Volume and water tanks · applications

Applications: Volume and Water Tanks

10 question types · Model Method and algebra, side by side

PSLE · SAT · GCSE Higher

01

Dual Taps (Inflow & Outflow) with Staggered Starts

heuristicSegmented Time-Rate Allocation / Net Rate Flow

A rectangular tank measuring 60 cm by 40 cm by 50 cm was initially empty. Tap A was turned on at 09:00 to fill the tank at a rate of 4.5 liters/min. At 09:08, Tap B was turned on to drain water from the bottom of the tank at a rate of 1.5 liters/min. Both taps remained turned on until the tank was 34 full. (a) How many liters of water were in the tank at 09:08? (b) At what time was the tank 34 full of water?

60 by 40, 50 tall: 120 L
The tank holds 60 × 40 × 50 = 120 000 cm³, 120 liters.
Total tank capacity bar = 120 liters.
step 1 of 8

Draw a capacity bar of 120 liters. Partition into a target block of 90 liters, split between Phase 1 (Tap A only) and Phase 2 (combined net rate).

  1. Total tank capacity bar = 120 liters.
  2. Shade 34 of the bar = 90 liters.
  3. Phase 1 (first 8 min): Tap A fills 8 × 4.5 ℓ = 36 liters.
  4. (a) Water at 09:08 is 36 liters.
  5. Unfilled portion of target bar: 90 − 36 = 54 liters.
  6. Phase 2: Every minute, Tap A adds 4.5 ℓ while Tap B removes 1.5 ℓ, leaving a net gain of 3.0 ℓ/min.
  7. Divide remaining target by net unit rate: 54 ÷ 3 = 18 minutes.
  8. Clock time: 09:08 + 18 min = 09:26.

answer(a) 36 liters; (b) 09:26

techniqueVolume of a Cuboid · Rates

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Using total capacity (120 liters) instead of 34 capacity (90 liters) to calculate the remaining fill time.
  • Applying the net rate (3.0 ℓ/min) from the very beginning at 09:00 instead of after 09:08.
  • Adding 1.5 ℓ/min to 4.5 ℓ/min instead of subtracting, treating the drain tap as a second filling tap.
02

Connecting Tanks with an Equalizing Valve (Hydrostatic Equilibrium)

heuristicCommon Height Equilibrium / Base Area Ratio Partitioning

Tank A is a rectangular tank measuring 40 cm by 30 cm with a height of 45 cm, filled with water to a depth of 30 cm. Tank B is another rectangular tank measuring 40 cm by 20 cm with a height of 40 cm, and is initially empty. A narrow pipe with a closed valve connects the bases of both tanks. When the valve is opened, water flows from Tank A into Tank B until the water levels in both tanks reach the same height. (a) What was the final height of water in both tanks? (b) How many liters of water flowed from Tank A into Tank B?

30 cmA: 40 by 30B: 40 by 20valve
Water flows until both surfaces are level, so the two depths end equal.
Find ratio of base areas: Base A : Base B = 1200 : 800 = 3 : 2.
step 1 of 8

Since final depth is identical, water distributes strictly in proportion to the base areas: Base A : Base B = 1200 : 800 = 3 : 2.

  1. Find ratio of base areas: Base A : Base B = 1200 : 800 = 3 : 2.
  2. Total water units at equilibrium: 3 units (in A) + 2 units (in B) = 5 units.
  3. Total water volume = 1200 × 30 = 36000 cm3.
  4. 1 unit of water = 36000 ÷ 5 = 7200 cm3.
  5. Water in Tank B (2 units) = 2 × 7200 = 14400 cm3.
  6. (b) Volume transferred = 14400 cm3 = 14.4 liters.
  7. Calculate common height using Tank B: Height = 14400 ÷ 800 = 18 cm.
  8. (a) Final height = 18 cm.

answer(a) 18 cm; (b) 14.4 liters

techniqueVolume of a Cuboid · Ratio

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Halving the initial depth (30 ÷ 2 = 15 cm), assuming water splits equally in depth regardless of differing base areas.
  • Averaging the heights of the two containers rather than dividing total volume by combined base area.
03

Completely Submerged Solid Object (Height Rise)

heuristicLiquid Displacement Principle / Equivalent Base Height Transfer

A rectangular glass tank with a base measuring 50 cm by 30 cm and a height of 40 cm contains water to a depth of 16 cm. Three identical solid metal cubes, each of edge length 10 cm, are lowered gently into the tank until they are completely submerged resting on the bottom. (a) What was the increase in the height of the water level? (b) What was the new depth of the water in the tank?

