Pouring and Recasting

When water is poured into another container, or metal is melted and recast into a new shape, its volume stays the same. Use this to find a new depth, whether water overflows, how many glasses or solids can be filled or made, and what is left over.

Before you start

The volume of a cuboid is length × breadth × height. The volume of any solid with a uniform cross-section is area of cross-section × length, or base area × height when it stands on its end (Volume of Prisms). A cylinder of radius $r$ and height $h$ has volume $\pi r^{2}h$ (Volume of Cylinders).
$1 \text{ cm}^{3} = 1$ mL, $1000 \text{ cm}^{3} = 1$ L and $1 \text{ m}^{3} = 1000$ L (Converting Units of Area and Volume).
To find an unknown length, substitute the numbers you know into the formula and solve the equation: $20 \times 15 \times h = 1500$ gives $300h = 1500$ , so $h = 1500 \div 300 = 5$ .

Water in a Container

Water takes the shape of its container. In a tank with a uniform cross-section standing on its base (a cuboid tank, or a cylinder standing upright), the water is itself a cuboid or a cylinder: its base is the inside base of the tank, and its height is the depth of the water.

Every question in this lesson rests on one fact: when water is poured, or a solid is melted down and made into a new shape, the volume stays the same. Nothing is made and, unless some spills, nothing is lost; only the shape changes.

What happens to the water level when a solid is lowered into the water, or lifted out, is the next lesson, Rising and Falling Water Levels.

Volume of water in a tank

Volume of water = inside base area of the tank × depth of the water

A cuboid tank: volume of water = internal length × internal breadth × depth.

A cylindrical tank of internal radius $r$ : volume of water = $\pi r^{2}$ × depth.

Use the internal (inside) measurements: the walls of a tank take up room, so the water fills only the inside.

Example 1: The depth from the volume

(a) Mrs Goh pours 6 L of water into an empty fish tank. The inside of the tank is 30 cm long, 25 cm wide and 20 cm high. Find the depth of the water.

(b) Siti pours 2 L of water into an empty cylindrical vase of internal radius 8 cm and height 25 cm. Using the $\pi$ button on your calculator, find the depth of the water, correct to 3 significant figures.

Solution

(a) Change the volume to $\text{cm}^{3}$ , to match the lengths: $6 \text{ L} = 6 \times 1000 = 6000 \text{ cm}^{3}$ .

Let the depth of the water be $d$ cm.

Volume of water = inside base area × depth

$6000 = 30 \times 25 \times d$

$6000 = 750d$

$d = 6000 \div 750 = 8$

The depth, 8 cm, is less than the height of the tank, 20 cm, so the water fits.

(b) $2 \text{ L} = 2000 \text{ cm}^{3}$ . Let the depth be $d$ cm.

Volume of water $= \pi r^{2} \times$ depth

$2000 = \pi \times 8^{2} \times d$

$64\pi d = 2000$

$d = 2000 \div (64\pi) = 9.947\ldots \approx 9.95$ (divide both sides by $64\pi$ )

On the calculator, put brackets round $64\pi$ : keyed as $2000 \div 64 \times \pi$ it would give 98.2, the wrong answer. The depth, 9.95 cm, is less than the vase's 25 cm, so the water fits.

(a) The water is 8 cm deep. (b) The water is about 9.95 cm deep.

Pouring from One Container into Another

When all the water in one container is poured into another, its volume does not change. So work in two steps: find the volume of the water in the first container, then use that same volume in the second container to find what you need, such as the new depth.

The water overflows if its volume is more than the capacity of the second container (the volume the container holds when it is full). The volume that spills is the difference.

Pouring

Volume of water poured out of the first container = volume of water in the second container (when none is spilt).

If the volume poured in is more than the capacity of the second container, the water overflows, and

volume that overflows = volume poured in − capacity of the container.

Example 2: From a jug into a rectangular container

A cylindrical jug has an internal radius of 7 cm and holds water to a depth of 15 cm. All the water is poured into an empty rectangular container whose inside is 22 cm long, 20 cm wide and 15 cm high. Find the depth of the water in the container. Take $\pi = \frac{22}{7}$ .

