Direct and Inverse Proportion

What you will learn

Before calculating, decide what stays fixed. Then use that fixed relationship to connect two situations. By the end you should be able to:

• Recognise a direct model from a constant unit rate or quotient
• Recognise an inverse model from a constant product
• Explain the one-unit method and the scale-factor method
• State the conditions needed for fixed work, stock and map-scale problems
• Reject a proposed proportion when its quotient or product changes
• Match units and check whether an answer has a sensible size and direction

This lesson follows Ratio and Proportion and uses earlier fraction and decimal arithmetic. We use positive quantities in our models. In particular, neither variable can be zero in an inverse model with a positive fixed product. A rate such as rupees per kilogram compares different kinds of quantities and carries units. A map ratio compares two lengths in matching units and is unitless. Do not cancel away the meaning of a rate.

1 Retrieve the tools you need

Try these four unscored checks before reading the feedback. Use 1 m = 100 cm.

1. Simplify 18/30.
2. Calculate (3/5) × 140.
3. Convert 1.25 m into centimetres.
4. Calculate 72 ÷ 0.6.

Feedback:

1. 18/30 = 3/5 after dividing numerator and denominator by 6. Their ratio is unchanged because both are divided by the same nonzero number.
2. (140 ÷ 5) × 3 = 28 × 3 = 84. Divide first when it keeps the arithmetic simple.
3. 1.25 × 100 = 125 cm. The numerical value changes because the unit is smaller; the length is unchanged.
4. 72 ÷ 0.6 = 720 ÷ 6 = 120. Multiplying both dividend and divisor by 10 preserves the quotient. Check: 120 × 0.6 = 72.

If the first two answers were difficult, revisit exact fractions. If the third failed, write the unit conversion beside the calculation. If the fourth failed, practise decimal division before using unit rates. These are tools for choosing and checking a model, not four new topics to memorise.

Coached block A: keep the same amount per unit

2 Identify the invariant before writing a formula

An invariant is a quantity or relationship that stays unchanged across the situations being compared. Suppose every kilogram of a product has the same price, with no fixed charge or quantity discount. Two kilograms then cost twice as much as one; half a kilogram costs half as much. Each extra kilogram contributes the same amount. The price per kilogram, not the total price, stays fixed.

Let x be the quantity bought and y be its total cost. Dividing y by x tells us the cost of one unit. If that same value k applies throughout the stated model, y/x = k, and multiplying by x gives y = kx. This is direct proportion. If x changes by a positive factor f, then the cost changes from kx to k(fx) = f(kx): y changes by the same factor f. Halving both quantities also preserves a direct proportion; it is not limited to increases.

For two corresponding pairs, y₁/x₁ = y₂/x₂. Rearranging gives y₂ = y₁ × (x₂/x₁). The two routes mean the same thing:

• One-unit route, also called the unitary method: divide the old total by the old quantity, then multiply by the new quantity
• Scale-factor route: multiply the old total by new quantity ÷ old quantity

Do not put unlike units into the two entries for the same variable. For example, use minutes in both time entries before comparing them. Also keep the quotient in a consistent order. Cost/quantity is rupees per kilogram; quantity/cost is kilograms per rupee. Both can be constant, but they are reciprocals with different meanings.

Mathematical basis: NCERT Class VIII: Direct and Inverse Proportions, §11.2, printed pp.129–135. This is the explicitly labelled legacy 2024–25 edition. Our explanations and questions are newly written.

Worked example 1: read a direct-model table

A stated direct model gives these corresponding values. Find the unit rate and y when x = 9.

x	y
2	30
4	60
7	105

Name the quantities: x and y are the two positive numerical variables. The model is stated to be direct; the table supplies matching pairs, not two unrelated lists.

Invariant: y/x = 30/2 = 60/4 = 105/7 = 15. Thus each one unit of x corresponds to 15 units of y.

One-unit route: x = 1 corresponds to y = 15, so x = 9 corresponds to y = 9 × 15 = 135. Scale-factor route: from x = 2 to x = 9 the factor is 9/2. Therefore y = 30 × 9/2 = 135. The factor need not be a whole number.

