Introduction by Rahul SirHello Aspirants! I am Rahul Sir. Pipes and Cistern is one of the most scoring chapters in IBPS PO and Clerk aptitude. It is based on work and efficiency concepts and can be solved quickly using the right approach. In this guide, we will cover important concepts, shortcuts, examples, and exam strategies to help you solve questions accurately under time pressure. 1. Understanding the Basics Pipes fill a tank while outlets empty it. Treat filling rates as positive and emptying rates as negative. If Pipe A fills in 10 hours, its rate is 1/10 of the tank per hour. Build confidence by converting time into work rates before solving. Example: A fills in 10 hours and B in 15 hours. Together they fill 1/10+1/15=1/6 of the tank per hour, so the tank fills in 6 hours. Practice identifying whether a pipe fills or empties before writing equations. Pipes fill a tank while outlets empty it. Treat filling rates as positive and emptying rates as negative. If Pipe A fills in 10 hours, its rate is 1/10 of the tank per hour. Build confidence by converting time into work rates before solving. Example: A fills in 10 hours and B in 15 hours. Together they fill 1/10+1/15=1/6 of the tank per hour, so the tank fills in 6 hours. Practice identifying whether a pipe fills or empties before writing equations. Pipes fill a tank while outlets empty it. Treat filling rates as positive and emptying rates as negative. If Pipe A fills in 10 hours, its rate is 1/10 of the tank per hour. Build confidence by converting time into work rates before solving. Example: A fills in 10 hours and B in 15 hours. Together they fill 1/10+1/15=1/6 of the tank per hour, so the tank fills in 6 hours. Practice identifying whether a pipe fills or empties before writing equations. 2. Work Rate Formula Always use Rate = Work/Time. Assume total work as one tank. Add rates for filling pipes and subtract outlet rates. Example: A=1/8, B=1/12, outlet=1/24. Net=1/8+1/12-1/24=1/6, therefore 6 hours. Repeated practice improves speed. Always use Rate = Work/Time. Assume total work as one tank. Add rates for filling pipes and subtract outlet rates. Example: A=1/8, B=1/12, outlet=1/24. Net=1/8+1/12-1/24=1/6, therefore 6 hours. Repeated practice improves speed. Always use Rate = Work/Time. Assume total work as one tank. Add rates for filling pipes and subtract outlet rates. Example: A=1/8, B=1/12, outlet=1/24. Net=1/8+1/12-1/24=1/6, therefore 6 hours. Repeated practice improves speed. 3. LCM Method Take the LCM of times to simplify calculations. If pipes take 12 and 18 hours, LCM is 36 units. Rates become 3 and 2 units/hour. Together they fill 5 units/hour, taking 36/5 hours. This reduces fractions and calculation errors. Take the LCM of times to simplify calculations. If pipes take 12 and 18 hours, LCM is 36 units. Rates become 3 and 2 units/hour. Together they fill 5 units/hour, taking 36/5 hours. This reduces fractions and calculation errors. Take the LCM of times to simplify calculations. If pipes take 12 and 18 hours, LCM is 36 units. Rates become 3 and 2 units/hour. Together they fill 5 units/hour, taking 36/5 hours. This reduces fractions and calculation errors. 4. Alternate Operation Questions often ask pipes operating alternately. Calculate work completed in one full cycle and divide remaining work carefully. Track partial cycles accurately and avoid assuming continuous operation. Questions often ask pipes operating alternately. Calculate work completed in one full cycle and divide remaining work carefully. Track partial cycles accurately and avoid assuming continuous operation. Questions often ask pipes operating alternately. Calculate work completed in one full cycle and divide remaining work carefully. Track partial cycles accurately and avoid assuming continuous operation. Questions often ask pipes operating alternately. Calculate work completed in one full cycle and divide remaining work carefully. Track partial cycles accurately and avoid assuming continuous operation. Questions often ask pipes operating alternately. Calculate work completed in one full cycle and divide remaining work carefully. Track partial cycles accurately and avoid assuming continuous operation. Questions often ask pipes operating alternately. Calculate work completed in one full cycle and divide remaining work carefully. Track partial cycles accurately and avoid assuming continuous operation. 5. Leakage Problems Treat leakage as a negative rate. Example: Fill=1/5, leak=1/20. Net=3/20 so total time=20/3 hours. Carefully read whether leakage starts immediately or later. Treat leakage as a negative rate. Example: Fill=1/5, leak=1/20. Net=3/20 so total time=20/3 hours. Carefully read whether leakage starts immediately or later. Treat leakage as a negative rate. Example: Fill=1/5, leak=1/20. Net=3/20 so total time=20/3 hours. Carefully read whether leakage starts immediately or later. Treat leakage as a negative rate. Example: Fill=1/5, leak=1/20. Net=3/20 so total time=20/3 hours. Carefully read whether leakage starts immediately or later. Treat leakage as a negative rate. Example: Fill=1/5, leak=1/20. Net=3/20 so total time=20/3 hours. Carefully read whether leakage starts immediately or later. Treat leakage as a negative rate. Example: Fill=1/5, leak=1/20. Net=3/20 so total time=20/3 hours. Carefully read whether leakage starts immediately or later. 6. Tank Already Filled If the tank is partially filled, calculate remaining fraction only. Multiply the remaining work by the net rate to find the required time. If the tank is partially filled, calculate remaining fraction only. Multiply the remaining work by the net rate to find the required time. If the tank is partially filled, calculate remaining fraction only. Multiply the remaining work by the net rate to find the required time. If the tank is partially filled, calculate remaining fraction only. Multiply the remaining work by the net rate to find the required time. If the tank is partially filled, calculate remaining fraction only. Multiply the remaining work by the net rate to find the required time. If the tank is partially filled, calculate remaining fraction only. Multiply the remaining work by the net rate to find the required time. 7. Exam Shortcuts Memorize common fractions, use LCM, avoid decimals unless necessary, and verify whether pipes are opened or closed at different times.

