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I completed MCA in 2009; I am working as a computer operator with seven years of experience; I have a knowledge of MS-Office and Desktop basics. Now I want to move to the IT field, is it possible?

      The present-day computer has evolved through several stages. The roots of the evolution are spread evenly in the prehistoric periods when the ancient man started to settle down in dwelling centres. In this period he started cultivation and other productive engagements. He faced the need for counting his products at this stage. In the early stages, he used his fingers for counting purposes. This was the reason for the establishment of a counting system based on the number ten, which is considered to be the most ancient system.

1. ABACUS

         The earliest counting machine, invented by man was a primitive one called ABACUS. It was developed by Chinese nearly 3000 years ago, the Abacus is known as 'Swan Pan' in China. This model is considered to be the best one among other Abacus models used in other countries.

         Abacus consists of a rectangular wooden frame with horizontal rods. These rods carry round beads. These beads are made out of stones, pearls, wooden blocks. Counting is done by shifting the beads from one side of the Abacus to another. One bead on a particular wire has the value of 1; two together have the value of 2. A bead on the on next line has the value of 10, and the bead on the third line would have the value of 100. Therefore, three properly placed beads - two with values of 1 and one with the value of 10 - could signify 12, and the addition of a fourth bead with the value of 100 could signify 112. Thus, the Abacus works on the principle of a place-value notation. The Abacus can be cleared or set to zero by moving all beads away from the crossbar.

1. JOHN NAPER

         John Napier was one of the earliest scientists who tried to invent some devices for calculations. He was A Scottish mathematician. Napier devised a set of rods for use in calculations involving multiplications. These rods were made from bones, and hence this device came to be known as NAPIER BONES. Napier introduced the use of notation to indicate a fractional position.

         The invention of odometer which -is now known as speedometer initiated the development of mechanical adders and multipliers. It was John Napier who developed the method of logarithm in 1617. The tables used in the logarithm represent analogue computing techniques. In logarithm multiplications and divisions can be done by adding and subtracting not the numbers themselves but with the help of related numbers.       

1. BLAISE PASCAL

         The credit for the invention of the first mechanical calculating machine goes to Blaise Pascal. He was a French Mathematician and Physicist, born on June 19, 1623, at Clermont-Ferrand, Auvergne. From the very childhood, Pascal was very much interested in the study of mathematics. When he was only sixteen Pascal published a book on the geometry of the conic sections that for the first time carried the subject well beyond the point at which Apollonius has left it nearly nineteen centuries before.

         In 1642, when Blaise Pascal was only nineteen, he invented the calculating machine called Pascaline, using cogged wheels, could add and subtract. It is considered as the ancestor of the mechanical devices that reached their culmination in the modern cash register.

         Blaise Pascal invented the calculating machine only to assist his father in tax calculations. His father was appointed as the Tax Superintendent of Royen.

         The calculating machine invented by Pascal consisted of gears, wheels and dials. Each wheel had ten segments, which can be compared to the wheel of a milometer. When one wheel completed 0 to 9 around their circumference, with this calculator addition and subtraction could be performed by dialling the series of wheels. He also invented the syringe, the hydraulic press, and Pascal's law of pressure. The basic principle of his calculator is still used today in water meters and modern-day odometers.

1. GOTTFRIED WILHELM VON LEIBNIZ

         Leibniz was a German philosopher and mathematician, born on July 1, 1646, at Leipzig, Saxony. He devised a calculating machine superior to that of Pascal. Leibniz modified the mechanical calculator invented by Pascal and perfected it to a certain context. The modified machine could multiply and divide as well as add and subtract. Leibniz was the first to recognize the importance of the binary system of notation, making the use of 1 and 0 only. This is important in connection with modern computers.

1. JOSEPH-MARIE JACQUARD

         He was a stonemason and later a weaver. He created the Punched Card Loom in 1801. This device was a new type of loom for weaving cloth—punched cards controlled its operation. Needles could pull threads through cards where there were holes and not where there were none. Thus patterns were stored on punched cards. It was adopted by Charles Babbage to control his Analytical Engine.

