Compound Measures

Worked example

1. 1

   $\text{mass}=31.5\text{ g},\text{density}=10.5{\text{ g/cm}}^3$ — A silver coin. We know the mass and the density, and we want the volume.

2. 2

   $\text{volume}=\frac{\text{mass}}{\text{density}}$ — Rearrange the formula so volume is alone: mass ÷ density.

3. 3

   $=\frac{31.5}{10.5}$ — Put the numbers in: the mass on top, the density underneath.

4. 4

   $=3{\text{ cm}}^3$ — 31.5 ÷ 10.5 = 3. Grams ÷ g/cm³ leaves cm³, so the answer is a volume.

Worked example

1. 1

   $V=4\times 4\times 4=64{\text{ cm}}^3$ — A metal cube of side 4 cm. Find its volume first, because density needs it: side × side × side.

2. 2

   $0.576\text{ kg}=0.576\times 1000=576\text{ g}$ — The density will be in g/cm³, so change the mass to grams. 1 kg = 1000 g.

3. 3

   $\text{density}=\frac{576}{64}$ — Density is mass ÷ volume.

4. 4

   $=9{\text{ g/cm}}^3$ — 576 ÷ 64 = 9. Each 1 cm³ of this metal weighs 9 grams.

Worked example

1. 1

   $2\text{ kg}=2000\text{ g}$ — A statue must weigh at least 2 kg. Change it to grams, to match the 576 g of one cube.

2. 2

   $2000÷576=3.47\dots$ — Divide the total mass by the mass of one cube to see how many cubes it takes.

3. 3

   $3\times 576=1728\text{ g}$ — Check 3 cubes: 1728 g is less than 2000 g, so 3 cubes is not enough.

4. 4

   $4\text{ cubes}$ — Round up, not to the nearest: you need a whole extra cube to reach 2 kg.

Worked example

1. 1

   $6\times 5=30,6\times 2=12,5\times 2=10{\text{ m}}^2$ — A box 6 m by 5 m by 2 m pushes with a force of 120 newtons. It has three different faces, so find the area of each.

2. 2

   $\frac{120}{10}=12{\text{ N/m}}^2$ — Greatest pressure: use the smallest face, because dividing by a smaller area gives a bigger answer.

3. 3

   $\frac{120}{30}=4{\text{ N/m}}^2$ — Least pressure: use the biggest face.

4. 4

   $12-4=8{\text{ N/m}}^2$ — Some questions ask for the greatest minus the least. Take one away from the other.

Worked example

1. 1

   $\text{area}=\frac{\text{force}}{\text{pressure}}$ — We want the area, so rearrange: force ÷ pressure.

2. 2

   $=\frac{90}{2.5}=36{\text{ cm}}^2$ — 90 ÷ 2.5 = 36. This is the base area, the face touching the table.

3. 3

   $36\times 10=360{\text{ cm}}^3$ — The solid is a prism 10 cm tall. A prism's volume is base area × height.

Worked example

1. 1

   $30\times 20=600{\text{ cm}}^2$ — A crate's base is 30 cm by 20 cm. Find the area on the floor first.

2. 2

   $\text{force}=\text{pressure}\times \text{area}$ — The pressure is 0.15 newtons per cm². We want the force, so multiply.

3. 3

   $0.15\times 600=90\text{ N}$ — 0.15 × 600 = 90. N/cm² × cm² leaves newtons.

Worked example

1. 1

   $\text{volume}=\frac{1068}{8.9}=120{\text{ cm}}^3$ — A cuboid of metal weighs 1068 g and its density is 8.9 g/cm³. Volume = mass ÷ density.

2. 2

   $8\times 5\times h=120$ — It is 8 cm long and 5 cm wide, with height h. Length × width × height is the volume.

3. 3

   $40h=120$ — 8 × 5 = 40, so 40 lots of h make 120.

4. 4

   $h=120÷40=3\text{ cm}$ — Divide both sides by 40 to get h alone.

Worked example

1. 1

   $\text{area}=\frac{48}{75}=0.64{\text{ m}}^2$ — A box pushes with 48 newtons and the pressure is 75 newtons per m². Area = force ÷ pressure.

2. 2

   $1.6\times x=0.64$ — The face on the table is 1.6 m by x m, so its area is 1.6x.

3. 3

   $x=0.64÷1.6=0.4\text{ m}$ — Divide by 1.6 to get x alone.

Worked example

1. 1

   $240\times 8=1920\text{ g}$ — Metal P: 240 cm³ with density 8 g/cm³. We need its mass, so mass = density × volume.

2. 2

   $1920+1080=3000\text{ g}$ — Metal Q weighs 1080 g. Add the masses to get the total mass.

3. 3

   $240+120=360{\text{ cm}}^3$ — Metal Q's volume is 120 cm³. Add the volumes to get the total volume.

4. 4

   $\frac{3000}{360}=8.33{\text{ g/cm}}^3$ — Total mass ÷ total volume = 8.333…, which is 8.33 to 3 s.f.

Worked example

1. 1

   $\frac{5.4\times 10^5}{1.2\times 10^3}$ — Town A has 540 000 people on 1200 km². Population ÷ area. Type the standard form into your calculator with the ×10ˣ key.

2. 2

   $=\frac{540000}{1200}=450$ — So 450 people per km².

3. 3

   $\text{Town B: }450\times 300$ — Town B has the same density and an area of 300 km². Rearrange: population = density × area.

4. 4

   $=135000\text{ people}$ — 450 × 300 = 135 000.

Worked example

1. 1

   $40\times 30\times 25=30000{\text{ cm}}^3$ — A tank is a cuboid 40 cm by 30 cm by 25 cm. Find its volume first.

2. 2

   $30000÷1000=30\text{ litres}$ — The tap gives 4 litres per minute, so change cm³ to litres: 1000 cm³ make 1 litre.

3. 3

   $30÷4=7.5\text{ minutes}$ — Time = volume ÷ rate of flow.

4. 4

   $0.5\times 60=30\text{ s, so }7\text{ min }30\text{ s}$ — 0.5 of a minute is not 50 seconds. Multiply by 60 to change it to seconds.

Worked example

1. 1

   $340÷100=3.4$ — A car uses 6.5 litres per 100 km and drives 340 km. How many lots of 100 km is that?

2. 2

   $3.4\times 6.5=22.1\text{ litres}$ — Each 100 km uses 6.5 litres, so multiply.

Worked example

1. 1

   $15\text{ m/s}$ — 15 metres every second. We want km in one hour.

2. 2

   $15\times 3600=54000\text{ m}$ — There are 60 × 60 = 3600 seconds in an hour, so in one hour it goes 3600 times as far.

3. 3

   $54000÷1000=54\text{ km}$ — 1000 metres make 1 km, so divide by 1000.

4. 4

   $15\text{ m/s}=54\text{ km/h}$ — × 3600 then ÷ 1000 is the same as × 3.6.

Worked example

1. 1

   $7.2\text{ kg per hour}$ — A machine makes 7.2 kg each hour. We want grams per second.

2. 2

   $7.2\times 1000=7200\text{ g per hour}$ — Change the top unit: 1 kg = 1000 g, so multiply.

3. 3

   $7200÷3600=2\text{ g per second}$ — Change the bottom unit: an hour is 3600 seconds, and the 7200 g is shared over them, so divide.