Hard GCSE Scale factors, diagrams and maps Questions

Write down what the map scale means

A scale of $1 : 25\,000$ means $1\,\mathrm{cm}$ on the map represents $25\,000\,\mathrm{cm}$ in real life. Both lengths must be measured in the same units.

Convert the first map length to a real length in centimetres

Multiply the map length by the scale factor.

Change that to metres

There are $100\,\mathrm{cm}$ in a metre, so the plot is $2000\,\mathrm{m}$ long.

Convert the second map length to a real length in centimetres

The same scale factor applies to every length on the map.

Change that to metres

The plot is $1250\,\mathrm{m}$ wide in real life.

Note that perimeter is a length, not an area

The fence runs along the edge, so it is a length: scale by $k$, not $k^2$.

Work out the real perimeter

Perimeter of a rectangle is twice the sum of the two different sides.

Multiply the perimeter by the cost per metre

Each metre of fencing costs $\pounds 12$.

Check with a second, independent method

Second method: find the perimeter on the map first, then scale it once by $k$; the answer is the same $6500\,\mathrm{m}$, because perimeter is one-dimensional.

Check the units

Metres multiplied by pounds per metre gives a cost in pounds.

Check the size of the answer is sensible

A plot two kilometres long needs kilometres of fence, so a bill in the tens of thousands of pounds is expected.

Avoid the usual mistake

The area of the plot is $2\,500\,000\,\mathrm{m}^2$, but the fence is priced per metre, not per square metre — do not use the area here.

Check the dimension of every quantity used

The perimeter is a length, so it takes only one factor of $25\,000$.

Note the rule this question is testing

A cost per metre must be multiplied by a length, never by an area.

Write down the final answer

The fence costs $\pounds 78\,000$.

Note that all spheres are mathematically similar

Spheres always have the same shape, so the scale-factor rules apply with the radius as the length.

Explain why mass follows the volume rule

The spheres are the same metal, so their densities match and mass is proportional to volume.

Work out the mass scale factor

Divide the mass of $B$ by the mass of A.

Take the cube root to get the length scale factor

Radii scale by $k = 3$, the cube root of the volume scale factor.

Multiply the radius by the length scale factor

Sphere $B$ has radius $6\,\mathrm{cm}$.

Check the masses

Scaling the mass of A by $k^3 = 27$ returns $2376\,\mathrm{g}$, the mass given.

Check with the sphere volume formula

Using $V = \frac{4}{3}\pi r^3$ directly, the volumes are in the ratio $27 : 1$ — the same factor, by an independent route.

Check with a second, independent method

A square root would give the area route, which does not apply to mass.

Check the units

Cube-rooting a mass ratio gives a pure number that scales the radius in centimetres.

Check the size of the answer is sensible

Sphere $B$ is much heavier, so it must have the larger radius.

Avoid the usual mistake

Using the MASS scale factor on the radius gives $54\,\mathrm{cm}$, an absurdly large sphere.

Check the dimension of every quantity used

The radius is a length; mass follows volume, which is three-dimensional.

Note one more wrong answer to avoid

Squaring the scale factor instead of using it directly gives $18\,\mathrm{cm}$, which is wrong.

Note the rule this question is testing

Equal densities are what let a mass ratio be used as a volume ratio.

Write down the final answer

Sphere $B$ has radius $6\,\mathrm{cm}$.

Write down what the model scale means

A scale of $1 : 15$ means every real length is $15$ times the matching length on the model.

Work out the area scale factor

Area is two-dimensional, so it is multiplied by $k^2 = 225$, not by $k$. This is the single most common mistake in this topic.

Decide whether to multiply or divide

The model is smaller than the real monument, so its area is the real area DIVIDED by $k^2 = 225$.

Write down the square-unit conversion

A square metre is a square of side $100\,\mathrm{cm}$.

Convert the real surface area to square centimetres

The real monument has surface area $450\,000\,\mathrm{cm}^2$.

Divide by the area scale factor

The model has surface area $2000\,\mathrm{cm}^2$.

Check by scaling the model area back up

Multiplying the model area by $k^2$ returns the real surface area, as required.

Check with a concrete face

A real face of $9\,\mathrm{m}$ by $5\,\mathrm{m}$ becomes a model face of $60\,\mathrm{cm}$ by $\frac{100}{3}$ cm, whose area is $2000\,\mathrm{cm}^2$ — the same factor of $225$ at work.