1. CHARLES BABBAGE

         He was an English mathematician born on December 26, 1791. He conceived of a machine that could be directed to work using punched cards, that could store partial answers to be performed upon them later, and that could print the results. Babbage developed a machine which he called the 'DIFFERENCE ENGINE'. This machine worked on a mathematical technique which repeatedly added differences between numbers to perform various types of calculations. Charles Babbage made only a working model of the difference engine. He did not make a full-scale difference engine because the necessary technology needed for its completion was not available at that time.

         Later on, he was able to develop another device called the 'ANALYTICAL ENGINE'. This was an automatic computing machine designed to make additions at the rate of sixty per minute. It had memory also. Babbage has introduced an input device also which was nothing but a collection of punched cards. He also introduced a control unit whose function would be to control the various operations of the engine. In short, the analytical engine was indeed a marvellous blueprint of a modern computer.

1. GEORGE SCOUTS

         Two Swedish printers, George Scheutz and his son Edward attracted towards the difference engine, and they were successful in making the first difference engine which became very popular among scientists as well as common people. Subsequently, they build several different engines. They got a Gold Medal at the Paris Exhibition in 1855 as the makers of the difference engine.

1. HERMAN HOLLERITH

         He built few difference engines in 1890 for use in the mathematical calculations of the census conducted there. These engines were proved to be highly efficient, and subsequently, several different engines were in use in the different parts of the world.

         The success story of his machine encouraged Hollerith to start the 'Tabulation Machine Company' in 1896. Later in 1911, this company became the 'Computing Tabulating Recording Company' This company is now known as the 'International Business Machines' - IBM Company.

In this lesson, we introduce the concept of probability with high school mathematics as a prerequisite. Before we start, I want to make you familiar with some standard terms in probability theory.

The first concept is the Random Experiment. If an experiment is conducted, and its result is unpredictable, i.e., the outcome can be one of many issues. Such an experiment is called a Random Experiment. Each outcome of a random experiment is an Event. Set of all possible outcomes is called the Sample Space. For example, when you are tossing a coin H(Head), and T(Tail) are two possible outcomes or event and { H, T } is the Sample Space of the Random Exprimenttossing coin.

There are two special kinds of events. Let A and B are two events. We call them Mutually Exclusive Events if A ∩ B = Φ, and we call them Mutually Exhaustive Events if A ∪ B = S. For example, in a random experiment, we are choosing a student to form a class. A is an event the student is a boy, and B is an event where the student is a girl. A and B are Mutually Exhaustive as a student is either a boy or girl there is no other option. And A and B are Mutually Exclusive because a student can not be a boy and girl both same time.

Probability of an event E = P(E) = (Number of outcomes in favor of E) / (Total number of outcomes)

For example, when you are tossing a coin probability of getting head is 1/2. Similarly, when you are throwing a dice probability of getting five is 1/6. As in the first case, the total number of outcome is 2 - head and tail, and the number of favourable results is one only head. Similarly, in the second case, the total number of results is 6 - 1 to 6, and the desired outcome is only 5.

So we have seen to calculate the probability we need to count the number of outcomes of an experiment. So it is crucial to know the theory of computing to work in probability. The fundamental theory of counting is that if you can do one work in m ways and another job in n ways, then you can do both tasks together in mxn ways. You can visualize this theory by imagining there is a closed room which has m doors to enter and n doors to exit. So a person can enter the room in m ways, and for each entry, he can exit in n ways. So he can enter and exit the room in mxn ways.