Hello Aspirants! I am Rahul Sir, and today we are going to master one of the most interesting and scoring topics in Quantitative Aptitude—Problems on Chain Rule. Although Chain Rule questions are not asked as frequently as Simplification or Arithmetic, they regularly appear in IBPS PO, IBPS Clerk, SBI PO, RBI Assistant, LIC AAO, and other banking examinations. The best part is that these questions follow a logical pattern and can often be solved without lengthy calculations if you understand the underlying concept. The Chain Rule is simply a method of comparing two or more quantities connected by multiple variables. Instead of solving lengthy equations, we create a chain of equivalent ratios to determine the required value. These questions usually involve men, machines, work, wages, money, goods, distance, or production where multiple factors influence the final result. Many students initially find Chain Rule confusing because they try to memorize formulas. However, there is no need to memorize dozens of formulas. Once you understand which quantities are directly proportional and which are inversely proportional, solving these questions becomes surprisingly easy. In this guide, I will explain every important concept, shortcut, and exam-oriented trick with examples. We’ll also discuss common mistakes, calculation techniques, and smart approaches that save valuable time during the exam. Let’s begin your journey towards mastering Chain Rule and increasing your Quant score. 1. What is the Chain Rule? The Chain Rule is a mathematical technique used when a problem contains multiple variables that affect one another. Instead of handling every variable separately, we compare them through proportional relationships. Suppose 10 men complete a work in 15 days. How many days will 15 men require? Here, Men ↑ → Days ↓ Since more workers finish the work faster, the quantities are inversely proportional. The relation becomes: 10 × 15 = 15 × x x = 10 days Now consider another example. If 8 machines produce 480 items in 6 days, how many items will 12 machines produce in 5 days? Here, Machines → Direct Days → Direct Items → Direct Required items = 480 × (12/8) × (5/6) = 600 items Notice that instead of solving complicated equations, we simply multiplied the proportional factors. Chain Rule questions commonly involve: The biggest advantage of understanding Chain Rule is that one concept solves dozens of question types. 2. Understanding Direct and Inverse Proportion Before solving Chain Rule questions, students must identify whether quantities are directly or inversely proportional. Direct Proportion If one quantity increases and the other also increases, they are directly proportional. Examples: Example: 5 workers make 200 bricks. 10 workers make? 200 × (10/5) = 400 bricks Inverse Proportion If one quantity increases while the other decreases, they are inversely proportional. Examples: Example: 8 men complete work in 15 days. 12 men complete work in? 15 × (8/12) = 10 days Shortcut Table Direct Increase → Increase Decrease → Decrease Inverse Increase → Decrease Decrease → Increase One of the biggest mistakes students make is applying the wrong proportional relationship. Always identify the nature of variables before calculation. Practice recognizing these relationships until they become automatic. 3. Step-by-Step Method to Solve Chain Rule Questions Many students panic when they see three or four variables together. The solution is to follow a fixed method. Step 1 Write all variables. Example: Workers Days Hours Production Step 2 Identify the required quantity. Step 3 Compare every variable with the required quantity. Step 4 Mark Direct (D) or Inverse (I). Step 5 Multiply all ratios. Example 15 men working 8 hours daily complete work in 12 days. How many days will 20 men working 6 hours require? Men → Inverse Hours → Inverse Days → Required Calculation 12 × (15/20) × (8/6) = 12 days Notice how both proportional relationships combine smoothly. Following these five steps reduces mistakes significantly. Never skip writing D or I beside every variable. It takes only a few seconds but saves many marks. 4. Shortcut Tricks for Faster Calculations IBPS exams demand speed. Instead of writing every step, learn these shortcuts. Trick 1 Cancel common factors. Example 360 × (12/18) Cancel 12 and 18 2/3 Result 360 × 2/3 = 240 Trick 2 Simplify before multiplication. 480 × 15/20 15/20 = 3/4 480 × 3/4 = 360 Trick 3 Multiply numerators first after simplification. Trick 4 Avoid decimal calculations. Convert decimals into fractions. 0.75 = 3/4 1.5 = 3/2 Trick 5 Keep ratios in lowest terms. Instead of 25/100 Use 1/4 Example 18 workers produce 720 toys in 10 days. 24 workers produce? 720 × (24/18) 720 × 4/3 = 960 These calculation tricks save nearly 30–40 seconds per question. 5. Chain Rule Questions on Men and Work The most common Chain Rule questions involve men, days, and working hours. Example 12 men working 8 hours complete work in 15 days. How many men are required to complete it in 12 days by working 10 hours daily? Required variable Men Days → Inverse Hours → Inverse Calculation 12 × (15/12) × (8/10) = 12 Thus, 12 men are required. Another Example 20 workers build 100 meters of road in 5 days. 30 workers build? 100 × (30/20) = 150 meters Remember Workers ↑ Production ↑ Workers ↑ Days ↓ Workers ↑ Hours remain constant unless mentioned. Exam Tip Never assume working hours unless provided. Always read the question carefully. 6. Chain Rule Questions on Cost and Quantity These questions test proportional thinking rather than arithmetic. Example If 12 kg sugar costs ₹720, 18 kg costs? 720 × (18/12) = ₹1080 Example 15 notebooks cost ₹450. 25 notebooks cost? 450 × (25/15) = ₹750 Sometimes discounts or rates change. Always compare unit prices carefully. Shortcut Cost = Quantity × Rate If rate remains constant, Cost is directly proportional to quantity. Students often overcomplicate these problems, but they usually involve simple direct proportions. Practice converting every question into ratio form before calculating. 7. Chain Rule Questions on Speed, Distance and Time Speed-Time problems also use Chain Rule. Example A

Hello Aspirants, I am Rahul Sir, and over the years I have trained thousands of banking aspirants preparing for IBPS PO, IBPS Clerk, SBI PO, RBI Assistant, and other competitive examinations. Among all aptitude topics, Ratio and Proportion is one of the most scoring yet misunderstood chapters. Students often memorize formulas without understanding the relationships between quantities, leading to unnecessary mistakes in the examination. Ratio and Proportion questions are frequently asked in IBPS exams directly and indirectly through topics like Partnership, Mixture & Allegation, Profit & Loss, Ages, Time & Work, and Data Interpretation. If your basics are clear, you can solve many questions mentally within seconds. The secret is not learning dozens of formulas but understanding the concept behind ratios. Once you know how quantities compare, you can easily solve even lengthy-looking problems using smart tricks. In this guide, I will explain every important concept, shortcuts, and examination strategies with solved examples that will help you maximize your score. Let’s master Ratio and Proportion the smart way. 1. Understanding Ratio and Proportion Basics Ratio represents the comparison between two quantities of the same kind using division. If two quantities are A and B, their ratio is written as A or A/B. Proportion states that two ratios are equal. For example, 3:5 = 12:20 Since both represent the same relationship, they are said to be in proportion. Important Formula If a = c then a × d = b × c This is called cross multiplication. Example The ratio of boys to girls is 4:5. If there are 36 girls, find the number of boys. Solution Girls = 5 parts 1 part = 36 ÷ 5 = 7.2 Boys = 4 × 7.2 = 28.8 Instead of decimals, use multiplication. 