Reasonableness check: 9 is greater than 7, so y should exceed 105. Also 135/9 = 15, matching every given quotient. Adding 7 to 30 would preserve a difference, not the stated unit rate.

A finite table can show that its listed pairs fit a model or contradict it. Matching a few rows alone does not prove that an unknown real-world process will follow that model at every other value. Here prediction is justified because the direct model is part of the question.

Worked example 2: find the cost of a fractional quantity

At a fixed price per kilogram, 4 kg of a product costs ₹176. There are no fees or quantity discounts. What does 6.5 kg cost?

What stays fixed: price per kilogram. What changes: mass and total cost. Invariant: cost/mass = ₹176/4 kg = ₹44 per kg.

First find one kilogram, then scale: 6.5 × ₹44 = ₹286. Equivalently, the mass factor is 6.5/4 = 13/8, so the cost is ₹176 × 13/8 = ₹286. These are the same multiplication and division in a different order.

Reasonableness check: 6 kg costs ₹264 and 7 kg costs ₹308, so ₹286 is in the correct interval. It is exactly halfway between those prices because 6.5 kg is halfway between 6 and 7 kg. Check the invariant: 286/6.5 = 44 rupees per kg. A result below ₹176 would have the wrong direction.

Worked example 3: convert the time before scaling

An object moves at a constant speed, covering 72 km in 90 minutes. There are no stops. How far does it move in 2.5 hours?

What stays fixed: speed. What changes: elapsed travelling time and distance. Convert first: 2.5 hours = 2.5 × 60 = 150 minutes. The decimal .5 hour means half an hour, or 30 minutes, not 5 minutes.

Invariant: distance/time = 72/90 = 0.8 km per minute. Distance in 150 minutes = 0.8 × 150 = 120 km. Alternatively, the time factor is 150/90 = 5/3, so distance = 72 × 5/3 = 120 km.

Reasonableness check: 150 minutes is longer than 90 minutes, so the distance must exceed 72 km. Both factors agree: 120/72 = 150/90 = 5/3. A second unit check gives 72 km in 1.5 hours, hence 48 km per hour; 48 × 2.5 = 120 km. Dividing 72 by 90 and then multiplying by 2.5 without converting units would mix minutes and hours.

Checkpoint A: explain, then calculate

Cover the feedback and answer in your own words.

1. In a direct model, x is multiplied by 3/2. What happens to y, and why?
2. At the same no-fee price, 2 kg costs ₹70. Find the cost of 0.5 kg and name the invariant.
3. Does knowing that both quantities increase establish direct proportion?

Feedback:

1. y is multiplied by 3/2, because y/x must stay fixed. It is not increased by the number 3/2.
2. The invariant is ₹70 ÷ 2 = ₹35 per kg. Therefore 0.5 kg costs ₹17.50. It is one quarter of the mass and one quarter of the ₹70 cost.
3. No. You need the same quotient under the stated conditions. A fixed starting fee can make both quantities increase while their quotient changes; we test this in worked example 8.

If you knew the direction but not the invariant, return to section 2. If the one-unit and factor routes disagree, inspect their units and arithmetic rather than choosing whichever answer looks convenient.

Coached block B: keep the same total, then challenge the model

3 Why a fixed product creates inverse proportion

Imagine a fixed number of labels packed with the same number in every full box. The total is labels per box × number of boxes. If each box holds twice as many labels, half as many boxes are needed, provided the packing divides exactly. The fixed total forces one factor to compensate for the other.

Let the two positive quantities be x and y. If their product has the fixed value k, then xy = k. Dividing by the nonzero x gives y = k/x. This is inverse proportion. If x is multiplied by f, y must be divided by f, because (fx) × (y/f) = xy. For two situations, x₁y₁ = x₂y₂, so y₂ = y₁ × (x₁/x₂). Notice the old/new factor here, whereas direct proportion uses new/old.

Do not memorise the reversed factor without the reason. Label the fixed product first. In a stock problem it may be person-days at an unchanged daily amount per person. In a fixed-work problem it may be worker-hours under identical steady rates. Neither label is an unconditional law about people.

