Explore topic-wise InterviewSolutions in Current Affairs.

Heiun – Tsang was a Chinese Buddhist traveller. He came to India to see the Buddhist places in India and to study Buddhist texts (granthas).

Ashoka followed the administrative system of Chandragupta Maurya, though he made some important changes and improvements in terms of implementation of policies and objectives. He called his subjects his children. He kept welfare of all people and valour as his primary duties. He appointed Rajuk, Yukta, and Pradeshik etc. They dealt with land, justice and accounts.

He created the post of Dhamma Mahamatra. His job was to create harmony between various communities. His job also included to provide aid and assistance to unjustly persectued people and their families. Ashoka set up such a system which enabled communication of the problems of common people to the king at all times and from all places. He appointed ‘Prativedaks’, the mention of whom is made in his 6th rock edict.

In order to bring uniformity in the judicial system, the Rajukas were given independent power in justice – related matters in the 25th year of Ashoka’s reign. He liberalised the penal code and abolished inhuman punishments. He released prisoners on his coronation day. Three days were given to death sentence awardee prisoners for repentence.

In non – violence oriented reforms, Ashoka gave up war policy. He emphasised medical care, roads, well and tree plantation for the welfare of all living beings. He made rural development his priority. Stradhyaksha, Vrajabhumik, Mahamantra, Nagar Vyavaharik, Antamahamatra etc. were appointed to oversee matters related to women, animal protection, justice and border regions.

These appointments helped Ashoka to connect administration with common people. By founding Dhamma’, Ashoka prepared an agenda for the king, subjects and bureaucracy. This improved mutual relations. Ashoka made an efficient foreign policy. Hence, Ashoka made the established Mauryan administration more efficient and capable.

नदी की यह विशेषता है कि वह अपना पानी स्वयं नहीं पीतीं।

सही विकल्प है C. 0

सही विकल्प है (ख) साइबेरिया का मैदानी भाग

पति ने ‘याद रखने के एक हजार तरीके’ नामक नयी किताब खरीदी।

पिन्टु ने 15 दिनों के बाद राजू को तमाचा मारा।

सही विकल्प है C. कम

कहीं तीली कम तो नहीं, यह जानने के लिए सुशील के पिताजी माचिस की सारी तीलियाँ गिनते हैं।

Avadh was declared as an independent state by Saadat Khan in 1739 A.D.

Correct option is (d) All of these

1. The colonial rule aimed at taking away raw material cheaply from colonies and selling away their finished products in colonies.

2. All colonial countries conducted sea trade. 

3. So they obviously connected different regions of raw material availability and markets to finished goods with these port cities. 

4. So the development was mostly concentrated in the port cities only. 

5. Apart from this, most of their factors lived there and their fortifications also present there only.

B) Fundamental rights

Correct option is B) Nepal

A) Fundamental duty

The Punjab played an important role in the history of India due to its special geographical location. It became the cradle of Indian civilization. The oldest ancient culture (Indus Valley Civilization) flourished in the Punjab. The Aryans made it the centre of their political sway. They composed their sacred books like the Vedas, Puranas, Mahabharta, Ramayana etc. in the Punjab. Punjab was the the Gateway of India.

All the invaders upto the medieval period came to India by passing through Punjab. Hence, people of Punjab had to fight numerous battles to hold back the advancing invaders, Apart from this, Punjab was the birthplace of Hinduism and Sikhism. Guru Nanak Dey Ji gave his divine message on this very land. It was here that Guru Gobind Singh created the Khalsa Panth and successfully resisted the Mughal oppression. Banda Singh Bahadur and Maharaja Ranjit Singh hold prominent places in the history of India.

The geographical features of Punjab can be divided into three parts, keeping in view the history of Punjab:

1. Himalayas and the North-West Mountain ranges.
2. The Terai region (foothills).
3. The Plains.

The mighty Himalayas form the boundary of Punjab in the north. The high rising peaks of the Himalayas are always covered with snow. The Himalayas hìve three ranges which run parallel to one another. There are numerous passes in the North-West ranges through which the invaders, traders and religious preachers had been corning to India since ancient times. The second geographical division of Punjab is Terai region. It is sandwiched between the mountains and the plains of Punjab. The population in this region is small.

