTER పదంలోని అక్షరాలను ప్రస్తావించడం వల్ల వచ్చే పదాలను నిఘంటువు క్రమంలో రాస్తే ఆ వరసలో MASTER పదము యొక్క కోటిని కనుక్కోండి.
సాధన:
REMAST పధంలోని అక్షరాల నిఘంటువు క్రమంలో వ్రాయగా A, E, M, R, S, T
A తో మొదలయ్యే పదాల సంఖ్య 5! = 120
E తో మొదలయ్యే పదాల సంఖ్య 5! = 120
MAE తో మొదలయ్యే పదాల సంఖ్య 3! = 6
MAR తో మొదలయ్యే పదాల సంఖ్య 3! = 6
MASE తో మొదలయ్యే పదాల సంఖ్య 2! = 2
MASR తో మొదలయ్యే పదాల సంఖ్య 2! = 2
తర్వాత పదం MASTER = 1
∴ MASTER పదం యొక్క కోటి = 120 + 1.20 + 6 + 6 + 2 + 2 + 1 = 257

ప్రశ్న 14.
\(\frac{4 n C_{2 n}}{2 n C_n}\) = \(\frac{1.3 .5 \ldots(4 n-1)}{\{1.3 .5 . \ldots(2 n-1)\}^2}\) అని చూపండి.
సాధన:

ప్రశ్న 15.
\(\frac{x^3}{(x-a)(x-b)(x-c)}\) ను పాక్షిక భిన్నాలుగా విడగొట్టండి.
సాధన:
\(\frac{x^3}{(x-a)(x-b)(x-c)}\)
= 1 + \(\frac{A}{x-a}\) + \(\frac{B}{x-b}\) + \(\frac{C}{x-a}\) అనుకొనుము
⇒ x³ = A (x – b) (x – c) + B (x – a) (x – c) + C (x – a) (x – b) + (x – a) (x – b) (x – c)
x = a వ్రాయగా,
a³ = A(a – b) (a – c)
⇒ A = \(\frac{a^3}{(a-b)(a-c)}\)
x = b వ్రాయగా
b³ = B(b – a) (b – c)
∴ B = \(\frac{b^3}{(b-a)(b-c)}\)
x = c వ్రాయగా
c³ = C(c – a) (c – b)
C = \(\frac{c^3}{(c-a)(c-b)}\)
∴ \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = 1 + \(\frac{x^3}{(a-b)(a-c)(x-a)}\) + \(\frac{b^3}{(b-a)(b-c)(x-b)}\) + \(\frac{c^3}{(c-a)(c-b)(x-c)}\)

ప్రశ్న 16.
ఒక కాంట్రాక్టరు రోడ్డు కాంట్రాక్టును పొందే సంభావ్యత \(\frac{2}{3}\), భవనం కాంట్రాక్టును పొందే సంభావ్యత \(\frac{5}{9}\) కనీసం ఒక కాంట్రాక్టునైనా పొందే సంభావ్యత \(\frac{4}{9}\) అతడు రెండు కాంట్రాక్టులను పొందే సంభావ్యతను కనుక్కోండి.
సాధన:
ఒక కాంట్రాక్టరు రోడ్డు కాంట్రాక్టును పొందే ఘటనను A అని మరియు భవనం కాంట్రాక్టుకు పొందే ఘటనను B అని అనుకొనుము.
∴ P(A) = \(\frac{2}{3}\), P(B) = \(\frac{5}{9}\) మరియు
P(A ∪ B) = \(\frac{4}{5}\)
సంభావ్యతల సంకలన సిద్ధాంతంననుసరించి
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= P(A ∩ B) = P(A) + P(B) = P(A ∪ B)
= \(\frac{2}{3}\) + \(\frac{5}{9}\) – \(\frac{4}{5}\)
= \(\frac{30 + 25 – 36}{45}\)
= \(\frac{19}{45}\)
∴ కాంట్రాక్టరు రెండు కాంట్రాక్టులను పొందే సంభావ్యత = \(\frac{19}{45}\)

ప్రశ్న 17.
75% సందర్భాలలో A నిజం మాట్లాడతాడు. 80% సందర్భాలలో B నిజం మాట్లాడ తాడు. ఒక సంఘటన గురించి వారు చెప్పే విషయం పరస్పరం విభేదించ డానికి సంభావ్యత ఎంత ?
సాధన:
A, B లు ఒక ఘటన గురించి నిజము చెప్పే ఘటనను వరుసగా E₁, E₂ అనుకొనుము.
∴ P(E₁) = \(\frac{75}{100}\) = \(\frac{3}{4}\)
P(\(E_1^c\)) = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
∴ P(E₂) = \(\frac{80}{100}\) = \(\frac{4}{5}\) P(\(E_2^c\)) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
ఒక ఘటన గురించి వారు చెప్పే విషయం పరస్పరం విభేదించే విధాలుగా ఘటనను E అనుకొనిన అవి రెండు విధాలుగా పరస్పరము వివర్జితాలై ఉంటాయి.
i) A నిజము చెప్పటం మరియు B అబద్ధం చెప్పటం
ii) A అబద్ధం చెప్పటం మరియు B నిజం చెప్పటం
∴ P(E) = P(E₁ ∩ \(E_2^c\)) + P(E\(E_1^c\) ∩ E₂)
= P(E₁). P(\(E_2^c\)) + P(\(E_1^c\)) . P(E₂)
= (\(\frac{3}{4}\)) (\(\frac{1}{5}\)) + (\(\frac{1}{4}\)) (\(\frac{4}{5}\))
= \(\frac{3}{20}\) + \(\frac{4}{20}\) = \(\frac{7}{20}\).

Section – C 5 × 7 = 35

III. దీర్ఘ సమాధాన తరహా ప్రశ్నలు.

1. ఏవైనా అయిదు ప్రశ్నలకు సమాధానాలు వ్రాయండి.
2. ప్రతి ప్రశ్నకు ఏడు మార్కులు.

ప్రశ్న 18.
n పూర్ణాంకం Z = cis θ, (θ ≠ (2n + 1)\(\frac{\pi}{2}\)) అయితే \(\) = i tan θ అని చూపండి.
సాధన:
Z = cosθ + i sinθ
Z²ⁿ = (cos θ + i sin θ)²ⁿ
= cos 2nθ + i sin 2nθ.
Z²ⁿ – 1 = cos 2nθ + i sin 2nθ – 1
= 1 – 2 sin² nθ + i . 2 sin nθ cos nθ
= -2 sin² nθ + 2i sin nθ cos nθ
= 2i² sin² nθ + 2i sin nθ cos nθ
= 2i sin nθ (i sin nθ + cos nθ)
= 2i sin nθ (cos nθ + i sin nθ).
Z²ⁿ + 1 = cos 2nθ + i sin 2n θ + 1
= 2 cos²nθ – 1 + i 2 sin nθ cos nθ + 1.
= 2 cos² nθ + i 2 sin nθ cos nθ
= 2 cos nθ (cos nθ + i sin nθ)
∴ \(\frac{z^{2 n}-1}{z^{2 n}+1}\) = \(\frac{2 i \sin n \theta(\cos n \theta+i \sin n \theta)}{2 \cos n \theta(\cos n \theta+i \sin n \theta)}\)
= \(\frac{i \sin n \theta}{\cos n \theta}\)
= i tan nθ.
∴ \(\frac{z^{2 n}-1}{z^{2 n}+1}\) = i tan nθ.

ప్రశ్న 19.
x⁴ + x³ – 16x² – 4x + 48 = 0 సమీకరణం రెండు మూలాల లబ్దం 6 అయితే, సమీకరణాన్ని సాధించండి.
సాధన:
x⁴ + x³ – 16x² – 4x + 48 = 0 సమీకరణం మూలాల α, β, γ, δ.
అనుకొనుము.
అనుకొనుము αβ = 6.
s₁ = α + β + γ + δ = -1 → (1)
s₂ = αβ + αγ + αδ + βγ + βδ + γδ = -16 → (2)
s₃ = αβγ + γβδ + αγδ + βγδ = 4 → (3)
s₄ = αβγδ = 48
⇒ 6 γδ = 48
⇒ γδ = 8
(3) నుండి ⇒ 6γ + 68 + δα + δβ = 4
⇒ 6(α + β + γ + δ) + 2α + 2β = 4
⇒ 6(-1) + 2α + 2β = 4
⇒ 2α + 2β = 10
⇒ α + β = 5
(1) నుండి ⇒ 5 + γ + δ = – 1
= γ + δ = -6
α + β = 5, αβ = 6 ⇒ α = 2, β = 3
γ + δ = -6, γδ = 8 ⇒ γ = -2, δ = −4
∴ 2, 3, -2, -4.

ప్రశ్న 20.
(1 + x)ⁿ విస్తరణలో 3 వరుస ద్విపద గుణకాలు 36, 84, 126 అయితే n విలువ కనుక్కోండి.
సాధన:
(1 + x)ⁿ విస్తరణలో 3 వరుస గుణకాలు ⁿC_{r – 1}, ⁿC_r ⁿC_{r + 1} అనుకొనుము.

⇒ 9r + 15 = 14r
⇒ 5r = 15
⇒ r = 3
(4) నుండి
\(\frac{n-3+1}{3}\) = \(\frac{7}{3}\)
3n – 6 = 21
3n = 27
∴ n = 9.

ప్రశ్న 21.
x = \(\frac{1}{5}\) + \(\frac{1.3}{5.10}\) + \(\frac{1.3.5}{5.10.15}\) + ……. ∞ అయితే 3x² + 6x విలువ కనుక్కోండి.
సాధన:

= (\(\frac{5}{3}\))^1/2
(x + 1)² = \(\frac{5}{3}\)
x² + 2x + 1 = \(\frac{5}{3}\)
3x² + 6x + 3 = 5
⇒ 3x² + 6x = 2.

ప్రశ్న 22.
క్రింది దత్తాంశానికి మధ్యమం నుంచి మధ్యమ విచలనాన్ని కనుక్కోండి.

సాధన:
ఇచ్చిన దత్తాంశానికి ఇప్పుడు క్రింది పట్టిక తయారు నిర్మిద్దాం.

ఇచ్చట N = 40 = Σf_i
Σf_ix_i = 320
అంక మధ్యమము \(\bar{x}\) = \(\frac{1}{N}\) Σ f_ix_i = \(\frac{320}{40}\) = 8
పట్టిక నుండి Σ f_i |x_i – \(\bar{x}\)| = 140
మధ్యమం నుంచి మధ్యమ విచలనం = \(\frac{1}{N}\) Σ f_i |x_i – \(\bar{x}\)|
= \(\frac{1}{40}\) (140)
= 3.5.

ప్రశ్న 23.
I, II, III అంకెలను కలిగిన మూడు పెట్టెలలో క్రింది విధంగా బంతులు ఉన్నాయి.

ఒక పెట్టెను యాధృచ్ఛికంగా ఎంపిక చేసి, దాని నుంచి ఒక బంతిని తీశారు. అది ఎర్రనిది అయితే, అది పెట్టె – II నుంచి తీయగల సంభావ్యతను కనుక్కోండి.
సాధన:
I, II, III అంకెలను కలిగిన పెట్టెలను ఎన్నుకొనే ఘటనలు వరుసగా B₁, B₂, B₃ అనుకొనుము.
∴ P(B₁) = P(B₂) = P(B₃) = \(\frac{1}{3}\)
P(\(\frac{R}{B_1}\)) = ఎన్నుకొన్న మొదటి పెట్టె నుండి ఎర్రబంతి కావడానికి సంభావ్యత
= \(\frac{3}{6}\)
P(\(\frac{R}{B_2}\)) = ఎన్నుకొన్న రెండవ పెట్టె నుండి ఎర్రబంతి కావడానికి సంభావ్యత
= \(\frac{1}{4}\)
P(\(\frac{R}{B_3}\)) ఎన్నుకొన్న మూడవ పెట్టె నుండి ఎర్రబంతి కావడానికి సంభావ్యత
= \(\frac{3}{12}\)
ఒక పెట్టె యాదృచ్ఛికంగా ఎంపిక చేసి, దాని నుంచి ఒకటి తీయగా అది ఎర్రనిది అయితే, అది పెట్టె II నుంచి తీయగల సంభావ్యత

ప్రశ్న 24.
ఒక పాచికను దొర్లించారు. దాని ముఖంపై కనబడే సంఖ్య X యొక్క అంకమధ్యమం విస్తృతిలను కనుక్కోండి.
సాధన:
శాంపిల్ ఆవరణము S అనుకొనుము మరియు ఒక పాచికను దొర్లించినపుడు దాని ముఖంపై కనబడే సంఖ్య X అనుకొనుము.