Mathematical basis: NCERT Ganita Prakash Grade 8 Part II: Proportional Reasoning–2, §3.6, printed pp.63–65, and NCERT Class VIII Mathematics, Chapter 11, §11.3, printed pp.137–140. We state the modelling assumptions explicitly before applying the relationship.

Worked example 4: use the product of each pair

A stated inverse model gives (x, y) = (3, 24), (4, 18), (6, 12). Find y when x = 9.

Invariant: xy = 3 × 24 = 4 × 18 = 6 × 12 = 72. All three given products agree. At x = 9, we need 9 × y = 72, so y = 72/9 = 8.

The factor route gives the same result: x changes from 3 to 9, a factor of 3. Therefore y changes from 24 to 24/3 = 8. Multiplying y by 3 would change the product instead of preserving it.

Reasonableness check: increasing x from 6 to 9 should reduce y below 12; 8 does. Substitution gives 9 × 8 = 72. The model is stated; three fitting rows alone would not establish an unknown law beyond the supplied data.

Worked example 5: state the work assumptions first

Six workers complete one fixed job in 12 hours. How long would nine workers take for the same job?

Required assumptions: all workers have the same steady work rate; all work together for the whole stated duration; the job can be shared without workers obstructing one another; there is no setup time, coordination overhead, pause or other additional delay. The amount of work is unchanged. These are the conditions of this question, not claims about every workplace.

Each worker contributes one worker-hour in one hour. Because their productivity is identical and steady, 6 workers × 12 hours delivers 72 worker-hours of the required work. Nine such workers deliver 9 worker-hours each hour.

Invariant: workers × hours = 72 worker-hours for this fixed job. Therefore 9 × t = 72, giving t = 8 hours. Factor check: workers increase by 9/6 = 3/2, so time is multiplied by 2/3: 12 × 2/3 = 8 hours.

Reasonableness check: more identical workers should need less than 12 hours. Also 9 × 8 = 6 × 12, so the required work is unchanged. The number 72 is a total in worker-hours, not the elapsed time for the group. With different work rates or a fixed setup delay, this whole-time inverse calculation would not be justified.

Worked example 6: preserve daily use as well as stock

A fixed food stock lasts 24 people for 15 days. Every person consumes the same daily amount, and that amount is unchanged in both situations. There is no new supply, waste or loss. How many days will the same stock last 30 people?

Think of one person's one-day ration as one unit of stock. The available stock contains 24 × 15 = 360 such daily rations. If 30 people each take one daily ration, the stock is used at 30 rations per day.

Invariant: people × days = 360 person-days, because the stock and daily amount per person are both fixed. Duration = 360/30 = 12 days. Equivalently, the number of people is multiplied by 30/24 = 5/4, so duration is multiplied by 4/5: 15 × 4/5 = 12.

Reasonableness check: more people use the stock faster, so 12 should be less than 15. Substitution gives 30 × 12 = 24 × 15 = 360. If each person changed the amount eaten per day, the same stock would no longer imply this same person-days product. Equal daily use within each group is insufficient unless it is also unchanged between the two groups.

4 Choose the relationship from what is fixed

A story about time is not automatically inverse, and a story about money is not automatically direct. At constant speed, distance/time is fixed, so distance and time are direct. For fixed work under the stated capacity assumptions, workers × time is fixed, so workers and time are inverse. Ask which pair of quantities is changing and which relationship is held constant.

Worked example 7: return to a direct model for a map

On a scale drawing marked 1:25,000, two points are 3.6 cm apart. Find the represented ground distance in metres. Use 100 cm = 1 m.

The scale means drawing length:represented ground length = 1:25,000 in matching units. Thus 1 cm on the drawing represents 25,000 cm on the ground. The scale is fixed while the two lengths change together.

Invariant: ground length/drawing length = 25,000 after both lengths are expressed in the same unit. This quotient is unitless because centimetres cancel with centimetres. Ground distance = 3.6 × 25,000 = 90,000 cm. Convert the answer: 90,000/100 = 900 m.

