### True Discount and Banker's Discount problems tricks in Hindi | fast track formulae for problem solving.

TRUE DISCOUNT AND BANKER'S DISCOUNT TRICKS IN HINDI

TRUE DISCOUNT 1). वास्तविक बट्टा = मिश्रधन - वर्तमान धन
2). यदि ब्याज की दर r% वार्षिक , समय t व वर्तमान धन (pw) है, तो वास्तविक बट्टा = PW×r×t/100
3). यदि r% तथा समय t के बाद देय धन A है, तो तत्काल धन pw = 100×A/(100+r.t)
4). यदि देय धन A पर r% व समय t दिए गए है, तो वास्तविक बट्टा = A.r.t /(100+r.t)
5). यदि किसी निश्चित समय के पश्चात निश्चित वार्षिक दर पर, देय धन पर वास्तविक बट्टा (TD) व समान समय व दर के लिए साधारण ब्याज (SI) हैं, तो देय धन A =      SI × TD/(SI - TD)
6). यदि t समय पश्चात r% वार्षिक दर पर, देय धन पर वास्तविक बट्टा (TD) व साधारण ब्याज (SI) है , तो -  SI - TD = TD×r×t/100
7). t वर्ष बाद r% चक्रवृद्धि दर पर देय धन A का तत्काल धन (PW) = A/(1+r/100)^t   व वास्तविक बट्टा = A - PW
BANKER'S DISCOUNT1). महाजनी बट्टा = शेष समय (समाप्त न हुए समय) के लिए बिल पर ब्याज = बिल की राशि × दर × शेष समय /100
2). महाजनी लाभ = महाजनी बट्टा - वास्तविक बट्टा
3). यदि बिल का मान / अंकित मूल्य A है, समय t व दर r% है, तो महाजनी बट्टा = A×r×t/100
4).…

### PROFIT and LOSS for competitive examination, shortcut tricks , fast track formulae

PROFIT AND LOSS
SOME IMPORTANT SUTRAS

*. Profit = selling price - cost price
*. Loss = cost price - selling price
*. Cost price = selling price - profit
*. Cost price = selling price + loss
*. Selling price = cost price + profit
*. Selling price = cost price - loss
*. Profit percent = profit × 100/cost price
*. Loss percent = loss ×100/cost price
*. Cost price = SP ×100/(100+P)
*. Cost price = SP ×100/(100 - L)
*. Selling price = CP(100+P)/100
*. Selling price = CP(100-L)/100

*. If a dishonest sellsman sell his goods on CP but he uses fake weight instead of original weight , then the profit percent =
100(original weight - fake weight)/fake weight

*. If a dishonest sellsman use x% less weight, but he also sells his goods at y% profit/loss, then net percentage profit/loss = (+-y+x)100/100-x

*. If a sellsman wants to earn b% profit after selling his goods at a% discount , then the sufficient increment in the marked price of the goods is = 100(b+a)/100-a

*. If P sells a object to Q at a% profit/loss , and then Q sells that object to R at b% profit/loss , if R gives A rupees, then the cost price for P is =
100×100×A/(100+-a)(100+-b)

*. If a% and b% are two successive profit/loss on any object then the resultant profit/loss =
(+-)a + (+-)b + [(+-)a(+-)b/100]

*. A sellsman sells his goods at a% profit , if he sells his goods at R rupees more, then he earn b% profit , then cost price of goods is = 100×R/(b-a)

*. A object is sells at a% profit, if CP and SP both are less by rupees R , then he earn b% more profit, then the CP of the object is = R(a+b)/b

*. If CP of (a) objects is equal to the SP of (b) objects , then the profit percent = 100(a-b)/b

*. If a part of a object is sells at x% profit/loss , b parts at y% profit/loss and c parts at z% profit/loss, this makes total profit/loss of R rupees, then the CP of the objects is =
100×R/(ax+by+cz)

*. If A th part of any object is sells with x% loss, if further there will be no loss or profit in whole transaction, for this, the profit percent for the remaining parts of the object should be =
Ax/(1-A)%

*. The selling price of two objects is Rs X , if one object is sold at r% loss and other object is sold at R% profit- then
**. The CP of the object which are sold at profit =
x(100+R)/(200-r+R)
**. CP at loss = x(100-r)/(200-r+R)

