Explore topic-wise InterviewSolutions in Current Affairs.

The first objective of education in India was to form a class of interpreters between the British rulers and the millions of Indians they governed. The second objective was to create a class of persons, Indian in blood and colour but British in taste, opinion, morals and intellect. The third objective was to obtain a cheap supply of clerks for holding subordinate posts in administration and British business concerns.

The Charter Act of 1813 directed the Company to spend 1 lakh rupees on the education of Indians. This was the first step taken by the British rulers towards the encouragement of the study of literature and science in India. The Charter Act, however, did not lay down any specific guidelines.

The new zamindars, unsure of retaining the contract at the next auction, had no permanent interest in the land and did nothing to improve it. The peasants were fleeced to meet the revenue targets.

Since revenue payments had to be paid in cash, the peasants began to grow cash crops like jute, cotton, sugarcane etc., which could be sold for ready cash in the markets.

Knowledge of contemporary nationalist movements in Europe fired the Indians with an intense desire to build a new India progressive, strong, prosperous and united.

1. The British system of education produced English-speaking Indian graduates who helped their British masters to run the empire. 

2. It also created a class of Indians who were Westernized to the extent that they rejected Indian culture and patronized anything and everything that was British including British goods.

The revenue from Bengal was used to cover as salaries of officials and to finance the trading activities of the company. Raw materials for England’s growing industries were bought with the revenues collected from Bengal.

B) E. Krishna Iyer

1. Subhash Chandra Bose became the President of Indian National Congress twice. 

2. He became the President of Congress when Second World War had commenced. He was of the opinion that as England was engaged in war, the Congress should intensify the agitation by taking advantage of this situation. 

3. If needed, India should seek the help of the enemies of England. 

4. Other senior leaders of the Congress did not agree with this view.

As a result, Subhash Chandra Bose resigned from the office of the Congress President.

(C) E. Krishna Iyer

(d) Madras Presidency

Answer is (A) Dappu

Devadasi was banned by an act in Bombay, Madras Presidency during 1934-1947.

The first 100 years, i.e., the first century of this era is written as years 1 – 100 CE or 1 – 100 AD.

The calendar we use today is based on the Christian era. Since it is commonly followed all over the world, now it is called as ‘Common Era’.

Prophet Muhammad migrated from Mecca to Medina, so, the ‘Hijri Era’ was started to commemorate this event.

Prophet Muhammad is the founder of Islam.

This era began in memory of Jesus Christ and so this era is called as Christian Era.

The first century before the Common Era is indicated as 100 – 1 BC.

The first millennium Before the Common Era is indicated as 1000 – 1 BCE.

The day is divided into two parts – day and night.

The Pre-historic period is the period for which no written records are available.

The period for which written records are available from which history is written is known as Historic period.

The time in history is divided into two main periods: Pre-historic and Historic period.

Our Earth is a planet in the solar system, so it is presumed that the earth was also formed 4.5 billion years ago.

1. Due to the earth’s rotation around its own axis, only a part of the earth’s surface faces the sun. 

2. It is only this part of the surface that becomes bright 

3. The part that is not facing the sun remains dark. 

4. Therefore we see light only in daytime even though the earth receives continuous light from the sun.

Chhatrapati Shivaji Maharaj started a new era or shaka known as ‘Rajyabhishek Shaka’ in 1674 to commemorate his coronation.

Our solar system came into being about 4.5 billion years ago

1. The span of 4.5 billion years, since the earth’s formation is a vast period of time. 

2. It is not easy to grasp this entire period all at оnсе. 

3. It is necessary to divide it into a number of stages in order to understand it better. 

4. Therefore, Time in history is divided into two main periods, i.e. Pre-historic and Historic period.

1. The calendar we use today is based on the Christian era.

2. This era is most widespread and commonly used all over the world. 

3. Therefore this era is now called the Common era.

i. There are different methods of reckoning time 

ii. Time is continuous, but for our convenience we divide time into periods.

It is an age-old custom to start a new era to commemorate a special event.

1. There are various methods of measuring time. 

2. These methods allow us to identify a particular day, month or year with respect to an earlier or later day, month or year. 

3. Thus, when we measure time, we actually measure its length.

1. Rajyabhishek Shaka 

2. Shalivahan Shaka, Vikram Samvat 

3. Prophet Muhammad 

4. Hijri 

5. history

The given equations are

11x + 15y = – 23 …(i)

and 7x -2y = 20 …(ii)

Multiplying (RBSESolutions.com) equation (i) by 2 and (ii) by 15 and then adding, we get

127x = 254

⇒ x = 2

Putting x = 2 in (ii), we get

7 × 2 – 2y = 20

⇒ 14 – 20 = 2y

⇒ 2y = – 6

⇒ y = – 3

Hence the required solution is x = 2 and y = – 3.