Reasonableness check: 1 drawing cm represents 250 m, so 3.6 drawing cm should represent 3.6 × 250 = 900 m. It lies between 750 m for 3 cm and 1,000 m for 4 cm. The ratio check uses 3.6:90,000 = 1:25,000; using 3.6:900 would wrongly compare centimetres directly with metres. This is the distance represented by the drawing, not a road-route length inferred from it.

Map-scale basis: NCERT Ganita Prakash, Grade 8 Part II, Chapter 3, §3.2, printed pp.56–57. The distinction between represented ground distance and road distance matters even when the arithmetic is correct.

Worked example 8: reject two tempting shortcuts

Claim A: A fee of ₹40 plus ₹12 per kilometre is directly proportional to the positive distance travelled.

What is fixed: the ₹40 starting charge and the ₹12 added per kilometre. Those facts do not say that total fee per kilometre is fixed. At 1 km the total is ₹52; at 2 km it is ₹64. Proposed invariant test: fee/distance is 52/1 = ₹52 per km, but 64/2 = ₹32 per km. It changes, so total fee and distance are not directly proportional. The correct fee relationship is C = 40 + 12d, with C in rupees and d in kilometres. The distance-dependent part C − 40 = 12d is direct, but the question concerns the whole fee C. A valid additive model is not an arithmetic mistake. Reasonableness check: doubling 1 km to 2 km does not double ₹52 to ₹104; the actual fee is ₹64. Only the distance charge doubles; the starting charge is paid once.

Claim B: The pairs (2, 10), (3, 9), (4, 8) are inverse because y decreases when x increases.

Proposed invariant test: xy = 2 × 10 = 20; 3 × 9 = 27; 4 × 8 = 32. The product changes, so these data are not inverse. They are not direct either: y/x is 5, 3 and 2. Here x + y stays at 12, showing that a fixed sum can also make one quantity decrease as the other increases. Reasonableness check: from x = 2 to x = 4, an inverse model would halve y from 10 to 5. The given y is 8, so even this factor check rejects the claim. Every supplied row matters when testing a proposed invariant.

Both proposed claims fail. Direction is a useful final check, but a constant quotient or product is the deciding test. The direct-proportion warning is supported by NCERT Class VIII Mathematics, Chapter 11, §11.2, printed pp.131–132; these fee and decreasing-table examples are our original counterexamples.

Checkpoint B: select, calculate and limit the claim

Try without using the quick reference.

1. In a positive inverse model, x is multiplied by 5/4. What factor multiplies y?
2. For an unchanged fixed job, why is “there are more workers” insufficient on its own to calculate the new time by inverse proportion?
3. A positive table has pairs (2, 15), (3, 10), (5, 6). Do its rows fit inverse proportion? Does the table alone prove an unknown process follows the rule for every x?
4. A stock lasts 10 people 18 days. Every person's daily use is equal and unchanged; there is no replenishment or loss. How long for 15 people?

Feedback:

1. The factor is 4/5. The two factors multiply to 1, leaving xy unchanged.
2. The workers must have identical steady rates, share the same fixed work without interference, and have no additional setup or coordination delay. Different rates or overhead can prevent workers × total elapsed time from being fixed.
3. Yes, the supplied products are all 30. No, a few matching rows do not prove an unknown process follows that law elsewhere. A prediction outside the table needs the model or conditions that justify it.
4. The invariant is 10 × 18 = 180 person-days. Duration = 180/15 = 12 days; 15 × 12 = 180, and more people means fewer days under these conditions.

If you chose the model only from the direction, revisit worked example 8. If your product was correct but the story assumptions were missing, revisit cases 5–6. If your map answer was 100 times too large or too small, revisit case 7 and write the conversion as a separate step.

5 Independent practice

Attempt all twelve questions before reading the key. Choose one option and write the invariant or the test that rejects the model. This is untimed learning practice: 1 mark per item, 0 negative marks and no pass cutoff. It is not a full RRB CBT simulation.

