Simplification
The topic of Simplification is an integral part of every exam. In bank and SSC exams, Simplification covers almost 4050% of the Quantitative Aptitude section. Simplification also covers the percentage, ratio, fractions topic. Simplification topic is asked to check the ability of students to find their ability to deal with the easy approach to solving complex problems. So, it’s important to have knowledge about simplification. Let’s try to understand Simplification
What is the simplification in Quantitative aptitude?
Simplification means to play with the questions by doing complex calculations with tricks such that you can solve these questions in just seconds. By using the following of the tricks and tactics, you can score well in your examination and save you some valuable time:
Before engaging with the topic, it is important to know about the topics and what are some basic approaches needed in order to solve the complex problems of the simplification
What are the types of simplifications?
Basically there two types of simplification problems.
Sometimes a number is missing we have to find out the number by performing some arithmetic’s operations
Secondly, we have just simplified or reduce the number by performing some common operation to attain a final value
In order to solve both kinds of problems various tool is needed which can ease our problems
 Basic multiplication techniques
 Memorizing tables up to 30 can help a lot. If you aren’t good at remembering things don’t worry there are many tips and tricks to learn the table.
 Learn tricks to understand and find cube roots and square roots of large numbers.
 Learn the concept of percentages, as we have already covered the topic earlier you can check our previous blogs. In this topic, we just need to learn (conversion of fractions to percentage & percentage to fractions)
 Memorize the reciprocals values.
 Learn BODMAS.
 Use the concept of the digital sum.
 Memorize cubes and squares of numbers up to 35.
 Learn tricks to know the squares and cubes of numbers that are greater than 35
While solving the problems of the simplification, we need to quick with the calculation in order to save time. Hence, we need some basic things to learn and some quick approaches to solving any question
Important Tips and Tricks for Simplifications
1. Digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9
2. Types of numbers.
(i) Natural numbers. = {1, 2, 3, 4 ...}
(ii) Whole numbers = {0, 1, 2, 3, 4, 5......}
(iii) Integers = {......, 3, 2, 1, 0, 1, 2, 3 ...}
(iv) Real numbers = {, 2.8, 2, 10, 1, 1.9,2, 3, 3.12, 3.13}
(v) Even numbers = {2, 4, 6, }
(vi) Odd numbers = {1, 3, 5, 7}
(vii) Prime numbers = {2, 3, 5, 7, 11, 13, 17, 19, }
3. Some basic formulas
While solving the mathematics problems, it is common that you will get a bunch of numbers, which will take a lot of time of yours. Hence, we just need to be keen and focused on the numbers such that we can multiply and divide the numbers, which can help us to get unit digits of the numbers, be 0.
What is BODMAS Rule?
B=Bracket
O=of
D=Divide
M=Multiplication
A=Addition
S=Subtraction.
While solving the problem of problems containing arithmetic operations and bracket we can simply solve the problems by using BODMAS rule. BODMAS rule simply helps us to solve the problem by steps. Firstly, we need to open brackets and solve the Brackets, then we can proceed further by solving OF, DIVIDE, MULTIPLY, ADDITION, SUBTRACTION problems.
Now let us study more about divisibility rule
Divisibility rules
Numbers 
IF A Number 
Examples 
Divisible by 2 
The numbers which end with 0,2,4,6,8 are divisible by 2 
352,70076 ,10000 ,96358 all are divisible by 2 
Divisible by 3 
Sum of its digits is divisible by 3 
381 are divisible by 3. [381=3+8+1][5+7+9=21]21 is divisible by 3. hence 579 is divisible by 3. 
Divisible by 4 
Last two digit divisible by 4 
7944 here last 2 digits are 44 which is divisible by 4 hence 7944 is divisible by 4. 
Divisible by 5 
Ends with 0 or 5 
1000 or 995 here the last digit is 0 or 5 that means 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, and 18 is divisible by 3. Hence number 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 
140, 2180, 111110 all numbers have the 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 the odd places)(Sum of its digits at the even places) (7+8+9+9)(4+8+7+3) (2312) 2312=11, 
What is Division & Remainder Rules? (Euclid’s Division lemma)
Suppose we divide 51 by 7
51=7*7+2
Hence, represent it as:
Dividend = (Divisor*Quotient) + Remainder
or
Divisor= [(Dividend)(Remainder] / quotient
Could be written as: a = bq + r
Where (a = dividend, b = divisor = quotient = remainder)
Some common fraction values used while solving problems
 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 × 0.1666 = 0.08333…
 1/13 = 0.076923
 1/14 = 0.071428
 1/15 = 1/3 × 1/5 = 0.333333 × 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 × 0.1111 = 0.055…
 1/19 = 0.052631
 1/20 = 0.05
Percentage – Ratio Equivalence:
Square & Cube
It is essential to learn the squares and perfect squares till 35 which can ease our problems and can save our time in the examinations
Numbers 
Squares 
Last digit 
Numbers 
Square 
1 
1 
1 
9 
81 
2 
4 
4 
8 
64 
3 
9 
9 
7 
49 
4 
16 
6 
6 
36 
5 
25 



Cube and cube Roots
Approximation
Mostly questions are asked from these topics in these we have to just put its approximate values and calculate an approximate answer
As the name suggests, if the given values are in points, then we can use its nearest value i.e., approximating the values to the nearest comfortable value which makes our calculation easier such that there is not much effect on the final answer
Examples on Approximations
1) 160.21 ✕ 6.976
Sol: 160 ✕ 7=1120
2. 25% of 2500 + 16.66% of 144 =?
Sol: (25/100)* 2500 + 1/6 *144 [1/6=16.66%]
⇒ 252 + 24 = 625 + 24
⇒ 649
3. ∛7.938 x∛999.9998– 4.9256 =?
Sol: Using the concept of approximation,
7.938 = 8
999.9998 = 1000
4.92856 = 5
⇒ 2✕(105)=205 =15
Tip:[you must know the cubes, and then you can calculate cube roots & squareroots easily and solve the problems]
4. 60% of 250 +25% of 600
Tip :Know the values of 60% =6/100 or 0.6 and 25 % = ¼ or 0.25
STEP 1: Now directly multiply 0.6×250 + 0.25×600
STEP 2: 0.6×250= 150 0.25×600=150
STEP 3: 150+ 150 = 300
STEP 4: Hence, the answer is 300
Solve mixed fraction – Multiplication
5. 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 38×(15/13)
4) Solve Mixed Fraction addition
6. 19×(3/5) + 23×(2/3) – 24×(1/5)
STEP 1: Take all the number outside i.e. 19+23 24 = 18
STEP 2: Add fractions 18×[(3/5) + (2/3) – (1/5)] = 18(16/15)
STEP 3: Hence the answer is 18(16/15)
7. (?)3+12×12= 62 ×5×2
STEP 1: Multiply 12 × 12 = 144
STEP 2: Square of 6 = 36
STEP 3: Multiply 36 ×5×2= 360
STEP 4: (a)power3 +144 = 360
STEP 5: (a)power3= 360144= 216
STEP 6: Therefore a = 12
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Author: Ashwani Kumar.