RRB NTPC/ASM/CA/TA Exam  >  RRB NTPC/ASM/CA/TA Notes  >  Mathematics for RRB NTPC / ASM  >  Simplification and Approximation

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA PDF Download

Introduction

  • The Quantitative Aptitude section in competitive exams assesses numerical skills and problem-solving abilities.
  • Common topics include Simplification, Number Series, Permutation & Combination, Quadratic Equations, Data Interpretation, and Data Analysis.
  • Around 10-12 questions often focus on basics like Percentages, Averages, Ratios and Proportions, Partnerships, Profit and Loss, and Interest Calculations (Simple and Compound Interest).
  • Simplification and Approximation are critical for quick, accurate solutions, helping candidates save time and maximize scores.

Here’s a quick guide to sharpen your skills in Simplification and Approximation, setting a strong foundation for success in Quantitative Aptitude.

What is simplification and approximation ?

Simplification means converting an expression into a more basic form using different operations, whereas approximation involves estimating the value of an expression by rounding its digits to the nearest value, but not exactly correct. 

Basic Rules for Simplification 

1. BODMAS  Rule

The BODMAS  rule dictates the order in which operations should be performed:

  • Brackets (Parentheses)
  • Order (Exponents and Roots)
  • Division and Multiplication (from left to right)
  • Addition and Subtraction (from left to right)

Example:

Simplify: 2 + 3 × (4 + 2)²

Step 1: Solve inside brackets: 4 + 2 = 6

Step 2: Apply the exponent: 6² = 36

Step 3: Multiply: 3 × 36 = 108

Step 4: Add: 2 + 108 = 110

2. Fraction Simplification 

When simplifying fractions, divide both the numerator and denominator by their HCF.

Example:

Simplify: 18 / 24

The HCFof 18 and 24 is 6. 

Divide both by 6: 18 ÷ 6 = 3 and 24 ÷ 6 = 4. 

The simplified fraction is 3 / 4.

3. Multiplying and Dividing Powers

  • Multiplying powers with the same base: Add the exponents.
  • Dividing powers with the same base: Subtract the exponents.
  • Power of a power: Multiply the exponents.

Examples:

2³ × 2⁴ = 2^(3+4) = 2⁷ = 128

x⁶ / x² = x^(6-2) = x⁴

(x²)³ = x^(2×3) = x⁶

4. Rationalizing the Denominator

  • If a fraction has a square root, cube root, or any irrational number in the denominator, we rationalize the denominator by multiplying both the numerator and the denominator by an appropriate term to eliminate the root.

Example:

Simplify: 5 / √2

Multiply the numerator and denominator by √2: (5 × √2) / (√2 × √2) = 5√2 / 2

5. Use of Algebraic Identities

Use algebraic identities to simplify expressions:

  • Square of a binomial: (a + b)² = a² + 2ab + b²
  • Difference of squares: a² - b² = (a - b)(a + b)
  • Cube of a binomial: (a + b)³ = a³ + 3a²b + 3ab² + b³

Examples:

(x + 3)² = x² + 6x + 9

x² - 16 = (x - 4)(x + 4)

(x + 2)³ = x³ + 6x² + 12x + 8

6. Simplifying Algebraic Expressions

  • Combine like terms and factor expressions wherever possible.

Example:

Simplify: 3x + 5x - 2x

Combine like terms: 3x + 5x - 2x = 6x

7. Simplification of Ratios and Proportions

  • Simplify ratios and proportions by dividing both terms by their HCF

Example:

Simplify the ratio 24:36

The HCF of 24 and 36 is 12, so divide both by 12: 24 ÷ 12 = 2 and 36 ÷ 12 = 3.

 The simplified ratio is 2:3.

8. Simplifying Exponential Expressions

  • When dealing with powers of powers, use the appropriate rules to simplify:

Example: Simplify (2³)²

Apply the power of a power rule: 2^(3×2) = 2⁶ = 64

9. Simplifying Square Roots

  • If the number under the square root is a perfect square, simplify it. If not, estimate the value of the square root.

Example: Simplify √50

Break 50 into 25 × 2, which simplifies to √25 × √2 = 5√2

10. Simplifying Cube Roots

  • Similarly, simplify cube roots by factoring the number into cubes of smaller numbers.

Example: Simplify ∛54

Break 54 into 27 × 2, which simplifies to ∛27 × ∛2 = 3∛2

Conclusion

Mastering these simplification rules is crucial for solving problems quickly and accurately in Quantitative Aptitude for government exams. By using these basic rules effectively, you can simplify expressions, reduce the complexity of calculations, and avoid making errors in your exam.

Example 1:  Solve 12 + 22 ÷ 11 × (18 ÷ 3)^2 - 10

Solution:  

= 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

Solution:

= 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)

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA


 

Basic Rules for Approximation 

1. Rounding Off Numbers:

Round numbers to the nearest value for easier calculations.

Example 1: Round 9854.678 to the nearest hundred: 9900.

Example 2: Round 657.293 to one decimal place: 657.3.

2. Estimation of Sums and Differences:

Round numbers in sums and differences for quick calculations.

Example: Estimate 8946 + 5723 - 2994

Round 8946 to 9000, 5723 to 5700, and 2994 to 3000.

Approximate result: 9000 + 5700 - 3000 = 8700.

