The volume (v) of water in a tank is propational to \( k \). where \( c \) is the rate at which water is kept into the tank and constant is \( k \). a) Find the value of \( k \) if \( v=60 \) and \( c \) is 36 . i) Use the value of \( k \) in i. to find the value of \( c \) when \( v=90 \)

Certainly! Let's break down the problem step by step. ### Understanding the Problem 1. **Proportional Relationship**: The volume of water \( v \) in a tank is proportional to the rate \( c \) at which water is added. Mathematically, this can be expressed as: \[ v = k \cdot c \] where: - \( v \) = volume of water in the tank - \( c \) = rate at which water is added - \( k \) = constant of proportionality ### Part a) Find the Value of \( k \) When \( v = 60 \) and \( c = 36 \) Given: \[ v = 60 \quad \text{and} \quad c = 36 \] Using the proportional relationship: \[ 60 = k \cdot 36 \] To solve for \( k \): \[ k = \frac{60}{36} = \frac{5}{3} \approx 1.6667 \] **So, the constant of proportionality \( k \) is \( \frac{5}{3} \).** ### Part i) Use the Value of \( k \) to Find \( c \) When \( v = 90 \) Given: \[ v = 90 \] \[ k = \frac{5}{3} \] Using the proportional relationship: \[ 90 = \frac{5}{3} \cdot c \] To solve for \( c \): \[ c = 90 \times \frac{3}{5} = 54 \] **Therefore, when the volume \( v \) is 90, the rate \( c \) is 54.** ### Summary - **a)** The constant \( k = \frac{5}{3} \). - **i)** When \( v = 90 \), the rate \( c = 54 \).