Arithmetic — Ratio, Averages, Time-Speed-Distance, and Profit-Loss

Why This Chapter Matters

Building on the number-system, percentage, and simplification foundation of the previous chapter, this chapter covers four closely related and consistently tested arithmetic topics — ratio and proportion, averages, time-speed-distance, and profit-loss — all of which appear in nearly every past Grama/Ward Sachivalayam paper and share a common thread: each topic is, at its core, a variation on the theme of comparing quantities and understanding how changing one quantity affects another in a predictable, formulaic way. These topics also have strong real-world relevance to the roles this examination recruits for, since dividing welfare benefits among eligible households in a fixed ratio, calculating average land holdings or beneficiary counts across villages, and verifying billing or subsidy amounts all draw on exactly the reasoning skills tested here, making this chapter valuable both for the exam itself and for the analytical thinking these posts genuinely require on the job.

Ratio and Proportion

A ratio is a comparison of two or more quantities of the same kind, expressed as one quantity divided by another, and written using a colon, such as 3:4. It is essential to understand that a ratio is a comparison of relative size and carries no unit of its own, and that a ratio remains unchanged if both its terms are multiplied or divided by the same non-zero number — so the ratio 3:4 is exactly equivalent to 6:8, 9:12, or 30:40, all of which represent the same underlying proportional relationship. When a ratio is expressed with the smallest possible whole numbers (such as 3:4 rather than 6:8), it is said to be in its simplest form, and this is generally the preferred form for stating a final answer unless the question specifically requires otherwise. A proportion is a statement that two ratios are equal, written as, for instance, 3:4 equals 6:8, and the fundamental rule for checking or solving a proportion is that the product of the extreme terms (the first and last terms) equals the product of the middle terms (the second and third terms) — commonly called the cross-multiplication rule. For example, given the proportion 3:4 equals x:12, cross-multiplying gives 4 multiplied by x equals 3 multiplied by 12, so 4x equals 36, and x equals 9.

Dividing a quantity in a given ratio is one of the most practically useful skills in this topic, since it appears not only as a direct question type but embedded within many word problems about sharing money, work, or resources. To divide a total quantity in a given ratio, the total is divided into a number of equal parts equal to the sum of the ratio's terms, and each party then receives a number of these parts corresponding to their term in the ratio. For example, if 720 rupees is to be divided between two people in the ratio 5:7, the sum of the ratio terms is 12, so each part is worth 720 divided by 12, which is 60 rupees; the first person then receives 5 parts, which is 5 multiplied by 60, giving 300 rupees, and the second person receives 7 parts, which is 7 multiplied by 60, giving 420 rupees — and as a useful check, 300 plus 420 does indeed equal the original 720 rupees. This same method extends naturally to three or more parties: if a sum of 1,800 rupees is divided among three people in the ratio 2:3:4, the sum of the terms is 9, so each part is worth 200 rupees, giving the three people 400, 600, and 800 rupees respectively, which together sum correctly to 1,800.

Compound ratios and combining two separate ratios that share a common term are frequently tested in slightly more advanced questions. If the ratio of A to B is 2:3, and the ratio of B to C is 4:5, these cannot be directly combined into a ratio of A to C without first making the B term identical in both ratios, since B is expressed as 3 in the first ratio and 4 in the second. To combine them, both ratios are scaled so that the B term matches: multiplying the first ratio by 4 gives A to B as 8:12, and multiplying the second ratio by 3 gives B to C as 12:15; since B is now 12 in both, the combined ratio of A to B to C is 8:12:15. This technique, often called the "chain ratio" or combining ratios through a common term, is a very common exam question type and is best mastered through repeated practice rather than memorising a rigid formula, since the specific numbers involved vary widely from question to question.

