Answer :
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.
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.
