# Mathematics Special: Tricks to Solve Simplification Questions

By Shashi Kant|Updated : February 7th, 2021

In the Mathematics section of most of the teaching exams, Simplification is considered as an important topic. The level of this section is increasing with every exam. The candidates must have clear understandings of concepts and practice a number of questions for quick calculations. Through this article, we will be sharing tips and tricks to crack questions of simplification quickly. These tips will be useful for the upcoming CTET, UPTET, DSSSB, KVS 2021 and other teaching exams.

In the Mathematics section of most of the teaching exams, Simplification is considered as an important topic. The level of this section is increasing with every exam. The candidates must have clear understandings of concepts and practice a number of questions for quick calculations. Through this article, we will be sharing tips and tricks to crack questions of simplification quickly. These tips will be useful for the upcoming CTET, UPTET, DSSSB, KVS 2020 and other teaching exams.

## Tips & Tricks to solve Simplification questions

This topic covers arithmetic concepts. The most important concept of simplification is VBODMAS. This rule governs the priority of operators when two or more operators are to be solved in an equation. The expended form of VBODMAS is:

V → Vinculum

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

O → Of

D → Division

M → Multiplication

S → Subtraction

The priority of operators can be modified using brackets. The use of VBODMAS rule is given below:-

856 + 986 x 65 ÷ 5 – 78 + (23 x 8 - 6)

Step 1: Start operations from left to right, as per the rule we will solve Bracket first and within bracket also apply the rule of BODMAS

856 + 986 x 65 ÷ 5 – 78 + (184 - 6)

856 + 986 x 65 ÷ 5 – 78 + (178)

Step 2: Now apply division operation.

856 + 986 x 13 – 78 + (178)

Step 3: Now apply multiplication operation.

856 + 12818 – 78 + (178)

Step 4: Now apply addition & subtraction operation.

13674 + 100

13774

In this way, we can solve sums very easily. The simplification questions at times become very complex and difficult to solve. For such questions, the candidates can apply the digital sum and unit digit method for quick solutions. These two methods are explained below:-

### Unit digit method

In simplification questions at times, we can eliminate options using the unit digit method. It means instead of solving question until the end, we can find the unit digit to solve the question.

For example, we are required to solve the following:-

2345 – 489 + 7890 + 45 x 16

1. 10466
2. 21054
3. 9867
4. 6548
5. 8769

In the question, 2345 – 489 + 7890 + 45 x 16, the unit digit of the answer should be:

5 – 9 + 0 + 0 = -4, since unit digit is negative we will add 10 to it (carry)

10 – 4 = 6, therefore the unit digit of this question must be 6. And 6 as a unit digit is in option A.

In this way, we can save our precious time. This method is very helpful when the options are not similar (not the same unit digit). If the options are similar in that case we can use this method with another method i.e digit sum method.

### Digit Sum Method

In this method, we keep on adding digits of the number until we get a single digit. According to this concept, the digit sum of an equation (addition, multiplication, subtraction) and its result remains the same. Using this method along with unit digit can save a huge amount of time.

Example: 2011×97+50123 =? × 743, what should come in place of question mark?

(A) 340 (B) 330 (C) 350 (D) 303 (E) 345

Solution:

In LHS 2011×97, unit digit will be 7 (7*1 = 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 multiply by 743 must consist 0 at its unit place. So, option (D) and (E) are eliminated.

Now let’s apply both Unit digit and the digit sum method together.

In LHS, 2011×97+ 50123

Adding digits we get, 2+0+1+1=4 and 9+7= 16 =1+6 (As 16 is a double digit number) and 5+0+1+2+3= 11

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.

### Types of Simplification questions

In exams, various types of simplification questions are asked. Some of the types are given below:

Type 1

135% of 342 - 342% of 13.5 =?

(1) 411.13

(2) 412.23

(3) 413.33

(4) 414.43

(5) 415.53

Using digital sum method, the digital sum of question is

135÷100 x 342 – 342÷100 x 13.5

9 x 9 – 9 x 9

81x81

(8+1)-(8+1)

9 – 9 = 0

Now in options digit sum of only option 5 is zero. Therefore the answer is 415.53

Type 2:

√1089 + 42 =? + √1024

• 17
• 19
• 18
• 20
• 22

In such questions, solve the surds and then solve the rest part.

33 + 16 = X + 32

49 = X + 32

X = 17

Type 3:

1210.5 * 43.5 * 33.5 = 12?

• 18
• 16
• 14
• 15
• 17

In such questions convert base of LHS and RHS numbers the same so as to compare for the answer.

1210.5 * 123.5 = 12x

1214 = 12x

X = 14

### Practice Quiz

The candidates are advised to practice questions from previous exams and from various platforms for maximum exposer to gain experience and score well in the exam. After studying thoroughly, try the below-given quiz to check your level of preparation:

## Mathematics Quiz on Simplification

Thanks

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