Question

A motor boat, whose speed is 18 km/hour in still water, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Hint:

To solve problems involving speed, time, and distance, use the relationship \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \) and apply the given conditions to set up equations.

Answer 1

Let the speed of the stream be \( x \) km/hour. The speed of the boat upstream will be \( 18 - x \) km/hour, and the speed of the boat downstream will be \( 18 + x \) km/hour. The time taken to go upstream is: \[ \text{Time upstream} = \frac{24}{18 - x}. \] The time taken to go downstream is: \[ \text{Time downstream} = \frac{24}{18 + x}. \] According to the problem, the time taken to go upstream is 1 hour more than the time taken to go downstream: \[ \frac{24}{18 - x} = \frac{24}{18 + x} + 1. \] Subtract \( \frac{24}{18 + x} \) from both sides: \[ \frac{24}{18 - x} - \frac{24}{18 + x} = 1. \] Simplify the left-hand side: \[ \frac{24(18 + x) - 24(18 - x)}{(18 - x)(18 + x)} = 1, \] \[ \frac{24(18 + x - 18 + x)}{(18 - x)(18 + x)} = 1, \] \[ \frac{24(2x)}{(18 - x)(18 + x)} = 1. \] Simplify further: \[ \frac{48x}{324 - x^2} = 1. \] Multiply both sides by \( 324 - x^2 \): \[ 48x = 324 - x^2. \] Rearrange the equation: \[ x^2 + 48x - 324 = 0. \] Solve this quadratic equation using the quadratic formula: \[ x = \frac{-48 \pm \sqrt{48^2 - 4(1)(-324)}}{2(1)} = \frac{-48 \pm \sqrt{2304 + 1296}}{2} = \frac{-48 \pm \sqrt{3600}}{2} = \frac{-48 \pm 60}{2}. \] So: \[ x = \frac{-48 + 60}{2} = \frac{12}{2} = 6 \quad \text{or} \quad x = \frac{-48 - 60}{2} = \frac{-108}{2} = -54. \] Since the speed of the stream cannot be negative, the speed of the stream is \( x = 6 \) km/hour.
Conclusion: The speed of the stream is \( 6 \) km/hour.