అంకమధ్యమము μ = 1 P(X = 1) + 2 P(X = 2) + 3 P(X = 3) + 4 P(X = 4) + 5 P(X = 5) + 6 P(X = 6)

Teachers often recommend practicing with AP 10th Class Biology Model Papers Set 6 in Telugu Medium to enhance exam readiness.

AP 10th Class Biology Model Paper Set 6 with Solutions in Telugu

Time: 2 Hours
Maximum Marks: 50

సూచనలు :

1. ఈ ప్రశ్నాపత్రంలో 4 విభాగాలు మరియు 17 ప్రశ్నలు ఉండును.
2. III వ విభాగం నందు కేవలం 12వ ప్రశ్నకు మాత్రమే మరియు IV వ విభాగం నందు గల అన్ని ప్రశ్నలకు అంతర్గత ఎంపిక కలదు.
3. 2 గంటల సమయంలో, 15 నిమిషాల సమయం ప్రశ్నపత్రం చదువుటకై కేటాయింపబడినది.
4. అన్ని సమాధానములు మీకివ్వబడిన సమాధాన పత్రంలోనే రాయవలెను.
5. అన్ని సమాధానములు స్పష్టంగాను, గుండ్రంగాను రాయవలెను.

విభాగము – I 8 × 1 = 8 మా.

1. అన్ని ప్రశ్నలకు సమాధానములు వ్రాయుము.
2. ప్రతి ప్రశ్నకు 1 మార్కు.

ప్రశ్న 1.
కిరణజన్య సంయోగ క్రియకు కావలసిన కారకాలు ఏవేవి ?
జవాబు:
కిరణజన్య సంయోగ క్రియకు నాలుగు కారకాలు అవసరమవుతాయి.

1. సూర్యకాంతి
2. నీరు
3. కార్బన్ డై ఆక్సైడ్
4. హరితరేణువు

ప్రశ్న 2.
దారువు, ప్రసరణ కణజాలానికి మధ్యగల సారూప్యతను రాయండి.
జవాబు:
దారువు, పోషక కణజాలాలు రెండూ మొక్కలలో నీరు, పోషకాల ప్రసరణకు సహాయపడతాయి. వీటిని ప్రసరణ కణజాలం అని అంటారు.

ప్రశ్న 3.
రెండు మూత్రపిండాలు చెడిపోతే జరిగే పర్యవసానాలు రాయండి.
జవాబు:
రెండు మూత్రపిండాలు పూర్తిగా చెడిపోతే శరీరంలోని వ్యర్థాలు విసర్జితం కావు. అవి రక్తంలో పేరుకొనిపోయి శరీరానికి హాని కలుగచేస్తాయి.

ప్రశ్న 4.
ప్రతీకార చర్యలో వెన్నుపాము పాత్రను ఎలా అభినందిస్తావు ?
జవాబు:
వెన్నుపాము శరీరానికి ఆధారాన్ని ఇవ్వడమే కాకుండా శరీరభాగాల నుంచి వచ్చే సమాచారాన్ని మెదడుకు చేరుస్తుంది. అకస్మాత్తుగా చర్యలకు వెంటనే ప్రతిస్పందనలను చూపుతుంది. అలా మనల్ని రక్షిస్తుంది.

ప్రశ్న 5.
మానవ శుక్రకణం యొక్క చిత్రం గీయండి.
జవాబు:

ప్రశ్న 6.
క్రింది పట్టికను దేనిని తెలియజేస్తుంది ?

జవాబు:
ఇది మెండల్ ఏక సంకర సంకరీకరణను తెలుపుతుంది.

విభాగము – II 4 × 2 = 8 మా.

సూచనలు :

1. క్రింది ప్రశ్నలన్నింటికి సమాధానములు వ్రాయుము.
2. ప్రతి ప్రశ్నకు 2 మార్కులు.

ప్రశ్న 7.
ఆహారపు గొలుసు నుంచి ఉత్పత్తిదారులను తొలగిస్తే ఏమవుతుంది ?
జవాబు:

1. ఆహారపు గొలుసులో ఉత్పత్తిదారులు మొదటి పోషకస్థాయిని చూపుతాయి.
2. ఉత్పత్తిదారులు స్వయంగా ఆహారాన్ని తయారుచేసుకునే స్వయంపోషకాలు. దీనిపై ఇతర జీవులు ఆధారపడతాయి.
3. ఉత్పత్తిదారులను తొలగిస్తే ఆహారగొలుసులోని మిగిలిన పోషకస్థాయిలు కనుమరుగవుతాయి.

ప్రశ్న 8.
ఈ క్రింది పై (Pie) చార్టును విశ్లేషించి దిగువనీయబడిన ప్రశ్నలకు జవాబులు రాయండి.

1) శిలాజ ఇంధనాలను గుర్తించండి.
జవాబు:
బొగ్గు, సహజవాయువు, నూనెలు.

2) వ్యర్థాలను ప్రధాన ఇంధన వనరులుగా భవిష్యత్తులో ఎందుకు అభివృద్ధి చేసుకోవాలి ?
జవాబు:
భవిష్యత్ , శిలాజ ఇంధనాలు తరిగిపోవచ్చు కనుక ప్రత్యామ్నాయ ఇంధనముగా వ్యర్థాలను భవిష్యత్తులో ఉపయోగించవచ్చును.

ప్రశ్న 9.
ఆహారవాహికలో శ్లేష్మం లేకపోతే ఏం జరుగుతుంది ?
జవాబు:

1. పెరిస్టాలిటిక్ చలనముకు అవరోధం ఏర్పడును.
2. ఆహారపు బోలస్, జారడంకు కష్టమగును.
3. ఆహారం మ్రింగడం కష్టతరమగును.
4. ఆహారవాహిక గోడలు దెబ్బతినును.
5. వివిధ గాఢతలు గల ఆహార పదార్థాల నుండి ఆహారవాహిక రక్షించబడదు.

ప్రశ్న 10.
అంటుకట్టుట వలన కలిగే ఉపయోగాలు ఏమిటి ?
జవాబు:

1. అంటుకట్టుట చాలా సులభమైన పద్ధతి. ఎలాంటి శాస్త్రీయ జ్ఞానం అవసరం లేదు.
2. రెండు వేర్వేరు లక్షణాలను కలిపి మరింత మేలైన లక్షణాలు గల మొక్కను ఏర్పరచవచ్చును.
3. ఈ పద్ధతి ద్వారా త్వరగా మొక్కలను ఏర్పరచవచ్చును.

విభాగము – III 5 × 4 = 20 మా.

సూచనలు :

1. 1. క్రింది ప్రశ్నలకు సమాధానములు వ్రాయుము.

1. ప్రతి ప్రశ్నకు 4 మార్కులు.

ప్రశ్న 11.
రంగయ్యకు ఆరోగ్యం సరిగా లేదు. డాక్టర్ నిర్వహించిన పరీక్షల్లో క్రింది ఫలితాలు వచ్చాయి. పట్టికను విశ్లేషించండి. కింది ప్రశ్నలకు సమాధానాలు రాయండి.

అ) రంగయ్యకు చక్కెర వ్యాధి ఉన్నది అని ఎలా చెప్పవచ్చు ?
ఆ) బైలిరూబిన్ గురించి తెలుసుకోవడానికి ఏ పరీక్షలు నిర్వహించాలి ?
ఇ) పై నివేదిక ఆధారంగా నీవేం గ్రహించావు ?
ఈ)పై నివేదిక ఆధారంగా డాక్టరును నీవు ఏ ఏ ప్రశ్నలు అడుగుతావు ?
జవాబు:
అ) ఆహారం తినకముందు సాధారణ చక్కెర స్థాయి (గ్లూకోజ్) 60 – 100 మధ్య ఉండాలి. కాని రంగయ్యకు చక్కెర స్థాయి ఆహారం తినకముందు 120 ఉన్నది. ఆహారం తిన్న తర్వాత గ్లూకోజు సాధారణ స్థాయి 160 180 మధ్య ఉండాలి. కాని రంగయ్యకు 220గా ఉన్నది. రెండు సందర్భాలలోను గ్లూకోజు స్థాయిలు రక్తం నందు ఎక్కువగా ఉండుట వలన రంగయ్యకు చక్కెర వ్యాధి ఉన్నట్లు చెప్పవచ్చు.

ఆ) బైలిరూబిన్ గురించి తెలుసుకోవడానికి రక్తపరీక్ష చేయాలి.

ఇ) పై నివేదిక ఆధారంగా రంగయ్యకు అధిక రక్త పీడనము, చక్కెర వ్యాధి ఉన్నదని, రక్తం నందు 24 గంటల ప్రోటీను స్థాయి కూడా ఎక్కువగా ఉన్నదని తెలియుచున్నది. మూత్రం నందు సోడియం స్థాయి సాధారణముగానే ఉన్నదని తెలియుచున్నది. రక్తం నందు పరిమాణం సాధారణ స్థాయి కంటే హెచ్చుగా నున్నది.

ఈ)

1. చక్కెర వ్యాధి వలన కలిగే నష్టాలు ఏమిటి ?
2. చక్కెర వ్యాధి కలుగుటకు కారణమేది ?
3. రక్తంలో గ్లూకోజు స్థాయిలను నియంత్రించు హార్మోను ఏది ? –
4. అధిక రక్త పీడనము ఎందువలన కలుగుతుంది ?
5. అధిక రక్త పీడనము వలన కలిగే అనర్థాలు ఏమిటి ?
6. మూత్రం నందు 24 గంటల ప్రోటీను ఎక్కువైతే ఏం జరుగుతుంది ?
7. మన శరీరానికి సోడియం ఏ విధంగా అవసరం అవుతుంది ?
8. బైలిరూబిన్ వర్ణక స్థాయి రక్తమునందు ఎక్కువ అయితే కలిగే అనర్థాలు ఏమిటి ?

ప్రశ్న12.
క్రింది పటములలో ఒక పటము గీసి, భాగాలను వ్రాయండి.
A) శ్వాసక్రియా మార్గాన్ని తెలియజేసే బొమ్మ గీసి, భాగాలు గుర్తించండి.
జవాబు:

(లేదా)
B) మానవుని విసర్జక వ్యవస్థ పటం గీచి భాగాలను గుర్తించండి.
జవాబు:

ప్రశ్న 13.
కేంద్రీయ నాడీవ్యవస్థ మరియు పరిధీయ నాడీవ్యవస్థ మధ్యగల భేదాలు ఏవి ?
జవాబు:

ప్రశ్న 14.
సహజ వనరులను సద్వినియోగం చేసుకోడమే దేశానికి మనం చేసే సేవ అని చెప్పవచ్చు. దీనిని నీవు సమర్థిస్తావా? ఎందుకు?
జవాబు:
సహజ వనరులను సద్వినియోగం చేసుకోవటమే దేశానికి మనం చేసే సేవ అని చెప్పటాన్ని నేను సమర్థిస్తున్నాను.

1. పెరుగుతున్న జనాభా వలన వనరుల వినియోగం విపరీతంగా పెరిగింది. ఈ పరిస్థితి కొంతకాలం కొనసాగితే, భవిష్యత్ తరాలకు వనరులు లభించవు.
2. ప్రధానంగా శిలాజ వనరుల కొరత ఏర్పడుతుంది. పెట్రోలు, బొగ్గు నిల్వలు అడుగంటిపోతాయి. మనం సాధించిన అభివృద్ధి అంతా శక్తిపైన ఆధారపడి ఉంది.
3. శక్తి రహిత ప్రపంచంలో ఏ పరిశ్రమలూ పనిచేయవు. ఏ రవాణా సాధనం నడవదు. అంటే మరలా మనం మధ్యయుగం నాటి పరిస్థితులకు వెళ్ళిపోతాము.
4. ఇటువంటి విపత్కర పరిస్థితులు రాకుండా ఉండాలంటే వనరుల సద్వినియోగం ప్రతి పౌరుడు తన బాధ్యతగా భావించాలి.
5. వనరుల వినియోగంలో విచక్షణ ఉపయోగించి భావితరాలకు వాటిని అందించాలి.
6. తరిగిపోయే వనరులను తగ్గించి వాడటం అలవాటు చేసుకోవాలి.
7. సౌరశక్తి, అలలశక్తి, పవనశక్తి వంటి ప్రత్యామ్నాయ వనరులకు ప్రాధాన్యం ఇవ్వాలి.
8. వనరుల వినియోగంలో సాంకేతిక జ్ఞానం పెంచాలి. అప్పుడే భవిష్యత్ తరాలకు మనం శక్తిమయ ప్రపంచం అందించగలం. అదే మనం మన దేశానికి, రేపటి తరానికి చేసే సేవ.