1. In a stated direct model, x = 3 gives y = 21. What is y when x = 8?

A. 26 B. 56 C. 168 D. 63/8

2. A positive inverse model has x = 4 and y = 27. What is its constant product k?

A. 27/4 B. 31 C. 108 D. 23

3. Three identical notebooks cost ₹81 at a fixed unit price, with no fees or quantity discount. What do five cost?

A. ₹48.60 B. ₹83 C. ₹405 D. ₹135

4. In a positive inverse model, x doubles. By what factor is y multiplied?

A. 1/2 B. 2 C. −2 D. 1

5. A fixed batch of 2,400 labels is packed with 30 labels in each full box. If the same batch is repacked with 40 labels in each full box and none left over, how many boxes are needed?

A. 80 B. 60 C. 106 2/3 D. 70

6. A copier prints at a fixed rate with no setup delay or interruption. It prints 84 pages in 3 minutes. How many pages does it print in 2.5 minutes?

A. 58 1/3 pages B. 56 pages C. 70 pages D. 100.8 pages

7. Eight identical machines, each working at the same steady rate, make one fixed batch in 15 hours. The work is shared without interference; there is no setup time, interruption or coordination overhead. How long do 12 such machines, all working together, need for the same batch under those conditions?

A. 10 hours B. 22.5 hours C. 11 hours D. 120 hours

8. A drawing has scale 1:40,000. Two points are 2.8 cm apart on it. What represented ground distance is this in metres? Use 100 cm = 1 m.

A. 112,000 m B. 112 m C. 1.12 m D. 1,120 m

9. Fixed supplies last 18 people for 20 days. Each person uses the same daily amount, unchanged in both situations; there is no replenishment or loss. How many days will the same supplies last 24 people?

A. 26 2/3 days B. 15 days C. 14 days D. 360 days

10. A fare is a fixed ₹35 plus ₹9 for each kilometre. Is the total fare directly proportional to positive distance? Choose the correct judgement and reason.

A. Yes; the added charge is ₹9 for each kilometre B. Yes; fare and distance both increase C. No; at 1 km and 2 km the total fare/distance values are ₹44/km and ₹26.50/km D. No; fare and distance are inverse, so more distance means less fare

11. The given pairs are (x, y) = (2, 18), (3, 15), (6, 6). A learner calls them inverse because y falls as x rises. Which assessment is correct?

A. They are inverse because y decreases B. They are inverse because the first and last products are both 36 C. They are direct because every y/x equals 9 D. They are not inverse because the products are 36, 45 and 36

12. Five workers finish one fixed job in 18 hours. For a prediction about nine workers doing the same job, assume all workers have identical steady rates, all work together, and work can be shared without interference, setup delay or coordination overhead. Which conclusion respects both the calculation and its assumptions?

A. 10 hours under the stated conditions; without them this prediction need not hold B. 10 hours for any nine workers, even when their rates or delays differ C. 32.4 hours under the stated conditions because time grows with worker count D. 14 hours because adding four workers subtracts four hours

6 Explained answer key

1. B is correct: the invariant is y/x = 21/3 = 7, so y = 7 × 8 = 56. Check: 56/8 = 7. A adds the change in x, 8 − 3 = 5, to 21; equal additions do not preserve this rate. C multiplies 21 by 8 without first dividing by the old x = 3. D uses 21 × 3/8 = 63/8, the inverse factor, and wrongly makes y smaller when x increases in this direct model.

2. C is correct: inverse proportion preserves xy, so k = 4 × 27 = 108. Substitution into y = k/x gives 108/4 = 27. A calculates y/x, the quotient used to test a direct model, not the requested product. B adds 4 + 27 = 31; an inverse model does not generally keep this sum fixed. D subtracts 27 − 4 = 23; a fixed difference is not the inverse invariant.

3. D is correct: one notebook costs 81/3 = ₹27, so five cost 5 × 27 = ₹135. The invariant is cost/notebook count = ₹27 per notebook. A uses the inverse calculation 81 × 3/5 = 48.60 and makes more notebooks cost less. B adds the two extra notebooks to the rupee amount as 81 + 2, mixing count with money. C uses 81 × 5 = 405, treating the cost of three notebooks as the cost of one.