5 parts = 36 4 parts = (36 × 4)/5 = 144/5 = 28.8 In competitive exams, values are generally chosen to give whole numbers. Suppose girls are 45. Then 1 part = 9 Boys = 36 Shortcut Always calculate one part first whenever total value of one side is given. Common Mistakes 2. Simplifying Ratios Quickly One of the fastest ways to solve questions is learning to simplify ratios mentally. Example 48:72 Find HCF HCF = 24 Ratio = 2:3 Shortcut Always divide by highest common factor. Decimal Ratio 0.8 : 1.2 Multiply by 10 8:12 Simplify 2:3 Fraction Ratio 1/2 : 3/4 Multiply both by LCM of denominators LCM = 4 Ratio 2:3 Example 25 kg : 500 g Convert units 25 kg = 25000 g Ratio 25000:500 50:1 Exam Trick Always convert units before taking ratio. Practice 150:210 Answer 5:7 Speed Tips The faster you simplify, the faster every question becomes. 3. Finding Unknown Values Using Ratio Many IBPS questions involve finding missing values. Suppose A = 7:9 If A = 56 Find B. Solution 7 parts = 56 1 part = 8 9 parts = 72 Answer = 72 Shortcut Formula Unknown = (Known × Required Ratio)/Known Ratio Example Ratio of pens and pencils = 5:8 Pens = 30 Pencils = ? = (30 × 8)/5 = 48 Reverse Question Ratio = 3:8 Second quantity = 80 First = (80 × 3)/8 =30 Trick Never calculate total unless asked. Go directly using proportional multiplication. Practice Ratio 8:11 Second quantity = 44 First? Answer 32 These direct questions should take less than 20 seconds. 4. Dividing Quantities According to Ratio Sharing money or objects according to ratio is a favorite IBPS topic. Example Divide ₹1800 in ratio 4:5. Solution Total parts 4+5=9 One part 1800/9=200 Shares 4×200=800 5×200=1000 Shortcut Formula Share (Total × Individual Ratio)/(Sum of Ratios) Example Divide ₹3900 in ratio 2:3:8 Total parts 13 Shares 600 900 2400 Speed Trick Always calculate one part first. Then multiply. Practice ₹960 Ratio 5:3 Shares 600 360 Common Mistake Students divide using difference instead of total parts. Always use addition of ratios. 5. Compound Ratio Made Easy Compound ratio combines two or more ratios. Example 2:3 and 4:5 Compound ratio 2×4 : 3×5 8:15 Example 3:7 5:9 Compound ratio 15:63 Simplify 5:21 Applications Compound ratio appears in Shortcut Multiply numerators together. Multiply denominators together. Simplify. Practice 6:5 2:9 Compound ratio 12:45 4:15 Easy marks if practiced properly. 6. Proportion Tricks and Cross Multiplication Cross multiplication saves enormous time. If x:15 = 12:20 Then 20x =15×12 180 x=9 Example 5 =10:18 Cross multiply 90=10x x=9 Direct Formula a=c Then ad=bc Practice 8:12 = x:18 8×18=12x 144=12x x=12 Shortcut Never expand ratios unnecessarily. Use cross multiplication directly. 7. Ratio Changes After Increase or Decrease These questions are very common. Example Ratio 3:5 Both increase by 6. New ratio ? Suppose original 3x and 5x (3x+6)/(5x+6) Use new condition if provided. Example Ratio 2:3 After adding 10 to each 3:4 Find original numbers. Solution 2x+10 3x+10 Cross multiply 4(2x+10) = 3(3x+10) 8x+40 9x+30 x=10 Numbers 20 30 Trick Represent original values as multiples of x. Never assume actual values. 8. Advanced Ratio Problems in IBPS Example Income ratio 5:7 Expenses ratio 4:5 Savings ₹2000 and ₹3000 Find incomes. Represent incomes 5x 7x Expenses 4y 5y Use savings equations. Solve simultaneously. Another Example Age ratio 3:5 After 10 years 5:7 Represent ages 3x 5x Apply condition. These questions appear difficult but become easy using variables. Tips 9. Ratio Questions in Data Interpretation Pie charts, bar graphs and tables often require ratio calculations. Example Company A 120 employees Company B 180 employees Ratio 2:3 Percentage to Ratio 40% 60% Ratio 2:3 Example Male 240 Female 360 Ratio 2:3 Shortcut Convert percentages into ratios. 25% 75% 1:3 20% 80% 1:4 30% 70% 3:7 These shortcuts save valuable seconds during DI. 10. Exam Strategies, Tricks and Practice Plan To master Ratio and Proportion: Daily Practice Solve 20 questions daily. Memorize Common Ratios 1:2 2:3 3:5 4:7 5:8 These appear repeatedly. Learn Percentage-Ratio Conversion 20%=1:4 25%=1:3 40%=2:3 50%=1:1 60%=3:2 75%=3:1 Time Target Easy Question 20 seconds Moderate 40 seconds Difficult 60 seconds Avoid These Mistakes Final Advice Concepts are more important than formulas. Practice enough

Hello Aspirants, Problems on Surds and Indices are important topics in the Quantitative Aptitude section of IBPS PO and Clerk examinations. Although the number of direct questions may be limited, concepts from surds and indices frequently appear in simplification, approximation, quadratic equations, and data interpretation questions. Candidates who understand the fundamental laws and shortcuts can solve these questions within seconds. Many students find roots and powers intimidating because of the symbols and mathematical expressions involved. However, once the basic rules are mastered, these questions become among the easiest scoring areas in banking exams. In this article, I will explain the concepts of surds and indices, important formulas, shortcut techniques, common mistakes, and exam-oriented examples that will help you improve both speed and accuracy. Let us begin our journey toward mastering Surds and Indices for IBPS PO and Clerk exams. — Rahul Sir 1. Understanding the Concept of Indices Indices, also known as exponents or powers, represent repeated multiplication of a number by itself. If a number ‘a’ is multiplied by itself ‘n’ times, it is written as aⁿ. For example: 2³ = 2 × 2 × 2 = 8 5² = 25 In IBPS exams, candidates are expected to apply exponent rules quickly rather than perform lengthy calculations. Understanding indices helps simplify complex expressions within seconds. Important Rules: aᵐ × aⁿ = aᵐ⁺ⁿ aᵐ ÷ aⁿ = aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ a⁰ = 1 a⁻ⁿ = 1/aⁿ Example: 2⁵ × 2³ = 2⁸ = 256 Example: 5⁷ ÷ 5⁴ = 5³ = 125 A strong understanding of exponent laws reduces calculation time significantly and helps solve simplification-based questions efficiently. 2. Understanding Surds and Their Importance A surd is an irrational root that cannot be simplified into a whole number. Examples: √2 √3 √5 √7 These numbers continue indefinitely without repeating. In banking exams, surds are generally used in simplification and approximation questions. Examples of Perfect Surds: √25 = 5 √49 = 7 √81 = 9 Examples of Non-Perfect Surds: √2 √6 √11 √17 Example: √50 = √(25 × 2) = 5√2 This process is known as simplifying surds. Candidates should practice identifying perfect squares because many questions can be solved instantly by breaking numbers into perfect square factors. Knowing squares from 1 to 50 and cubes from 1 to 20 is extremely beneficial for faster calculations. 