The number of nearest neighbours of a particle in solids is called Co-ordination number. 

C.No.of FCC structure–12

(i) \(\frac{dx}{dt}\) = K[A][B]²

(ii) \(\frac{dx}{dt}\) = K[A][2B]² 4 times

Faraday’s law states that “The amount of chemical reaction which occurs at any electrode during electrolysis by a current is directly proportional to the quanting of electricity passed through the electrolyte. 

To deposit one mole of zinc

2 × 96500 = 1,93,000 C of electricity is required.

In aniline, lone pair of electrons of nitrogen in not free for donation because it is involved in resonance but in cyclohexamine, lone pair of electrons on nitrogen is free for donation. So cyclohexamine is more basic than Aniline.

Gay Lussac’s law of gaseous volumes.

Roasting:

Heating of metal ore in presence of air is called roasting. In this step metal ore is converted to oxide.

2ZnS + 3O₂ → 2ZnO + 2SO₂

Calcination:

Heating of metal ore in absence of air is called calcination. In this step volatile impurities escape and some of the ore decomposes.

CaCO₃ → CaO + CO₂ ↑

A firm is a commercial enterprise, a company that buys and sells products and/or services to consumers with the aim of making a profit.

Vajapeya Yajna: Vajapeya is the highest form of soma Yajna, involving the pressing of “Soma” juice and performed in autumn season. It is called an 'AtiYajna' as it is ranked as great as Rajasuya or Aswamedha Yajna, which demanded immense man power and resources.

Kautilya

The Saptanga theory of the state was given by Kautilya.

Pharmacodynamics (PD) is the study of the biochemical and physiologic effects of drugs (especially pharmaceutical drugs).

Correct option is (d) Assertion (A) is false, but Reason (R) is True

When the new partner is not in a position to bring his share of goodwill in cash, then goodwill account is adjusted through the old Partners' Capital Account.

Correct option is D) changing

From statement I

If the content of the two containers containing mixture of cashews and raisins in the ratio a : b and c : d respectively is mixed then the ratio of cashews and raisins in the final mixture = [a/(a + b) + c/(c + d)] : [b/(a + b) + d/(c + d)]

a = 1, b = 3, c = 2 and d = 5

⇒ {[1/(1 + 3)] + [2/(2 + 5)]}: {[3/(1 + 3)] + [5/(2 + 5)]}

⇒ {[1/4 + 2/7]} : {[3/4] + [5/7]}

⇒ [15/28] : [41/28]

⇒ 15 : 41

From statement II

Only the weight of cashews is given, so we can’t get the answer from this information only.

∴ The data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question

From statements I and II

Let the total distance be x km

One-third of the distance = x/3 km

Remaining distance = 2x/3 km

Time taken to cover x/3 km = (x/3) ÷ 20 = x/60 hr

Time taken to cover (2x/3) km = (2x/3) ÷ 10 = 2x/30

According to question,

⇒ [(2x/30) + (x/60)] = 10

⇒ (4x + x)/60 = 10

⇒ 5x/60 = 10

⇒ x = 10 × 60/5

⇒ x = 120 km

∴ The data in both the statement I and II together are necessary to answer.

From statements I and II

Let the present ages of Salim and Arman be 6x and 8x years respectively.

⇒ (6x) × (8x) = 192

⇒ 48x² = 192

⇒ 4 = x²

⇒ x = 2

Present age of Salim = 6 × 2 = 12 years and after 4 years will be 16 years.

Present age of Arman = 8 × 2 = 16 years and after 4 years will be 20 years.

⇒ Ratio of their ages after 4 years = 16 : 20

⇒ 4 : 5

∴ The data in both the statement I and II together are necessary to answer.

From statements I and II

According to the question,

⇒ 2 pencils + 1 pen = Rs. 150

⇒ 2 × 40 + 1 pen = Rs. 150

⇒ 80 + 1 pen = Rs. 150

⇒ Cost of 1 pen = Rs. (150 – 80) = Rs. 70

From statement I

Let the ‘E’ is an event of probability

Number of favorable outcomes: ⁴C₂ = 4!/(2! × 2!) = 6

Total number of outcomes = ⁹C₂ = 9!/(2! × 7!) = 36

⇒ P(E) = (6/36) = 1/6

From statement II

The number of pencils of each colour is not given. So, we can’t get the answer.

