Hard GCSE Scale factors, diagrams and maps Questions

Challenging, exam-style GCSE Scale factors, diagrams and maps questions with worked solutions. Stretch yourself on the hardest map scales, cm to km, speed, area under scaling problems.

map scalescm to kmspeedarea under scalingk squaredscale drawings
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A map has a scale of 1:25 0001:25\,000. A rectangular plot of land measures 88 cm by 55 cm on the map. A fence is to be built all the way round the edge of the plot. Fencing costs £12 per metre. Work out the total cost of the fence.
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Worked solution

  1. Write down what the map scale means

    1:25 000  ⇒  1 cm→25 000 cm1 : 25\,000 \;\Rightarrow\; 1\text{ cm} \to 25\,000\text{ cm}

    A scale of 1:25 0001 : 25\,000 means 11 cm on the map represents 25 00025\,000 cm in real life. Both lengths must be measured in the same units.

  2. Convert the first map length to a real length in centimetres

    8×25 000=200 000 cm8 \times 25\,000 = 200\,000\text{ cm}

    Multiply the map length by the scale factor.

  3. Change that to metres

    200 000÷100=2000 m200\,000 \div 100 = 2000\text{ m}

    There are 100100 cm in a metre, so the plot is 20002000 m long.

  4. Convert the second map length to a real length in centimetres

    5×25 000=125 000 cm5 \times 25\,000 = 125\,000\text{ cm}

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

  5. Change that to metres

    125 000÷100=1250 m125\,000 \div 100 = 1250\text{ m}

    The plot is 12501250 m wide in real life.

  6. Note that perimeter is a length, not an area

    perimeter scales by k, not k2\text{perimeter scales by } k, \text{ not } k^2

    The fence runs along the edge, so it is a length: scale by kk, not k2k^2.

  7. Work out the real perimeter

    2×(2000+1250)=2×3250=6500 m2 \times (2000 + 1250) = 2 \times 3250 = 6500\text{ m}

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

  8. Multiply the perimeter by the cost per metre

    6500×12=78 0006500 \times 12 = 78\,000

    Each metre of fencing costs £12.

  9. Check with a second, independent method

    map perimeter 2(8+5)=26 cm,26×25 000=650 000 cm=6500 m\text{map perimeter } 2(8+5) = 26\text{ cm}, \quad 26 \times 25\,000 = 650\,000\text{ cm} = 6500\text{ m}

    Second method: find the perimeter on the map first, then scale it once by kk; the answer is the same 65006500 m, because perimeter is one-dimensional.

  10. Check the units

    m×£ per m=£\text{m} \times £\text{ per m} = £

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

  11. Check the size of the answer is sensible

    6500 m of fence at £126500\text{ m of fence at } £12

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

  12. Avoid the usual mistake

    area=2000×1250=2 500 000 m2\text{area} = 2000 \times 1250 = 2\,500\,000\text{ m}^2

    The area of the plot is 2 500 0002\,500\,000 m2^2, but the fence is priced per metre, not per square metre — do not use the area here.

  13. Check the dimension of every quantity used

    perimeter×k,area×k2\text{perimeter} \times k, \quad \text{area} \times k^2

    The perimeter is a length, so it takes only one factor of 25 00025\,000.

  14. Note the rule this question is testing

    price per metre⇒use the perimeter\text{price per metre} \Rightarrow \text{use the perimeter}

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

  15. Write down the final answer

    £78 000£78\,000

    The fence costs £78 000.

Answer
£78 000\pounds 78\,000
Question 2
5 markschallenging
Two solid spheres are made from the same metal. Sphere A has radius 22 cm and mass 8888 g. Sphere B is made from the same metal and has mass 23762376 g. Work out the radius of sphere B, in centimetres.
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Worked solution

  1. Note that all spheres are mathematically similar

    every sphere is similar to every other sphere\text{every sphere is similar to every other sphere}

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

  2. Explain why mass follows the volume rule

    mass=density×volume\text{mass} = \text{density} \times \text{volume}

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

  3. Work out the mass scale factor

    k3=237688=27k^3 = \frac{2376}{88} = 27

    Divide the mass of B by the mass of A.

