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Hello students, I am Rahul Sir from OdTutor, and over the years of training aspirants for IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, SSC, and Railway examinations, I have seen one topic consistently intimidate students who are otherwise quite well-prepared — Logarithm. The moment students see the word “log” in a question, something switches off in their brain. They skip it, mark it for later, and often never return to it during the exam, losing easy marks in the process. Here is the truth I tell every batch of students I teach: Logarithm is one of the most formula-friendly chapters in the entire quantitative aptitude syllabus. It does not require lengthy calculations, it does not require complex reasoning, and once you understand what a logarithm actually means, the formulas feel logical rather than arbitrary. You stop memorizing and start understanding, and that makes all the difference. At OdTutor, I have developed a step-by-step teaching approach for Logarithm that takes even the most hesitant student from absolute confusion to confident problem-solving within a matter of days. In this article, I am going to walk you through every important concept, formula, and question type that IBPS PO and Clerk exams test, with clear solved examples at every stage. Read it from beginning to end, practice the examples alongside, and I promise you — Logarithm will become one of your favorite scoring topics in the exam. Let’s get started. 1. What Exactly Is a Logarithm? The Concept Explained Simply Before any formula, any shortcut, or any trick, you must understand what the word “logarithm” actually means. I spend a full 20 minutes on this in my live classes, because every single formula that comes afterward is rooted in this one definition. Consider this simple question: “To what power must we raise 2 to get 8?” You already know the answer — 2 raised to the power 3 gives 8, so the answer is 3. Logarithm is simply a way of writing this question and its answer in a formal mathematical notation. We write it as: log₂(8) = 3 This is read as “log base 2 of 8 equals 3,” and it means exactly the same thing as 2³ = 8. The general definition is: logₐ(b) = c means aᶜ = b Here, ‘a’ is called the base, ‘b’ is the number whose logarithm is being taken, and ‘c’ is the answer — the exponent to which ‘a’ must be raised to get ‘b’. I give my students this mental anchor: the logarithm answers the question “what is the exponent?” That’s all it is. Every time you see logₐ(b), your brain should immediately ask: “a raised to WHAT power gives b?” The answer to that question is your logarithm value. This conversion between logarithmic form and exponential form is the very first skill you must master, because every property and formula of logarithm is simply the exponential rule rewritten in log language. Get this conversion absolutely smooth, and the rest of the chapter flows naturally. 2. Essential Properties of Logarithm You Must Know by Heart Now that the definition is clear, let’s build the formula toolkit. These properties appear directly or indirectly in almost every Logarithm question in IBPS exams. I want you to not just memorize them, but understand why each one makes sense based on what you already know about exponents. Property 1 — Product Rule: log(m × n) = log m + log n When you multiply two numbers, their logs add. This mirrors the exponent rule: aˣ × aʸ = aˣ⁺ʸ. Property 2 — Quotient Rule: log(m / n) = log m − log n When you divide two numbers, their logs subtract. This mirrors: aˣ / aʸ = aˣ⁻ʸ. Property 3 — Power Rule: log(mⁿ) = n × log m An exponent inside a log comes out as a multiplier in front. This is extremely frequently tested. Property 4 — Change of Base Rule: logₐ(b) = log(b) / log(a) This lets you convert any log to base 10, which makes calculations far easier. Property 5 — Log of 1: logₐ(1) = 0 for any base a Because a⁰ = 1 always, the log of 1 is always 0 regardless of the base. Property 6 — Log of Base Itself: logₐ(a) = 1 Because a¹ = a always, the log of the base equals 1. Property 7 — Log Base Reciprocal: logₐ(b) = 1 / logᵦ(a) Switching the base and the number gives you the reciprocal. This is a lifesaver in competitive exams. Write all seven properties on a single sheet. Review them every morning for one week until you can write all seven from memory in under two minutes. 3. Solved Examples: Applying Basic Properties Let’s immediately put those properties to work with examples that mirror the style of actual IBPS questions. Example 1: Find the value of log₂(32). Solution: We need to find: 2 raised to what power gives 32? Since 32 = 2⁵, we have log₂(32) = 5. That’s it. Example 2: Simplify: log(6) + log(5) − log(3) Solution: Using Product Rule first: log(6) + log(5) = log(6 × 5) = log(30) Then using Quotient Rule: log(30) − log(3) = log(30/3) = log(10) = 1 (Since log base 10 of 10 = 1, and when no base is written, base 10 is assumed.) The answer is 1. Example 3: Simplify: log₃(81) − log₃(9) Solution: log₃(81) = log₃(3⁴) = 4 log₃(9) = log₃(3²) = 2 Answer = 4 − 2 = 2 Alternatively using Quotient Rule: log₃(81/9) = log₃(9) = 2. Same answer. Example 4: If log(2) = 0.3010, find log(8). Solution: log(8) = log(2³) = 3 × log(2) = 3 × 0.3010 = 0.9030 This is a hugely important question type. IBPS almost always gives you a standard log value and asks you to calculate another one by expressing it as a power of a given number. I will cover this pattern extensively in the coming sections. 4. The Standard Log Values You Must Memorize

