Basic Rules of Simplification
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BODMAS Rule
Easiest way to choose simplification questions:
STEP 1: Know about BODMAS Rule. Following are the list of priority given for brackets and signs.
STEP 2: If an expression Contains brackets, the expression within thebrackets should be simplified first.
STEP 3: If an expression contains ‘Of’, multiplication, division, addition and subtraction, thenof should be performed first then followed by multiplication or division.
Proceeding from left to right, addition and subtraction are carried out in the order in which the sign of addition and subtraction are given.
If expression contains‘Of’ and Division – always do ‘Of’ and then do division
STEP 4: If expression involves all thefouroperations,then multiplication and divisionis carried outfirst in the order in which they are given from left to right. The same rules are carried out for addition and 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 
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)
2312=11, which is divisible by 11. So 39798847 is divisible by 11. 
Division & Remainder Rules
Sum Rules
Arithmetic Progression (A.P.)
a) nth term = a + ( n – 1 ) d
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
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

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. 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: 
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 decimal places among the given numbers. 
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
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. 1076Multiplication 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. Pure Recurring Decimal Mixed Recurring Decimal 
Surds & Indices
 Some Rules of Indices
 Some Rules of Surds
Learnsquares and cubes of number:
Squares(1^{2} to 30^{2}):
 1^{2 }– 1
 2^{2 }– 4
 3^{2 }– 9
 4^{2 }– 16
 5^{2 }– 25
 6^{2}– 36
 7^{2}– 49
 8^{2}– 64
 9^{2}– 81
 10^{2}100
 11^{2 }–121
 12^{2}144
 13^{2}– 169
 14^{2}– 196
 15^{2}– 225
 16^{2}– 256
 17^{2}– 289
 18^{2}– 324
 19^{2}– 361
 20^{2}– 400
 21^{2}– 441
 22^{2}– 484
 23^{2}– 529
 24^{2}– 576
 25^{2}– 625
 26^{2}– 676
 27^{2}– 729
 28^{2}– 784
 29^{2}– 841
 30^{2}– 900
Cubes (1^{3}to 15^{3}):
 1^{3}– 1
 2^{3}– 8
 3^{3}– 27
 4^{3}– 64
 5^{3}– 125
 6^{3}– 216
 7^{3}– 343
 8^{3}– 512
 9^{3}– 729
 10^{3}– 1000
 11^{3}– 1331
 12^{3}– 1728
 13^{3}– 2197
 14^{3}– 2477
 15^{3}– 3375
Fast and Easy Method to Take Square Root Math Tricks:
Fast and Easy Method to Take Square Root is given here, candidates those who are preparing for banking and all other competitive exams can use this trick in the examination to save time. We have given the link to download in PDF.
POINTS TO REMEMBER:
 When 2^{2}= 4, then √4 = 2
 Here 4 is thesquareof 2
 2 is thesquare rootof 4
 ASquareof a number cannever end with2, 3, 7and 8
Table 1:
One’s digit of a square  One’s digit of the square root 
1  1 or 9 
4  2 or 8 
5  5 
6  4 or 6 
9  7 or 3 
To find the square of a number which is amultiple of ‘5’
25^{2} = [2×3] 5^{2}
= [6] 25 = 625
i.e., AB^{2}whereB=5
AB^{2}= [A× next number] B^{2}
For example, 85^{2}= [8×9] 25 =7225
115^{2}= [11×12] 25 =13225
155^{2}= [15×16] 25 =24025
This method can be followed for all numbers divisible by 5
TYPE 1:
To find the square root of a 3digit number
EXAMPLE:√841
STEP 1: Consider the one’s digit of the given number i.e., 1
From Table 1, if the one’s digit of the square is ‘1’ then the square root would either end with‘1’ or‘9’
STEP 2:Alwaysignore the ten’s digit of the given number
STEP 3:Now the remaining number other than the one’s and the ten’s digit in the given number is‘8’
Consider asquareroot of a square which isnearer to as well aslesser than ‘8’.
Here it is ‘4’ which is nearer to as well as lesser than ‘8’. Hence the square root of 4 i.e.,‘2’is taken
STEP 4: we already know the one’s digit of the square root to be either 1 or 9 from STEP 1
