Let's break down the problem step by step and solve each part in detail.
### Part (a): Relationship between [tex]\( t, p \)[/tex], and [tex]\( v \)[/tex]
Given: The time [tex]\( t \)[/tex] (in minutes) taken to buy fuel varies directly as the number of vehicles [tex]\( v \)[/tex] in a queue and inversely as the number of pumps [tex]\( p \)[/tex]. This relationship can be expressed as:
[tex]\[ t = k \frac{v}{p} \][/tex]
Where [tex]\( k \)[/tex] is a proportionality constant. We will determine [tex]\( k \)[/tex] using the given information:
- [tex]\( t = 10 \)[/tex] minutes
- [tex]\( v = 20 \)[/tex] vehicles
- [tex]\( p = 5 \)[/tex] pumps
Substituting these values into the relationship, we get:
[tex]\[ 10 = k \frac{20}{5} \][/tex]
[tex]\[ 10 = k \cdot 4 \][/tex]
[tex]\[ k = \frac{10}{4} \][/tex]
[tex]\[ k = 2.5 \][/tex]
So, the relationship between [tex]\( t \)[/tex], [tex]\( p \)[/tex], and [tex]\( v \)[/tex] is:
[tex]\[ t = 2.5 \frac{v}{p} \][/tex]
### Part (b): Time it takes to fuel 50 vehicles at a station with 2 pumps
We need to find the time [tex]\( t \)[/tex] when:
- [tex]\( v = 50 \)[/tex] vehicles
- [tex]\( p = 2 \)[/tex] pumps
Using the relationship:
[tex]\[ t = 2.5 \frac{v}{p} \][/tex]
[tex]\[ t = 2.5 \frac{50}{2} \][/tex]
[tex]\[ t = 2.5 \cdot 25 \][/tex]
[tex]\[ t = 62.5 \][/tex]
So, it takes 62.5 minutes to fuel 50 vehicles at a station with 2 pumps.
### Part (c): Number of pumps to fuel 50 vehicles in 20 minutes
We need to find the number of pumps [tex]\( p \)[/tex] when:
- [tex]\( v = 50 \)[/tex] vehicles
- [tex]\( t = 20 \)[/tex] minutes
Using the relationship:
[tex]\[ t = 2.5 \frac{v}{p} \][/tex]
[tex]\[ 20 = 2.5 \frac{50}{p} \][/tex]
Rearranging to solve for [tex]\( p \)[/tex]:
[tex]\[ 20 = 125 \frac{1}{p} \][/tex]
[tex]\[ 20p = 125 \][/tex]
[tex]\[ p = \frac{125}{20} \][/tex]
[tex]\[ p = 6.25 \][/tex]
So, 6.25 pumps are required to fuel 50 vehicles in 20 minutes.
