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

Introduction: Applying Your Foundation

Having built a solid foundation in the number system, percentages, and simplification in the previous chapter, you are now ready for four topics that appear with great regularity in the arithmetic/aptitude portion of the DSC exam: Ratio and Proportion, Averages, Time-Speed-Distance, and Profit-Loss. These four topics are, in a real sense, "applied" arithmetic — each one models a common real-world situation using the mathematical tools you have already learned, particularly fractions and percentages. You will notice as you work through this chapter that these topics also connect richly to each other: averages problems often use ratio concepts, time-speed-distance problems often use ratio and proportion, and profit-loss problems are, at their heart, percentage problems wearing a business-context costume.

As with the previous chapter, engage actively with every worked example: attempt each problem yourself before reading the solution, and revisit any technique that does not click immediately. Arithmetic mastery, for exam purposes, is built through repetition of correct method until that method becomes fast and automatic — there are no true shortcuts around this practice, though there are, as you will see, many legitimate calculation shortcuts that make the practice itself faster and more efficient.

Section 1: Ratio and Proportion

1.1 Understanding Ratios

A ratio compares two or more quantities of the same kind, showing their relative sizes. The ratio of a to b is written a:b, and is equivalent to the fraction a/b. Ratios should always be expressed in their simplest form — divide both terms by their HCF. For example, the ratio 24:36 simplifies to 2:3 (dividing both terms by their HCF, 12).

Important properties to remember: A ratio has no units — it is a pure number comparison, so both quantities being compared must be in the same unit before forming the ratio (you cannot directly form a ratio between a length in metres and a length in centimetres without first converting one to match the other). Multiplying or dividing both terms of a ratio by the same non-zero number does not change the ratio's value — this is exactly analogous to the fraction-simplification principle from the previous chapter.

1.2 Dividing a Quantity in a Given Ratio

A very common question type asks you to divide a total quantity according to a given ratio.

Worked example: Divide ₹720 between two people, Ravi and Suresh, in the ratio 5:4.
Total parts = 5 + 4 = 9
Ravi's share = (5/9) × 720 = ₹400
Suresh's share = (4/9) × 720 = ₹320
Check: 400 + 320 = 720 ✓

Worked example (three-way division): Divide 1,800 mangoes among three schools in the ratio 3:5:7.
Total parts = 3+5+7 = 15
School A: (3/15) × 1,800 = 360
School B: (5/15) × 1,800 = 600
School C: (7/15) × 1,800 = 840
Check: 360+600+840 = 1,800 ✓

1.3 Proportion

A proportion is an equation stating that two ratios are equal: a:b = c:d, often written as a:b :: c:d. In this proportion, a and d are called the "extremes," and b and c are called the "means." A fundamental property: the product of the extremes equals the product of the means — that is, a × d = b × c. This is the basis for solving for an unknown term in a proportion.

Worked example: Find x if 4:9 = x:36.
Using the cross-multiplication property: 4 × 36 = 9 × x
144 = 9x
x = 16

Direct and Inverse Proportion: Two quantities are in direct proportion if an increase in one causes a proportional increase in the other (e.g., more working hours generally means more work completed, other things equal). Two quantities are in inverse proportion if an increase in one causes a proportional decrease in the other (e.g., more workers on a fixed job generally means less time needed to complete it, other things equal). Recognising which type of relationship a word problem describes is essential before setting up the correct equation.

Worked example (direct proportion): If 8 pens cost ₹120, find the cost of 15 pens.
Cost per pen = 120/8 = ₹15
Cost of 15 pens = 15 × 15 = ₹225

Worked example (inverse proportion): If 12 workers can complete a task in 18 days, how many days will 9 workers take to complete the same task, assuming equal work rate per worker?
Total work (in worker-days) = 12 × 18 = 216 worker-days
With 9 workers: days needed = 216/9 = 24 days

Section 2: Averages

2.1 The Basic Formula

The average (or arithmetic mean) of a set of numbers is found by dividing the sum of all the numbers by the count of numbers: Average = Sum of all values ÷ Number of values. This is likely the single most instinctively familiar arithmetic concept for a teacher-candidate, since calculating class averages is a routine professional task — precisely why this chapter's Excel section in the earlier computer awareness chapter emphasised the AVERAGE() function.

