PSLE Mathematics Learning Guide · Guide 48
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A rectangular tank problem becomes easier when the learner stops thinking first about litres and starts thinking about three-dimensional space.
For a rectangular tank with a flat rectangular base:
liquid volume = base area × liquid height.
This one relationship can find the amount of liquid, the height of the liquid, the base area, the rise after more liquid is added, or the final level after liquid is poured into another rectangular tank.
This guide builds one habit: identify the tank’s internal base area, convert the liquid amount into a compatible volume unit, use volume = base area × height, and preserve total liquid when the problem states that none is lost.
Guide 7 owns general cube/cuboid volume. Guide 31 owns capacity and unit relationships. This page takes the narrower liquid-level application and connects those two owners without replacing either.
The MOE Primary Mathematics syllabus updated October 2025 provides curriculum context. All examples below are original eduKate teaching material.
Find base area before working with liquid height
A rectangular tank has internal base dimensions 20 cm by 15 cm.
Base area = 20 × 15 = 300 cm².
If the liquid is 8 cm deep, liquid volume = 300 × 8 = 2400 cm³.
The tank’s full height is irrelevant unless the problem asks about full capacity or overflow.
Connect cubic centimetres and millilitres
In the metric system:
1 cm³ = 1 mL.
Therefore 2400 cm³ = 2400 mL = 2.4 L.
This conversion makes it possible to move between tank dimensions in centimetres and liquid amounts stated in litres or millilitres.
Find liquid height from volume
A tank has base area 250 cm² and contains 2000 cm³ of liquid.
Height = volume ÷ base area = 2000 ÷ 250 = 8 cm.
Check: 250 × 8 = 2000 cm³.
Use length × width × liquid height when base area is not given
A tank has base 25 cm by 16 cm and contains 3200 mL.
3200 mL = 3200 cm³.
Base area = 25×16 = 400 cm².
Liquid height = 3200 ÷ 400 = 8 cm.
An added volume creates a predictable rise
A tank has base area 300 cm². Another 900 mL is poured in.
Added volume = 900 cm³.
Rise in liquid level = 900 ÷ 300 = 3 cm.
If the original level was 5 cm, the new level is 8 cm, provided the tank is tall enough and no liquid spills.
Removing liquid lowers the level by removed volume ÷ base area
A rectangular tank has base area 240 cm². 720 mL is removed.
Drop = 720 ÷ 240 = 3 cm.
If the original depth was 10 cm, the remaining depth is 7 cm.
Pouring between rectangular tanks conserves volume
Tank A contains 3600 cm³ of liquid. All of it is poured into empty Tank B whose base area is 450 cm².
Final height in B = 3600 ÷ 450 = 8 cm.
The shape changed, so the height changed. The liquid volume remained 3600 cm³ because the problem states that all liquid was transferred without loss.
A wider base gives a lower level for the same volume
Two rectangular tanks contain the same 2400 cm³ of liquid.
Tank X base area = 300 cm² → height = 8 cm.
Tank Y base area = 600 cm² → height = 4 cm.
Doubling the base area halves the height when volume stays fixed.
This inverse relationship is a useful reasonableness check.
The same added liquid causes different rises in different tanks
Add 1200 mL to Tank A with base area 200 cm²:
rise = 1200 ÷ 200 = 6 cm.
Add the same 1200 mL to Tank B with base area 400 cm²:
rise = 3 cm.
The narrower tank rises more because the same volume is spread across a smaller base.
A partial transfer requires the transferred volume first
Tank A contains 5 L. Three fifths is poured into Tank B.
Transferred volume = 3/5 × 5 L = 3 L = 3000 cm³.
If Tank B base area is 500 cm², its liquid level rises to:
3000 ÷ 500 = 6 cm, assuming it began empty.
Tank A retains 2 L.
A fraction-full statement relates current volume to capacity
A rectangular tank is 3/4 full. Its full capacity is 8 L.
Current volume = 3/4 × 8 = 6 L.
If the tank’s base area is 500 cm², 6 L = 6000 cm³, so current height = 6000 ÷ 500 = 12 cm.
The full internal height would be 8000 ÷ 500 = 16 cm under the same rectangular dimensions.
Find base area from volume and height
A rectangular tank contains 4500 cm³ at a liquid depth of 9 cm.
Base area = 4500 ÷ 9 = 500 cm².
If one base dimension is 25 cm, the other is 500 ÷ 25 = 20 cm.
Capacity limits the final state
A tank has base area 300 cm² and internal height 12 cm.
Full capacity = 300×12 = 3600 cm³ = 3.6 L.
It currently contains 3 L. Another 1 L is poured toward it.
Available capacity = 0.6 L.
Therefore 0.6 L can remain in the tank and 0.4 L would overflow under the idealised problem model.
Do not calculate a liquid height above the stated tank height and leave it uninterpreted.
Multi-stage tank problems need state tracking
A tank with base area 250 cm² starts at 4 cm depth.
Initial volume = 250×4 = 1000 cm³.
500 mL is added: new volume = 1500 cm³, height = 6 cm.
Then 250 mL is removed: remaining volume = 1250 cm³, height = 5 cm.
Write each state instead of applying all changes to the original height.
Main worked workshop: sixteen original tank problems
1.
Base 20×10 cm, depth 6 cm. Volume?
Answer: 1200 cm³ = 1.2 L.