ప్రశ్న 15.
డార్విన్ యొక్క ‘ప్రకృతివరణం’ సిద్ధాంతాన్ని ఒక ఉదాహరణతో వివరించండి.
జవాబు:
ప్రకృతివరణ సిద్ధాంతాన్ని ‘చార్లెస్ డార్విన్’ ప్రతిపాదించాడు. దీని ప్రకారం ప్రకృతి మాత్రమే ఒక జీవి మనుగడ సాగించాలా లేక నశించాలా అని నిర్ణయిస్తుంది. ఏ జీవి మనుగడకు అవసరమైన లక్షణాలు కలిగి ఉంటాయో అవి మాత్రమే ప్రకృతిలో జీవించగలిగి మిగిలినవి మరణిస్తాయి. దీనిని ‘ప్రకృతి వరణం’ (Natural Selection) అంటారు.

ఉదా : ఒక కుందేలుకు పుట్టిన ఐదు పిల్లలలో మూడు వేగంగా పరిగెత్తే ధర్మం కలిగి ఉన్నాయి. రెండు వేగంగా పరిగెత్తలేవు. శత్రువు దాడి చేసినపుడు వేగంగా పరిగెత్తే కుందేళ్లు తప్పించుకొని జీవించగలుగుతాయి. పరిగెత్తలేనివి మరణిస్తాయి.

ఈ సందర్భంలో వేగంగా పరిగెత్తటం వాటి మనుగడకు తోడ్పడింది. ఈ అనుకూలనం ఉన్న జీవులను ప్రకృతి మనుగడకు ఎన్నిక చేసింది.

విభాగము – IV 2 × 8 = 16 మా.

సూచనలు :

1. అన్ని ప్రశ్నలకు సమాధానములు వ్రాయుము.
2. ప్రతి ప్రశ్నకు 8 మార్కులు.
3. ప్రతి ప్రశ్నకు అంతర్గత ఎంపిక కలదు.

ప్రశ్న 16.
ఎ) క్రింది భాగాలలో పెరిస్టాలిసిస్ విధులను తెల్పండి.
ఎ) ఆహారవాహిక
బి) జీర్ణాశయం
సి) చిన్నప్రేగు
డి) పెద్దప్రేగు
జవాబు:
ఎ) ఆహారవాహిక : ఆహారవాహికలోని ‘పెరిస్టాల్సిస్’ చలనం వలన ఆహారం నోటి నుండి జీర్ణాశయంకు చేరుతుంది.
బి) జీర్ణాశయం : జీర్ణాశయంలోని పెరిస్టాల్సిస్ చలనం వలన ఆహారం జఠర రసంతో బాగా కలపబడి జీర్ణక్రియ వేగవంతం అవుతుంది. ఆహారాన్ని చిలకటానికి, కదపటానికి జీర్ణాశయంలో ‘పెరిస్టాల్సిస్’ తోడ్పడుతుంది.
సి) చిన్నప్రేగు : చిన్నప్రేగులో పెరిస్టాల్సిస్ వలన ఆహారం నెమ్మదిగా ముందుకు కదిలి పెద్దప్రేగును చేరుతుంది.
డి) పెద్దప్రేగు : పెద్దప్రేగులో పునఃశోషణ జరిగిన పిదప, వ్యర్థపదార్థాలు క్రమేణా ముందుకు జరిగి పాయువు ద్వారా విసర్జింపబడతాయి.
(లేదా)
బి) సమ విభజన, క్షయకరణ విభజనకు గల తేడాలేమిటి ?
జవాబు:
సమవిభజన – క్షయకరణ విభజన :

ప్రశ్న17.
ఎ) శ్వాసక్రియలో కార్బన్ డై ఆక్సైడ్ వెలువడునని అర్థము చేసుకొనుటకు నీవు ప్రయోగశాలలో నిర్వహించిన ప్రయోగ
జవాబు:
విధానము తెలుపుము.
ఉద్దేశం : శ్వాసక్రియలో CO₂ వెలువడుతుందని నిరూపించుట.
పరికరాలు : వెడల్పు మూతిగల రెండు గాజు సీసాలు, మొలకెత్తుతున్న శనగగింజలు, పొడి శనగగింజలు, సున్నపునీరు ఉన్న బీకర్లు.
ప్రయోగ విధానం :

1. వెడల్పు మూతిగల రెండు గాజుసీసాలు తీసుకోవాలి. ఒకదానిలో మొలకెత్తుతున్న శనగగింజలు ఉంచాలి. రెండవ దానిలో పొడి శనగగింజలు ఉంచవలెను.
2. రెండు గాజుసీసాలలో సున్నపుతేటతో నింపిన బీకర్లు ఉంచాలి. తరువాత రెండు సీసాల రబ్బరు బిరడాలను గట్టిగా బిగించాలి.
3. సీసామూతి చుట్టూ గాలి చొరబడకుండా వేజిలైన్ పూయవలెను. సీసాలను కదపకుండా రెండు రోజులు ఉంచవలెను.
   పరిశీలన :
4. ఒకటి రెండు రోజులు గమనించినట్లయితే మొలకెత్తుతున్న శనగగింజలు ఉండే సీసాలోని బీకరులో ఉన్న సున్నపు తేట ఎక్కువగా తెల్లటి పాలవలె మారుతుంది.
5. దీనికి కారణము మొలకెత్తే విత్తనాలు శ్వాసక్రియ జరపడం వల్ల వెలువడిన కార్బన్ డై ఆక్సైడ్ వల్లనే సున్నపు తేట పాల వలె మారిందని చెప్పవచ్చు.
6. పొడిగింజలు గల సీసాలోని సున్నపు తేట తెల్లగా పాలవలె అంతగా మారదు.
   ఫలితం :
7. కనుక శ్వాసక్రియలో కార్బన్ డై ఆక్సైడ్ విడుదల అవుతుందని నిరూపించడమైనది.

(లేదా)
బి) జాన్ కాగితం కప్పు, ప్లాస్టిక్ గొట్టాలను ఉపయోగించి స్టెతస్కోపును తయారుచేశాడు. అతడు అనుసరించిన విధానాన్ని రాయండి.
జవాబు:
జాన్ స్టెతస్కోప్ నిర్మించటానికి ఈ క్రింది విధానం అనుసరించాడు.

1. ఒక కాగితం కప్పు తీసుకొని దాని మధ్యన రంధ్రం చేసి, ఒక చిన్న గొట్టం అమర్చాడు.
2. రెండు ప్లాస్టిక్ గొట్టాలను తీసుకొని వాటిని రబ్బరు ట్యూబ్లో కలిపాడు.
3. రబ్బరు ట్యూబ్ను కాగితం కప్పుకు అమర్చిన గొట్టానికి కలిపాడు.
4. అందువలన Y ఆకారంలో ప్లాస్టిక్ గొట్టాలు అమర్చబడ్డాయి.
5. దానికి క్రిందుగా కాగితం కప్పు వ్రేలాడుతూ ఉంది.
6. ప్లాస్టిక్ గొట్టాలను చెవిలో ఉంచుకొని గుండెపై కాగితం కప్పు ఆన్చి హృదయ స్పందనను వినవచ్చు.
7. ఇది హృదయస్పందనను పరిశీలించే స్టెతస్కోప్లో పని చేస్తుంది.

Students can use TS 10th Class English Model Papers Set 9 as a tool for exam preparation.

TS 10th Class English Model Paper Set 9 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

1. Read the question paper carefully.
2. Answer the questions under Part – A in the answer booklet provided.
3. Write the answer to the questions under Part – B on the question paper itself.
4. Avoid overwriting.

Part – A (60 Marks)

(Questions 1 – 4) : Read the following passage carefully.

Father had provided for my education, and I had been able to realize his dreams. My parents were truly proud of me. It was through me that they had earned a greater degree of admiration and respect from the villagers. My father would not like to see me carrying a trunk on my back and would be very hurt if I did so. I concluded that it would be better to let him carry it. Father was used to carrying luggage anyway. He was stronger and more skilled than me in these matters.

I had never got used to physical labour having stayed in hostels right from my childhood. So, in sSpite of my youth and strength, I was physically useless. I continued walking silently with father. We rested at tvyo places on the way and had our tiffin but we hardly talked. Finally, we reached Dirang. The bus from Tawang had not yet reached Dirang and so we had some time in hand. We entered a tea shop and sat facing each other. Father appeared tired. I felt sorry for him but couldn’t bring myself to say anything. I asked the waiter to get us two cups of tea. Just as I was going to take my first sip, I heard father’s voice,’Do you have a pair of old shoes ?’
‘Why?’ I asked.
The road is uneven and full of pebbles, it hurts while walking.’ (The Journey)

Answer each of the following questions in three to four sentences. 4 × 3 = 12 M
Question 1.
How was the narrator able to realize the dreams of his father ?
Answer:
The narrator’s father had provided for the narrator’s education. The narrator worked hard and got a white-collar job. He became a government officer. Thus the narrator was able to realize the dreams of his father.

Question 2.
‘It was through me that they had earned a greater degree of admiration.’ Do you think it is true ? Explain.
Answer:
No, it’s not true. In my view, the narrator had earned a greater degree of admiration and respect from the villagers because of his father. If his father had not provided for his education, he would not have been able to get a good position in life.

Question 3.
Why did the narrator let his father carry luggage ? If you were in his place, what would you do?
Answer:
The narrator let his father carry luggage because of his false prestige. He had the feeling that if he carried the luggage, the whole world would laugh at him. He thought that his education had made him shun physical labour.
If I where in his place, I would not feel so. I would carry my luggage without looking for others’ help.

Question 4.
Why do you think the narrator felt sorry for his father ?
Answer:
In spite of his youth and strength, the narrator was physically useless. His father had to carry the luggage all the way to Dirang. The narrator only followed his father. So, the narrator felt sorry for his father when he appeared tired.

(Questions 5 – 8) : Read the following passage.

Falling for her beauty and intelligence, as also the dowry which her father offered, a young man tied the three sacred knots around her heck, made her the housewife to a household and said to her, ‘Look, ammadu, this is your home.’ Then the hqusewife immediately pulled the end of her sari and tucked it in at the waist and swabbed the entire house and decorated the floor with muggulu designs. The young man promptly praised her work. ‘You are dexterous at swabbing the floor – even more dexterous in drawing the muggulu. Sabash, keep it up.’ He said it in English, giving her a pat on the shoulder in appreciation. Overjoyed, the housewife began living with swabbing as the chief mission in her life. She scrubbed the house spotlessly clean at all times and beautifully decorated it with multi-coloured designs. That’s how her life went on, with a sumptuous and ceaseless supply of swabbing cloths and muggu baskets. (What is My Name ?)

Answer each of the following questions in three to four sentences. 4 × 3 = 12 M
Question 5.
According to the passage, how does a woman require her household ?
Answer:
According to the passage, a woman requires her household to be safe and happ^. She also requires her household to be lovable, affectionate, caring, supporting and secure.

Question 6.
What activities of the woman made her husband appreciate her ?
Answer:
The woman swabbed the entire house and kept it clean and neat. She decorated the floor with muggulu designs and made it look attractive. These activities made her husband appreciate her.

Question 7.
What ‘chief mission’ is mentioned in the passage ? Is it true ? Explain.
Answer:
The ’chief mission’ mentioned in the passage was swabbing. Yes, it is true. After the marriage, the woman kept the house always neat and clean. Her husband promptly praised her work. When her husband appreciated her, she was overjoyed and completely immersed herself in swabbing the house as if it was the chief mission in her life.

Question 8.
Pick Three TRUE sentements according to the passage.
A) The young man married the ugly woman because he was offered a lot of dowry by her father.
B) The housewife was skilful at swabbing the floor.
C) The young man praised hi9 wife for skilfully drawing muggulu and cleaning the floor spot lessly
B) The innocent woman overjoyed with the appreciation of her husband made swabbing as thechiefmissionofherlife.
E) The young woman cannot understand English.
Answer:
The following statements are TRUE.
A) The housewife was skilful at swabbing the floor.
B) The young man praised his wife for skilfully drawing muggulu and cleaning the floor spotlessly.
C) The innocent woman overjoyed with the appreciation of her husband made swabbing as the chief mission of her life.