Therefore the square root of ‘841’ lies between21and29
STEP 5:
Take a numberdivisible by ‘5’ between 21 and 29, that is‘25’
25^{2}= [2×3] 25 =625
Now625< 841
25^{2}is itself lesser than 841. Then21^{2}will bemuch lesser than841.
Therefore, the remaining option is ‘29’
√841 = 29
TYPE 2:
To find the square root of a 4digit number
EXAMPLE:√8464
STEP 1: Consider the one’s digit of the given number i.e., 4
From Table 1, if the one’s digit of the square is ‘4’ then the square root would either end with‘2’ or‘8’
STEP 2:Alwaysignore the ten’s digit of the given number
STEP 3:Now the remaining numbers other than the one’s and the ten’s digit in the given number is‘84’
Consider asquareroot of a square which isnearer to as well aslesser than ‘84’.
Here it is‘81’ which is nearer to as well as lesser than ‘84’. Hence the square root of 81 i.e.,‘9’ is taken
STEP 4: we already know the one’s digit of the square root to be either 2 or 8 from STEP 1
Therefore the square root of ‘8464’ lies between92and98
STEP 5:
Take a numberdivisible by ‘5’ between 92 and 98, that is‘95’
95^{2}= [9×10] 25 =9025
Now9025> 8464
95^{2}is itself greater than 8464. Then98^{2}will bemuch greater than8464
Therefore, the remaining option is ‘92’
√8464 = 92
TYPE 4:
To find the square root of a 5digit number
EXAMPLE:√18769
STEP 1: Consider the one’s digit of the given number i.e., 9
From Table 1, if the one’s digit of the square is ‘9’ then the square root would either end with‘3’ or‘7’
STEP 2:Alwaysignore the ten’s digit of the given number
STEP 3:Now the remaining numbers other than the one’s and the ten’s digit in the given number is‘187’
Consider asquareroot of a square which isnearer to as well aslesser than ‘187’
Here it is‘169’ which is nearer to as well as lesser than 187. Hence the square root of 169 i.e.,‘13’ is taken
STEP 4: we already know the one’s digit of the square root to be either 3 or 7 from STEP 1
Therefore the square root of ‘18769’ lies between133and137
STEP 5:
Take a numberdivisible by ‘5’ between 133 and 137, that is‘135’
135^{2}= [13×14] 25 =18225
Now18225< 18769
135^{2}is itself smaller than 18769. Then133^{2}will bemuch lesser than18769
Therefore, the remaining option is ‘137’
√18769 = 137
Shortcut Tricks to find the Square of Numbers
Simple Shortcut Tricks to find the Square of Numbers is given here, which was very helpful for the speedy calculations. Candidates those who are preparing for banking and all other competitive exams can use also download this in PDF.
SQUARE OF NUMBERS
 Square of a number is the product obtained by multiplying a number by itself.
2×2 = 4
11×11 = 121
 To find the square of two digit numbers(1099)we can consider the following steps.
1) Let the 2digit number be = AB
2) Now to find AB^{2}
3) Unit digit of square = B^{2}
4) Ten’s digit of square = 2×A×B (+ Carry if any from the previous step)
5) The rest of the digits of square = A^{2}(+ Carry if any from the previous step)
FOR EXAMPLE
To find of Square of 67,
67^{2} = ?
AB^{2} = 67^{2}
STEP 1:
B^{2}^{=}7^{2}=49
From Step 1, here “9” is the unit digit and “4” is carry
à 67^{2}= _ _ _ 9
STEP 2:
[2×A×B]+ Carry from previous step i.e.,4
2×A×B = 2×6×7 = 84
Add Carry ‘4’ with the above ‘84’ we get [84+4=88]
Therefore at the end of the 2^{nd} Step
Here “8” is the unit digit and “8” is carry
à67^{2}= _ _ 8 9
STEP 3:
A^{2}= 6^{2} = 36
There is a carry of ‘8’ from the previous step
Therefore, [36+8] = 44
The final answer is67^{2}=4489
 To find the square of a number, which is amultiple of ‘5’
AB^{2}= [A× next number] B^{2}
25^{2} = [2×3] 5^{2}
= [6] 25 = 625
i.e., AB^{2}whereB=5
AB^{2}= [A× next number] B^{2}
For example, 85^{2}= [8×9] 25 =7225
115^{2}= [11×12] 25 =13225
155^{2}= [15×16] 25 =24025
This method can be followed for all numbers divisible by 5
Fast and Easy Method to Take Cube Root Math Tricks:
Fast and Easy Method to Take Cube Root is given here, candidates those who are preparing for banking and all other competitive exams can use this trick in the examination to save time.
