Solving a problem does not mean merely performing calculations: it means understanding the question, selecting the necessary information, and recognising the relationship between the data and the unknowns. This pathway, originally prepared for a 2011-2012 induction course for new teachers, moves gradually from natural language to algebra and also shows when an equation is not the most immediate route.
The original material was printed on .
1. Summary
When learning mathematics, students often show difficulty or uncertainty in solving problems. It is therefore useful to propose activities that help them internalise a guided pathway until they can solve a problem situation correctly and independently.
Discomfort with problems appears early. Four fundamental operations must be practised in order to interpret a text:
- understand the text through deliberate reasoning;
- recognise and identify the essential data and information;
- recognise the question, or questions, being asked;
- grasp the relationships between the data and the required result.
2. I have a problem: how can I solve it?
Whenever we face a problem, the first question should be: what kind of problem is this?
| Kind of problem | Example and observation |
|---|---|
| A problem that cannot be solved | Luke plays cards with Mark and wins the game. By how many points did he win? The problem cannot be solved because essential information is missing. Knowing that Luke played well does not determine the score. |
| A problem that is not really a problem | Paula is 8 years old. Her friend Anna has exactly the same date of birth. How old is Anna? The answer is already in the text: the two friends were born on the same day of the same year, so Anna is 8. |
| A problem solved by taking action | Andrew must do a school assignment but discovers that his notebook is full. How can he complete the assignment? The problem is solved by acting: Andrew gets a new notebook. |
| A problem solved by arithmetic | Philip owns 35 collectible cards and his grandfather gives him 12 more. How many cards does Philip have altogether? An addition is needed: |
| A problem solved by insight | A complete standard 52-card deck, with every card equally thick, measures 1.3 cm when viewed from the side. How thick would it be after removing all the kings? A ruler is available, and the answer must be given in one minute. Removing the four kings leaves 48 cards. Thickness is proportional to the number of cards: |
3. Where should I start?
Problem solving requires several skills, but understanding the question is fundamental. Otherwise, we risk solving a different problem and all our calculation skills become useless.
A medical-school admissions test once included this question: “Fourteen people go on a trip. All but nine die. How many survivors are there?”
Possible answers: a) 14; b) 9; c) 5; d) 23.
More than half of the candidates chose 5: the wording prompts the brain to perform 14 − 9 automatically. Let us analyse the text instead.
| Who are we talking about? | People. |
|---|---|
| How many are there? | 14. |
| How many people die? | All except 9. |
| How many survive? | 9. |
The first subproblem is therefore always: what is the problem asking? If the request is unclear, the reasoning that follows cannot lead to the correct answer.
4. From natural language to algebraic language
Many problems are solved by translating the data into algebra, but this transition is not immediate. Algebra makes visible a rule that can remain hidden in numerical cases.
Consider the statement: “Together, you and I are 40 years old.” Knowing either age lets us determine the other:
I am 30 years old → you are 40 − 30 = 10
I am 35 years old → you are 40 − 35 = 5
I am x years old → you are 40 − xSaying only that 30 + 10 = 40 and 35 + 5 = 40 verifies the two examples but does not yet state the rule. The numerical rule is “subtract the known age from 40”; its algebraic formulation is x ↦ 40 − x.
When we know how to perform a procedure with numbers, we should ask which rule we used and try to express it algebraically.
5. Guided problems
To help students translate natural language into algebra, it is useful to guide them with carefully chosen questions. In the following problems, we pretend that we know the answer, represent it with a letter, and use the information in the text to construct an equation.
5.1. Guided problem 1: Claudia’s age
If we asked Claudia how old she was, she would answer: “If I were 4 years older, I would be over 20 by as many years as I am now under 20.” How old is she?
| Assumed current age | x |
|---|---|
| Age if she were 4 years older | x + 4 |
| Years she currently lacks to reach 20 | 20 − x |
| Equation | x + 4 = 20 + (20 − x) |
x + 4 = 20 + 20 − x
x + x = 40 − 4
2x = 36
x = 18Claudia is 18. If she were four years older, she would be 22, which is 2 years over 20 - the same number of years she is currently under 20.
