Question

Катер прошел 10км по течению и 20 км по озеру,затратив на весь путь 1 ч. Скорость течения реки равна 7км/ч. Найдите скорость катера по течению.

Answer 1

Ответ:  35 км/ч .

Собственная скорость катера равна  х км/ч .

Время, затраченное на движение по озеру, равно  20/х ч .

Скорость катера по течению равна  (х+7) км/ч .

Время, затраченное на движение по течению равно  10/(х+7) ч .

На весь путь ушёл 1 час, тогда

$\dfrac{10}{x+7}+\dfrac{20}{x}=1\ \ \ ,\ \ \ \dfrac{10x+20(x+7)}{x(x+7)}=1\ \ \ ,\ \ \ 10x+20(x+7)=x(x+7)\ \ ,\\\\\\30x+140=x^2+7x\ \ ,\ \ \ x^2-23x-140=0\ \ ,\ \ \ D=1089=33^2\ \ ,\\\\x_1=-5$

Отрицательное значение скорость катера не может принимать , поэтому  скорость катера равна  28 км/ч , а скорость катера по течению равна  28+7=35 км/ч .

Answer 2

Problem Analysis

We are given that a boat traveled 10 km downstream and 20 km on a lake, taking a total of 1 hour for the entire journey. The speed of the river current is given as 7 km/h. We need to find the speed of the boat in still water.

Solution

Let's assume the speed of the boat in still water is x km/h. Since the boat is traveling downstream, it will benefit from the speed of the river current. Therefore, the effective speed of the boat downstream will be (x + 7) km/h.

We can calculate the time taken to travel downstream using the formula: time = distance / speed. The distance traveled downstream is 10 km, and the effective speed is (x + 7) km/h. Therefore, the time taken to travel downstream is 10 / (x + 7) hours.

Similarly, when the boat is traveling on the lake, it will not be affected by the river current. Therefore, the effective speed of the boat on the lake will be x km/h. The distance traveled on the lake is 20 km. Using the formula for time, the time taken to travel on the lake is 20 / x hours.

The total time taken for the entire journey is given as 1 hour. Therefore, the sum of the times taken for downstream and on the lake should be equal to 1 hour:

10 / (x + 7) + 20 / x = 1

To solve this equation, we can multiply through by the least common multiple (LCM) of the denominators to eliminate the fractions. The LCM of (x + 7) and x is x(x + 7):

10x + 70 + 20(x + 7) = x(x + 7)

Simplifying the equation:

10x + 70 + 20x + 140 = x^2 + 7x

Combining like terms:

30x + 210 = x^2 + 7x

Rearranging the equation:

x^2 - 23x - 210 = 0

Now we can solve this quadratic equation for x.