About Rahul Sir Rahul Sir is a renowned Aptitude and Reasoning trainer with extensive experience preparing students for competitive examinations such as IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, SSC, Railways, and other government exams. Known for his simple teaching style and practical shortcut techniques, he helps students solve complex aptitude problems quickly and accurately. His teaching focuses on building strong fundamentals, improving calculation speed, and mastering exam-oriented tricks that save valuable time during competitive exams. Through structured lessons, real exam questions, and regular practice sessions, Rahul Sir has guided thousands of aspirants toward achieving their career goals. In this article, Rahul Sir breaks down one of the most scoring yet often misunderstood topics in Quantitative Aptitude — Alligation and Mixture — into simple, exam-ready techniques that you can apply within seconds during your IBPS PO and Clerk exams. 1. What is Alligation and Why It Matters in Banking Exams Alligation is a mathematical technique used to solve problems involving the mixing of two or more ingredients, quantities, or values that have different properties — such as price, concentration, or ratio — to find a resultant mixture with a specific average value. In IBPS PO and Clerk exams, Alligation questions typically appear in the Quantitative Aptitude section and can involve mixing liquids of different prices, milk-water solutions, different grades of items, or even average-based problems disguised as mixtures. This topic is important because it usually takes candidates a long time to solve using conventional algebraic methods, but with the Alligation rule, the same question can be solved in under 20 seconds. Banking exams are highly time-sensitive, with candidates having roughly a minute or less per question, so mastering a fast method for mixture problems gives you a real edge over other aspirants. Alligation questions are also popular because examiners can twist them in many ways — replacement of mixtures, mixing more than two components, or combining alligation with ratio-proportion and profit-loss concepts. Understanding the core logic thoroughly, rather than memorizing formulas blindly, ensures you can adapt to any variation the exam throws at you, making this one of the highest-value topics to master for scoring well in the Quant section. 2. The Basic Alligation Rule (Formula and Logic) The foundation of solving any Alligation question lies in one simple rule: the ratio in which two quantities at different values must be mixed to produce a mixture at a given average value is inversely proportional to the difference of the extreme values from the average value. The standard formula is written as: Quantity of Cheaper : Quantity of Dearer = (Dearer Value − Mean Value) : (Mean Value − Cheaper Value) This is best visualized using the well-known Alligation Cross Method, where the cheaper value (C) and dearer value (D) are placed at the top two corners, the mean value (M) is placed in the middle, and the differences (D − M) and (M − C) are cross-multiplied diagonally to get the ratio. For example, if a shopkeeper mixes tea worth ₹60/kg with tea worth ₹80/kg to get a mixture worth ₹68/kg, the ratio of cheaper to dearer tea is (80−68):(68−60) = 12:8 = 3:2. This means for every 3 units of the ₹60 tea, you need 2 units of the ₹80 tea. Rahul Sir emphasizes that students should not just memorize this formula but understand why it works — the mean value always divides the mixture in inverse proportion to how far each component’s value is from the average. Once this logic is internalized, even unfamiliar or twisted questions become manageable within seconds, without needing to write a single equation. 3. Alligation Cross Method — Step-by-Step Visual Trick The Alligation Cross Method is the single most powerful visual shortcut for solving mixture problems quickly, and Rahul Sir recommends practicing it until it becomes second nature. Here’s how to apply it step by step: Step 1: Draw a cross (X) diagram. Write the cheaper quantity’s value at the top-left corner and the dearer quantity’s value at the top-right corner. Step 2: Write the mean (average) value of the final mixture in the center of the cross. Step 3: Subtract diagonally — subtract the mean value from the dearer value, and write this result at the bottom-left corner (this becomes the ratio of the cheaper quantity). Then subtract the cheaper value from the mean value, and write this at the bottom-right corner (this becomes the ratio of the dearer quantity). Step 4: The bottom-left and bottom-right numbers, simplified, give the required ratio of cheaper to dearer quantities. Example: A trader mixes two varieties of rice costing ₹40/kg and ₹60/kg to make a mixture worth ₹52/kg. Using the cross method: (60−52):(52−40) = 8:12 = 2:3. So the cheaper rice and dearer rice must be mixed in a 2:3 ratio. This visual method eliminates the need for algebraic equations entirely. Rahul Sir advises students to practice drawing this cross quickly on rough paper during mock tests until they can compute the ratio mentally within 10-15 seconds, which is essential given the strict time constraints of IBPS PO and Clerk exams. 4. Solving Mixture Problems Involving Milk and Water One of the most frequently tested variations of Alligation in IBPS exams involves milk-and-water mixtures, where a vessel contains milk mixed with water in a certain ratio, and candidates must find the ratio needed to achieve a desired concentration, or determine how much water must be added or removed. For pure milk-water problems, treat pure milk as having a “value” of 100% (or 1) and water as having a “value” of 0%, then apply the same alligation cross rule using percentages instead of prices. Example: In what ratio should a milkman mix pure milk with water to get a mixture that is 80% pure milk? Using alligation: pure milk = 100%, water = 0%, mean = 80%. Ratio = (100−80):(80−0) = 20:80 = 1:4. So milk and water should be mixed in a 1:4 ratio. A trickier variant involves mixing two different milk solutions,

