The same problem can be expressed through objects, drawings, tables, equations, or parametric formulas. These are not competing solutions: they are languages suited to different ages and learning goals. This article is a complementary exploration of the pathway from natural language to algebra; here the focus is on instructional design and on how a general solution can suggest an elementary explanation.
Advanced mathematics can remain backstage: it is a tool for the teacher to check the data, uncover a structure, and choose an effective example—not something to impose on a child before it is needed.
For each of the eight situations, we distinguish four moments: the problem, a concrete primary-school approach, a lower-secondary formalisation, and a general upper-secondary or teacher-level view. The final check is not an afterthought: it always places the answer back into the original situation.
1. One problem, several representations
“Simpler” does not mean less rigorous. In primary school, the entire mathematical structure can be preserved while letters are replaced by strips, counters, systematic trials, or questions such as “what stays the same?”. In the final year of primary school or the beginning of lower secondary, there is no need to impose formal algebra: the unknown can appear later, when it gives a short name to a relationship that students already understand. At upper-secondary level, parameters show which data produce a possible answer and which properties hold for a whole family of problems.
| Level | Guiding question | Typical tools |
|---|---|---|
| Primary | What can I see, build, or compare? | Drawings, materials, tables, counting, invariants. |
| Lower secondary | How can I translate the relationship into a checkable calculation? | Ratios, unknowns, equations, graphs. |
| Upper secondary or teacher | What happens when the data change? | Functions, parameters, existence conditions, proofs. |
2. Chickens and rabbits
There are 35 animals in a yard, all chickens or rabbits, and 94 legs altogether. How many rabbits and how many chickens are there?
Primary: pretend they are all chickens
If all 35 animals were chickens, they would have 35 × 2 = 70 legs. The observed total has 24 extra legs. Replacing one chicken with one rabbit adds exactly two legs, so there are 24 / 2 = 12 rabbits. The remaining 35 − 12 = 23 animals are chickens. The transformation can be modelled with 35 two-legged counters, adding one pair of legs to twelve of them.
Lower secondary: a system
Let r be the number of rabbits and c the number of chickens:
r + c = 35
4r + 2c = 94
2(r + c) = 70
(4r + 2c) − (2r + 2c) = 94 − 70
2r = 24
r = 12, c = 23Upper secondary and teacher: the parametric model
With H animals and L legs, the general formulas are:
rabbits = (L − 2H) / 2
chickens = (4H − L) / 2The product 2 legs/animal × H animals = 2H legs also provides a dimensional reading: 2 is not an abstract number; it carries the unit “legs per animal”, which cancels with “animals”. A solution in non-negative whole numbers requires 2H ≤ L ≤ 4H and an even L. The teacher can therefore design valid data by choosing the two populations first, or immediately test whether a problem statement is coherent. The rabbit formula already contains the elementary idea: subtract the number of legs all the animals would have as chickens, then divide the surplus by two.
Check: 12 + 23 = 35 and 12 × 4 + 23 × 2 = 48 + 46 = 94. The correct answer is therefore 12 rabbits and 23 chickens.
3. The full, half-full, and empty barrel
A full barrel weighs 80 kg; the same barrel when half full weighs 50 kg. How much does the empty barrel weigh?
Primary: remove the same half-quantity
Going from full to half full removes 80 − 50 = 30 kg: this is the mass of half the contents. The half-full barrel still contains another 30 kg of product. Removing it from 50 kg leaves the barrel alone: 50 − 30 = 20 kg. A drawn balance and two identical blocks for the two halves of the contents make the step visible.
Lower secondary: distinguish container from contents
Let e be the mass of the empty barrel and f the mass of all its contents:
e + f = 80
e + f/2 = 50
f/2 = 80 − 50 = 30
e = 50 − 30 = 20Upper secondary and teacher: use the data as parameters
Let P be the mass of the full barrel and M that of the half-full barrel. Then:
half the contents = P − M
empty barrel = M − (P − M) = 2M − P
all the contents = 2(P − M)The subtractions are dimensionally sound because P and M are both masses. A physically meaningful case requires P/2 ≤ M < P. The formula 2M − P directly suggests the equation-free solution: remove from the half-full barrel the same mass that was removed when moving from full to half full.
Check: the empty barrel weighs 20 kg and the complete contents 60 kg; full it weighs 20 + 60 = 80 kg, and half full 20 + 30 = 50 kg.
