A method for turning compound proportionality into a sequence of simple steps
A classic problem
12 hens lay 48 eggs in 12 days. How many eggs do 3 hens lay in 3 days?
The numbers are simple, but the structure is interesting because it involves three quantities simultaneously: production, resources and time.
1. First of all: the simple rule of three
The simple rule of three involves two quantities, while any third quantity remains fixed.
For example: 12 hens lay 48 eggs in 12 days. How many eggs would 3 hens lay in the same 12 days?
Time does not change. Only hens and eggs change. Going from 12 hens to 3 means dividing by 4; production is therefore also divided by 4:
48 : 4 = 12
Thus 3 hens lay 12 eggs in 12 days. A single proportion is sufficient here: this is a simple rule of three problem.
2. Why the initial problem uses the compound rule of three
In the initial problem, however, both the number of hens and the number of days change. We must move simultaneously from 12 hens to 3 hens and from 12 days to 3 days.
Production is directly proportional to both quantities: fewer hens mean fewer eggs; fewer days mean fewer eggs again. This is why we speak of the compound rule of three.
3. The classic solution
The traditional method uses the two factors of change:
48 × (3/12) × (3/12) = 3
The first ratio accounts for the change in resources; the second for the change in time. The answer is therefore 3 eggs.
The general formula, denoting production by P, resources by R and time by T, is:
P2 = P1 × (R2/R1) × (T2/T1)
The procedure is correct and fast. But a student may still wonder: why exactly do two ratios appear?
4. The triple method
We can represent each situation by the triple:
(P, R, T)
where P is Production, R is Resources and T is Time. The important feature is to show the requested triple with its question mark immediately as well, so that we can clearly see where we need to arrive.
| Initial situation | Requested situation |
|---|---|
| (48, 12, 12) | (?, 3, 3) |
We can now proceed one quantity at a time.
First transformation: change the resources
We need to reduce the hens from 12 to 3: we divide by 4. If time remains unchanged, we also divide production by 4.
(48, 12, 12) -> (12, 3, 12)
Second transformation: change the time
There are already 3 hens. We need to reduce the days from 12 to 3: we divide by 4 again, this time dividing production and time.
(12, 3, 12) -> (3, 3, 3)
The requested triple was:
(?, 3, 3)
and we have reached:
(3, 3, 3)
The required value is therefore 3.
5. The rule of the triple
In linear production problems, we always work on one pair at a time: (P,R) or (P,T).
Rule: multiply or divide production and just one of resources or time by the same number, leaving the other quantity unchanged.
(P, R, T) -> (kP, kR, T)
(P, R, T) -> (kP, R, kT)
6. Why it works
If production is linear, the following relationship holds:
P = kRT
and therefore:
P / (R × T) = k
The quantity P/(RT) remains constant. In our example:
48/(12×12) = 12/(3×12) = 3/(3×3) = 1/3
The triple method is therefore not a trick: it preserves the invariant of the problem.
7. The reverse problem
Three hens lay 3 eggs in 3 days. How many eggs do 12 hens lay in 12 days?
| Initial situation | Requested situation |
|---|---|
| (3, 3, 3) | (?, 12, 12) |
We multiply production and resources by 4:
(3, 3, 3) -> (12, 12, 3)
Then we multiply production and time by 4:
(12, 12, 3) -> (48, 12, 12)
The answer is therefore 48 eggs.
8. The teaching advantage
The classic formula compresses everything into one line. The triple method instead makes the two steps contained in the formula visible. In this way, the compound rule of three becomes a sequence of two simple proportionality problems.
(48,12,12) -> (12,3,12) -> (3,3,3)
The central idea is: one change at a time.
Five classic versions of the problem
Let us now look at five typical problems. In each, we immediately write the known triple and the requested triple with its question mark, then apply the method quickly.
1. Workers and parts
8 workers produce 120 parts in 5 days. How many parts do 12 workers produce in 10 days?
| Initial situation | Requested situation |
|---|---|
| (120, 8, 5) | (?, 12, 10) |
From 8 to 12 workers: × 3/2 on P and R
(120, 8, 5) -> (180, 12, 5)
From 5 to 10 days: × 2 on P and T
(180, 12, 5) -> (360, 12, 10)
Answer: 360 parts.
2. Machines and components
6 machines produce 900 components in 3 hours. How many components do 4 machines produce in 6 hours?
| Initial situation | Requested situation |
|---|---|
| (900, 6, 3) | (?, 4, 6) |
From 6 to 4 machines: × 2/3 on P and R
(900, 6, 3) -> (600, 4, 3)
From 3 to 6 hours: × 2 on P and T
(600, 4, 3) -> (1200, 4, 6)
Answer: 1200 components.
