The core idea
A ratio compares quantities by division. Equivalent ratios scale every part by the same factor. A unit rate states the amount for one unit. Direct proportion keeps that rate constant as the quantities change.
1. Ratios describe a relationship
An illustrative art kit contains 6 red beads and 9 blue beads. The red-to-blue ratio is 6:9, read “six to nine”; the colon separates the quantities. Dividing both numbers by 3 gives 2:3. There are two red beads for every three blue beads. This does not mean the kit contains only five beads: it contains three groups of that five-bead pattern.
Order matters: blue-to-red is 3:2. The whole contains 15 beads, so the red fraction is 6/15 = 2/5, not 2/3. Label quantities before simplifying. For three colours, a ratio of 2:3:1 has six total parts; a collection of 30 beads would contain 10, 15 and 5 beads in that order. Multiplying all parts by five preserves the relationship.
2. Align the units
For a pure ratio of two lengths, use the same unit. A 1.5-metre strip compared with a 50-centimetre strip is not 1.5:50. One metre equals 100 centimetres, so the first strip is 150 centimetres. The ratio is 150:50 = 3:1: the first strip is three times as long. The units cancel because centimetres are divided by centimetres.
A rate compares different kinds of quantities and keeps a compound unit. Travelling 18 kilometres in 1.5 hours gives 18 ÷ 1.5 = 12 kilometres per hour. The sign ÷ means division and “per” means for each. This is an average over the trip, not a claim of identical speed at every moment. Always state the unit above and below the division.
3. Solved example: scale a mixture
An illustrative drink recipe mixes concentrate and water in a volume ratio of 2:5. To make 1.4 litres, count 2 + 5 = 7 total parts. One litre contains 1000 millilitres, so 1.4 litres is 1400 millilitres. One part is 1400 ÷ 7 = 200 millilitres. Use 2 × 200 = 400 millilitres of concentrate and 5 × 200 = 1000 millilitres of water; × means multiplication.
Check the sum and ratio: 400 + 1000 = 1400, and 400:1000 simplifies to 2:5. Doubling the mixture requires 800 and 2000 millilitres. Adding 200 to each original quantity instead produces 600:1200 = 1:2, changing the proportions. Equal additions generally do not preserve a ratio. This classroom model assumes the ingredient volumes add.
Sources: OpenStax: Prealgebra 2e: Solve Proportions and their Applications ↗
4. Solved example: compare unit prices
Illustrative packet A contains 750 grams of a material for ₹90; packet B contains 1.2 kilograms of the same material for ₹138. The symbol ₹ means rupees, and a kilogram is 1000 grams. A costs ₹90 ÷ 0.75 = ₹120 per kilogram. B costs ₹138 ÷ 1.2 = ₹115 per kilogram. B is cheaper by ₹5 per kilogram despite its higher packet price.
Compare a common 3-kilogram mass as a check: A's rate gives ₹360 and B's rate gives ₹345. The difference is ₹15. Unit price measures one aspect of value. If only 500 grams can be used before the rest becomes waste, the larger packet may not be preferable. Keep the mathematical comparison separate from assumptions about what is useful.
Sources: OpenStax: Prealgebra 2e: Ratios and Rate ↗ · OpenStax: Prealgebra 2e: Solve Proportions and their Applications ↗
5. Test whether a rate stays constant
Let n be the number of identical notebooks and C their total cost in rupees. If each costs ₹18 without additional charges, C = 18n; symbols placed together mean multiplication. For n = 1, 2 and 5, the costs are 18, 36 and 90. Dividing cost by count always gives 18. This is direct proportion; zero notebooks cost zero in the model.
Adding a ₹30 delivery charge changes the rule to C = 30 + 18n. Two notebooks cost ₹66 and four cost ₹102, not twice ₹66. Cost rises by a constant amount per notebook, but total cost per notebook changes. A constant increase differs from a constant ratio. Discounts and changing speeds likewise require checking before extending a proportional model.
