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

By | November 6, 2019

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.

इस खंड से निपटने के लिए महत्वपूर्ण बोडामास नियम को याद रखना है और सभी टेबल, क्यूब्स, वर्ग पूरी तरह से याद होने चाइये । जिससे आपकी प्रश्न को सॉल्व करने में आसानी होगी और समय की भी बचत होगी

## 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

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. Simplification Tips & Tricks

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

 Types Description 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 Numbers which 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

 Numbers IF A Number 5 Divisible by 2 End with 0,2,4,6,8 are divisible by 2 124,278,4578,98254 all are divisible by 2 Divisible by 3 Sum of its digits  is divisible by 3 375,4251,78123 all are divisible by 3.  [852=8+5+2=15] and 15 is divisible by 3  hence 852 is divisible by 3. Divisible by 4 Last 2 digit divisible by 4 2544 here last 2 digits are 44 which is divisible by 4 hence 2544 is also divisible by 4. Divisible by 5 Ends with 0 or 5 120 or 125 or 280 or 390 here last digit is 0 or 5 that mean both the numbers are divisible by 5. Divisible by 6 Divides by Both 2 & 3 3636 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 8 Last 3 digit divide by 8 574888 here last 3 digit 888 is divisible by 8 hence 574888 is also divisible by 8. Divisible by 10 End with 0 1200,480,780,145020 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 11 Consider 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 2018  ### 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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