Теплоход прошёл 28 км против течения реки и вернулся обратно, потратив на обратный путь на 4 мин меньше. Найти скорость теплохода в стоячей воде, если скорость течения реки составляет 1 км/ч.

Let's denote the speed of the boat in still water as v km/h. Since the velocity of the current is 1 km/h, the speed of the boat against the current is (v - 1) km/h, and the speed of the boat with the current is (v + 1) km/h.

On the forward journey, the boat traveled 28 km against the current, so the time taken is given by:

time = distance / speed = 28 / (v - 1)

On the return journey, the boat traveled the same distance of 28 km with the current, but it took 4 minutes less time. We need to convert these minutes into hours by dividing by 60:

4 minutes = 4 / 60 = 1 / 15 hours

So, the time taken on the return journey is:

time - 1/15 = 28 / (v + 1)

Now, we can set up an equation using the above two equations:

28 / (v - 1) = 28 / (v + 1) - 1/15

To solve this equation, we can multiply both sides by (v - 1)(v + 1) * 15 to eliminate the denominators:

28 * 15 = 28(v - 1) - (v - 1)(v + 1)

420 = 28v - 28 - v^2 + 1

Rearranging the equation and simplifying:

v^2 - 28v + 349 = 0

Using the quadratic formula, we can solve for v:

v = (-(-28) ± sqrt((-28)^2 - 41349)) / (2*1) = (28 ± sqrt(784 - 1396)) / 2 = (28 ± sqrt(-612)) / 2

Since the square root of a negative number is not a real number, there are no real solutions for v. Therefore, there is no speed for the boat in still water that satisfies the given conditions.