(Questions 9 – 12) : The Pie-chart given below provides information about enrolment of students in a college. Study the chart carefully and aswer the following questions. (4 × 2 = 8 M)

Answer each of the following questions in three to four sentences.
Question 9.
What is the pie-chart about ? How many sectors are given in the pie-chart ?
Answer:
The Pie-chart is about enrolment of students in different secotrs. Six sectors are given in the pie-chart.

Question 10.
How many students are enrolled for MBBS ? How many more students are in MBBS than in Hotel Management ?
Answer:
Five hundred students are enrolled for MBBS. Two hundred and fifty more students are in MBBS than in Hotel Management.

Question 11.
How many sectors have the same number of students? What are they?
Answer:
Three sectors have the same number of students.
Those sectors are :
1) Computer Science,
2) Commerce,
3) Science.

Question 12.
How many students are enrolled in total ? Name the sector that has the least number of students.
Answer:
One. thousand and twenty five students are enrolled in total. The sector that has the least number of students is ‘Dentisting’.

(Question 13) :
Read the passage given below focussing on the parts that are underlined. Answer any 4 of the ques-tions as directed and write them in the answer booklet. 4 × 2 = 8 M

Once there lived a woman in a city, (i) She was fond of penguin birds, (ii) She bought a penguin, (iii) “It’s very nice.” She said to her doughter.
“Yes, mon,” said the daughter, (iv) “It is the most beautiful penguin in the world.”
“Yes,” said the mother.
The next day, (v) the neighbours came to their house. They wanterLto see the penguin.
One of them got a doubt.
“Do penguins live here ?” he said, (vi) “Penguins live in cold places. The temperature will be verylow.”

i) Add a suitable question tag.
ii) Rewrite the sentence beginning with ‘a penguin’.
iii) Rewrite the sentence beginnign with ‘She said to her daughter that …………………’
iv) Rewrite the sentence using ‘more beautiful’.
v) Combine the two sentences using ‘not ‘in order to’.
vi) Combine the two sentences using ‘where’.
Answer:
i) wasn’t she ?
ii) A penguin was bought by her.
iii) She said to her daughter that it was very nice.
iv) It is more beautiful than any other penguin in the world.
v) The neighbours came to their house in order to see the penguin.
vi) Penguins live in cold places where the temperature will be very low.

(Question 14) : 10 M
Wangari Maathai, was an Environmental Activist and Nobel Prize Winner. She started the Green Belt Movement and also fought for the equal rights for women in Africa.
Imagine you are one of the famous journalists – You are asked to interview her. Write a script of interview based on the above information.
You may include ideas :
♦ Basic needs of the people
♦ Degradation of the environment
♦ Activities taken up by the Green Belt Movement
♦ Protection of the forests
♦ Rehabilitation of the environment
♦ Responsibilities of the people
(OR)
The local authorities decided to conduct an elocution competition on the topic, ‘Company of bad friends lead to bad habits’ on ANTI-DRUGS DAY. The youth is going on wrong path.
Prepare a SPEECH SCRIPT on the topic.

You can include the following ideas :
♦ Importance of the day
♦ Age group addicted
♦ Present status
♦ Harmful effects
♦ Remedies ………………. etc.
Answer:
Interview with Wangari Maathai
Journalist : Hello, ma’am. Glad to meet you.
Wangari Maathai : Hello, thank you very much.
Journalist : Ma’am tell us why. the people in rural areas are asking for protecting basic needs ?
Wangari Maathai : The basic needs in rural areas are clean drinking water, food, fire wood, building materials and fodder for their animals. All those come from the land. But unfortunately they did not have those things because the environment was degraded.
Journalist : What do you think is the main reason for the degradation of environment?
Wangari Maathai : Most of the people who’are in power, want to control lot of natural resources and want to keep away many other out of it. The resources are not shared equitably. And this has resulted in environment degradation due to mismanagement.
Journalist : Tell us about the Green Belt movement, that you have started ?
Wangari Maathai : Green Belt Movement aims at organizing women especially rural women to plant trees, combat environment degradation, deforestation, re¬store their main sources of fuel for cooking, generate income and stop soil erosion.
Journalist : Did you see any transformation in the people to protect forests ?
Wangari Maathai : Of course, there is a huge transformation. I saw the willingness of the people to fight for their rights, to have a good and clean environment to fight for their rights on forests, to protect them and not allow any corrupt leaders to take their lands.
Journalist : How do you think you have succeeded in making the people rehabitate the environment and understand their responsibilities ?
Wangari Maathai : Because we are dealing with environment, we gave them education both in civics and also in environment. That made them understand clearly why they should take up the responsibility of protecting their environment and it was not only the responsibility of somebody else to come and rehabilitate their environment, it’s them, and it’s their responsibility.
Journalist : So, what is the one thing you expect from everyone of us ?
Wangari Maathai : I think it is the responsibility of everyone of us to plant a tree. If we plant a tree, we plant the future for ourselves, for our children and for everyone.
Journalist : Thank you, very much Ma’am for giving your time and all the best for your great work.
(OR)
Honorable judges and my dear students a very good morning. I would like to share some of my views on the topic “Company of bad friends lead to bad habits” on this “Anti-Drugs Day.”

This Anti – Drugs Day is celebrated every year to create awareness and to educate the people against the usage of drugs. It’s very alarming situation that at the age of high school, children are taking and addicting to the drugs.

The drug sellers are attracting the children through food items and grabbing their future into their hands. It’s a very dangerous situation. The addicted children are becoming puppets in the hands of suppliers and they are using these innocent children to their illegal and un-ethical activities.

The addicted children will become inactive, passive. They don’t like to mingle with friends and family members. Their mood swings and happens very frequently. The academic stand¬ards become very low and become nervous and anxious for every thing.

To overcome this problem parents and teacher should take care of them. Parents should always observe and have an eye on the friends and company to whom their children are mov¬ing. Always good company leads to good thoughts and habits. One can say one’s nature by seeing their friends. So, we should maintain good company. If our friends are bad, we also become bad and habituate to bad things. So, beware of bad friends and be away froni bad habits.
Thank you one and all for this opportunity.

(Question 15) : 5M
In the interview with Wangari Maathai, she explained the NHK Radio, how she had taught the women in Africa to plant trees.
On the Green Day, in your school you might have planted a sapling and adopted it to take care of it.
Now, make a diary entry about your feelings on this memorable day.
– Transformations brought.
– Message spread to the world
Answer:
21st July, 2018
Saturday.

Today is a memorable day in my life. Being the part of the program Harithaharam our school management decided to plant saplings in and around our school premises. Our class teacher had allotted each student with a sapling and said that they are adopting that to us. And our Head Master told us to look after the plants as of our own child as our parents to us. At that moment I felt a very strange and unique feeling in my heart. Then I took an oath to keep up the plant carefully and to safeguard it from all types of dangers that may occur to it. After planting the tree, when I touched the leaves of that sapling I thought that it was smiling and saying thanks. I should say thanks to my school management and the government for giving me such a good opportunity. I am very eager to go to school to see my new little friend on tomorrow morning.

(Question 16) : 5 M
Plastic waste is one of many types of wastes that takes too long to decompose and pollutes the environment. The toxic chemicals of plastic affects human health and many cause cancers, impaired immunity etc.

As a Secretary of “Health Club” of your school you want to organize a compaign against plastics.
Now, prepare ‘a poster’ stating the harm done by the plastics and some alternative steps to be taken to reduce plastic usage.

You may, include :
♦ Effects of the plastic
♦ Choose paper bags over plastic bags
♦ Stop buying bottled water
♦ Carry your own shopping bags
♦ Store food in non plastic containers.
Answer:
Poster writing:

Part – B (20 Marks)

Instructions :

1. Answer the questions on the question paper itself and attach it to the answer booklet of Part – A.
2. Avoid overwriting.

(Questions 17 – 21) : Read the following poem.

Glossary :
Float : put on the surface of water
Blooms : flowers

(Questions 17 – 21) :
Now answer the questions. Each question has four choices. Choose the correct answer and write A, B, C or D in the brackets given. (5 × 1 = 5 M)

Question 17.
The poet leaves the paper boats ……………..
A) in the upstream
B) in the lake
C) in the downstream
D) in the sea
Answer:
C) in the downstream

Question 18.
‘I write my name on them.’ The word Them’ refers to ………………
A) flowers
B) paper boats
C) villages
D) houses
Answer:
B) paper boats

Question 19.
The mood of the poet is ……………….
A) serious
B) apathetic
C) angry
D) joyful
Answer:
D) joyful

Question 20.
The poet believes the paper boats …………………
A) will reach his village
B) will reach his friends
C) will reach people who may not know about him
D) will join the sea
Answer:
C) will reach people who may not know about him

Question 21.
The poet floats the paper boats ………………
A) in the morning
B) in the afternoon
C) in the evening
D) on a holiday.
Answer:
A) in the morning

(Questions 22 – 26):
In the following passage, five sentences are numbered and 6ach of them has an error. Correct them and rewrite them in the given space. 5 × 1 = 5 M

(22) “Then Second World War was over and India’s freedom was imminent. (23) “Indians will built their own India,” declared Gandhiji. (24) The whole country was filled with an unprecedented optimism. (25) I asked to my father for permission to leave Rameswaram and study at the district headquarters in Ramanathapuram. (26) He told me as if thinking aloud, “Abul! I know you have to go away to grew…..”
22) ……………… 23) …………….. 24) ………….. 25) …………….. 26) ……………….
Answer:
22. Then the Second World War was over and India’s freedom was imminent.
23. “Indians will build their own India,” declared Gandhiji.
24. The whole country was filled with an unprecedented optimism.
25. I asked my father for permission to leave Rameswaram and study at the district headquarters in Ramanathapuram.
26. He told me as if thinking aloud, “Abdul! I know you have to go away to grow…..”

(Questions 27 – 31): Complete the passage choosing the right words from those given below. Each blank is numbered and each blank has four choices (A), (B), (C) and (D). Choose the correct answer and write (A), (B), (C) or (D) in the brackets given. 5 × 1 = 5 M

About 70 percent (27) ………………… a person’s body is made up of water. The water percentage (28) ……………….. as a person gets older. Infants and children have higher content of water (29) ………………… adults. It is important (30) …………………. everyone to maintain proper water level in (31) ……………… a body.

Question 27.
A) of
B) off
C) on
D) in
Answer:
A) of

Question 28.
A) decrease
B) decreasing
C) decreases
D) decreased
Answer:
C) decreases

Question 29.
A) than
B) up
C) upon
D) from
Answer:
A) than

Question 30.
A) from
B) for
C) up
D) upto
Answer:
B) for

Question 31.
A) a
B) an
C) the
D) one
Answer:
C) the

(Questions 32 – 36) :
Read the following passage with focus on the underlined parts. Answer them as directed in the space given. 5 × 1 = 5M
The next morning the boys were there again, and a man was standing at the gate watching them. I stopped and looked, just to see what the white boy was making his little servant,(32) do. For (33) my utter astonishment (34) the little dark boy was striding imperiously up and under (35) the lawn, while the white youngster walked abjectly behind him.
‘Get me a banana !’ The little boy ran into the house and reappeared shortly with a banana. ‘Peel it for me !’ a (36) little white boy peeled the banana and handed it to his dark master.

Questions :
32. Write the word that is the opposite in meaning of the underlined word.
Answer:
master

33. Replace the underlined word with a correct one.
Answer:
To

34. Write the word that has the same meaning of the underlined word.
Answer:
surprise

35. Replace the underlined word with a correct one in the context.
Answer:
for

36. Replace the underlined word with the correct article.
Answer:
the

Access to a variety of TS Inter 1st Year Economics Model Papers and TS Inter 1st Year Economics Question Paper May 2022 helps students overcome exam anxiety by fostering familiarity.

TS Inter 1st Year Economics Question Paper March 2022

Sestion – A

Note : i) Answer ANY THREE of the following questions in 40 lines each.
ii) Each question carries TEN marks. (3 × 10 = 30)

Question 1.
What are the various methods of calculating National Income?

Question 2.
Explain the meaning of Perfect Competition. Illustrate the mechanism of Price Determination under Perfect Com-petition.

Question 3.
Describe the law of diminishing marginal utility, its limita-tions and importance.

Question 4.
Critically examine the Law of Variable Proportions.

Question 5.
Explain the Keynesian Theory of Employment.

Question 6.
Explain the law of demand and examine its exceptions.

Question 7.
What is a demand function? What are the factors that deter-mine the demand for a good?