∴ The data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question.

From statements I and II

Let the total work be 1

Work done by Rahul and Amit = 1/4

⇒ Work done by Anil = 1 – (1/4) = 3/4

Share of Anil will be proportional to work done by him

⇒ Share of Anil = (3/4) × Rs. 24,000 = Rs. 18,000

∴ The data in both the statement I and II together are necessary to answer.

Formula Used:

For a cylinder,

Curved Surface Area = 2πrh

Volume = πr²h

where, r → Radius of cylinder, h → Height of cylinder

Statement I.

The sum of the height and radius of the cylinder is 10 cm.

⇒ r + h = 42

Value of the r and h are not given

So, statement I alone is not sufficient

Statement II.

The curved surface area of cylinder is 370π cm2

⇒  2πrh = 370π 

⇒ rh = 370π/2π = 185

Value of the r and h are not given

So, statement I alone is not sufficient

Combining Statement I and Statement II 

r + h = 42      ----(1)

rh = 185

⇒ r = 185/h

Putting r in (1), we get

r + 185/r = 10

⇒ r² – 42r + 185 = 0

⇒ r2 – 37r – 5r + 185 = 0

⇒ (r – 37) (r – 5) = 0

⇒ r = 5, 37

We need maximum value of r

⇒ r = 37 cm

rh = 185

⇒ h = 185/37 = 5 cm

⇒ h/r = 5/37

∴ Both statements together are needed to answer the question.

Calculation:

Statement I.

15 years ago Rohit was 24 years younger than his father.

Rohit is 24 years younger than his father.

Let Rohit’s present age be x

Age of Rohit’s father = x + 24

We get a relation between their ages but not the exact value

So, Statement I alone is not sufficient

Statement II.

Let Rohit’s age be 2x

Age of Rohit’s father = 5x

Again we get a relation between their ages but not the exact values

So, Statement II alone is not sufficient

Combining Statement I and Statement II

2x + 24 = 5x

So, solving it we will get the exact values of their present ages and sum of their ages after 10 years could be calculated

∴ Both statements together are needed to answer the question

Formula Used:

Profit = Investment × Time

From statement I 

Aniket's investment = Rs. 5050

Bhanu's investment = (120/100) × 5050

⇒ Bhanu's investment = Rs. 6060

Profit = Investment × Time

⇒ Aniket's profit = Rs. 5050 × 8

Bhanu's profit = Rs. 6060 × 10

Ratio of their profit shares = 5050 × 8 : 6060 × 10

⇒ Ratio of their profit shares = 2 : 3

Total profit = Rs. 5050/5 = Rs. 1010

Profit share of Aniket = Rs. (2/5) × 1010

⇒ Profit share of Aniket = Rs. 404

From statement II

The ratio of their total investment time is 1 : 4.

We cannot find profit of Aniket form here.

So, statement II is not sufficient to answer the question.

∴ The statement I alone is sufficient to answer the question but, statement II alone is not sufficient.

Concept Used:

Total work is LCM of work done by all the persons.

Formula Used:

Total Work = Efficiency × Time

Calculations: 

Statement I.

X, Y and Z together can do a piece of work in 19 days.

X and Y together can complete the work in 28.5 days

Y and Z together can complete the work in 38 days.

Total work = LCM of (19, 28.5, 38) = 114

Total Work = Efficiency × Time

114 = (X + Y + Z) × 19

⇒ (X + Y + Z) = 114/19 = 6      ----(1)

Similarly, 114 = (X + Y) × 28.5

⇒ (X + Y) = 114/28.5 = 4      ----(2)

Similarly, 114 = (Y + Z) × 38

⇒ (Y + Z) = 114/38 = 3      ----(3)

Adding (2) and (3) and subtracting (1) from it, we get

(X + 2Y + Z) – (X + Y + Z) = (4 + 3) – 6

⇒ Y = 1

Putting in Y (3), we get

Z = 3 – 1 = 2

Now, Total Work = Efficiency × Time

⇒ 114 = Z × Time

⇒ Time = 114/2 = 57 days

So, statement I alone is sufficient to answer.

