Kid's homework: 4 equations 5 unknowns? Going crazy!
I'm new here, and I'm hoping someone can help out. My 10 year old son has been set a maths problem, which I can't solve. I've got a PhD in neuroscience and do a fair amount of matlab stuff (data analysis, image processing) on a daily basis, but I can't work this out.
The problem is expressed in words, but I've read it through a dozen times and I'm sure it boils down to the following:
a + b = 55
b + c = 43
c + d = 42
d + e = 37
They are asked to find the value of e. But this is 4 equations with 5 unknowns. Is there really a unique solution for this system of equations? Where am I going wrong?
If you set one of the variable to 0 you can solve for the rest, of course, but I'm pretty sure this is not what they are meant to do. The hint says it's easiest to start by working out the value of c.
I'm lost, any help would be most appreciated!
The exact question is:
The following people take part in a school trip: 55 boys and girls; 43 girls and fathers; 42 fathers and mothers and 37 mothers and teachers. How many teachers took part in the school trips?
Assuming the classes are mutually exclusive (i.e. no teachers are also parents), I'm pretty sure that is the set of equations I posted. The other problems in the same homework are similar in form but all have 1 additional piece of information: the total number (e.g., a + b + c + d + e = 100). Those ones are solvable no problem.
Solution 1:
Edit. By back substitution, one can easily express $a,b,c,d$ in terms of $e$: \begin{cases} d=37-e,\\ c=e+5,\\ b=38-e,\\ a=e+17. \end{cases} Therefore $a+b+c+d+e=97+e$. It is very likely that they have simply forgotten the constraint that there are $100$ participants.
Solution 2:
We add these equalities in this manner $$a+b+43+c+d+37=55+b+c+42+d+e$$ now we cancel we find $$\require{cancel}a+\cancel{b}+43+\cancel{c}+\cancel{d}+37=55+\cancel{b}+\cancel{c}+42+\cancel{d}+e$$ hence $$a+80=e+97\iff e=a-17$$ so each time you take a value of $a$ we find a value of $e$. Can you now answer your son?
Solution 3:
The following people take part in a school trip: 55 boys and girls; 43 girls and fathers; 42 fathers and mothers and 37 mothers and teachers. How many teachers took part in the school trips? Assuming the classes are mutually exclusive (i.e. no teachers are also parents)
If you assume there were zero teachers, you get 37 mothers, 5 fathers, 38 girls and 17 boys.
If, on the other hand, you assume there were 37 teachers, you get 0 mothers, 42 fathers, 1 girl and 54 boys.
Everything between 0 and 37 teachers should admit a solution too, so the solution is rather non-unique, even if everything is constrained to be a positive integer.
Solution 4:
We have:
$$b=55-a$$ $$c=43-b=43-(55-a)=a-12$$ $$d=42-c=42-(a-12)=54-a$$ $$e=37-d=37-(54-a)=a-17$$ For all these numbers to be non-negative we therefore need $$17\le a\le 54$$with the inequalities strict if we require all the numbers to be positive.
It was just possible that the constraints that all the numbers are non-negative (or positive) integers would have fixed a value for $a$, but they don't, and any $a$ which satisfies the constraints gives a consistent solution.
Solution 5:
Not really an answer, but I couldn't post an image as a comment. Looks like you already have some nice answers, but a visual always helps me :D
Edit: Again, this wasn't intended to solve the question, just offer another point of view to play with. Here is a rough Legend for my comically grotesque visual :)
Teachers Mothers Fathers
boys girls
Each number is the quantity of persons in the group(s) it overlaps:
37 Teachers and Mothers
42 Mothers and Fathers
54 Mothers and boys
97 Mothers, Fathers, boys, and girls (all members except Teachers.)
43 Fathers and girls
55 boys and girls
100? = OP suggests this may be the 'total' accidentally omitted from the problem.
3? = rough guess at an answer (based on 100 total.)