5.2. Guided problem 2: Alberto’s answers
Alberto’s grandmother asks him twenty questions. She gives him 3 euros for every correct answer and takes away 2 euros for every wrong answer. In the end Alberto receives 10 euros. How many questions did he answer correctly?
| Assumed number of correct answers | x |
|---|---|
| Wrong answers | 20 − x |
| Earnings from correct answers | 3x |
| Loss from wrong answers | 2(20 − x) |
| Balance | 3x − 2(20 − x) = 10 |
3x − 40 + 2x = 10
5x = 50
x = 10Alberto gave 10 correct and 10 wrong answers: he earned 30 euros and lost 20, leaving a balance of 10 euros.
5.3. Guided problem 3: coins in a vending machine
A hot-drinks vending machine accepts 0.50, 1, and 2 euro coins. Its coin box contains 100 euros altogether. There are three times as many 1 euro coins as 0.50 euro coins, while the number of 2 euro coins is 45 less than the number of 1 euro coins. How many coins of each denomination are there?
| Assumed number of 0.50 euro coins | x |
|---|---|
| Number of 1 euro coins | 3x |
| Number of 2 euro coins | 3x − 45 |
| Total value | (1/2)x + 3x + 2(3x − 45) = 100 |
x + 6x + 12x − 180 = 200
19x = 380
x = 20There are 20 half-euro coins, 60 one-euro coins, and 15 two-euro coins. The check is 10 + 60 + 30 = 100 euros.
5.4. Guided problem 4: the square of students
This problem is slightly harder: we must ask what remains unchanged in the two situations? Moreover, the unknown is not directly the requested answer.
When the students in a class form a square, 8 are left over. If the side of the square is increased by one, there is instead a shortage of 1 student. How many students are in the class?
| Assumed side of the first square | x |
|---|---|
| Number of students in the first arrangement | x² + 8 |
| Side of the second square | x + 1 |
| Number of students in the second arrangement | (x + 1)² − 1 |
| Equation | x² + 8 = (x + 1)² − 1 |
x² + 8 = x² + 2x + 1 − 1
8 = 2x
x = 4The number of students is 4² + 8 = 24, or equivalently 5² − 1 = 24.
6. Formalising the chickens-and-rabbits problem
Problems can often be solved without equations, but the necessary reasoning is not always obvious. Something interesting happens if we first solve a problem in general form, using parameters - known data represented by symbols rather than numbers: the resulting formula reveals the arithmetic reasoning that may initially have escaped us.
A farmer raises chickens and rabbits. He has 55 animals with 160 legs altogether. How many are rabbits and how many are chickens?
| Quantity | Rabbits | Chickens |
|---|---|---|
| Number of animals | x | 55 − x |
| Number of legs | 4x | 2(55 − x) |
The statement “there are 160 legs altogether” produces the equation:
4x + 2(55 − x) = 160
4x + 110 − 2x = 160
2x = 50
x = 25 rabbits
55 − x = 30 chickensGeneralisation
Now suppose that the farmer has T animals with Z legs altogether. If x is the number of rabbits, the number of chickens is T − x. The rabbits have 4x legs and the chickens have 2(T − x) legs.
4x + 2(T − x) = Z
4x + 2T − 2x = Z
2x = Z − 2T
x = (Z − 2T) / 2 rabbits
T − x = (4T − Z) / 2 chickensFor the problem to yield non-negative whole-number counts, T and Z must be integers, Z must be even, and 2T ≤ Z ≤ 4T. Under these conditions, both formulas return integers between 0 and T.
The two formulas have a concrete meaning. 2T is the number of legs there would be if all the animals were chickens. The difference Z − 2T leaves two extra legs for every rabbit; dividing by 2 therefore gives the number of rabbits.
Similarly, 4T is the number of legs there would be if all the animals were rabbits. The difference 4T − Z leaves two legs for every chicken; dividing by 2 gives the number of chickens. Algebra has made an arithmetic procedure explicit.