Statement II.

Y takes 29 days to complete the work alone.

We cannot find efficiency of Z from this statement alone.

So, statement II alone is not sufficient to answer the question.

∴ The statement I alone is sufficient to answer the question but, statement II alone is not sufficient.

Given:

Time taken by A to complete a work = 30 hours

Amount earned through the work = Rs 60,000

The penalty on delaying work by each hour = Rs 1,000

Formula used:

Efficiency = Total work/Total time

Calculations:

Let the total work be 90 units.

⇒ Efficiency of A = 90/30

⇒ 3 units/hour.

According to question,

A completed one-third work at usual pace (3 units/hour),

⇒ Time to complete one-third work = 30/3

⇒ 10 hours

A completed another one-third work at two-third of the usual pace (2 units/hour),

⇒ Time to complete another one-third work = 30/2

⇒ 15 hours

A completed remaining work at one-third of the usual pace (1 units/hour),

⇒ Time to complete another one-third work = 30/1

⇒ 30 hours

Total time taken by A to complete whole work = (10 + 15 + 30)

⇒ 55 hours

As there is delay of 25 hours,

⇒ Penalty = Rs 25,000

Amount earned after penalty = (60,000 - 25,000)

⇒ Rs 35,000

∴ The total time taken by A to complete the work and wages earned after penalty are 55 hours and Rs 35,000 respectively.

Formula Used:

SP = MRP × (1 – D%/100)

Profit = SP – CP

Profit% on SP = (P/SP) × 100

where, MRP → Marked price, SP → Selling price, CP → Cost price, D% → Discount percent,

P% → Profit percent

Calculations:

Statement I.

Relation between MRP and CP is given, but the CP is not given

⇒ Statement I alone is not sufficient 

Statement II.

CP = Rs. 250

MRP = Rs. (115/100) × 250

⇒ MRP = Rs. 287.5

SP = MRP × (1 – D%/100)

⇒ SP = 287.5 × (1 – 4/100)

⇒ SP = Rs. 276

Profit = Rs. (276 – 250)

⇒ Profit = Rs. 26

Now, Profit% on SP = (P/SP) × 100

⇒ Profit% on SP = (26/276) × 100

⇒ Profit% on SP = 9.4%

∴ The statement I and statement II both are needed to answer the question.

Given:

Days taken by  A and  B together to do total work = 25 days

Days taken by B to do \(66\frac{2}{3}\)% of total work = 30 days

Formula used:

No of days taken = (Total work)/(Efficiency)

Calculation:

Days taken by B to do (\(66\frac{2}{3}\)% = 2/3) ​of total work = 30

⇒ Days taken by B to do total work = 30 × 3/2 = 45 days

Total work = LCM of (30, 45) = 225 

The efficiency of A and B together = 225/25 = 9

The efficiency of  B  = 225/45 = 5

The efficiency of  A = 9 – 5 = 4

Days taken by A to do (4/15) part of total work = (225/4) × (4/15) = 15 days

∴ Time taken by A to do (4/15) part of the total work is 15 days

Given:

28 men can do the work in 15 days

Formula used:

(M1 × D1 × T1)/W1 = (M2 × D2 × T2)/W2

Calculation:

We know that,

Work is the same W1 =W2

Let M1 be 28

Let D1 be 15

Let M2 be x

Let D2 be 42

28 × 15 = x × 42

⇒ x = (28 × 15)/42

⇒ x = 10

∴ Men required to complete the work in 42 days is 10.