Worked example: Find the average of 45, 62, 78, 51, and 84.
Sum = 45+62+78+51+84 = 320
Average = 320/5 = 64

2.2 Using the Average to Find a Missing Value

Worked example: The average marks of 5 students in a test is 72. If four of the students scored 65, 70, 75, and 80, find the fifth student's score.
Sum of all 5 scores = 72 × 5 = 360
Sum of known 4 scores = 65+70+75+80 = 290
Fifth student's score = 360 − 290 = 70

2.3 Weighted Average

When different groups within a combined set have different sizes, a simple average of the group averages is incorrect — you must use a weighted average, which accounts for the relative size of each group.

Worked example: Section A of a class has 30 students with an average score of 70. Section B has 20 students with an average score of 80. Find the combined average score of both sections.
Total marks in Section A = 30 × 70 = 2,100
Total marks in Section B = 20 × 80 = 1,600
Combined total marks = 2,100 + 1,600 = 3,700
Combined total students = 30 + 20 = 50
Combined average = 3,700/50 = 74
Common trap: A student who incorrectly averages 70 and 80 directly (getting 75) ignores the different group sizes — this is wrong whenever group sizes differ, since the larger group (Section A, with more students at the lower average) should pull the combined average closer to its own value, and indeed 74 is closer to 70 than a naive (70+80)/2=75 would suggest.

2.4 Average Speed — A Frequent Trap

This deserves special mention here because it connects averages directly to the next section's topic, and because it is one of the most consistently mis-solved problem types in this entire chapter. When a journey is covered at different speeds for EQUAL DISTANCES (not equal time), the average speed is NOT the simple average of the two speeds — it requires the harmonic mean formula:

Average speed (for two equal distances at speeds s1 and s2) = (2 × s1 × s2) / (s1 + s2)

Worked example: A car travels from town A to town B at 60 km/h and returns from B to A at 40 km/h (same distance both ways). Find the average speed for the entire round trip.
Average speed = (2 × 60 × 40) / (60 + 40) = 4,800/100 = 48 km/h
Note the trap: The naive simple average of 60 and 40 would give 50 km/h, which is WRONG. The correct answer, 48 km/h, is lower than the simple average because the car spends more TIME travelling at the slower speed (since the distances, not the times, are equal) — and average speed is fundamentally total distance divided by total time, which naturally weights more heavily toward the slower speed when distances are equal. If instead the EQUAL TIME (not equal distance) were spent at each speed, the simple average WOULD be correct — always identify whether the problem specifies equal distances or equal times before choosing your method.

Section 3: Time, Speed, and Distance

3.1 The Fundamental Relationship

The core relationship connecting these three quantities: Distance = Speed × Time, which can be rearranged as Speed = Distance/Time and Time = Distance/Speed. Nearly every problem in this topic is built on correctly applying and rearranging this single relationship, so ensure you can instantly rearrange it in your head in either direction.

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

This is an essential, extremely frequently tested skill. To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5. Memorise these two conversion factors precisely, as they appear in a very large proportion of time-speed-distance questions.

Worked example: Convert 72 km/h to m/s.
72 × (5/18) = 72×5/18 = 360/18 = 20 m/s

Worked example: Convert 25 m/s to km/h.
25 × (18/5) = 25×18/5 = 450/5 = 90 km/h

3.3 Problems Involving Trains

Train problems are a classic sub-category within this topic, distinguished by the fact that the LENGTH of the train itself matters, since a train must cover its own length in addition to the distance to a point or object it is passing.

