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Simplification and Approximation Shortcut Tricks & Tips

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Basic Rules of Simplification


The Quantitative Aptitude section of the Bank exam  consists of questions such as Simplification, Number Series, Permutation & Combination, Quadratic Equation, Data Interpretation, Data Analysis and other Miscellaneous questions.
 
Here is a short study-guide to help you crack questions on “Simplification and Approximation“

BODMAS Rule

It defines the correct sequence in which operations are to be performed in a given mathematical expression to find the correct value. This means that to simplify an expression, the following order must be followed –

 

B = Bracket,

= Order (Powers, Square Roots, etc.)

D = Division

M = Multiplication

A = Addition

S = Subtraction

⇒To solve simplification questions correctly, you must apply the operations of brackets first. Further, in solving for brackets, the order – (), {} and [] – should be stricly followed. 

⇒Next you should evaluate exponents (for instance powers, roots etc.)

⇒Next, you should perform division and multiplication, working from left to right. (division and multiplication rank equally and are done left to right).

⇒Finally, you should perform addition and subtraction, working from left to right. (addition and subtraction rank equally and are done left to right).

EXAMPLE :  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)
 
 
                                                                                            BRACKETS
The various kind of brackets are:
(i) ‘–’ is known as line (or bar) bracket or vinculum.
(ii) ( ) is known as parenthesis, common bracket or small bracket.
(iii) { } is known as curly bracket, brace or middle bracket.
(iv) [ ] is known as rectangular bracket or big bracket.
The order of eliminating brackets is:
(i) line bracket
(ii) small bracket (i.e., common bracket)
(iii) middle bracket (i.e., curly bracket)
(iv) big bracket (i.e., rectangular bracket)
 
 
 
Note:
(i) Even + Even = Even
Even – Even = Even
Even × Even = Even
Even ÷ Even = Even
 
(ii) Odd + Odd = Even
Odd – Odd = Even
Odd × Odd = Odd
Odd ÷ Odd = Odd
 
(iii) Even + Odd = Odd
Even – Odd = Odd
Even × Odd = Even
Even ÷ Odd = Even

                                                                                       NUMBER SYSTEM 

Types of Numbers:
1. Natural Numbers: A number n > 0 where n is counting number; [1,2,3…]
2. Whole Numbers: A number n ≥ 0 where n is counting number; *0,1,2,3…+.
0 is the only whole number which is not a natural number.
Every natural number is a whole number.
3. Integers: A number n ≥ 0 or n ≤ 0 where n is counting number;…,-3,-2,-1,0,1,2,3… are integers.
4. Positive Integers: A number n > 0; [1,2,3…]
5. Negative Integers: A number n < 0; [-1,-2,-3…]
6. Non-Positive Integers: n ≤ 0; *0,-1,-2,-3…]
7. Non-Negative Integers: A number n ≥ 0; *0,1,2,3…+
0 is neither positive nor negative integer.
8. Even Numbers: A number divisible by 2; [for example 0,2,4,…]
9. Odd Numbers: A number not divisible by 2; [for example 1,3,5,…]
10. Prime Numbers: A number numbers which is divisible by themselves only apart from 1.
1 is not a prime number.

 

   DIVISIBILITY

1. Divisibility by 2:A number is divisible by 2 if its unit digit is 0,2,4,6 or 8.
Example: 64578 is divisible by 2 or not?
Step 1 – Unit digit is 8.
Result: 64578 is divisible by 2.

 

2. Divisibility by 3: A number is divisible by 3 if sum of its digits is completely divisible by 3.
Example: 64578 is divisible by 3 or not?
Step 1 – Sum of its digits is 6 + 4 + 5 + 7 + 8 = 30
which is divisible by 3.
Result: 64578 is divisible by 3.

 

3. Divisibility by 4: A number is divisible by 4 if number formed using its last two digits is completely
divisible by 4.
Example: 64578 is divisible by 4 or not?
Step 1 – number formed using its last two digits is 78
which is not divisible by 4.
Result: 64578 is not divisible by 4.