Given:

2 men and 3 women can do a piece of work in 4 days

6 men and 1 woman can do the same work in 2 days

Concept used:

Efficiency ∝ (1/Time)

Calculation:

Let men be m

Let women be w

Let x be the days 4 men and 6 women take to complete the work

(2m + 3w) × 4 = (6m + 1w) × 2

⇒ (2m + 3w) × 2 = (6m + 1w) × 1

⇒ (4m + 6w) = (6m + 1w)

⇒ (6w – 1w) = (6m – 4m)

⇒ (5w) = (2m)

⇒ m : w = 5 : 2

Total work = [(2m + 3w) × 4] or [(6m + 1w) × 2]

Taking one of them

[(2m + 3w) × 4]

⇒ [{(2 × 5) + (3 × 2)} × 4]

⇒ [(10+ 6) × 4]

⇒ Total work = 64

(4m + 6w) × x = 64

⇒ [(4 × 5)  + (6 × 2)] × x = 64

⇒ 32 × x = 64

⇒ x = 2

∴ 4 men and 6 women take to complete the work in 2 days.

Given:

Two years ago, a village had 16,000 people. In the coming year the population increased by 10%. But in the next year, if some people had left the village and the population had decreased by 18%.

Concept used:

Ratio and Proportion

Calculation:

Population increased by 10%

⇒ 10 : 11

Population decreased by 18%

⇒ 50 : 41

Total population 

⇒ 10 × 50 : 11 × 41

⇒ 500 : 451

Two years ago a village had a population of 16000

500 unit = 16000

1 unit = 320

451 unit = 451 × 320

Present population will be 14432.

Given:

The total population of village A in 2010 = 20,000

The ratio of men and women in 2010 = 3 ∶ 2 

The percentage increase in the population of the village in 2011 = 20% 

The men population of the village in 2011 = 15000

Calculations:

Let the men and women in 2010 be 3x and 2x 

According to the question 

⇒ 3x + 2x = 20,000

⇒ 5x = 20,000

⇒ x = 4,000

The women population of the village in 2010 = 8,000

Total population of the village in 2011 = 20,000 × (120/100) = 24,000

Total women population of the village in 2011 = Total population of the village in 2011 – Total men population of the village in 2011 

⇒ 24,000 – 15,000 = 9,000

The ratio of female population in 2010 and 2011 = 8000 ∶ 9000 = 8 ∶ 9

∴ The ratio of female population in 2010 and 2011 is 8 ∶ 9 

Given:

Ratio of carbon and oxygen = 1 : 4.

Formula used:

Percentage of carbon = (Value of carbon/ sum of ratio) × 100

Calculation:

Sum of the terms of ratio = 1 + 4 = 5

Percentage of carbon = (1/5) × 100

⇒ 20%

∴ The percentage of carbon is 20%

Solving for Quantity I:

Given:

Three years ago, the ratio of age of son and father = 1 : 8

After three years, the ratio of age of son and father = 5 : 19

Calculations:

Let three years ago from present, the age of son and father be x and 8x years respectively.

So, After 6 years ratio of their age will be 5 : 19

⇒ (x + 6)/(8x + 6) = 5 : 19

⇒ 19x + 114 = 40x + 30

⇒ 21x = 84

⇒ x = 4

Present age of son = x + 3

⇒ 4 + 3

⇒ 7

∴ Present age of son is 7 years.

Solving for Quantity II:

Given:

Three years ago, the ratio of age of son and mother = 1 : 9

After three years the ratio of age of son and mother = 1 : 4

Calculations:

Let three years ago from present, the age of son and mother be x and 9x years respectively.

So, After 5 years ratio of their age will be 1 : 4

⇒ (x + 5)/(9x + 5) = 1 : 4

⇒ 4x + 20 = 9x + 5

⇒ 5x = 15

⇒ x = 3

Present age of son = x + 3

⇒ 3 + 3

⇒ 6

∴ Present age of son is 6 years.

∴ Quantity I > Quantity II

The correct answer is Mauritania.

• The Equator is an imaginary line notionally drawn on the earth equidistant from the poles.
• It divides the earth into the northern and southern hemispheres.
• It constitutes the parallel of latitude 0°.
• It passes through 13 countries.

