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

Why This Chapter Matters

If the previous chapter built your foundation in numbers, percentages, and simplification, this chapter puts that foundation to work on the four topic areas that generate the largest share of applied arithmetic word problems on almost every competitive exam, including AP VRO/VRA-level papers: ratio and proportion, averages, time-speed-distance, and profit-loss. These four topics share a common character — each one models a real-world situation using a small set of relationships, and once you have internalised those relationships, a huge variety of differently worded questions all reduce to the same underlying calculation. The candidates who struggle with this chapter's material are almost never candidates who cannot do arithmetic; they are candidates who have not yet learned to recognise which underlying relationship a given word problem is testing. This chapter is built to fix exactly that gap, through careful explanation paired with worked examples at every step.

These topics also have genuine everyday relevance to the work you are training for. A revenue official regularly deals with proportional divisions of land or produce, average yields across plots, and cost-and-value assessments that mirror profit-loss reasoning directly. Learning these topics well is therefore not merely an exam requirement but a form of practical numeracy that will serve you throughout your career.

Ratio and Proportion

Understanding Ratio

A ratio compares two or more quantities of the same kind, expressing how many times one quantity contains another. The ratio of a to b is written a:b, and it is fundamentally a fraction, a/b, expressed in its simplest form. If a class has 20 boys and 15 girls, the ratio of boys to girls is 20:15, which simplifies (dividing both terms by their HCF, 5) to 4:3. It is essential to always express ratios in their lowest terms and to always keep the order of terms consistent with how the question states them — the ratio of boys to girls (4:3) is different in meaning, though not in value, from the ratio of girls to boys (3:4), and questions are precise about which order they want.

Dividing a Quantity in a Given Ratio

A very common question type asks you to divide a total quantity according to a given ratio. If a sum of ₹4,800 is to be divided between two people in the ratio 5:7, first find the total number of parts: 5 + 7 = 12 parts. Each part is worth 4800/12 = ₹400. The first person gets 5 parts = 5 × 400 = ₹2,000, and the second gets 7 parts = 7 × 400 = ₹2,800. Always verify by checking that the shares sum back to the original total: 2,000 + 2,800 = 4,800, confirming the calculation is correct.

This same technique extends naturally to three or more parties. If ₹18,000 is divided among three people in the ratio 2:3:4, total parts = 2+3+4 = 9, each part = 18000/9 = ₹2,000, giving shares of ₹4,000, ₹6,000, and ₹8,000 respectively.

Combining Ratios

When a question gives you the ratio of A to B and separately the ratio of B to C, and asks for the combined ratio A:B:C, you must make the B term common across both ratios before combining. For example, if A:B = 2:3 and B:C = 5:7, the B terms (3 and 5) need to be made equal by finding their LCM, which is 15. Scale A:B = 2:3 by a factor of 5 to get A:B = 10:15, and scale B:C = 5:7 by a factor of 3 to get B:C = 15:21. Now that both ratios share B = 15, we can combine directly: A:B:C = 10:15:21.

Proportion

A proportion states that two ratios are equal: a:b = c:d, often written a:b :: c:d. The fundamental property of a proportion is that the product of the extremes equals the product of the means: a × d = b × c. This is the basis of the widely used "rule of three" or unitary-method cross-multiplication technique for solving proportion problems. For example, if 8 workers can complete a task and the question asks how many workers are needed under a proportionally scaled scenario, setting up a proportion and cross-multiplying resolves it directly.

A particularly important application is direct and inverse proportion. Two quantities are in direct proportion when an increase in one causes a proportional increase in the other (for example, cost and quantity purchased — more items cost proportionally more). Two quantities are in inverse proportion when an increase in one causes a proportional decrease in the other (for example, speed and time taken to cover a fixed distance — higher speed means proportionally less time). Recognising whether a word problem describes a direct or inverse relationship before setting up your equation is essential, because setting it up backward gives an answer that is the reciprocal of the correct one.

Averages

The Basic Formula

The average (or arithmetic mean) of a set of values is the sum of all the values divided by the number of values: Average = (Sum of values) / (Number of values). This simple formula underlies every average-based question, though the wording of exam questions often disguises the calculation needed to apply it. For example, if a student scores 72, 85, 68, 91, and 74 in five subjects, the average score is (72+85+68+91+74)/5 = 390/5 = 78.