The most important geographical division of Punjab is its plains, which are very fertile. They extend from river Indus in the north-west to river Yamuna in the southeast. It is formed by the fertile soil deposited by the rivers from the Himalayas. It is the cause of the prosperity of Punjab since ancient times.

Each geographical feature of Punjab had influenced differently the history of Punjab:

1. The passes of the North-West Mountain Ranges of Himalayas allowed passage to numerous invaders. Hence, the security of the North-west region remained a major problem for every ruler. On the other hand, the snow-capped high rising peaks of the Himalayas in the north of Punjab provided perfect security from all the possible invaders from the northern side.
2. The distinctive culture of Punjab is a gift of the Himalayas standing in the North and the West of Punjab.
3. The prosperity and wealth of Punjab had always attracted greedy invaders. Consequently, Punjab had to face their invasions frequently.
4. The Terai region provided shelter to the Sikhs during their hard times. The Sikhs saved themselves and courageously faced the oppressive rulers.

The rivers acted as a great hurdle in the path of invaders of Punjab.

Our National Emblem is used at the following places:

1. On the letter head of the President, the Vice-President and the Governor of India. 

2. On the currency of India. 

3. On the uniform of Indian Navy, Army and Air Force. 

4. On judicial and non judicial stamp papers. 

5. On postal stamps.

1. If the migration takes place on person’s free will, initiative and desire to live in a better place.

2. To improve their financial status, the migration is said to be voluntary

Urbanization is driven by three factors. They are, 

1. Natural population growth.

2. Rural to Urban Migration and 

3. Reclassification bf rural areas into urban areas.

Demographic causes of Migration:

1. In demographic sense, the population composition like age and sex, over population and under population are the major causes of migration.

2. It is well known fact that adults are more migratory than any other age-groups.

3. Women mostly migrate after their marriage.

4. Generally over population is considered as a push factor and under population to be Pull factor in the context of migration.

Political cause of Migration:

1. Various political causes like colonization, wars, government policies etc. have always been playing important role in human migration from time to time.

2. Wars have been one of the significant causes of migration since ancient time.

Yes, globalization has led to polarization of classes. This has led to segregation of people in the society that may emerge from income inequality, economic restructuring etc. It leads to differentiation of groups on the basis of high income and low income. Skilled people manage to get high paying jobs while the less educated/skilled people receive low wages.

The unfavorable factors which make the people to move out from a location are called push factors.

1. Economy is one of the most important causes of human migration from one area to another.

2. The availability of fertile agricultural land, employment opportunities, development of technology etc. are some of the economic causes that attract the migration.

(b) Ancient time

(c) 936

The number of 4-letter words which can be formed from 6 letters when one or more of the letters is repeated = 6 × 6 × 6 × 6 = 1296 

Number of 4-letter words which can be formed from the given 6 letters when none of the letters is repeated 

= Number of arrangements of 6 letters taken 4 at a time 

= ⁶P₄ = \(\frac{6!}{(6-4)!} = \frac{6!}{2!}\) = 6 × 5 × 4 × 3 = 360 

∴ Number of 4 letter words which have at least one of their letters repeated = 1296 – 360 = 936.

If any letter can be used any number of times, then the number of words of 4 Letters with 8 different letters is 8 × 8 × 8 × 8 = 8⁴ = 4096 

Number of words of 4 letters with at least one letter repetition not allowed = ⁸P₄ = 8 × 7 × 6 × 5 = 1680 

∴ Number of 4 letter words with at least one letter repeated is 8⁴ – ⁸P₄ = 4096 – 1680 = 2416.

(d) 720

The first place will always be filled by C and the last place will always be filled with Y. The remaining six places can be filled by the remaining 6 letters in ⁶P₆ ways. 