Problem 1: A polygon has 44 diagonals. What is the number of sides?
Solution:
If the polygon has n sides, we can choose the first end of a diagonal in n ways and second point in (n-3) ways excluding the point, and it's adjacent two which form side of the polygon. So both can be selected in n*(n-3) ways. Now in choosing the point, we are selecting each diagonal for twice. Each end is counted as fist and second end. So the number of diagonal = n*(n-3)/2 = 44 => n*(n-3) = 88 = 11x8.
So n = 11 (Ans)

Problem 2: Find the number of devisor in the form 4n + 2 (n>1) of integer 210.
Solution:
The divisor is in the form = 2*(2n+1) form. Factor of 210 in 2n+1 (n>1) form is 1,3,5. So we have to find the all possible combination of 3, 5, and 7.
Let generalized the problem assuming we have n different elements x1, x2 .. Xn and we have to find the all possible set can form by xi. So we can choose the first element in 2 ways either we can choose it or not choose it. Similarly, the second element we can also choose in 2 ways. So the first two elements we can choose in 2x2 ways. Similarly the first 3 elements we can choose 2^3 ways. So all element we can choose it in 2^n ways. So with n distinct elements, we can form 2^n different sets. There is one set which has no element ie, an empty set. So the number of the nonempty set is 2^n -1.
In our case, n=3 so answer is 7. Numbers are 3, 5, 7, 15, 21, 35, 105. (Ans)

Now we introduce the concept of Permutation and Combination. The permutation is the number of arrangement we can make by some elements, and the combination is the number of sets we can make by those elements. Difference between permutation and combination is that permutation consider the order or arrangement of the members which combination does not. Suppose a & b are two elements then {ab} and {ba} are two different permutations, but the same combination as the combination does not consider the arrangement.

Let, nPr is the number of permutations of r objects selected from n different objects. We can imagine there are n different balls and r basket and nPr is the number of ways we can fill the boxes by the given balls. So the first box we can fill in n ways as there are n balls. Then the second balls can be filled in (n-1) ways as one ball is already allocated in the first box. So according to the theory of counting 2 boxes can be filled in n*(n-1) ways. In same way r box can be filled in n*(n-1)*(n-2)...(n-r+1) ways.
So nPr = n*(n-1)*(n-2)....(n-r+1)
Now we are defining factorial of n as n!=n*(n-1)*(n-2).....3*2*1
Then nPr = n!/(n-r)!
The thing to be noted, nPn = n! as 0! =1 as it indicates an empty set or permutation.
Similarly, nCr is the number of combination of r objects selected from n different objects. Now you take one combination. It has r differents. So you can r! Permutation from it. Now if you take each combination and create r! Permutation from it, you get all possible permutation of r elements from n distinct elements.
So nCr * (r!) = nPr
=> nCr = (n!)/(n-r)! * (r!)

Problem 3: Find the sum of the digits in the unit place of all numbers forms with the help of 3,4,5,6.

Solution:

So the number of all the number's last digit where 3 is the last is the number of all permutation of rest three numbers = 3P3
So the sum of all the number's last digit where 3 is the last = 3P3 x 3
Similarly, the sum of all the number's last digit where 4 is the last = 3P3 x 4
So the sum of the digits in the unit place of all numbers forms with the help of 3,4,5,6 = 3P3x(3+4+5+6)(Ans)

Problem 4: Find the number of parallelograms that can form from a set of 4 parallel lines intersecting another 3 parallel lines.

Solution:

Answer is 4C2 x 3C2 ( An explanation will be provided on Request )

Problem 5: If two verticals are selected form a polygon of n side in random, what is the probability that it is diagonal?

The number of pairs of vertices = nC2 (number of ways 2 points selected from n points )
The number of diagonal = n(n-3)/2 ( see problem 1)

So, the answer = n(n-3)/2*4C2

Some properties of combinations:

1. nCr = nCn-r

Every time we choose r points from a set of n points there is a set of n-r points remain in the original set.  So nCr = nCr-1

2. (n+1)Cr = nCr + nCr-1

(n+1)Cr = number of ways r elements can be choose from n+1 elements.

Let, one element keeps apart from n+1 elements. Then number of combination = nCr

So (n+1)Cr = nCr + number of set can be formed with one seperated elements.