Finding a Missing Value

A common question type gives you the average and asks you to find a missing individual value. For example: "The average weight of 6 people is 58 kg. If a seventh person joins and the new average becomes 60 kg, find the weight of the seventh person." The total weight of the original 6 people = 6 × 58 = 348 kg. The total weight after the seventh person joins = 7 × 60 = 420 kg. The seventh person's weight = 420 - 348 = 72 kg. This "total before and after" technique — converting averages back into sums, performing the arithmetic on sums, and only converting back to an average at the end if needed — is the single most useful technique for average word problems and should become your default approach whenever a question involves someone joining or leaving a group.

Change in Average When Replacing a Value

Another very common pattern: "The average age of a group of 10 people is 32 years. One person aged 40 leaves and is replaced by a new person, and the new average becomes 31 years. Find the age of the new person." Original total age = 10 × 32 = 320. New total age = 10 × 31 = 310. The change in total = 320 - 310 = 10 (a decrease). Since the person who left was 40, and the total decreased by 10 upon replacement, the new person's age = 40 - 10 = 30 years.

Average Speed — A Special Case

A frequent error occurs when candidates compute average speed for a journey covered at different speeds by simply averaging the two speeds arithmetically. This is incorrect when the distances (not the times) covered at each speed are equal, and the correct formula in that specific case is the harmonic mean: Average Speed = (2 × S1 × S2)/(S1 + S2), where S1 and S2 are the two speeds, applicable specifically when equal distances are covered at each speed. 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 (2 × 40 × 60)/(40+60) = 4800/100 = 48 km/h — notably not the simple average of 50 km/h. This distinction, between simple average and this special harmonic-mean case, is a classic exam trap and deserves careful attention; the simple arithmetic average of speeds is only correct when equal time (not equal distance) is spent at each speed.

Time, Speed, and Distance

The Core Relationship

The foundational relationship governing this entire topic is: Distance = Speed × Time, which can be rearranged as Speed = Distance/Time, or Time = Distance/Speed, depending on which two quantities are known and which is to be found. Every problem in this topic, however elaborately worded, ultimately reduces to correctly identifying which two of these three quantities are given and applying the appropriate rearrangement.

Unit Conversion — km/h to m/s and Back

Speed is commonly given in kilometres per hour but questions sometimes require metres per second, or vice versa, and this conversion is tested frequently enough that it deserves to be entirely automatic. To convert km/h to m/s, multiply by 5/18 (because 1 km = 1000 m and 1 hour = 3600 seconds, so 1000/3600 simplifies to 5/18). To convert m/s to km/h, multiply by 18/5. For example, 72 km/h converted to m/s: 72 × 5/18 = 20 m/s. And 15 m/s converted to km/h: 15 × 18/5 = 54 km/h.

Relative Speed — Trains and Meeting Problems

When two objects move toward each other, their relative speed (the rate at which the gap between them closes) is the sum of their individual speeds. When two objects move in the same direction, their relative speed (the rate at which one gains on the other) is the difference of their individual speeds. This single principle underlies almost all "two trains" or "two people walking toward or away from each other" problems.

Worked example (opposite directions, meeting problem): Two trains, 300 km apart, start moving toward each other at speeds of 60 km/h and 90 km/h respectively. How long until they meet? Their relative speed (moving toward each other, so speeds add) = 60 + 90 = 150 km/h. Time to meet = Distance/Relative Speed = 300/150 = 2 hours.

Worked example (same direction, catching up): A person walking at 5 km/h starts 2 hours before a cyclist who travels at 15 km/h in the same direction. How long after the cyclist starts will they meet? By the time the cyclist starts, the walker has a head start distance of 5 × 2 = 10 km. The cyclist's relative speed with respect to the walker (same direction, so speeds subtract) = 15 - 5 = 10 km/h. Time for the cyclist to close a 10 km gap at a relative speed of 10 km/h = 10/10 = 1 hour.

Problems Involving Trains Crossing Objects

A train crossing a stationary point (a pole, or a person standing still) covers a distance equal to its own length in the crossing time. A train crossing a platform or bridge covers a distance equal to the sum of its own length and the platform's length. For example: a train 150 metres long crosses a platform 250 metres long in 20 seconds; find the train's speed. Total distance covered = 150 + 250 = 400 metres, covered in 20 seconds, so speed = 400/20 = 20 m/s, which converts to 20 × 18/5 = 72 km/h.

Boats and Streams

This is a specialised sub-topic of relative speed involving water currents. If a boat's speed in still water is "b" and the stream's (current's) speed is "s," then the boat's effective speed downstream (moving with the current) is b + s, and its effective speed upstream (moving against the current) is b - s. Given downstream and upstream speeds, you can recover the boat's still-water speed and the stream's speed: still-water speed = (downstream speed + upstream speed)/2, and stream speed = (downstream speed - upstream speed)/2. For example, if a boat covers 30 km downstream in 2 hours and the same 30 km upstream in 3 hours, downstream speed = 15 km/h and upstream speed = 10 km/h; still-water speed = (15+10)/2 = 12.5 km/h, and stream speed = (15-10)/2 = 2.5 km/h.

