The speed of a boat in still water is x km/hr and the speed of the stream is 3 km/hr.

(i) Write the speed of the boat upstream, in terms of x.

(ii) Write the speed of the boat downstream, in terms of x.

(iii) If the boat goes 15 km upstream and 22 km downstream in 5 hours, write an equation in x to represent the statement.

(iv) Solve the equation to evaluate x.

(i) Given,

Speed of a boat in still water = x km/hr

Speed of stream = 3 km/hr

Upstream speed = (x − 3) km/hr.

Hence, the speed of the boat upstream = (x - 3) km/hr.

(ii) Given,

Speed of a boat in still water = x km/hr

Speed of stream = 3 km/hr

Downstream speed = (x + 3) km/hr.

Hence, the speed of the boat downstream = (x + 3) km/hr.

(iii) By formula,

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Given,

The boat goes 15 km upstream and 22 km downstream in 5 hours.

$\therefore\dfrac{15}{x - 3} + \dfrac{22}{x + 3} = 5$

Hence, the required equation : $\dfrac{15}{x - 3} + \dfrac{22}{x + 3} = 5$.

(iv) Solving,

$\Rightarrow \dfrac{15}{x - 3} + \dfrac{22}{x + 3} = 5 \\[1em] \Rightarrow \dfrac{15(x + 3) + 22(x - 3)}{(x + 3)(x - 3)} = 5 \\[1em] \Rightarrow \dfrac{15x + 45 + 22x - 66}{x^2 - 3^2} = 5 \\[1em] \Rightarrow \dfrac{37x - 21}{x^2 - 9} = 5 \\[1em] \Rightarrow 37x - 21 = 5(x^2 - 9) \\[1em] \Rightarrow 37x - 21 = 5x^2 - 45 \\[1em] \Rightarrow 0 = 5x^2 - 45 - 37x + 21 \\[1em] \Rightarrow 5x^2 - 37x - 24 = 0 \\[1em] \Rightarrow 5x^2 - 40x + 3x - 24 = 0 \\[1em] \Rightarrow 5x(x - 8) + 3(x - 8) = 0 \\[1em] \Rightarrow (5x + 3)(x - 8) = 0 \\[1em] \Rightarrow (5x + 3) = 0 \text{ or }(x - 8) = 0 \text{….[Using zero-product rule]} \\[1em] \Rightarrow x = -\dfrac{3}{5} \text{ or } x = 8.$

Since, speed must be positive,

∴ x = 8.

Hence, x = 8 km/hr.