Question

a boat travels with the current at 15km/h and travels against the current at 7km/h. what is the speed of the boat in still water, and what is the rate of the current?

Explanation:

Step1: Define variables

Let \( b \) be the speed of the boat in still water (in km/h) and \( c \) be the speed of the current (in km/h). When the boat travels with the current, its effective speed is \( b + c \), and when it travels against the current, its effective speed is \( b - c \). We know that \( b + c = 15 \) (with the current) and \( b - c = 7 \) (against the current).

Step2: Solve for \( b \) (speed of boat in still water)

We can add the two equations together to eliminate \( c \). So, \( (b + c) + (b - c) = 15 + 7 \). Simplifying the left side: \( b + c + b - c = 2b \). The right side is \( 22 \). So, \( 2b = 22 \), which means \( b=\frac{22}{2}=11 \) km/h.

Step3: Solve for \( c \) (speed of the current)

Now that we know \( b = 11 \), we can substitute back into one of the original equations. Let's use \( b + c = 15 \). So, \( 11 + c = 15 \). Subtract \( 11 \) from both sides: \( c = 15 - 11 = 4 \) km/h.

Answer:

The speed of the boat in still water is \( 11 \) km/h, and the rate of the current is \( 4 \) km/h.