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A mathematical theory is a numerical model that depends on aphorisms (an explanation that is taken to be valid, to fill in as a reason or beginning stage for further thinking and contentions). (Wikipedia Numerical Hypothesis). Parts of science, for example, bunch hypothesis and number hypothesis and some other more hypothesis


Two words that usually contradict each other in mathematics are the words “theory” and “theorem”.

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Theorems are what arithmetic is all about. A theorem is a statement which has been proved true by a special kind of logical argument called a rigorous proof. A rigorous proof is simply a sound deductive argument, meaning that it starts with statements which we know to be true and then makes small steps, each step following from the initial steps, until we reach our conclusion.

As our hypotheses are being tested and refined until our level of confidence in them is very high, we seek a set of principles which provide a coherent explanation for the various laws and facts which we’ve assembled. This kind of detailed explanation of some aspect of reality, incorporating all of the various well-tested hypotheses and mathematical models and explaining the various facts and laws that we’ve observed, is what we call a scientific theory.

The main distinguishable factor about theorems and theory is that theorems are proved while theories are not proved.

A theorem which is proved as a step to proving another theorem is called a lemma, while theorems that results as a consequence of another theorem is called a corollary.

A Hypothesis is associated with a theory in that it has not been proven, it is an unproven statement supported by redundant results and little or no available data. An unproven mathematical statement is called a conjecture.


Math problems could be very tedious and tasking a times. Some require little or no effort to get a solution; whereas, others need formulae, laws and techniques to break the problem into easier bits. Truth be told, some math word problems look really difficult and some people find a hard time finding a solution, but believe it or not, these questions aren’t any harder to solve than non-word problems. Mindset is really important in the way an individual would try to resolve a problem. A lot of people fidget when they come across a word problem, but really, the main challenge is turning the word problem into an algebraic equation. Once this is done, the rest is a piece of cake.

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Many strategies and techniques exist in the world of mathematics. These possible strategies might differ in steps and time taken to solve the problem, but what they have in common, is the answer. Deciphering a question takes the appropriate technique and brain power. A simple starting point in solving a word problem is a four-step approach. Below, are the basic steps required to solve a word problem in mathematics:


Rushing into a question leads to a lot of issues during calculations. This would eventually lead to an incorrect answer. However, this doesn’t mean you should overthink about the question and sleep on it. When faced with a word problem, stop working on it. Although, it sounds paradoxical, but people make mistakes by trying to solve the problem too soon. Ask yourself: What do I really what to do? Make sure you understand exactly what the question is asking of you make sure you understand what you are solving for.


Just as English is a language, mathematics is a universal language. Everything human beings do, involve the manipulation of numbers to get a result. The second step in solving a problem involves turning the words into one or more mathematical expressions or equations. This is to achieve a simpler and cleaner solving ground. Letters could be assigned to the subjects in the given question. It also encourages the manipulation of the equations and the use of the laws and theory to conquer a question. Knowing how to write an equation helps to break down the problem into subcomponents.


The third step in solving a word problem is to solve for the variable you are interested in. The equation gotten from the second step could look abstract, but the equation would need to be simplified using any procedure of choice. Now this procedure entails converting the abstract solution into a numerical solution by substituting in integer values for the already put alphabets(variables). An equality sign might be included in the equation as many questions might vary. The main aim of this procedure is to look for a specific answer to a specific problem.


Solutions come in different ways and different manners. Back solving is another way people seek for solution which might require number play. Your solution should indicate that you actually gave a solution to the question given. Check if the solution you gave is plausible and makes common sense. Thinking about your results helps for future references if presented with the question again. The fourth step is closely related to the first. Understanding is really important during the process of problem solving. Take a minute to think about your result. Use the What-if-should-have method to check and clarify your answers. When done with the solution, carry out this simple checklist:

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 Does the solution seem probable?

 Does it answer the given question?

 Did the solution some in the same unit?

 Did you answer using the language in the question?

If your solution answers all the questions positively, then consider the problem solved correctly, if being judged by steps.


Speed maths is solving mathematics with clever and intuitive techniques in order to cut down time significantly. Speed maths is a basic skill in which quizzers must learn to beat their opponents or have the upper hand in performing lightening quick calculations. This is doing mental mathematics with speed.

Equipped with the skill, enables one do complex calculations without the use of a calculator. It is entirely based on the use of brain power and the ability to circuit equational trends and short cuts. It can also be referred to as hands-free maths. With a little practise, an individual can get the hang of the tricks and use them at will.

Having knowledge in speed math would help boost confidence in handling numbers. When presented with a problem, look for clue words. This is one of the built-in traits you need to unlock to become a speed math guru. As you begin to look for words, certain arithmetic operations begin to stick out.

The secret of speed math is going against the conventional way of solving math problems. In conventional pen on paper calculations, it is easier to solve from right to left, but speed math demands that one solve from left to right when doing calculations in your head. Doing otherwise, leads to difficulty in mental math’s. Mental maths is solved this way because the answer starts calling out before actually completing the problem resolution . As stated earlier on, regular practice is mandatory if you want to get comfortable solving math from left to right.

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Common words in addition problems include:

• Total

• Sum

• Perimeter

• In all

Common words in subtraction problems include:

• How much more

• Exceed

• Difference

Common words for division problems:

• Distribute

• Quotient

• Average

• Share

Common words for multiplication problems:

• Product

• Total

• Area

• Times


Before we get into the nitty-gritty, let us look into what a multiplicand and multiplier is. Let us take the multiplication problem 32 x 42. Here, 32 is the multiplicand; that is: the value being multiplied and 42 is the multiplier- the value multiplying the leading number. Taking a sample problem, let’s say: 324 x 11. This would surely take you a while if you don’t know the simple and quick technique.

First, the initial number of the multiplicand is put down as the first number of the answer.

324 x 11 = 3_ _ _

Then each successive number of the multiplicand is added to the consecutive number at the right.

3 (3+2) (2+4) = 456¬_

To get the last number, simply make the last number of the multiplicand the last number of the solution which gives: 3564

Some problems would require carrying over. Let us take 628 x 11 for example. The first value of the multiplicand is always the first number of the solution. That is:

628 x 11 = 6_ _ _

Each successive number of the multiplicand is added to the neighboring number at the left as done in the previous example. Addition of the last two digits of the multiplicand results in a two-digit sum which would require carrying over. Therefore:

6 (6+2) (2+8) = 6 8 10

So now, we drop the zero in 10 and carry over 1 and add it to 8. This gives: 690_

This last number is simply just put there as the last number of the multiplicand. This gives a solution of: 6908.


This is a simple rule of addition from left to right. Take this example: 9881 + 1234

9 8 8 1

+1 2 3 4

This gives (9+1), (8+2), (8+3), (1+4)

10 10 11 5

Three of the first sums are double digits. This demands a carry-over of the numbers to the front of the number. Therefore, 10

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