Explore topic-wise InterviewSolutions in Current Affairs.

Organisms show moments and locomotion.

Land tax is recovered by the king in exchange for the use of agricultural land. The land tax was the main source of state income. The people of the society used to take the land in their possession and produce crops on it, in exchange for this, the people the had to pay 1/5^th of the agricultural produce as the tax to the king. The tax was paid in cash or in the form of farm produce.

When a person takes more essential things from another person in return for his less essential thing, then it is called barter exchange.

Kautilya was of the opinion that the king should not levy taxes arbitrarily, but the rate of tax should be fair and equitable. It is not necessary to levy heavy taxes on infertile land.

Marcuse advocates collective control to be essential in community life as a principle of positive liberty and hence supports state interference in the individual’s life.

‘Swa’ (self) and ‘Raj’ (rule) hence Swaraj implies ‘self-rule’ or self-governance.

The two rules of taxation are 

1. Timely tax collection 

2. Taxation in accordance with capability

Correct Answer is: (c) 3

(d) All of these

(b) 1/6^th of Production 

Correct Answer is: (b) 10%

The definition of economics by Kautilya is as follows- “Man’s behaviour or livelihood is called Artha. The name of the land possessed by man is Artha. The science that prescribes the measures to develop or nurture such land, is called economics.” Kautilya explains that this economics uses wealth in Dharma, Artha and Kaam and protects it.

Four key goals of state given by Kautilya are : 

1. Acquiring the unobtained 

2. Protecting the acquired 

3. Ensuring growth of the protected and 

4. Utilising the accumulated for welfare of the subjects, or spending the earnings from taxes for welfare of citizens.

1/6^th part of the production of the farm had to be given to the king.

The definition of economics by Kautilya is as follows- “Man’s behaviour or livelihood is called Artha. The name of the land with man is Artha. The science that prescribes the measures to develop or nurture such land, is called economics.” Kautilya explains that this economics uses Artha in religion, economy and sensual pleasure and protects it.

According to Kautilya, a penalty of 12 times the loss should be imposed on the officials who harm the treasury by embezzlement.

1. Land Tax, House Tax and Incidental Tax 

2. Sales Tax, Import- Export Tax

The act was passed as a result of people’s mobilization as well as in recognition of the provisions in the Constitution that provide Rights to citizens.

Kautilya says embezzlement is the irregularity done in management of the state treasury. According to him, not entering the income received in the register, not spending the regular tax entered in the register and refusing to acknowledge the savings, are the three types of embezzlement. Through embezzlement, the person who has been harming the treasury, adhyakshya has right to recover 12 times of the embezzled amount from them.

The following sources of state income are described by Kautilya :

1. Different types of land tax, house tax in town, incidental tax, etc. 

2. Tax on items sold in market, Import-Export Tax, 

3. Road Tax, Canal Tax and Tax on heavy loading vehicles, 

4. Artist Tax, Fisheries Tax,

5. Excise and Taxes on intoxicants, 

6. Property Tax, Forest Produce Tax, Mining Tax, Monopolistic Tax on Salt, etc. 

7. Labour Tax 

8. Contingent Income Tax

9. Interest on Loans 

10.Charitable Tax 

11.Penalties 

12.State Profits 

13.Taxes on horses, wool, elephants, fruits and trees.

The period (about 270 – 280 days) a baby is born is called the gestation period.

The correct answer is B) i only correct.

Correct option is (d) All of the above

Wrong Pair: 

Kautilya – Samajshashtra

Diameter of the roller = 1.4 m 

Radius (r) = 1.4/2 = 0.7 m 

and length (h) = 2m

Curved surface area = 2πrh = 2 x 22/7 x 0.7 x 2 cm² = 8.8 m² 

Area covered in 50 complete revolutions = 8.8 x 50 m² = 440 m²  revolutions = 8.8 x 50 m² = 440 m² 

Area of the playground = 440 m² 

x = 180 - 75

= 105°       [Co-interior angles]

y = 180 - x

= 180 - 105

= 75°     [Co-interior angles]

z = 75°   [Corresponding angles]

Correct option is (A) 3

(b) 16

Let the required four parts be (a – 3d), (a – d), (a + d), (a + 3d) 