∴ Total number of words beginning with C and ending with Y = 1 × 1 × ⁶P₆ = 6 ! = 720.

(b) 4320

There are eight letters in the word “DAUGHTER” including three vowels (A, U, E) and 5 consonants (D, G, H, T, R) If the vowels are to be together, we consider them as one letter, so the 6 letters now (5 consonants and 1 vowels entity) can be arranged in ⁶P₆ = 6 ! ways. Also corresponding to each of these arrangements, the 3 vowels can be arranged amongst themselves in 3 ! ways. 

∴ Required number of words = 6 ! × 3 ! 

= 6 × 5 × 4 × 3 × 2 × 1 × 3 × 2 = 4320.

We have Number of red balls = 6 

Number of White balls = 6 

Number of blue balls = 6 

Since each selection consists of 3 balls of each = 11  colours, so 3 red balls can be selected out of 6 balls in ⁶C₃ ways, 3 white balls out of 5 can be selected in ⁵C₃ ways and 3 blue balls out of 5 can be selected in ⁵C₃ ways 

∴ Required number of ways = ⁶C₃ x ⁵C₃ x ⁵C₃ 

= 20 x 10 x 10 = 2000 

In a dictionary, the words are arranged in an alphabetical order.

 (i) Starting with A, the remaining 4 letters G, A, I, N can be arranged in 4! = 24 ways. These are the first 24 words. 

(ii) Then, starting with G, the remaining letters A, A, I, N can be arranged in \(\frac{4!}{2!}\) = 12 ways. Thus, there are 12 words starting will G. 

(iii) Now, the words will start with I. Starting with I, the remaining letters A, G, A, N can be arranged in \(\frac{4!}{2!}\) = 12 ways. So, there are 12 words, which start with I. 

(iv) Thus, so far, we have constructed 24 + 12 + 12, i.e., 48 words. The 49th word will start with N and is NAAGI. Hence, the 50th word is NAAIG.

In the given word EXAMINATION, there are 11 letters out of which, A, I, and N appear 2 times and all the other letters appear only once. 

The words that will be listed before the words starting with E in a dictionary will be the words that start with A only. 

Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, and then the remaining 10 letters taken all at a time are rearranged. Since there are 2 Is and 2 Ns in the remaining 10 letters, 

Number of words starting with A=10!/2!2!=907200

Thus, the required numbers of words is 907200.

The projection of vector(a on b) = vector(a.b)/|vector b|

Here, vector a = i + j + k and vector b = j

So, vector(a.b) = 1 and |vector b| = 1

So, projection of i + j + k along, j = 1/1 = 1 unit.

∵  ⁿC₈ = ⁿC₂,n = 8 + 2 = 10

∴ ⁿC₂ = ¹⁰C₂ = (10 x 9)/(2 x 1) = 45

A million is a 7-digit number. So any number greater than 1 million will contain all the seven digits. Since the digit 2 occurs twice and digit 3 occurs thrice and the rest are different, therefore, number of possible numbers which can be formed with the given seven digits = \(\frac{7!}{(2!)(3!)}=420.\)

These possible numbers include those which have 0 at the millions place. Keeping 0 fixed at the millions place, the remaining 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are different can be arranged in = \(\frac{6!}{(2!)(3!)}\) = 60. ways. 

∴ Number of numbers greater than 1 million made from the given digits = 420 – 60 = 360.

Since no two teachers are together and teachers must be seated in between students. 7 students can be permuted in 7! way, there are 8 places in which the 4 teachers can be seated in ⁸P₄ ways. 

∴ Total ways is 7! × ⁸P₄

Given Curve is y = A.e^x + B.e^-x ...(1)

Differentiating both sides w.r.t. 'x', we have 

dy/dx = Ae^x - Ae^-x ...(2)

Again differentiating, w.r.t. x

d²y/dx² = Ae^x + Be^-x

i.e., d²y/dx² = y 

So, (d²y/dx²) - y = 0 

which is require differential Equation.