Direct and inverse proportion, often tested through word problems, describe how two quantities change in relation to each other. Two quantities are in direct proportion when an increase in one causes a proportional increase in the other (and a decrease causes a proportional decrease) — for example, the cost of a certain item and the quantity purchased are directly proportional, since buying twice as much costs twice as much, assuming a constant price per unit. Two quantities are in inverse proportion when an increase in one causes a proportional decrease in the other — for example, the number of workers assigned to a task and the time taken to complete that task are inversely proportional, since more workers complete the same task in proportionally less time, assuming all workers work at the same constant rate. Recognising correctly whether a given word problem describes a direct or inverse relationship before setting up the proportion is essential, since applying the wrong type of proportion to a word problem produces an answer that is often wildly incorrect, and examiners frequently phrase questions in ways designed to test whether a candidate can correctly identify which type of relationship is at play.

Averages

The average (or arithmetic mean) of a set of numbers is found by adding all the values together and dividing the sum by the count of values. For instance, the average of the numbers 12, 15, 18, and 19 is found by adding them to get 64, and dividing by 4 (the count of numbers), giving an average of 16. A very useful reformulation of this basic idea, which speeds up many exam questions considerably, is that the average of a set of numbers can also be thought of as a "central" value such that the sum of the positive deviations of numbers above the average exactly equals the sum of the negative deviations of numbers below it — this deviation method is often much faster than direct summation for numbers that are close together, since a candidate can pick any convenient reference number, find how far each value deviates from it (positively or negatively), average those deviations, and add that average deviation back to the reference number to get the true average, without ever summing the original large numbers directly.

A frequently tested category involves the average of consecutive numbers, where a useful shortcut applies: the average of a set of consecutive numbers (whether consecutive integers, consecutive even numbers, or consecutive odd numbers) is always simply the average of the first and last numbers in the set, since consecutive sequences are symmetric around their midpoint. For example, the average of all integers from 11 to 25 is found by averaging the first and last terms, 11 and 25, which gives 18, without needing to sum all fifteen numbers individually.

Questions involving a change in average when a value is added, removed, or replaced are extremely common and reward a systematic approach. If the average of a group changes when a new member joins, the key relationship is that the change in the total sum equals the change in average multiplied by the new total count of members (or, depending on how the problem is framed, the original count). For example, if the average age of a group of 20 people is 30 years, and a new person joins making the average rise to 31 years for the now 21-member group, the new total sum of ages must be 31 multiplied by 21, which is 651, while the original total sum was 30 multiplied by 20, which is 600; the age of the new person joining must therefore be the difference between these two totals, 651 minus 600, which is 51 years. Similarly, if a value is removed from a group and the average changes, or if one member of a group is replaced by another and the average changes by a specified amount, the same underlying principle applies: work out the total sum before and after the change using the average-multiplied-by-count relationship, and the difference between the two totals reveals the value that was added, removed, or the net effect of the replacement.

Weighted averages appear when the items being averaged do not carry equal importance or equal count, and the formula requires multiplying each value by its corresponding weight (often a count or a proportion), summing these products, and dividing by the sum of the weights. For example, if a class of 30 students has an average score of 70 and another class of 20 students has an average score of 80 on the same test, the combined average is not simply the average of 70 and 80 (which would incorrectly give 75), because the two classes have different sizes; the correct combined average is found by adding the total score of both classes — 30 multiplied by 70, giving 2,100, plus 20 multiplied by 80, giving 1,600, for a combined total of 3,700 — and dividing by the combined count of 50 students, giving a correct combined average of 74. This distinction between a simple average and a weighted average is a common trap in exam questions, and candidates should always check whether the groups being averaged are of equal size before applying a simple averaging shortcut.

Time, Speed, and Distance

The foundational relationship in this topic is that distance equals speed multiplied by time, from which the other two useful rearrangements follow directly: speed equals distance divided by time, and time equals distance divided by speed. Candidates must be fluent in converting between the two most common speed units used in these problems: kilometres per hour (km/h) and metres per second (m/s). To convert from km/h to m/s, multiply by 5 and divide by 18 (since one kilometre is 1,000 metres and one hour is 3,600 seconds, and 1,000 divided by 3,600 simplifies to 5 over 18); to convert from m/s to km/h, multiply by 18 and divide by 5, the exact reverse. For example, a speed of 72 km/h converts to metres per second by multiplying 72 by 5 and dividing by 18, giving 20 m/s, and this particular conversion — 72 km/h being exactly 20 m/s — is worth memorising directly since it recurs so often in exam problems as a round, convenient number.