Given, a – 3d + a – d + a + d + a + 3d = 20 

⇒ 4a = 20 ⇒ a = 5

Also, \(\frac{(a-3d)(a+3d)}{(a-d)(a+d)}\) = \(\frac{2}{3}\)

⇒ 3(a² – 9d²) = 2(a² – d²) 

⇒ 3(25 – 9d²) = 2(25 – d²) 

⇒ 75 – 27d² = 50 – 2d² ⇒ 25 = 25d² 

⇒ d = ± 1 

Taking d = 1, the numbers are 5 – 3, 5 – 1, 5 + 1, 5 + 3, i.e., 2, 4, 6, 8 

Taking d = –1, the numbers are 5 + 3, 5 + 1, 5 – 1, 5 – 3, i.e., 8, 6, 4, 2 

∴ Required product in either case = 2 × 8 = 16.

(c) - (p + q)

Since S_n = \(\frac{n}{2}\)[2a + (n - 1) d] for an A.P. whose first term 

= a, common difference = d, number of terms = n.

∴ S_p = q = \(\frac{p}{2}\) (2a + (p – 1)d)

⇒ 2q = 2ap + p (p – 1)d                   ...(i)

S_q = p = \(\frac{q}{2}\) (2a + (q – 1) d) 

⇒ 2p = 2aq + q(q – 1)d                   ...(ii) 

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

2(q – p) = 2a (p – q) + (p² – q²)d – (p – q) d 

⇒ – 2(p – q) = 2a (p – q) + (p – q) (p + q) d – (p – q) d 

– 2 = 2a + [(p + q) – 1]d                     ...(iii)

Now, S_{p + q} = \(\frac{p+q}{2}\) [2a + (p + q - 1) d ]

= \(\frac{p+q}{2}\) x - 2            (From (iii))

= - (p + q).

Correct option is (C) 4

Let three terms be \(\frac{a}r, a,ar\) 

According to the given conditions,

\(\frac{a}r+a+ar=21/4\).......(i)

and \(\frac{a}r\) x a x ar = 1,

i.e., a³ = 1

∴ a = 1

∴ from equation (i), we get

\(\frac1r+1+r=21/4\) 

By solving this, we get r = 4.

Answer : (b) 2

Sum of first 10 terms = \(\frac{1}{2}\) (Sum of next 10 terms)

⇒ S₁₀ = \(\frac{1}{2}\) (S₂₀ – S₁₀) ⇒ 3S₁₀ = S₂₀ 

⇒ 3 × \(\frac{1}{2}\)[2a + 9d] = \(\frac{1}{2}\)[2a + 19d]  \(\big[\because S_n = \frac{n}{2}[2a+(n-1)d]\big]\) 

⇒ 3(2a + 9d) = 2(2a + 19d) 

⇒ 6a + 27d = 4a + 38d 

⇒ 2a = 11d 

⇒ 2(13 + d) = 11d (∵ Second term = 13 and 13 – a = d) 

⇒ 26 – 2d = 11d 

⇒ 13d = 26 

⇒ d = 2.

(d) \(\frac{3}{5}\)

Let a be the first term and r the common ratio of the G.P. 

Then, \(\frac{S_3}{S_6}\) = \(\frac{125}{152}\) ⇒ \(\frac{\frac{a(r^3-1)}{(r-1)}}{\frac{a(r^6-1)}{(r-1)}}\) = \(\frac{125}{152}\)

\(\frac{(r^3-1)}{(r^6-1)}\) = \(\frac{125}{152}\) ⇒ \(\frac{(r^3-1)}{(r^3+1)(r^3-1)}\) = \(\frac{125}{152}\)

⇒ r³ + 1 = \(\frac{125}{152}\) ⇒ r³ = \(\frac{152}{125}\) - 1 = \(\frac{27}{125}\)

⇒ r³ = \(\big(\frac{3}{5}\big)^3\) ⇒ r = \(\frac{3}{5}.\)

Given, r = -3, t₆ = 1701

t_n = ar^n-1

\(\therefore\) t₆ = a(-3)^6-1

\(\therefore\) 1701 = a(-3)⁵

\(\therefore\) a = \(\frac{1701}{-243}\)