We can put the special element in one place of r place then

number of set can be formed with special element = from n element how to choose element for r-1 place = nCr-1

So  (n+1)Cr = nCr + nCr-1

3. nC0 + nC1 + nC2 + ....... + nCn = 2^n

See problem 2

Problem 6: Tn = number of triangles drawn from a n side polygon. Tn+1 - Tn =21. n=?

Solution:

(n+1)C3 - nC3 = 21

=> nC3 + nC2 -nC3 = 21

=> nC2 = 21

=> n(n-1) = 42 = 6x7 => n = 7 (Ans)

Back to Probability:

If A` is event for not happenning A then  P(A`) = 1- P(A)

If A and B are independent event then

                P(A∩B) = P(A)*P(B)

                P(A∪B) = P(A) + P(B) - P(A∩B)

Problem 7: Probability of A and B will die within a year is p and q. What is the probability that only one of them will alive at the end of the year.

Ans: p*(1-q) + q*(1-p)

Problem 8: Probabilty that a person will hit a target is 0.3. He shoots 10 times. What is the probability, he hits the target?

Ans: 1- (0.7)^10

Problem 9: In a throw of dice find the probability of getting one in even number of throw.

Ans:

(1/6)*(5/6) + (1/6)*(5/6)^3 + (1/6)*(5/6)^5 +   .... ∝

=(5/36) [ 1+ (5/6)^2 + (5/6)^4 + (5/6)^6 + .... ∝ ]

= (5/36)[ 1/(1-25/36)] = 5/36 * 36/11 = 5/11 ( Ans )

Problem 10: What is the probability that a leap year selected in random has 53 Sunday.

Ans: 364 days has 52 Sundays. 

So the probability of there is one Sunday in remaining two consecutive days =1/7 + 1/7 = 2/7 ( Ans )

Problem 11: E and F are two independent events. Probability of both events happen togethre is 1/12 and neither will happen is 1/2. P(E) = ? and P(F) = ?

Ans: P(E) * P(F) = 1/12

       1 - P(E or F) = 1/2 => P(E) + P(F) -1/12 = 1/2 => P(E) + P(F) = 7/12

You have two equation p+q = 7/12 and pq = 1/12. Solve them and get p and q.

Baye's Theorem:

Problem 12: A box contains N coins, m of which are fair and rest are biased. The probability of getting head of biased coin is 2/3. A coin is drawn from the box at random and tossed twice. First time it shows head and second time tail. What is the probability the coin is fair.

Problem 13: In a multiple choice question there are m options. A student know the answer, that's probability is p. He answered the question correctly. What is the probability that he knows the answer?

Some harder problems:

Problem 14: X is a set of n distinct elements. If A and B are two subsets of X, picked at random. What is the probability that A and B have same number of element?

Problem 15: A box contains ticket number from 1 to N. n tickets are drawn with replacement. Find the probability that the largest number in on drawn ticket is k.

 Now I will explain how a bank calculates the probability of a cardholder to be a fraud. They select some crucial features like age, income .. of the people. Let say a card holder's age is 40 and income INR 50,000 from the historical data they calculate the probability of a person become fraud when his age is 40 from the ratio of the total number of person of age. 40, and the number of people among them are a fraud. Similarly, they can calculate the probability of becoming fraud with income 40000. The final probability is the product of each feature probability. This is known as the Naive Bayes Classifier.

Online money making is hot topi now these days. Everyone is looking for scam free online money making ways on Google. Many experts are running training courses on this topic across the web, But very few money making tips and guides are legit and effective for newbies.

Today I am going to share a lesson about top website when it comes to make money online using skills and expertise. I am talking about FIVERR.

What is Fiverr.com ?

Fiverr is an online platform and it works are aggregator between skilled people and businesses those are looking for skilled people to do Business/Personal Gigs. 

Gigs means a job usually for a specified time. On Fiverr, people create Gigs ( Services ) based on their skills and needy people buy those gigs and pay.