16 cm50 by 30, water 16 cm
The base is 50 × 30 = 1500 cm².
Draw the tank base: Area = 50 × 30 = 1500 cm2.
step 1 of 6

Treat the 3 cubes as a volumetric slab of water spread across the full 1500 cm² base area.

  1. Draw the tank base: Area = 50 × 30 = 1500 cm2.
  2. Calculate solid volume introduced: 3 × (10 × 10 × 10) = 3000 cm3.
  3. Imagine flattening this 3000 cm3 volume into a water layer over the 1500 cm2 floor.
  4. Layer thickness = 3000 ÷ 1500 = 2 cm.
  5. (a) Rise in water level = 2 cm.
  6. (b) Total depth = 16 + 2 = 18 cm.

answer(a) 2 cm; (b) 18 cm

techniqueVolume of a Cuboid

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Adding the cube edge length (10 cm) directly to the water height (16 + 10 = 26 cm).
  • Multiplying cube volume by 3 and dividing by the tank height instead of the tank base area.
04

Partially Submerged Solid Object (Effective Base Reduction)

heuristicEffective Cross-Sectional Area / Annular Base Volume

A rectangular tank has a base measuring 30 cm by 20 cm and a height of 35 cm. It contains water to a depth of 10 cm. A tall solid rectangular metal bar with a square base of 10 cm by 10 cm and a height of 25 cm is placed vertically upright onto the bottom of the tank. The water level rises but the top of the bar remains above the water surface. (a) What was the new depth of the water in the tank? (b) What fraction of the volume of the metal bar was submerged underwater?

10 cm30 by 20, water 10 cm
The floor is 600 cm².
Total tank floor area = 600 cm2.
step 1 of 8

Model the container base as 6 equal blocks of 100 cm². The bar occupies 1 block, leaving 5 blocks to hold the full 6000 cm³ of water.

  1. Total tank floor area = 600 cm2.
  2. Bar base occupies 100 cm2.
  3. Remaining free floor area = 600 − 100 = 500 cm2.
  4. The original water volume (600 × 10 = 6000 cm3) is squeezed into this 500 cm2 area.
  5. Depth = 6000 ÷ 500 = 12 cm.
  6. (a) New depth = 12 cm.
  7. Compare submerged portion (12 cm) to total bar height (25 cm): 1225.
  8. (b) Fraction submerged = 1225.

answer(a) 12 cm; (b) 1225

techniqueVolume of a Cuboid · Area and Perimeter

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Assuming the full volume of the bar is submerged (10 × 10 × 25 = 2500 cm3) and calculating 2500 ÷ 600 = 4.17 cm rise, ignoring that the top of the bar is dry.
  • Dividing 6000 cm³ by 600 cm² and adding 10 cm.
05

Submerged Object Causing Overflow

heuristicCapacity Headroom Comparison / Net Retained Volume

A rectangular container with a base of 25 cm by 20 cm and a height of 20 cm was filled with water to a depth of 18 cm. A solid metal block measuring 15 cm by 10 cm by 12 cm was lowered into the container and rested completely on the bottom, causing water to overflow. (a) How many cubic centimeters of water overflowed from the container? (b) After the metal block was carefully removed from the container, what was the depth of the remaining water?

18 cm25 by 20, 20 tall
Above the 18 cm of water is 2 cm of room: 500 × 2 = 1000 cm³.
Container base = 500 cm2. Empty space at top = 2 cm deep = 1000 cm3.
step 1 of 8

Model the empty space as 2 cm of height (1000 cm³). The block displaces 1800 cm³, pushing out 800 cm³ of overflow, which lowers the water level by 800 ÷ 500 = 1.6 cm from the initial level.

  1. Container base = 500 cm2. Empty space at top = 2 cm deep = 1000 cm3.
  2. Block volume = 15 × 10 × 12 = 1800 cm3.
  3. Block fills the 1000 cm3 headspace and pushes the rest over the brim.
  4. Overflow = 1800 − 1000 = 800 cm3.
  5. (a) Overflow = 800 cm3.
  6. When the block is removed, the 800 cm3 lost causes the water level to drop from its initial 18 cm.
  7. Height drop = 800 ÷ 500 = 1.6 cm.
  8. New depth = 18 − 1.6 = 16.4 cm.

answer(a) 800 cm3; (b) 16.4 cm

techniqueVolume of a Cuboid · Milliliters and Liters

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Assuming all 1800 cm³ of the block overflows, ignoring that 1000 cm³ was accommodated in the empty air space.
  • Calculating remaining water depth by dividing total capacity (10,000 cm³) by 500 cm² minus block height.
06

Water Transfer Between Tanks with Unequal Base Areas

heuristicRatio of Heights and Base Areas / Volume Invariance

Tank A has a base of 40 cm by 25 cm and contains water to a depth of 36 cm. Tank B has a base of 30 cm by 20 cm and is empty. Some water is poured from Tank A into Tank B until the depth of water in Tank A is three times the depth of water in Tank B. (a) What was the depth of water in Tank B? (b) How many liters of water were poured from Tank A into Tank B?