4. The age riddle
Anna is 6 years older than Bruno and says to him: “Today I am twice the age you were when I was the age you are today.” How old are they?
Primary: move back by the age difference
Anna is six years older. When Anna was Bruno’s current age, it was therefore six years ago; at that time Bruno was also six years younger than he is now. The statement says that Anna’s present age is twice Bruno’s age at that earlier time. Using six-year strips, we can see that Bruno’s past age contains two strips, his current age three, and Anna’s current age four. Their ages are therefore 18 and 24.
Lower secondary: translate the timeline
Let x be Bruno’s present age. Anna is x + 6. Six years ago Bruno was x − 6:
x + 6 = 2(x − 6)
x + 6 = 2x − 12
x = 18
Bruno = 18, Anna = 24Upper secondary and teacher: use the difference as a parameter
Let d be the positive difference between the older and younger ages, and let y be the younger age. When the older person was y, the younger person was y − d. The riddle gives:
y + d = 2(y − d)
y = 3d
older age = y + d = 4d
younger age / older age = 3/4The relationship does not determine unique ages until d is known; it determines the ratio 3 : 4. The teacher can choose a suitable difference and immediately obtain a consistent example. With d = 6, the ages are 3d = 18 and 4d = 24.
Check: six years ago Anna was 18, Bruno’s current age, while Bruno was 12. Anna is now 24, twice 12.
5. The rectangle with fixed perimeter
A rectangle is made with a 40 cm length of string. Which dimensions maximise its area?
Primary: build and compare
The perimeter is 40 cm, so two adjacent sides add up to 20 cm. Rectangles with whole-number sides can be drawn on squared paper and their areas ordered:
sides: 1 and 19 2 and 18 3 and 17 ... 9 and 11 10 and 10
area: 19 36 51 ... 99 100As the two side lengths become closer, the area grows. The 10 cm square is the best candidate. A closed loop of string or a hinged strip makes it visible that the perimeter stays fixed while the enclosed space changes.
Lower secondary: sides equally distant from 10
Write the side lengths as 10 − d and 10 + d. Their sum remains 20:
A = (10 − d)(10 + d)
A = 100 − d²Every non-zero value of d subtracts something from 100, so the maximum occurs at d = 0.
Upper secondary and teacher: a family of parabolas
For a general perimeter P, one side is x and the other is P/2 − x:
A(x) = x(P/2 − x)
= P²/16 − (x − P/4)²The area is greatest when the subtracted term is zero, namely when x = P/4. The same result follows from the vertex of the parabola or from differentiation. Parameter P shows that this is not a coincidence caused by the number 40.
Check: four 10 cm sides give a 40 cm perimeter, and the area is 10 × 10 = 100 cm².
6. Mixing two concentrations
How many litres of a 40% drink and how many litres of a 5% drink are needed to make 21 litres of a 25% drink? Assume that the volumes are additive.
Primary: balance with equal parts
Forty per cent is 15 percentage points above 25%, while 5% is 20 points below. One litre of the first drink brings a surplus of 15 points and one litre of the second a deficit of 20. To balance them, the quantities must be in the inverse ratio 20 : 15 = 4 : 3. The seven parts total 21 litres, so each part is 3 litres: 4 × 3 = 12 litres at 40% and 3 × 3 = 9 litres at 5%. Hundred-square grids or coloured water can make the weighted average visible.
Lower secondary: conserve the amount of concentrate
0.40x + 0.05(21 − x) = 0.25 × 21
0.35x + 1.05 = 5.25
x = 12The second quantity is 21 − 12 = 9 litres.
Upper secondary and teacher: weighted average
Given concentrations c₁ > c > c₂ and total volume V, the quantity of the first mixture is:
x = V(c − c₂) / (c₁ − c₂)
V − x = V(c₁ − c) / (c₁ − c₂)The model also identifies impossible cases: without additions or reactions, the final concentration must lie between the two initial concentrations. To create manageable data, the teacher can begin with the desired ratio and work backwards to the total volume.
Check: 12 × 0.40 + 9 × 0.05 = 4.80 + 0.45 = 5.25 equivalent litres of concentrate, which is 21 × 0.25.
7. Handshakes
At a meeting, every pair of people exchanges exactly one handshake. There are 45 handshakes altogether. How many people attend?