Sestion – B

Note : i) Answer ANY EIGHT of the following questions in 20 lines.
ii) Each question carries Five marks.
iii) ALL are of short answer type questions. (8 × 5 = 20)

Question 8.
Explain the differences between free goods and economic goods.

Question 9.
What is Utility? What are its types?

Question 10.
What is an Indifference Curve? What are its assumptions?

Question 11.
Analyse the characteristics of wants.

Question 12.
Explain the concept of Indifference Curve. Discuss its prop¬erties.

Question 13.
Illustrate the reasons for negative sloping demand curve.

Question 14.
Explain the relationship between Average Cost and Marginal Cost.

Question 15.
What is Monopoly ? What are its characteristics?

Question 16.
Discuss the concept of Income Demand.

Question 17.
Write a note on teh classification of Markets based on Time and Area.

Question 18.
What are the factors that determine National Income?

Question 19.
What are the determining factors of Real Wages?

Question 20.
Point out the redemption methods of Public Debt.

Question 21.
What are the sources of Public Revenue?

Question 22.
What is Barter System ? What are its difficulties?

Question 23.
Calculate the median for the following data :

Question 24.
Explain the Primary and Secondary functions of Money.

25. What is Statistics? Explain its relationship with Economics.

Section – C

Note : i) Answer ANY FIFTEEN of the following questions in 5 lines.
ii) Each Question carries TWO marks. (15 × 2 = 30)

Question 26.
Explain the capital goods.

Question 27.
What is Wealth?

Question 28.
What is Microeconomics?

Question 29.
Explain Ordianl Utility.

Question 30.
What is Price?

Question 31.
What is Price demand?

Question 32.
Draw the indifference Map.

Question 33.
Explain Substitute goods.

Question 34.
What is Cross Demand?

Question 35.
What is an Opportunity Cost?

Question 36.
Define the Production Function.

Question 37.
Define Market.

Question 38.
What is Net Profit?

Question 39.
What is Contract Rent?

Question 40.
Expand C.S.O What is its responsibility?

Question 41.
What ae Transfer Payments ? Give examples.

Question 42.
What is Budget?

Question 43.
Define Public Finance.

Question 44.
What is Net Banking ? Explain the merits of it.

Question 45.
What is Clearance House?

Question 46.
What is Currency?

Question 47.
What are the uses of Overdrafts?

Question 48.
What are the Merits of Median?

Question 49.
Explain the concept of Mode.

Question 50.
Find the mode from the following data.

Self-assessment with TS Inter 2nd Year Sanskrit Model Papers Set 8 allows students to take charge of their own learning.

TS Inter 2nd Year Sanskrit Model Paper Set 8 with Solutions

Time : 3 Hours
Max Marks : 100

Note :

• All questions should be attempted.
• Question Nos. 1, 2 and 3 should be answered either in the medium of instructions of the candidate or in Sanskrit (Devanagari Script) only.
• The remaining questions should be answered in Sanskrit (Devanagari Script) only.

सूचना : प्रथम, द्वितीय, तृतीय प्रश्नान् विहाय अन्ये सर्वेऽपि प्रश्नाः संस्कृतभाषायामेव समाधातव्याः ।

1. एकस्य श्लोकस्य प्रतिपदार्थं भावं च लिखत । (1 × 6 = 6)

अ) यथा यथा हि पुरुषः कल्याणे कुरुते मनः ।
तथा तथास्य सर्वार्थाः सिद्ध्यन्ते नात्र संशयः ॥
समाधान:
पदच्छेदः (Word Division) यथा यथा हि पुरुषः कल्याणे कुरुते मनः तथा तथा+अस्य सर्व+अर्थाः सिद् ध्यन्ते न + अत्र संशयः ।

अर्थाः (Meanings) पुरुषः = A man; यथा यथा = whenever; कल्याणे = in auspicious things; मनः = mind; कुरुते हि = fixes; अस्य = his; `सर्व + अर्थाः = all desires; तथा तथा = then; सिद्ध्यन्ते = are fulfilled; अत्र = here; संशय: = doubt; न = is not there.

भाव: (Substance) Whenever man fixes his mind on doing good things, then all his desires will be fulfilled. There is no doubt in it.

आ) चक्रमेकं निरालम्बो मार्गोऽनूरुश्च सारथिः ।
तथाप्येति सदा भानुः महतां किं नु दुष्करम् ॥
समाधान:
पदच्छेदः (Word Division)चक्रम् एकम्, निरालम्बः, मार्गः + अनूरुः+च सारथिः, तदा+अपि एति सदा भानुः महताम्, किम्, नु दुष्करम् |

अर्थाः (Meanings) एकम् = one; चक्रम् = wheel; निरालम्ब: without support; मार्गः = path; सारथिः च = and the driver is; अनूरुः = Anuru, one having no legs; तदा अपि = Even then ; भानुः = the sun;सदा = always; एति = moves; महताम् = to the great; दुष्करम् = impossible; किम् नु = what indeed ?

भावः (Substance) There is only one wheel to the chariot, the path has no support, and the driver has no legs. Still, the sun always goes round. What is impossible to the great?

2. एकं निबन्धप्रश्नं समाधत्त । (पद्यभागात्) । (1 × 6 = 6)

अ) कुशस्य सरयूविहारं तत्परिणामं च विवृणुत ।
Answer:
Introduction: The lesson Kumudvati parinaya is an extract from the sixteenth canto of Raghuvamsa. The poet was Kalidasa, who lived in 1st century B.C. Kalidasa wrote three dramas and four poems.

Kusa’s water sport: Kusa became the king of Ayodhya after Sri Rama ascended to heaven. When summer set in with days becoming hotter, and nights becoming shorter, he desired to go for swimming in the river Sarayu along with his wife. He got tents erected on the riverbank and crocodiles removed from the waters. He enjoyed in the river just as Indra in the Akasa Ganga.

Loss of the bracelet: While Kusa was swimming, his bracelet of victory, passed on to him along with the kingdom by Sri Rama slipped into the waters unnoticed. When Kusa came out of the river, he immediately noticed the loss of the bracelet even before he finished dressing. Kusa considered flowers and ornaments in the same way. स तुल्यपुष्पाभरणो हि धीरः । Still, he was dejected as it was a charm of victory, and was worn by his father. He ordered skillful divers to search for it. Their search was in vain. They told him that the greedy serpent king Kumuda must have taken it. Kusa got angry and aimed Garuda arrow to destroy the serpents. The river got agitated at that.

Kumuda’s offer : Kumuda then came out of the river along with a young woman. He had the ornament in his hand. On seeing him, Kusa withdrew his arrow, and calmed down. प्रहेष्वनिर्बन्धरुषो हि सन्तः । He offered it to Kusa. मुर्धाभिषिक्तं कुमुदो बभाषे । He said that his sister Kumudvati threw a ball up while playing. Then she saw this bracelet falling down like a shooting star, and took it. He requested Kusa to accept his sister. When Kusa agreed, their marriage was performed.

आ) देवगिरिनाथः महादेवः रुद्रमाम्बया कथं पराजितः ?
Answer:
Introduction: The lesson Rudrama Devi was written by Sri D. Venkataramacharya. It is taken from the author’s poem Silpasri. The present lesson describes Rudrama’s victory over Mahadeva, the ruler of Devagiri.

Queen of Kakati Kingdom: When the king of Kakati kingdom became old, his minister advised him to install his daughter Rudrama as the queen. It would make the kingdom along with the subjects pleased. निजधरां मुदितां कुरु सप्रजाम् ।

The king felt happy at that suggestion, and made announcement to that effect. On an auspicious day, Rudrama was made the queen while the priests chanted mantras, and water from holy rivers and oceans was sprinkled on her. The pious queen became successful as people and feudatory lords paid their taxes willingly. The Jealous Enemies’: However, Harihara and Murari became jealous of her success. The king of Devagiri lured them by his crooked ways. He said that a weak.woman insulted them. The ruler of Devagiri, who was proud of his army attacked Rudrama. समभियान्त्य्यरयो विभियोऽधियः ।

At that time, Rudrama went to worship Siva. On getting the news that the city was under siege, she returned with limited forces to the city. When she blew her conch, the soldiers and feudatory kings understood that it was their queen, and blew their conches.

The Warrior Queen: The queen and her soldiers attacked the enemy from opposite sides. Rudrama threw her sword and killed the mahout of Mahadeva’s elephant. The king of Devagiri jumped in to another chariot. Rudrama killed the charioteer and the horses, and broke his string and flag. नृपाग्रणी तंविश्यं चकार | The Devagiri ruler and his followers ran away from the battlefield. He understood that women were not weak.

3. एकं निबन्धप्रश्नं समाधत्त । (गद्यभागात्) (1 × 6 = 6)

अ) हर्षवर्धन राज्यश्रियं कथम् अरक्षत् ?
Answer:
Introduction: The lesson Bhratruvatsalyam is an extract from Harshacharita written by Bana. Bana was the court poet of king Harshavardhana who ruled Sthaneswar in the seventh century AD. The present lesson describes how Harshavardhana saved his sister, Rajyasri, who was about to enter fire.

Rajyasri’s Plight: When Harshavardhana enquired the messenger Bhandi about his sister Rajyasri, he latter told him that after the death of the king’s elder brother Rajyavardhana, Gupta prisoned Rajyasri. However, she escaped from there, and entered the Vindhya forest. Then Harsha decided to go in search of her.

Meeting with Divakaramitra:While Harsha was searching for her, he came to know from a tribal boy Nirghata that the Buddhist teacher Divakaramitra might know about her. Harsha thought that he must be the friend of Rajyasri’s husband who became a monk at a young age. He met him and told him that he lost all his relatives except his younger sister. She lost her husband and while wandering, entered the Vindhya forest. He asked him whether he knew anything about her.

Saving Rajyasri: At that very moment one of the monk’s disciples came there and reported that he met a woman in the forest who was about to enter the fire as she lost her father and husband. The king understood that it was his sister, and asked the monk to show the way. They reached the place where Rajyasri was, and the king stopped her. On seeing her brother, she wept for a long time. Divakaramitra gave water for washing her face, and advised her to go to home. Harsha spent the night there, and later accompanied by his sister, reached his camp.

आ) वृक्षाणां रक्षणे केसरसिंहस्य श्रद्धां विशदयत ।
Answer:
Introduction: The lesson Vriksharakshika Pitamahi was written by Padma Sastri. It is taken from the author’s Sanskrita kathasatakam Part 1. This lesson narrates the importance of planting the trees.

Kesara Simha’s Grandmother: Kesara Simha lived with his parents in one of the valleys of Kasmir. His grandmother was interested in planting trees. She would cover the trees with grass during winter to protect them from snow. She advised Kesara Simha to plant trees if he wanted to eat fruits. By that, he would get merit. Mother Earth would be pleased. पृथिवी अनेन प्रसन्ना भवति | They would get flowers, fruits and wood. Influenced by the words of his grandmother, Kesara Simha also planted trees, and took care of them. Five years later grandmother died.

Kesara Simha’s Dreams: One day, Kesara’s grandmother appeared in his dream, and consoled him. She looked young. She advised him to plant trees everyday so that he could forget her. प्रत्यहं नूतनान् वृक्षानारोपय | Kesara Simha woke up and started to take care of trees. His grandmother again appeared in his dream and offered him delicious fruits in a silver bowl.

Grandmother’s Advice: After many days grandmother appeared in Kesara Simha’s dream sitting on a throne like a celestial damsel. She was pleased as he planted many trees. She said that from that day onwards there would be no disease in the village. As trees would give pure air, there would be the benefit of health. The nymphs who appeared there took him in chariot to see the clear streams and trees full of fruits and flowers. His grandmother advised him that trees alone were the most important ones in this world. अस्मिन् संसारे वृक्षाणामेव प्राधान्यं वर्तते । They would give pure air, water and soil. If the trees were cut indiscriminately, earth would heat up, snow would melt, and oceans would flood the earth. People would suffer. She also advised that education was necessary to know the secrets. अतः पठनमावश्यकं वर्तते ।

Kesara Simha took a firm decision that he would protect the world by growing trees.