Worked example: A train 150 metres long is running at a speed of 54 km/h. Find the time it takes to cross a stationary pole.
Convert speed to m/s: 54 × 5/18 = 15 m/s
To cross a stationary point (like a pole), the train needs to cover a distance equal to its own length: 150 metres.
Time = Distance/Speed = 150/15 = 10 seconds

Worked example (crossing a platform): The same 150-metre train at 54 km/h (15 m/s) needs to cross a platform that is 250 metres long. Find the time taken.
When crossing a platform, the train must cover its own length PLUS the platform's length: 150+250 = 400 metres.
Time = 400/15 = 26.67 seconds (approximately)
Key concept to remember: crossing a stationary point/pole/man requires covering only the train's own length; crossing a platform, bridge, or another stationary object of some length requires covering the train's length PLUS that object's length.

Worked example (two trains crossing each other): Two trains, 120 m and 180 m long, running in opposite directions at 45 km/h and 63 km/h respectively, cross each other. Find the time taken to cross completely.
When two objects move in OPPOSITE directions, their relative speed is the SUM of their individual speeds: 45+63 = 108 km/h. Converting to m/s: 108 × 5/18 = 30 m/s.
Total distance to be covered = sum of both train lengths = 120+180 = 300 metres.
Time = 300/30 = 10 seconds
Note on direction: If the two trains were moving in the SAME direction instead, their relative speed would be the DIFFERENCE of their individual speeds, not the sum — this distinction (sum for opposite directions, difference for same direction) is a fundamental and frequently tested principle for all relative-speed problems, not just trains.

3.4 Boats and Streams

This is a related sub-category using the same relative-speed principle. Let the speed of the boat in still water be b, and the speed of the stream/current be s.

• Downstream speed (boat moving in the same direction as the current, so the current assists it) = b + s
• Upstream speed (boat moving against the current, so the current opposes it) = b − s

Worked example: A boat's speed in still water is 12 km/h, and the speed of the stream is 3 km/h. Find the boat's downstream and upstream speeds, and the time taken to travel 45 km downstream.
Downstream speed = 12+3 = 15 km/h
Upstream speed = 12−3 = 9 km/h
Time for 45 km downstream = 45/15 = 3 hours

A useful related pair of formulas, derivable from the two above: boat's still-water speed b = (downstream speed + upstream speed)/2, and stream speed s = (downstream speed − upstream speed)/2. These are useful when the downstream and upstream speeds are given directly and you need to find b and s individually.

3.5 Problems Involving Two Moving Bodies — Meeting and Catching Up

Worked example (moving toward each other): Two towns A and B are 300 km apart. A car starts from town A toward town B at 50 km/h, and simultaneously another car starts from town B toward town A at 40 km/h. Find the time after which they will meet.
Since they move toward each other, relative speed = 50+40 = 90 km/h
Time to meet = Total distance/Relative speed = 300/90 = 10/3 hours = 3 hours 20 minutes

Worked example (catching up, same direction): A starts cycling from a point at 12 km/h. Two hours later, B starts from the same point in the same direction at 18 km/h. Find how long it takes B to catch up with A from the time B starts.
By the time B starts, A has already covered: 12 × 2 = 24 km head start.
Since B is chasing A in the same direction, relative speed = 18−12 = 6 km/h
Time for B to close the 24 km gap = 24/6 = 4 hours

Section 4: Profit and Loss

4.1 Core Terminology

Master these terms precisely, as questions frequently hinge on correctly identifying which figure is which:

• Cost Price (CP): The price at which an article is purchased or produced.
• Selling Price (SP): The price at which an article is sold.
• Profit (Gain): Occurs when SP > CP. Profit = SP − CP.
• Loss: Occurs when SP < CP. Loss = CP − SP.
• Profit Percentage: (Profit/CP) × 100 — note carefully that profit/loss percentage is ALWAYS calculated with Cost Price as the base, never Selling Price. This is one of the most important and most frequently tested facts in this entire topic.
• Loss Percentage: (Loss/CP) × 100
• Marked Price (MP) (also called List Price): The price displayed or listed on an article before any discount is applied.
• Discount: A reduction from the Marked Price, given to arrive at the actual Selling Price. Discount = MP − SP. Discount Percentage = (Discount/MP) × 100 — note that, unlike profit/loss percentage (which is based on CP), discount percentage is based on Marked Price.