3. Important Laws of Indices for Competitive Exams The laws of indices form the backbone of many simplification problems. Law 1: aᵐ × aⁿ = aᵐ⁺ⁿ Example: 3² × 3⁴ = 3⁶ = 729 Law 2: aᵐ ÷ aⁿ = aᵐ⁻ⁿ Example: 7⁵ ÷ 7² = 7³ = 343 Law 3: (aᵐ)ⁿ = aᵐⁿ Example: (2³)² = 2⁶ = 64 Law 4: (ab)ⁿ = aⁿbⁿ Example: (2 × 3)² = 2² × 3² = 36 Law 5: (a/b)ⁿ = aⁿ/bⁿ Example: (4/5)² = 16/25 These laws frequently appear in IBPS simplification questions and should be memorized thoroughly. 4. Simplifying Surds Quickly One of the easiest ways to simplify surds is factorization. Example: √72 = √(36 × 2) = 6√2 Example: √108 = √(36 × 3) = 6√3 Example: √200 = √(100 × 2) = 10√2 Shortcut: Always look for the largest perfect square factor. Common Perfect Squares: 1² = 1 2² = 4 3² = 9 4² = 16 5² = 25 6² = 36 7² = 49 8² = 64 9² = 81 10² = 100 In the examination hall, quickly identifying perfect square factors can reduce lengthy calculations and save valuable time. 5. Rationalization of Surds Rationalization means removing the surd from the denominator. Example: 1/√2 Multiply numerator and denominator by √2. = √2/2 Example: 3/√5 = 3√5/5 For expressions like: 1/(√3 + √2) Multiply by the conjugate. = (√3 − √2)/(3 − 2) = √3 − √2 This technique is often tested indirectly through simplification questions. Remember: Conjugate of (a + b) is (a − b) Conjugate of (a − b) is (a + b) Mastering rationalization improves your ability to solve advanced simplification problems rapidly. 6. Fractional and Negative Indices Fractional indices represent roots. a¹/² = √a a¹/³ = ∛a Example: 64¹/² = √64 = 8 Example: 125¹/³ = ∛125 = 5 Negative indices indicate reciprocals. a⁻² = 1/a² Example: 2⁻³ = 1/2³ = 1/8 Example: 5⁻² = 1/25 These concepts frequently appear in simplification and approximation questions in IBPS exams. Candidates should practice converting roots into exponents and vice versa. 7. Shortcut Techniques for IBPS Exams The key to scoring well is applying shortcuts rather than lengthy calculations. Technique 1: Break powers into manageable parts. Example: 2¹⁰ = 1024 Technique 2: Use exponent laws instead of expansion. Example: 3⁴ × 3² = 3⁶ Technique 3: Memorize squares up to 50. Technique 4: Memorize cubes up to 20. Technique 5: Recognize common roots. √144 = 12 √169 = 13 √196 = 14 Example: √288 = √(144 × 2) = 12√2 Such techniques can save 15–20 seconds per question, making a significant difference in overall exam performance. 8. Common Exam Questions with Examples Example 1: 2⁴ × 2⁵ = 2⁹ = 512 Example 2: √98 = √(49 × 2) = 7√2 Example 3: (3²)³ = 3⁶ = 729 Example 4: 81¹/² = 9 Example 5: 125¹/³ = 5 Example 6: 4⁻² = 1/16 Example 7: √147 = √(49 × 3) = 7√3 These examples demonstrate the direct application of formulas and are similar to questions commonly asked in banking examinations. 9. Common Mistakes Students Should Avoid Mistake 1: Ignoring exponent laws. Mistake 2: Expanding expressions unnecessarily. Mistake 3: Forgetting negative exponent rules. Mistake 4: Incorrect simplification of surds. Example: √18 ≠ 9√2 Correct: √18 = 3√2 Mistake 5: Confusing square roots with cube roots. Mistake 6: Not remembering perfect squares and cubes. Mistake 7: Calculation errors during rationalization. Regular practice and careful application of formulas help avoid these mistakes and improve accuracy. 10. Practice Strategy to Master Surds and Indices To become proficient in Surds and Indices: Day 1–2:Learn all exponent laws. Day 3–4:Practice simplification of surds. Day 5:Learn rationalization

Hello Aspirants, Square Root and Cube Root questions are among the most scoring topics in IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, RBI Assistant, and other banking examinations. These questions frequently appear in Simplification, Approximation, and Number Series sections. Candidates who master the shortcuts for identifying square roots and cube roots can solve questions within seconds and significantly improve their speed. Many students fear lengthy calculations, but the reality is that banking exams rarely require traditional methods. Instead, they test your understanding of number patterns, unit digits, perfect squares, perfect cubes, and estimation techniques. In this article, I will explain the most effective tricks and strategies for solving Square Root and Cube Root problems quickly. Along with conceptual understanding, you will find practical examples similar to those asked in IBPS examinations. By practicing these techniques regularly, you can increase your accuracy and reduce calculation time during the exam. Let us begin our journey toward mastering Square Root and Cube Root problems. — Rahul Sir 1. Understanding Square Roots and Cube Roots Before learning shortcuts, it is important to understand the basic concepts. A square root of a number is a value that, when multiplied by itself, gives the original number. Example: √64 = 8 Because: 8 × 8 = 64 Similarly, a cube root of a number is a value that, when multiplied three times by itself, gives the original number. Example: ∛125 = 5 Because: 5 × 5 × 5 = 125 In banking exams, square roots and cube roots are commonly tested through simplification and approximation questions. Candidates often waste time performing long calculations when a simple pattern-based approach can produce the answer instantly. Understanding perfect squares and perfect cubes is the foundation of these questions. Every aspirant should memorize squares from 1 to 50 and cubes from 1 to 30. This helps in quickly recognizing answers during examinations. Examples: 25 = 5² 36 = 6² 49 = 7² 64 = 8² 125 = 5³ 216 = 6³ 343 = 7³ 512 = 8³ When solving banking aptitude questions, recognizing these values instantly can save valuable seconds. Once students become comfortable with these fundamentals, advanced shortcut methods become much easier to apply. Regular practice of perfect squares and cubes develops mental calculation abilities and boosts confidence during exams. Most toppers rely on these basics before applying advanced techniques. 2. Memorizing Important Perfect Squares One of the biggest advantages in IBPS exams comes from memorizing perfect squares. The majority of square root questions involve numbers whose roots lie between 1 and 50. Therefore, candidates should memorize these values thoroughly. Important Squares: 11² = 121 12² = 144 13² = 169 14² = 196 15² = 225 16² = 256 17² = 289 18² = 324 19² = 361 20² = 400 Instead of memorizing randomly, create patterns. Observe: 21² = 441 22² = 484 23² = 529 24² = 576 25² = 625 Such patterns help recall values quickly. Example Question: √529 = ? Since 23² = 529 Answer = 23 Another Example: √784 = ? Since 28² = 784 Answer = 28 IBPS simplification questions often contain square roots of perfect squares. If you know these values instantly, solving the entire question becomes much faster. Daily revision of square tables takes only a few minutes but provides huge benefits in competitive exams. Candidates who memorize squares properly can solve several questions without using rough work. 3. Shortcut Method to Find Square Roots of Large Perfect Squares Large perfect square questions appear frequently in bank exams. Example: √1764 Step 1: Look at the last two digits. 64 can come from: 8² = 64 2² = 04 Since the unit digit is 4, possibilities are 2 or 8. Step 2: Ignore the last two digits. Remaining number = 17 Find the largest square less than 17. 4² = 16 5² = 25 Therefore, first digit = 4. Possible answers: 42 or 48 Step 3: Choose larger unit digit because 17 is closer to 16. Answer = 42 Check: 42 × 42 = 1764 This method eliminates lengthy calculations and allows candidates to solve large square root questions in seconds. Practice with numbers such as: 2025 2401 3136 4096 5184 These values frequently appear in competitive examinations. 