Problems Involving Transferring Water
We calculate how the depth changes when water in container A is transferred to container B.
Key Points for Water Transfer Problems
- Even after water is transferred, the total volume of water does not change
- Volume before transfer = base area of A × depth of A
- Depth after transfer = volume ÷ base area of B
Example 1: Container A has a base area of 30 cm² and contains water with a depth of 12 cm. If all of this water is transferred to container B, which has a base area of 18 cm², what will the depth of the water be in cm?
Solution:
① Volume of water = 30 × 12 = 360 cm³
② New depth = 360 ÷ 18 = 20 cm
Solution:
① Volume of water = 30 × 12 = 360 cm³
② New depth = 360 ÷ 18 = 20 cm
Example 2: Container X has a base area of 60 cm² and a depth of 15 cm, while container Y has a base area of 40 cm² and a depth of 5 cm. They are connected at the bottom, and the water levels are made equal. What is the common depth in cm?
Solution:
① Total volume of water = 60 × 15 + 40 × 5 = 900 + 200 = 1100 cm³
② Total base area = 60 + 40 = 100 cm²
③ Common depth = 1100 ÷ 100 = 11 cm
Solution:
① Total volume of water = 60 × 15 + 40 × 5 = 900 + 200 = 1100 cm³
② Total base area = 60 + 40 = 100 cm²
③ Common depth = 1100 ÷ 100 = 11 cm
Problems in which “the water levels are made equal” can be solved by calculating “total volume of water ÷ total base area.”