4. द्वयोः प्रश्नयोः समाधानानि लिखत (उपवाचकात्) । (2 × 4 = 8)

अ) अमात्यराक्षसः किं कृत्वा नगरादपक्रान्तः ?
समाधान:
परिचय : मुद्राराक्षसम् इति पाठ्यभागः विशाखदत्तस्य मुद्राराक्षसनाटकस्य प्रथमाङ्कात् स्वीवृत्तः । अत्र नन्दवंशनिर्मूलनस्य अनन्तरं चाणक्येन चन्द्रगुत्पस्य कृते कृताः प्रयत्नाः वर्णिताः ।
अमात्यराक्षसः मित्रस्य चन्दनदासस्य गृहे कलत्रं न्यासीकृत्य नगरादपक्रान्तः ।

आ) चाणक्येन किमर्थं यथाशक्ति प्रयत्नः क्रियते ?
समाधान:
परिचय : मुद्राराक्षसम् इति पाठ्यभागः विशाखदत्तस्य मुद्राराक्षसनाटकस्य प्रथमाङ् कात् स्वीकृतः । अत्र नन्दवंशनिर्मूलनस्य अनन्तरं चाणक्येन चन्द्रगुप्तस्य कृते कृताः प्रयत्नाः वर्णिताः ।
अमात्यराक्षसः कथं चन्द्रगुप्तस्य साचिव्यग्रहणम् अङ्गीकरोति इति चाणक्येन प्रयत्नः क्रियते ।

इ) ब्रह्मचारी कुत्र उषितवान् ? किमर्थम् ?
समाधान:
परिचय : न्यासरक्षां इति पाठ्यभागः स्वप्नवासवदत्तम् इति नाटकस्य प्रथमाङ्कः । अस्य कविः भासः ।
ब्रह्मचारी राजगृहवासी । वेदाध्ययनार्थं सः लावणकम् इति ग्रामे उषितवान् । सः ग्रामः वत्सभूमौ आसीत् । तत्र ग्रामदाहेन वासवदत्ता दग्धा इति सः अवदत् ।

ई) कश्यपेन किमिति निश्चितम् ?
समाधान:
परिचय : आनूरवम् इति रूपकम् श्रीमता कोगण्टि सीतारामाचार्येण विरचितम् । अस्मिन् सूर्यस्य सारथेः अनुरोः जन्मादिकं वर्णितम् ।
कुमारयुगले एकः सवित्रे देयः इति कश्यपेन निश्चितम् । स एव द्युतिमान् अनूरुः ।

5. द्वयोः ससंदर्भ व्याख्यानं लिखत । (पद्यभागात्) (2 × 3 = 6)

अ) प्रवेष्वनिर्बन्धरुषो हि सन्तः ।
समाधान:
परिचय : एतत् वाक्यं कुमुद्वतीपरिणयः इति पाठ्यभागात् स्वीकृतम् । एषः पाठ्यांशः रघुवंशे षोडशसर्गात् गृहीतः । अस्य कविः कालिदासः ।
सन्दर्भ : सरयूनद्याः पतितं जैत्राभरणम् अलभमानः कुशः भुजङ्गवधाय गारुत्मतम् अस्त्रं समाददे । तदा नागराजः कुमुदः आभरणहस्तः सन्, कुमुद्वतीसहितः बहिः अगच्छत् । तं दृष्ट्वा कुशः अस्त्रं प्रतिसञ्जहार । विनीतेषु सत्पुरुषाः कोपं न प्रदर्शयन्ति ।
भाव : विनीतेषु जनेषु सत्पुरुषाः कोपं न प्रदर्शयन्ति ।

आ) आस्तामयं मे युवराजभावः ।
समाधान:
परिचय : एतत् वाक्यं विक्रमस्य औदार्यम् इति पाठ्यभागात् स्वीकृतम् । अस्य कविः बल्हणः । अयं पाठ्यभागः विक्रमाङ्कदेवचरितमिति काव्यात् गृहीतः । . सन्दर्भ : चालुक्यराजः आहवमल्लः स्वस्य द्वितीयपुत्रं विक्रमाङ्कं युवराजं कर्तुम् ऐच्छत् । परन्तु कुमारः न अङ्गीकृतवान् । मम आज्ञां सर्वे राजानः पालयन्ति । त्यागभोगयोः संपद् व्ययीकरोमि। अहं युवराजो न भवामि इति उक्तवान् ।
भाव : मम युवराजत्वम् आस्ताम् ।

इ) समभियान्त्यरयो विभियोऽधियः ।
समाधान:
परिचय : एतत् वाक्यं रुद्रमदेवी इति पाठ्यभागात् स्वीकृतम् । अस्य रचयिता श्री डी. वेङ्कटरामाचार्यः ।
सन्दर्भ : काकतीयराजः स्वपुत्रीं रुद्रमदेवीम् राज्ये अभिषिक्तवान् । तदसहमानाः देवगिरिराजादयः वयम् अबलया अवमानिताः इति संचिन्त्य ताम् अभियौ ।
भाव : भयहीनाः मूर्खाः शत्रवः अभिद्रवन्ति ।

ई) स्तम्भेऽस्ति श्रीपतिर्विष्णुः ।
समाधान:
परिचय : एतत् वाक्यं नृसिंहाविर्भावः इति पाठ्यभागात् स्वीकृतम् । अस्य कविः डा. के. सुधाकररावः ।
सन्दर्भ : हिरण्यकशिपुः स्वपुत्रं प्रह्लादम् अवदत् यत् अस्मिन् स्तम्भे हरिं दर्शय । तं गदया हनिष्यामि । सः कुत्रास्ति इति । प्रह्लादः अवदत् यत् सः स्तम्भे अस्ति इति ।
भाव : विष्णुः अस्मिन् स्तम्भे अस्ति ।

6. द्वयोः ससंदर्भ व्याख्यानं लिखत । (गद्यभागात्) (2 × 3 = 6)

अ) असौ दस्युः ।
समाधान:
परिचय : एतत् वाक्यं सोमदत्तचरितम् इति पाठ्यभागात् स्वीकृतम् । अस्य कविः दण्डी । एषः पाठ्यभागः कवेः दशकुमारचरितात् गृहीतः ।
सन्दर्भ : विप्राय माणिक्यं दत्वा सोमदत्तः निद्रावशं गतः । तदनु सः विप्रः कशाघातचिह्नितः सन् भटैः सह आगत्य सोमदत्तम् अयमेव चोरः इति अदर्शयत् ।
भाव : अयम् चोरः ।

आ) महानेवासौ देवं द्रष्टुमिच्छामि ।
समाधान:
परिचय : एतत् वाक्यं सुहृद्भेदः इति पाठ्यभागात् स्वीकृतम् । अस्य कविः नारायणपण्डितः । एषः पाठ्यभागः हितोपदेशात् स्वीकृतः ।
सन्दर्भ : दमनककरटको सञ्जीवकं पिङ्गलसमीपम् अनीतवन्तौ । तौ दृष्ट्वा पिङ्गलकः अपृच्छत् यत् सः अपूर्वः जन्तुः दृष्टः वा इति । तदा दमनकः उक्तवान् यत् स एव भवन्तं द्रष्टुम् इच्छति इति ।
भाव : महान् अयं जन्तुः स्वामिनं द्रष्टुम् इच्छति ।

इ) अरे मुर्खाः, सर्वदा स्वसुखचिन्तैव युष्माकम् ।
समाधान:
परिचय : एतत् वाक्यं भिषजः भैषज्यम् इति पाठ्यभागात् स्वीकृतम् । अस्य रचयिता श्रीमान् पुल्लेल श्रीरामचन्द्रमहोदयः ।
सन्दर्भ : एकस्मिन् दिने केचन ग्रामीणाः वेङ्कटरावस्य सकाशं निर्धारि समयानन्तरं आगताः । मार्गे यानघातेन मूर्छितं बालकं ते आनयन् । परन्तु समयनियमो नाम समयनियमः एव, वैद्यस्यापि विश्रान्तिसुखादिकम् आवश्यकम् इत्युक्त्वा वेङ्कटरावः तान् प्रेषयामास ।
भाव : सर्वदा यूयं स्वसुखमेव पश्यथ ।

ई) अहं वीरपितुः कन्या अस्मि ।
समाधान:
परिचय : एतत् वाक्यं भारतभूषा वीरयोषा इति पाठ्यभागात् स्वीकृतम् । अस्य रचयिता गलगरीरामाचार्यः ।
सन्दर्भ : चिन्तादेव्याः पिता युद्धे वीरस्वर्गं गतः । अनाथायाः तस्यां स्थितिं परिचिन्त्य पितुः सहचराः अशोचन् । तदा चिन्तादेवी उक्तवती यत् अहं वीरपितुः कन्या । पितरम् अनुचिकीर्षामि इति ।
भाव : अहं वीरस्य पितुः पुत्री अस्मि |

7. द्वौ लघुप्रश्नौ समाधत्त । (पद्यभागात्) (2 × 3 = 6)

अ) विक्रमाङ्कदेवस्य आज्ञा किं करोति ?
समाधान:
विक्रमाङ्कदेवस्य आज्ञा पार्थिवानां शिरः चुम्बति ।

आ) विष्णोः सङ्कल्पेन शक्त्या च किं किं भविष्यति ?
समाधान:
विष्णोः सङ्कल्पेन विश्वसृष्टिः भविष्यति । तस्य शक्त्या विश्वविनाशः भविष्यति ।

इ) पाषाणैः प्रहरतेऽपि कः न कुप्यति ?
समाधान:
पाषाणैः प्रहरते स्वम् अपि सहकारः न कुप्यति ।

ई) कस्य जीवितं निष्फलम् ?
समाधान:
यः गुणधर्मविहीनः अपि जीवति, तस्य जीवितं निष्फलम् ।

8. द्वौ लघुप्रश्नौ समाधत्त । (गद्यभागात्) (2 × 3 = 6)

अ) हर्षवर्धनः भण्डिं किमुवाच ?
समाधान:
हर्षवर्धनः भण्डिं राज्यश्रीव्यतिकरः कः इति उवाच ।

आ) युद्धे कः केन व्यापादितः ?
समाधान:
युद्धे सञ्जीवकः पिङ्गलेन व्यापादितः ।

इ) एकाकिन्यपि चिन्ता कथं खेलति स्म ?
समाधान:
एकाकिन्यपि चिन्ता मृदा दुर्गं निर्माय, पुनः तद्भङ्गलीलां खेलति स्म ।

ई) केसरसिंहः कुत्र न्यवसत् ?
समाधान:
केसरसिंहः काश्मीरस्य उपत्यकायां न्यवसत् ।

9. सम्पूर्णेन वाक्येन समाधत्त । (पद्यभागात्) (5 × 1 = 5)

अ) कुशः किम् अस्त्रं समाददे ?
समाधान:
गारुत्मतम्

आ) विक्रमाङ्कस्य औदार्यं कः अरचयत् ?
समाधान:
बिल्हणः

इ) कस्य वचः निशम्य नृपतिः मुदम् अवाप ?
समाधान:
सचिवस्य वचः

ई) मुरारिः कीदृशः इतीरितम् ?
समाधान:
सर्वगतः

उ) कस्य गुणः अवर्ण्यः ?
समाधान:
साङ्गत्यस्य

10. सम्पूर्णेन वाक्येन समाधत्त । (गद्यभागात्) (5 × 1 = 5)

अ) हर्षवर्धनस्य का अवशिष्टा ?
समाधान:
हर्षवर्धनस्य यवीयसी स्वसा राज्यश्रीः अवशिष्टा ।

आ) पिङ्गलकसिंहस्य मन्त्रिणौ कौ ?
समाधान:
पिङ्गलकसिंहस्य मन्त्रिणौ दमनककरटकौ 1

इ) वेङ्कटरावः पितुः नाम किम् ?
समाधान:
वेङ्कटरावः पितुः नाम सुब्बय्यशास्त्री |

ई) शीतत पितामही वृक्षाणाम् उपरि किम् आच्छादयति स्म ?
समाधान:
शीतर्तौ पितामही वृक्षाणामुपरि घासम् आच्छादयति स्म ।

उ) कन्यानुसरणे कः नियुक्तः ?
समाधान:
कन्यानुसरणे मानपालः नियुक्तः ।

11. अधोनिर्दिष्टयोः एकं पत्रं लिखत । (1 × 5 = 5)

अ) धनयाचनार्थं पितरं प्रति पत्रम्
समाधान:

*छात्रावासात्
दिनाङ्कः 22-7-2019

आदरणीय पितृवर्य !
सादरं नमामि । अत्र कुशलं तत्रास्तु | अहं सश्रद्धं विद्याभ्यासं करोमि । अहं कानिचन पुस्तकानि क्रेतुम् इच्छामि । एतदर्थं रूप्यकाणां सहस्त्रं कृपया प्रेषयतु भवान् ।
मातृचरणयोः मम प्रणामाः ।