Average speed problems, particularly those involving a journey covered at two different speeds over two equal distances (rather than two equal times), are a very frequent source of error because candidates instinctively want to simply average the two speeds, which is incorrect whenever the two speeds apply over equal distances rather than equal time durations. When equal distances are covered at two different speeds, the correct average speed for the entire journey is the harmonic mean of the two speeds, calculated as twice the product of the two speeds divided by their sum. For example, if a person travels a certain distance at 40 km/h and returns the same distance at 60 km/h, the average speed for the entire round trip is not 50 km/h (the simple average) but is instead found by taking twice the product of 40 and 60, which is 4,800, and dividing by their sum, which is 100, giving an average speed of 48 km/h. This result being lower than the simple average makes intuitive sense once explained: the traveller spends more time travelling at the slower speed (since covering the same distance at a lower speed always takes longer), so the slower speed has a proportionally greater influence on the overall average, pulling the true average speed below the midpoint of 50 km/h. Candidates should note carefully, however, that if a problem instead describes two different speeds maintained for two equal time periods (rather than two equal distances), the simple average of the two speeds is indeed the correct approach in that case, so correctly identifying whether the problem holds distance or time constant is the essential first step before choosing the appropriate averaging method.

Problems involving trains, boats and streams, and relative speed extend the basic time-speed-distance framework and are worth understanding conceptually rather than as isolated formulas. When two objects move in the same direction, their relative speed (the rate at which the gap between them changes) is the difference between their individual speeds; when they move in opposite directions (or toward each other), their relative speed is the sum of their individual speeds. This principle explains why two trains approaching each other from opposite directions cover the distance between them faster (at their combined speed) than one train would take to catch up to another moving in the same direction ahead of it (at only the difference of their speeds). For boats and streams problems, a boat's speed while moving downstream (with the current) equals the boat's own speed in still water plus the speed of the current, while its speed moving upstream (against the current) equals the boat's own speed in still water minus the speed of the current; from these two relationships, the boat's speed in still water can be found as half the sum of the downstream and upstream speeds, and the speed of the current can be found as half the difference between the downstream and upstream speeds. For example, if a boat's downstream speed is 18 km/h and its upstream speed is 12 km/h, the boat's speed in still water is half of 18 plus 12, which is 15 km/h, and the current's speed is half of 18 minus 12, which is 3 km/h.

Problems involving trains crossing platforms, bridges, or other trains require the additional recognition that the distance to be covered equals the length of the train plus the length of the stationary or moving object being crossed, since a train is only considered to have fully "crossed" an object when its entire length, from front to back, has passed that object. For instance, if a train 150 metres long crosses a platform that is 100 metres long, the train must cover a total distance of 250 metres (the sum of both lengths) for the crossing to be complete, and this combined-length principle, once understood conceptually, resolves what otherwise looks like a confusing category of problem into a straightforward application of the basic distance-speed-time formula.

Profit and Loss

Profit and loss problems revolve around two central terms: the cost price (CP), which is the price at which an item is purchased or produced, and the selling price (SP), which is the price at which it is sold. When the selling price exceeds the cost price, the difference is profit, and when the cost price exceeds the selling price, the difference is loss. Profit percentage is calculated as the profit divided by the cost price, multiplied by 100, and loss percentage is calculated analogously as the loss divided by the cost price, multiplied by 100 — in both cases, it is critical to remember that the cost price, not the selling price, is always the base for calculating profit or loss percentage unless a question explicitly states otherwise, since this is one of the most common sources of error in this topic.

