Proportion is about what remains linked while quantities change. In direct proportion, two quantities grow or shrink together by a constant multiplier. In inverse proportion, one grows while the other shrinks so that their product remains constant.
This twenty-sixth Secondary 4 Mathematics Learning Guide develops direct and inverse proportion, scale and rate modelling as one relationship system. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
Current syllabus connection: ratio, direct and inverse proportion, scales, average rates, speed and unit conversion form part of the Number and Algebra progression in the 2026 O-Level Mathematics 4052 and 2027 SEC G3 Mathematics K310 syllabuses.
Direct proportion has a constant quotient
If y is directly proportional to x, write y=kx. The ratio y/x remains constant and equals k.
If x doubles, y doubles. If x is multiplied by 5, y is multiplied by 5, provided the same model remains valid.
Worked Example 1 | Build a direct-proportion rule
y is directly proportional to x. When x=8, y=20. Find y when x=14.
Write y=kx. Using 20=8k gives k=2.5.
y=2.5x, so at x=14, y=35.
Check using scale: x changed by factor 14/8=1.75. So y should change from 20 by the same factor: 20×1.75=35.
Inverse proportion has a constant product
If y is inversely proportional to x, write y=k/x for x≠0. The product xy remains constant and equals k.
If x doubles, y halves. If x becomes three times as large, y becomes one third as large.
Worked Example 2 | Build an inverse-proportion rule
y is inversely proportional to x. When x=6, y=15. Find y when x=10.
Write y=k/x. Then 15=k/6, so k=90.
y=90/x, so at x=10, y=9.
Check the product: 6×15=90 and 10×9=90.
Do not classify proportion from words alone
“More workers means less time” may suggest inverse proportion, but only if total work and productivity per worker remain fixed. If workers interfere with each other or tasks cannot be divided evenly, the simple model may not apply.
Mathematics models relationships under stated or implied conditions. A proportional rule is powerful because it is simple, but its assumptions still matter.
Worked Example 3 | Workers and time
Six equally productive workers complete a fixed task in 15 hours. Assuming time is inversely proportional to the number of workers, how long would 10 workers take?
Let n be number of workers and t the time. Since t=k/n:
15=k/6, so k=90 worker-hours.
For 10 workers, t=90/10=9 hours.
The result depends on the equal-productivity and perfectly divisible-work assumptions.
Rates compare two quantities with units
A rate describes one quantity per unit of another: kilometres per hour, dollars per kilogram, litres per minute, or words per minute.
Units are part of the mathematics. A numerical answer without the correct unit can hide a conversion error.
Worked Example 4 | Average speed
A car travels 150 km in 2.5 hours. Find its average speed.
Average speed = distance/time = 150/2.5 = 60 km/h.
This does not mean the car travelled at exactly 60 km/h at every instant. Average speed compresses the whole journey into total distance divided by total time.
Average speed is not the average of two speeds unless conditions justify it
If a journey has different speeds over different durations or distances, first find total distance and total time.
Worked Example 5 | Equal distances at different speeds
A cyclist travels 30 km at 20 km/h and another 30 km at 30 km/h. Find the average speed for the whole journey.
First part time=30/20=1.5 h.
Second part time=30/30=1 h.
Total distance=60 km and total time=2.5 h.
Average speed=60/2.5=24 km/h.
The ordinary average of 20 and 30 is 25, which is wrong here because more time is spent at the slower speed.
Unit conversion should happen before rates are combined
To convert m/s to km/h, multiply by 3.6. To convert km/h to m/s, divide by 3.6.
This follows from 1 km=1000 m and 1 hour=3600 s.
Worked Example 6 | Convert speed units
Convert 18 m/s to km/h.
18×3.6=64.8 km/h.
Scale is a proportional relationship
A map scale of 1:50,000 means every drawing length is 1/50,000 of the corresponding actual length when both use the same unit.
This is direct proportion: doubling map distance doubles actual distance.
Worked Example 7 | Scale and real distance
On a 1:40,000 map, two points are 6.5 cm apart. Find the actual distance in kilometres.
Actual distance=6.5×40,000=260,000 cm.
260,000 cm=2.6 km.
The separate Congruence, Similarity, Scale Drawings and Area-Volume Ratios guide develops geometric scaling further.
Rates can form equations
When distance is fixed, time=d/s. Comparing two speeds therefore creates a fractional relationship that may lead to an equation.
Worked Example 8 | Time saved by travelling faster
A journey of 120 km is completed at speed v km/h. At v+20 km/h, the journey takes one hour less. Find v.
Time difference:
120/v − 120/(v+20)=1.
Multiply by v(v+20):
120(v+20)−120v=v(v+20).
2400=v²+20v.
v²+20v−2400=0.
(v+60)(v−40)=0.
Speed must be positive, so v=40 km/h.
Check: 120/40=3 h and 120/60=2 h, exactly one hour less.
Cost per unit creates a comparison model
If one pack costs $7.20 for 600 g and another costs $9.50 for 1 kg, compare on a common unit such as dollars per kilogram.
First pack: $7.20/0.6 kg=$12/kg.
Second pack: $9.50/kg.
The second pack has the lower unit price, assuming product quality and other relevant factors are comparable.
Worked Example 9 | Flow rate and time
A tank needs 840 litres. Water enters at 35 litres per minute. How long does filling take at this constant rate?
Time=840/35=24 minutes.
If two taps operate together, their flow rates can be added only if both rates are measured in compatible units and remain constant.
Direct and inverse graphs look different
For y=kx, the graph is a straight line through the origin. For y=k/x with positive k, the graph is reciprocal-shaped and excludes x=0.
A table can help classify the relationship: check whether y/x is constant for direct proportion or xy is constant for inverse proportion.
Worked Example 10 | Classify from data
| x | y |
|---|---|
| 2 | 24 |
| 3 | 16 |
| 4 | 12 |
| 6 | 8 |
The products xy are 48 throughout. Therefore y is inversely proportional to x with k=48:
y=48/x.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Uses y=k/x for direct proportion | Two models memorised without invariants | Check whether quotient or product stays constant |
| Averages two speeds directly | Total distance and time ignored | Return to average speed=total distance/total time |
| Combines km/h with m/s | Units treated as decoration | Convert before calculating |
| Assumes worker-time inverse proportion in every context | Model assumptions ignored | State fixed-work and equal-productivity assumptions |
| Uses percentage change as simple addition over repeated periods | Multiplicative change missed | Track repeated multipliers |
Independent practice
- y is directly proportional to x. If y=18 when x=6, find y when x=15.
- y is inversely proportional to x. If y=20 when x=4, find y when x=10.
- Eight workers take 12 hours for a fixed task. Under an inverse-proportion model, how long would 6 workers take?
- Convert 25 m/s to km/h.
- A map scale is 1:80,000. A road measures 4.2 cm on the map. Find its actual length in km.
- A car travels 40 km at 40 km/h and 60 km at 60 km/h. Find the overall average speed.
Explained answers
1. k=18/6=3, so y=3(15)=45.
2. k=xy=80, so y=80/10=8.
3. Worker-hours=8×12=96, so time=96/6=16 hours.
4. 25×3.6=90 km/h.
5. 4.2×80,000=336,000 cm=3.36 km.
6. Times are 1 h and 1 h, so total distance=100 km and total time=2 h. Average speed=50 km/h.
Final thought
Proportion is not a keyword hunt. It is a search for the invariant relationship connecting changing quantities. Rate problems add units; scale problems add representation; inverse models add reciprocal change.
Find what stays constant. Then build the equation around it.
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