4.2 Basic Profit and Loss Calculations

Worked example: A shopkeeper buys an article for ₹800 and sells it for ₹920. Find the profit and profit percentage.
Profit = 920 − 800 = ₹120
Profit percentage = (120/800) × 100 = 15%

Worked example: An item purchased for ₹1,500 is sold at a loss of 12%. Find the selling price.
Loss = 12% of 1,500 = 180
SP = CP − Loss = 1,500 − 180 = ₹1,320
Alternative faster method: SP = CP × (1 − loss%/100) = 1,500 × 0.88 = ₹1,320

Worked example (finding CP from SP and profit%): An article is sold for ₹690, yielding a profit of 15%. Find the cost price.
Since SP = CP × (1 + profit%/100):
690 = CP × 1.15
CP = 690/1.15 = ₹600
Common trap: Do not calculate 15% of 690 and subtract it from 690 to "find" the cost price — this is incorrect for the same underlying reason discussed in the percentage-reversal trap in the previous chapter: the 15% profit was calculated on the CP, not on the SP, so you must set up the correct proportional equation rather than naively reversing the percentage on the wrong base.

4.3 Marked Price and Discount

Worked example: A shopkeeper marks an article at ₹2,000 and offers a discount of 10%. If his cost price for the article is ₹1,620, find his actual profit percentage after the discount.
Discount = 10% of 2,000 = 200
SP = MP − Discount = 2,000 − 200 = ₹1,800
Profit = SP − CP = 1,800 − 1,620 = 180
Profit percentage = (180/1,620) × 100 = approximately 11.11%

4.4 Successive Discounts

Just as with successive percentage changes in the previous chapter, successive discounts do NOT simply add together. A "20% and then 10%" successive discount is NOT the same as a flat 30% discount.

Worked example: A shop offers successive discounts of 20% and 10% on an item marked at ₹1,000. Find the final selling price and the effective single discount percentage.
After first discount (20%): 1,000 × 0.80 = ₹800
After second discount (10% on the new price of 800): 800 × 0.90 = ₹720
Final SP = ₹720
Effective discount = 1,000 − 720 = 280, which as a percentage of the original marked price = (280/1,000) × 100 = 28%
Note: 28% is LESS than the naive 20+10=30% a hasty student might assume — this is directly analogous to the successive percentage change formula from the previous chapter, and indeed you can verify it using that same formula with a=−20, b=−10: net change = −20−10+(−20×−10)/100 = −30+2 = −28%, confirming the 28% effective discount calculated directly above.

4.5 A Common Trick Question Type: Equal Percentage Profit and Loss on Two Items

Worked example: A shopkeeper sells two items, each for ₹1,200. On one, he makes a profit of 20%, and on the other, he incurs a loss of 20%. Find his overall profit or loss percentage on the entire transaction.
This is one of the most consistently mis-solved question types in this entire topic, because it seems intuitive that equal-percentage profit and equal-percentage loss should cancel out to a net zero — but this is INCORRECT, precisely because the same 20% is being calculated on two DIFFERENT cost prices (since the selling price, not the cost price, is what is equal in this scenario).
For the item sold at 20% profit: SP = CP × 1.20, so CP = 1,200/1.20 = ₹1,000
For the item sold at 20% loss: SP = CP × 0.80, so CP = 1,200/0.80 = ₹1,500
Total CP = 1,000 + 1,500 = ₹2,500
Total SP = 1,200 + 1,200 = ₹2,400
Since Total SP < Total CP, there is an overall LOSS of 2,500 − 2,400 = ₹100
Overall loss percentage = (100/2,500) × 100 = 4%
Useful shortcut formula for this exact scenario (equal selling price, with profit x% on one and loss x% on the other): the seller always incurs a net loss, and the loss percentage equals (x/10)², expressed as a percentage. Here, x=20, so loss% = (20/10)² = 2² = 4%, matching our detailed calculation above exactly. This shortcut is worth memorising given how frequently this exact question pattern appears.