Understanding of Fiverr 

There are few things you need to know before starting working on fiverr like platforms ( eg- People Per Hours).

• How to deide rates for your freelancing services
• How to post Gigs on Fiverr to get maximum exposure
• How to optimize Your Gigs to get orders
• How to write title, description of your services to inrease visibility in Fiverr Searches for relevant keywords.
• How to create Graphics for your services for better CTR ( Click through Rate)
• How to upsell on Fiverr to Increase your revenue & earn more money on fiverr.

All this Questions need practical approach so i created 2 videos on fiverr tutorial. This video training is long training session but it will help you to understand fiverr algorithms and ways to earn money through fiverr.

Fiverr Tutorial Part -1 

Fiverr Tutorial in Hindi Part 2 

Check this lesson to get all answers related to fiverr money making topic if you have any query please feel free to comment here on urban pro.

Happy learning

To convert Binary to hexadecimal:

1. Starting from the right, divide the number in the binary into groups of 4.
2. Write down the hexadecimal value from the following table:

For Example:

Convert  11111011101110110

Solution:

1.Starting from the right divide into 4 groups as 

2.You would notice that the last group has only 1 binary digit i.e 1. To make it a group of 4 we can add three 0’s in front of it to make it 0001.

3. Using the table above we can write it as

0110 as 6, 0111 as 7, 1111 as F and 0001 as 1 to get 1F776.

To convert Hexadecimal to Binary:

1. Write down the number in the Hexadecimal System.
2. Write down the Binary value of each from the following table:

                For Example,

                Convert the Hexadecimal number FACB23.

                Solution:

1. Starting from the left , we have F A             C             B             2              3
2. Using the table above, write F as 1111, A as 1010, C as 1100, B as 1011, 2 as 0010 and 3 as 0011
3. We get the Binary number as 111110101100101100100011.

Exercise 1.

Convert the following binary numbers into hexadecimal:

1. 111001111100
2. 111001111100
3. 100110101100
4. 110010101001
5. 101011000011

Exercise 2. Convert the following from hexadecimal number system to binary number system:

1. A193
2. BCF8
3. ABC4
4. BEF4
5. CDAB

Answers to the Exercise:

1. CE7
2. E7C
3. 9AC
4. CA9
5. AC3

Answers to Exercise2:

1. 1010000110010011
2. 1011110011111000
3. 1010101111000100
4. 1011111011110100
5. 1100110110101011

For students seeking education in school, all you need to know is that knowledge is more important than your marks. So, whenever you have your exams on the head, you start cramming up everything and end up forgetting things in the exam. Please start early, and don't cram up topics for the exam but learn because you need to, learn for knowledge, ask questions and enquire the topic. Develop an interest in the subject, and you see how your knowledge, as well as your marks, upgrade!

Hi everyone here I am going to solve a fundamental question related to quant the answer of which probably most of you know, but the method is critical that applies to other topics like time and distance, time and work compound interest etc.

Q. There are ten cats and dogs. Each cat eats five biscuits, and each dog eats six, and if there are total 56 biscuits find the number of cats and dogs separately?

First approach

Let the number of cats is c, and the number of dogs is d.

So according to question.

c+d=10; 5×c+6×d=56

Now multiply first equation with 5

5c+5d=50

And subtract the second by a third equation

5xc+6xd-5xc-5×d= 56-50

d=6

And apply this value of d in the very first equation

c= 10-6=4

So finally cats are 4 and dogs are 6.

but this approach is old and obsolete let's see

The cat is taking 5, and we can rearrange dog is taking 5+1=6 biscuits

So five is common for all ten animals.

And then 5×10=50

Subtract this from 56-50=6

Now we can see one is left in dog group so divide six by 1

We get six dogs and since total.animals are ten so 10-6=4 cats.

At least 6 marks we can get from the quant section by this method, and it applies to different topics

Stay happy and be selected.

Thanks