36 cmA: 40 by 25B: 30 by 20
At the end A is three times as deep as B.
Ratio of depth: Tank A : Tank B = 3 : 1.
step 1 of 9

Express volume as base area units multiplied by height units: Tank A has 1000 cm² × 3 parts = 3000 volume units, Tank B has 600 cm² × 1 part = 600 volume units.

  1. Ratio of depth: Tank A : Tank B = 3 : 1.
  2. For every 1 cm of height, Tank A holds 1000 cm3 and Tank B holds 600 cm3.
  3. When Tank B has 1 unit of depth and Tank A has 3 units of depth:
  4. Volume in Tank A = 3 × 1000 = 3000 cm3 per unit height.
  5. Volume in Tank B = 1 × 600 = 600 cm3 per unit height.
  6. Total volume per unit height = 3000 + 600 = 3600 cm3.
  7. Total water = 36000 cm3 ⟹ 36000 ÷ 3600 = 10 cm of depth.
  8. (a) Depth in Tank B = 10 cm.
  9. Volume in Tank B = 10 × 600 = 6000 cm3 = 6 liters.

answer(a) 10 cm; (b) 6 liters

techniqueVolume of a Cuboid · Ratio

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Setting the ratio of volumes directly as 3 : 1 without accounting for the different base areas (1000 vs 600).
  • Dividing the initial depth (36 cm) into a 3 : 1 ratio (27 cm and 9 cm), which violates volume conservation.
07

Stepped / L-Shaped Cross-Section Tanks

heuristicSegmented Geometric Sections / Piecewise Filling Rates

A stepped container consists of a wide lower cuboid section and a narrower upper cuboid section. The lower section measures 40 cm by 25 cm with a height of 15 cm. The upper section measures 20 cm by 25 cm with a height of 20 cm, open at the top. A tap delivers water into the container at a constant rate of 2.5 liters/min. (a) How many minutes did it take to fill the lower section completely? (b) How many minutes did it take for the total water level to reach a height of 27 cm from the base?

lower 40 by 25, 15 tallupper 20 by 25
The lower section holds 40 × 25 × 15 = 15 000 cm³, 15 liters.
Tier 1 (Lower block): 40 × 25 × 15 = 15000 cm3 = 15 liters.
step 1 of 8

Split the container into two stacked horizontal tiers: Tier 1 (15 liters) and Tier 2 up to target height (6 liters).

  1. Tier 1 (Lower block): 40 × 25 × 15 = 15000 cm3 = 15 liters.
  2. At 2.5 ℓ/min, Tier 1 takes 15 ÷ 2.5 = 6 minutes.
  3. (a) 6 minutes.
  4. Tier 2 (Upper block up to 27 cm total height):
  5. Height in upper block = 27 − 15 = 12 cm.
  6. Upper volume = 20 × 25 × 12 = 6000 cm3 = 6 liters.
  7. Upper filling time = 6 ÷ 2.5 = 2.4 minutes.
  8. (b) Total time = 6 + 2.4 = 8.4 minutes (8 min 24 s).

answer(a) 6 minutes; (b) 8.4 minutes, that is 8 min 24 s

techniqueVolume of a Cuboid · Rates

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Using the lower section's base area (1000 cm2) for the upper height calculation.
  • Calculating total volume to 27 cm as if the base was uniform throughout.
08

Tilted Rectangular Container Spillage (Prism Geometry)

heuristicRight-Angled Triangular Prism / Air Void Volume Deduction

A rectangular container measuring 30 cm in length, 20 cm in breadth, and 18 cm in height was completely filled with water. The container was then tilted slowly about one of its bottom edges of breadth 20 cm. Water spilled out until the remaining water formed a right-angled triangular cross-section whose legs along the length and height measured 30 cm and 12 cm respectively. (a) How many liters of water were spilled? (b) When the container was returned to sit flat horizontally on its base, what was the depth of the remaining water?

side 30 by 18 = 540 cm²30 by 12: 180
Seen from the side, the full container is 30 by 18.
Side face area of container = 30 × 18 = 540 cm2.
step 1 of 8

View the triangular face as a fraction of the rectangular side face (30 × 18).

