Ratios without algebra? Surely not trial and error? A 10 year old wants to learn about proportions but has no algebra background:
Their questions is:

30 Kids go on a school trip 
Cost of the trip (t) + a lunch (l) =5pounds 
Kids who go on the trip but who don't want lunch pay =3pounds 
If the total paid comes to 110 pounds how many don't want a lunch?

METHOD 1: Is trial and error the only solution without algebra?
METHOD2: Algebra
Let $y$ be the number of kids not wanting lunch then we have
\begin{eqnarray}
t+l&=&5\\
30t+yl&=&110\\
\end{eqnarray}
but we know if t=3 then l=2 from the first two bits of info then:
\begin{eqnarray}
\Longrightarrow\quad 90+2y&=&110\\
\text{so}\quad y&=&10 kids
\end{eqnarray}
They don't know anything about algebra or multiplication with the (x) sign. What should I do? How can you solve this problem easily?
 A: "Trial and error" can be done intelligently. e.g. if all 30 want lunch, it's 150 pounds. If none do, it's 90 pounds. If 15 do, it's 120 pounds. And so forth.
(Or you might have picked 10 to try as it's an easier number and you can visually see the answer is closer to 0 than 30, at which point you get lucky and hit the answer dead on)

The algebra can be done in words. Starting from none buying lunch works out to 90 pounds and each additional student who buys lunch increases it by two points, that means you need 10 students to make the 20 pound difference.
In fact, even knowing algebra, this is probably how I would have done it anyways if I wanted to do a quick calculation in my head.

I'm not sure why you mention proportions specifically; have you looked at their text to see the sorts of methods they are solving with ratios? I can imagine a solution here where they can recognize they have to work out the "proportion" between

90 - 110 - 150

is the same as the one between

0 - # of students - 30

i.e. that they are looking for the ratio 20 : 60, i.e. 1:3. So one third of the 30 students want a lunch.