4. Important Perfect Cubes to Memorize Cube root questions become extremely easy when perfect cubes are memorized. Important Cubes: 1³ = 1 2³ = 8 3³ = 27 4³ = 64 5³ = 125 6³ = 216 7³ = 343 8³ = 512 9³ = 729 10³ = 1000 Continue memorization up to 30³. Examples: 11³ = 1331 12³ = 1728 13³ = 2197 14³ = 2744 15³ = 3375 Example Question: ∛3375 = ? Since 15³ = 3375 Answer = 15 Many banking exams directly ask cube root values within simplification questions. Candidates who know cube tables can answer instantly without calculation. Daily practice of cubes strengthens numerical ability and improves overall speed in quantitative aptitude sections. 5. Unit Digit Trick for Cube Roots Cube root questions can be solved using unit digit patterns. Observe: 1³ ends in 1 2³ ends in 8 3³ ends in 7 4³ ends in 4 5³ ends in 5 6³ ends in 6 7³ ends in 3 8³ ends in 2 9³ ends in 9 This pattern is extremely important. Example: ∛2197 Last digit = 7 Only 3³ ends with 7. So unit digit = 3. Remaining digits = 2 Largest cube below 2 is 1³. Therefore first digit = 1. Answer = 13 Check: 13³ = 2197 This shortcut is commonly used in IBPS and SBI exams because it allows instant identification of cube roots. 6. Estimation Techniques for Approximation Questions Approximation questions often include square roots and cube roots. Example: √998 ≈ ? Nearest perfect square: 31² = 961 32² = 1024 998 is closer to 1024. Answer ≈ 32 Similarly: ∛990 ≈ ? 10³ = 1000 Answer ≈ 10 Approximation helps solve questions quickly without exact calculations. IBPS exams frequently test estimation

Hello Aspirants, Simplification is one of the most scoring topics in IBPS PO and Clerk examinations. Every year, questions from simplification appear in the Quantitative Aptitude section, allowing candidates to secure quick marks within a short time. Since these questions are calculation-based, mastering shortcuts, approximation techniques, and BODMAS rules can significantly improve both speed and accuracy. Many students spend too much time performing lengthy calculations and lose valuable marks. However, with the right strategies, simplification questions can be solved within seconds. The key lies in understanding mathematical operations, memorizing squares and cubes, improving calculation speed, and practicing regularly. In this guide, I will share important tips, tricks, and examples that will help you tackle simplification questions efficiently in IBPS PO and Clerk exams. Whether you are a beginner or an advanced learner, these methods will help you improve your score and confidence in the examination hall. — Rahul Sir 1. Understanding Simplification Questions Simplification refers to finding the correct value of a mathematical expression involving arithmetic operations such as addition, subtraction, multiplication, division, percentages, square roots, and exponents. In IBPS exams, simplification questions are designed to test calculation speed and numerical accuracy. Candidates usually encounter 10–15 questions from this topic. Since the difficulty level is moderate, these questions provide an excellent opportunity to maximize scores. For example: 48 ÷ 6 × 3 + 12 Using BODMAS: 48 ÷ 6 = 8 8 × 3 = 24 24 + 12 = 36 Answer = 36 Many students mistakenly solve operations from left to right without following the correct order. Understanding the structure of simplification questions is therefore essential. Regular practice helps candidates identify patterns and avoid common mistakes. Since IBPS exams are highly competitive, solving simplification questions accurately within a few seconds can create a significant advantage over other candidates. Aspirants should begin by strengthening basic arithmetic skills and gradually move towards advanced simplification problems involving fractions, roots, percentages, and powers. 2. Mastering the BODMAS Rule BODMAS stands for: B – Brackets O – Orders (Powers and Roots) D – Division M – Multiplication A – Addition S – Subtraction This rule determines the sequence of operations in simplification problems. Example: 20 + (15 ÷ 3) × 4 Bracket first: 15 ÷ 3 = 5 Multiplication: 5 × 4 = 20 Addition: 20 + 20 = 40 Answer = 40 Many errors occur because students ignore the order of operations. Learning BODMAS thoroughly prevents unnecessary mistakes. In IBPS exams, questions often include multiple operations in a single expression. Candidates must quickly identify which operation to perform first. Practice questions involving nested brackets, powers, roots, and fractions. With repeated exposure, the application of BODMAS becomes automatic. A strong command of BODMAS is the foundation of simplification and directly improves accuracy and speed in examinations. 3. Importance of Fast Calculation Skills Fast calculations play a major role in simplification. Candidates who perform arithmetic quickly can save valuable time for difficult questions. Important calculations to memorize: Squares up to 50 Cubes up to 30 Fraction-to-percentage conversions Tables up to 25 Example: 25² = 625 35² = 1225 45² = 2025 Similarly: 12.5% = 1/8 20% = 1/5 25% = 1/4 Rapid recall eliminates the need for lengthy calculations during the exam. Daily practice of mental arithmetic improves numerical agility and confidence. Candidates should spend 15–20 minutes every day practicing multiplication, division, squares, and percentage calculations. Using calculation shortcuts reduces dependence on rough work and enhances speed significantly. 4. Approximation Techniques Approximation helps solve lengthy calculations quickly. Example: 498 × 21 Approximate: 500 × 20 = 10000 Actual answer will be close to 10000. Approximation is particularly useful when answer choices are widely separated. Common approximation rules: 498 ≈ 500 199 ≈ 200 9.98 ≈ 10 4.99 ≈ 5 Candidates should use approximation only when answer options allow it. If options are very close, exact calculation becomes necessary. Approximation saves time and is one of the most important techniques used by toppers in banking examinations. 5. Solving Questions with Fractions Fractions frequently appear in simplification problems. Important Fraction Values: 1/2 = 50% 1/3 = 33.33% 1/4 = 25% 1/5 = 20% 1/8 = 12.5% Example: (3/4 × 200) + (1/5 × 100) = 150 + 20 = 170 Understanding fraction conversions simplifies calculations dramatically. Practice converting percentages into fractions and vice versa. This technique helps solve questions involving ratios, percentages, and simplification much faster. 6. Square Roots and Cube Roots Tricks Questions involving roots are common in IBPS exams. Example: √144 = 12 √225 = 15 ∛125 = 5 ∛216 = 6 Memorizing perfect squares and cubes enables instant solutions. Candidates should learn: Squares up to 50 Cubes up to 30 Square roots of common perfect squares Cube roots of common perfect cubes Strong knowledge of roots reduces calculation time and improves accuracy. 7. Percentage-Based Simplification Percentages are often combined with arithmetic operations. Example: 25% of 400 + 20% of 250 = 100 + 50 = 150 Useful conversions: 50% = 1/2 25% = 1/4 75% = 3/4 12.5% = 1/8 Converting percentages into fractions makes calculations easier and faster. Candidates should regularly practice percentage calculations because they appear in multiple aptitude topics. 8. Common Mistakes to Avoid Frequent mistakes include: Ignoring BODMAS Incorrect decimal calculations Wrong percentage conversions Calculation errors Rushing through questions Example: 16 ÷ 4 × 2 Correct: 16 ÷ 4 = 4 4 × 2 = 8 Many students mistakenly write: 16 ÷ 8 = 2 Avoiding such mistakes can improve scores significantly. Accuracy should always accompany speed. 9. Time Management Strategies Simplification questions should ideally be solved within 20–30 seconds. Tips: Attempt simplification first. Skip lengthy calculations initially. Use approximation wherever possible. Avoid spending more than 40 seconds on a single question. Regular timed practice develops examination temperament and improves overall performance. Candidates should solve mock tests regularly to identify weak areas and improve speed. 10. Practice Examples for IBPS Exams Example 1: 45 + 15 × 2 = 45 + 30 = 75 Example 2: 225 ÷ 15

