Simplification Techniques and Tricks

Simplification Techniques and Tricks

 Simplification is one of the most important part of Quantitative Aptitude section of any competitive exam.

Rules of Simplification

V → Vinculum

B → Remove Brackets - in the order ( ) , { }, [ ] 

O → Of

D → Division

M → Multiplication

A → Addition

S → Subtraction

Important Parts of Simplification

  • Number System
  • HCF & LCM
  • Square & Cube
  • Fractions & Decimals
  • Surds & Indices

 Number System

  • Classification
  • Divisibility Test
  • Division& Remainder Rules
  • Sum Rules

  Classification 

Types

Description

Natural Numbers:

all counting numbers ( 1,2,3,4,5....∞)

Whole Numbers:

natural number + zero( 0,1,2,3,4,5...∞)

Integers:

All whole numbers including Negative number + Positive number(∞......-4,-3,-2,-1,0,1,2,3,4,5....∞)

Even & Odd Numbers :

All whole number divisible by 2 is Even (0,2,4,6,8,10,12.....∞) and which does not divide by 2 are Odd (1,3,5,7,9,11,13,15,17,19....∞)

Prime Numbers:

It can be positive or negative except 1, if the number is not divisible by any number except the number itself.(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61....∞)

Composite Numbers:

Natural numbers which are not prime

Co-Prime:

Two natural number a and b are said to be co-prime if their HCF is 1.

 Divisibility

Numbers

IF A Number

Examples

Divisible by 2

End with 0,2,4,6,8 are divisible by 2

254,326,3546,4718 all are divisible by 2

Divisible by 3

Sum of its digits  is divisible by 3

375,4251,78123 all are divisible by 3.  [549=5+4+9][5+4+9=18]18 is divisible by 3  hence 549 is divisible by 3.

Divisible by 4

Last two digit divisible by 4

5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4.

Divisible by 5

Ends with 0 or 5

225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5.

Divisible by 6

Divides by Both 2 & 3

4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6.

Divisible by 8

Last 3 digit divide by 8

746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8.

Divisible by 10

End with 0

220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10.

Divisible by 11

[Sum of its digit in
 odd places-Sum of its digits
in even places]= 0 or multiple of 11

Consider the number 39798847

 (Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3)

(23-12)

23-12=11, which is divisible by 11. So 39798847 is divisible by 11.

Division & Remainder Rules

Suppose we divide 45 by 6

 hence ,represent it as:

dividend = ( divisor✘quotient ) + remainder

or

divisior= [(dividend)-(remainder] / quotient

 Sum Rules

(1+2+3+.........+n) = 1/n(n+1)

(12+22+32+.........+n2) = 1/n (n+1) (2n+1)

(13+23+33+.........+n3) = 1/4 n2 (n+1)2

Arithmetic Progression (A.P.)

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

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

 H.C.F. & L.C.M.

  • Factorization & Division Method
  • HCF & LCM of Fractions & Decimal Fractions

 Methods

 

On Basis

H.C.F. or G.C.M

L.C.M.

Factorization Method

Write  each number as the product of the prime factors. The product of least powers of common prime factors gives H.C.F.
Example:
Find the H.C.F. of 108, 288 and 360.

108 = 22✘33, 288 = 25✘32 and 360 = 23✘5✘32

H.C.F. = 22✘32=36

Write each numbers into a product
of prime factors. Then, L.C.M is
the product of highest powers of
 all the factors.
Examples:
Find the L.C.M. of 72, 108 and 2100.
72=23✘32,108=33✘22,
2100=22✘52✘3✘7.
L.C.M.=23✘33✘52✘7=37800

Division Method

Let we have two numbers .Pick the smaller one and divide it by the larger one. After that divide the divisor with the remainder. This process of dividing the preceding number by the remainder will repeated until we got the zero  as remainder.The last divisor is the required H.C.F.
Example:

 

H.C.F. of given numbers = 69

Let we have set of numbers.
First of all find the number
 which divide at least two of
the number in a given set of
 number.remainder and
not divisible numbers
will carry forward as it is.
Repeat the process till
at least  two number is
not divisible by any number
except 1.The product of
the divisor and the
undivided numbers is the required
L.C.M.

Example:
Find the L.C.M. of 12,36,48,72

H.C.F. & L.C.M. of Fractions

H.C.F. =  H.C.F. of Numerator / L.C.M. of Denominators

L.C.M. = L.C.M. of Numerator /H.C.F. of Denominators

Product of H.C.F. & L.C.M.

H.C.F * L.C.M. = product of two numbers

Decimal numbers

H.C.F. of Decimal numbers
Step 1. Find the HCF of the given
numbers without decimal.
Step 2.Put the decimal point ( in the
HCF of Step 1) from right to left
according to the MAXIMUM
 deciaml places among the given numbers.

L.C.M. of Decimal numbers
Step 1. Find the LCM of the given
numbers without decimal.
Step 2.Put the decimal point ( in the
LCM of Step 1) from right to left
according to the MINIMUM
 deciaml places among the given numbers.

 

 

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