Let A = (3, 4, 1) and B = (5, 1, 6)

Then direction ratio of line AB are 3 - 5, 4 - 1, 1 - 6,

= (-2, 3 ,5)

 So, Equation of any line AB will be

(x - 3)/-2 = (y - 4)/3 = (z - 1)/-5 = r (Let) 

Co-ordinate of any point P of above line is

P(-2r + 3, 3r + 4, -5r + 1) .

If p lies on xy plane, then -5r + 1 = 0

r = 1/5

∴ co-ordinate require is (13/5,23/5,0)

Answer is (a) x - α = y - β = z - γ

Answer is (a) 0

Given planes are :

- x + y + 2z = 9 ...(i)

and x + 2y + z = 5 ...(ii)

The direction cosine of the normal to the plane (1) are (-1/√6,1/√6,2/√6) and that of plane (2) are (1/√6,1/√6,2/√6). Let be the angle between given plane, then

cos = (-1/√6)(1/√6) + (1/√6)(2/√6) + (2/√6)(1/√6)

= (-1/6) + (2/6) + (2/6) = 1/2 

∴ θ = 60°

Let vector a = 2i - 3j + 4k and vector b = -4i + 6j - 8k

= 2(2i - 3j + 4k) = -2 vector a

Hence, vector(a and b) are co-linear

Answer is (d) 0

By distance formula, 

AB²  = (10 – 3)²  + (20 – 6)²  + (30 – 9)² 

= 49 + 196 + 441 = 686 

BC²  = (25 -10)²  + (- 41 – 20)²  + (5 – 30)² 

= 225 + 3721 + 625 = 4571 

CA²  = (3 – 25)²  + (6 + 41)²  + (9 – 5)² 

= 484 + 2209 + 16 = 2709 

We find that AB²  + CA²  ≠ BC²

(∵ 3395 ≠ 4571) 

∴ ABC is not a right angled triangle.

We have to show that the three points are colinear , i.e. they all lie on the same line, 

If we define a line which is having a parallel line to AB and the points A and B lie on it, if point C also satisfies the line then, the three points are colinear, 

Given A(2, 3, 4) and B(-1, -2, 1), AB = -3i – 5j -3k 

The points on the line AB with A on the line can be written as,

R = (2, 3, 4) +a(-3, -5, -3) 

Let C = (2-3a, 3-5a, 4-3a)

(5, 8, 7) = (2-3a, 3-5a, 4-3a)

If a = -1, then L.H.S = R.H.S, thus 

The point C lies on the line joining AB, 

Hence, the three points are colinear.

Answer is (a) a^x log a

Given differential equation may be written as

(dy/dx) + (1/x log x)y = 2/x² (which is linear diff. equation)

Here, P = 1/x log x and Q = 2/x²

Now, ∫Pdx = ∫(1/x log x) dx [put log x = z, (1/x) dx = dz]

= ∫(1/z) dz = log z = log(log x)

I..F = e^∫Pdx = e^{log(log x)} = log x 

Hence, solution is given by 

y x log x = ∫(2/x²).log x dx = 2∫x^-2.log x dx

= 2[(log x).(x^{-2 + 1})/(-2 + 1) - ∫(1/x)(-1/x)] dx

= 2[(-1/x)log x + ∫(1/x²) dx]

= 2[(-1/x)log x - (1/x)] + k = (-2/x)(log x) + k

Here, x = a((1 + t²)/(1 - t²)) = a(-1 + (2/(1 - t²)))

∴ dy/dt = a(0 + 2 x(-1/(1 - t²)²).(-2t)) = 4at/(1 - t²)²

and y = 2t/(1 - t²)

∴ dy/dx = ((1 - t)².2 - 2t.(-2t))//(1 - t²)² = 2(1 + t²)/(1 - t²)²

∴ dy/dx = (dy/dt)/(dx/dt) = (2(1 + t²)/(1 - t²)²) x ((1 - t²)²/4at)

= (1 + t²)/2at