Decimal Fraction is one of the most important topics in the Quantitative Aptitude section of IBPS PO and Clerk examinations. Questions based on decimal fractions frequently appear in simplification, approximation, percentage, ratio, profit and loss, and data interpretation sections. Candidates who understand decimal concepts thoroughly can solve many aptitude questions quickly and accurately. I am Rahul Sir, and over the years of mentoring banking aspirants, I have observed that students often lose marks due to simple mistakes in decimal calculations. The good news is that decimal fraction questions are generally easy if approached with the right techniques and shortcuts. By mastering decimal operations such as addition, subtraction, multiplication, division, conversion to percentages, and approximation methods, candidates can significantly improve their speed. In banking examinations, every second matters. Spending excessive time on calculations can affect your overall performance. Therefore, learning smart methods to handle decimal fractions becomes essential. Through consistent practice and understanding of core concepts, aspirants can confidently tackle even the most challenging decimal-based problems. In this guide, we will discuss important concepts, practical shortcuts, exam-oriented tricks, and solved examples that will help you score better in IBPS PO and Clerk exams. Let us begin our journey toward mastering decimal fractions. 1. Understanding Decimal Fractions A decimal fraction is a fraction whose denominator is a power of 10 such as 10, 100, 1000, etc. These fractions are represented using decimal points. Examples: Decimal fractions are widely used in aptitude examinations because they simplify complex calculations. Understanding place values is crucial: Example: In 23.456 Many students make mistakes by ignoring place values during calculations. Examiners often design questions that test this basic understanding. Shortcut: To convert a fraction into decimal form, divide the numerator by the denominator. Example: 3/8 = 3 ÷ 8 = 0.375 To convert decimal to fraction: 0.75 = 75/100 = 3/4 IBPS questions frequently involve conversions between fractions, percentages, and decimals. Therefore, candidates should memorize common equivalents: A strong understanding of decimal fractions forms the foundation for solving many advanced aptitude topics efficiently. 2. Addition and Subtraction of Decimal Fractions Adding and subtracting decimal numbers is simple when decimal points are aligned properly. Example: 12.45 + 7.36 Example: 25.80 – 13.45 Important Rule: Always place decimal points exactly below each other. Common Error: Students sometimes add numbers without aligning decimals, leading to incorrect answers. IBPS Shortcut: Break numbers into whole and decimal parts. Example: 15.75 + 4.25 = (15 + 4) + (0.75 + 0.25) = 19 + 1 = 20 Practice mental calculations: Exam Example: A shopkeeper earns ₹12.75 on one item and ₹18.25 on another. Find total profit. Solution: 12.75 + 18.25 = 31.00 Answer = ₹31 Regular practice improves speed and reduces calculation mistakes during examinations. 3. Multiplication of Decimal Fractions Decimal multiplication is a favorite area in banking aptitude. Rule: Multiply numbers normally and place decimal points according to total decimal digits. Example: 2.5 × 1.2 Step 1: 25 × 12 = 300 Step 2: Total decimal places = 2 Answer: 3.00 Shortcut Multiplications: Example: 0.5 × 84 = 42 0.25 × 80 = 20 0.75 × 40 = 30 Exam Example: Find: 2.4 × 1.5 24 × 15 = 360 Two decimal places Answer = 3.60 Approximation Trick: 2.48 × 4.02 ≈ 2.5 × 4 = 10 Useful for approximation questions in IBPS. Mastering decimal multiplication can save significant time in quantitative aptitude sections and data interpretation sets. 4. Division of Decimal Fractions Decimal division often appears difficult but becomes easy with proper practice. Example: 6.3 ÷ 0.9 Multiply numerator and denominator by 10: 63 ÷ 9 = 7 Shortcut: Remove decimals by multiplying both numbers by powers of 10. Example: 4.25 ÷ 0.25 425 ÷ 25 = 17 Important Concepts: Examples: 40 ÷ 0.5 = 80 50 ÷ 0.25 = 200 Exam Example: Find: 12.6 ÷ 0.3 126 ÷ 3 = 42 This shortcut helps solve questions much faster than long division methods. 5. Converting Decimal Fractions into Percentages Percentage questions dominate banking exams. Conversion Rule: Decimal × 100 Example: 0.75 × 100 = 75% 0.25 × 100 = 25% 0.125 × 100 = 12.5% Quick Conversion Table: Exam Example: Convert 0.68 into percentage. 0.68 × 100 = 68% Understanding these conversions helps in profit and loss, simple interest, and DI questions. 6. Converting Percentages into Decimals Reverse conversion is equally important. Rule: Percentage ÷ 100 Example: 25% = 25 ÷ 100 = 0.25 Examples: Exam Example: A bank offers 7.5% interest. Decimal form: 7.5 ÷ 100 = 0.075 This conversion is frequently used in interest and growth-rate calculations. 7. Approximation Tricks Using Decimals Approximation questions are scoring opportunities in IBPS exams. Example: 49.8 × 10.2 Approximate: 50 × 10 = 500 Example: 198.7 ÷ 4.9 ≈ 200 ÷ 5 = 40 Benefits: Practice estimating answers before performing exact calculations. This habit often helps eliminate incorrect options quickly. 8. Common Decimal Values to Memorize Memorization reduces calculation time drastically. Important Values: Example: If profit is 3/5 of cost price, Use 0.6 directly. This significantly speeds up problem-solving in examinations. 9. Solving Decimal Fraction Word Problems Word problems combine decimal fractions with real-life situations. Example: A customer buys goods worth ₹250.75 and ₹149.25. Total cost: 250.75 + 149.25 = ₹400 Example: A worker completes 0.75 of a task. Remaining work: 1 – 0.75 = 0.25 Exam Example: A student scored 0.82 fraction of total marks. Percentage score: 0.82 × 100 = 82% Understanding the language of word problems is essential for converting statements into mathematical expressions. 10. Exam Strategy for Decimal Fraction Questions Success in decimal fraction questions depends on speed and accuracy. Key Strategies: Memorize Common Conversions Learn important fraction-decimal-percentage equivalents. Use Approximation Approximate whenever exact answers are not required. Practice Mental Calculations Reduce dependency on rough sheets. Avoid Decimal Placement Errors Double-check decimal positions. Solve Daily Practice at least 15–20 decimal questions every day. Example: 2.5 × 4 = 10 0.25 × 80 = 20 0.75 × 120 = 90 Frequent revision strengthens concepts and improves confidence during examinations.

