(**Part V****I**) consists of shortcut methods, tips and tricks to solve question from Numbers and Decimal Fraction, Logarithm and Height and Distance. Now (**Part VI****I**) includes another three very important section from quantitative aptitudes including *“**Percentage**”, “**Simplification”** and “**Ration and Proportion**”*.

Shortcut methods and tricks to solve questions from **‘****Percentage****‘** are given below:

__Percentage Concept__

When we say certain percentage, it means many hundredths. e.g. p percent means p hundredths, denoted by p% = (p/100) as fraction.

Thus 25% = (25/100) = (¼)

(x/y) as percent means (x/y) = {(x/y) x 100}%

Thus ¼ = (¼ x 100) = 25%

__Increase or decrease of percentage:__

When the price of a particular commodity is increased by A%, then to keep the amount of expenditure constant, the reduction in consumption required is:

[{A / (100 + A)} x 100]%

When the price of a particular commodity is decreased by A%, then to keep the amount of expenditure constant, the increase in consumption required is:

[{A / (100 – A)} x 100]%

__Results on Population:__

Consider that the population of a particular town is X as of now and suppose it increases at the rate of S% per annum, then:

1. Population after n years = X { 1 + (S / 100)}^{n}

2. Population n years ago = X / { 1 + (S / 100)}^{n}

__Results on Depreciation:__

Consider that the present value of a particular machine be Y. Suppose it depreciates at the rate of D% per annum.

Then:

1. Value of the machine after n years = Y { 1 – (D / 100)}^{n}

2. Value of the machine years ago = Y / { 1 – (D / 100)}^{n}

3. If X is P% more than Y, then Y is less than X by [{P / (100 + P)} x 100}] %

4. If X is P% less than Y, then Y is more than X by [{P / (100 – P)} x 100}] %

If there is always increase of I% and then decrease of D%, then there is always decrease or loss in that particular condition.

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Shortcut methods and tricks to solve questions from **‘Simplification’** are given below:

**‘B O D M A S’** is very important part of this section. Many problems could be solved by using this rule. This is explained below:

B = Brackets () {} []

O = Order power of √ ()2

D = Divide / ÷

M = Multiply * x

A = Addition +

S = Subtraction –

Most of the questions could easily be solved if you remember these easy rules.

__Sum Rules:__

(1+2+3+………+n) =1/2 [n(n+1)]

(1^{2}+2^{2}+3^{2}+………+n^{2}) =1/6 [n (n+1) (2n+1)]

(1^{3}+2^{3}+3^{3}+………+n^{3}) =1/4 [n^{2}(n+1)^{2}]

Arithmetic Progression (A.P.)

x, x+ a, x+ 2a, x+ 3a, ….are said to be in A.P. in which first term = x and common difference = a.

Let the nth term be t_{n} and last term = l, then

a) nth term = a + ( n – 1 ) d

b) Sum of n terms =n/2 [2a + (n-1)d]

c) Sum of n terms =n/2 [(a+l)] where l is the last term

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Shortcut methods and tricks to solve questions from **‘Ration and Proportion’** are given below:

The ratio of two quantities a and b in the same units, is the fraction a/b and is denoted by a:b where a is called ‘antecedent’, and b is called ‘consequent’.

The equality of two ratios is called proportion.

If a:b = p:q, then we can write a:b :: p:q, and we say that a, b, p and q are in the proportion where a and q are called extremes, b and p are called mean terms.

Product of means = product of extremes.

Thus, a:b :: p:q => (b*p) = (a*p)

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