 

4. Divisibility by 5: A number is divisible by 5 if its unit digit is 0 or 5.
Example: 64578 is divisible by 5 or not?
Step 1 – Unit digit is 8.
Result: 64578 is not divisible by 5.

 

5. Divisibility by 6: A number is divisible by 6 if the number is divisible by both 2 and 3.
Example: 64578 is divisible by 6 or not?
Step 1 – Unit digit is 8. Number is divisible by 2.
Step 2 – Sum of its digits is 6 + 4 + 5 + 7 + 8 = 30
which is divisible by 3.
Result: 64578 is divisible by 6.

 

 

6. Divisibility by 7:A number of 2 digits is divisible by 7 because 3 × 6 + 3 = 21. 21 is divisible by 7.
A number of three or more digits is divisible by 7 if the sum of the numbers formed by the last two digits and twice the number formed by the remaining digits is divisible by 7.
For Example:
(i) 574 is divisible by 7 because 74 + 5 × 2 = 74 + 10 = 84 is divisible by 7.
(ii) 2268 is divisible by 7 because 68 + 22 × 2 = 68 + 44 = 112 is divisible by 7.

7. Divisibility by 8: A number is divisible by 8 if number formed using its last three digits is completely
divisible by 8.
Example: 64578 is divisible by 8 or not?
Step 1 – number formed using its last three digits is 578
which is not divisible by 8.
Result: 64578 is not divisible by 8.

8. Divisibility by 9: A number is divisible by 9 if sum of its digits is completely divisible by 9.
Example: 64579 is divisible by 9 or not?
Step 1 – Sum of its digits is 6 + 4 + 5 + 7 + 9 = 31
which is not divisible by 9.
Result: 64579 is not divisible by 9.

 

 

 SQUARE ROOT AND CUBE ROOT

Main Concepts and Results

• A natural number is called a perfect square if it is the square of some natural number. i.e., if m = n2, then m is a perfect square where m and n are natural numbers.

• A natural number is called a perfect cube if it is the cube of some natural number. i.e., if m = n3, then m is a perfect cube where m and n are natural numbers.

• Number obtained when a number is multiplied by itself is called the square of the number.

• Number obtained when a number is multiplied by itself three times are called cube number.

• Squares and cubes of even numbers are even.

• Squares and cubes of odd numbers are odd.

• A perfect square can always be expressed as the product of pairs of prime factors.

• A perfect cube can always be expressed as the product of triplets of prime factors.

 

                                                                               Square Root

Example: Square root of 3481

Step 1 :- Split number in two parts i.e. 34 and 81(last two digits)
Step 2 :-We know that square of number ends in 1, so square root ends either in 1 or 9.
Step 3:-Check, 34 lies between 25 (square of 5) and 36 (square of 36). Take smaller number.
So, our answer is either 51 or 59.
but we know 502 = 2500 and 602 = 3600, 3481 is nearest to 3600. So the answer is 59.
 
 
 
Ex4: 1000
Step 1 :- 961(312) < 1000 < 1024(322)
Now, 1000 is nearest to 1024
So, 32 – ((1024-1000)/(2× 32))
32 – (24/64)
32-.375 = 31.625
or 31+((1000-961)/(2× 31))
 31 + (39/62)
31+.629 ≈ 31.63
 
 

                                                                                            Cube root

Example: cube root of 3112136
Split in two parts  3112    136
Number will end with 6
143 (2744) < 3112 < 153 (3375)
Choose the smaller number and answer will be 146.
 
 
Square Root Upto 25:-
 
 
                                                                                    Percentage Type
 

Common Percentage method:-

12 % of 555 + 15 % of 666= ?