Rahul Sir, a renowned aptitude trainer associated with OdTutor, has helped thousands of banking aspirants master quantitative aptitude concepts through logical methods and shortcut techniques. Among the most important topics in IBPS PO and Clerk examinations are Problems on H.C.F (Highest Common Factor) and L.C.M (Least Common Multiple). These concepts form the foundation of number systems and frequently appear in prelims and mains examinations. Many candidates find H.C.F and L.C.M questions easy in theory but struggle with speed and accuracy during exams. Rahul Sir emphasizes understanding the relationship between factors and multiples instead of memorizing formulas blindly. Once students understand divisibility rules, prime factorization, and shortcut methods, they can solve complex questions within seconds. In IBPS PO and Clerk exams, questions on H.C.F and L.C.M are often combined with word problems involving bells, events, workers, distribution, grouping, and repetitive cycles. A strong command of these concepts not only helps in direct questions but also improves performance in Number Series, Simplification, and Data Interpretation sections. This guide covers practical tips, tricks, formulas, and examples that Rahul Sir recommends for mastering H.C.F and L.C.M. By following these techniques consistently and practicing regularly, aspirants can significantly improve their speed and confidence while attempting banking aptitude questions. 1. Understanding the Basics of H.C.F and L.C.M Before learning shortcuts, students must clearly understand what H.C.F and L.C.M represent. H.C.F (Highest Common Factor):The greatest number that divides two or more numbers exactly. L.C.M (Least Common Multiple):The smallest number that is exactly divisible by two or more numbers. Example: Find H.C.F and L.C.M of 12 and 18. Factors of 12 = 1, 2, 3, 4, 6, 12 Factors of 18 = 1, 2, 3, 6, 9, 18 Common factors = 1, 2, 3, 6 Therefore: H.C.F = 6 Multiples of 12 = 12, 24, 36, 48, 60… Multiples of 18 = 18, 36, 54… Common multiple = 36 L.C.M = 36 Understanding this basic concept is crucial because many aspirants jump directly to formulas without grasping the logic behind factors and multiples. Once the concept is clear, solving advanced banking questions becomes significantly easier. Rahul Sir advises students to spend adequate time mastering these fundamentals before attempting shortcut techniques. Strong basics ensure fewer mistakes and faster calculations during competitive examinations. 2. Prime Factorization Method for Quick Solutions Prime factorization is one of the most reliable methods for solving H.C.F and L.C.M questions. Example: Find H.C.F and L.C.M of 24 and 36. Prime factors: 24 = 2 × 2 × 2 × 3 36 = 2 × 2 × 3 × 3 For H.C.F: Take common factors with smallest powers. H.C.F = 2² × 3 = 4 × 3 = 12 For L.C.M: Take all factors with highest powers. L.C.M = 2³ × 3² = 8 × 9 = 72 This method is highly accurate and works well even for large numbers. Banking examinations frequently include questions where prime factorization provides the fastest path to the solution. Students should memorize prime numbers up to at least 100 and practice factorizing numbers quickly. With sufficient practice, identifying prime factors becomes almost automatic. Rahul Sir often recommends writing prime factors vertically because it reduces confusion and calculation errors. This systematic approach ensures maximum accuracy under exam pressure. 3. Product Formula Trick One of the most important formulas in aptitude exams is: H.C.F × L.C.M = Product of the Numbers For two numbers: HCF × LCM = First Number × Second Number Example: Two numbers are 24 and 36. HCF = 12 LCM = ? LCM = (24 × 36)/12 = 864/12 = 72 This formula is frequently used in IBPS examinations. Example: H.C.F = 8 LC.M = 120 One number = 24 Find the other number. 8 × 120 = 24 × Other Number 960 = 24 × Other Number Other Number = 40 Questions based on this concept are very common because they test conceptual understanding rather than lengthy calculations. Rahul Sir advises students to immediately identify opportunities to use this formula, as it can save valuable exam time and eliminate unnecessary calculations. 4. Shortcut for Finding H.C.F Using Division Method The division method is extremely useful when dealing with large numbers. Example: Find H.C.F of 84 and 126. 126 ÷ 84 = 1 remainder 42 84 ÷ 42 = 2 remainder 0 Therefore, H.C.F = 42 This method is much faster than writing all factors. Example: Find H.C.F of 198 and 252. 252 ÷ 198 = remainder 54 198 ÷ 54 = remainder 36 54 ÷ 36 = remainder 18 36 ÷ 18 = 0 Therefore, H.C.F = 18 The division method is particularly useful in banking exams where speed matters. Students should practice this technique regularly because it can solve large-number questions within seconds. Rahul Sir recommends performing rough calculations mentally while using the division method to reduce dependence on written work and increase speed during examinations. 5. Solving Word Problems Based on L.C.M Many IBPS questions involve repetitive events. Example: Two bells ring every 12 minutes and 18 minutes. If they ring together now, after how many minutes will they ring together again? L.C.M of 12 and 18 12 = 2² × 3 18 = 2 × 3² L.C.M = 36 Answer = 36 minutes Such questions appear frequently in banking exams. Common applications include: The key is recognizing that whenever events repeat at different intervals and meet again, L.C.M is generally required. Rahul Sir teaches students to immediately identify keywords like “together again,” “simultaneously,” and “after how long” as indicators that L.C.M may be involved. 6. Distribution and Grouping Problems Using H.C.F H.C.F is commonly used when items must be distributed into maximum equal groups. Example: 96 chocolates and 144 candies are to be packed into identical packets. Find the maximum number of packets. H.C.F of 96 and 144: 96 = 2⁵ × 3 144 = 2⁴ × 3² H.C.F = 2⁴ × 3 = 48 Answer = 48 packets Each packet contains: 96/48 = 2 chocolates 144/48 = 3 candies Whenever questions ask