POINTS TO REMEMBER:
 When 2^{3}= 8, then (8)^{1/3}= 2
 Here 8 is thecubeof 2
 2 is thecube rootof 8
Table 1:
One’s digit of a cube  One’s digit of the cube root 
1  1 
2  8 
3  7 
4  4 
5  5 
6  6 
7  3 
8  2 
9  9 
0  0 
TYPE 1:
To find the cube root of a 4digit number
EXAMPLE:(9261)^{1/3}
STEP 1: Consider the one’s digit of the given number i.e.,1
From Table 1, if the one’s digit of the cube is ‘1’ then the cube root would also end with‘1’
STEP 2:Alwaysignore the ten’sand the hundredth digits of the given number i.e., ignore‘2’ and‘6’ in the given number
STEP 3:Now the remaining number other than the one’s, ten’s and the hundredth digit in the given number is‘9’
Consider acuberoot of a cube which isnearer to as well aslesser than ‘9’.
Here it is ‘8’ which is nearer to as well as lesser than ‘9’. Hence the cube root of 8 i.e.,‘2’ is taken
STEP 4: we already know the one’s digit of the cube root is1 from STEP 1
Therefore the cube root of ‘9261’ is21
Therefore(9261)^{1/3} = 21
TYPE 2:
To find the cube root of a 5digit number
EXAMPLE:(32768)^{1/3}
STEP 1: Consider the one’s digit of the given number i.e., 8
From Table 1, if the one’s digit of the cube is ‘8’ then the cube root would end with‘2’
STEP 2:Alwaysignore the ten’s and the hundredth digitsof the given number i.e., ‘6’ and ‘7’ here
STEP 3:Now the remaining numbers other than the one’s, ten’s and the hundredth digits in the given number is‘32’
Consider acuberoot of a cube which isnearer to as well aslesser than ‘32’.
Here it is‘27’ which is nearer to as well as lesser than ‘32’. Hence the cube root of 27 i.e.,‘3’ is taken
STEP 4: we already know the one’s digit of the cube root to be2 from STEP 1
Therefore the cube root of ‘32768’ is32
Therefore(32768)^{1/3} = 32
Example 1: 21^{2} / 49 × 6
Solution:From the above question if we know the square value of 21^{2}, then this question will be easily solved
STEP 1:21^{2}= 441
STEP 2:441/49= 9
STEP 3:9×6 = 54
STEP 4:Hence the answer for above series is54
2) REMEMBER FREQUENTLY ASKED FRACTION VALUES
 5% = 0.05
 6 ¼ % = 0.0625
 10% = 0.1
 12 ½ = 0.125
 16 × (2/3)% = 0.166
 20 % = 0.2
 25 % = 0.25
 33 × (1/3)%= 0.33
 40 % = 0.4
 50% = 0.5
 60% = 0.6
 66 × (2/3) =0.66
 75 %= 0.75
 80 %= 0.8
 90 % = 0.9
 100% = 1
 125 % = 1.25
 150% = 1.5
 200 % = 2
 250 % =2.5
EXAMPLE 2: 60% of 250 +25% of 600
STEP 1: Know the values of 60% =0.6 and 25 % = 0.25
STEP 2: Now directly multiply 0.6×250 + 0.25×600
STEP 3:0.6×250= 150
0.25×600=150
STEP 4: 150+ 150 = 300
STEP 5:Hence the answer for above series is300
3) Solve mixed fraction – Multiplication
EXAMPLE 3: 2×(3/5) × 8×(1/3) + 7 ½ × 2×(2/3)
STEP 1:2×(3/5) × 8×(1/3) = (13/5) × (25/3) = 65/3
STEP 2: + 7 ½ ×2×(2/3)= 43/6 × 12/5 = 86/5
STEP 3:65/3 + 86/5 = 38×(15/13)
STEP 4:hence the answer for above series is38×(15/13)
4) Solve Mixed Fraction addition
Example 4:19×(3/5) + 23×(2/3) – 24×(1/5)
STEP 1:Take all the whole number outside the bracket i.e. 19+23 24 = 18
STEP 2:Add fractions within bracket 18×[(3/5) + (2/3) – (1/5)] = 18(16/15)
STEP 3: Hence the answer for above series is18(16/15)
Example 5: (?)^{2}+18×12= 6^{2}×5×2
STEP 1:Multiply 18 × 12 = 216
STEP 2:Square of 6 = 36
STEP 3:Multiply 36 ×5×2= 360
STEP 4:(X)^{2 }+216 = 360
STEP 5:(X)^{2} = 360216 = 144
STEP 6:Therefore X = 12
Multiplication Tricks – Find solution within 20 seconds
Today I am going to share quick multiplication tricks that will help you find the solution within 20 seconds. Must read – 10 Coolest Maths tricks
Multiply by 9,99,999,etc…
Multiply by 125
Step 1. Each time you just need to pick 125 multiply it by 8 will get 1000
Multiply two digits numbers ending in 1
Multiply numbers between 11 and 19
Multiply two digit number by 11
Step 3: 649
Multiply by 5
Step 2: Multiply the result from Step 1 by 10 : 617*10=6170
Multiply by 25
18*25=450
Step 1: Divide the number by 4:18/4
Step 2: Multiply the number from Step 1 by 100: 4.5 * 100 = 450
