A motorboat is capable of traveling at a speed of 15 miles per hour in still water. On a particular day, it took 30 minutes longer to travel a distance of 10 miles upstream than it took to travel the same distance downstream. What was the rate of current in the stream on that day?
Let's say the rate of the current is x miles per hour.
When the motorboat travels upstream, it has to overcome the current, so its effective speed is decreased by x mph.
Therefore, the time it takes to travel 10 miles upstream is 10 / (15 - x) hours.
When the motorboat travels downstream, it benefits from the current, so its effective speed is increased by x mph.
Therefore, the time it takes to travel 10 miles downstream is 10 / (15 + x) hours.
The problem states that it took 30 minutes longer to travel upstream than downstream, which can be expressed mathematically as:
10 / (15 - x) = 10 / (15 + x) + 0.5
Multiplying both sides of the equation by (15 - x)(15 + x), we get:
10(15 + x) = 10(15 - x) + 0.5(15 - x)(15 + x)
150 + 10x = 150 - 10x + 112.5 - 0.5x^2
0.5x^2 + 20x - 112.5 = 0
Dividing the equation by 0.5, we get:
x^2 + 40x - 225 = 0
Using the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = 1, b = 40, and c = -225, we get:
x = (-40 ± sqrt(40^2 - 4(1)(-225))) / (2(1))
x = (-40 ± sqrt(1600 + 900)) / 2
x = (-40 ± sqrt(2500)) / 2
x = (-40 ± 50) / 2
x = (50 - 40) / 2 or x = (-50 - 40) / 2
x = 10 / 2 or x = -90 / 2
x = <<10/2=5>>5 or x = -45
Since the rate of the current cannot be negative, the rate of the current in the stream was 5 miles per hour. Answer: \boxed{5}.