How count numbers of containers in filing containers of three different sizes with known exchange rates using algebra rather than via my shortcut?

Question

A manufacturer packages soap powder in containers of three different sizes. The amount of soap powder in a full large container could fill exactly 3 of the medium containers or exactly 5 of the small containers. If an equal number of small and large containers are to be filled with the amount of soap powder that would fill 90 medium containers, how many small containers will be filled?

$90$ medium containers is equivalent to $30$ large containers.

If you divided $30$ into $25$ and five, the five large containers could fill $25$ small containers, so the answer is $25$ large containers, and $25$ small containers.

There's a longer algebraic way to solve this problem, and I'm seeing too much of a shortcut to do it. How to do it the algebraic way?

I don't know what to set as the "unknown" in the problem. Do you set the number of containers of small($z$),medium($y$), and large($x$) to the three unknowns or do you express everything in one variable such as the number of large containers?

I also don't know how to express there must be an equal number of small and large containers in algebra.

$x=y?$

$x=5 \cdot y$?

$x+1/5x =?$

Answer 1

Since the small and medium size containers are characterized in terms of the large containers, let $x$ be the size of the large container. Since three medium containers are equivalent to a large container, the size of a medium container is $x/3$. Since five small containers are the size of a large container, the size of a small container is $x/5$.

We are told that an equal number of small and large containers are to be filled with the amount of soap powder that would fill $90$ medium containers. Let $n$ be the number of small containers to be filled and the number of large containers to be filled. Then the condition means $$90 \cdot \frac{x}{3} = n\left(x + \frac{x}{5}\right)$$ Solving that equation for $n$ yields \begin{align*} 30x & = n \cdot \frac{6}{5}x\\ 150x & = 6nx\\ 25x & = nx\\ (25 - n)x & = 0 \end{align*} Since a product is equal to zero if and only if one of its factors is equal to zero, \begin{align*} 25 - n & = 0 & \text{or} & & x & = 0\\ 25 & = n \end{align*} If $x = 0$, then all the containers are the same size, contrary to assumption. Hence, $n = 25$, as you found.

Answer 2

I think you could let $x,y,z$ be the 'size' of the large, medium, and small containers respectively. Let $n$ be the same numbers of containers that small and large shares. According to the question, we could write down \begin{align*} &x=3y\\ &x=5z\\ &90y=n(x+z), \end{align*} and the rest is to solve $n,$ which is obviously 25 by expressing $y$ and $z$ into $x$ in the final equation.