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 11 cm to 55 km. Two villages are 66 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\text{ cm} \to 5\text{ km}

    Every 11 cm measured on the map is 55 km on the ground.

  2. Multiply the map distance by five

    6×5=30 km6 \times 5 = 30\text{ km}

    Six centimetres on the map is six lots of 55 km.

  3. State the real distance

    30 km30\text{ km}

    The real distance is 3030 km, which is far bigger than the 66 cm on the map — as it must be.

Answer
30 km30 \text{ 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\text{ cm} \times 5\text{ cm} = 15\text{ cm}^2

    Take a 33 cm by 55 cm rectangle, whose area is 1515 cm2^2.

  2. Double every length

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

    Doubling both sides gives a 66 cm by 1010 cm rectangle with area 6060 cm2^2.

  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 1010 cm tall and holds 250250 ml. The larger jug holds 20002000 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\text{ cm}

    The larger jug is 2020 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\text{ ml}

    Scaling the smaller capacity by k3=8k^3 = 8 returns 20002000 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\text{ml} \to \text{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\text{ cm tall, } 2\text{ litres}

    A 2020 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 8080 cm, an absurdly tall jug.

  10. Write down the final answer

    20 cm20\text{ cm}

    The larger jug is 2020 cm tall.

Answer
20 cm20 \text{ cm}
Question 5
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

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