Volume and water tanks · applications

Applications: Volume and Water Tanks

10 problems

PSLE · SAT · GCSE Higher

01

Filling a 120-Liter Tank While a Drain Runs from 09:08

heuristicWork in stages: one tap alone, then both taps, which together gain the difference of their rates

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.
The tank holds 60 × 40 × 50 = 120000 cm3, which is 120 liters.
step 1 of 8

Draw the tank's 120 liters as a bar and mark three quarters of it. Tap A fills the first part alone; after 09:08 the drain takes away part of each minute's water.

All steps
  1. The tank holds 60 × 40 × 50 = 120000 cm3, which is 120 liters.
  2. Three quarters of it is 34 × 120 = 90 liters.
  3. From 09:00 to 09:08 only Tap A runs: 8 × 4.5 = 36 liters.
  4. (a) At 09:08 the tank holds 36 liters.
  5. To reach 90 liters it needs 90 − 36 = 54 liters more.
  6. From 09:08 each minute Tap A adds 4.5 liters and Tap B drains 1.5, so the tank gains 4.5 − 1.5 = 3 liters a minute.
  7. The 54 liters take 54 ÷ 3 = 18 minutes.
  8. (b) The tank is three quarters full 18 minutes after 09:08, at 09:26.

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

techniqueVolume of a Cuboid · Rates

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Filling to the full 120 liters. The taps stop at three quarters, 90 liters, so only 90 − 36 = 54 liters are left at 09:08.
  • Using the net 3 liters a minute from 09:00. The drain starts at 09:08; before that, Tap A fills at the full 4.5 liters a minute.
  • Adding the drain's 1.5 liters to Tap A's 4.5. Tap B takes water out, so the tank gains 4.5 − 1.5 = 3 liters a minute.
02

Two Tanks Joined at the Base: the Water Levels Out

heuristicWhen the levels are equal, the water divides between the tanks in the ratio of their base areas

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.
Water flows through the pipe until both surfaces are level, so the two tanks end with the same depth.
step 1 of 8

With equal depths, each tank holds water in proportion to its base area, so share the water out in units of 3 to 2.

All steps
  1. Water flows through the pipe until both surfaces are level, so the two tanks end with the same depth.
  2. At the same depth, each tank holds water in proportion to its base area: 1200 : 800 = 3 : 2.
  3. So the water ends as 3 units in Tank A and 2 units in Tank B, 5 units in all.
  4. The water is 1200 × 30 = 36000 cm3, so one unit is 36000 ÷ 5 = 7200 cm3.
  5. Tank B gets 2 units: 2 × 7200 = 14400 cm3.
  6. (b) 14400 cm3 = 14.4 liters flowed into Tank B.
  7. Its depth is 14400 ÷ 800 = 18 cm. Check: Tank A keeps 3 units, 21600 cm3, and 21600 ÷ 1200 = 18 cm too.
  8. (a) The water ends 18 cm deep in both tanks.

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

techniqueVolume of a Cuboid · Ratio

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Halving the 30 cm. The tanks have different bases, so equal depths do not mean equal halves of the water: the 36000 cm3 spreads over 2000 cm2, 18 cm deep.
  • Giving the answer to (b) in cubic centimeters when liters are asked: 1000 cm3 is 1 liter, so 14400 cm3 is 14.4 liters.
03

Three Metal Cubes Lowered into a Tank of Water

heuristicA sunken solid pushes the water up by its own volume, spread over the base

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.
The base of the tank is 50 × 30 = 1500 cm2.
step 1 of 6

Picture the volume of the cubes as an extra layer of water lying over the whole floor of the tank.

All steps
  1. The base of the tank is 50 × 30 = 1500 cm2.
  2. Three cubes of edge 10 cm bring 3 × 10 × 10 × 10 = 3000 cm3 under the water.
  3. The water they push aside has to go up, and it spreads over the whole base as a layer.
  4. The layer is 3000 ÷ 1500 = 2 cm thick.
  5. (a) The water rises 2 cm.
  6. (b) The new depth is 16 + 2 = 18 cm. That is above the cubes' 10 cm, so they are fully under water.

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

techniqueVolume of a Cuboid

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Adding the cubes' 10 cm to the depth. The cubes take up only part of the floor, so their volume, spread over the whole base, raises the water only 2 cm.
  • Dividing the 3000 cm3 by the tank's height. A rise in depth is volume divided by the base area, 1500 cm2.
04

A Tall Bar Stood in a Tank: the Water Has Less Floor

heuristicA solid that stands out of the water takes floor away, and the same water stands deeper on the floor that is left

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.
The floor of the tank is 30 × 20 = 600 cm2.
step 1 of 8

Look at the floor. The bar stands on 100 cm2 of it and pokes out of the water, so the same water must stand on the 500 cm2 that is left.

