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1933 and 1934 more

The word Maasai means my people.

The chiefs appointed by the colonial government often accumulated wealth over time. They had a regular income with which they could buy animals, goods and land. They lent money to poor neighbours who needed cash to pay taxes. Many of them began living in towns, and became involved in trade. Their wives and children stayed back in the villages to look after the animals. 

These chiefs managed to survive the devastations of war and drought. They had both pastoral and non-pastoral income, and could buy animals when their stock was depleted. 

(i) Drought affected the life of pastoralists everywhere. When rains fail and pastures are dry, cattle are likely to starve unless they can be moved to areas where forage is available. 

(ii) Since they could not shift their cattle to places where pastures were available, large numbers of Maasai cattle died of starvation and disease in the years of drought. 

(iii) The Maasai in Kenya possessed 720,000 cattle, 820,000 sheep and 171,000 donkeys. In just two years of severe drought, 1933 and 1934, over half the cattle in the Maasai Reserve died. 

1. Peaceful Co-existence
2. Not to attack each other

First conference of Non-Aligned movement was held at Belgrade in 1961 A.D.

Lakhs of people of Bengal and Punjab died due to division of India and lakhs had to leave their homes. Around 80 lakh people of eastern and western Punjab had to leave their lands, shops and other properties.

The three problems that the newly independent nation of India faced are as follows:

• Rehabilitation of a large number of refugees.
• Assimilation of princely states.
• Ensuring the unity of a country which is full of diversity.

Afro-Asian Conference of 1955 was held at Bandug in Indonesia. Indian Prime Minister Pt. Jawahar Lai Nehru, Chau-in-lai of China and Sukarno of Indonesia took part in it.

This commission had three members.

After the Second World War, the entire world was divided into two opposite alliances. America was the leader of one alliance which was known as Western Bloc. U.S.S.R. was the leader of second alliance which was known as Eastern Bloc. Serious Cold War started between them. Military treaties and pacts like Nato and Warsah Pact have made situation more tense. India wanted to keep its own sovereignty and didn’t become the member of any group. That’s why India started Non-Aligned Movement with the help of other countries. Founders of this movement were Pt. Jawahar Lai Nehru, Tito of Ugoslavia and Nasir of Egypt.

Non-Aligned Movement started in 1961 A.D. It was based on principles of Panchsheel. All the members of this movement didn’t want to include themselves in any of the alliance. Its first conference held at Belgrade in 1961 A.D. It was started with 25 members but now it has more than 100 members.

Peaceful co-operation.

India, after independence, adopted the foreign policy based on the concept of peaceful co-existence. Its main features are given below:

1. India respects the sovereignty and freedom of all the countries of the world.
2. India believed that people of all the religions, nations and races are equal.
3. India strongly oppose those countries which discriminate the people on the basis of colour, race or class. For example, India had opposed the racial policy of South African Government and its discriminational policy with Asian people and original inhabitants of Africa.
4. India believes that all international disputes should be resolved through peaceful methods.

The Planning Commission was to formulate policies which would guide the economic development. Productivity and employment opportunities were to be increased through proper implementation of those policies.

Pt. Jawahar Lai Nehru of India, President of Ugoslavia Tito and President of Egypt Nasir were the founders of Non-Aligned Movement.

Correct option is (d) All of these

• Subjects that were placed on the Union List were _________, _________ and _________.
  
  Answer: taxes, defence and foreign affairs
• Subjects on the Concurrent List were _________ and _________.
  
  Answer: forests and agriculture
• Economic planning by which both the state and the private sector played a role in development was called a _________ _________ model.
  
  Answer: Mixed economy
• The death of _________ sparked off such violent protests that the government was forced to give in to the demand for the linguistic state of Andhra.
  
  Answer: Potti Sriramulu

Some leaders believed that English should be done away with and Hindi should be promoted as the national language. But this idea was opposed by the leaders from non-Hindi areas. They did not want an imposition on Hindi on the people of those areas. Finally, it was decided that while Hindi would be the ‘official language’; English would be used for communication among various states.

Non-Aligned Movement started in 1961 A.D. It was based upon the principles of Panchsheel.

Lahore agreement took place between Indian Prime Minister Atal Behari Vajpai and Pakistani Prime Minister, Nawaz Sharif. Its objective was to resolve mutual disputes of India and Pakistan peacefully.

Removing poverty and building a modern technical and industrial base were important objectives for the new nation. The Planning Commission was set up in 1950 to plan and execute policies for economic development.

States were reorganized in November 1956 A.D. 14 States and 6 Union Territories were formed at that time.

Communalism, casteism, linguism, poverty, unemployment, illiteracy, population explosion, etc.

Answer: (a) True, (b) False, (c) False, (d) True

War between India and China started in 1962 A.D. due to border dispute.