7. It is not always easy
The more problems we examine, the clearer it becomes that a mechanical translation from natural language is not enough. We must process the information, reflect on it, and derive further information. There is no single fixed scheme beyond the general process of working with information.
Two questions, however, are often decisive:
- where is the equality?
- what has not changed in the situations described?
Three situations to explore
There are 158 students, with more girls than boys. One eleventh of the girls wears glasses, while one seventh of the boys wears glasses. How many boys and girls are there?
Brothers and sisters sit at a table. Aldo says, “The number of my brothers equals the number of my sisters.” Claudia says, “The number of my brothers is twice the number of my sisters.” How many brothers and sisters are there?
Marco is 6 years older than his wife Mara. Marco says, “Four years ago, I had been married for a number of years equal to half my age at that time.” Mara says, “Ten years from now, I will have spent two thirds of your lifetime as a married woman.” How long have they been married, and how old is each of them?
The original document develops the first example in detail. The other two show how demanding it can be to identify the equality hidden in family and time relationships.
7.1. Boys and girls: using divisibility
In the first example, an equation is not immediately apparent, but the text implies two constraints:
- if one eleventh of the girls wears glasses, the number of girls must be a multiple of 11;
- if one seventh of the boys wears glasses, the number of boys must be a multiple of 7.
Because the girls are more numerous, their number F is greater than half of 158: F > 79. The first multiple of 11 to test is therefore 88. For each multiple, compute M = 158 − F and check whether M/7 is an integer.
| Multiplier k | F = 11k | M = 158 − F | M/7 |
|---|---|---|---|
| 8 | 88 | 70 | 10 |
| 9 | 99 | 59 | 8.428571 |
| 10 | 110 | 48 | 6.857143 |
| 11 | 121 | 37 | 5.285714 |
| 12 | 132 | 26 | 3.714286 |
| 13 | 143 | 15 | 2.142857 |
| 14 | 154 | 4 | 0.571429 |
| 15 | 165 | −7 | −1 |
The only admissible solution is 88 girls and 70 boys. This method, which is easy to reproduce in a spreadsheet, turns the information about fractions into divisibility conditions.
7.2. What does not change?
A ship carries 20 men and enough water for 33 days. Two shipwreck survivors come aboard, and the captain reduces the daily water ration to
5/6of the original ration. How many days will the water now last?
Let us suppose that we know the new duration and call it x. The decisive question is: what remains unchanged in the two situations? The total amount of water does not change.
| First situation | 20 · 33 · 1 rations |
|---|---|
| Second situation | 22 · x · 5/6 rations |
20 · 33 · 1 = 22 · x · 5/6
x = (6/5) · (20 · 33 / 22)
x = 36With 22 people and the ration reduced to five sixths, the water lasts 36 days.
7.3. Working backwards
Three friends, A, B, and C, have 48 euros altogether. First A gives B enough money to double B’s amount; then B gives C enough money to double C’s amount; finally C gives A enough money to double A’s amount. At the end, all three have the same amount. How much did they have at the beginning?
Work backwards, one step at a time. At the end each person has 48/3 = 16 euros:
final situation: A = 16, B = 16, C = 16In the last transaction, A doubled his money: immediately beforehand he had 8 euros, and C gave him another 8. Before giving the money away, C therefore had 16 + 8 = 24.
before the last move: A = 8, B = 16, C = 24In the preceding transaction, C doubled his money: beforehand he had 12 euros, and B gave him 12. Before giving it away, B therefore had 16 + 12 = 28.
before the second move: A = 8, B = 28, C = 12In the first transaction, B doubled his money: beforehand he had 14 euros, and A gave him 14. Before giving it away, A therefore had 8 + 14 = 22.
initial situation: A = 22, B = 14, C = 12The required initial situation is therefore A = 22 euros, B = 14 euros, C = 12 euros. A system with three unknowns would be possible but much more laborious than working backwards.
Curated by Salvatore Mosaico.