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Basics

Basic concepts on a subject form the foundation. Basics are for building the foundational concepts in concise but comprehensive manner.

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 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...

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...

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