Hello students, I am Rahul Sir, and for many years now I have been training aspirants for IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, SSC, and Railway exams. Over thousands of classroom hours, I have noticed one thing again and again — students fear Boats and Streams not because the concept is hard, but because they never sat down and understood the logic behind the formulas. They memorize blindly, get confused under exam pressure, and lose marks on questions that are actually quite simple once the basics are clear. At OdTutor, my teaching philosophy has always been the same: understand first, then practice, then speed up. In this article, I am going to break down Boats and Streams exactly the way I teach it in my live batches — step by step, with real exam-style examples, so that by the end, you will be solving these questions faster than the person sitting next to you in the exam hall. Let’s begin. 1. Understanding the Basic Concept Before touching any formula, you must understand what is actually happening physically. Imagine a boat moving in a river. The river itself has a current, which we call the “stream.” This stream has its own speed, and it affects how fast the boat appears to move. When the boat moves in the same direction as the stream, the river is pushing the boat forward, so the boat’s effective speed increases. This is called moving “downstream.” When the boat moves against the direction of the stream, the river is pushing against it, slowing it down. This is called moving “upstream.” Here is the key insight I always give my students: the boat has its own speed in still water, and the stream has its own speed. These two speeds simply add up or subtract from each other depending on direction. That’s it. There is no rocket science here. So we define two terms clearly: Once you fix these two values in your mind, every single Boats and Streams question becomes a simple speed-time-distance problem, just like the ones you’ve already solved in basic motion chapters. The trick is not in new mathematics — it is in correctly identifying which speed to use in which situation. I tell my students: get this concept rock solid first, because every formula ahead is built on top of it. 2. The Two Core Formulas You Must Memorize Now that the concept is clear, let’s lock in the two formulas that form the backbone of this entire chapter. I want you to write these on a sticky note and paste it on your study table. Downstream Speed = (b + s) When the boat moves with the current, the current’s speed adds to the boat’s own speed. So if a boat’s speed in still water is 10 km/hr and the stream’s speed is 2 km/hr, the downstream speed becomes 10 + 2 = 12 km/hr. Upstream Speed = (b − s) When the boat moves against the current, the current’s speed gets subtracted from the boat’s speed. Using the same numbers, the upstream speed would be 10 − 2 = 8 km/hr. Now, here is something extremely important for exams: once you know the downstream speed (let’s call it D) and the upstream speed (let’s call it U), you can directly find the boat’s speed and the stream’s speed using these reverse formulas: Speed of boat in still water = (D + U) / 2 Speed of stream = (D − U) / 2 I cannot stress enough how often these two reverse formulas appear in IBPS PO and Clerk papers. Many students only memorize the first set and panic when the question gives downstream and upstream speeds and asks for the boat’s speed or stream’s speed. Don’t be that student. Master both directions of these formulas so you can move from any given information to any required answer without hesitation. Practice writing these four formulas from memory at least ten times until they become instinct, not something you have to think about during the exam. 3. Solved Example: Finding Downstream and Upstream Speed Let’s apply what we just learned with a typical exam question. Question: A boat’s speed in still water is 15 km/hr, and the speed of the stream is 3 km/hr. Find the downstream and upstream speeds of the boat. Solution: Here, b = 15 km/hr and s = 3 km/hr. Downstream speed = b + s = 15 + 3 = 18 km/hr Upstream speed = b − s = 15 − 3 = 12 km/hr That’s all there is to it. I know this looks almost too simple, and that’s exactly the point — in the actual exam, IBPS rarely asks such a direct question alone. Instead, they wrap this basic concept inside a distance or time problem, which is why understanding this foundational step deeply matters. If you fumble here, the entire question collapses. A small variation IBPS loves to ask: Question: The downstream speed of a boat is 20 km/hr and the upstream speed is 12 km/hr. Find the speed of the boat in still water and the speed of the stream. Solution: Speed of boat in still water = (D + U)/2 = (20 + 12)/2 = 32/2 = 16 km/hr Speed of stream = (D − U)/2 = (20 − 12)/2 = 8/2 = 4 km/hr Notice how this is simply the reverse application of the same formula. I always tell my students in class: don’t treat “find b and s from D and U” as a different question type. It’s the same coin, just flipped. Once this clicks, you’ll never get confused between which formula to apply. 4. Solved Example: Distance, Speed and Time Combined Now let’s bring in the classic distance-speed-time relationship, since most real exam questions combine Boats and Streams with this formula: Distance = Speed × Time Question: A boat covers a distance of 36 km downstream in 3 hours. The speed of the boat

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