3. Printers and pages
5 printers print 2000 pages in 4 hours. How many pages do 10 printers print in 6 hours?
| Initial situation | Requested situation |
|---|---|
| (2000, 5, 4) | (?, 10, 6) |
From 5 to 10 printers: × 2 on P and R
(2000, 5, 4) -> (4000, 10, 4)
From 4 to 6 hours: × 3/2 on P and T
(4000, 10, 4) -> (6000, 10, 6)
Answer: 6000 pages.
4. Looms and fabric
9 looms produce 540 metres of fabric in 6 hours. How many metres do 3 looms produce in 12 hours?
| Initial situation | Requested situation |
|---|---|
| (540, 9, 6) | (?, 3, 12) |
From 9 to 3 looms: ÷ 3 on P and R
(540, 9, 6) -> (180, 3, 6)
From 6 to 12 hours: × 2 on P and T
(180, 3, 6) -> (360, 3, 12)
Answer: 360 metres.
5. Pumps and litres of water
4 pumps transfer 2400 litres in 2 hours. How many litres do 6 pumps transfer in 5 hours?
| Initial situation | Requested situation |
|---|---|
| (2400, 4, 2) | (?, 6, 5) |
From 4 to 6 pumps: × 3/2 on P and R
(2400, 4, 2) -> (3600, 6, 2)
From 2 to 5 hours: × 5/2 on P and T
(3600, 6, 2) -> (9000, 6, 5)
Answer: 9000 litres.
Conclusion
The compound rule of three remains perfectly valid, but the triple method makes its structure explicit. First we show the initial situation, then the requested situation with its question mark:
(P1, R1, T1) -> (?, R2, T2)
At that point, we change just one quantity at a time, accompanying the change with the same change in production.
(P1, R1, T1) -> (P', R2, T1) -> (P2, R2, T2)
The classic formula:
P2 = P1 × (R2/R1) × (T2/T1)
thus becomes the compact expression of two elementary transformations. The main advantage is educational: students are not merely asked to apply a formula; the path leading to the solution becomes visible.
When the question mark is on Resources or Time
So far, we have mainly sought production. But the triple (P,R,T) works in the same way when the unknown is the number of resources or the time.
The rule does not change: transform one pair at a time, choosing (P,R) or (P,T), until the requested triple is reached.
1. Unknown resources - workers
6 workers produce 240 parts in 5 days. How many workers are needed to produce 360 parts in the same 5 days?
| Initial situation | Requested situation |
|---|---|
| (240, 6, 5) | (360, ?, 5) |
Time already matches the requested value. Production must go from 240 to 360: × 3/2. We work on the pair (P,R).
(240, 6, 5) -> (360, 9, 5)
9 workers are needed.
2. Unknown resources - machines
8 machines produce 400 parts in 10 hours. How many machines are needed to produce 300 parts in 5 hours?
| Initial situation | Requested situation |
|---|---|
| (400, 8, 10) | (300, ?, 5) |
First I reduce time from 10 to 5 hours: ÷ 2 on P and T.
(400, 8, 10) -> (200, 8, 5)
Now I need to increase production from 200 to 300: × 3/2 on P and R.
(200, 8, 5) -> (300, 12, 5)
12 machines are needed.
3. Unknown resources - printers
10 printers produce 5000 pages in 4 hours. How many printers are needed to produce 7500 pages in 6 hours?
| Initial situation | Requested situation |
|---|---|
| (5000, 10, 4) | (7500, ?, 6) |
I increase time from 4 to 6 hours: × 3/2 on P and T.
(5000, 10, 4) -> (7500, 10, 6)
10 printers are needed.
4. Unknown time - machines
5 machines produce 600 parts in 4 hours. In how many hours do they produce 900 parts, still using 5 machines?
| Initial situation | Requested situation |
|---|---|
| (600, 5, 4) | (900, 5, ?) |
Resources already match the requested value. Production goes from 600 to 900: × 3/2. We work on the pair (P,T).
(600, 5, 4) -> (900, 5, 6)
6 hours are needed.
5. Unknown time - workers
6 workers produce 900 parts in 3 days. In how many days do 4 workers produce 1200 parts?
| Initial situation | Requested situation |
|---|---|
| (900, 6, 3) | (1200, 4, ?) |
First I reduce the workers from 6 to 4: × 2/3 on P and R.
(900, 6, 3) -> (600, 4, 3)
Now production must go from 600 to 1200: × 2 on P and T.
(600, 4, 3) -> (1200, 4, 6)
6 days are needed.
The triple is symmetric with respect to the unknown
The question mark can therefore occupy any position:
(?, R, T) or (P, ?, T) or (P, R, ?)
There is no need to learn three different procedures. We start from the known triple, write the triple we want to reach, and carry out successive transformations on the permitted pairs only:
(P,R) or (P,T)
This is one of the most useful features of the method: the procedure remains identical even when the quantity to be determined changes.