Solution

Volume of water in the jug $= \pi r^{2} \times$ depth

$= \frac{22}{7} \times 7^{2} \times 15$

$= 22 \times 7 \times 15$ (one 7 cancels)

$= 2310 \text{ cm}^{3}$

The volume stays the same, so the water in the container is also $2310 \text{ cm}^{3}$ . Let its depth be $d$ cm.

$22 \times 20 \times d = 2310$

$440d = 2310$

$d = 2310 \div 440 = 5.25$

Check: 5.25 cm is less than the container's height of 15 cm, so none overflows.

The water in the container is 5.25 cm deep.

Example 3: Does it overflow?

A cylindrical container of internal radius 10 cm and height 30 cm is full of water. All of it is poured into an empty rectangular basin whose inside is 40 cm long, 25 cm wide and 9 cm high.

(a) Show that the water overflows. (b) Find the volume of water that overflows, in millilitres. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

Solution

(a) Volume of water $= \pi r^{2}h = \pi \times 10^{2} \times 30 = 3000\pi \text{ cm}^{3} = 9424.77\ldots \text{ cm}^{3}$

Capacity of the basin $= 40 \times 25 \times 9 = 9000 \text{ cm}^{3}$

The water, $9424.77\ldots \text{ cm}^{3}$ , is more than the basin holds, $9000 \text{ cm}^{3}$ , so it overflows.

(b) Volume that overflows = volume poured in − capacity

$= 3000\pi - 9000$

$= 424.77\ldots \text{ cm}^{3} \approx 425 \text{ cm}^{3}$

Since $1 \text{ cm}^{3} = 1$ mL, this is about 425 mL. (Keep the full value of $3000\pi$ until the last line: rounding it to 9420 first would give 420 mL, which is wrong.)

(a) The water ( $9424.77\ldots \text{ cm}^{3}$ ) is more than the capacity of the basin ( $9000 \text{ cm}^{3}$ ), so it overflows. (b) About 425 mL of water overflows.

Example 4: How many glasses can be filled?

A rectangular container whose inside is 25 cm long and 18 cm wide holds water to a depth of 12 cm. The water is poured into identical cylindrical glasses, each of internal radius 3.5 cm and height 10 cm, filling each glass to the brim before starting the next. Take $\pi = \frac{22}{7}$ .

(a) How many glasses can be filled? (b) How much water is left over?

Solution

Volume of water $= 25 \times 18 \times 12 = 5400 \text{ cm}^{3}$

Volume of one glass $= \pi r^{2}h = \frac{22}{7} \times 3.5^{2} \times 10 = \frac{22}{7} \times 12.25 \times 10 = 385 \text{ cm}^{3}$

(a) Number of glasses $= 5400 \div 385 = 14.025\ldots$

Only a full glass counts, so round down: 14 glasses are filled. There is not enough water left for a 15th.

(b) Water used $= 14 \times 385 = 5390 \text{ cm}^{3}$

Water left over $= 5400 - 5390 = 10 \text{ cm}^{3}$ , which is 10 mL.

(a) 14 glasses can be filled. (b) 10 mL of water is left over.

Melting and Recasting

A solid made of metal (or wax, or clay) can be melted and recast: poured into a mould to set as a new solid of a different shape. As with water, none of the metal is made or lost, so the total volume stays the same.

When a large solid is recast into many identical small ones, the number made is a whole number: divide the volumes, then round down, since part of a solid is not a solid. The metal left over is the volume of the large solid less the volume of the small solids made.

Melting and recasting

Total volume of the new solid or solids = volume of the solid melted down.

Number of identical small solids made = the whole number part of (volume of the large solid ÷ volume of one small solid).

Metal left over = volume of the large solid − (number made × volume of one small solid).

Example 5: How many cubes, and how much left over?

A metal cuboid 20 cm long, 12 cm wide and 9 cm high is melted and recast into as many cubes of edge 4 cm as possible.

(a) How many cubes are made? (b) What volume of metal is left over?

Solution

Volume of the cuboid $= 20 \times 12 \times 9 = 2160 \text{ cm}^{3}$

Volume of one cube $= 4^{3} = 64 \text{ cm}^{3}$

(a) $2160 \div 64 = 33.75$

There is enough metal for 33 whole cubes but not for a 34th, so round down: 33 cubes.