Hello Aspirants, I am Rahul Sir, and over the years I have guided hundreds of banking aspirants preparing for IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, RRB, and other competitive examinations. One topic that consistently appears in aptitude sections is Problems on Numbers. Many students find number-based questions tricky because they involve concepts from divisibility, remainders, factors, multiples, digit properties, and logical number patterns. The good news is that these questions become much easier when you understand a few fundamental rules and shortcut techniques. In IBPS examinations, number problems are generally designed to test logical thinking rather than lengthy calculations. Therefore, students who know the right tricks can solve questions quickly and accurately. In this article, I will explain the most important concepts, shortcuts, and exam-oriented strategies for solving Problems on Numbers. Each section includes practical examples similar to those asked in banking examinations. If you master these techniques and practice regularly, you can significantly improve your speed and accuracy in the quantitative aptitude section. Let us begin with the essential concepts that every banking aspirant must know. 1. Understanding Number Systems Problems on Numbers begin with understanding the number system. Every question is built around basic properties of numbers. The main categories include: Understanding these categories helps identify patterns quickly. Example Which of the following is a prime number? A) 39B) 51C) 53D) 57 Check divisibility: 53 is not divisible by 2, 3, 5, or 7. Therefore, 53 is prime. IBPS Tip Memorize prime numbers up to 100. Many number-based questions become easier when you instantly recognize prime numbers. Shortcut A number greater than 3 is prime only if it is not divisible by any prime number less than its square root. Strong understanding of number classifications creates the foundation for advanced number problems in banking examinations. 2. Divisibility Rules You Must Know Divisibility rules save valuable exam time and reduce calculations. Important rules: Example Is 35784 divisible by 9? 3 + 5 + 7 + 8 + 4 = 27 27 is divisible by 9. Therefore, 35784 is divisible by 9. Example Is 6524 divisible by 4? Last two digits = 24 24 is divisible by 4. Hence 6524 is divisible by 4. IBPS Tip Most remainder and factor questions can be solved quickly using divisibility rules without actual division. Practice applying these rules mentally to improve speed. 3. Factors and Multiples Techniques Factors and multiples form the basis of many aptitude questions. Important Formula If N = pᵃ × qᵇ × rᶜ Number of factors = (a+1)(b+1)(c+1) Example Find total factors of 72. 72 = 2³ × 3² Factors = (3+1)(2+1) = 4 × 3 = 12 factors Multiple Concept A multiple is obtained by multiplying a number by an integer. Multiples of 12: 12, 24, 36, 48, 60… Example Find the least number divisible by both 12 and 18. LCM(12,18) 12 = 2² × 3 18 = 2 × 3² LCM = 2² × 3² = 36 IBPS Tip Questions involving common divisibility often require LCM and HCF concepts. Learning prime factorization speeds up calculations significantly. 4. Remainder Theorem Tricks Remainder questions are extremely common in IBPS examinations. Basic Formula: Dividend = Divisor × Quotient + Remainder Example What is the remainder when 257 is divided by 6? 6 × 42 = 252 257 − 252 = 5 Remainder = 5 Shortcut Example Find remainder when 2⁵ divided by 3. 2⁵ = 32 32 ÷ 3 Remainder = 2 Pattern Method Find remainder when 7⁵ divided by 5. Units digit pattern: 7¹ = 7 → 2 remainder 7² = 49 → 4 remainder 7³ → 3 remainder 7⁴ → 1 remainder Cycle repeats. 7⁵ gives remainder 2. IBPS Tip Many large power questions can be solved using remainder cycles instead of calculating actual values. This technique saves considerable time during exams. 5. Finding Unit Digits Quickly Unit digit questions are favorites in banking exams. Example Find unit digit of 8⁵³. Observe pattern: 8¹ = 8 8² = 64 → 4 8³ → 2 8⁴ → 6 Pattern repeats every 4 terms. 53 ÷ 4 gives remainder 1. Therefore unit digit = 8. Example Unit digit of 3⁴² Pattern: 3, 9, 7, 1 42 ÷ 4 gives remainder 2 Second digit in pattern = 9 Answer = 9 IBPS Tip Memorize unit digit cycles for: 2, 3, 4, 7, 8, 9 These appear frequently in competitive examinations. Mastering this concept allows solving questions within seconds. 6. HCF and LCM Shortcuts HCF and LCM are essential topics in number problems. Relationship Formula HCF × LCM = Product of Numbers Example Two numbers are 12 and 18. HCF = 6 LCM = 36 6 × 36 = 216 12 × 18 = 216 Verified. Example HCF of 48 and 72 48 = 2⁴ × 3 72 = 2³ × 3² Common factors: 2³ × 3 = 24 IBPS Tip Prime factorization is usually faster than repeated division. Questions involving bells, clocks, intervals, or repeated events often use LCM concepts. Regular practice makes these calculations effortless. 7. Digit-Based Number Problems Many IBPS questions involve digits and their relationships. Example A two-digit number has digits whose sum is 9 and difference is 3. Find the number. Let digits be x and y. x + y = 9 x − y = 3 Adding: 2x = 12 x = 6 y = 3 Number = 63 Example If digits of a two-digit number are reversed, the number decreases by 27. Difference between digits? Difference = 27 ÷ 9 = 3 Shortcut Reversing a two-digit number changes value by: 9 × Difference of digits IBPS Tip Memorizing such relationships saves time during algebraic number questions. 8. Consecutive Number Problems Questions involving consecutive numbers are common. Formulas Sum of first n natural numbers: n(n+1)/2 Example Sum of first 20 natural numbers 20 × 21 / 2 = 210 Example Three consecutive numbers have sum 72. Let numbers be: x−1, x, x+1 3x = 72 x = 24 Numbers = 23, 24, 25