The topic Simplification is asked in every exam. In bank exams, Simplification covers almost 3040% of the Quantitative Aptitude section. Simplification also covers the percentage chapter. Simplification topic is asked to check the ability of students to find their ability to deal with the numbers. So, it’s important to have knowledge about simplification. Let’s try to understand Simplification
Simplification means to play with the questions by doing complex calculations. By using following of the tricks and tactics you can score good by saving your valuable time:
If we don’t omit ‘9’, then also the digital sum remains same.
Example: 293 = 2 + 9 + 3 = 14, 1 + 4 = 5 [answer remains same]
Let’s discuss one more example
b. 290100
c. 290140
d. 293990
Sol: We can find out the answer by option method without doing multiplication. This is only possible with the help of Digital sum.
Now, Digital sum, 326 ✕ 890 = (3 + 2 +6) ✕ ( 8 +9 + 0 )
⇒ 11 ✕ 17
⇒ (1+1) ✕ (1+7)
⇒ 2 ✕ 8 = 16
⇒ digital sum (16) = 7
Now find out the digital sum of the given options
Simplification means to play with the questions by doing complex calculations. By using following of the tricks and tactics you can score good by saving your valuable time:
 Learn BODMAS.
 Use the concept of digital sum.
 Memorize tables up to 30.
 Memorize cubes and squares of numbers up to 35.
 Learn tricks to find squares and cubes of numbers greater than 35.
 Learn tricks to find cube roots and square roots of large number.
 Learn the concept of percentages (conversion of fractions to percentage & percentage to fractions)
 Memorize the reciprocals.
Learn Fast Maths Tricks here
Rules of Simplification
V → Vinculum
B → Remove Brackets  in the order ( ) , { }, [ ]
O → Of
D → Division
M → Multiplication
A → Addition
S → Subtraction
⇒ 15✕(12)^{2} [brackets are solved first and table of 19 and 15 must be on tips]
⇒ 120+144 [must know the squares]
⇒ 264
Let’s discuss an example based on BODMAS 
Example: 152 × 2^{3} + (228 ÷ 19)^{2} =?
Sol:⇒ 15✕(12)^{2} [brackets are solved first and table of 19 and 15 must be on tips]
⇒ 120+144 [must know the squares]
⇒ 264
Important Parts of Simplification
 Digital Sum
 Number System
 Reciprocals
 HCF & LCM
 Percentages
 Square & Cube
 Approximation
 Fractions & Decimals
 Surds & Indices
Digital Sum
Digital sum is the sum obtained after adding all the digits of any given number successively. Example: 568 = 5+6+8 = 19, 1 + 9 = 10.
 Example: 6 ✕ 9 = 54, 5+4 = 9
 Example: 293 = 2 + 9 + 3 = 2 + 3 = 5 [ ‘9’ is omitted ]
If we don’t omit ‘9’, then also the digital sum remains same.
Example: 293 = 2 + 9 + 3 = 14, 1 + 4 = 5 [answer remains same]
Let’s discuss one more example
Example: 326 ✕ 890 =?
a. 291140b. 290100
c. 290140
d. 293990
Sol: We can find out the answer by option method without doing multiplication. This is only possible with the help of Digital sum.
Now, Digital sum, 326 ✕ 890 = (3 + 2 +6) ✕ ( 8 +9 + 0 )
⇒ 11 ✕ 17
⇒ (1+1) ✕ (1+7)
⇒ 2 ✕ 8 = 16
⇒ digital sum (16) = 7
Now find out the digital sum of the given options
 DS (291140) = 8
 DS (290100) = 3
 DS (290140) = 7
 DS (293990) = 5
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 
CoPrime:
 Two natural number a and b are said to be coprime 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 placesSum 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)
(2312)
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
could be write it as
x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)
Example:
On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ?
Number = 342k + 47
( 18 ✘19k ) + ( 18 ✘2 ) + 11
18 ✘( 19k + 2 ) +11.
Remainder = 11
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.)
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 t_{n} and last term = l, then
b) Sum of n terms = ^{n}/_{2} [2a + (n1)d]
c) Sum of n terms = ^{n}/_{2} (a+l) where l is the last term
Reciprocals
Up to 5, the reciprocals are easy to remember. But after 5 are explained as below 1/7 = 0.142857
 2/7 = 0.285714
 3/7 = 0.42857
 5/7 = 0.714285
 6/7 = 0.857142
 1/8 = 0.125
 2/8 = ¼ = 0.25
 3/8 = 3 × 1/8 = 0.375
 4/8 = ½ = 0.5
 5/8 = 4/8 + 1/8 = 0.5 + 0.125 = 0.625
 6/8 = ¾ = 0.75
 7/8 = 6/8 + 1/8 = 0.75 + 1.25 = 0.875
 1/9 = 0.1111…
 2/9 = 0.2222…
 3/9 = 0.3333…
 1/11 = 0.090909…
 2/11 = 0.181818…
 3/11 = 0.272727…
 10/11 = 0.909090…
 1/12 = ½ × 1/6 = ½ × 0.166666 = 0.083333…
 1/13 = 0.076923
 1/14 = 0.071428
 1/15 = 1/3 × 1/5 = 0.33333 × 0.2
 1/15 = 0.0666…
 1/16 = ½ × 1/8 = ½ × 0.125 = 0.0625
 1/17 = 0.058823
 1/18 = ½ × 1/9 = ½ × 0.111111 = 0.0555…
 1/19 = 0.052631
 1/20 = 0.05
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 = 2^{2}✘3^{3}, 288 = 2^{5}✘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
 Example:
H.C.F. of given numbers = 69