Given a cost price and a desired profit or loss percentage, the corresponding selling price can be found directly: selling price equals cost price multiplied by (100 plus profit percentage) divided by 100 in the case of a profit, or cost price multiplied by (100 minus loss percentage) divided by 100 in the case of a loss. For example, if an item costing 800 rupees is sold at a 15 percent profit, the selling price is 800 multiplied by 115 and divided by 100, giving 920 rupees; if instead it is sold at a 15 percent loss, the selling price is 800 multiplied by 85 and divided by 100, giving 680 rupees. Conversely, when a selling price and a profit or loss percentage are given and the cost price needs to be found, the same relationship is simply rearranged: cost price equals selling price multiplied by 100, divided by (100 plus profit percentage) for a profit case, or by (100 minus loss percentage) for a loss case. A common error here is to subtract or add the percentage directly to the selling price to "reverse" the calculation (for instance, incorrectly assuming that if an item was sold at a 15 percent loss for 680 rupees, the cost price must simply be 680 plus 15 percent of 680); this is wrong precisely because the profit or loss percentage was originally calculated on the cost price, not the selling price, so working backward from the selling price requires the full formula involving 100 in the denominator rather than a direct percentage addition on the selling price itself.

Marked price and discount problems add a further layer: the marked price (MP) is the label or listed price of an item before any discount is applied, and the discount is typically expressed as a percentage of the marked price, so the selling price equals the marked price multiplied by (100 minus discount percentage) divided by 100. A frequently tested scenario combines a discount with a desired profit margin over cost price: for example, if a shopkeeper wants to mark up an item so that even after offering a 20 percent discount on the marked price, a 20 percent profit over the cost price is still achieved, the marked price must be calculated by first finding the required selling price (cost price multiplied by 120 and divided by 100), and then working backward from the discount relationship (selling price equals marked price multiplied by 80 and divided by 100) to isolate the marked price — if the cost price is, say, 500 rupees, the required selling price for a 20 percent profit is 600 rupees, and since this selling price equals 80 percent of the marked price, the marked price is 600 multiplied by 100 and divided by 80, giving 750 rupees. Successive discounts, where two or more discounts are applied one after another rather than simultaneously, behave exactly like the successive-percentage-change situation described in the previous chapter: they are never simply added together, and the combined effect must be calculated by applying each discount in turn (or by using the same "net change" shortcut formula), since two successive discounts of 10 percent each, for instance, produce a combined discount of 19 percent rather than 20 percent, precisely because the second discount is applied to the already-reduced price rather than to the original marked price.

False weight problems, where a dishonest trader uses an inaccurate weighing measure while still claiming to sell at cost price, are a specific and recurring profit-and-loss variant worth understanding on its own terms. If a trader claims to sell at cost price but uses a weight that is lighter than what is claimed — for instance, using a 900-gram weight while claiming it to be a full kilogram — the trader is effectively gaining a profit equal to the shortfall relative to the false weight used. The profit percentage in such cases is calculated as the error (the difference between the true and false weight) divided by the false weight used, multiplied by 100 — in the example given, the error is 100 grams (1,000 minus 900) and the false weight used is 900 grams, so the profit percentage is 100 divided by 900, multiplied by 100, which gives approximately 11.11 percent. This category of problem is really just the basic profit-and-loss framework applied to a quantity discrepancy rather than a price discrepancy, and recognising this underlying similarity helps candidates apply the same systematic, cost-price-as-base approach they already use for standard profit-and-loss questions.

Integrating These Topics — Exam Strategy

Because ratio, averages, time-speed-distance, and profit-loss so often appear together within a single, longer word problem — for instance, a question might describe a journey with a given ratio of speeds, then ask for an average speed, or a business scenario involving a ratio of cost prices between two items alongside a combined profit percentage — candidates benefit enormously from practising problems that deliberately blend two or more of these topics rather than drilling each topic in complete isolation. When faced with such a blended problem, the most reliable strategy is to break it into clearly labelled steps on scratch paper: first extract and simplify the given ratio or set up the given relationship, then apply the relevant averaging, speed, or profit-loss formula to each piece separately, and finally combine the results according to what the question is actually asking, rather than attempting to solve the entire problem in one mental leap. Consistent, timed practice across genuinely mixed problem sets — not just topic-by-topic drills — is what ultimately builds the pattern-recognition speed that distinguishes a strong performer in this section, since the underlying mathematics in each individual topic covered across this chapter and the previous one is not inherently difficult, but the ability to correctly identify which formula or method a given word problem calls for, quickly and under time pressure, is a skill built only through sustained and varied practice.