Decimal:

Do you ever thought that if there were no any decimal system in early age then if some want to buy only a part of whole then how they did that?

Decimal system helped us to simplify the number system and elevate it. With the help of decimal system you now can take any amount of the whole and pay accordingly by using decimal system.

Dot (∙) represents decimal.

Tenth:

When a whole object is described in 10 units one part of the whole is said to be tenth of that object.

Similarly 2 part of the whole is said to be 2 tenths and so on.

And the most important thing is when we have to describe 3 part and 2 objects, we can write

3 part of whole = \(\frac{3}{10}\) also written as 0.3

2 whole = 2

Total = 2 + \(\frac{3}{10}\) = 2.3 read as two point three.

Representing Decimals on number line:

We have to represent 0.6 on number line which is 6 part in 10.

Fractions as decimals:

We can represent any fraction as decimal.

As \(\frac{8}{5}\) can be written as 10^th order.

We multiply its numerator and denominator by 2.
Thus, \(\frac{8}{5}\) can be written as \(\frac{8x2}{5x2}\) = \(\frac{16}{10}\), which can be easily represented in decimal system as 1.6.

Decimals as fractions:

We can represent any decimal as fraction.

As 3.5 when written in fraction is \(\frac{35}{10}\) which when divided by equal numbers to numerator and denominator i.e. divide it by 5 we get \(\frac{7}{2}\).

Hundredth:

If a whole object contains 100 part and then 1 part of it is called hundredth.

As 5 part of the whole is \(\frac{5}{100}\) = 0.05

Comparing Decimals:

When we have to compare two decimal terms as 5.3 and 5.37.

First, we will check the whole part of the two numbers which in these number is same. Then we check its tenth part which is also the same in this case then we will check its hundredth part which in first number is zero and in 2^nd number is 7 which is greater than zero.

Thus, second number is greater than the first.

“ We use Decimals in many ways in our lives. For example, in representing units of money, length and weight.”

Decimal in terms of

Money:

We know that 100 paise = Rs.1

So, 1 paise = Rs. \(\frac{1}{100}\) = Rs. 0.01

Also, 55 paise = Rs. \(\frac{55}{100}\) = Rs. 0.55

And Rs1 and 32 paisa = Rs 1.32

Length:

We know that 1cm = \(\frac{1}{100}\) m or 0.01 m.

56cm = \(\frac{56}{100}\) m = 0.56 m

1m and 30 cm =1m + \(\frac{30}{100}\) = 1.3m

Weight:

We know that 1000 g = 1 kg

Therefore, 1 g =\(\frac{1}{1000}\)kg = 0.001 kg

The correct answer is 9, September.

• Telangana State will celebrate its language day known as ‘Telangana Bhasha Dinosthavam’ on September 9 every year to coincide with the birth anniversary of noted Telangana poet Kaloji Narayana Rao.
• As a poet and freedom fighter, Kaloji had played a crucial role in the promotion of the culture of Telangana. 

Monopoly is the addition to two words, i.e. ‘Mono’ + ‘Poly’, i.e. single seller in the market. Being the single seller in on the supply of the commodity. Impure monopoly even no close substitute of In product is available in the market. In monopoly, no distinction arises between ‘firm’ and ‘industry’ i.e. firm is industry and industry is firm.

Definitions:- 

1. According to Mc councell, “Pure monopoly exists when a single firm is the sole producer of a product for which there are no close substitutes 

Features of Monopoly: The main features of monopoly are as follows:- 

1. Single seller and large Number of buyers: Monopoly market consists single seller of the product but the number of buyers stands very large. No buyer can influence the price of the product due to his large number in the market. 

2. No close substitutes: Monopoly firm produces such commodity which has no close substitute and as a result the cross elasticity of demand becomes zero. 

3. Monopolist as a price- maker: Being the single seller in the market. The monopolist firm in itself a price- maker A monopolist firm can determine both price and quantity but not simultaneously (i.e either price or quantity at a particular time.) 

