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Tutorials

In work life and personal life often we need to learn the basics on a subject quickly. Usually it is not easy to acquire the basic concepts on any subject reliably in a short time from a concise material.

Basic concepts on any subject form the foundation, but generally we have found it is not easy to build it in a short time. Our focus here is on base concepts in a concise form that can be learned quickly and be used as a launch pad for further learning and subject specific problem solving.

Often we added a rich concept layer derived from the base concepts that helps to solve more complex problems with great ease. For us, it is a two layer concept model: Base concepts and Rich concepts derived from base concepts. Objective is efficient problem solving.

Subcategories

Basic maths: Basic concepts on maths. The content is primarily targeted for competitive test requirements.

Reasoning and logic analysis: Tutorials, problems and puzzles on competitive test reasoning and logic analysis.

How to factorize 25 selected quadratic equations quickly by factor analysis

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In school exams or in competitive exams, how quickly and accurately you can factorize a quadratic equation is always a problem we have to grapple with. In this session a comprehensive method to factorize quadratic equations of various difficulties is presented through an exercise of 25 selected problems followed by explanatory solutions...

Algebraic proof of Least value of sum of reciprocals for any number of positive variables

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The result of the least value of sum of reciprocals, used in Algebra represents an important principle of value sharing.

For two positive real variables $a$ and $b$ for example, if $a+b=1$, the least value of sum of reciprocals of $a$ and $b$ will occur when the value of the sum 1 is shared equally between the two variables. This remarkable equal value sharing principle hold true for any number of variables as well as for any positive sum of the variables...

How to solve difficult fraction comparison problems in a few steps

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Comparison of fractions is a tricky process, but if you learn to use patterns inherent in a set of fractions, you should be able to quickly solve even knotty fraction comparisons by applying a suitable method exploiting the pattern. In this session, as many as six different techniques of fraction comparison are presented with examples...

How to solve Clock problems

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In an analog clock the minutes hand moves faster than the hour hand and crosses it 22 times in a day, both going round and round over the clock face. In a typical clock problem, we are asked to find the angle between the two clock hands at a specific time...

How to find perfect and approximate square root of integers or decimals

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For most needs of finding square root, if we remember by heart squares and cubes of a few common integers, using quick factorization we can figure out the square roots or cube roots of numbers be it integer or decimal. But occasionally, in a few specific types of arithmetic problems, there is no other option than to actually evaluate the square root of a number following a method specifically tailored for the purpose...

Partially hidden Componendo dividendo revealed to solve a not so difficult Algebra problem in minimum time 7

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In a problem in which Componendo dividendo is partially hidden, expression with distinctive pattern exists, but its inseparable pair, the fraction of two variables involved in the distinctive expression is not visible. Uncovering and using Componendo dividendo in such a problem invariably gives you the solution faster than any other method. But revealing even partially hidden pattern is possible only when you look for it...

Hidden Componendo dividendo uncovered to solve difficult Algebra problems quickly 6

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Occasionally the distinctive expression and its inseparable companion fraction of two componendo dividendo variables, both exist in given and target expressions, but in hidden form. With a closer look and component expression substitution, once possibility of using componendo dividendo is revealed, the problem can be solved in mind quickly, faster than any other method. These form another important class of hidden componendo dividendo problems that need uncovering...

Componendo dividendo uncovered to solve difficult Algebra problems quickly 5

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Occasionally the distinctive expression appears in one of the expressions, but its inseparable pair, the fraction of the two variables, is not readily available in the problem statement. We say in these cases that componendo dividendo appears in partial form, and the problem is of hidden componendo dividendo type. We need to change the given expression to match the target expression in such cases...

Componendo dividendo adapted to solve difficult Algebra problems quickly 4

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When the problem appears in a form exactly suitable for applying the method, problem solving is trivial and very easy. The real challenge lies in problems where, there is some pattern similar to Componendo dividendo in the problem, but only partially, so at first glance we don't consider Componendo dividendo as the suitable approach...

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