4. A is correct: (2x) × (y/2) = xy, so the factor is 1/2. B also doubles y and makes the product four times its old value. C makes positive y negative; multiplying by −2 is neither halving nor subtracting 2. Even subtracting a fixed 2 would not in general preserve the product. D leaves y unchanged, which would double xy instead of keeping it fixed.

5. B is correct: labels per box × boxes = 2,400, so boxes = 2,400/40 = 60. The old count was 2,400/30 = 80; 80 × 30/40 = 60 gives the inverse scaling. A keeps the old 80 boxes despite the larger packing size. C uses the direct factor 80 × 40/30 = 106 2/3; it fails the fixed-total test and cannot be the count of full boxes here. D subtracts the packing increase, 40 − 30 = 10, from 80; 70 boxes at 40 labels would hold 2,800, not 2,400.

6. C is correct: the rate is 84/3 = 28 pages per minute, so 28 × 2.5 = 70 pages. Check: a shorter time gives fewer than 84 pages, and 70/2.5 = 28. A results from misreading 2.5 minutes as 2 minutes 5 seconds: 84 × 125/180 = 58 1/3. In fact 2.5 minutes is 150 seconds, or 2 minutes 30 seconds. B counts only 2 minutes, omitting the half-minute. D uses the inverse calculation 84 × 3/2.5 = 100.8 even though output at a fixed rate is direct with time.

7. A is correct: the fixed batch requires 8 × 15 = 120 machine-hours, so 12t = 120 and t = 10 hours. Check: 12 × 10 = 120 and more identical machines take less time. B uses the direct factor 15 × 12/8 = 22.5, increasing time instead. C subtracts the four extra machines from 15 hours, mixing different quantities and failing 12 × 11 = 132. D mistakes the total 120 machine-hours for the group's elapsed hours; divide by 12 machines to obtain elapsed time.

8. D is correct: the scale compares matching length units, so ground distance = 2.8 × 40,000 = 112,000 cm = 1,120 m. Check: one drawing centimetre represents 400 m; 2.8 × 400 = 1,120. A relabels centimetres as metres without dividing by 100. B divides the centimetre value by 1,000 instead of 100. C obtains the number 1.12 by converting to kilometres but incorrectly labels it metres. This represented distance does not by itself give a road-route length.

9. B is correct: the invariant is people × days = 18 × 20 = 360 person-days at the unchanged daily amount per person. Thus days = 360/24 = 15, and 24 × 15 = 360. A uses the direct factor 20 × 24/18 = 26 2/3 and wrongly makes the stock last longer for more people. C subtracts the six extra people from 20 days, mixing units; 24 × 14 = 336, not 360. D labels the total 360 person-days as elapsed days without dividing by 24 people.

10. C is correct: F = 35 + 9d gives F = ₹44 at 1 km and ₹53 at 2 km. Thus F/d changes from 44 to 26.50 rupees per km, so it is not direct. A confuses the constant added charge ₹9/km with the average total fare per kilometre, which includes the fixed ₹35. B uses direction alone, an insufficient test. D correctly starts with 'No' but falsely calls the relation inverse: the fare increases, and the products 1 × 44 = 44 and 2 × 53 = 106 differ. Only F − 35 is directly proportional to distance.

11. D is correct: 2 × 18 = 36, 3 × 15 = 45 and 6 × 6 = 36. One mismatching row is enough to reject a constant-product model for the supplied data. A checks only direction, not the product. B skips the middle row; agreement at the endpoints cannot cancel a contradiction. C miscalculates the quotients: 18/2 = 9, 15/3 = 5 and 6/6 = 1, so the table is not direct either.

12. A is correct: under the stated assumptions the fixed job needs 5 × 18 = 90 worker-hours. Nine such workers need 90/9 = 10 hours, and 9 × 10 = 5 × 18. B has the same numerical answer but wrongly extends it to unequal rates or additional delays; worker count alone then does not establish the fixed product. C uses the direct factor 18 × 9/5 = 32.4 instead of the inverse factor. D uses the additive change 18 − 4 = 14; 9 × 14 = 126 worker-hours does not match 90.

