Simplification & Approximation Rules & Smart Tricks for RBI Grade B Exam

The Quantitative Aptitude section in RBI Grade B Phase 1 has 30 questions. Simplification & Approximation usually takes 5–7 of them. That’s around 17 to 23% of the section. Today, we’ll discuss the most important concepts. The types of questions that appear again and again. Short tricks to solve them fast. And practice questions with full solutions. By the end, you’ll know how to attempt this part quickly. Read on to know the tricks needed to score these marks with confidence.

ALSO READ: How to Score Maximum Marks in Cloze Test & Ratio & Proportion Concepts and Shortcuts for RBI Grade B Exam

What is Simplification & Approximation?

In questions on Simplification & Approximation, you are given a numerical expression. It may involve addition, subtraction, multiplication, division, percentages, fractions, or square roots.

• Simplification means solving the expression exactly and finding the correct answer.
• Approximation means replacing numbers with close values to solve faster, without losing much accuracy.

Both types are designed to test your speed and calculation skills.

Why is This Topic Important for RBI Grade B?

• You may get 5–6 questions from this topic in the Quant section.
• It’s direct and scoring – no complex reasoning steps needed.
• If you are good with basic maths, these questions can be solved in less than a minute each.

Example – Simplification

Question:
25 × 4 + 120 ÷ 6 – 15 = ?

Solution:

• Step 1: Follow BODMAS (Brackets → Orders → Division/Multiplication → Addition → Subtraction)
• Step 2: 25 × 4 = 100
• Step 3: 120 ÷ 6 = 20
• Step 4: Now 100 + 20 – 15 = 105

Answer = 105

Example – Approximation

Question:
(49.8 × 20.2) ÷ 9.9 ≈ ?

Solution:

• Round 49.8 to 50, 20.2 to 20, and 9.9 to 10
• Now (50 × 20) ÷ 10 = 1000 ÷ 10 = 100

Much faster, right?

Rules You Must Remember

Rule 1: Follow BODMAS strictly

• Always do brackets first, then powers/square roots, then multiplication/division, then addition/subtraction.

Rule 2: Keep rounding simple for approximation

• Round to nearest 10, 100, or simple fraction.
• Example: 19.8 → 20, 199 → 200, 3.49 → 3.5

Rule 3: Use fraction–percentage links

• 1/2 = 50%, 1/3 ≈ 33.33%, 1/4 = 25%, 3/4 = 75%, 1/5 = 20%

Rule 4: Avoid over-rounding

• If you round too much, the answer may shift far from the options.

Smart Tricks to Save Time

Trick 1: Split numbers cleverly
Example: 25 × 16 = (25 × 4) × 4 = 100 × 4 = 400

Trick 2: Use multiplication shortcuts

• 11 × 32 = 352 (Add digits in the middle: 3 + 2 = 5 → 3 5 2)
• Squaring numbers ending with 5: 35² = (3×4) | 25 = 1225

Trick 3: Cancel before multiplying
Example: (48 × 25) ÷ 50 → Cancel 25 and 50 first → (48 × 1) ÷ 2 = 24

Trick 4: For approximation, match with options early

• Sometimes, you don’t need the exact value. Estimate and compare with choices.

How to Practise Effectively

1. Start with accuracy – Solve slowly at first to avoid silly mistakes.
   
2. Then work on speed – Set a timer: 1 minute per question.
3. Mix both types – Practice both simplification and approximation so you can switch easily in the exam.
4. Note your errors – Keep a small notebook for calculation mistakes and keep revising.

Practice Time

Try this:

(32.5 × 8.1) + (250 ÷ 5) ≈ ?

• Round 32.5 → 33, 8.1 → 8, so 33 × 8 = 264
• 250 ÷ 5 = 50
• Add: 264 + 50 = 314

Daily Mini Task

• Solve 10 simplification problems
• Solve 5 approximation problems
• Learn 1 new shortcut
• Revise fraction–percentage values

Repeat this for 10 days, and you’ll see your speed improve.

Sample Questions Simplification & Approximation

Solve the following questions. Each question carries one mark. Use smart calculation tricks wherever possible to save time.

Q1. Find the value of:

1250÷0.25+3.6×2501250 \div 0.25 + 3.6 \times 250

A. 5400
B. 5800
C. 5900
D. 6000

Answer: C

Solution: 1250÷0.25=50001250 \div 0.25 = 5000, 3.6×250=9003.6 \times 250 = 900. Adding: 5000+900=59005000 + 900 = 5900.

Q2. Simplify:

711 of 198 ÷ 38 of 160\frac{7}{11} \text{ of } 198 \ \div \ \frac{3}{8} \text{ of } 160

A. 1.8
B. 2.0
C. 2.1
D. 2.2

Answer: C

Solution: 711\frac{7}{11} of 198 = 126; 38\frac{3}{8} of 160 = 60. Now 126÷60=2.1126 \div 60 = 2.1.

Q3. Approximate to 2 decimal places:

15.99×24.0215.99 \times 24.02

A. 384.05
B. 384.08
C. 384.12
D. 384.15

Answer: B

Solution: Use (16−0.01)×(24+0.02)(16 – 0.01) \times (24 + 0.02) = 384+0.32−0.24−0.0002=384.0798384 + 0.32 – 0.24 – 0.0002 = 384.0798 → 384.08.

Q4. Find the exact value:

(125×24)÷(0.5×75)(125 \times 24) \div (0.5 \times 75)

A. 75
B. 80
C. 85
D. 90

Answer: B

Solution: 125×24=3000125 \times 24 = 3000; 0.5×75=37.50.5 \times 75 = 37.5; 3000÷37.5=803000 \div 37.5 = 80.

Q5. Calculate quickly:

2987×5032987 \times 503

A. 1,500,461
B. 1,501,461
C. 1,502,461
D. 1,503,461

Answer: C

Solution: 2987×500=1,493,5002987 \times 500 = 1,493,500, 2987×3=8,9612987 \times 3 = 8,961; total = 1,502,4611,502,461.

Takeaway

Simplification & Approximation is a high-return topic for RBI Grade B Phase 1. It’s all about strong basics, quick thinking, and clean calculation. The rules are simple, and the tricks are easy to master with practice. Spend a few minutes daily, and you’ll turn this into a guaranteed scoring area.

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