Simplification Tips & Tricks – How to Solve Simplification – Approximation Problems & Quiz

Simplification Tips & Tricks | How to Solve Simplification Problems & Quiz How to Solve Simplification & Approximation Questions? Tips & Tricks, simplification tricks in Hindi, simplification problems with solutions pdf

Simplification Tips & Tricks: – At least five questions are asked from the topic ” Simplification & Approximate” in every bank exam, SSC Exam, Railway, CDS, CHSL, LDC, MTS and others Exam. Here we provide the simple toips and tricks to solved the Simplification question.

Check Tricks to save time & solve questions in few second. To save the time in the exam hall, you must check these tips and practice the previous paper, model paper to improve your speed and accuracy. These tricks can even be helpful while doing Data Interpretation {DI} questions also. This will cover all the theory needed to master Number System in Numerical Abilitys for Banking & SSC and other competitive Exams.

Simplification Tips & Tricks - How to Solve Simplification - Approximation Problems & Quiz

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Tips and Tricks to Solve Simplification and Approximation Problem

Rules of Simplification

V → Vinculum

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

O → Of

D → Division

M → Multiplication

A → Addition

S → Subtraction

Digital Sum

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

Example: 129 = 1+2+9 = 12, 1 + 2 = 3.

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

Example: 6 ✕ 9 = 54, 5+4 = 9

Trick: In order to save time if we find digit 9 or multiples of 9, then 9 or its multiple can be neglected.

Example: 193 = 1 + 9 + 3 = 13 = 1 + 3 = 4  [ ‘9’ is omitted ]

‘9’ is omitted to reduce the calculation.

If we don’t omit ‘9’, then also the digital sum remains same.

Example: 293 = 2 + 9 + 3 = 14, 1 + 4 = 5 [answer remains same]

Let’s discuss one more example-

Classification 

TypesDescription
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 Numbers :All whole number divisible by 2 is Even (0,2,4,6,8,10,12…..∞)
Odd Numberswhich 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
Co-Prime:Two natural number a and b are said to be co-prime if their HCF is 1.

Simplification Divisibility Rules

NumbersIF A Number5
Divisible by 2End with 0,2,4,6,8 are divisible by 2124,278,4578,98254 all are divisible by 2
Divisible by 3Sum of its digits  is divisible by 3375,4251,78123 all are divisible by 3.  [852=8+5+2=15] and 15 is divisible by 3  hence 852 is divisible by 3.
Divisible by 4Last 2 digit divisible by 42544 here last 2 digits are 44 which is divisible by 4 hence 2544 is also divisible by 4.
Divisible by 5Ends with 0 or 5120 or 125 or 280 or 390 here last digit is 0 or 5 that mean both the numbers are divisible by 5.
Divisible by 6Divides by Both 2 & 33636 here last digit is 6 so it divisible by 2 & sum of its digit (like 3+6+3+6=18) is 18 which is divisible by 3.Hence 3636 is divisible by 6.
Divisible by 8Last 3 digit divide by 8574888 here last 3 digit 888 is divisible by 8 hence 574888 is also divisible by 8.
Divisible by 10End with 01200,480,780,145020 all numbers has a last digit zero it means all are divisible by 10.
Divisible by 11[Sum of its digit in  odd places-Sum of its digits in even places]= 0 or multiple of 11Consider 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)

(23-12)

23-12=11,
which is divisible by 11. So 39798847 is divisible by 11.

Percentage – Ratio

Common Percentage method

Example:)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

Example: 1.2 % of 1225 + 2.4 % of 975=?

Read 12 % of 122.5 + 24 % of 97.5

Here 10 % is common 10 % of (122.5+97.5)=10% of 220=22

First part (10+2)=2% of 122.5=2.450

Second part (10+10+4)=10% 97.5 + 4 % of 97.5=9.75 + 3.900

So (22+2.45+9.75+3.90)= 38.10

B)Breaking Method

Example🙂  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

Example: –  92% of 4890 -86 % of 5200 +42 % of 4150 = ?

=(90+2) % of 4890 + (40+2)% of 4150- (80+6) % of 5200

=4401+97.80+1660+83-4160-312

=1769.80

Base method

Example: – 95% of 4860 = ?

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

Whole Number method

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

Solution: – (140 *1300)/100 + (160 *1360)/100

=1820+2176=3996

Fraction method

Example: – 125 of 488 + .625 of 824 = ?

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

Example: –  875 of 872 + .625 of 448=?

=87.5 % of 872 + 62.5 % of 448 =7/8 * 872 + 5/8 * 448 =763 + 280 =1043

Square & Cube Simplification Rule

  • Square & Cube
  • Square Root & Cube Root
  • Factorization Method

Quantitative Aptitude Exam Syllabus pdf 2022

Approximation

As the name suggests, 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.

Example: 159.99 ✕ 6.011

Solution: – 160 X 6 = 960

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