(b) Metal used $= 33 \times 64 = 2112 \text{ cm}^{3}$

Metal left over $= 2160 - 2112 = 48 \text{ cm}^{3}$

Melting is what lets you divide the volumes. Cutting 4 cm cubes from the solid block without melting it is a different question: you count along each edge ( $5 \times 3 \times 2 = 30$ cubes, since only 2 fit in the 9 cm height), as in Converting Units of Area and Volume.

(a) 33 cubes are made. (b) $48 \text{ cm}^{3}$ of metal is left over.

Example 6: The height of the new solid

A solid metal cylinder of radius 7 cm and height 12 cm is melted and recast into a cuboid whose base is 22 cm long and 14 cm wide. Find the height of the cuboid. Take $\pi = \frac{22}{7}$ .

(not drawn to scale)

Solution

Volume of the cylinder $= \pi r^{2} \times \text{height} = \frac{22}{7} \times 7^{2} \times 12 = 22 \times 7 \times 12 = 1848 \text{ cm}^{3}$

The volume stays the same, so the cuboid has volume $1848 \text{ cm}^{3}$ . Let its height be $h$ cm.

$22 \times 14 \times h = 1848$

$308h = 1848$

$h = 1848 \div 308 = 6$

The cuboid is 6 cm high.

Common mistakes

Using the outside measurements. Water fills only the inside of a container: use its internal length, breadth or radius.
Using the diameter as the radius. A dispenser of diameter 24 cm has radius 12 cm.
Mixing litres and cubic centimetres. Change litres to $\text{cm}^{3}$ ( $1 \text{ L} = 1000 \text{ cm}^{3}$ ) before you use lengths in cm.
Forgetting the water already there. When water is added to a tank that is not empty, add the volumes (or find the rise from the water added only).
Rounding a count up. 33.75 cubes or 14.025… glasses means 33 cubes or 14 glasses: part of a cube is not a cube, and a part-filled glass is not full.
Dividing by $64\pi$ without brackets. $2000 \div (64\pi)$ means 2000 divided by the whole of $64\pi$ . Keyed in as $2000 \div 64 \times \pi$ it gives 98.2, not 9.95: put brackets round $64\pi$ on your calculator.
Rounding too early. Keep values such as $3000\pi$ in full until the last line.

Practice

Try each question before you open its solution. Each question says which value of π to use; give answers that are not exact correct to 3 significant figures.

Question 1

Hui Min pours 1.5 L of water into an empty cylindrical jug of internal radius 6 cm and height 20 cm. Find the depth of the water in the jug. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

$1.5 \text{ L} = 1500 \text{ cm}^{3}$ . Let the depth be $d$ cm.

Volume of water $= \pi r^{2} \times$ depth

$1500 = \pi \times 6^{2} \times d$

$36\pi d = 1500$

$d = 1500 \div (36\pi) = 13.26\ldots \approx 13.3$

Check: 13.3 cm is less than the jug's 20 cm, so the water fits.

The water is about 13.3 cm deep.

Question 2

The inside of a fish tank is 50 cm long, 30 cm wide and 35 cm high, and it holds water 18 cm deep. Mei Ling pours in another 6 L of water. Find the new depth of the water.

ShowHide solution

$6 \text{ L} = 6000 \text{ cm}^{3}$ . The water added spreads over the whole base of the tank.

Inside base area $= 50 \times 30 = 1500 \text{ cm}^{2}$

Let the rise be $x$ cm: $1500x = 6000$ , so $x = 4$ .

New depth $= 18 + 4 = 22$ cm (below the top, 35 cm).

Or: water now $= 1500 \times 18 + 6000 = 33\,000 \text{ cm}^{3}$ , and $1500d = 33\,000$ gives $d = 22$ .

The water is now 22 cm deep.

Question 3

Water 12 cm deep in a cylindrical jug of internal radius 6 cm is all poured into an empty cylindrical vase of internal radius 4 cm and height 30 cm. Find the depth of the water in the vase.

ShowHide solution

Volume of water $= \pi \times 6^{2} \times 12 = 432\pi \text{ cm}^{3}$

Let the depth in the vase be $d$ cm.

$\pi \times 4^{2} \times d = 432\pi$

$16\pi d = 432\pi$

$16d = 432$

$d = 27$

Check: 27 cm is less than the vase's height of 30 cm, so none overflows.

The water in the vase is 27 cm deep.

