GCSE Scale factors, diagrams and maps Practice Questions

Free GCSE Scale factors, diagrams and maps practice questions with full step-by-step worked solutions. Covers map scales, multiplying by a scale, reversing a scale, scale models. Practise exam-style problems and check your method.

map scalesmultiplying by a scalereversing a scalescale modelsunit conversionratio scale
GCSE Foundation70 questionsStep-by-step solutions
Question 1
2 markseasy
A map has a scale of 1 cm1\,\mathrm{cm} to 5 km5\,\mathrm{km}. Two villages are 6 cm6\,\mathrm{cm} apart on the map. Work out the real distance between the villages, in kilometres.
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Worked solution

  1. Write down what the scale means

    1 cm→5 km1\,\mathrm{cm} \to 5\,\mathrm{km}

    Every 1 cm1\,\mathrm{cm} measured on the map is 5 km5\,\mathrm{km} on the ground.

  2. Multiply the map distance by five

    6×5=30 km6 \times 5 = 30\,\mathrm{km}

    Six centimetres on the map is six lots of 5 km5\,\mathrm{km}.

  3. State the real distance

    30 km30\,\mathrm{km}

    The real distance is 30 km30\,\mathrm{km}, which is far bigger than the 6 cm6\,\mathrm{cm} on the map — as it must be.

Answer
30 km30\,\mathrm{km}
Question 2
1 markeasy
A model of a car is made to a scale of 1:301:30. Which statement is correct?
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Worked solution

  1. Interpret the scale for lengths

    1:30  ⇒  real=30×model1 : 30 \;\Rightarrow\; \text{real} = 30 \times \text{model}

    Every real length is 3030 times the matching length on the model.

  2. Work out the area scale factor

    k2=302=900k^2 = 30^2 = 900

    Area is two-dimensional, so real areas are 900900 times the model areas.

  3. Compare the statements

    k=30,k2=900k = 30,\qquad k^2 = 900

    Only the statement with a length factor of 3030 and an area factor of 900900 is correct.

Answer
real length=30×model length,real area=900×model area\text{real length} = 30 \times \text{model length},\quad \text{real area} = 900 \times \text{model area}
Question 3
2 marksintermediate
Ben says, “If I double every length of a rectangle, then the area of the rectangle also doubles.” Which statement correctly explains Ben’s mistake?
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Worked solution

  1. Try a rectangle to test the claim

    3 cm×5 cm=15 cm23\,\mathrm{cm} \times 5\,\mathrm{cm} = 15\,\mathrm{cm}^2

    Take a 3 cm3\,\mathrm{cm} by 5 cm5\,\mathrm{cm} rectangle, whose area is 15 cm215\,\mathrm{cm}^2.

  2. Double every length

    6 cm×10 cm=60 cm26\,\mathrm{cm} \times 10\,\mathrm{cm} = 60\,\mathrm{cm}^2

    Doubling both sides gives a 6 cm6\,\mathrm{cm} by 10 cm10\,\mathrm{cm} rectangle with area 60 cm260\,\mathrm{cm}^2.

  3. Compare the two areas

    6015=4\frac{60}{15} = 4

    The area is multiplied by 44, not by 22, so Ben is wrong.

  4. Check with a second, independent method

    k2=22=4k^2 = 2^2 = 4

    The general rule agrees: doubling every length multiplies the area by k2=22=4k^2 = 2^2 = 4.

  5. Avoid the usual mistake

    area≠k×area\text{area} \ne k \times \text{area}

    Area is two-dimensional. Both the length and the width double, so the area gains a factor of 22 twice.

  6. Write down the final answer

    area×4\text{area} \times 4

    Doubling every length multiplies the area by 44.

Answer
The area is multiplied by 22=4\text{The area is multiplied by } 2^2 = 4
Question 4
3 markshard
Two jugs are mathematically similar. The smaller jug is 10 cm10\,\mathrm{cm} tall and holds 250 ml250\,\mathrm{ml}. The larger jug holds 2000 ml2000\,\mathrm{ml}. Work out the height of the larger jug, in centimetres.
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Worked solution

  1. Work out the volume scale factor

    k3=2000250=8k^3 = \frac{2000}{250} = 8

    Divide the larger capacity by the smaller capacity.

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

    k=83=2k = \sqrt[3]{8} = 2

    Heights scale by kk, the cube root of the volume scale factor.

  3. Multiply the height by the length scale factor

    10×2=20 cm10 \times 2 = 20\,\mathrm{cm}

    The larger jug is 20 cm20\,\mathrm{cm} tall.

  4. Check the cube

    23=82^3 = 8

    Cubing 22 gives back the volume scale factor 88.

  5. Check the capacities

    250×8=2000 ml250 \times 8 = 2000\,\mathrm{ml}

    Scaling the smaller capacity by k3=8k^3 = 8 returns 2000 ml2000\,\mathrm{ml}, as given.

  6. Check with a second, independent method

    8≈2.83≠2\sqrt{8} \approx 2.83 \ne 2

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

  7. Check the units

     ml→ cm\,\mathrm{ml} \to\,\mathrm{cm}

    Cube-rooting a capacity ratio gives a pure number, which then scales the height in centimetres.

  8. Check the size of the answer is sensible

    20 cm tall, 2 litres20\,\mathrm{cm}\text{ tall, } 2\text{ litres}

    A 20 cm20\,\mathrm{cm} jug holding 22 litres is a realistic kitchen jug.

  9. Avoid the usual mistake

    10×8=80≠2010 \times 8 = 80 \ne 20

    Using the VOLUME scale factor on a height gives 80 cm80\,\mathrm{cm}, an absurdly tall jug.

  10. Write down the final answer

    20 cm20\,\mathrm{cm}

    The larger jug is 20 cm20\,\mathrm{cm} tall.

Answer
20 cm20\,\mathrm{cm}
Question 5
6 markschallenging
A map has a scale of 1:25 0001:25\,000. A rectangular plot of land measures 8 cm8\,\mathrm{cm} by 5 cm5\,\mathrm{cm} on the map. A fence is to be built all the way round the edge of the plot. Fencing costs £12\pounds 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\,\mathrm{cm} \to 25\,000\,\mathrm{cm}

    A scale of 1:25 0001 : 25\,000 means 1 cm1\,\mathrm{cm} on the map represents 25 000 cm25\,000\,\mathrm{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\,\mathrm{cm}

    Multiply the map length by the scale factor.

  3. Change that to metres

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

    There are 100 cm100\,\mathrm{cm} in a metre, so the plot is 2000 m2000\,\mathrm{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\,\mathrm{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\,\mathrm{m}

    The plot is 1250 m1250\,\mathrm{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\,\mathrm{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\pounds 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\,\mathrm{cm}, \quad 26 \times 25\,000 = 650\,000\,\mathrm{cm} = 6500\,\mathrm{m}

    Second method: find the perimeter on the map first, then scale it once by kk; the answer is the same 6500 m6500\,\mathrm{m}, because perimeter is one-dimensional.

  10. Check the units

     m×£ per  m=£\,\mathrm{m} \times £\text{ per }\,\mathrm{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\,\mathrm{m}\text{ 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\,\mathrm{m}^2

    The area of the plot is 2 500 000 m22\,500\,000\,\mathrm{m}^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\pounds 78\,000.

Answer
£78 000\pounds 78\,000

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