All steps
  1. The floor of the tank is 30 × 20 = 600 cm2.
  2. The bar stands on 10 × 10 = 100 cm2 of it, and its top stays above the water.
  3. So the water stands on the floor around the bar: 600 − 100 = 500 cm2.
  4. The water is still the 600 × 10 = 6000 cm3 there was at first.
  5. Spread over 500 cm2, it stands 6000 ÷ 500 = 12 cm deep.
  6. (a) The new depth is 12 cm. Check: the water and the bar's wet part fill 600 × 12 = 7200 cm3, and 6000 + 100 × 12 = 7200.
  7. The bar is under water up to 12 cm of its 25 cm height, and its cross-section is the same all the way up.
  8. (b) 1225 of the bar's volume is under water.

answer(a) 12 cm; (b) 1225

techniqueVolume of a Cuboid · Area and Perimeter

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Adding the whole bar's volume, 2500 cm3, to the water. The top of the bar stays dry; only the part below the surface pushes water aside.
  • Dividing the 6000 cm3 by the whole 600 cm2 floor. The bar takes 100 cm2 of it, so the water has only 500 cm2.
05

A Block Lowered into a Nearly Full Container Makes It Overflow

heuristicOnly the part of the block's volume beyond the empty space at the top spills over

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.
The base is 25 × 20 = 500 cm2. Above the 18 cm of water there are 2 cm of space: 500 × 2 = 1000 cm3.
step 1 of 8

Draw the 2 cm of space above the water. The block fills that space first, and whatever of its volume does not fit goes over the brim.

All steps
  1. The base is 25 × 20 = 500 cm2. Above the 18 cm of water there are 2 cm of space: 500 × 2 = 1000 cm3.
  2. The block is 15 × 10 × 12 = 1800 cm3. It is 12 cm tall, so it ends up fully under water.
  3. As the block goes down, the water rises and fills the 1000 cm3 of space up to the brim.
  4. The rest of the block's volume pushes water over the brim: 1800 − 1000 = 800 cm3.
  5. (a) 800 cm3 of water overflowed.
  6. Check: the container now holds 9000 + 1800 − 800 = 10000 cm3, exactly 500 × 20, full to the brim.
  7. Lift the block out. The water is the 500 × 18 = 9000 cm3 there was at first, less the 800 that overflowed: 8200 cm3.
  8. (b) It stands 8200 ÷ 500 = 16.4 cm deep, 800 ÷ 500 = 1.6 cm below the first 18 cm.

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

techniqueVolume of a Cuboid · Milliliters and Liters

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Counting all 1800 cm3 as overflow. The first 1000 cm3 only fills the space above the water.
  • Giving 18 cm for (b), as if no water had been lost. The 800 cm3 that overflowed is gone, so the water stands 1.6 cm lower.
06

Pouring from One Tank to Another Until One Is Three Times as Deep

heuristicCount the water in units of depth: one unit in Tank B goes with three in Tank A

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.
At the end, Tank A is three times as deep as Tank B: call Tank B's depth 1 unit and Tank A's 3 units.
step 1 of 9

Measure depth in units. Tank A ends 3 units deep when Tank B is 1 unit deep, so find the volume that one such set of depths holds.

All steps
  1. At the end, Tank A is three times as deep as Tank B: call Tank B's depth 1 unit and Tank A's 3 units.
  2. Each centimeter of depth in Tank A holds 40 × 25 = 1000 cm3.
  3. Each centimeter of depth in Tank B holds 30 × 20 = 600 cm3.
  4. Try a unit of 1 cm. Tank A, 3 cm deep, would hold 3 × 1000 = 3000 cm3.
  5. Tank B, 1 cm deep, would hold 600 cm3.
  6. Together that is 3000 + 600 = 3600 cm3 of water for every centimeter in a unit.
  7. All the water, 1000 × 36 = 36000 cm3, is still in the two tanks, so a unit is 36000 ÷ 3600 = 10 cm.
  8. (a) Tank B is 10 cm deep, and Tank A 3 × 10 = 30 cm.
  9. (b) Tank B holds 600 × 10 = 6000 cm3, so 6 liters were poured. Check: Tank A went from 36 cm to 30 cm, 1000 × 6 = 6000 cm3.

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

techniqueVolume of a Cuboid · Ratio

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Splitting the water 3 : 1 by volume. The 3 : 1 is about depths; the tanks hold different amounts per centimeter, 1000 and 600 cm3.
  • Splitting the 36 cm into 27 cm and 9 cm. That ignores the different bases and does not keep the water's volume: 27000 + 5400 = 32400, not 36000.
07

Filling a Stepped Container: Wide Below, Narrow Above

heuristicFill one section at a time, each with its own base area

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.
The lower section holds 40 × 25 × 15 = 15000 cm3, which is 15 liters.
step 1 of 8

Split the container into its two sections. Water fills the wide lower section first, then rises into the narrow upper one; each has its own base.

All steps
  1. The lower section holds 40 × 25 × 15 = 15000 cm3, which is 15 liters.
  2. At 2.5 liters a minute it fills in 15 ÷ 2.5 = 6 minutes.
  3. (a) 6 minutes.
  4. Above 15 cm the water rises in the narrow upper section, whose base is 20 × 25 = 500 cm2.
  5. A level of 27 cm is 27 − 15 = 12 cm into the upper section.
  6. Those 12 cm hold 20 × 25 × 12 = 6000 cm3, which is 6 liters.
  7. They take 6 ÷ 2.5 = 2.4 minutes more.
  8. (b) 6 + 2.4 = 8.4 minutes, and 0.4 minute is 0.4 × 60 = 24 seconds: 8 minutes 24 seconds.