भवदीयः पुत्र / भवदीया पुत्री
श्रीनिवासः / अहल्या

आ) अवकाशप्रदानाय प्राचार्यं प्रति पत्रम्
समाधान:

दिनाङ्कः 22-7-2019

सेवायाम्
श्रीमान् प्राचार्यमहोदयः
उच्च माध्यमिकविद्यालयः
भाग्यनगरम्
महोदय !
सविनयं निवेदयामि यत् अहम् ज्वरग्रस्तः अस्मि । शिरोवेदना च मां पीडयति । अतः अद्य विद्यालयम् आगन्तुं न शक्रोमि । कृपया 23-7-2019 तः 26-7- 2019 पर्यन्तं दिनत्रयस्य अवकाशं प्रदाय माम् अनुगृह्णातु ।

भवतः शिष्यः / भवदीया शिष्या
श्रीनिवासः / अहल्या
द्वितीयवर्षम्
उच्च माध्यमिक विद्यालयः
भाग्यनगरम्

Note : नगरनाम न परिवर्तितव्यम् ।

12. चत्वारि सन्धिनामनिर्देशसहितं विघटयत ।

1) हरिश्शेते
समाधान:
हरिस् + शेते = श्चुत्व सन्धिः

2) रामोऽयम्
समाधान:
रामः + अयम् = विसर्ग सन्धिः

3) इष्टः
समाधान:
इष् + तः = ष्टुत्व सन्धिः

4) सद्धर्मः
समाधान:
सत् + धर्मः = जश्त्वं सन्धिः

5) चिन्मयम्
समाधान:
चित् + मयम् = अनुनासिक सन्धिः

6) पाचकः पचति
समाधान:
पाचकः + पचति = विसर्ग सन्धिः

7) रामश्चिनोति
समाधान:
रामस् + चिनोति = श्चुत्व सन्धिः

8) अजन्तः
समाधान:
अच् + अन्तः = जश्त्व सन्धिः

13. चत्वारि सन्धिनामनिर्देशसहितं सन्धत्त ।

1) सद् + जनः
समाधान:
सज्जनः = श्चुत्व सन्धिः

2) रामस् + टीकते.
समाधान:
रामष्टीकते = ष्टुत्व सन्धिः

3) अप् + जम्
समाधान:
अब्जम् = जश्त्व सन्धिः

4) षट् + मासाः
समाधान:
षण्मासाः = अनुनासिक संन्धिः

5) वाक् + मयम्
समाधान:
वाङ्मयम् = अनुनासिक सन्धिः

6) नमः + ते
समाधान:
नमस्ते = विसर्ग सन्धिः

7) शरत् + चन्द्रः
समाधान:
शरच्चन्द्रः = श्चुत्व सन्धिः

8) रामः + खादति
समाधान:
रामः खादति = विसर्ग सन्धिः

14. द्वयोः शब्दयोः अन्त-लिङ्ग – वचनमात्रनिर्देशसहितं रूपाणि लिखत ।

अ) गुणिन्
नकारान्तः पुंलिङ्गो ‘गुणिन्’ शब्दः (MERITORIOUS)
समाधान:

आ) सरित्
तकारान्तः स्त्रीलिङ्गः ‘सरित्’ शब्दः ( RIVER)
समाधान:

इ) नामन्
नकारान्तः नपुंसकलिङ्गो ‘नामन्’ शब्दः (NAME)
समाधान:

15. त्रयाणां समासनामनिर्देश सहितं विग्रहवाक्यानि लिखत।
1) यथाशक्ति
समाधान:
शक्तिम् अनतिक्रम्य = अव्ययीभाव समास:

2) दुःखातीतः
समाधान:
दुःखं अतीतः = द्वितीयातत्पुरुष समास:

3) भोज्योष्णम्
समाधान:
भोज्यं च तत् उष्णम् $=$ कर्मधारय समास:

4) पञ्चवटी
समाधान:
पंचानां वटानां समाहारः = द्विगु समासः

5) पार्वतीपरमेश्वरौ
समाधान:
पार्वती च परमेश्वर: च = द्वन्द्व समास:

6) पीताम्बर:
समाधान:
पीतं अम्बरं यस्य स: = बहुव्रीहि समासः

16. अधोरेखाङ्कितपदानि शुद्धीकृत्य वाक्यानि लिखत । (5 × 1 = 5)

1. बालका: फलानि खादति ।
समाधान:
बालक: फलानि खादति।

2. वनं वायुप्रदूषणं परिहरन्ति ।
समाधान:
वनानि वायुप्रदूषणं परिहरन्ति।

3. वारीणि मलिनं प्रक्षालयति।
समाधान:
वारि मलिनं प्रक्षालयति।

4. गुणिन: लोकम् उपकरोति ।
समाधान:
गुणी लोकम् उपकरोति।

5. यूयं विदूषकं परिहससि।
समाधान:
त्वं विदूषकं परिहससि।

Collaborative study sessions centered around TS Inter 1st Year Commerce Model Papers and TS Inter 1st Year Commerce Question Paper March 2018 can enhance peer learning.

TS Inter 1st Year Commerce Question Paper March 2018

Section – A
(2 × 10 = 20)

Answer any two of the following questions in not exceeding 40 lines each.

Question 1.
Define the Co-operative Society. Explain its features.
Answer:
Meaning : The term “co-operation” is derived from the Latin word “co-operari”. The work “Co” means “with” and “operari” means “to work”. Thus the term co-operation means working together. Cooperative society is a voluntary association of persons who work together to promote their economic interest.

Definition : The Indian co-operative societies Act 1932, section (4) defines co-operatives society as “a society which has its objectives for the promotion of economic interests of its members in accordance with co-operative principles”.

Features :

1. Voluntary Association: A co-operative society is a voluntary association of persons. That means, persons can join and leave the society when they want.
2. Open Membership : The Membership is open to all persons having a common economic interest. Any person can became a member irrespective of his/her caste, creed, religion, colour, sex etc.
3. Number of Members : A Minimum of TO’ members are required to form a co-operative Society. In case of multistate co-operative societies, the minimum number of members should be ’50’ from each state.
4. State control : Every co-operative society comes under the control and supervision of the Government. Every society has to get its accounts audited from the co-operative Department of the government.
5. Capital: The capital of the co-operative society is contributed by its members, it after depends on the loans and grants from state and central Government.
6. Democratic set up : The co-operative societies are managed in a democratic manner. The Management of a co-operative society is control by managing committee elected on the basis of “one-man one-vote” irrespective of the member of shares held by any member.
7. Service Motive : The primary objective of all co-operative societies is to provide services to its members, rather than to earn profits.
8. Return on capital Investment : The members of co-operative society get returns on their capital investment in the form of dividend.
9. Distribution of surplus : After giving dividends to the members of the society, the surplus profit is distributed among the members in the form of bonus.
10. Registration of the society : In India, co-operative societies are registered under the co-operative Societies Act.

Question 2.
Distinguish between a Private Company and a Public Company.
Answer:
Introduction :
On the basis of Public Interest, companies may be classified into two types. They are

(1) Private company and
(2) Public company

(1) Private company : According to companies Act, private company is a company which is formed with the association of at least two members but not exceed fifty and with minimum paid up capital of one lakh rupees. It is prohibited from issue of share to the public and also transfer of shares.

(2) Public Company : Public company is a company which is formed with the Association of minimum seven members and there is no limit on maximum number of members. Public companies are required to issue a prospecters for inviting people to purchase their shares. Public company minimum paid up capital is five Lakhs rupees.

Difference between private company and public company :

Question 3.
What is Business Finance ? Explain its needs and significance in the business organisations.
Answer:
Meaning : The requirement of funds by business firm to accomplish its various activities is called “Business finance”.

Finance is considered as the life blood of any organization. The success of any organization depends on the availability of adequate finance.

Definition :
According to B.O. Wheeler : “Finance is that business activity which is concerned with the acquisition and conservation of capital funds in meeting the financial needs and overall objectives of a business enterprise”. ,

Need and Significance :

1. To meet fixed capital requirement of business : Business requires finance to meet fixed capital requirements such as purchase fixed assets like land and building, plant and Machinery, furniture and fixtures etc.
2. To meet working capital requirements : Working capital is used for holding current assets such as purchase of material, payment of wages, transportation expenses, etc. business reequires finance.
3. For growth and expansion : For growth and expansion activities, a business requires finance. It may be required increase production, to install more machines, to set up a R & D centre, etc.
4. For diversification : Entering into new business and new lines of activities is known as diversification. The Business finance is needed to start any new activity in business.
5. For Survival : Without the required finance, organizations can not survive for long time. To carry out the various business operations in continuity, business finance is needed.
6. Liabilities : To meet liabilities of business, may be long-term or short, a business requires sufficient finance, e.g. for payment of loan installments, creditors etc.
7. For payment of expenses : Business finance is needed for payment of expenses like paying salaries, wages, taxes, rent, advertisements etc.

Section -B
(4 × 5 = 20)

Answer any four of the following questions not exceeding 20 lines each :

Question 4.
What is meant by Industry ? Explain four types of Industries.
Answer:
Industry is a business activity, which is related to the extract producing, processing or manufacturing of goods.

Classification / Types of Industries :

1. Primary Industry : Primary Industry is concerned with production of goods with the help of nature and which requires very little human effort. For example : Agriculture ; Farming ; Fishing, etc.
2. Genetic Industry : Genetic Industries are engaged, in re-production and multiplication of certain species of plants and animals with, the object of sale to earn profit Eg : plant Nurseries, fish hatcheries, cattle breeding etc.
3. Extractive Industry : Extractive Industry is concerned with extraction or drawing out goods from the spil, air or water. Generally products of this industry are in raw form and they are used by manu-facturing and construction Industries.
   Eg : Mining Industry, Coal, Mineral, oil ; Iron ore industries etc.
4. Manufacturing Industry: Manufacturing Industry is engaged in the conversion of raw material into semifinished or finished goods. This industry creates form utility in goods by making them suitable for human use.
   Eg : Cement Industry, sugar Industry, Cotton textile industry etc.
5. Construction Industry : construction Industry engaged in the construction of buildings, bridges, roads, dams, canals, etc. These industries are engaged in the creation of infrastructure for the smooth development of the economy.
6. Service Industry : In modem times, service sector plays an important role in the development of the nation and therefore it is named as service Industry.
   Eg : Hotel Industry ; Tourism Industry, entertainment industry etc.

Question 5.
State three advantages and two disadvantages of Hindu Undivided Family business.
Answer:
Hindu undivided family business is a form of business where the family members of three successive generations own the business jointly. The head of the family known as “karta” manages the business. The other members are called “co-parceners”. All of them have equal ownership right over the properties of the business.

Advantages :

1. Continuity : It is not dissolved by the death of insanity of a coparcener.
2. Centralized and efficient management: The management of J.H.E firm is vested in the hands of karta only. This results in the unity of command and disciplined management.
3. No limit to membership : By birth every coparcener will automatically became a number in joint hindu family. Hence, there is no limit to membership.
4. Better credit : It credit worthiness is better than of sole Trader.

Disadvantages :

1. No Direct correlation between Efforts and Rewards: Kartha alone looks after the business; but benefits are shared among all coparceners. There is no rewards for their work.
2. Limited managerial ability: In Joint Hindu undivided family business, the kartha alone to manage, so in expansion and growth of business- the management and control becomes difficult.
3. Limited capital and Financial Resources : compared with partnership and joint stock company the capital and financial resources of the joint Hindu undivided family firm.

Question 6.
What is Partnership Deed ? Also write its main contents.
Answer:
Partnership deed is a document containing the terms and conditions of a partnership. It is a agreement in writing signed by all the partners duly stamped and registered. The partnership deed defines certain rights, duties and obligations of partners and governs relationship among them in the conduct of business. This deed has to be stamped in accordance with the Indian stamp Act 1899.

The following points are generally be included in the deed.

Contents :

1. Name of the firm.
2. Nature of the business.
3. Name and address of partners.
4. Location of business.
5. Duration of partnership.
6. Amount of capital to be contributed by each partner.
7. profit and loss sharing ratio.
8. Duties, powers and obligations of partners.
9. Salaries and withdrawals of the partners.
10. Preparation of accounts and their auditing.
11. Procedure for dissolution of the firm.
12. Procedure for settlement of disputes.

Question 7.
Explain the’advantages of Equity Shares.
Answer:
Equity shares are also known as “ordinary shares” represent the ownership of a company. A company can raised the capital by issue of equity shares.