Profit and Loss

Core Definitions and Formulas

Cost Price (CP) is the price at which an item is purchased or produced. Selling Price (SP) is the price at which it is sold. When SP exceeds CP, the seller makes a profit, equal to SP - CP; when CP exceeds SP, the seller incurs a loss, equal to CP - SP. Profit and loss are typically expressed as percentages of the cost price (this is the standard convention unless a question explicitly states otherwise): Profit % = (Profit/CP) × 100, and Loss % = (Loss/CP) × 100.

These formulas can be rearranged to find SP directly when CP and the profit or loss percentage are known: SP = CP × (100 + Profit%)/100 for a profit scenario, or SP = CP × (100 - Loss%)/100 for a loss scenario. For example, if an item costing ₹800 is sold at a 15% profit, SP = 800 × 115/100 = ₹920. If instead it is sold at a 15% loss, SP = 800 × 85/100 = ₹680.

Conversely, when SP and the profit or loss percentage are known and CP is required, the formulas rearrange to: CP = SP × 100/(100 + Profit%) for a profit scenario, or CP = SP × 100/(100 - Loss%) for a loss scenario. For example, if an item is sold for ₹690 at a 15% loss, CP = 690 × 100/85 = ₹811.76 (approximately).

Marked Price and Discount

The Marked Price (MP), also called the list price, is the price displayed on an item before any discount is applied. Discount is typically expressed as a percentage of the marked price, and the Selling Price after discount is: SP = MP × (100 - Discount%)/100. For example, an item marked at ₹1,200 with a 20% discount has SP = 1200 × 80/100 = ₹960.

A frequently tested variation combines marked price, discount, and profit or loss together in a single problem: "A shopkeeper marks an item 40% above its cost price and then offers a 10% discount on the marked price. Find the shopkeeper's profit percentage." Let CP = 100 (using 100 as a convenient base value simplifies percentage-based calculations enormously, and is a technique worth adopting whenever a question involves only percentages, since the final answer will itself be a percentage). Marked Price = 100 × 140/100 = 140. Selling Price after 10% discount on MP = 140 × 90/100 = 126. Since CP was 100 and SP is 126, the profit = 26, and since CP = 100, the profit percentage is directly 26%.

Successive Discounts

When two discounts are applied successively (for example, "30% off, and an additional 10% off on the reduced price"), they cannot simply be added together — this is another instance of the successive percentage change principle introduced in the previous chapter. Using the formula a + b + (ab/100) with a = -30 and b = -10: net change = -30 + (-10) + [(-30)(-10)/100] = -40 + 3 = -37%. So successive discounts of 30% and 10% are equivalent to a single discount of 37%, not 40% as a hasty candidate might assume.

False Weight Problems

A specialised profit-loss variant involves a dishonest trader who claims to sell at cost price but uses a false (reduced) weight to actually give the customer less than they paid for, thereby earning a hidden profit. The formula for the profit percentage earned through this method is: Profit% = (True Weight - False Weight)/(False Weight) × 100. For example, a trader claims to sell at cost price but uses a weight of 900 grams for every claimed kilogram (1000 grams). The trader's profit percentage = (1000-900)/900 × 100 = 100/900 × 100 = 11.11% (approximately). This topic combines profit-loss reasoning with ratio thinking and is worth practising as its own distinct pattern once the basic profit-loss formulas are secure.

Bringing the Four Topics Together

Ratio, averages, time-speed-distance, and profit-loss might look like four separate topics in a syllabus list, but in practice they share deep structural overlaps: ratio thinking underlies boats-and-streams and mixture problems; the "assume a convenient base value of 100" technique from profit-loss applies equally well to percentage-heavy average and ratio problems; and average-speed calculations are themselves a direct application of the harmonic-mean special case within a broader time-speed-distance framework. As you practise, resist treating these as four disconnected checklists of formulas to memorise, and instead notice the recurring techniques — converting between sums and averages, using a convenient base value of 100, applying the successive-change formula, and setting up relative-speed equations — that carry across all four areas. This is what genuine arithmetic fluency looks like, and it is exactly the fluency that will let you solve unfamiliar-looking word problems confidently on exam day, because you will recognise the underlying pattern rather than searching for a memorised formula that matches the exact wording in front of you.