Multiply by 9
Factorization
Step 5:7*500=3500
Some Special type
When sum of unit digit is 10 and remaining digit is same.
43×47 = 4 × ( 4+1 ) / 3×7
Ans = 202172×78 = 7×8 / 2×8
= ^{56}/16
Ans = 5616104 × 106 = 10× ( 10+1 ) / 4×6
= 10 × ^{11}/24
= ^{110}/24
Ans = 11024
When sum of ten’s digit is 10 and unit digit is same
= ( 4×6 ) +6 / 6×6
= 24 + ^{6}/36
= ^{30}/ 36
Ans = 303683 × 23
= ( 8×2 ) +3 / 3×3
= ^{19}/09
Ans = 190992 × 12
= ( 9×1 ) + 2/2 × 2
= ^{11}/04
Ans = 1104
When unit digit is 5 in both the numbers and difference between both number is 10.
75 × 65
= 6 × ( 7+1 ) / 75
= ^{48}/75
Ans = 4875
45 × 35
= 3 × ( 4+1 )/75
= 15 /75
Ans = 1575
105 × 95
= 9 × ( 10+1 ) / 75
= 99/75
Ans = 9975
Find the unit digit of 147^{128 }* 138^{148} ?
Multiply by 11,111,1111….so on
Sol:
No of digits in multiplier = 9
Write in ascending order from left side like this:
987654321
and now 91=8
write it in descending order just after it
12345678
now you will get like this:
12345678987654321
hence
111111111✘111111111 = 12345678987654321
Ques 2. 1111111111 ✘ 1111111111 = ?
Sol:
No of digits in multiplier =10
Write in ascending order from left side like this:
10 9 8 7 6 5 4 3 2 1
and now 101=9
write it in descending order just after it
1 2 3 4 5 6 7 8 9
and after it just add the carry
1 2 3 4 5 6 7 8/ 9/ 10 9 8 7 6 5 4 3 2 1
8+1/ 9+1 / 0
1 2 3 4 5 6 7 9 0 0 9 8 7 6 5 4 3 2 1
now you will get like this:
1234567900987654321
hence
1111111111✘1111111111 = 1234567900987654321
Ques 3. 1111111✘2222222 = ?
Sol:
No of digit in the multiplier is 7 then let n=7;
Now Just multiply the digit 2 from 1 to 7 time & arrange them from extreme left to right in ascending order,you will get like this:
14 12 10 8 6 4 2
and now just subtract one from n.like this n=7,so n1=6.
Multiply the digit 2 from 1 to 6 time & arrange them from just right after it,you will get like this:
2 4 6 8 10 12
Now placing both outcome like this & add the carry
2 4 6 8 10 12 14 12 10 8 6 4 2
8+1/0+1/2+1/4+1/2+1
You will get the answer:
2 4 6 9 1 3 5 3 0 8 6 4 2
Ques 4. 1111111✘5555555 = ?
Sol:
No of digit = 7
Now Just multiply the digit 5 from 1 to 7 time & arrange them from extreme left to right in ascending order,you will get like this:
35 30 25 20 15 10 5
Just right after it perform same action but in descending order & till 6 times only.like this:
5 10 15 20 25 30
Now placing together ,just add the carry
5 10 15 20 25 30 35 30 25 20 15 10 5
6 1 7 2 8 3 8 2 7 1 6 0 5
1111111✘5555555=6172838271605
Tricks to find Square Root and Cube Roots
# Division Method
 Step 1. Make Pair of digits of given number from left to right
 Step 2. Pick first pair, like here 6 find the square which is equals to 6 or less than it.Like 2
 Step 3. So Place it to in the section of Quotient as well as in the divisor.
 Step 4: then subtract from square of no which is equals to 6 or less than it with 6
 Step 5. Now comes to second pair bring it down like here 40 ,double the quotient like 2 = 4 and write the result on the left of 240 .It is just like division.Now repeat From Step 2 until you got the remainder zero.
# Prime Factor Method
# Square Root of a Decimal Fraction
 Step 1. Make the pair of integer part first.
 Step 2. Now find whether the decimal part is odd or even if it is odd then make it odd by placing at the end of it zero.
 Now just find the square root by the division method as discussed above and don’t forget to put the decimal point in the square root as the integer part is over.
# Method of Finding Cube Root of Perfect Cube
Fast Math and Shortcut Rules for Multiplication:
Useful Shortcut rules for Multiplication which was more helpful in the fast math to solve the aptitude questions were given below. Candidates those who are preparing for the banking and all competitive exams can use this
Type1: If the unit figure is same and the sum of the tens figure is 10, then follow the below method.
General Shortcut Method:
