Question

Поливочная машина двигается с постоянной скоростью.и каждую минуту из нее вытекает одно и то же кол-во воды.Если увеличить скорость движения в 2 паза,а скорость вытекания воды увеличить в 3 раза.то содержащейся в машине воды хватит на то,чтобы полить 4 км дороги.Сколько километров дороги удастся полить,если начальную скорость движения увеличить в 3 раза,а начальную скорость вытекания воды увеличить в 2 раза?

Answer 1

Пусть х км/ч первоначальная скорость машины, у л - скорость вытекания воды, А л - воды вмещается в машину.Тогда А/у ч - время расхода воды, А*х/у км - длина дороги, которую можно полить.
Тогда при увеличении скорости движения в 2 раза, а скорости вытекания воды в 3 раза получим, А/(3у) ч - время расхода воды, (А*2х)/(3у) =4 км - длина дороги.
Если начальную скорость движения увеличить в 3 раза, а скорость вытекания воды увеличить в 2 раза, получим А/(2у) ч - время расхода воды, (А*3х)/(2у) км - длина дороги, которую можно полить.
Из выражения (А*2х)/(3у)=4 выразим А=(4*3у)/(2х)
подставим А  в выражение (А*3х)/(2у)=(4*3у*3х)/(2х*2у)=(4*3*3)/(2*2)=9 км
Ответ: 9 км

Answer 2

Problem Analysis

We are given a scenario where a watering machine moves at a constant speed and releases the same amount of water every minute. If we increase the speed of the machine by 2 times and the rate of water release by 3 times, we need to determine how much road can be watered if we increase the initial speed of the machine by 3 times and the initial rate of water release by 2 times.

Solution

Let's assume the initial speed of the machine is S km/min and the initial rate of water release is R liters/min. We are also given that the machine can water 4 km of road with the initial settings.

To solve this problem, we can use the concept of ratios.

Let's calculate the amount of water released by the machine in 1 minute with the initial settings: - Speed of the machine = S km/min - Rate of water release = R liters/min

Therefore, the amount of water released in 1 minute = R liters.

Now, if we increase the speed of the machine by 2 times and the rate of water release by 3 times, the new speed of the machine will be 2S km/min and the new rate of water release will be 3R liters/min.

We are given that with these new settings, the machine can water 4 km of road.

To find the amount of water required to water 4 km of road, we can set up the following ratio:

Amount of water released in 1 minute / Distance covered in 1 minute = Amount of water required to water 4 km of road / 4 km

Using the initial settings, the ratio becomes:

R / S = Amount of water required to water 4 km of road / 4 km

Now, let's calculate the amount of water required to water 4 km of road using the new settings:

3R / (2S) = Amount of water required to water 4 km of road / 4 km

To find the distance that can be watered with the new settings, we can rearrange the equation:

Distance covered = (3R / (2S)) * 4 km

Let's substitute the values and calculate the distance covered:

Distance covered = (3R / (2S)) * 4 km

Calculation

Given that the initial speed of the machine (S) is increased by 3 times and the initial rate of water release (R) is increased by 2 times, we can substitute the values into the equation:

Distance covered = (3 * R) / (2 * S) * 4 km

Answer

The distance that can be watered with the increased speed and rate of water release is (3 * R) / (2 * S) * 4 km.

Please note that the specific values for the initial speed (S) and rate of water release (R) were not provided in the question, so we cannot calculate the exact distance that can be watered.