Here 10 % is common 10 % of (555+666)=10 % of 1221=122.10
First part (10+2)=2% of 555= 11.10
Second part (10+5)=5% of 666= 33.30
So (122.10+11.10+33.30)= 166.50

Breaking Method:-

68 % of 4096 +72% of 5120 -23 % of 6931 -17 % of 1341= ?(Approximate)

(60+8)%of 4096 +(70+2) % of 5120 –(20+3)% of 6931 –(10+7)% of 1341
=2456.00 +327.68 +3584.00+102.40 -1386.20-207.93-134.10-93.87
=6470.08 -1821.47
=4648.61
≈450

 

Unit digit Method:-

75 % of 4860 = ?
=(300/4)% of 4860
=3*1215
=3645

Base method:-

95% of 4860 = ?

=(100-5)% of 4860
=4860-243
=4617

Whole Number method:-

139.001 % of 1299.99 + 159.99 % of 1359.99= ? (Approximate)

(140 *1300)/100 + (160 *1360)/100
=1820+2176=3996

Fraction method:-

125 of 488 + .625 of 824 = ?

=12.5 % of 488 + 62.5 % of 824
=1/8 *488 + 5/8 * 824
=61 + 515
=576

 

   ADDITION AND SUBTRACTION

I. Addition of smaller number to larger number is easier than addition of larger number to smaller number

Example :- addition in the order 5817 + 809 + 67 + 8 is easier than the addition in the order 8 + 67 + 809 + 5817. Hence to add the numbers, it is better to first arrange them in decreasing order and then add them.

 

II. To find the sum like 6345 + 2476 + 802, first add the thousands and then hundreds, tens and once in order.

Thus
6345 + 2476 + 802

= 6000 + 2000 ( = 8000) + 300 (= 8300)+ 400 (= 8700) + 800 (= 9500)+ 40 (= 9540) + 70 (= 9610) + 5 (= 9615) + 6 (9621) + 2

= 9623

 

                                                                                     Addition Shortcut

Example:- 116 + 39

(Here we can write this 39 as 40-1)
= 116 + (40 – 1)
= 116 + 40 – 1
= 156 – 1 (Instead of adding 39 to 116, we just add 40 to 116 (because we can do this without using pen and paper) and later we subtract one from it)
= 155

 

Example:- 116 + 97
= 116 + (100 – 3)
= 116 + 100 – 3 (Here, instead of adding 97 to 116, we are just adding a 100 to 116 and then subtracting 3 from it .
= 216 – 3
= 213

 

 

  H.C.F and L.C.M

H.C.F is the highest common factor or also known as greatest common divisor, the greatest number
which exactly divides all the given numbers.

Follow the steps below to find H.C.F of given numbers by prime factorization method.

1. Express the given numbers as product of their prime factors
2. Check for the common prime factors and find least index of each common prime factor in the given numbers
3. The product of all common prime factors with the respective least indices is H.C.F of given numbers.

Example-
H.C.F of 12, 36, 48

Solution-

1. Express the numbers as product of prime factors

12= 3 * 2^2

36=3^2 * 2^2

48=3 * 2^4

2. The common prime factors are 2 and 3 and the corresponding least indices are 2 and 1 respectively
3. The product of all the common prime factors with the respective least indices

H.C.F of 12, 36, 48 =2^2 *3= 12

 

 

Least Common Multiple (L.C.M)
L.C.M is least common multiple, the smallest number which is exactly divisible by all the given numbers.

Follow the steps below to find L.C.M of given numbers .
1. Express the given numbers as product of their prime factors.
2. Find highest index in all the prime factors of given numbers.
3. The product of all the prime factors with respective highest indices is the L.C.M of given numbers.

 

Example:
L.C.M of 14, 42, 36

Solution:-

1.Express the numbers as product of prime factors.

14= 2*7
36= 3^2 *2^2
42 = 2*3*7

2. The highest index of 2, 3, 7 are 2, 2, 1 respectively
3. The product of all the prime factors with the respective highest indices.

L.C.M of 14, 36, 42 =2^ 2  *3^2 *7 = 252

 

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