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Home / Aptitude / Quantitative Aptitude Shortcut Tricks For IBPS Exam – Part VII

Quantitative Aptitude Shortcut Tricks For IBPS Exam – Part VII

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

Aptitude Tricks BODMAS Rule

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)]

(12+22+32+………+n2) =1/6 [n (n+1) (2n+1)]

(13+23+33+………+n3) =1/4 [n2(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 tn 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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                                                                                                                (……to be continued in Part VIII)

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