Advantages :

1. Equity shares do not create any obligation to pay a fixed rate of dividend.
2. Equity shares can be issued without creating any charge over the assets of the company.
3. It is a permanent source of capital and the company need not to repay it except under liquidation.
4. Equity share holders are the real owners of the company who have the voting right.
5. Incase of profits, equity shareholders are the real gainers by way of increased dividends and appreciation in the value of shares.

Question 8.
Differentiate between a share and a debenture.
Answer:
The capital of a company is divided into small units called “share”. “Debentures” are important instruments for raising long term debt capital.

The following are the differences between shares and debentures.

Question 9.
Explain the features of MNCs.
Answer:
MNCs means Multinational Corporations. The word “Multinational” is a combination of words “Multi” which refers to many and” national” refers to a country. So multinationals means a company which operates in more than one country. MNC refers to an enterprise whose managerial headquarters are located in one country, while it carries out operations in a number of other countries.

According to Pavid E. Liliental, defines the MNCs as “Corpo-rations which have their home in one country but operate and live under the laws and customs of other countries as well”.

Features of MNCs :

1. Giant size : The assets and sale of MNCs are very large. These companies operate on large scale as they trade in more than one country. Some times their sales turnover exceeds the Gross National product (GNP) of developing countries.
2. International Operation : A MNC operates and sells their products in different countries. It operates through a network of branches and subsidiaries in host countries.
3. Professional Management : A Multinational corporation employees professional experts, specialized people. MNCs try to keep their employees updated by training from time to time to handle the advance in technology effectively.
4. Oligo polistic powers : Oligopoly means power in the hands of few companies only. Due to their gaint size, the MNCs occupy dominating position in the market.
5. Centralized control : MNC’s will have managerial headquarters in the home country. All the subsidiary and branches work under the supervision, guidance and control of the head office.
6. Sophisticated Technology : Multinational companies make use of latest, advanced and innovative technology to produce quality goods.

Section – C
(5 × 2 = 10)

Answer any five of the following questions not exceeding 5 lines each :

Question 10.
What is Employment ?
Answer:

1. An employment is a contract of service between employee and employer. An employee works under an agreement as per the rules of service assigned to him by the employer.
2. For performing such activities employee receives remuneration from employer in the form of “wage” or salary.

Question 11.
What is Sole Proprietorship ?
Answer:

1. Sole proprietorship is a form of business organisation in which a single individual introduces his own capital; skill in the management and he only responsible for the results of its operations.
2. According to kimball and Kimball: “sole proprietorship is a form of business where the individual proprietor is the supreme judge of all matters to his business”.

Question 12.
Define Active Partner.
Answer:

1. The partner who actively participate in the day-to-day operations of the business is called as “Active partners”.
2. Active partner is also called as “working partner”.

Question 13.
Define M.O.A.
Answer:

1. M.O.A. Means memorandum of Association. It is the constitution of the company. It defines the scope of company’s activities as well as its relation with the out siders of the company.
2. According to companies Act, “The memorandum of Association of a company as originally framed or as altered from time to time in pursurance of ahy previous company laws or of this Act”.

Question 14.
Define preference shares.
Answer:
1) As per sec 85 of the Indian companies Act 1956, preference shares are those shares which carry special rights over the equity shares.
2) preference shares have the following two rights :

1. Right to receive the fixed rate of dividend irrespective of profits.
2. Right to get back the capital prior to equity shareholders at the time of company’s liquidation.

Question 15.
Define Service Enterprise.
Answer:
1) The enterprises involve in providing or rendering services are called “service enterprises”.

2) There are 3 types of services enterprises based on Investment in equipment.
a) Micro service enterprise : Investment in equipment does not exceed ₹ 10 lakhs.
b) Small service enterprise : Investment in equipment is more than ₹ 10 lakhs but not exceeds ₹ 2 crore.
c) Medium service Enterprise : Investment and equipment is more than ₹ 2 crore but not exceed ₹ 5 crore.

Question 16.
What is Fixed Capital ?
Answer:

1. The capital which is used to acquire fixed assets such as land and buildings, plant and machinery etc., is called “Fixed capital”.
2. Business enterprises use fixed capital for longterms requirements.

Question 17.
What is E-Commerce’?
Answer:

1. E – Commerce is short for “Electronic commerce”.
2. Transacting or facilitating business through Internet is called “E – commerce”.
3. Popular examples of E – Commerce revolve around buying and selling online.

Section – D (1 × 20 = 20)
(50 Marks)

Answer the following question :

Question 18.
From the following Trial Balance, prepare Final accounts of Sathish as on 31.12.2012.
Trial Balance

Adjustments:
(1) Closing Stock ₹ 2,100.
(2) Outstanding Stationery ₹ 600.
(3) Depreciation on Machinery ₹ 10%.
(4) Bad debts ₹ 500.
(5) Prepaid Wages ₹ 500.
Answer:
Trading and profit & loss Account of sathish for the year ended 31-12-2017

Balance sheet of Sathish as on 31-12-2017

Section – E (1 × 10 = 10)

Answer any one of the following questions.

Question 19.
Prepare a three-column cash Book from the following particulars

Answer:

Question 20.
From the following information, prepare a Bank Reconciliation Statement of M/s. Sathish Brothers as on 30th April, 2017 :
(1) Credit balance of Pass Book ₹ 45,000.
(2) Cheque amounting to ₹ 10,500 were deposited in the Bank but only cheques of ₹ 4,500 were cleared upto 30^th April.
(3) Cheques amounting ₹ 15,000 were issued but cheques worth ₹ 5,100 not been Presented for payment in the Bank upto 30^th April.
(4) In the Pass Book there was a credit ₹ 300 for interest on investments and debit of ₹ 75 for bank charges.
Answer:
Bank reconciliation statement of M/s. Sathish Brothers as on 30^th Apr.2017

Section – F (2 × 5 = 10)

Answer any two of the following questions :

Question 21.
Explain different types of Accounts along with their debit-credit rules.
Answer:
Accounts are classified into 3 types. Which are

1) Personal Accounts
2) Real Accounts and
3) Nominal Accounts.

1) Personal Accounts : The accounts which are related to persons, firms and companies are called “personal accounts”. Ex : Ramu a/c , State Bank of India a/c etc.
Rule : Debit The receiver
Credit the giver.

2) Real Accounts : Accounts related to properties and assets of
business are called “Real accounts”. Ex : Cash a/c, Furniture a/c, land & building a/c ; goodwill a/c etc

Rule : Debit what comes in
Credit what goes out.

3) Nominal Accounts: The accounts related to expenses, losses, incomes and gains are called as “Nominal Accounts”. Ex : Wages a/c , salaries a/c, commission a/c, rent a/c, discount received a/c etc.
Rule : Debit all expenses and losses
Credit all incomes and gains.

Question 22.
Prepare Shirisha’s account from the following particulars :

Answer:

Question 23.
Prepare Sales Book and Sales Return Book from the following :

Answer:

Question 24.
What-are the Errors disclosed by the Trial Balance ?
Answer:
The errors which affect the agreement of Trial Balance are called Errors disclosed by the Trial Balance.
The following are the errors disclosed by Trial Balance :
1) Posting of Transaction to wrong side of an account: Posting of transaction to wrong side of account affect the agreement of Trial Balance.
Eg : Discount allowed posted to credit side of discount account.
2) Posting of wrong amount to an account: Posting of wrong amount affect the agreement of Trial Balance.
Eg : Sales of ₹ 5000 posted as ₹ 500 to sales account.
3) Errors in totaling : Errors in totaling it may be over cast or undercast affects the agreement of trial Balance.
Eg: Sales Returns Book Overcast by ₹ 100.
4) Errors of carrying forward: If a mistake in carrying forward a total of one page to the next page. This error affect the agreement of Trial Balance.
Eg: purchase Book total is carried forward as ₹ 1,500 instead of ₹ 150.
5) Posting of only one aspect of Journal Entry into ledger :
Some times accountant may post only one aspect of entry to the ledger account, it affect the Agreement of Tril Balance. Eg: Sale of goods to Ramesh ₹ 200 Posted to sales book only.
6) Recording one aspect twice :
An accountant may be record one aspect twice it affect the Agreement of Trial Balance.
Eg : Paid salaries ₹ 600 debited twice in salaries account.

Section – G (5 × 2 = 10)

Answer any five of the following questions :

Question 25.
What is an Accounting Cycle ?
Answer:
An Accounting Cycle is a complete sequence of accounting process that begins with the recording of business transactions and ends with the preparation of final accounts.
It involves Journal, ledger, Trial Balance and financial statements such as Trading account, profit and loss account and Balance sheet.

Question 26.
Explain Business Entity Concept.
Answer:
As per this concept, Business organisation are treated as a separated entity which can be distinguished from the “owners” or stakeholders who provide capital to it.
This concept helps in keeping private affairs of the owners and stakeholders separate from the business affairs.

Question 27.
Journalise the following transactions :

Answer:
Journal Entries

Question 28.
Write opening Journal Entry from the following as on 1^st April, 2017 :

Answer:
Opening Entry as on 1^st April 2017

Question 29.
What iš Debit Note?
Answer:
Debit Note is a document sent to the supplier while returning the goods purchased on credit from him.
This Debit Note intimating that the supplier account is debited to the extent of goods returned and reasons for returning them.

Question 30.
What is Bank Overdraft’
Answer:
Bank overdraft is an arrangement made by the customer with banker to drawn an excess amount of what the customer has in his bank account balance subject to a specific predetermined limit.

Question 31.
Give the meaning of Bad debts.
Answer:
The debts which are not collected and irrecoverable are knowns as “Bad Debts”.
Baddebts are treated as loss to the business.

Question 32.
Prepare Trial Balance of Sathish:

Answer:
Trial Balance of Sathish

Utilizing AP Inter 1st Year Chemistry Model Papers Set 2 helps students overcome exam anxiety by fostering familiarity.

AP Inter 1st Year Chemistry Model Paper Set 2 with Solutions

Time: 3 Hours
Maximum Marks: 60

Note : Read the following instructions carefully.

1. Answer all questions of Section – A. Answer ANY SIX questions in Section – B and ANY TWO questions in Section – C.
2. In Section – A, questions from Sr. Nos. 1 to 10 are of Very short answer type. Each question carries TWO marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
3. In Section – B, questions from Sr. Nos. 11 to 18 are of Short answer type. Each question carries FOUR marks. Every answer may be limited to 75 words.
4. In Section – C, questions from Sr. Nos. 19 to 21 are of Long answer type. Each question carries EIGHT marks. Every answer may be limited to 300 words.
5. Draw labelled diagrams, wherever necessary for questions in Section – B and Section – C.

Section – A

Note : Answer ALL questions.

Question 1.
Define receptor, sink.

Question 2.
Which oxides cause the acid rain? What is the pH value of acid rain?

Question 3.
Calculate kinetic energy of three moles of CO₂ at 27°C (in calories only)

Question 4.
What is disproportionation reaction? Give one example.

Question 5.
Calculate pH of a 1.0 × 10 8 M solution of HCl.

Question 6.
Lithium salts are mostly hydrated. Why? Give one example.

Question 7.
Why is gypsum added to cement?

Question 8.
What is synthesis gas? How is it prepared?

Question 9.
Define catenation. Write two allotropic crystalline forms of carbon.

Question 10.
Write IUPAC names of the following compounds :
a) (CH₃)₃C CH₂ C (CH₃)₃

Section – B

Note : Answer ANY SIX questions.

Question 11.
Write the postulates of kinetic molecular theory of gases.

Question 12.
Define normality. Calculate the normality of oxalic acid solution containing 6.3 g of H₂C₂O₄.2H₂O in 500 ml of solution.

Question 13.
Define heat capacity. Derive C_p – C_v = R.

Question 14.
State Lechatelier’s principle. Explain the application of Le Chatelier’s principle on the synthesis of sulphur trioxide (write the effects of temperature and pressure only).

Question 15.
Which salts are responsible for the hardness of water ? How the hardness of water is removed by Calgon method?

Question 16.
Explain the structure of diborane.

Question 17.
How is acetylene prepared from the following compounds :
a) Calcium carbide
b) 1, 2-dibromoethane

Question 18.
How does ethylene react with the following :
(a) Cl₂
(b) HBr
(c) H₂SO₄
(d) O₃

Section – C

Note : Answer ANY TWO of the following questions.

Question 19.
Explain the Quantum numbers and their significances briefly.

Question 20.
Write an essay on s, p, d and f block elements.

Question 21.
Define hybridization. Explain the types of hybridization involving s and p orbitals with one example each.