Question 4

A solid metal cube of edge 10 cm is melted and recast into a solid cylinder of radius 5 cm. Find the height of the cylinder. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

(not drawn to scale)

ShowHide solution

Volume of the cube $= 10^{3} = 1000 \text{ cm}^{3}$

Let the height of the cylinder be $h$ cm. The volume stays the same:

$\pi \times 5^{2} \times h = 1000$

$25\pi h = 1000$

$h = 1000 \div (25\pi) = 12.73\ldots \approx 12.7$

The cylinder is about 12.7 cm high.

Question 5

A cylindrical container of internal radius 7 cm and height 20 cm is full of water. Ali pours all of it into an empty rectangular tank whose inside is 25 cm long, 16 cm wide and 9 cm high. Does the water overflow? If not, how far below the top of the tank is the water? Take $\pi = \frac{22}{7}$ .

ShowHide solution

Volume of water $= \frac{22}{7} \times 7^{2} \times 20 = 22 \times 7 \times 20 = 3080 \text{ cm}^{3}$

Capacity of the tank $= 25 \times 16 \times 9 = 3600 \text{ cm}^{3}$

$3080 < 3600$ , so the water does not overflow.

Let the depth be $d$ cm: $25 \times 16 \times d = 3080$ , $400d = 3080$ , $d = 7.7$ .

Distance below the top $= 9 - 7.7 = 1.3$ cm

No, it does not overflow: the water is 7.7 cm deep, 1.3 cm below the top of the tank.

Question 6

A cylindrical pot of internal radius 14 cm and height 20 cm is full of soup. All of it is poured into an empty cylindrical container of internal radius 10.5 cm and height 30 cm. Show that the soup overflows, and find how much overflows, in litres. Take $\pi = \frac{22}{7}$ .

ShowHide solution

Volume of soup $= \frac{22}{7} \times 14^{2} \times 20 = 22 \times 28 \times 20 = 12\,320 \text{ cm}^{3}$

Capacity of the container $= \frac{22}{7} \times 10.5^{2} \times 30 = 22 \times 1.5 \times 10.5 \times 30 = 10\,395 \text{ cm}^{3}$

$12\,320 > 10\,395$ , so the soup overflows.

Volume that overflows $= 12\,320 - 10\,395 = 1925 \text{ cm}^{3}$

$1925 \text{ cm}^{3} = 1925 \div 1000 = 1.925$ L

The soup overflows, and 1.925 L of it spills.

Question 7

(a) A metal cuboid 16 cm long, 8 cm wide and 4 cm high is melted and recast into a cube. Find the length of an edge of the cube.

(b) A solid metal cylinder of radius 3 cm and length 10 cm is melted and drawn out into a wire, a long thin cylinder of radius 0.1 cm. Find the length of the wire, in metres.

ShowHide solution

(a) Volume of the cuboid $= 16 \times 8 \times 4 = 512 \text{ cm}^{3}$

Let the edge of the cube be $x$ cm: $x^{3} = 512$ , so $x = \sqrt[3]{512} = 8$ .

(b) Volume of the cylinder $= \pi \times 3^{2} \times 10 = 90\pi \text{ cm}^{3}$

Let the length of the wire be $l$ cm. The volume stays the same:

$\pi \times 0.1^{2} \times l = 90\pi$

$0.01\pi l = 90\pi$

$0.01l = 90$

$l = 90 \div 0.01 = 9000$

$9000 \text{ cm} = 90$ m

(a) The edge of the cube is 8 cm. (b) The wire is 90 m long.

Question 8

A metal cuboid 30 cm long, 20 cm wide and 10 cm high is melted and recast into as many solid cylinders of radius 2 cm and height 5 cm as possible. (a) How many cylinders are made? (b) What volume of metal is left over? Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

Volume of the cuboid $= 30 \times 20 \times 10 = 6000 \text{ cm}^{3}$

Volume of one cylinder $= \pi \times 2^{2} \times 5 = 20\pi = 62.83\ldots \text{ cm}^{3}$

(a) $6000 \div (20\pi) = 95.49\ldots$ , so round down: 95 cylinders (not enough metal for a 96th).