Section 5: An Integrated Practice Set

These problems deliberately combine concepts across this chapter's four topics, since real exam papers frequently do the same. Attempt each before checking the method notes.

• Q1. The ratio of ages of two teachers is 4:5. After 6 years, the ratio becomes 5:6. Find their present ages. (Method: let ages be 4x and 5x. After 6 years: (4x+6)/(5x+6) = 5/6. Cross-multiply: 6(4x+6)=5(5x+6) → 24x+36=25x+30 → x=6. Present ages: 4x=24 and 5x=30.)
• Q2. The average age of a class of 25 students is 15 years. When the teacher's age is included, the average becomes 16 years. Find the teacher's age. (Method: total age of students = 25×15=375. Total including teacher = 26×16=416. Teacher's age = 416−375=41 years.)
• Q3. A man covers a certain distance at 40 km/h and returns over the same route at 60 km/h. Find his average speed for the whole journey. (Method: use the equal-distance average speed formula: (2×40×60)/(40+60) = 4800/100 = 48 km/h.)
• Q4. A trader marks his goods 40% above cost price and then allows a discount of 25%. Find his profit or loss percentage. (Method: let CP=100. MP=140. SP after 25% discount on MP = 140×0.75=105. Profit = 105−100=5, so profit percentage = 5%.)

Common Exam Traps

• Calculating profit/loss percentage on Selling Price instead of Cost Price. This is, without exaggeration, one of the single most common errors across all profit-loss questions in competitive exams generally. Always anchor profit and loss percentage calculations to Cost Price.
• Assuming successive percentage changes or successive discounts simply add together. As demonstrated repeatedly in this chapter and the previous one, they do not — always calculate successive changes step by step, or use the correct successive-change formula, remembering to use a negative value for any decrease/discount within that formula.
• Using a simple arithmetic average for average speed when distances (not times) are equal. Remember to use the harmonic-mean-based formula (2×s1×s2)/(s1+s2) specifically for the equal-distance case, and reserve the simple average only for the equal-time case.
• Confusing relative speed rules for objects moving in the same direction versus opposite directions. Remember: SUM of speeds for opposite directions (approaching or separating), DIFFERENCE of speeds for the same direction (one chasing another).
• Forgetting to add the length of a train (or other extended-length moving object) when it must cross a platform, bridge, or another train, as opposed to a stationary point like a pole or a person, where only the object's own length needs to be covered.
• Assuming equal-percentage profit and equal-percentage loss on two items (with equal selling prices) cancel out to zero net effect. As demonstrated in Section 4.5, this scenario always produces a net LOSS, and the shortcut formula (x/10)² is worth memorising for speed.
• Averaging group averages directly without weighting by group size when the groups being combined are of different sizes — always use the weighted average approach shown in Section 2.3 whenever group sizes differ.
• Errors in km/h ↔ m/s conversion, particularly forgetting which fraction (5/18 or 18/5) applies in which direction. A reliable memory check: km/h is a "bigger" unit relationship numerically for the same physical speed than m/s in most everyday contexts (e.g., 36 km/h = 10 m/s, a smaller number), so converting FROM km/h TO m/s should produce a SMALLER number, meaning you multiply by the fraction less than 1 (5/18); the reverse conversion multiplies by the fraction greater than 1 (18/5).

With the completion of this chapter, you now have a comprehensive, exam-ready arithmetic toolkit spanning the number system, percentages, simplification, ratio and proportion, averages, time-speed-distance, and profit-loss. The path from here to genuine exam-day confidence is simple to describe, even if it requires sustained discipline to execute: revisit these worked examples regularly, attempt fresh practice problems in each category without looking at the methods first, time yourself as your exam date approaches to build speed alongside accuracy, and revisit any specific trap listed in this chapter's and the previous chapter's "Common Exam Traps" sections whenever you notice yourself falling into it during practice. Arithmetic, more than almost any other section of your DSC syllabus, rewards steady, consistent practice with real, measurable improvement — trust that process, and you will see it pay off both in your exam result and in the quiet confidence you will carry into every future classroom conversation involving numbers, marks, and data.