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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.


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.

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


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 Clock problems


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


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


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...

How to solve Surds part 3, Surd expression comparison and ranking


In this third tutorial on how to solve surd problems, we will take up explaining two related problem classes, namely, two-term surd expression comparison and ranking. First we will state and explain two important rich concepts used for quick solution and then apply the rich concepts in solving chosen SSC CGL level problems...


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