People speaking different languages live over here in India. Problem in this relation is that some people consider their language as superior than the other languages.

1. Increasing population is the basic reason of poverty and unemployment.
2. Development plans of government either fail or they move on a slow pace.

Correct option is (d) Linguistic

India had to face many problems after getting independence. One of these problems was of local kingdoms. They were 562 jp number and were ruled by Indian kings or rulers. According to Act of 1947 A.D., these kingdoms were free to keep their own freedom or were free to be included in the countries of either India or Pakistan. That’s why many local kingdoms liked to remain free. But first Home Minister of free India Sardar Vallabh Bhai Patel handled the matter with great intelligence and asked the rulers of local kingdoms to become a part of Indian union. Small kingdoms were included into provinces.

Some kingdoms were sharing cultural values and even they were sharing common boundries. They were joined and made state. For example kingdoms of Kathiawar were included in Saurashtra and the kingdoms of Patiala, Nabha, Faridkot, Jind and Malerkotla were joined to make Pepsu state. Now only three kingdoms remained which were not ready to be included in India and these were Hyderabad, Junagarh and Kashmir.

Hyderabad: Nizam of the kingdom of Hyderabad Usman Ah Khan refused to merge his kingdom into union of India. So Indian police was sent to Hyderabad on 13th September 1948 A.D. and on 17th September 1948 A.D. kingdom of Hyderabad was merged into union of India.

Junagarh:. Nawab of Junagarh wanted to go with Pakistan. But plebiscite (Public Survey) took place in Junagarh on 20th February 1948 A.D. and public wished to be merged in Union of India. Therefore Junagarh was merged into union of India.

Kashmir: Ruler of Kashmir Maharaja Hari Singh also wanted to remain free. But Pakistan wanted to control kingdom of kashmir. So ruler of Kashmir called for Indian help and proposed to merge his state into Indian Union. Indian government accepted the request of ruler of Kashmir and send its army to Kashmir. War started between India and Pakistan but a large part of Kashmir was occupied by Pakistan.

Other Kingdoms: Except these kingdoms, there were certain other small kingdoms which were included in the nearby states. Baroda was included in the province of Bombay. Unified state was founded by joining few small states. For example one union was made in March, 1948 by joining Bharatpur, Dholpur, Alwar and Karavli. After this Rajasthan union was formed in which kingdoms of Boondi, Talwara, Pratapgarh, Shahpur, Banswara, Kota, Kishangarh, etc. were included in it.

Sardar Vallabh Bhai Patel was instrumental in the unification of Princely States.

Correct option is (c) 8 million

Correct option is (a) 345

Correct option is (d) 565

1. In the absence of atmosphere, there will not be any scattering of light and so light will reach our eye, i.e. the sky will appear black instead of blue at night in the absence of light. 

2. On the moon, since there is no atmosphere, there is no scattering of sunlight reaching the moon surface. Hence to an observer on the surface of moon, no light reaches except the light directly from sun. Thus sky will have no colour and will appear black to an observer on the moon surface. This is applicable for any planet which does not have atmosphere. 

3. When an astronaut goes above the atmosphere of the earth in rocket he sees the sky black.

1. The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of size bigger than the wavelength of visible light. 

2. Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light does not reach our eye, the clouds seem black.

Whether we toss a coin 6 times or toss 6 coins at a time, the number of arrangements will be the same. 

∴ The number of arrangements of 4 heads and 2 tails out of 6 is \(\frac{6!}{4!2!}=15.\)

1 + \(\frac{3}{2}+\frac{5}{4}+\frac{7}{8}\) + ..... is an arithmetico-geometric series with corresponding A.P. and G.P as: 

A.P. : 1 + 3 + 5 + 7 + ......

G.P. : 1 + \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\) + ...........

nth term of A.P. = (1 + (n – 1) 2) = (2n – 1)

nth term of G.P. = 1. \(\big(\frac{1}{2}\big)^{n-1}\) = \(\frac{1}{2^{n-1}}\)

∴ nth term of the given A.G.P. = \(\frac{2n-1}{2^{n-1}}\)

∴ S_n = 1 + \(\frac{3}{2}+\frac{5}{4}+\frac{7}{8}\) + ........ + \(\frac{2n-1}{2^{n-1}}\)

⇒ \(\frac{1}{2}\)S_n = \(\frac{1}{2}+\frac{3}{4}+\frac{5}{8}\) + ......+........+ \(\frac{2n-3}{2^{n-1}}\) + \(\frac{2n-1}{2^n}\)

On subtraction, we get

\(\big(1-\frac{1}{2}\big)\)S_n = 1 + \(\frac{2}{2}+\frac{2}{4}+\frac{2}{8}\) + ........... + \(\frac{2}{2^{n-1}}\) - \(\frac{2n-1}{2^n}\)

⇒ \(\frac{1}{2}\)S_n = 1 + 2 \(\bigg\{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)+ ........+ \(\frac{1}{2^{n-1}}\bigg\}\) - \(\frac{2n-1}{2^n}\)

⇒ \(\frac{1}{2}\)S_n = 1 + 2 \(\bigg\{\frac{\big(\frac{1}{2}\big(1-\big(\frac{1}{2}\big)^n\big)}{1-\frac{1}{2}}\bigg\}\) - \(\frac{2n-1}{2^n}\) = 1 + 2 = \(\frac{4}{2^n}\) - \(\frac{2n-1}{2^n}\) = 3 - \(\frac{2n-3}{2^{n}}\)

∴ S_n = 6 - \(\frac{2n-3}{2^{n-1}}\).