Check with a second, independent method

Second method: divide first, then convert. $0.2\,\mathrm{m}^2$ is $0.2 \times 10\,000 = 2000\,\mathrm{cm}^2$.

Check the units

Multiplying by $100$ instead of $10\,000$ is the usual slip here.

Check the size of the answer is sensible

A model with about a fifth of a square metre of surface is a sensible desk-sized model.

Avoid the usual mistake

Dividing an area by $k$ instead of $k^2$ leaves the answer $15$ times too big.

Check the dimension of every quantity used

Surface area is two-dimensional, so reducing it needs $k^2 = 225$, not $k = 15$.

Note the rule this question is testing

Check the direction of the scale factor: the model is the smaller object, so divide.

Write down the final answer

The model has surface area $2000\,\mathrm{cm}^2$.

Write the two lengths as a ratio

The map length comes first, then the real length.

Convert the real length to metres

There are $1000\,\mathrm{m}$ in a kilometre.

Convert the real length to centimetres

There are $100\,\mathrm{cm}$ in a metre, so the real road is $450\,000\,\mathrm{cm}$ long.

Write the ratio with both parts in centimetres

A map scale compares like with like, so both parts must be in the same unit.

Divide both parts by the map length

Dividing both parts by $18$ puts the ratio in the form $1 : n$.

State the scale

The map has a scale of $1 : 25\,000$.

Check what one centimetre now represents

On this scale $1\,\mathrm{cm}$ is $0.25\,\mathrm{km}$.

Check the whole road with that scale

Eighteen centimetres at $0.25\,\mathrm{km}$ per centimetre is $4.5\,\mathrm{km}$, the real length given.

Check with a second, independent method

Second method: one centimetre of map covers $450\,000 \div 18 = 25\,000\,\mathrm{cm}$ of road, which is the scale directly.

Check the units

A ratio scale has no units, so the conversion to centimetres is essential.

Check the size of the answer is sensible

A map scale must have a large $n$, because a real distance is far bigger than the map distance.

Avoid the usual mistake

Dividing kilometres by centimetres gives $0.25$, not the scale; both parts must be in the same unit first.

Check the dimension of every quantity used

Both parts of the ratio are lengths in centimetres, so the scale itself is a pure number.

Note the rule this question is testing

To reach the form $1 : n$, divide both parts of the ratio by the first part.

Write down the final answer

The scale of the map is $1 : 25\,000$.

Work out the volume of the original cube

A cube of side $4\,\mathrm{cm}$ has volume $64\,\mathrm{cm}^3$.

Scale the side length

Every length is multiplied by the scale factor $1.5$.

Work out the volume of the enlarged cube directly

A cube of side $6\,\mathrm{cm}$ has volume $216\,\mathrm{cm}^3$.

Work out the volume scale factor from the two volumes

The volume has been multiplied by $3.375$.

Check against the cube of the scale factor

Cubing the length scale factor gives the same $3.375$, confirming the rule $V \times k^3$.

Rule out the length factor

Multiplying the volume by $1.5$ gives $96\,\mathrm{cm}^3$, which is not the volume of a $6\,\mathrm{cm}$ cube.

Rule out the area factor

Multiplying the volume by $k^2 = 2.25$ gives $144\,\mathrm{cm}^3$, which is also wrong.

Rule out a factor of $27$

A factor of $27$ would apply to a scale factor of $3$, not $1.5$.

Check with a second, independent method

The ratio of the volumes is exactly the cube of the ratio of the sides.

Check the units

Volumes of these cubes are measured in cubic centimetres.

Check the size of the answer is sensible

The enlarged cube must have the bigger volume.

Avoid the usual mistake

The enlarged side is $4 \times 1.5 = 6\,\mathrm{cm}$, not $5.5\,\mathrm{cm}$.

Check the dimension of every quantity used

The side is one-dimensional and the volume is three-dimensional.

Note the rule this question is testing

A scale factor need not be a whole number: cubing $1.5$ gives $3.375$.

Write down the final answer

The enlarged cube has side $6\,\mathrm{cm}$ and volume $216\,\mathrm{cm}^3$, which is $3.375$ times the original $64\,\mathrm{cm}^3$.