7 Let an error choose what to revisit

• Questions 1, 3, 6: revisit block A for unit rates and one-unit reasoning
• Questions 2, 4, 5: revisit the start of block B for fixed products and reciprocal factors
• Questions 7, 9, 12: revisit cases 5–6 for fixed work/stock and rate conditions
• Question 8: revisit case 7 for matching length units and scale
• Questions 10–11: revisit case 8 to test every quotient/product rather than direction alone

After revisiting, retry the relevant item without the key and explain why each of its other three options fails.

Sources and lesson boundary

The official RRB mathematics list includes Ratio and Proportion: RRB CEN 09/2025, §14.1, printed p.28. This establishes topic scope; it is not a claim about question frequency or a prediction of a future exam.

The mathematical bases for direct and inverse relationships are NCERT Class VIII Mathematics, Chapter 11 §11.2 pp.129–137 and §11.3 pp.137–143, and NCERT Ganita Prakash, Grade 8 Part II, Chapter 3 §3.6 pp.63–65. For map scale, see NCERT Ganita Prakash, Grade 8 Part II, Chapter 3 §3.2 pp.56–57. All exposition, worked cases, practice and counterexamples here are independently written around the cited concepts; no textbook question or scan is reproduced. Identical steady rates, fixed work, unchanged daily use per person and absence of overhead are explicit conditions of our models. Combined unequal-worker rates, pipes, relative speed and compound proportion with several changing quantities are outside this lesson.

Analogy

Think of a stack of identical rectangular cards laid flat in a single row, with no gaps or overlaps. If each card is 6 cm wide, twice as many cards make twice the total width. The fixed width per card is a quotient: total width/card count = 6 cm per card. This pictures a direct relationship.

Now keep one 60 cm ribbon fixed and cut it into equal pieces with no material lost. More equal pieces mean shorter pieces: 5 pieces are 12 cm each, whereas 10 pieces are 6 cm each. Piece count × length per piece remains 60 cm. This pictures an inverse relationship because the whole ribbon, not the piece length, is fixed.

The useful question in both pictures is “What are we keeping unchanged?” The pictures have limits: unequal cards, gaps, unequal ribbon pieces or cutting loss change the relationships. They also do not prove that any larger group of workers is proportionally faster; that needs its own equal-rate and no-overhead assumptions. A picture helps you name the invariant, but does not replace checking the story's conditions.

Quick reference

Choose before calculating

1. Label the two changing quantities and write what stays fixed.
2. Put both entries for the same quantity in the same unit.
3. Test the invariant, then calculate; substitute the answer and check its direction.

Direct proportion: same quotient

• y/x = k; y = kx, for positive values used here
• y₂ = y₁ × (x₂/x₁): x multiplied by f means y multiplied by f
• One-unit method: old y ÷ old x, then multiply by new x
• Typical condition: unchanged unit price with no fee/discount, or unchanged output rate with no setup delay
• Keep rate units: ₹/kg, pages/minute, km/hour

Inverse proportion: same product

• xy = k; y = k/x; x and y are positive and x ≠ 0
• y₂ = y₁ × (x₁/x₂): x multiplied by f means y multiplied by 1/f
• Fixed work: workers × time only for identical steady workers, unchanged divisible work, no interference or extra delay
• Fixed stock: people × days only when each person's equal daily use is unchanged in both situations, with no replenishment/loss
• Fixed packing: items per full box × box count = total items; respect whole-box feasibility

Units and traps

• 1 hour = 60 minutes; 2.5 minutes = 2 minutes 30 seconds
• 1 m = 100 cm; 1 km = 1,000 m = 100,000 cm
• Scale 1:n means drawing length:represented ground length in matching units = 1:n; multiply drawing length by n, then convert the answer's unit
• Map distance is represented ground distance, not an assumed road-route length
• Both rising does not prove direct; opposite movement does not prove inverse
• A fixed fee in total cost usually defeats the direct model: test total cost/quantity
• Check every supplied row. Matching finite rows shows compatibility, not a universal real-world law
• If neither the quotient nor product is fixed, do not force either model

Error repair

Wrong units: rewrite conversions. Wrong direction: recheck the chosen model. Correct direction but wrong value: test the actual quotient/product. Missing assumptions: the numerical prediction is not yet justified.