2.
Base area 300 cm², volume 2400 cm³. Height?
Answer: 8 cm.
3.
Base 25×12 cm, liquid 3 L. Height?
Answer: 3000÷300 = 10 cm.
4.
Base area 400 cm². Add 1600 mL. Rise?
Answer: 4 cm.
5.
Base area 250 cm². Remove 1000 mL. Drop?
Answer: 4 cm.
6.
3600 cm³ poured into tank base 600 cm².
Answer: 6 cm.
7.
Same 3600 cm³ into base 300 cm².
Answer: 12 cm.
8.
A 10 L tank is 65% full.
Answer: 6.5 L.
9.
6.5 L in a rectangular tank with base area 500 cm². Depth?
Answer: 6500÷500 = 13 cm.
10.
Volume 4200 cm³, depth 7 cm. Base area?
Answer: 600 cm².
11.
Base area 600 cm², one side 30 cm. Other side?
Answer: 20 cm.
12.
A tank 20×15×10 cm is full. Capacity?
Answer: 3000 cm³ = 3 L.
13.
The tank in Problem 12 contains 2.4 L. Empty capacity?
Answer: 0.6 L.
14.
A learner divides 2400 mL by tank height 8 cm to find liquid height.
Repair: divide volume by base area, not by height. If base area is 300 cm², height = 8 cm.
15.
A learner pours 2 L into a wider tank and expects the same height.
Repair: height depends on base area. Same volume can produce different depths.
16.
Tank base 500 cm² starts at 3 cm. Add 2 L. New height?
Reasoning: initial 1500 cm³ + 2000 = 3500; height = 7 cm.
Conservation is a condition, not an assumption in every story
If the problem says all liquid is poured from one tank to another without spilling, transferred volume is conserved.
If the problem says some spills, evaporates, remains behind or is removed, adjust the volume before calculating the new height.
Do not automatically set starting and ending volumes equal when the story explicitly contains a loss or addition.
Area units and volume units must stay distinct
Base area is measured in cm².
Liquid volume is measured in cm³ or a capacity equivalent such as mL.
Liquid height is measured in cm.
In the relationship:
cm² × cm = cm³.
The units themselves check the structure.
Litres belong to volume through conversion
1 L = 1000 mL = 1000 cm³.
Therefore a 2.5 L liquid volume is 2500 cm³ before it is combined with a base area measured in cm².
Use Guide 31 if this conversion is the main difficulty.
Use the general volume owner for missing cuboid dimensions
When the problem is about a solid cuboid or full container dimensions rather than a liquid surface level, use Guide 7: Rebuild Cube and Cuboid Volume From Base Area and Height.
Common rectangular-tank errors
Error 1: use full tank height instead of current liquid height.
Error 2: divide volume by one base dimension instead of the full base area.
Error 3: mix litres directly with cm² without converting litres to cm³.
Error 4: assume equal volume gives equal height in tanks with different bases.
Error 5: conserve volume even when the story states a spill or removal.
Independent transfer check
- Tank base 18×20 cm, liquid depth 5 cm. Find volume in mL.
- Base area 450 cm², volume 3600 mL. Find depth.
- Base area 250 cm², add 750 mL. Find rise.
- A 4.5 L volume is poured into an empty tank with base 500 cm². Find depth.
- A tank base is 30×20 cm and contains liquid 8 cm deep. How much liquid is present in litres?
- The tank in Question 5 loses 1.2 L. Find new depth.
- Explain why the same 3 L produces a lower level in a tank with double the base area.
- Explain why a final calculated depth above the tank’s internal height signals an overflow issue.
Independent-check answers
1. 18×20×5 = 1800 cm³ = 1800 mL.
2. 3600÷450 = 8 cm.
3. 750÷250 = 3 cm.
4. 4500÷500 = 9 cm.
5. 30×20×8 = 4800 cm³ = 4.8 L.
6. Remaining 3.6 L = 3600 cm³; depth = 3600÷600 = 6 cm.
7. Height = volume ÷ base area, so doubling the denominator halves the height for fixed volume.
8. The physical container cannot hold liquid above its stated top; excess volume must overflow or the assumptions need rechecking.
Parent and tutor guide
Draw a top-view rectangle first and write the base area inside it. Then draw a side view with the liquid height. Keeping these two measurements visually separate reduces the tendency to multiply or divide by the wrong dimension.
For pouring problems, write a “before volume → transferred volume → after volume” chain. Ask explicitly whether any liquid is lost.
After a correct direct problem, reverse it. Give volume and base area to find height, then give height and volume to recover base area. This tests whether the relationship is understood rather than memorised in one direction.
The learner’s final card
What is the internal base area? What liquid volume is present now? Are the volume and area units compatible? Am I finding current height, rise, drop, capacity or transferred volume? Is liquid conserved in this story?
Complete Batch 12
Use Guide 45: Division as Fraction or Decimal Quotients, Guide 46: GST and Annual Interest, and Guide 47: Unit Cubes and Isometric Solids.
Return to the PSLE Learning Guide Mathematics route.
Sources and boundaries
MOE Primary Mathematics syllabus, updated October 2025; SEAB 2026 PSLE formats.
All tank dimensions, liquid amounts and solutions are original teaching material. The examples assume ideal rectangular internal dimensions and explicitly state when volume is conserved or when overflow must be considered.