Question 1170042: Tank A and Tank B are rectangular prisms and are sitting on a flat table.
Tank A is 10 cm × 8 cm × 6 cm and is sitting on one of its 10 cm × 8 cm faces.
Tank B is 5 cm × 9 cm × 8 cm and is sitting on one of its 5 cm × 9 cm faces.
Initially, Tank A is full of water and Tank B is empty.
The water in Tank A drains out at a constant rate of 4 cm3/s.
Tank B fills with water at a constant rate of 4 cm3/s.
Tank A begins to drain at the same time that Tank B begins to fill.
(i) Determine after how many seconds Tank B will be exactly 1
3
full.
(ii) Determine the depth of the water left in Tank A at the instant when Tank
B is full.
(iii) At one instant, the depth of the water in Tank A is equal to the depth of
the water in Tank B. Determine this depth.
(b) Tank C is a rectangular prism that is 31 cm × 4 cm × 4 cm.
Tank C sits on the flat table on one of its 31 cm × 4 cm faces.
Tank D is in the shape of an inverted square-based pyramid, as shown. It is
supported so that its square base is parallel to the flat table and its fifth vertex
touches the flat table.
The height of Tank D is 10 cm and the side length of its square base is 20 cm.
Initially, Tank C is full of water and Tank D is empty.
Tank D begins filling with water at a rate of 1 cm3/s.
Two seconds after Tank D begins to fill, Tank C begins to drain at a rate of
2 cm3/s.
At one instant, the volume of water in Tank C is equal to the volume of water
in Tank D.
Determine the depth of the water in Tank D at that instant.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem into parts and solve each one.
**(a) Tank A and Tank B:**
**(i) Time for Tank B to be 1/3 full:**
* Volume of Tank B = 5 cm × 9 cm × 8 cm = 360 cm³
* 1/3 of Tank B's volume = 360 cm³ / 3 = 120 cm³
* Tank B fills at 4 cm³/s.
* Time = Volume / Rate = 120 cm³ / 4 cm³/s = 30 seconds.
**(ii) Depth of water in Tank A when Tank B is full:**
* Time to fill Tank B = 360 cm³ / 4 cm³/s = 90 seconds.
* Volume drained from Tank A in 90 seconds = 90 s × 4 cm³/s = 360 cm³.
* Volume of Tank A = 10 cm × 8 cm × 6 cm = 480 cm³.
* Volume remaining in Tank A = 480 cm³ - 360 cm³ = 120 cm³.
* Area of the base of Tank A = 10 cm × 8 cm = 80 cm².
* Depth of water in Tank A = Volume / Area = 120 cm³ / 80 cm² = 1.5 cm.
**(iii) Depth when water depths are equal:**
* Let 't' be the time in seconds.
* Depth of water in Tank B = (4t) / (5 × 9) = 4t / 45 cm.
* Volume of water remaining in Tank A = 480 - 4t cm³.
* Depth of water in Tank A = (480 - 4t) / (10 × 8) = (480 - 4t) / 80 cm.
* Equate the depths: 4t / 45 = (480 - 4t) / 80.
* Cross-multiply: 320t = 45(480 - 4t).
* 320t = 21600 - 180t.
* 500t = 21600.
* t = 21600 / 500 = 43.2 seconds.
* Depth = (4 × 43.2) / 45 = 172.8 / 45 = 3.84 cm.
**(b) Tank C and Tank D:**
* Volume of Tank C = 31 cm × 4 cm × 4 cm = 496 cm³.
* Volume of a pyramid = (1/3) × base area × height.
* Volume of Tank D (when full) = (1/3) × (20 cm)² × 10 cm = 4000/3 cm³.
* Let 't' be the time in seconds after Tank D starts filling.
* Volume of water in Tank D = t cm³.
* Tank C begins draining at t = 2 seconds.
* Volume of water in Tank C at time t = 496 - 2(t - 2) = 500 - 2t cm³.
* We need to find when the volumes are equal:
* t = 500 - 2t
* 3t = 500
* t = 500/3 seconds
* Volume of water in Tank D at this time = 500/3 cm³.
* Volume of water in a square based pyramid when water depth is h is: V = (1/3)*(20*h/10)^2 * h = 4/3*h^3.
* 500/3 = 4/3*h^3
* 500/4 = h^3
* 125= h^3
* h = 5 cm.
**Answers:**
**(a)**
* **(i) 30 seconds**
* **(ii) 1.5 cm**
* **(iii) 3.84 cm**
**(b)**
* **5 cm**
|
|
|