Keep the rate constant
| Notebooks | Cost | Cost per notebook |
|---|---|---|
| 3 | ₹90 | ₹90 ÷ 3 = ₹30 |
| 5 | ₹150 | ₹150 ÷ 5 = ₹30 |
| 8 | ₹240 | ₹240 ÷ 8 = ₹30 |
Sources: NCERT Class VIII Mathematics Exemplar: Direct and Inverse Proportions ↗ · OpenStax: Prealgebra 2e: Solve Proportions and their Applications ↗ · OpenStax: Prealgebra 2e: Graphing Linear Equations ↗
6. Find an unknown quantity
Eight identical sheets weigh 60 grams. For the same size and material, 20 sheets weigh 20 × (60 ÷ 8) = 150 grams. The unit rate is 7.5 grams per sheet. Alternatively, the sheet count grows by 20/8 = 2.5, so the mass grows by the same factor: 60 × 2.5 = 150. Both routes preserve the relationship.
Write 60/8 = m/20, where m is the unknown mass in grams. Multiplying both sides by 20 isolates m. Keeping mass above sheet count on both sides makes the proportion meaningful. The arrangement 60/8 = 20/m reverses only one ratio and breaks the unit agreement. Cross-multiplication helps only after you have set up a valid proportion.
A calculated quantity and a usable whole-item count may differ. Suppose an illustrative vehicle holds 12 passengers and a group contains 45 people. Division gives 45/12 = 3.75 vehicle-loads, but three vehicles provide only 36 places. Four vehicles provide 48 places and can carry everyone in one trip. The capacity ratio helped calculate the requirement; the whole-vehicle constraint determines the final choice. Always ask whether fractional units are possible and whether the task asks for an exact amount, a minimum capacity or a maximum within a limit.
Sources: OpenStax: Prealgebra 2e: Solve Proportions and their Applications ↗
7. Recognise inverse proportion and its limits
For a fixed 120-kilometre journey, constant speeds of 30 and 60 kilometres per hour give times of 4 and 2 hours. Doubling speed halves time because speed multiplied by time stays 120 kilometres. This is inverse proportion. It assumes fixed distance and ignores stops. Do not automatically apply it to group work: doubling the workers can introduce coordination delays or leave too few tools to keep everyone working.
Sources: OpenStax: Prealgebra 2e: Ratios and Rate ↗ · OpenStax: Prealgebra 2e: Solve Proportions and their Applications ↗
PUT IT INTO PRACTICE
Prepare an illustrative seed mixture
- A paper exercise uses a sunflower-to-millet mass ratio of 3:7 for a 2-kilogram mixture. Convert the total to grams and find one part before calculating both amounts.
- Calculate the amounts for 3 kilograms using a scale factor. Compare your method with finding each ingredient's amount per kilogram.
- Check: 2 kilograms needs 600 and 1400 grams; multiplying both by 1.5 gives 900 and 2100 grams. Explain why adding 500 grams to both original amounts changes the ratio.
Check your understanding
A group has 12 beginners and 18 experienced learners. What fraction are beginners?
12/(12 + 18) = 2/5. The part-to-part ratio 2:3 uses a different reference.
Which has the lower unit price: 6 pens for ₹54 or 10 for ₹85?
The rates are ₹9 and ₹8.50 per pen, so the ten-pen packet has the lower unit price.
A map uses 1 centimetre for 2 kilometres. What does 3.5 centimetres represent?
3.5 × 2 = 7 kilometres, assuming the same scale along the measured route.
A service charges ₹40 plus ₹5 per item. Do 8 items cost twice as much as 4?
No. Four cost ₹60 and eight ₹80. The fixed charge prevents direct proportion.
A 90-kilometre trip takes 3 hours. At the same average rate, how far in 5 hours?
The rate is 30 kilometres per hour, giving 150 kilometres if that average rate continues.