PSLE Volume and Water-Level Problems test whether pupils can connect three-dimensional space to area, height and displacement. The core formulas are simple: volume of a cuboid = length × breadth × height, and volume = base area × height. The difficulty comes from deciding what changed, what stayed fixed and whether the problem is asking about container volume, liquid volume or change in water level.
Students searching for PSLE volume questions, water level problems, cube and cuboid volume, liquid in tanks or Primary 6 volume often remember the formula but still struggle when a solid is added, liquid is transferred or one dimension is unknown. The reliable method is identify the solid/container → mark dimensions → find base area → identify volume change → translate that change into height.
This page is the canonical Volume and Water-Level owner under the PSLE Mathematics Heuristics hub. Under Singapore’s current Primary 6 Mathematics syllabus, pupils work with volume of cubes and cuboids, including finding an unknown dimension from volume and the other dimensions.
Quick answer: the three equations
- Volume of cuboid = length × breadth × height
- Volume = base area × height
- Change in water level = displaced / added volume ÷ base area
Worked example 1: find volume
Problem: A tank is 12 cm long, 8 cm wide and 10 cm high. Find its volume.
Volume = 12 × 8 × 10 = 960 cm³.
Worked example 2: find an unknown height
Problem: A cuboid has volume 720 cm³ and base area 90 cm². Find its height.
Height = Volume ÷ Base area = 720 ÷ 90 = 8 cm.
This is why base area × height is often more useful than memorising length × breadth × height only.
Water-level problems are volume problems in disguise
If water rises in a rectangular tank, the added volume occupies a rectangular prism whose base is the tank’s base and whose height is the rise in water level.
Added volume = tank base area × rise in water level.
Worked example 3: added water
Problem: A rectangular tank has base dimensions 20 cm by 15 cm. Water is added and the level rises by 4 cm. How much water was added?
Base area = 20 × 15 = 300 cm².
Added volume = 300 × 4 = 1200 cm³.
Since 1000 cm³ = 1 litre, this is 1.2 litres.
Worked example 4: water-level rise
Problem: 1800 cm³ of water is poured into a tank with base area 450 cm². By how much does the water level rise?
Rise = 1800 ÷ 450 = 4 cm.
Displacement: an immersed object takes up space
When a solid object is completely immersed, the water-level rise corresponds to the volume displaced by the submerged part of the object.
Displaced volume = tank base area × rise in water level.
If the object is only partly submerged, only the submerged volume contributes to displacement.
Worked example 5: fully immersed cuboid
Problem: A solid cuboid is placed completely into a tank. The tank base area is 250 cm² and the water rises by 3.2 cm. Find the cuboid’s volume.
Displaced volume = 250 × 3.2 = 800 cm³.
Because the cuboid is completely immersed, its volume is 800 cm³.
Transfer between containers
When liquid is transferred from one rectangular container to another, the liquid volume remains constant unless some is spilled or removed.
This creates a useful invariant:
Original liquid volume = new liquid volume.
Worked example 6: transfer
Problem: A tank with base 12 cm by 10 cm contains water to a depth of 15 cm. All the water is poured into a tank with base 18 cm by 10 cm. Find the new water depth.
Original volume = 12 × 10 × 15 = 1800 cm³.
New base area = 18 × 10 = 180 cm².
New depth = 1800 ÷ 180 = 10 cm.
When water-level problems become multi-step
- object added, then removed
- water transferred between containers
- only part of a solid is submerged
- tank dimensions change between stages
- water is added after an object is already inside
- difference between original and final water levels is given
Use a before-and-after table to keep stages separate.
The before-and-after tank table
- Before: base area, water height, water volume.
- Change: added liquid, removed liquid, object inserted, object removed.
- After: new water height, displaced volume or remaining liquid.
The table prevents pupils from mixing container volume with liquid volume.
Volume units
- 1 cm³ = 1 ml
- 1000 cm³ = 1000 ml = 1 litre
- 1 m³ = 1,000,000 cm³
Do not convert unless the question requires a different unit. Keep units consistent through the calculation.
Common errors
- uses perimeter instead of base area
- forgets one dimension
- uses the tank’s full height instead of actual water depth
- assumes container volume equals liquid volume
- adds water-level heights instead of volumes
- forgets only submerged volume displaces water
- mixes cm² and cm³
- converts litres and cubic centimetres incorrectly
The diagnostic checklist
- What is the base area?
- What volume is actually changing?
- Is the liquid volume constant?
- Is an object fully or partly submerged?
- What height belongs to the water, and what height belongs to the container?
- Are the units consistent?
Frequently asked questions
Why use base area × height?
Because water-level changes are directly linked to the same base area across the tank.
Does a floating object displace its full volume?
Not necessarily. Only the submerged portion displaces water by volume.
If all water is transferred, what stays constant?
The liquid volume, assuming none is lost.
What is the most common mistake?
Using the wrong height or confusing liquid volume with the container’s total capacity.
Where does this sit in Atlas?
This is the canonical Volume and Water-Level owner under PSLE Mathematics Heuristics.
The final Volume rule
Think in layers: base area first, then height. Water-level change is simply a volume change spread across a fixed base. Once the correct volume is identified, the rest is usually one division or multiplication.