*. If a object is sold at Rs A with r% profit/loss, then the selling price of the object, if it is sold with R% profit/loss is = A(100+-R)/(100+-r)

*. If a person sold a object with R% profit/loss instead of r% profit/loss and gains the profit of Rs A, then the CP of the object = 100A/R-r

*. If the selling price of two objects are same, one object sold at r% profit and other object sold at r% loss, in this type of transaction always occurred loss, then the loss percent is = r^2/100

*. If CP of x objects is equal to the SP of y objects, if x>y then always be profit then profit percent = 100(x-y)/y , but if x<y then always be loss, then loss percent = 100(y-x)/y

*. A buys a object in Rs x, and sells it to B at r% profit/loss, again B sells it to A at R% profit/loss , then profit of A in this transaction =
x(100+-r/100)(1-[100+-R/100])

*. r% profit/loss been if x objects are bought in 1 rupees , then the numbers of objects selling in 1 rupees will be = 1×100/100+-r

*. A person buy y objects in x rupees, and then he sells x objects in y rupees-
If x>y then loss percent = (x^2 - y^2)100/x^2
If x<y , then profit percent = (y^2 - x^2)100/x^2

*. A person sells a object in x% profit/loss , if he bought at y% less/more, and sell at Rs A more/less , then he gains the profit of z% then the CP of the object =
A[(100+-y/100)(100+-z/100) - (100+-x/100)]

*. After selling a object in x rupees, the gained profit/loss is equal to the cost price of that object then the cost price is = +-50+-10√(25+-x)

*. If a object is bought at the rate of (a) in x rupees, and sells at the rate of (b) in y rupees-then the percentage profit = (ay-bx)100/bx

*. A sellsman cheat while buying object with x% and while selling object he again cheat with x% , then the profit percent = 2x + x^2/100

*. A object with having CP of x rupees, is sold with y% profit , and by another person that object is sold again with z% profit/loss , then the final selling price of that object will be =
x(100+-y/100)(100+-z/100)

*. A shopkeeper sells his object at cost price with a% profit/loss , but he use b grams instead of c grams  , then profit/loss percent =
[(c/a)(100+-a) - 100]%

*. The decrease/increases in the usage of the object such that expenditure only increase/decrease by y%, if x% gets increases/decrease in the value of the object is =
[1 - (100+-y/100+-x)]100%

*. After selling a object in x rupees , the gained profit is equal to the loss obtained while selling the object in y rupees, then the CP of the object is = x+y/2

### Boat and stream short-tricks in Hindi | Fast track arithmetic formulae for competitive examination.

BOAT AND STREAM (नाव एवं धारा)
1). यदि शांत जल में नाव या तैराक की चाल x किमी/घंटा व धारा की चाल y किमी/घंटा है, तो धारा के अनुकूल नाव अथवा तैराक की चाल = (x+y) किमी/घंटा
2). धारा के प्रतिकूल नाव अथवा तैराक की चाल = (x-y) किमी /घंटा
3). नाव की चाल = (अनुप्रवाह चाल + उद्धर्वप्रवाह चाल)/2
4). धारा की चाल =  (अनुप्रवाह चाल - उद्धर्वप्रवाह चाल)/2
5). यदि धारा की चाल a किमी/घंटा है, तथा किसी नाव अथवा तैराक को उद्धर्वप्रवाह जाने में अनुप्रवाह जाने के समय का n गुना समय लगता है,(समान दूरी के लिए), तो शांत जल में नाव की चाल = a(n+1)/(n-1) किमी/घंटा
6). शांत जल में किसी नाव की चाल x किमी/घंटा व धारा की चाल y किमी/घंटा है, यदि नाव द्वारा एक स्थान से दूसरे स्थान तक आने व जाने में T समय लगता है, तो दोनो स्थानों के बीच की दूरी = T(x^2 - y^2)/2x km
7). कोई नाव अनुप्रवाह में कोई दूरी a घंटे में तय करती है, तथा वापस आने में b घंटे लेती है, यदि नाव कि चाल c किमी/घंटा है, तो शांत जल में नाव की चाल = c(a+b)/(b-a) km/h
8). यदि शांत जल में नाव की चाल a किमी/घंटा है, तथा वह b किमी/घंटा की चाल से बहती हुई नदी में गत…