The given equations are

0.4x + 0.3y = 1.7 …(i)

0.7x – 0.2y = 0.8 …(ii)

Multiplying equation (i) by 2 and (ii) by 3 and then adding, we get

2.9x = 5.8

⇒ x = 2

Substituting the value of x in (i), we get

0.4 x 2 + 0.3y = 1.7

⇒ 0.3y = 1.7 – 0.8

⇒ 0.3y = 0.9

⇒ y = 3

We have,

((x+1)/(x+2))² = (x+2) / (x + 4)

(x+1)² / (x+2)² – (x+2) / (x + 4) = 0

By taking LCM as (x+2)² (x+4)

((x+1)² (x+4) – (x+2) (x+2)²) / (x+2)² (x+4) = 0

By cross-multiplying we get,

(x+1)² (x+4) – (x+2) (x+2)² = 0

Let us expand the equation

(x² + 2x + 1) (x + 4) – (x + 2) (x² + 4x + 4) = 0

x³ + 2x² + x + 4x² + 8x + 4 – (x³ + 4x² + 4x + 2x² + 8x + 8) = 0

x³ + 2x² + x + 4x² + 8x + 4 – x³ – 4x² – 4x – 2x² – 8x – 8 = 0

-3x – 4 = 0

x = -4/3

Now let us verify the given equation,

((x+1)/(x+2))² = (x+2) / (x + 4)

By substituting the value of ‘x’ we get,

(x+1)² / (x+2)² = (x+2) / (x + 4)

(-4/3 + 1)² / (-4/3 + 2)² = (-4/3 + 2) / (-4/3 + 4)

((-4+3)/3)² / ((-4+6)/3)² = ((-4+6)/3) / ((-4+12)/3)

(-1/3)² / (2/3)² = (2/3) / (8/3)

1/9 / 4/9 = 2/3 / 8/3

1/4 = 2/8

1/4 = 1/4

Hence, the given equation is verified.

We have,

((2x+3) – (5x-7))/(6x+11) = -8/3

By cross-multiplying we get,

3((2x+3) – (5x-7)) = -8(6x+11)

3(2x + 3 – 5x + 7) = -48x – 88

3(-3x + 10) = -48x – 88

-9x + 30 = -48x – 88

-9x + 48x = -88 – 30

39x = -118

x = -118/39

Now let us verify the given equation,

((2x+3) – (5x-7))/(6x+11) = -8/3

By substituting the value of ‘x’ we get,

((2(-118/39) + 3) – (5(-118/39) – 7)) / (6(-118/39) + 11) = -8/3

((-336/39 + 3) – (-590/39 – 7)) / (-708/39 + 11) = -8/3

(((-336+117)/39) – ((-590-273)/39)) / ((-708+429)/39) = -8/3

(-219+863)/39 / (-279)/39 = -8/3

644/-279 = -8/3

-8/3 = -8/3

Hence, the given equation is verified.

We have,

((x+1)/(x-4))² = (x+8)/(x-2)

(x+1)² / (x-4)² – (x+8) / (x-2) = 0

By taking LCM as (x-4)² (x-2)

((x+1)² (x-2) – (x+8) (x-4)²) / (x-4)² (x-2) = 0

By cross-multiplying we get,

(x+1)² (x-2) – (x+8) (x-4)² = 0

Upon expansion we get,

(x² + 2x + 1) (x-2) – ((x+8) (x² – 8x + 16)) = 0

x³ + 2x² + x – 2x² – 4x – 2 – (x³ – 8x² + 16x + 8x² – 64x + 128) = 0

x³ + 2x² + x – 2x² – 4x – 2 – x³ + 8x² – 16x – 8x² + 64x – 128 = 0

45x – 130 = 0

x = 130/45

= 26/9

Now let us verify the given equation,

((x+1)/(x-4))² = (x+8)/(x-2)

(x+1)² / (x-4)² = (x+8) / (x-2)

By substituting the value of ‘x’ we get,

(26/9 + 1)² / (26/9 – 4)² = (26/9 + 8) / (26/9 – 2)

((26+9)/9)² / ((26-36)/9)² = ((26+72)/9) / ((26-18)/9)

(35/9)² / (-10/9)² = (98/9) / (8/9)

(35/-10)² = (98/8)

(7/2)² = 49/4

49/4 = 49/4

Hence, the given equation is verified.

The general form of a linear equation in two variables x and y is ax + by + c = 0, where a, b, c are real numbers and a ≠ 0, b ≠ 0. 

Five linear equations in two variables are as follows: 

i. 3x + 4y – 12 = 0 

ii. 3x – 4y + 12 = 0 

iii. 5x + 5y – 6 = 0 

iv. 7x + 12y – 11 = 0 

v. x – y + 5 = 0

The given equations are

2x + y = 13 …(i)

and 5x – 3y= 16 …(ii)

Multiplying equation (i) by 3 and then adding in (ii),

we get

11x = 55

\(x=\frac { 55 }{ 11 }=5\)

Substituting the value of x in (i), we get

2 × 5 + y = 13

⇒ 10 + y = 13

⇒ y = 13 – 10 = 3

Hence the required solution is x = 5, y = 3.

i. m + 3 = 5 

m = 5 – 3 

∴ m = 2 

ii. 3y + 8 = 22 

∴ 3 y = 22 – 8 

∴ 3 y = 14 

∴ y = 14/9 

iii. x/3 = 2 

∴ x = 2 × 3 

∴ x = 6 

iv. 2p = p + (4/9)

