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Simplification and Approximation

While preparing for your exams, you are stuck at simplification and approximation? This article is going to help you with simplification and approximation questions.

Simplification is the most scoring section in the exam. In this section, you have to solve expressions using basic simplification rules. While solving these questions, you must follow the principle of the BODMAS rule. For solving simplification questions, we need to find the exact output by simplifying the mentioned expression. Whereas in approximation questions, the approximate output is required. In approximation questions, there is no need to calculate the exact answer. However, it should be close to the exact answer; basically, the most appropriate answer is considered while others are eliminated. Let’s know more about the critical concepts of Simplification and Approximation, along with the various types of questions and tips and tricks.

Techniques and Tricks to Solve Simplification Questions

Simplification means to ease the questions by doing complex calculations. By using following tricks and tips you can solve Simplification questions in no time.

BODMAS rule
The concept of digital sum should be on your tips.
Try to memorize tables up to 30.
Try to memorize cubes and squares of numbers up to 25.
Use tricks to find squares and cubes of bigger values.
Use tricks to find cube roots and square roots of large numbers.
Concept of percentages
Try to memorize the reciprocals.

Basic Rules of Simplification

V → Vinculum

B → Remove Brackets – in the order ( ) , { }, [ ]

O → Of

D → Division

M → Multiplication

A → Addition

S → Subtraction

Important Techniques of Simplification

Digital Sum
Number System
Reciprocals
HCF & LCM
Percentages
Square & Cube

Digital Sum

Digital sum is the sum obtained after adding each digits of any given number.

Example: 554 = 5+5+4 = 14, 1 + 4 = 5.

Important Note: if any number multiplied by 9, then the digital sum is always 9.

Number System
Classification
Divisibility Test
Division & Remainder Rules
Sum Rules

Classification

Natural Numbers: ( 1,2,3,4,5,6,7,8,9….∞)

Whole Numbers: (0,1,2,3,4,5,6,7,8,9,10,11…∞)

Integers: (∞……-4,-3,-2,-1,0,1,2,3,4,5….∞)

Even & Odd Numbers :

Even: (0,2,4,6,8,10,12…..∞)

Odd: (1,3,5,7,9,11,13,15,17,19….∞)

Prime Numbers: A number having only factors 1 and itself. (2,3,5,7,11,13,17,19,23,29,31,37,41…∞)

Composite Numbers:

All natural numbers excluding prime numbers.

(4,6,8,9,10,12,14,15,16,18,20……∞)

Co-Prime:

Two natural numbers a and b are said to be co- prime if and only if their HCF is 1.

Divisibility

Divisible by 2: No. Which end with 0,2,4,6,8 are divisible by 2

Divisible by 3: Digital sum is divisible by 3

Divisible by 4: Last two digits are divisible by 4

Divisible by 5: No. Which ends with 0 or 5 are divisible by 5

Divisible by 6: No. divisible by Both 2 & 3 are divisible by 6

Divisible by 8: Last 3 digit divide by 8

Divisible by 10: No. Which end with 0

Divisible by 11: (Sum of its digit in odd places)-(Sum of its digits in even places)= 0 or multiple of 11

Division & Remainder Rules

represent it as:

dividend = ( divisor*quotient ) + remainder

divisor= [(dividend)-(remainder] / quotient

could be write it as

x = kq + r

where, x = dividend,

k = divisor,

q = quotient,

r = remainder

Sum Rules

(1+2+3+………+n) = ½* n*(n+1)
(12+22+32+………+n2) = ⅙*n*(n+1)*(2n+1)
(13+23+33+………+n3) = ¼*n2*(n+1)2

Arithmetic Progression (A.P.)

a, a + d, a + 2d, a + 3d, ….are said to be in A.P. in where, a = first term,

d = common difference

Let the nth term = an and last term = l, then

a) nth term = a + ( n – 1 )*d
b) Sum of n terms = (n/2)*[2*a + (n-1)*d]
c) Sum of n terms = (n/2)*(a+l) where l is the last term.

Reciprocals

The reciprocals are easy to memorize upto 10. Reciprocals after that along with more are below-

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 = 0.11…
2/9 = 0.222…
3/9 = 0.333…
1/11 = 0.0909…
2/11 = 0.1818…
10/11 = 0.909090…
1/12 = ½ *1/6 = ½ * 0.1666 = 0.08333…
1/13 = 0.076923
1/15 = 1/3 * 1/5 = 0.333 * 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 = 0.0555…
1/19 = 0.052
1/20 = 0.05

HCF and LCM

The highest common factor (HCF) of two or more numbers is the largest number which divides each of the given numbers without remainder = 0.

The lowest common multiple (LCM) of two or more numbers is the smallest of the multiples of those numbers.

Points to note

If the HCF of 2 numbers is 1, then their LCM will be simply their product.
For coprime numbers, the HCF is always 1.

Percentages

Percent literally means “for every hundred” per means “upon ” and cent means”100 “. % is read as percentage and y % is read as “y percent”.

Calculation of a % of b

(a/100)* b = (a* b)/100

a% of b = b% of a

Square, Cubes, square roots and Cube roots

Squares, cubes, square roots and cube roots for numbers ranging 1 to 25:

NoSquareCubeSquare RootCube Root

1 1 1 1.000 1.000

2 4 8 1.414 1.260

3 9 27 1.732 1.442

4 16 64 2.000 1.587

5 25 125 2.236 1.710

6 36 216 2.449 1.817

7 49 343 2.646 1.913

8 64 512 2.828 2.000

9 81 729 3.000 2.080

10 100 1000 3.162 2.154

11 121 1331 3.317 2.224

12 144 1728 3.464 2.289

13 169 2197 3.606 2.351

14 196 2744 3.742 2.410

15 225 3375 3.873 2.466

16 256 4096 4.000 2.520

17 289 4913 4.123 2.571

18 324 5832 4.243 2.621

19 361 6859 4.359 2.668

20 400 8000 4.472 2.714

21 441 9261 4.583 2.759

22 484 10648 4.690 2.802

23 529 12167 4.796 2.844

24 576 13824 4.899 2.884

25 625 15625 5.000 2.924

Approximation

As the name stands its definition, 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 but calculations are a bit simplified.

Example: 150.02 * 5.976

Sol: 150.0 * 6 =900.0

Tips to solve approximation questions

Conversion of decimal numbers to the nearest number to minimum decimal places
To solve approximation questions, first, convert the decimal to the nearest value. Then simplify the given equation using the approximated values that you obtained.

Examples on Simplification and Approximation

Example 1.

Find an approximate value of what should come in place of the question mark (?)

39.012 × 14.98 – 28.013 × 9.999 = (20 + ?) × 5.23

Solution→

As per BODMAS, we first multiply,

39.012 × 14.98 = 39 × 15 = 585

28.013 × 9.999 = 28 × 10 = 280

So, 585 – 280 = (20 + x) × 5.23

⇒ 585 – 280 = 100 + 5x

⇒ 305 – 100 = 5x

⇒ 205 = 5x

⇒ x = 41

Example2

What would be the Square of 35?