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 for the upper 12 cm. Above 15 cm the water rises in the narrow section, 500 cm2, not 1000 cm2.
  • Reading 8.4 minutes as 8 minutes 40 seconds. A tenth of a minute is 6 seconds, so 0.4 minute is 24 seconds.
08

A Full Container Tilted, Then Lowered Back Until the Water Just Covers the Base

heuristicSeen from the side, the water left is a fraction of the side; the same fraction of the water is left

A rectangular container 30 cm long, 20 cm wide and 18 cm high was full of water. It was tilted about one of its 20 cm bottom edges, and water spilled out. It was then lowered back a little, until the water just covered the whole base: seen from the side, the water formed a right triangle with legs of 30 cm along the base and 12 cm up the lower end. (a) How many liters of water were spilled? (b) The container was then set flat on its base. How deep was the water?

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

Every slice of the container across its width looks the same, so compare areas on the side. The fraction of the side that is wet is the fraction of the water that is left.

All steps
  1. Seen from the side, the full container is a rectangle 30 × 18 = 540 cm2.
  2. The water left is a right triangle with legs 30 and 12: 12 × 30 × 12 = 180 cm2.
  3. Every slice across the 20 cm width is the same, so the water left is the same fraction of the full container as the triangle is of the rectangle: 180540 = 13.
  4. So 1 − 13 = 23 of the water spilled.
  5. The full container held 30 × 20 × 18 = 10800 cm3, which is 10.8 liters.
  6. (a) 23 × 10.8 = 7.2 liters spilled.
  7. Set flat, the third of the water that is left fills a third of the container's height, because the base is the whole base again.
  8. (b) 13 × 18 = 6 cm deep.

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

techniqueVolume of a Cuboid · Area of a Triangle

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Stopping at the triangle's area, 180 cm2. The water is a prism 20 cm wide: 180 × 20 = 3600 cm3, which is 3.6 liters left.
  • Taking half the water as left, as if the surface ran corner to corner. That needs legs of 30 and 18; with 12 up the lower end, a third is left.
09

Reading Two Taps' Rates from a Graph of Water Depth

heuristicTurn each rise in depth into liters with the base area, then take one rate from the other

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.
The base is 50 × 30 = 1500 cm2, so each centimeter of depth is 1500 cm3, 1.5 liters.
step 1 of 8

Read the graph in two stretches, Tap X alone and both taps together. A rise in depth times the base area is the water that came in.

All steps
  1. The base is 50 × 30 = 1500 cm2, so each centimeter of depth is 1500 cm3, 1.5 liters.
  2. From 0 to 10 minutes only Tap X runs, and the depth goes from 6 cm to 16 cm: 10 cm in 10 minutes, 1 cm a minute.
  3. 1 cm a minute is 1.5 liters a minute.
  4. (a) Tap X gives 1.5 liters a minute.
  5. From 10 to 18 minutes both taps run, and the depth goes from 16 cm to 36 cm: 20 cm in 8 minutes, 2.5 cm a minute.
  6. That is 2.5 × 1.5 = 3.75 liters a minute for the two taps together.
  7. Tap X still gives 1.5 of them, so Tap Y gives 3.75 − 1.5 = 2.25.
  8. (b) Tap Y gives 2.25 liters a minute.

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

techniqueLine Graphs · Rates

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Reading 16 cm in 10 minutes as Tap X's rise. The tank already held 6 cm at the start, so Tap X added 16 − 6 = 10 cm.
  • Taking the second stretch's 3.75 liters a minute as Tap Y's rate. Tap X is still running then; Tap Y's share is 3.75 − 1.5 = 2.25.
10

A Tank That Overflows into the Tank Below

heuristicThe upper tank must fill before any water reaches the lower one

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.
In 11 minutes at 3 liters a minute the tap gives 11 × 3 = 33 liters.
step 1 of 8

Draw the 11 minutes of water as one bar. The first part fills Tank P; only what comes after that reaches Tank Q.

All steps
  1. In 11 minutes at 3 liters a minute the tap gives 11 × 3 = 33 liters.
  2. Tank P holds 30 × 20 × 25 = 15000 cm3, which is 15 liters.
  3. It fills in 15 ÷ 3 = 5 minutes, and only then does water spill over.
  4. (a) Water starts to overflow into Tank Q after 5 minutes.
  5. The other 33 − 15 = 18 liters, 18000 cm3, spill into Tank Q.
  6. Tank Q's base is 40 × 30 = 1200 cm2.
  7. The water in Tank Q stands 18000 ÷ 1200 = 15 cm deep, below its 20 cm height, so none is lost.
  8. (b) Tank Q is 15 cm deep after 11 minutes.

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

techniqueVolume of a Cuboid · Milliliters and Liters

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Letting Tank Q fill from the start. No water reaches it until Tank P is full, after 5 minutes.
  • Dividing all 33 liters by Tank Q's base. Tank P keeps 15 of them, so Tank Q gets 18.