Primary: grow the picture
When a new person arrives, that person greets everyone already present. Starting with one person, the total therefore grows by successively adding 1, 2, 3, and so on:
people: 1 2 3 4 5 6 7 8 9 10
handshakes: 0 1 3 6 10 15 21 28 36 45The total reaches 45 with 10 people. Dots joined by lines or name cards help prevent the same pair from being counted twice.
Lower secondary: count and divide by two
With n people, each can greet n − 1 others. Product n(n − 1) counts every pair twice:
n(n − 1) / 2 = 45
n(n − 1) = 90
10 × 9 = 90
n = 10Upper secondary and teacher: triangular numbers
H = n(n − 1) / 2
n = (1 + √(1 + 8H)) / 2A total H is admissible only when 1 + 8H is a perfect square. This condition prevents an accidentally impossible number of handshakes and connects the problem with complete graphs and binomial coefficients.
Check: 10 × 9 / 2 = 45.
8. Two taps and a tank
One tap fills a tank in 6 hours and another fills it in 4 hours. How long do they take when working together?
Primary: divide the tank into twelve parts
Choose 12 parts because 12 is divisible by both 6 and 4. In one hour, the first tap fills 2 parts and the second fills 3. Together they fill 5 parts per hour, so 12 parts require 12/5 = 2.4 hours, or 2 hours 24 minutes. A tank divided into twelve strips prevents the common error of adding or averaging the completion times.
Lower secondary: add fractions of the job
t/6 + t/4 = 1
2t/12 + 3t/12 = 1
5t = 12
t = 12/5 hoursUpper secondary and teacher: add rates
If two taps take a and b hours separately, their joint time is:
1/T = 1/a + 1/b
T = ab / (a + b)For more taps, all rates are added. A drain is represented by a negative rate, provided the total rate remains positive. Choosing times with a small least common multiple makes the concrete solution more accessible.
Check: in 2.4 hours the first tap fills 2.4/6 = 0.4 of the tank and the second 2.4/4 = 0.6; together they fill exactly one tank.
9. The basket of eggs and remainders
When some eggs are grouped in twos, 1 remains; in threes, 2 remain; in fours, 3 remain; in fives, 4 remain; and in sixes, 5 remain. When grouped in sevens, none remain. What is the smallest possible number of eggs?
Primary: a systematic search
In each of the first five groupings, one more egg would complete all the groups. The required number plus one must therefore be divisible by 2, 3, 4, 5, and 6. The candidates are one less than common multiples of these numbers: 59, 119, 179, and so on. The first one divisible by 7 is 119.
Lower secondary: use the least common multiple
Since lcm(2, 3, 4, 5, 6) = 60, we can write:
N + 1 = 60k
N = 60k − 1
k = 1 → N = 59, not divisible by 7
k = 2 → N = 119, divisible by 7Upper secondary and teacher: congruences
60k − 1 ≡ 0 (mod 7)
4k ≡ 1 (mod 7)
k ≡ 2 (mod 7)
N = 119 + 420t, for every non-negative integer tThe solution is unique modulo 420. To create variants of the same type, the teacher can choose a new divisor coprime to the preceding least common multiple; the Chinese remainder theorem then guarantees one class of solutions.
Check: dividing 119 by 2, 3, 4, 5, and 6 gives remainders 1, 2, 3, 4, and 5 respectively; moreover, 119 = 7 × 17.
10. Conclusion: algebra as a lens, not a barrier
In all eight examples, the advanced solution does not erase the intuitive one. It often explains it: subtraction L − 2H suggests pretending that every animal is a chicken; formula 2M − P says to remove the same half-content twice; the square in the area formula indicates that equal sides are best; and the least common multiple turns many guesses into a systematic search. The teacher may follow the general path while preparing and offer the class only the portion appropriate to its level.
The goal is not to introduce difficult symbols prematurely, but to offer a concrete experience today that can be recognised inside a formula tomorrow.
A practical checklist for teachers
- Solve the general model: identify parameters, constraints, and existence conditions.
- Check the data: choose whole-number results, consistent units, and realistic situations.
- Look for the invariant: what stays unchanged while the situation changes?
- Choose a representation: objects, a drawing, a table, a timeline, or systematic trials.
- Allow several strategies: compare methods before formalising them.
- Introduce symbols when useful: an equation should name a relationship that is already understood.
- Check in context: always substitute the result back into the problem data.
- Generalise: finally ask what would change if one item of data were altered.
A cura di Salvatore Mosaico.