When two rectangular tanks reach the same water depth, multiply that common depth by the sum of their base areas to get the total volume. Divide the total volume by the combined area to find the common depth. Then compare the water in either tank before and after to find the amount transferred.
MOE’s Primary 6 Mathematics syllabus includes using the volume and base area of a cuboid to find its height. The original problems below apply that relationship to two tanks with different base areas and no spillage.
- Find each tank’s starting volume.
- Use the combined base area only after the depths are equal.
- Find the transfer by subtracting the starting and final volume in one tank.
Worked example | Both tanks contain water at the start
Tank A has a base of 20 cm by 10 cm and starts with water 7 cm deep. Tank B has a base of 15 cm by 10 cm and starts with water 2 cm deep. Both tanks can hold water at the final depth. How much water must you pour from A into B so the depths match?
A starts with 200 × 7 = 1,400 cm³. B starts with 150 × 2 = 300 cm³. Together they hold 1,700 cm³. At a common depth, each centimetre of water depth across the two tanks holds 200 + 150 = 350 cm³.
The common depth is 1,700 ÷ 350 = 34/7 cm. B ends with 150 × 34/7 = 5,100/7 cm³. Subtract its starting 300 cm³. The amount transferred is 3,000/7 cm³, or about 429 cm³ to the nearest cubic centimetre.
Check using Tank A. At the common depth it holds 200 × 34/7 = 6,800/7 cm³. Its loss is 1,400 − 6,800/7 = 3,000/7 cm³, the same as B’s gain.
Why averaging the starting depths can give the wrong answer
The average of 7 cm and 2 cm is 4.5 cm. At that depth, the tanks would hold 350 × 4.5 = 1,575 cm³ altogether. That would leave 125 cm³ unaccounted for. The tanks have different base areas, so one centimetre of depth represents different amounts in each tank.
The common depth should lie between the starting depths when the tanks have uniform bases and all water stays in the tanks. The exact depth is closer to the starting depth of the tank with the larger base.
Try a version with a whole-number answer
Two tanks have base areas of 120 cm² and 80 cm². Their starting depths are 8 cm and 3 cm respectively. The total volume is 960 + 240 = 1,200 cm³. The common depth is 1,200 ÷ 200 = 6 cm.
The second tank rises by 3 cm and gains 80 × 3 = 240 cm³. The first falls by 2 cm and loses 120 × 2 = 240 cm³. Ask a study partner to explain how different depth changes can represent the same transferred volume.
Frequently asked questions
Does equal depth mean equal volume?
With different base areas, the volumes at the same depth are different.
Can I use this method for tanks that widen towards the top?
The calculation here assumes constant horizontal cross-sectional areas. A widening tank requires more information about its shape.
Official information
MOE’s Primary Mathematics syllabus, Primary Six, Volume of Cube and Cuboid. Checked 9 November 2026.