The number of signals = \(\frac{9!}{2!2!5!}\) = \(\frac{9\times8\times7\times6\times5\times4\times3\times2\times1}{2\times2\times5\times4\times3\times2\times1}\) = 756.

The word GARDEN contains 6 letters — 4 consonants (G, R, D, N) and 2 vowels A, E. The 4 consonants can be arranged in 6 places in ⁶P₄ ways. In each of these arrangements two places will remain blank in which the first place will be filled by A and the place following it by E as vowels have to be in alphabetical order. This can always be done in only 1 way, i.e., E following A 

∴ Required number of ways = ⁶P₄ x 1 = \(\frac{6!}{(6-4)!}=\frac{6!}{2!} \times1\) = 6 × 5 × 4 × 3 = 360.

Let S_∞ = 1 + \(\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+.....\infty\)                     .....(i)

\(\frac{1}{5}\)S_∞  = \(\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+.....\infty\)              .....(ii)

Subtracting eqn (ii) from eqn (i), we get

\(\big(1-\frac{1}{5}\big)S_\infty\) = 1 + \(\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+.....\infty\)

\(\frac{4}{5}\)S_∞ = 1 + 3\(\big(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.....\infty\big)\)

= 1 + 3 \(\bigg(\frac{\frac{1}{5}}{1-\frac{1}{5}}\bigg)\)            (∵ Sum of infinite G.P. = \(\frac{a}{1-r}\)) 

= 1 + \(\frac{\frac{3}{5}}{\frac{4}{5}}\) = 1 + \(\frac{3}{4}\) = \(\frac{7}{4}\)

S_∞ = \(\frac{7}{4}\) x \(\frac{5}{4}\) = \(\frac{35}{16}.\)

3 + 5r + 7r² + ...... ∞ is an infinite arithmetico-geometric series, where

a = 3, d = 2, common ratio (r) = r.

Sum to infinity of an A.G.P., with first term of A.P, as a, common difference d and common ratio r is

S_∞ = \(\frac{a}{1-r}\) + \(\frac{dr}{(1-r)^2}\)

∴ \(\frac{44}{9}\) = \(\frac{3}{1-r}\) + \(\frac{2r}{(1-r)^2}\) ⇒ \(\frac{44}{9}\) = \(\frac{3(1-r)+2r}{(1-r)^2}\)

⇒ 44 (1 – r)² = 9 (3 – r) ⇒ 44 (1 – 2r + r²) = 27 – 9r 

⇒ 44 – 88r + 44r² = 27 – 9r ⇒ 44r² – 79r + 17 = 0 

⇒ (4r – 1) (11r – 17) = 0 ⇒ r = \(\frac{1}{4}\) or \(\frac{17}{11}\)

r ≠ \(\frac{17}{11}\) as it is not possible to find the sum of an infinite G.P. with | r | > 1. So r = \(\frac{1}{4}\).

The radiations which are associated with electrical and magnetic fields are called electromagnetic radiations. When an electrically charged particle moves under acceleration, alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in the form of waves. These waves are called electromagnetic waves or electromagnetic radiations.

Properties of electromagnetic radiations: 

Oscillating electric and magnetic field are produced by oscillating charged particles. These fields are perpendicular to each other and both are perpendicular to the direction of propagation of the wave.

They do not need a medium to travel. That means they can even travel in vacuum. 

Let \(x\) = \(2^{\frac{1}{4}}\) x  \(4^{\frac{1}{8}}\) x \(8^{\frac{1}{16}}\) x \(16^{\frac{1}{32}}\)......∞

∴ log \(x\) = \(\frac{1}{4}\) log 2 + \(\frac{1}{8}\) log 4 + \(\frac{1}{16}\) log 8 + \(\frac{1}{32}\) log 16 + ..........∞

=   \(\frac{1}{4}\) log 2 + \(\frac{1}{8}\) log 2² + \(\frac{1}{16}\) log 2³ + \(\frac{1}{32}\) log 2⁴ + ..........∞

=   \(\frac{1}{4}\) log 2 + \(\frac{2}{8}\) log 2 + \(\frac{3}{16}\) log 2 + \(\frac{4}{32}\) log 2 + ..........∞

= \(\big(\)\(\frac{1}{4}\) + \(\frac{2}{8}\) + \(\frac{3}{16}\) + \(\frac{4}{32}\) + .........∞\(\big)\) log 2               .....(i)

Now, \(\frac{1}{4}\) + \(\frac{2}{8}\) + \(\frac{3}{16}\) + \(\frac{4}{32}\) + ....∞ is an Arithmetico-Geometric series.