  4. Take the cube root to get the length scale factor

    k=273=3k = \sqrt[3]{27} = 3

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

  5. Multiply the radius by the length scale factor

    2×3=6 cm2 \times 3 = 6\text{ cm}

    Sphere B has radius 66 cm.

  6. Check the masses

    88×27=2376 g88 \times 27 = 2376\text{ g}

    Scaling the mass of A by k3=27k^3 = 27 returns 23762376 g, the mass given.

  7. Check with the sphere volume formula

    43π×6343π×23=2168=27\frac{\frac{4}{3}\pi \times 6^3}{\frac{4}{3}\pi \times 2^3} = \frac{216}{8} = 27

    Using V=43πr3V = \frac{4}{3}\pi r^3 directly, the volumes are in the ratio 27:127 : 1 — the same factor, by an independent route.

  8. Check with a second, independent method

    27≈5.2≠3\sqrt{27} \approx 5.2 \ne 3

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

  9. Check the units

    g→cm\text{g} \to \text{cm}

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

  10. Check the size of the answer is sensible

    6>26 > 2

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

  11. Avoid the usual mistake

    2×27=54≠62 \times 27 = 54 \ne 6

    Using the MASS scale factor on the radius gives 5454 cm, an absurdly large sphere.

  12. Check the dimension of every quantity used

    radius×k,mass×k3\text{radius} \times k, \quad \text{mass} \times k^3

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

  13. Note one more wrong answer to avoid

    2×32=18≠62 \times 3^2 = 18 \ne 6

    Squaring the scale factor instead of using it directly gives 1818 cm, which is wrong.

  14. Note the rule this question is testing

    same metal⇒mass∝volume\text{same metal} \Rightarrow \text{mass} \propto \text{volume}

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

  15. Write down the final answer

    6 cm6\text{ cm}

    Sphere B has radius 66 cm.

Answer
6 cm6 \text{ cm}
Question 3
6 markschallenging
A model of a monument is made so that model length : real length =1:15= 1 : 15. The real monument has a surface area of 4545 square metres. Work out the surface area of the model, in square centimetres.
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Worked solution

  1. Write down what the model scale means

    1:15  ⇒  1 cm→15 cm1 : 15 \;\Rightarrow\; 1\text{ cm} \to 15\text{ cm}

    A scale of 1:151 : 15 means every real length is 1515 times the matching length on the model.

  2. Work out the area scale factor

    k2=(15)2=225k^2 = (15)^2 = 225

    Area is two-dimensional, so it is multiplied by k2=225k^2 = 225, not by kk. This is the single most common mistake in this topic.

  3. Decide whether to multiply or divide

    model=real÷k2\text{model} = \text{real} \div k^2

    The model is smaller than the real monument, so its area is the real area DIVIDED by k2=225k^2 = 225.

  4. Write down the square-unit conversion

    1 m2=100×100=10 000 cm21\text{ m}^2 = 100 \times 100 = 10\,000\text{ cm}^2

    A square metre is a square of side 100100 cm.

  5. Convert the real surface area to square centimetres

    45×10 000=450 000 cm245 \times 10\,000 = 450\,000\text{ cm}^2

    The real monument has surface area 450 000450\,000 cm2^2.

  6. Divide by the area scale factor

    450 000÷225=2000 cm2450\,000 \div 225 = 2000\text{ cm}^2

    The model has surface area 20002000 cm2^2.

  7. Check by scaling the model area back up

    2000×225=450 000 cm2=45 m22000 \times 225 = 450\,000\text{ cm}^2 = 45\text{ m}^2

    Multiplying the model area by k2k^2 returns the real surface area, as required.