\(\therefore\) a = -7

Here, t₁ = a = 5, r = \(\frac{t_2}{t_1}=\frac{25}5=5,t_n = 5^{10}\)

t_n = ar^n-1

\(\therefore\) 5¹⁰ = 5 x 5(n - 1)

\(\therefore\) 5¹⁰ = 5 x 5^(n-1)

\(\therefore\) 5¹⁰ = 5^{(1 + n - 1)}

\(\therefore\) 5¹⁰ = 5ⁿ 

\(\therefore\) n = 10

\(\therefore\) 5¹⁰ = 5ⁿ

\(\therefore\) n = 10

\(\therefore\) 5¹⁰ is the 10^th term of the G.P.

Alternate Method:

t₁ = 5, t₂ = 25 = 5², t₃ = 125 = 5³, t₄ = 625 = 5⁴

\(\therefore\) t₁₀ = 5¹⁰

Given, a = 7/243, r = 3

t_n = ar^n-1

\(\therefore\) t₆ = \(\frac7{243}\times(3)^{6-1}\) 

\(=\frac7{243}\times3^5\)

= 7

Let a be the first term of the GP

Given,

r = 2 and

a₃ + a₅ = 90

⟹ ar² + ar⁴ = 90

⟹ a(2)² + a(2)⁴ = 90

⟹ 4a + 16a = 90

⟹ 20a = 90

⟹ ^{\(a \frac{9}{2}\)}

Given,

G.P. is 2,1, \(\frac{1}{2}\) , \(\frac{1}{4}\) …

First term a = 2

and common ratio = \(\frac{1}{2}\)

Let n^th term of G.P. be \(\frac{1}{1024}\)

∴ T_n = arⁿ⁻¹

⇒ \(\frac{1}{1024}\) = 2(\(\frac{1}{2}\))ⁿ⁻¹

⇒ \(\frac{1}{2048}\) = (\(\frac{1}{2}\))ⁿ⁻¹

⇒ (\(\frac{1}{2}\))¹¹ = (\(\frac{1}{2}\))ⁿ⁻¹

∴ n − 1 = 11

Or n = 12

i. Fourth root of 81. 

ii. Square root of 49. 

iii. Fifth root of 15. 

iv. Ninth root of 512. 

v. Nineteenth root of 100. 

vi. Seventh root of 6.

Complete table:

 Using laws of indices, write proper numbers in the following boxes.

i. 3⁵ × 3² = 3⁽⁷⁾

^ii. 3⁷ × 3⁹ = 3^(-2)

iii. (3⁴)⁵ = 3⁽²⁰⁾

iv. 5^-3 = 1/5⁽³⁾

v. 5⁰ = (1)

vi. 5¹ = (5)

vii. (5 × 7)² = 5⁽²⁾ × 7⁽²⁾

viii. \((\frac{5}{7})^3\) = \(\frac{(5)^3}{(7)^3}\)

ix. \((\frac{5}{7})^{-3}\) = \(\left(\frac{(7)}{(5)}\right)^3\)

Consider, 6³ = 6 × 6 × 6 = 216 and (-4)³ = (- 4) × (- 4) × (- 4) = – 64

Thus, cube of a positive number is positive and cube of a negative number is negative. 

∴ Sign of a number = sign of its cube.

i. (121)^5/2

ii. (324)^3/4

iii. (264)^2/5

iv. (3)^3/3

It is a group which itself does not act as chromophore but when attached to the chromophore, it shifts the adsorption towards longer wavelength along with an increase in the intensity of adsorption.
Examples:
-OH, -NH₂ , -NHR, -NR₂ are auxochromes.

Soft soaps: Soft soaps are potassium salts of fatty acids.

Hard soaps: Hard soaps are sodium salts of fatty acids.

Answer is (A) Farmers

Ashoka is a unique leader.

The term ‘gana’ means people of equal status. Sangha means ‘assembly’. Gana Sangha means an assembly of equal status people. They cover a small area that was ruled by a superior group among them. These gana sanghas practiced “all are equal” traditions. A kingdom means a territory and was ruled by a king or queen. In a kingdom, a family which rules for a long, period becomes a dynasty.

(B) Chandragupta Maurya

(B) Blacksmith

1. B 

2. F 

3. G 

4. E 

5. C