To solve this problem, we start with the given mathematical relationship:
1. The time [tex]\( t \)[/tex] taken to buy fuel at a filling station varies directly as the number of vehicles [tex]\( \omega \)[/tex] in a queue. This implies:
[tex]\[
t \propto \omega
\][/tex]
2. The time [tex]\( t \)[/tex] varies inversely as the number of pumps [tex]\( p \)[/tex] available at the station. This implies:
[tex]\[
t \propto \frac{1}{p}
\][/tex]
Combining these two relationships, we get:
[tex]\[
t \propto \frac{\omega}{p}
\][/tex]
We can express this as:
[tex]\[
t = k \cdot \frac{\omega}{p}
\][/tex]
where [tex]\( k \)[/tex] is the constant of proportionality.
Next, we need to find the value of [tex]\( k \)[/tex] using the given information:
- Number of pumps, [tex]\( p = 5 \)[/tex]
- Number of vehicles, [tex]\( \omega = 20 \)[/tex]
- Time taken, [tex]\( t = 10 \)[/tex] minutes (noting that "comm" appears to be a typo for "10 minutes")
Using these values, we substitute into the equation:
[tex]\[
10 = k \cdot \frac{20}{5}
\][/tex]
Simplifying the right-hand side:
[tex]\[
10 = k \cdot 4
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
k = \frac{10}{4}
\][/tex]
[tex]\[
k = 2.5
\][/tex]
Therefore, the relationship between the time [tex]\( t \)[/tex], the number of vehicles [tex]\( \omega \)[/tex], and the number of pumps [tex]\( p \)[/tex] is:
[tex]\[
t = 2.5 \cdot \frac{\omega}{p}
\][/tex]
So, the proportional relationship is:
[tex]\[
t = 2.5 \cdot \frac{\omega}{p}
\][/tex]
This equation succinctly describes the time required to fuel vehicles at the station, considering it varies directly with the number of vehicles and inversely with the number of pumps available.