Let  S =   \(\frac{1}{4}\) + \(\frac{2}{8}\) + \(\frac{3}{16}\) + \(\frac{4}{32}\) + .......∞             .....(ii)

\(\frac{1}{2}\)S = \(\frac{1}{8}\) +\(\frac{2}{16}\) + \(\frac{3}{32}\) + .......∞                ........(iii)

On subtracting eqn (iii) from eqn (ii), we get

 \(\frac{1}{2}\)S = \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\) + \(\frac{1}{32}\) + ........∞ = \(\frac{\frac{1}{4}}{1-\frac{1}{2}}\) = \(\frac{\frac{1}{4}}{\frac{1}{2}}\) = \(\frac{1}{2}\) ⇒ S = 1

∴ log \(x\) = 1 × log 2                     (From (i)) 

⇒ log \(x\) = log 2 ⇒ \(x\) = 2.

Correct option (c) –2500 

Explanation:

(x–1)(x–3)(x–5) ................ (x–99)

= x⁵⁰ – S₁ x⁴⁹ + S₂ x ⁴⁸ .............

Coefficient of x⁴⁹ is – S₁

= –(1+3+5+......+99) 

= –50² = – 2500 

(c) 99.2¹⁰⁰ +1 

1 + 2.2 + 3.2² + 4.2³ + ..... + 100.2⁹⁹ is 

clearly an AGP with A.P. : 1 + 2 + 3 + ..... 100 

and  G.P. : 1 + 2 + 2² + ..... + 2⁹⁹. 

Let S = 1 + 2.2 + 3.2² + 4.2³ + ..... + 100.2⁹⁹ 

∴ 2S = 2 + 2.2² + 3.2³ + ..... + 99.2⁹⁹ + 100.2¹⁰⁰ 

∴ S – 2S = (1 + 2 + 2² + 2³ + ..... 2⁹⁹) – 100.2¹⁰⁰

⇒ -S = \(\frac{(2^{100}-1)}{2-1}\) - 100.2¹⁰⁰

(∵ S_n = \(\frac{a(r^n-1)}{r-1}\) and this is a G.P. with 100 terms, a = 1, r = 2)

= 2¹⁰⁰ – 1 – 100.2¹⁰⁰ 

S = 2¹⁰⁰ . 100 – 2¹⁰⁰ + 1 = 2¹⁰⁰ . (100 – 1) + 1 

= 99.2¹⁰⁰ + 1.

The value of a house is Rs. 15 Lac

Appreciation rate = 5% = \(\frac5{100}\) = 0.05

Value of house after 1st year = 15(1 + 0.05) = 15(1.05)

Value of house after 6 years = 15(1.05) (1.05)⁵

= 15(1.05)⁶

= 15(1.34)

= 20.1 lac. 

Amount invested = Rs. 10000

Interest rate = \(\frac8{100}=0.08\) 

amount after 1st year = 10000(1 + 0.08)

= 10000(1.08)

Value of the amount after n years

= 10000(1.08) × (1.08)^n-1

= 10000(1.08)ⁿ

= 20000

∴ (1.08)ⁿ = 2

∴ (1.08)⁵= 1.47 …..[Given]

∴ n = 10 years, (approximately)

1/2, 1/4, 1/8, 1/16,....

Here, a = 1/2, r = \(\frac{1/4}{1/2}=1/2, |r|<1\)

\(\therefore\) Sum to infinity exists.

\(\therefore\) Sum to infinity = \(\frac{a}{1-r}\)

 = \(\frac{1/2}{1-1/2}=1\)

(b) 2

S = \(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+......\) upto ∞

\(\frac{1}{3}S\) = \(\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+......\)upto ∞

⇒ S - \(\frac{1}{3}S\) = \(1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+......\)upto ∞

= \(\frac{4}{3}\) + 4 \(\bigg[\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.......\infty\bigg]\)

= \(\frac{4}{3}\) + 4 \(\bigg[\frac{\frac{1}{3^3}}{1-\frac{1}{3}}\bigg]\)\(\big(\because\,S_\infty=\frac{a}{1-r}\big)\)

= \(\frac{4}{3}\) + 4 \(\bigg[\frac{\frac{1}{9}}{\frac{2}{3}}\bigg]\) = \(\frac{4}{3}\) + \(\frac{4}{6}\) = \(\frac{12}{6}\) = 2.