As shown in a figure, there are three water tank of different sizes. Every tank is a rectangular prism.
In addition, A and B, B and C are connected by a thin pipe with a cock X, Y respectively.
When the cock is opened, water in connected two water tanks moves through the pipe until height of the water surface in each tank is same.
Answer the following questions.
Pay attention that the amount of the water in the pipe shall not be considered.
(1) At first, water is contained up to a height of 100cm from bottom in A and no water in B and C.
When the cock X is opened while cock Y being closed, the height of the water surface of A and B comes to be 40 cm.
Indicate the ratio of the amount of water in the tank of A and B as the integral ratio at this moment.
(2) Next, when the cock X is closed and the cock Y is opened, the height of the water surface of B and C comes to be 25cm.
Indicate the ratio of the amount of water in the tank of B and C as the integral ratio.
(3) Next, with cock Y being opened, cock X is opened again, the height of the water surface of the water in the three tanks comes to be same.
Find the height of the water surface at this moment.
In addition, A and B, B and C are connected by a thin pipe with a cock X, Y respectively.
When the cock is opened, water in connected two water tanks moves through the pipe until height of the water surface in each tank is same.
Answer the following questions.
Pay attention that the amount of the water in the pipe shall not be considered.
(1) At first, water is contained up to a height of 100cm from bottom in A and no water in B and C.
When the cock X is opened while cock Y being closed, the height of the water surface of A and B comes to be 40 cm.
Indicate the ratio of the amount of water in the tank of A and B as the integral ratio at this moment.
(2) Next, when the cock X is closed and the cock Y is opened, the height of the water surface of B and C comes to be 25cm.
Indicate the ratio of the amount of water in the tank of B and C as the integral ratio.
(3) Next, with cock Y being opened, cock X is opened again, the height of the water surface of the water in the three tanks comes to be same.
Find the height of the water surface at this moment.
Answer
Solution
(1) The height of the water surface of A is to 40cm down by 60cm from 100cm, the height of the water surface of B becomes 40cm from 0cm.
The height of the surface of the water of A having fallen 60cm, 60/100=3/5 of the amount of the water in A moved to B, and 1 - 3/5 = 2/5 was left in A.
Therefore, the ratio of the quantity of the water of A and B of this time is 3/5 : 2/5 = 3 : 2.
(2) When the cock Y is opened, the height of the surface of the water of B is down by 15cm from 40cm and becomes to 25cm.
15/40 = 3/8 of the amount of the water in B moved to C and 1 - 3/8 = 5/8 was left in B.
Therefore, the ratio of the amount of the water of B and C is 5/8 : 3/8 = 5 : 3.
(3) When cock X and Y are opened, water in A goes into B and C and some remain in A.
Then the height of the water surface of the water in the three tanks comes to be same.
The height of the water surface of this case is same as the height of the water surface when water is transferred to the water tank with the combined bottom area of A, B and C.
That is , to find the height of the water level at this time, the amount of water in A first may be divided by the total of the bottom area of A, B, and C.
The ratio of the base of A and B is same as the ratio of the volume of A and B found in (1), it is 2 : 3.
The ratio of the base of B and C is 5 : 3 as (2).
2 sets of these ratios are made into a continuous ratio, it is A : B : C = 10 : 15 : 9.
Because the height of the surface of the water of A of the beginning was 100cm, the volume of the water becomes 10 × 100 = 1000.
Total of base area of the three is 10 + 15 + 9 = 34.
Therefore the height is 1000/34 = 500/17.
(1) 3 : 2
(2) 5 : 3
(3) 500/17 cm
(2) 5 : 3
(3) 500/17 cm
(1) The height of the water surface of A is to 40cm down by 60cm from 100cm, the height of the water surface of B becomes 40cm from 0cm.
The height of the surface of the water of A having fallen 60cm, 60/100=3/5 of the amount of the water in A moved to B, and 1 - 3/5 = 2/5 was left in A.
Therefore, the ratio of the quantity of the water of A and B of this time is 3/5 : 2/5 = 3 : 2.
(2) When the cock Y is opened, the height of the surface of the water of B is down by 15cm from 40cm and becomes to 25cm.
15/40 = 3/8 of the amount of the water in B moved to C and 1 - 3/8 = 5/8 was left in B.
Therefore, the ratio of the amount of the water of B and C is 5/8 : 3/8 = 5 : 3.
(3) When cock X and Y are opened, water in A goes into B and C and some remain in A.
Then the height of the water surface of the water in the three tanks comes to be same.
The height of the water surface of this case is same as the height of the water surface when water is transferred to the water tank with the combined bottom area of A, B and C.
That is , to find the height of the water level at this time, the amount of water in A first may be divided by the total of the bottom area of A, B, and C.
The ratio of the base of A and B is same as the ratio of the volume of A and B found in (1), it is 2 : 3.
The ratio of the base of B and C is 5 : 3 as (2).
2 sets of these ratios are made into a continuous ratio, it is A : B : C = 10 : 15 : 9.
Because the height of the surface of the water of A of the beginning was 100cm, the volume of the water becomes 10 × 100 = 1000.
Total of base area of the three is 10 + 15 + 9 = 34.
Therefore the height is 1000/34 = 500/17.
