Question

В трёх сосудах налита вода. Если 60% воды из первого сосуда перелить во второй затем 50% воды, оказавшейся во втором сосуде перелить в третий и, 60% воды из третьего сосуда перелить в первый, то в каждом сосуде окажется по 1л. Сколько было воды в каждом сосуде?

Answer 1

Подробное решение:
3 сосуда
х л- в 1-ом сосуде
y л- в 2-ом сосуде
z л- в 3-ем сосуде

х/3                             1/3 1-го
х-x/3=2х/3                  осталось в 1-ом(1/3 отлили из 1-го)
y+х/3                           стало во 2-ом ,1/3 из 1-го перелили во 2-ой
(y+х/3)/4                       1/4  2-го
(y+х/3)-(y+х/3)/4               осталось во 2-м , так как 1/4 перелили в 3-й
z+(y+х/3)/4                     cтало в 3-м (добавили 1/4 из 2-го)
(z+(y+х/3)/4)/10                1/10 3-го
z+(y+х/3)/4 - (z+(y+х/3)/4)/10   осталось в 3-м (отлили 1/10)
2х/3+(z+(y+х/3)/4)/10           стало в 1-м (добавили 1/10 из 3-го)

2х/3+(z+(y+х/3)/4)/10 = 9       *3*10 (умножим левую и правую части)
(y+х/3)-(y+х/3)/4 =9                               (y+x/3)(1-1/4)=9
(z+(y+х/3)/4) - (z+(y+х/3)/4)/10 =9         (z+(y+x/3)/4)(1-0,1)=9

20х+3z+3(y+х/3)/4=270     *4 (умножим левую и правую части)
3(y+х/3)/4 =9             *4 (умножим левую и правую части)
0,9(z+(y+х/3)/4) =9            

80х+12z+3y+х=1080
3y+x=36
z+(y+х/3)/4 =10           *4*3 (умножим левую и правую части)

81х+12z+3y=1080
3y+x=36              x=36-3y
12z+3y+x=120    12z+3y+36-3y=120     12z=84     z=7
 
81х+12z+3y=1080 81(36-3y)+12*7+3y=1080  2916-243y+84+3y=1080 -240y=-1920 y=8
x=36-3y
z=7

y=8
x=36-3y  x=36-3*8=36-24=12  x=12
z=7

x=12 л было воды в 1-м сосуде
y= 8 л было воды в 2-м сосуде
z= 7 л было воды в 3-м сосуде

Проверка:
12+8+7=27л
9*3=27 л
Фуууууууууух дописал))) Надеюсь правильно!))))

Answer 2

Problem Analysis

We have three vessels filled with water. We are given a series of operations where a certain percentage of water is transferred from one vessel to another. After these operations, each vessel contains 1 liter of water. We need to determine the initial amount of water in each vessel.

Solution

Let's assume the initial amount of water in the first vessel is x liters, in the second vessel is y liters, and in the third vessel is z liters.

According to the given information, the following operations are performed: 1. 60% of the water from the first vessel is transferred to the second vessel. 2. 50% of the water from the second vessel (which now contains 60% of the water from the first vessel) is transferred to the third vessel. 3. 60% of the water from the third vessel (which now contains 50% of the water from the second vessel) is transferred back to the first vessel.

Let's calculate the amount of water in each vessel after these operations.

Calculation

1. After transferring 60% of the water from the first vessel to the second vessel, the second vessel contains 60% of x liters, which is 0.6x liters. 2. After transferring 50% of the water from the second vessel to the third vessel, the third vessel contains 50% of 0.6x liters, which is 0.3x liters. 3. After transferring 60% of the water from the third vessel back to the first vessel, the first vessel contains 60% of 0.3x liters, which is 0.18x liters.

According to the given information, after these operations, each vessel contains 1 liter of water. Therefore, we can set up the following equations:

0.18x = 1 (equation for the first vessel) 0.3x = 1 (equation for the third vessel)

Solving these equations will give us the initial amount of water in each vessel.

Solution

Let's solve the equations:

0.18x = 1 x = 1 / 0.18 x ≈ 5.56

0.3x = 1 x = 1 / 0.3 x ≈ 3.33

Therefore, the initial amount of water in the first vessel is approximately 5.56 liters, in the second vessel is 0.6 * 5.56 ≈ 3.33 liters, and in the third vessel is 0.3 * 3.33 ≈ 1 liter.

Answer

The initial amount of water in each vessel is approximately: - First vessel: 5.56 liters - Second vessel: 3.33 liters - Third vessel: 1 liter