### Emergence of British East India Company as an Imperialist Political Power in India

EMERGENCE OF BRITISH EAST INDIA COMPANY AS AN IMPERIALIST POLITICAL POWER IN INDIA
Dynamically changing India during early eighteenth century had a substantially growing economy under the authority of Mughal emperor Aurangzeb. But after his demise in 1707, several Mughal governors established their control over many regional kingdoms by exerting their authority. By the second half of eighteen century, British East India Company emerged as a political power in India after deposing regional powers and dominating over Mughal rulers. The present article attempts to analyze the reasons for emergence of British East India Company as an imperial political power in India and their diplomatic policies of territorial expansion. In addition to this, I briefly highlighted the Charter Acts (1773, 1793, 1813, 1833 and 1853) to trace its impact on the working process of Company.Establishment of East India Company in India

In 1600, British East India Company received royal charter or exclusive license…

### Three dimensional geometry (part-1) | study material for IIT JEE | concept booster , chapter highlights.

THREE DIMENSIONAL GEOMETRY

ORIGIN
In the following diagram X'OX , Y'OY and Z'OZ are three mutually perpendicular lines , which intersect at point O. Then the point O is called origin.
COORDINATE AXES
In the above diagram X'OX is called the X axes, Y'OY is called the Y axes and Z'OZ is called the Z axes.
COORDINATE PLANES
1). XOY is called the XY plane. 2). YOZ is called the YZ plane. 3). ZOX is called the ZX plane.
If all these three are taken together then it is called the coordinate planes. These coordinates planes divides the space into 8 parts and these parts are called octants.
COORDINATES  Let's take a any point P in the space. Draw PL , PM and PN perpendicularly to the XY, YZ and ZX planes, then
1). LP is called the X - coordinate of point P. 2). MP is called the Y - coordinate of point P. 3). NP is called the Z - coordinate of the point P.
When these three coordinates are taken together, then it is called coordinates of the point P.
SIGN CONVENTI…

### A detailed unit conversion table in Hindi.

UNITS CONVERSION TABLE
CENTIMETRE GRAM SECOND SYSTEM (CGS)1). MEASUREMENT OF LENGTH (लंबाई के माप) 10 millimeter = 1 centimetres10 centimetre = 1 decimetres  10 decimetre = 1 metres 10 metre = 1 decametres 10 decametres = 1 hectometres 10 hectometres = 1 kilometres 10 kilometres = 1 miriametresMEASUREMENTS OF AREAS ( क्षेत्रफल की माप )  100 millimetre sq. = 1 centimetre sq.
100 centimetre sq. = 1 decimetres sq. 100 decimetres sq. = 1 metre sq. 100 metre sq. = 1 decametres sq  100 decametres sq. = 1 hectometres sq. 100 hectometres sq. = 1 kilometres sq. 100 kilometres sq. = 1 miriametres sq.
MEASUREMENTS OF VOLUME ( आयतन की माप) 1000 millimetre cube. = 1 centimetre cube.
1000 centimetre cube. = 1 decimetres cube. 1000 decimetres cube. = 1 metre cube. 1000 metre cube. = 1 decametres cube. 1000 decametres cube. = 1 hectometres cube. 1000 hectometres cube. = 1 kilometres cube. 1000 kilometres cube. = 1 miriametres cube.
MEASUREMENTS OF VOLUME OF LIQUIDS  (द्रव्य के आयतन का माप) 10 millilitre=…

### THE GENERAL THEORY OF RELATIVITY | A Unique way to explain gravitational phenomenon.

Today we are going to talk about a very important and revolutionary concept that is THE GENERAL THEORY OF RELATIVITY.
This theory came into existence after 10 years of special theory of relativity (1905), and published by Albert Einstein in 1915.
This theory generalise the special theory of relativity and refines the Newton's laws of universal gravitation.
After coming this theory people's perspective about space and time has been changed completely. And this theory give a new vision to understand the spacetime geometry.
This theory gives a unified description of gravity as a geometrical properties of space and time.
This theory helps us to explain some cosmological phenomenon that is ,

* why small planets revolve around the big stars?
* Why everything in this universe is keep moving?
* Why mostly planets and stars are spherical in shape?
* Why does gravity create?
* Why does time become slow near the higher gravitating mass. Ie. Gravitational time dilation.
And gravitational…