# How to Solve Simplification - Approximation (Tips and Tricks)

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.
There are almost 10 to 12 questions in the paper which deal with topics such as Percentage and Average, ratio & proportions, partnerships, Profit & loss, calculations of simple & compound interests etc.
Here is a short study-guide to help you crack questions on “Simplification and Approximation“,

## Basic Rules of Simplification

### 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,

O = Order (Powers, Square Roots, etc.)

D = Division
M = Multiplication

= Addition
S = Subtraction
1. Hence, 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.
2. Next you should evaluate exponents (for instance powers, roots etc.)
3. Next, you should perform division and multiplication, working from left to right. (division and multiplication rank equally and are done left to right).
4. Finally, you should perform addition and subtraction, working from left to right. (addition and subtraction rank equally and are done left to 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

The Modulus (or the absolute value) of x is always either positive or zero, but never negative. For any real number x, the absolute value or modulus of x is denoted by |x| and is defined as

|x|x {if ≥ 0} and x {if 0}
EXAMPLE 1: Solve |8|
|8| = |-8|  = 8

## Tips to Crack Approximation

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.

Now we are going to share some important tips and tricks that will help you prepare the Simplification - Approximation topic better.

## Simplification / Approximation: Tips and Tricks

We strictly recommend you to learn square (up to 30) and cube (up to 20).We will discuss here methods to solve and types of problems which are generally asked in exams.

## Unit Digits and its applications

Ex: 298: 8 is the unit place in 298.
Ex: 1947: 7 is the unit place in 1847.

Ex: 2345×6789

(A)15920206 (B)15920208 (C) 15920205 (D) 15920204

Solution: When unit place of 5 in 2345 and unit place of 9 in 6789 multiplies we will get 45. So when both numbers are multiplies it should have 5 at its unit place which is only in option C.

Ex: 43 × 36 + 57 × 89

(A)6380 (B)5728 (C)6782 (D)6621

The unit digit will be the sum of the individual unit digits.

(3×6)+(7×9) = 18+63 = 81

So the resultant number must have 1 at its unit place.

### Digit Sum

It is the sum of all digits of the number used in making the number and keep adding till we have only one digit left.

Ex: 2345
Digit sum = (2+3+4+5) = 14 = 1+4 = 5

Ex: 123456789

Digit sum = (1+2+3+4+5+6+7+8+9) = 45 = (4+5) = 9
Note: In this case our assumption is that 9 should be treated as 0.

Ex: 123 × 456 × 781

(A)43804728 (B) 53804728 (C) 53804528 (D)33804958
LHS (Digit sum)= (1+2+3)×(4+5+6)×(7+8+1)= 6× 6× 7 = 36× 7 = 9 × 7 = 63 = 0
RHS (Digit sum):
(A) (4+3+8+0+4+7+2+8)= 36 =(3+6)= 9 = 0
(B) (5+3+8+0+4+7+2+8) = 37 = 10 = (1+0) = 1
(C) = 35 = (3+5) = 8
(D) =31 = (3+1) = 4
So, Option A is the answer.

Ex: 2011×97+50123 = ? × 743
(A) 340 (B) 330 (C) 350 (D) 303 (E) 345

Solution:

In LHS 2011×97, unit digit will be 7
In 50123, the unit digit is 3, So when we add these, the addition will have ‘0’ at its unit place.
In RHS, we also need ‘0’ at the unit place, the number which has to multiplied by 743 must consist 0 at its unit place. So, option (D) and (E) are eliminated.
Now Let’s apply Unit digit and digit sum
In LHS, 2011×97+ 50123
4 × 7 + 11 = 28+11 = 10+2 = 1+2 = 3
In RHS if option is (A)
then 340 × 743 = 7×14 = 7 × 5 = 35 = 8
LHS ≠ RHS
In RHS if option is (B)
then 330 × 743 = 6× 14 = 6× 5 = 30 = 3
LHS = RHS, It is the answer. If you check other options it will not satisfy this.

Ex: 6269+0.75× 4444+0.8×185 =?

(A)9759 (B)9750 (C)9740 (D)9755 (E)9655

Solution:

6269+ (3/4)×4444+148.0

6269+3333+148
We can see that unit digit is Zero. So options remained are B and C.
Now, (23)+(12)+(13)
5+3+4 = 12 = 3
Applying digit sum for (C) = 2 and (B) = 3
So, answer is B

## 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
When a number is given, split it in two parts, in such a way that 2nd part has last two digits of number and first part will have remaining digits.

Ex 1:  Square root of 3481
Split number in two parts i.e. 34 and 81(last two digits)
We know that square of number ends in 1, so square root ends either in 1 or 9.
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.
or 34 is more close to 36 than 25, so the answer is 59.

Ex 2: 76176
Split: 761  76
Number will end in either 4 or 6,
729(272) < 761 < 784 (282), So the answer may be 274  or 276. 761 is more close to 784, so the answer is 276.

Ex 3: square root of 75076
Split: 750  76
Number will end in either 4 or 6
729(272) < 750 < 784 (282), So the answer may be 274  or 276. 750 is more close to 729 than 784, so the answer is 274.

Non-Perfect Square: This gives approximate value not an exact value.

Ex4: 1000

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

## 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
When a number is given, split it in two parts, in such a way that 2nd part has last three digits of number and first part will have remaining digits.

Ex 1: cube root of 3112136
Split in two parts  3112    136
Number will end with 6
14(2744) < 3112 < 153 (3375)
Choose the smaller number and answer will be 146.

Ex 2: cube root of 2406104

split in two parts 2406   104

Number will end with 4

133 (2197) < 2406 < 143(2744)

So the answer will be 134.
###### Share To: 