  8. Check with a concrete face

    real face 9 m×5 m  ⇒  model 60 cm×50015 cm\text{real face } 9\text{ m} \times 5\text{ m} \;\Rightarrow\; \text{model } 60\text{ cm} \times \frac{500}{15}\text{ cm}

    A real face of 99 m by 55 m becomes a model face of 6060 cm by 1003\frac{100}{3} cm, whose area is 20002000 cm2^2 — the same factor of 225225 at work.

  9. Check with a second, independent method

    45÷225=0.2 m2=2000 cm245 \div 225 = 0.2\text{ m}^2 = 2000\text{ cm}^2

    Second method: divide first, then convert. 0.20.2 m2^2 is 0.2×10 000=20000.2 \times 10\,000 = 2000 cm2^2.

  10. Check the units

    m2×10 000=cm2\text{m}^2 \times 10\,000 = \text{cm}^2

    Multiplying by 100100 instead of 10 00010\,000 is the usual slip here.

  11. Check the size of the answer is sensible

    2000 cm2=0.2 m22000\text{ cm}^2 = 0.2\text{ m}^2

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

  12. Avoid the usual mistake

    450 000÷15=30 000≠2000450\,000 \div 15 = 30\,000 \ne 2000

    Dividing an area by kk instead of k2k^2 leaves the answer 1515 times too big.

  13. Check the dimension of every quantity used

    area÷k2\text{area} \div k^2

    Surface area is two-dimensional, so reducing it needs k2=225k^2 = 225, not k=15k = 15.

  14. Note the rule this question is testing

    model<real⇒divide\text{model} < \text{real} \Rightarrow \text{divide}

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

  15. Write down the final answer

    2000 cm22000\text{ cm}^2

    The model has surface area 20002000 cm2^2.

Answer
2000 cm22000 \text{ cm}^2
Question 4
5 markschallenging
On a map, a straight road of real length 4.54.5 km measures 1818 cm. Write the scale of the map in the form 1:n1 : n, where nn is a whole number.
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Worked solution

  1. Write the two lengths as a ratio

    18 cm:4.5 km18\text{ cm} : 4.5\text{ km}

    The map length comes first, then the real length.

  2. Convert the real length to metres

    4.5 km=4500 m4.5\text{ km} = 4500\text{ m}

    There are 10001000 m in a kilometre.

  3. Convert the real length to centimetres

    4500 m=450 000 cm4500\text{ m} = 450\,000\text{ cm}

    There are 100100 cm in a metre, so the real road is 450 000450\,000 cm long.

  4. Write the ratio with both parts in centimetres

    18:450 00018 : 450\,000

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

  5. Divide both parts by the map length

    1818:450 00018=1:25 000\frac{18}{18} : \frac{450\,000}{18} = 1 : 25\,000

    Dividing both parts by 1818 puts the ratio in the form 1:n1 : n.

  6. State the scale

    1:25 0001 : 25\,000

    The map has a scale of 1:25 0001 : 25\,000.

  7. Check what one centimetre now represents

    25 000 cm=250 m=0.25 km25\,000\text{ cm} = 250\text{ m} = 0.25\text{ km}

    On this scale 11 cm is 0.250.25 km.

  8. Check the whole road with that scale

    18×0.25=4.5 km18 \times 0.25 = 4.5\text{ km}

    Eighteen centimetres at 0.250.25 km per centimetre is 4.54.5 km, the real length given.

  9. Check with a second, independent method

    1 cm→450 00018=25 000 cm1\text{ cm} \to \frac{450\,000}{18} = 25\,000\text{ cm}

    Second method: one centimetre of map covers 450 000÷18=25 000450\,000 \div 18 = 25\,000 cm of road, which is the scale directly.

  10. Check the units

    cm:cm\text{cm} : \text{cm}

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

  11. Check the size of the answer is sensible

    n=25 000 is largen = 25\,000 \text{ is large}

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

  12. Avoid the usual mistake

    4.518=0.25≠25 000\frac{4.5}{18} = 0.25 \ne 25\,000

    Dividing kilometres by centimetres gives 0.250.25, not the scale; both parts must be in the same unit first.