[Tens fig. × Tens fig. + Unit fig.] [Unit fig × Unit fig]Example:
86 × 26 = [8 × 2 + 6] [6 × 6] = [22] [36]; so answer is: 2236.
Note: Here, the unit figure denotes the number that present in the ones digit (6, 6), and tens figure denotes the number that present in the tens digit (8, 2).
Type2: If the sum of the unit figure is 5 and the tens figure are equal. Then follow the below method.
General Shortcut Method:
[(Tens figure)^{2}+ ½ × Tens figure] [Unit fig. × Unit fig]Example:
83 × 82 = [8^{2}+ ½ × 8] [ 3 × 2 ] = [68] [06] So the answer is: 6806
Type3: If the unit figures are same and the sum of tens figures is 5.
General Shortcut Method:
[Tens fig × Tens fig + ½ × Unit fig] [(Unit fig.)^{2}]Example:
36 × 26 = [3 × 2 + ½ × 6] [6^{2}]
= [9] [36]; Answer is 936.
Type4: If the unit figures are 5 and difference between the tens figures is 1 then the rule is,
General Shortcut Method:
[(Larger tens fig + 1) × (Smaller tens fig)] [75]Example:
35 × 45 = [(4 + 1) × 3] [75]
= [15] [75]; So the Answer is, 1575.
More Shortcuts Will be Updated Soon.
Fast Math and Shortcut Rules for Division:
Useful Shortcut rules for Division which was more helpful in the fast math to solve the aptitude questions were given below. Candidates those who are preparing for the banking and all competitive exams can use this.
1.)DIVISIBLE BY 2:
A number will be divisible by 2, if the unit digit in the number is 0, 2, 4, 6 and 8.
Example: Numbers like, 56456, 32658, 89846 are divisible by 2.
2.)DIVISIBLE BY 4:
A number will be divisible by 4, if the last two digits of the number is divisible by 4.
Example: Numbers like 56536 is divisible by 4, because the last two digits of this number is divisible by 4 and the number 546642 is not divisible by 4 because the last two digits of this number is not divisible by 4.
3.)DIVISIBLE BY 6:
A number will be divisible by 6, if that number is divisible by both 2 and 3.
Example: 36 is divisible by 6 because 36 is divisible by both 2 and 3.
4.)DIVISIBLE BY 8:
A number will be divisible by 8, if the last three digits of that number are divisible by 8.
Example: 565144 is divisible by 8 because the last three digits 144 is divisible by 8. And the number 554314 is not divisible by 8 because the last three digits 314 is not divisible by 8.
5.)DIVISIBLE BY 5:
A number will be divisible by 5 if the unit digit is either 0 or 5.
Example: Numbers like 565520 and 898935 are divisible by 5.
6.)DIVISIBLE BY 3:
A number will be divisible by 3, if the sum of the digits in the number is divisible by 3.
Example: 658452 is divisible by 3 because the sum of the numbers is divisible by 3, 6+5+8+4+5+2= 30, which is divisible by 3.
The number 456455 is not divisible by 3, because the sum of the number is not divisible by 3. 4+5+6+4+5+5= 29 this is not divisible by 29.
7.)DIVISIBLE BY 9:
A number will be divisible by 9, if the sum of the digits in the number is divisible by 9.
Example: 898686 is divisible by 9 because the sum of the numbers is divisible by 9,
8+9+8+6+8+6= 45, this is divisible by 9.
8.)DIVISIBLE BY 11:
A number will be divisible by 11, if the difference of the sum of the digits in the Odd places and Sum of the digits in the Even places, is either zero or divisible by 11.
Example: 502678 is divisible by 11 because, the sum of the digits of the odd places, 5+2+7= 14, sum of the digits in the even places, 0+6+8=14, the difference is 1414=0, so this number is divisible by 11.
9.)DIVISIBLE BY 12:
A number is divisible by 12, if the number is divisible by both 3 and 4.
Example: 144 is divisible by 12, because it is divisible by both 3 and 4.
10.)DIVISIBLE BY 10:
Any number that ends with zero will be divisible by 10.
Fast Math and Shortcut Rules for Subtraction RuleI:
Useful Shortcut rules for Subtraction which was more helpful in the fast math to solve the aptitude questions were given below. Candidates those who are preparing for the banking and all competitive exams can use this.
RuleI:Borrowing and Paying Back Method:
This method is the quickest method of subtraction. This method is also called equal additions method.
Example (1): Suppose we have to subtract 55 from 91. Mentally we have to increase the number to be subtracted to the nearest multiple of 10 i.e., increase 55 to 60 by adding 5 to it. Mentally increase the other quantity by the same amount i.e., by 5. Therefore, the problem is 96 minus 60 i.e., our answer is 96 – 60= 36.
Example (2): Sometimes it is useful to increase the number to be subtracted to the nearest multiple of 100 for example 442 – 179. Therefore 179 becomes 200 by adding 21 and 442 becomes 463 by adding 21. Then the problem becomes 463 – 200= 263. Now we see that 463 – 200 is easier than 442 – 179. The result is same as 263.
Example (3): Another example is 2326 – 1875. Here 1875 becomes 2000 by adding 125 and 2326 becomes 2451 by adding 125. The number becomes 2451 – 2000= 451. Here the subtraction 2451 – 2000 is easier than the subtraction 2326 – 1875. The answer of both is same 451.
Example (4): The subtraction of 3786 – 2998. Here 2998 becomes 3000 by adding 2 and 3786 becomes 3788 by adding 2. The problem of 3788 – 3000 is easier than 3786 – 2998 and our answer is 788. This answer is same for both the problems.
Fast Math and Shortcut Rules for Subtraction RuleII:
Useful Shortcut rules for Subtraction which was more helpful in the fast math to solve the aptitude questions were given below. Candidates those who are preparing for the banking and all competitive exams can use this.
Rule: II. Double Column Addition and Subtraction Method:
This following method is works when there is a series of additions and subtractions are to be performed in a line.
Example (1):
1026
– 4572
+ 5263
– 2763
+ 8294
_____________
Explanation: We have to look the signs given before the numbers and then start adding and adding and subtracting from the top right position.
Step I: First Double Column
26 – 72= – 46, – 46 + 63= 17,
17 – 63= – 46, – 46 + 94= 48.
Answer is 48à StepI
Step II: Second Double Column
10 – 45= – 35, – 35 + 52= 17,
17 – 27= – 10, – 10 + 82= 72.
Answer is 72à StepII
Now Combine the Step II and Step I.
Answer will be 7248.
You will get the same answer if you also use the normal method.
Example (2):
7676
– 1431
+ 5276
– 3489
+ 1546
_____________
Explanation: We have to look the signs given before the numbers and then start adding and adding and subtracting from the top right position.
Step I: First Double Column
76 – 31= 45, 45 + 76= 121 (here in the 121 take the last two digits from 121 i.e., 21)
21 – 89= – 68, – 68 + 46= 22 (Here the answer comes in minus so add 100 with the answer)
100 + ( 22) = 78.
Answer is 78àStep I
Step II: Second Double Column
76 – 14= 62, 62 + 52= 114 (Take the last two digits from 114 i.e., 14)
14 – 34= – 20, – 20 + 15= – 5 (Here the answer comes in minus so add 100 with the answer)
100 + (5) = 95
Answer is 95àStep II.
Now Combine the Step II and Step I.
Answer will be 9578.
To Solve Modulus of a Real Number
Tips to Crack Approximation
4433.764 = 4434
2211.993 = 2212
1133.667 = 1134
3377.442 = 3377
Now, simplify 530 x 20% + 225 x 17%
= 106 + 38.25 = 144.25
 To simplify a square root, you can follow these steps:
 Factor the number inside the square root sign.
 If a factor appears twice, cross out both and write the factor one time to the left of the square root sign. If the factor appears three times, cross out two of the factors and write the factor outside the sign, and leave the third factor inside the sign. Note: If a factor appears 4, 6, 8, etc. times, this counts as 2, 3, and 4 pairs, respectively.
 Multiply the numbers outside the sign.
 Multiply the numbers left inside the sign.
 To simplify the square root of a fraction, simplify the numerator and simplify the denominator.
Simplification / Approximation: Tips and Tricks
Unit Digits and its applications
Digit Sum
How to calculate Square Root?
Perfect Square
If the square ends in

