How do I solve this Math problem

Question

Hi guys. Does this problem involve the use of proportions?

Answer 1

If the tank has a capacity of $50$ liters and $2.5$ liters is added each second, it takes $$\frac{50}{2.5}=20$$ seconds for it to be filled.

Answer 2

You solve it by thinking, and if that fails then by trial and error. If you don't see the answer immediately, then try figuring it out "the hard way". How much water has been filled in 1 second? In 2 seconds? In 10 seconds? Now how long do you think it would take to fill 50 liters?

That will give you the answer much faster than posting on an online forum, waiting for an answer. And you will hopefully learn enough to cope with the next exercise without having to ask anyone for help. After doing a few exercises like this, maybe you can even spot how to do them quickly, without trial and error. At the very least you will be much better equipped to understand and remember the quick solution once it's presented to you.

Thinking about whether it "involves the use of proportions" will in my opinion only help you when the answer to that question is obvious to you. Before that time, it's just in the way of understanding what's actually going on, what proportionality means, and why it's a useful concept. Also it won't save you any time, or work.

Answer 3

A tank has a capacity of $50$ litres

This is called a volume in mathematics. We write $$ V_\text{tank}=50 \,\text{l} $$ in physics, attaching the unit to the number.

How many seconds will it be filled if the flow rate into it is $2.5$ liters per second.

A flow rate, or more general: a rate, describes the change over time of a quantity.

The changing quantity here is the volume of the body of in flowing water $$ V_\text{water}(t) $$

E.g. it could be $$ \begin{align} V_\text{water}(1\,\text{s}) &= 7\,\text{l} \\ V_\text{water}(2\,\text{s}) &= 10\,\text{l} \\ V_\text{water}(5\,\text{s}) &= 2\,\text{l} \end{align} $$ which is obviously not our problem, as that tank seems to have a different growth (what would we expect for $V_\text{water}(1\,\text{s})$?) and it even seems to leak water after some time.

How many seconds will it be filled if the flow rate into it is $2.5$ liters per second.

That bit of text gives you this information $$ \frac{dV_\text{water}}{dt}(t) = 2.5 \,\frac{\text{l}}{s} \quad\quad (*) $$ This is a very simple differential equation. It has many solutions. To pick the proper solution one needs an extra bit of information. Here it is the initial condition $$ V_\text{water}(0\,\text{s}) = 0 \,\text{l} \quad\quad (**) $$ It is not given explicitly in your task, but it is reasonably to assume that the tank is empty at the start time $t=0\,\text{s}$.

The idea is now to solve equation $(*)$ and then use the initial condition $(**)$ to find the solution function $V_{water}(t)$.

Forgetting about linear differential equations for a second, we read $(*)$ like a physicist $$ dV_{water} = 2.5 \frac{l}{s} \, dt $$ this means the change in volume $dV_{water}$ of the accumulating water body is $2.5 \,\text{l}/\text{s}$ times the change in time $dt$. As the rate is constant regarding time we can infer $$ V_\text{water}(t) = 2.5 \frac{l}{s} \, t $$ and this also fulfills the initial condition $(**)$.

Then one poses the equation $$ V_{water}(t) = V_{tank} $$ and solves for $t$, to answer the question when the tank is full.

I leave this to you.