# Percentages – Important Concepts and Shortcuts

## Percentages – Important Concepts and Shortcuts One of the most important subjects in bank exam is Quantitative Aptitude. So it is very important for the candidates to practice and score well in this section. Percentage is one such chapter which is related to the real world problems too. To solve the sums accurately there should be certain strategy and shortcut tricks so that the time is well managed during exams.

The Percentage is that arithmetic chapter which includes a lot of calculation so to be efficient and fast in calculation all one needs is practice. The previous year question papers give an idea about the type of questions and their difficulty level of the problems so that the aspirants become well acquainted with the problems based on percentages. In the written examination, sectional cutoff plays a vital role along with overall cutoff. Therefore, no sections can be taken lightly.

There are 2-3 questions which are likely to come in the exam and those are basic percentage problems mainly calculation oriented. Percentage basically means a fraction whose denominator is considered to be 100. If it is said that x percentage indicates x%.

Some concepts related to percentages are:

 x%  will be represented as a fraction: x% = x/100 ; 10% = 10/100(10 part out of 100

parts)

 x/y is to be assumed as a percentage:$\frac{x}{y}= \left ( \frac{x}{y}\times 100 \right )$ ;

 If there is an increase on the price of a good by R% then the consumption will be

reduced by not increasing the expenditure: $\left [ \frac{R}{100+R}\times 100 \right ]$  %

 If there is an decrease on the price of a good by R% then the consumption will be

more by not decreasing the expenditure: $\left [ \frac{R}{100-R}\times 100 \right ]$%

Population related problems: Population be P which increases at the rate of R% per

annum-

1. After n years the population will be-  $P\left ( 1+\frac{R}{100} \right )^{n}$

2. Before n years ago the population was- $\frac{P}{\left ( 1+\frac{R}{100} \right )^{n}}$

Depreciation related problems: The present value of a machine be P and the rate

depreciates at the rate of R%  per annum-

1. After n years the value of the machine will be- $P\left ( 1-\frac{R}{100} \right )^{n}$

2. Before n years the value of the machine was- $\frac{P}{\left ( 1- \frac{R}{100} \right )^{n}}$

There are certain tricks which can be explained through some examples:

 40% of Ram’s income is Rs. 1200. To find 75% of Ram’s income?

40% =Rs 1200

Trick:  $\frac{1200}{40}\times 75=2250$

 ⅓ part of Rams income

Trick: 40% means ⅖

$\frac{1200}{2}\times \frac{5}{3}=1000$

The problems related to percentage are best to solve using these tricks. Candidates can

themselves see that the time taken if solved by shortcut method or tricks is much less than the

time taken to solve it through the basic formula. As the exams like IBPS, the type of questions

are objective type so all they count is the ultimate result and if the number of question a

candidate attempts is more than the chances of getting better marks increases. Therefore

percentages should be solved using maximum tricks.