Question 2 The time (t) taken to buy a fuel at a petrol station variation varies directly as the number of vechice \( v \) on a quene and inwersely as the number of pumps (p) available in a station. In a station of 5 pumps, it took 10 minutes to fuel 20 vechiles Find; (a) The relationship betweet \( t \) \( p \) and \( U \). (b) The time it will taple to fuel 50 vechiles in Station usithe 2 pumps

Let's solve the problem step by step. ### Given: - **Direct Variation:** Time \( t \) varies directly with the number of vehicles \( v \). - **Inverse Variation:** Time \( t \) varies inversely with the number of pumps \( p \). ### (a) Establishing the Relationship Between \( t \), \( p \), and \( v \) When a variable varies **directly** with one variable and **inversely** with another, the relationship can be expressed as: \[ t = k \times \frac{v}{p} \] Where: - \( t \) = Time taken to fuel the vehicles - \( v \) = Number of vehicles in the queue - \( p \) = Number of pumps available - \( k \) = Constant of proportionality **So, the relationship is:** \[ t = \frac{k \cdot v}{p} \] ### (b) Calculating the Time to Fuel 50 Vehicles with 2 Pumps **Step 1: Find the Constant \( k \) Using Given Data** Given: - \( p = 5 \) pumps - \( v = 20 \) vehicles - \( t = 10 \) minutes Plug these values into the relationship: \[ 10 = \frac{k \cdot 20}{5} \] \[ 10 = 4k \] \[ k = \frac{10}{4} = 2.5 \] **Step 2: Use \( k \) to Find the Time for \( v = 50 \) Vehicles and \( p = 2 \) Pumps** Plug \( k = 2.5 \), \( v = 50 \), and \( p = 2 \) into the relationship: \[ t = \frac{2.5 \cdot 50}{2} \] \[ t = \frac{125}{2} = 62.5 \text{ minutes} \] **Therefore, it will take **62.5 minutes** to fuel 50 vehicles using 2 pumps.** ### Summary - **(a)** The relationship is \( t = \frac{k \cdot v}{p} \). - **(b)** It will take **62.5 minutes** to fuel 50 vehicles with 2 pumps.