1

4

5

6

9

0

The number would end in

1,9

2,8

5

4,6

3,7

0

How to calculate Cube root?
If the cube ends in

1

2

3

4

5

6

7

8

9

0

The number would end in

1

8

7

4

5

6

3

2

9

0

Approximation Techniques for speed calculations
While dealing with calculation intensive sections like Quant and DI, it is very important to pick the right questions use the best methods. As we grow, we tend to get habituated in our day to day calculations and employ conventional methods without thinking whether it is the best for a given scenario. Many a time there will be an approach that is much easier than the conventional methods to solve a given problem. Most of the so called calculation intensive questions are not that scary if we think a bit before solving them. As a rule of thumb, always spend few seconds to identify the best approach before start solving.
Some useful methods are given below which can help in our calculations.
To approximate Actual values
If (and only if) we need to find the actual value of a given fraction, represent the numerator as sum or difference of terms related to denominator.
1449/132 =
1449 = 1320 + 132 – 3
1449/132 = 10 + 1 – a small value ≈ little less than 11 (actual value is 10.977)
36587 / 123 =
36587 = 36900 – 246 – 61.5 – …
36587 / 123 = 300 – 2 – 0.5 – a small value ≈ little less than 297.5 (actual is 297.455)
1569 / 12 =
1569 = 1200 + 360 + 8.4 + 0.6
1569 / 12 = 100 + 30 + 0.7 + 0.05 = 130.75
This method should suffice for the level of accuracy expected in our exams.
Another method is to reduce the complexity of fraction and then solve. Complexity of a fraction can be directly related to the complexity of its denominator. If we simplify denominator, we simplify the fraction. Add to or subtract from the denominator to make it an easier value (like add 2 to 1998 to get 2000 or subtract 16 from 116 to get 100).
While adjusting the denominator always remember to BALANCE the fraction. Balancing fraction is not just adding/subtracting the same number to/from the numerator that we used to change the denominator.
Consider a fraction p/q = n; then p = qn.
If we add a number x to q, we need to add nx to p to balance the fraction. Also if q is reduced by a number x, p needs to be reduced by nx.
Here the approximation comes while fixing n. If the given options are separated well enough from each other and simplification of denominator is pretty obvious, then this method can be employed. If we have closer options it is better to stick with the method we discussed first.
1569 / 12 = ?
Here if we make the denominator as 10 we can tell the value in no time. To do so, we need to subtract 2 from denominator. Numerator is more than 130 times the denominator (n ≈ 130). Hence to balance the fraction we need to subtract 2 * 130 from numerator.
1569 / 12 ≈ 1309 / 10 ≈ 130.9 (actual value is 130.75)
To Approximate relative values
Most of the DI questions revolves around sorting the given numbers/fractions or finding its relative position (lesser/greater than) based on a reference value. If we don’t need the actual value, DON’T find the actual value.
Find the largest and smallest value among the below fractions
56/298, 46/374, 138/493, 37/540, 670/2498
We will do the first level approximation by guesstimating the given fractions. Try to represent the given numbers in 1/x format. While arranging fractions we usually try to represent the given fractions with the same denominator after finding the LCM of all denominators. But we are here to solve faster using approximation. We will take an easier route, Make the numerator same, i.e. one.
56/298, we know 56 * 6 > 298 = > 56/298 > 1/6. Note that we didn’t find the actual value of 56 * 6; we just want to get the closest multiple of 56 to the number 298.
56/298 = Greater than 1/6
46/374= Less than 1/8
138/493 = Greater than 1/4
37/540 = Greater than 1/15
670/2498 = Greater than 1/4
We don’t have any confusion in finding the smallest which is 37/540 (1/15 is less than other values). But we have 2 candidates fighting for the largest fraction title, 138/493 and 670/2498. We will consider only those two and try to get an approximate value. We will try both methods discussed before for finding the actual value.
Method 1:
138 = 98.6 + 24.65 + 12.325 + …
138/493 ≈ 0.2 + 0.05 + 0.025 + small value ≈ greater than 0.275