  13. Check the dimension of every quantity used

    cm:cm⇒no units\text{cm} : \text{cm} \Rightarrow \text{no units}

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

  14. Note the rule this question is testing

    divide both parts by the map length\text{divide both parts by the map length}

    To reach the form 1:n1 : n, divide both parts of the ratio by the first part.

  15. Write down the final answer

    1:25 0001 : 25\,000

    The scale of the map is 1:25 0001 : 25\,000.

Answer
1:25 0001 : 25\,000
Question 5
5 markschallenging
A cube has side length 44 cm. The cube is enlarged by a scale factor of 1.51.5. Which statement about the enlarged cube is correct?
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Worked solution

  1. Work out the volume of the original cube

    43=64 cm34^3 = 64\text{ cm}^3

    A cube of side 44 cm has volume 6464 cm3^3.

  2. Scale the side length

    4×1.5=6 cm4 \times 1.5 = 6\text{ cm}

    Every length is multiplied by the scale factor 1.51.5.

  3. Work out the volume of the enlarged cube directly

    63=216 cm36^3 = 216\text{ cm}^3

    A cube of side 66 cm has volume 216216 cm3^3.

  4. Work out the volume scale factor from the two volumes

    21664=3.375\frac{216}{64} = 3.375

    The volume has been multiplied by 3.3753.375.

  5. Check against the cube of the scale factor

    1.53=3.3751.5^3 = 3.375

    Cubing the length scale factor gives the same 3.3753.375, confirming the rule V×k3V \times k^3.

  6. Rule out the length factor

    64×1.5=96≠21664 \times 1.5 = 96 \ne 216

    Multiplying the volume by 1.51.5 gives 9696 cm3^3, which is not the volume of a 66 cm cube.

  7. Rule out the area factor

    64×2.25=144≠21664 \times 2.25 = 144 \ne 216

    Multiplying the volume by k2=2.25k^2 = 2.25 gives 144144 cm3^3, which is also wrong.

  8. Rule out a factor of 27

    64×27=1728≠21664 \times 27 = 1728 \ne 216

    A factor of 2727 would apply to a scale factor of 33, not 1.51.5.

  9. Check with a second, independent method

    6343=21664=(64)3=1.53\frac{6^3}{4^3} = \frac{216}{64} = \left(\frac{6}{4}\right)^3 = 1.5^3

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

  10. Check the units

    cm3\text{cm}^3

    Volumes of these cubes are measured in cubic centimetres.

  11. Check the size of the answer is sensible

    216>64216 > 64

    The enlarged cube must have the bigger volume.

  12. Avoid the usual mistake

    side 5.5≠6\text{side } 5.5 \ne 6

    The enlarged side is 4×1.5=64 \times 1.5 = 6 cm, not 5.55.5 cm.

  13. Check the dimension of every quantity used

    side×1.5,volume×1.53\text{side} \times 1.5, \quad \text{volume} \times 1.5^3

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

  14. Note the rule this question is testing

    k=1.5⇒k3=3.375k = 1.5 \Rightarrow k^3 = 3.375

    A scale factor need not be a whole number: cubing 1.51.5 gives 3.3753.375.

  15. Write down the final answer

    side 6 cm,  V=216 cm3=3.375×64 cm3\text{side } 6\text{ cm}, \; V = 216\text{ cm}^3 = 3.375 \times 64\text{ cm}^3

    The enlarged cube has side 66 cm and volume 216216 cm3^3, which is 3.3753.375 times the original 6464 cm3^3.

Answer
side 6 cm,V=216 cm3=3.375×64 cm3\text{side } 6\text{ cm}, \quad V = 216\text{ cm}^3 = 3.375 \times 64\text{ cm}^3

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