(b) Metal used $= 95 \times 20\pi = 1900\pi \text{ cm}^{3}$

Left over $= 6000 - 1900\pi = 30.97\ldots \approx 31.0 \text{ cm}^{3}$

(a) 95 cylinders are made. (b) About $31.0 \text{ cm}^{3}$ of metal is left over.

Question 9

A solid metal cylinder of radius 6 cm and height 20 cm is melted and recast into as many small solid cylinders of radius 1.5 cm and height 3.5 cm as possible. (a) How many small cylinders are made? (b) What volume of metal is left over? Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

Volume of the large cylinder $= \pi \times 6^{2} \times 20 = 720\pi \text{ cm}^{3}$

Volume of one small cylinder $= \pi \times 1.5^{2} \times 3.5 = 7.875\pi \text{ cm}^{3}$

(a) Number $= \frac{720\pi}{7.875\pi} = \frac{720}{7.875} = 91.42\ldots$ , so round down: 91 small cylinders (not enough metal for a 92nd).

(b) Metal used $= 91 \times 7.875\pi = 716.625\pi \text{ cm}^{3}$

Left over $= 720\pi - 716.625\pi = 3.375\pi = 10.60\ldots \text{ cm}^{3} \approx 10.6 \text{ cm}^{3}$

(a) 91 small cylinders are made. (b) About $10.6 \text{ cm}^{3}$ of metal is left over.

Question 10

At Siti's class party, a cylindrical drinks dispenser of internal diameter 24 cm holds a drink to a depth of 30 cm. The drink is poured into cylindrical cups of internal radius 3.5 cm, each filled to a depth of 8 cm. (a) How many cups can be filled to that depth? (b) How much drink is left over, in millilitres? Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

Radius of the dispenser $= 24 \div 2 = 12$ cm

Volume of drink $= \pi \times 12^{2} \times 30 = 4320\pi \text{ cm}^{3}$

Volume in one cup $= \pi \times 3.5^{2} \times 8 = 98\pi \text{ cm}^{3}$

(a) Number of cups $= \frac{4320\pi}{98\pi} = \frac{4320}{98} = 44.08\ldots$ , so round down: 44 cups. The drink left over is not enough to fill a 45th cup to 8 cm.

(b) Left over $= 4320\pi - 44 \times 98\pi = 4320\pi - 4312\pi = 8\pi = 25.13\ldots \text{ cm}^{3} \approx 25.1 \text{ cm}^{3}$ , which is about 25.1 mL.

(a) 44 cups can be filled. (b) About 25.1 mL of drink is left over.

Question 11

A rectangular tank whose inside is 60 cm long and 40 cm wide holds water 25 cm deep. The water is used to fill identical containers, each a cuboid 15 cm long, 12 cm wide and 20 cm high inside, one after another. (a) How many containers can be filled completely? (b) What depth of water is left in the tank?

ShowHide solution

Volume of water $= 60 \times 40 \times 25 = 60\,000 \text{ cm}^{3}$

Capacity of one container $= 15 \times 12 \times 20 = 3600 \text{ cm}^{3}$

(a) $60\,000 \div 3600 = 16.66\ldots$ , so round down: 16 containers are filled. What is left cannot fill a 17th.

(b) Water used $= 16 \times 3600 = 57\,600 \text{ cm}^{3}$ , so water left $= 60\,000 - 57\,600 = 2400 \text{ cm}^{3}$ .

Let the depth left be $d$ cm: $60 \times 40 \times d = 2400$ , so $2400d = 2400$ and $d = 1$ .

(a) 16 containers can be filled completely. (b) 1 cm of water is left in the tank.

Key points

Water poured from one container into another, and metal melted and recast, keep the same volume.
Volume of water in a tank = inside base area × depth; use internal measurements.
To find a depth or a height, substitute the numbers into the volume formula and solve the equation.
Water overflows when its volume is more than the capacity of the container it is poured into; the overflow is the difference.
Number of glasses filled or solids made: divide the volumes and round down; what is left = total − number × one.
Recasting by melting divides the volumes; cutting a block without melting counts along each edge instead.
Change litres to $\text{cm}^{3}$ before working with lengths in cm: $1 \text{ L} = 1000 \text{ cm}^{3}$ , $1 \text{ cm}^{3} = 1$ mL.
Use the value of $\pi$ the question gives; bracket a divisor with $\pi$ in it; round only at the end.