670 = 499.6 + 124.9 + 49.96 – 4.46
670/2498 ≈ 0.2 + 0.5 + 0.02 – small value ≈ less than 0.27
Hence 138/493 is the largest.
Method 2:
138/493,
We can see denominator is close to 3.5 times numerator. Hence if we increase denominator by x, we need to balance the fraction by increasing numerator by x/3.5. We will get an easier fraction if we can write denominator as 500 by adding 7. We also need to add 7/3.5 = 2 to the numerator.
138/493 ≈ 140/500 ≈ 0.28
Similarly for 670/2498, here we can get a neat fraction by adding 2 to the denominator. And here as 2 is negligible compared to the denominator we can very well skip the balancing part and write fraction as 670/2500 = 0.268
Hence, 138/493 is the largest.
Here we wrote 670/2500 = 0.268. How?
670/2500 = 67/250, we can get denominator as 1000 by multiplying both sides by 4. Hence 67/250 = 268/1000 = 0.268
We used the same logic while ‘cleaning up’ 140/500. Multiply both sides with 2 to get denominator as 1000. Fraction becomes 280/1000 = 0.028
Here, instead of finding actual values of all five fractions and comparing them we just played with the relative values of the fractions and found actual values only for two cases which were required to get the answer.
Another usual DI question type is to find the relative position of a given value based on a reference value. This question comes like ‘How many students scored marks more than class average (Reference value)’ , ‘How many players has strike rate higher than Sachin (Reference value)’ etc…
How many of the given values are greater than 0.7
11/13, 25/34, 33/46, 44/65, 56/81
As we are asked to find only the relative values (with respect to 0.7) don’t jump into finding actual values. Take few seconds to write the below statement which will help us in solving faster.
If x/y > 0.7, x > 0.7 y, 10x > 7y
So we need to find all fractions where 10 times numerator is greater than 7 times y. multiplying both sides with 10 is to ease the calculation and simplify the comparison
Take fractions one by one
Three fractions (11/13, 25/34 and 33/46) are greater than 0.7
Most of us have higher comfortable level with multiplication than division. To find relative values based on a reference point, convert division into multiplication. This way we can get our answers faster without messing with our accuracy.
In our example 56/81 = 0.69, still we were able to find it is lesser than 0.7 without doing any complicated or time consuming stuff. Sweet, right!
EXAMPLE 1: Solve 12 + 22 ÷ 11 × (18 ÷ 3)^2 – 10
= 12 + 22 ÷ 11 × 6^2 – 10 (Brackets first)
= 12 + 22 ÷ 11 × 36 – 10 (Exponents)
= 12 + 2 × 36 – 10 = 12 + 72 – 10 (Division and multiplication, left to right)
= 84 – 10 = 74 (Addition and Subtraction, left to right)
EXAMPLE 2: Solve 4 + 10 – 3 × 6 / 3 + 4
= 4 + 10 – 18/3 + 4 = 4 + 10 – 6 + 4 (Division and multiplication, left to right)
= 14 – 6 + 4 = 8 + 4 = 12 (Addition and Subtraction, left to right)
To Solve Modulus of a Real Number
Conversion of decimal numbers to nearest number
To solve such questions, first convert the decimal to nearest value. Then simplify the given equation using the new values that you have obtained.
EXAMPLE 1: Solve 4433.764 – 2211.993 – 1133.667 + 3377.442
Here,
4433.764 = 4434
2211.993 = 2212
1133.667 = 1134
3377.442 = 3377
Now simplify, 4434 – 2212 – 1134 + 3377 = 4466
EXAMPLE 2: Solve 530 x 20.3% + 225 x 16.8%
Here, 20.3% becomes 20% and 16.8% becomes 17%
Now, simplify 530 x 20% + 225 x 17%
= 106 + 38.25 = 144.25
Approximation of Square Roots
(1) To simplify a square root, you can follow these steps:
(2) Factor the number inside the square root sign.
(3) If a factor appears twice, cross out both and write the factor one time to the left of the square root sign. If the factor appears three times, cross out two of the factors and write the factor outside the sign, and leave the third factor inside the sign. Note: If a factor appears 4, 6, 8, etc. times, this counts as 2, 3, and 4 pairs, respectively.
(4) Multiply the numbers outside the sign.
(5) Multiply the numbers left inside the sign.
(6) To simplify the square root of a fraction, simplify the numerator and simplify the denominator.
NOTE: Check that the outside number squared times the inside number should equal the original number inside the square root.
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