4. No Entry of new firm : The entry of new firm into the industry is strictly prohibited. There is no competitor of monopoly firm in the market. 

5. Demand curve Negatively sloped : Monopolist demand curve in negatively sloped and marginal revenue (MR) is lesion than average revenue (AR). The slope of demand curve depends on elasticity of demand.

We have A = \(\hat i\) - 2\(\hat j\) + 3\(\hat k\)

and 3 = 3\(\hat i\) - 2\(\hat j\) + \(\hat k\)

2A + 5B = 2(\(\hat i\) -  2\(\hat j\) + 3\(\hat k\)) + 5 (3\(\hat i\) - 2\(\hat j\) + \(\hat k\))

= (2\(\hat i\) - 4\(\hat j\) + 6\(\hat k\) ) + (15\(\hat i\) - 10\(\hat j\) + 5\(\hat k\))

= (2 + 15)\(\hat i\) - (4 + 10)\(\hat j\) + (6 + 5)\(\hat k\)

= 17 \(\hat i\) - 14 \(\hat j\) + 11\(\hat k\)

Ratio and Proportion Formulas & Notes:

In everyday life, we always compare something of the same type. The ratio is also a mode of such comparisons. In this section, we will read ratio and proportion notes. Just an example you brought a bicycle for Rs 2000 and one of your friend brought the same bicycle in another shop for Rs. 1500. Now you compare the price as you had been thug by the shopkeeper by an amount of Rs.500.

This amount of 500 came by taking the difference between the price of same quantity. This is the one way of comparison, but the method of taking difference is not good in case when we compare a much smaller amount to much higher amount. Such as we compare the price of a Tata Nano car to Lamborghini car. Read further in ratio and proportion notes.

For this, the most accurate system was given as ratio which is comparison by division which is termed as Ratio.

Ratio:

When two quantities of same units are compared by division, this method of comparison is called ratio. It is said in this mode of comparison that one quantity is how many times the other.

• The symbol for expression of ratio is “ : ”
• For comparison by ratio, the two quantities must be in the same unit. If they are not, they should be expressed in the same unit.
• we can get equivalent ratios by multiplying or dividing the numerator and denominator by the same number.
• Ratio and proportion notes will provide you with the complete guide of chapter of class 6.

Examples of ratio:

• Length of a room is 50 m and its breadth is 40 m. So, the ratio of length of the room to the breadth of the room = \(\frac{50}{40}\) = \(\frac{5}{4}\) = 5 : 4.
• There are 32 girls and 20 boys going for a picnic. Ratio of the number of girls to the number of boys = \(\frac{32}{20}\) = \(\frac{8}{5}\) = 8 : 5.
• Whereas that of the breadth of a table to the length of the table is 2 : 3.

Proportion:

If two ratios are equal, we say that they are in proportion and use the symbol ‘::’ or ‘=’ to equate the two ratios. This equation of ratio is said to be proportion.

If two ratios are not equal, then we say that they are not in proportion. Also take a look at examples given in ratio and proportion notes.

For representation of proportion: we can say 2, 5, 14 and 35 are in proportion which is written as 2 : 5:: 14 : 35 and is read as 2 is to 5 as 14 is to 35 or it is written as 2 : 5 = 14 : 35.

Terms related to proportion:

• Extreme terms: In the statement of proportion which contains 4 terms, in that we denote 1^st and 4^th term as extreme terms.
  Example: In proportion 2 : 5:: 14 : 35, 2 and 35 are extreme terms.
• Middle Terms: In the statement of proportion which contains 4 terms, in that we denote 2^nd and 3^rd term as middle terms.
  Example: In proportion 2 : 5 :: 14 : 35, 5 and 14 are middle terms.

Now read the other section of ratio and proportion notes i.e. unitary method.

Unitary method:

The method in which first we find the value of one unit and then the value of required number of units is known as Unitary Method.

• Suppose the cost of 5 cans is Rs. 175. To find the cost of 4 cans, using the unitary method, we first find the cost of 1 can. It is Rs.175/5 or Rs.35. From this, we find the price of 4 cans as 35 × 4 or Rs.140.