3. Estimation of Products:

Round each factor in a product to simplify multiplication.

Example: Estimate 378 × 52 × 18

Round 378 to 400, 52 to 50, and 18 to 20.

Approximate result: 400 × 50 × 20 = 400000.

4. Estimation of Quotients:

Round both the dividend and divisor for simple division.

Example: Estimate 5289 ÷ 17

Round 5289 to 5300 and 17 to 20.

Approximate result: 5300 ÷ 20 = 265.

5. Using the Nearest Tens, Hundreds, or Thousands:

Round large numbers to the nearest ten, hundred, or thousand.

Example: Estimate 29482 + 17563 - 9281 by rounding to the nearest thousand.

Round 29482 to 29000, 17563 to 18000, and 9281 to 9000.

Approximate result: 29000 + 18000 - 9000 = 38000.

6. Square Root Approximation:

Approximate the square root by choosing a nearby perfect square.

Example: Approximate √567

Since 567 is close to 576, approximate: √567 ≈ √576 = 24.

7. Cube Root Approximation:

Approximate the cube root by choosing a nearby perfect cube.

Example: Approximate ∛997

Since 997 is close to 1000, approximate: ∛997 ≈ ∛1000 = 10.

8. Percentage Approximation:

Round values to convenient numbers for quick percentage calculations.

Example: Estimate 22% of 793

Round 793 to 800.

Approximate result: 22% of 800 = 176.

9. Using Approximations in BODMAS Calculations:

Apply BODMAS with rounded values to simplify complex expressions.

Example: Approximate 375 + 243 × (96 + 19) ÷ 18

Round 375 to 380, 243 to 240, (96 + 19) to 115, and 18 to 20.

Approximate result: 380 + 240 × 115 ÷ 20 = 1760.

Following are some solved examples to help you prepare well for the upcoming exams:

Ques1. A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes, and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?
A.    45
B.    60
C.    75
D.    90
 

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Answer: Option D

Explanation:

Let the number of notes of each denomination be x.

Then x + 5x + 10x = 480

=> 16x = 480

Therefore, x = 30.

Hence, total number of notes = 3x = 90.

 Ques2. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
A.    20
B.    80
C.    100
D.    200

 

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Answer: Option C

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10      x - y = 20 .... (i)

     and x + 20 = 2(y - 20)      x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

Therefore, The required answer A = 100.

 Ques3. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:
A.    Rs. 3500
B.    Rs. 3750
C.    Rs. 3840
D.    Rs. 3900

 

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Answer: Option D

Explanation:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

Then, 10x = 4y   or   y =    5/2 x.
Therefore, 15x + 2y = 4000

=> 15x + 2*(5/2)x = 4000

=> 20x = 4000

Therefore, x = 200.

So, y = (5/2) * 200 = 500.

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

    = Rs. (2400 + 1500)

    = Rs. 3900.

Ques4. If a - b = 3 and a2 + b2 = 29, find the value of ab.
A.    10
B.    12
C.    15
D.    18

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Answer: Option A

Explanation:

2ab = (a2 + b2) - (a - b)2

   = 29 - 9 = 20

   => ab = 10.

Ques5. The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
A.    Rs. 1200
B.    Rs. 2400
C.    Rs. 4800
D.    Cannot be determined
E.    None of these

 

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Answer: Option B

Explanation:

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 .... (i)

    and x + 6y = 1600 .... (ii)


Divide equation (i) by 2, we get the below equation. 

=> x +  2y =  800. --- (iii)

Now subtract (iii) from (ii)

 x +  6y = 1600  (-)
 x +  2y =  800  
----------------
      4y =  800
----------------

Therefore, y = 200.

Now apply value of y in (iii)

=>  x + 2 x 200 = 800

=>  x + 400 = 800

Therefore x = 400

Solving (i) and (ii) we get x = 400, y = 200.

Therefore, Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

Example 6: 64 × 99

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Solution 6:
⇒Step 1: 64 – 1 = 63
⇒Step 2: Complement of 64 = 100 – 64 = 36
Ans: 6336.

Example 7: 678 × 999 = ?

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Solution 7:
⇒Step 1: 678 – 1 = 677
⇒Step 2: Compliment of 678 = 1000 – 678 = 322
Ans: 677322.

Example 8: 78 × 999 = ?

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Solution 8: Take 78 as 078 and solve normally.
⇒Step 1: 078 – 1 = 077
⇒Step 2: Compliment of 078 = 1000 – 078 = 922
Ans: 77922

Example  9: Square of number 988?

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Solution 9: Nearest best to 988 = 1000. 988 is less than 100 by 12
⇒Step 1: Subtract 988 by 12 = 988 – 12 = 976.
⇒Step 2: Square of 12 = 144 (Number of digits should be equal to number of zeros in base)
Ans: 976144 

Example 10: Square of number 102?

Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA  View Answer

Solution: Nearest best to 102 = 100. 102 is more than 100 by 2
⇒Step 1: Add 102 by 2 = 102 + 2 = 104.
⇒Step 2: Square of 2 = 04 (Number of digits should be equal to number of zeros in base)
Ans: 10404. 

The document Simplification and Approximation | Mathematics for RRB NTPC / ASM - RRB NTPC/ASM/CA/TA is a part of the RRB NTPC/ASM/CA/TA Course Mathematics for RRB NTPC / ASM.
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