∴ 2p – p = (4/9)

∴ p = 4/9 

We have,

((x+2) (2x-3) – 2x² + 6)/(x-5) = 2

By cross-multiplying we get,

(x+2) (2x-3) – 2x² + 6) = 2(x-5)

2x² – 3x + 4x – 6 – 2x² + 6 = 2x – 10

x = 2x – 10

x – 2x = -10

-x = -10

x = 10

Now let us verify the given equation,

((x+2) (2x-3) – 2x² + 6)/(x-5) = 2

By substituting the value of ‘x’ we get,

((10+2) (2(10) – 3) – 2(10)² + 6)/ (10-5) = 2

(12(17) – 200 + 6)/5 = 2

(204 – 194)/5 = 2

10/5 = 2

2 = 2

Hence, the given equation is verified.

i. 2x + y = 5 …(i) 

3x – y = 5 …(ii) 

Adding equations (i) and (ii), 

2x + y = 5 + 3x – y = 5 

5x = 10 

∴ x = 10/5 

∴ x = 2 

Substituting x = 2 in equation (i), 

2(2) + y = 5 

4 + y = 5 

∴ y = 5 – 4 = 1 

∴ (2, 1) is the solution of the given equations.

ii. x – 2y = -1 

∴ x = 2y – 1 … .(i) 

∴ 2x – y = 7 ….(ii) 

Substituting x = 2y – 1 in equation (ii), 

2(2y – 1) – y = 7 

∴ 4y – 2 – y = 7 

∴ 3y = 7 + 2 

∴ 3y = 9 

∴ y = 9/3 

∴ y = 3 

Substituting y = 3 in equation (i), 

x = 2y – 1 

∴ x = 2(3) – 1 

∴ x = 6 – 1 = 5 

∴ (5, 3) is the solution of the given equations.

iii. x + y = 11 

∴ x = 11 – y …(i) 

2x – 3y = 7 …….(ii) 

Substituting x = 11 - y in equation (ii),

2(11 – y) – 3y = 7 

∴ 22 – 2y – 3y = 1 

∴ 22 – 5y = 7 

∴ 22 – 7 = 5y 

∴ 15 = 5y 

∴ y = 15/5 

∴ y = 3 

Substituting y = 3 in equation (i), 

x = 11 – y 

∴ x = 11 – 3 = 8 

∴ (8, 3) is the solution of the given equations.

iv. 2x + y = -2 …(i) 

3x – y = 7 …(ii) 

Adding equations (i) and (ii), 

2x + y = -2 + 3x – y = l 

5x = 5 

∴ x = 5/5

∴ x = 1 

Substituting x = 1 in equation (i), 

2x + y = -2 

∴ 2(1) + y = -2 

2 + y = -2 

∴ y = – 2 – 2 

∴ y = -4 

∴ (1, -4) is the solution of the given equations.

v. 2x – y = 5 

∴ -y = 5 – 2x 

∴ y = 2x – 5 …(i) 

3x + 2y = 11 ……(ii) 

Substituting y = 2x – 5 in equation (ii), 

3x + 2(2x – 5) = 11 

∴ 3x + 4x - 10 = 11 

∴ 7x = 11 + 10 

∴ 7x = 21 

∴ x = 21/7

∴ x = 3

Substituting x = 3 in equation (i), 

y = 2x – 5 

∴ y = 2(3) – 5 

∴ y = 6 – 5 = 1 

∴ (3,1) is the solution of the given equations.

vi. x – 2y = -2 

∴ x = 2y – 2 …(i) 

x + 2y = 10 …..(ii) 

Substituting x = 2y – 2 in equation (ii), 

2y – 2 + 2y = 10 

∴ 4y = 10 + 2 

∴ 4y = 12 

∴ y = 12/4

∴ y = 3 

Substituting y = 3 in equation (i), 

x = 2y – 2 

∴ x = 2(3) – 2 

∴ x = 6 – 2 = 4

∴ (4, 3) is the solution of the given equations.

We have,

(y² + 4)/(3y² + 7) = 1/2

By cross-multiplying we get,

2(y² + 4) = 1(3y² + 7)

2y² + 8 = 3y² + 7

3y² – 2y² = 7 – 8

y² = -1

y = √-1

= 1

(A) 2

Adding the given equations, 

3x+ 5y = 9 

5x + 3y = 7

8x + 8y = 16 

∴ x + y = 2 .. [Dividing both sides by 8]

i. x + 4 = 9 

∴ x + 4 – 4 = 9 – 4 

… [Subtracting 4 from both the sides]

∴ x = 5

ii. x – 2 = 7 

∴ x – 2 + 2 = 7 + 2 

… [Adding 2 on both the sides] 

∴ x = 9

iii. 4x = 24 

∴ \(\cfrac{4x}{4} =\cfrac{24}{4}\)

… [Dividing both the sides by 4] 

∴ x = 6

(i), (ii), (iv), (viii) are the simple equations. Since these are all in the form of ax + b = 0.