  1. Side face area of container = 30 × 18 = 540 cm2.
  2. Side face area occupied by remaining water = 12 × 30 × 12 = 180 cm2.
  3. Fraction of water remaining = 180540 = 13.
  4. Fraction of water spilled = 1 − 13 = 23.
  5. Total capacity = 30 × 20 × 18 = 10800 cm3 = 10.8 liters.
  6. (a) Spilled water: 23 × 10.8 = 7.2 liters.
  7. Remaining depth on flat base is simply 13 of the original height (18 cm):
  8. (b) Remaining depth = 13 × 18 = 6 cm.

answer(a) 7.2 liters; (b) 6 cm

techniqueVolume of a Cuboid · Area of a Triangle

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Forgetting to multiply the triangular cross-section by the breadth (20 cm) to obtain 3D volume.
  • Assuming the remaining water is half of the container (5.4 liters), confusing this with the diagonal corner-to-corner tilt (18 cm leg).
09

Water Height Against Time Graph Interpretation

heuristicGradient-to-Flow Rate Conversion / Multi-Stage Inflow

A rectangular tank with a base measuring 50 cm by 30 cm contained some water at first. At t = 0 min, Tap X was turned on to fill the tank. At t = 10 min, Tap Y was also turned on to assist Tap X. A line graph tracking the water depth shows that the water height was 6 cm at t = 0 min, 16 cm at t = 10 min, and 36 cm at t = 18 min. (a) What was the rate of flow of Tap X in liters per minute? (b) What was the rate of flow of Tap Y in liters per minute?

0101861636minutesdepth, cm
Each centimeter of depth is 1500 cm³, 1.5 liters.
Base area = 1500 cm2. Each 1 cm depth corresponds to 1500 cm3 = 1.5 liters.
step 1 of 8

Model the height increases on a timeline: 10 cm in 10 min gives 1 cm/min, while 20 cm in 8 min gives 2.5 cm/min.

  1. Base area = 1500 cm2. Each 1 cm depth corresponds to 1500 cm3 = 1.5 liters.
  2. Phase 1: Water level rises 10 cm in 10 min ⟹ 1 cm/min.
  3. Tap X delivery rate = 1 cm/min × 1.5 ℓ/cm = 1.5 liters/min.
  4. (a) Tap X rate = 1.5 liters/min.
  5. Phase 2: Water level rises 20 cm in 8 min ⟹ 2.5 cm/min.
  6. Combined delivery rate = 2.5 × 1.5 = 3.75 liters/min.
  7. Tap Y delivery rate = 3.75 − 1.5 = 2.25 liters/min.
  8. (b) Tap Y rate = 2.25 liters/min.

answer(a) 1.5 liters per minute; (b) 2.25 liters per minute

techniqueLine Graphs · Rates

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing 16 cm by 10 min, forgetting to subtract the initial 6 cm water height.
  • Dividing the 36 cm final height by 18 min to find Tap Y's rate directly.
10

Cascading Overflow Across Multi-Tier Tanks

heuristicSequential Capacity Saturation / Multi-Tier Spillway

A fountain has two rectangular tanks, Tank P and Tank Q. Tank P is mounted directly above Tank Q. Tank P measures 30 cm by 20 cm by 25 cm. Tank Q measures 40 cm by 30 cm by 20 cm. Both tanks are initially empty. A tap fills Tank P at a constant rate of 3 liters/min. When Tank P is completely full, excess water overflows from its rim into Tank Q. The tap runs continuously for 11 minutes. (a) How many minutes did it take before water first started overflowing into Tank Q? (b) What was the depth of water in Tank Q at the end of 11 minutes?

P: 30 by 20, 25 tallQ: 40 by 30, 20 tall
Eleven minutes at 3 liters a minute: 33 liters.
Total water delivered in 11 minutes: 11 × 3 = 33 liters.
step 1 of 8

Draw a total water delivery bar for 11 minutes (33 liters) and partition it into Tank P's 15-liter bucket and Tank Q's 18-liter overflow.

  1. Total water delivered in 11 minutes: 11 × 3 = 33 liters.
  2. Tank P capacity: 30 × 20 × 25 = 15000 cm3 = 15 liters.
  3. Time to fill Tank P: 15 ÷ 3 = 5 minutes.
  4. (a) Overflow starts at 5 minutes.
  5. Water overflowing into Tank Q: 33 − 15 = 18 liters = 18000 cm3.
  6. Base of Tank Q: 40 × 30 = 1200 cm2.
  7. Depth in Tank Q: 18000 ÷ 1200 = 15 cm.
  8. (b) Depth = 15 cm.

answer(a) 5 minutes; (b) 15 cm

techniqueVolume of a Cuboid · Milliliters and Liters

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Assuming water enters Tank Q immediately from minute 0, ignoring the 5-minute pre-fill delay of Tank P.
  • Dividing the entire 33 liters by the base area of Tank Q.
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