Examples from ratio and proportion notes:

1. There are 40 persons working in an office. If the number of females is 15 and the remaining are males, find the ratio of the number of females to number of males and the number of males to number of females.

Solution:
Number of females = 15

Total number of workers = 40

Number of males = 40 – 15 = 25

Therefore, the ratio of number of females to the number of males
= 15 : 25 = 3 : 5

And that of number of males to the number of females
= 25 : 15 = 4 : 5.

2. Give two equivalent ratios of 12 : 4.
   Solution:
   Ratio 12:4 = \(\frac{12x2}{4x2}\) = \(\frac{24}{8}\)
   Therefore, 24 : 8 is an equivalent ratio of 12 : 4
   similarly, \(\frac{12/4}{4/4}\) = \(\frac{3}{1}\) = 3 : 1
   3 : 1 is also an equivalent of 12 : 4.

Learn and revise ratio and proportion notes to score better in examinations.

The correct answer is a - IV, b - III, c - II, d - I.

• Ghori first invaded Multan in 1175 AD. At this time, Karamati, the followers of the Shiites was ruling. These Karamati were Buddhists before becoming Muslims. Ghori won the Multan.
• Ghori did the second invasion on Gujarat in 1178 AD, but Moolraj II defeated him in the foothills of the Abu Mountains. This was the first defeat of Muhammad Ghori in India. This war was conducted by Nayika Devi, the wife of Moolraj.
• After Ghori’s death, his kingdom was divided into three of his main slaves.
  • Qutubuddin Aibak – Indian Area. Aibak made Delhi the center of the Islamic Empire.
  • Tajuddin Yaldoz – Ghazni area
  • Nasiruddin Kubacha – Uchchh and Sindh (Pakistan)
  • Muhammad ibn Bakhtyar - Indus Area

• The period between 1206 A.D. and 1526 A.D. in India's history is known as the Delhi Sultanate period.
• During this period of over three hundred years, five dynasties ruled in Delhi. These were:
  • The Slave dynasty (1206-90), Khilji dynasty (1290-1320), Tughlaq dynasty (1320-1413), Sayyid dynasty (1414-51), and Lodhi dynasty (1451-1526).
•  Qutub-ud-din Aibak, a slave of Muhammad Ghori, who became the ruler after the death of his master, founded the Slave Dynasty.
• He was a great builder who built the majestic 238 feet high stone tower known as Qutub Minar in Delhi.

International trade and the flow of foreign investments are adversely affected by fluctuation of exchange rates. For solving this problem, Smithsonian Agreement was made on December 18,1971. This agreement raised the fluctuation range of exchange parity at both ends by 1 percent to 2.25 percent. 

On February 12,1973, after the devaluation of American dollar Smithsonian Agreement was broken. After Jamaica Agreement in January 1976, floating exchange rate system was launched under International monetary fund. 

In fact, floating exchange rate system was not a free flexible exchange rate system but is a managed floating exchange rate system in which the monetary authority of the nations were given the responsibility to eliminate short-run fluctuations of exchange rates without affecting the long-run trends of the exchange rates. Thus, in this given opportunity to enjoy the benefits of fixed exchange rates on one hand and also to maintain flexible exchange rate for adjusting imbalances in balance of payments on the other.

In managed floating exchange-rate short-term fluctuations of exchange rate are controlled with the use of foreign exchange funds without affecting the long-term inclination of exchange rates. 

Clean Floating and Dirty Floating systems. 

i. In clean floating system, exchange rate is allowed to be determined by two free forces of demand and supply in the market. In this system, monetary authority is not allowed to interfere and foreign exchange market re-adjusts itself by demand and supply forces without any intervention. 

ii. In Dirty Floating system, the monetary authority of the country marks intervention to check fluctuations of exchange rate. The basic aim of this intervention is to favour domestic nation. 

Thus, in clear floating system exchange rte stability is obtained with flexibility to same extent but in dirty floating system, there is no flexibility in exchange rate.