Let we have set of numbers.
Example: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. 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. 
Percentages
Percent implies “for every hundred”. % is read as percentage and y % is read as y per cent. Calculation of q % of z
 (q/100) ✕ z = (q ✕ z)/100
 q % of z = z % of q
Percentage – Ratio Equivalence:
Example:12.5% of 800
Sol: 1/8 ✕ 800 [12.5% = 1/8, Learn the ratio equivalence table]⇒100
Square & Cube
 Square & Cube
 Square Root & Cube Root
 Factorization Method
Perfect Square

NonPerfect Square

last digit is 1, 4, 9, 6, 5  last digit is 2, 3, 7, 8 
Square Root & Cube Root
Approximation
As the name suggests, if the given values are in points, then approximate the values to the nearest comfortable value so that there is not much effect on the final results.
Example: 150.4 ✕ 5.876
Sol: 1500 ✕ 6 =9000
Fractions & Decimals
On Basis  Explanation 

Decimal Fractions
 A number with a denominator of power of 10 is a decimal fractions. ^{1}/_{10}= 1 tenth; ^{1}/_{100}= 0.1;^{38}/_{100}=0.38 
Vulgar Fractions
 Conversion of 0.64(decimal number) into a Vulgar Fraction.First of all write the numeric digit 1 in the denominator of a number (like here 0.64) and add as many numeric zeros as the digit in the number after decimal point.After that removes the decimal point from the given number.At last step just reduce the fraction to its lowest terms. So, 0.64 = ^{64}/_{100}=^{16}/_{25};25.025 = ^{25025}/_{1000} = ^{1001}/_{4} 
Operations  Addition & Subtraction To perform the addition and subtraction of a decimal fraction could be done through placing them right under each other that the decimal points lie in one column. 3.424+3.28+.4036+6.2+.8+4 3. 424 3. 28 . 4036 6. 2 . 8 +4______ 18. 1076 Multiplication of a Decimal Fraction To find the multiplication of decimal fraction , first of all you need to remove the decimal point from the given numbers and then perform the multiplication after that assign the decimal point as many places after the number as the sum of the number of the decimal places in the given number. Step 1. 0.06*0.3*0.40 Step 2. 6*3*40=720 Step 3. 0.00720 Multiplication of a decimal fraction by power of 10 A multiplication of a decimal fraction by a power of 10 can be perform through shifting the decimal point towards right as many places as is the power of 10. like 45.6288*100=45628.8, 0.00452*100=0.452 Division 
Comparison of Fractions  To compare the set of fractions numbers,first of all you need to convert each fraction number or value into a equal decimal value and then it will be became easy for you to assign them ( the numbers or value) in a particular way( ascending or descending order). ^{3}/5,^{4}/7,^{8}/9 and ^{9}/11 Arranging in Ascending Order ^{3}/5= 0.6, ^{4}/7 = 0.571, ^{8}/9 = 0.88, ^{9}/11 = 0.818. Now, 0.88 > 0.818 > 0.6 > 0.571 ^{8}/9>^{9}/11>^{3}/5>^{4}/7 
Recurring Decimal  Recurring Decimal A decimal number in which after a decimal point a number or set of number are repeated again and again are called recurring decimal numbers.It can be written in shorten form by placing a bar or line above the numbers which has repeated. A decimal number in which all digits are repeated after a decimal point. A decimal number in which certain digits are repeated only. 
Surds & Indices
 Some Rules of Indices
 Some Rules of Surds
Now let’s discuss some questions based on the above tricks:
1. (9.979)^{3} – (23.99)^{2} + (1.99)^{5} =?
Sol: Using the concept of approximation,⇒ 10^{3} – 24^{2} + 2^{5} = 1000 – 576 – 32
⇒ 456
2. 25% of 2500 + 16.66%% of 144 =?
Sol: 25/100 ✕ 2500 + 1/6 ✕ 144 [16.66% =1/6]⇒ 25^{2} + 24 = 625 + 24
⇒ 649
3. ∛7.938 x∛999.9998– 4.9256 =?
Sol: Using the concept of approximation,⇒ 2✕105 [must know the cubes, then you can also find out cube roots & squareroots easily]
⇒ 15
4. (125)^{1.3} × (25)^{1.8} ÷ (5)^{0.5} = (5)^{?}
Sol: (5)^{3×1.3} ✕ (5)^{2×1.8}(5)^{0.5} = (5)^{?}⇒ (5)^{3.9} ✕ (5)^{3.6} ÷ (5)^{0.5} = (5)^{?}
⇒ (5)^{3.9+3.60.5} = (5)^{?}
⇒ ? = 7.5 – 0.5 = 7
5. 13 × 2 (65 – 15) + 168 ÷ 2 ÷ 2 (11)2 =?
Sol: Using the concept of BODMAS,⇒ 13 ✕ 2 ✕ 50 + 168/2 2 – 121 =?
⇒ 1300 + 84/2 – 121 =?
⇒ 1300 + 42 – 121
⇒ 1221
Use the above tricks with accuracy in order to increase your speed. So, practice as much as you can. Set a timer and make sure yourself that you have to attempt 1012 questions within 45 minutes while practising and if you get failed in doing this, then practice again and again.
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