[discia] GCSE Maths › Ratio and Proportion › Writing and Simplifying Ratio › Practise
  1. Either paper 2 marks

    There are 96 members in a swimming club. 56 of the members are adults. The rest are children.

    Write down the ratio of adults to children. Give your answer in its simplest form.

    Show answer
    \(7 : 5\)

    First work out how many children there are:

    \(96 - 56 = 40\)

    So the ratio of adults to children is \(56 : 40\).

    The highest common factor of 56 and 40 is 8, so divide both parts by 8:

    \(56 : 40 = 7 : 5\)

    • M1children found, \(96 - 56 = 40\)
    • A1ratio in simplest form \(7 : 5\)
  2. Either paper 2 marks

    Write each of these ratios in its simplest form.

    (a)

    \(24 : 60\)

    (b)

    \(7 : 63\)

    Show answer
    (a) \(2 : 5\)
    (b) \(1 : 9\)

    The highest common factor of 24 and 60 is 12.

    \(24 \div 12 = 2\) and \(60 \div 12 = 5\)

    \(24 : 60 = 2 : 5\)

    7 divides exactly into 63, so the highest common factor of the two numbers is 7 itself.

    \(7 \div 7 = 1\) and \(63 \div 7 = 9\)

    \(7 : 63 = 1 : 9\)

  3. Either paper 3 marks

    Shape \(ABCDE\) is drawn on a centimetre grid.

    xyABCDE
    (a)

    Complete this statement by writing the name of one side of the shape.

    \(AB : \_\_\_\_\_\_ = 4 : 1\)

    (b)

    On the centimetre grid below, draw a rectangle with the same area as shape \(ABCDE\).

    xy

    You cannot draw on this page. Draw (b) on paper, then show the answer and check your drawing against the model.

    Show answer
    (a) CD (or DC)
    (b) Any rectangle of area \(18\) cm\(^2\), for example \(6\) cm by \(3\) cm
    (b)xy

    Counting squares on the grid, \(AB = 8\) cm.

    For \(AB : \_\_\_\_\_\_ = 4 : 1\), the missing length is \(8 \div 4 = 2\) cm.

    The side that is \(2\) cm long is \(CD\).

    Shape \(ABCDE\) is a rectangle \(8\) cm by \(3\) cm with a right-angled triangle cut from one corner. The triangle has legs \(6\) cm and \(2\) cm.

    Area of the rectangle \(= 8 \times 3 = 24\) cm\(^2\)

    Area of the triangle \(= \frac{1}{2} \times 6 \times 2 = 6\) cm\(^2\)

    Area of shape \(ABCDE = 24 - 6 = 18\) cm\(^2\)

    Any rectangle whose length times width is \(18\) will do, for example \(6\) cm by \(3\) cm.

    • (b) M1area of shape \(ABCDE\) \(18\) cm\(^2\), shown by a shape of that area
    • (b) A1rectangle of area \(18\) cm\(^2\) drawn, for example \(6\) by \(3\)
  4. Either paper 2 marks

    A necklace is made with red beads, green beads and silver beads in the ratio \(36 : 54 : 30\).

    Write this ratio in its simplest form.

    Show answer
    \(6 : 9 : 5\)

    The highest common factor of 36, 54 and 30 is 6, so divide every part of the ratio by 6:

    \(36 \div 6 = 6\), \(54 \div 6 = 9\), \(30 \div 6 = 5\)

    \(36 : 54 : 30 = 6 : 9 : 5\)

    • M1every part divided by the common factor 6
    • A1simplest form \(6 : 9 : 5\)
  5. Either paper 2 marks

    Tomas is asked to write the ratio \(12 : 60\) in its simplest form.

    He writes \(12 : 60 = 4 : 20\)

    Explain what Tomas has done wrong, and write \(12 : 60\) in its simplest form.

    Show answer
    Tomas divided both parts by 3, which is not the highest common factor. \(12 : 60 = 1 : 5\)

    Tomas has divided both parts by 3. That is a common factor of 12 and 60, but it is not the highest one — 4 and 20 can both be divided by 4 again.

    The highest common factor of 12 and 60 is \(3 \times 4 = 12\).

    \(12 \div 12 = 1\) and \(60 \div 12 = 5\)

    \(12 : 60 = 1 : 5\)

    • C1error: divided by 3 only, not the highest common factor \(12\)
    • A1simplest form \(1 : 5\)
  6. Either paper 2 marks

    Omar says that \(5 : 2\) is an equivalent ratio to \(15 : 6\).

    Ella says that \(2.5 : 1\) is an equivalent ratio to \(15 : 6\).

    Who is correct? Tick one box.

    - Both of them - Omar only - Ella only - Neither of them

    Give reasons for your answer.

    Show answer
    Both of them. Omar: dividing 15 and 6 by 3 gives 5 : 2, so Omar is right. Ella: 15 ÷ 6 = 2.5, so 15 : 6 = 2.5 : 1 and Ella is right.

    Test Omar's claim by dividing both parts of \(15 : 6\) by the same number, 3:

    \(15 \div 3 = 5\) and \(6 \div 3 = 2\), so \(15 : 6 = 5 : 2\)

    Omar wrote \(5 : 2\), so Omar is right.

    Test Ella's claim by dividing both parts by 6 so that the second part is 1:

    \(15 \div 6 = 2.5\) and \(6 \div 6 = 1\), so \(15 : 6 = 2.5 : 1\)

    Ella wrote \(2.5 : 1\), so Ella is right.

    Putting the two tests together, the box to tick is: Both of them.

    • A1Both of them
    • M1tests both claims: 15 : 6 divided by 3 is 5 : 2, and 15 ÷ 6 = 2.5
  7. Either paper 2 marks

    Write the ratio 9 millimetres to 4 centimetres in its simplest form.

    Show answer
    \(9 : 40\)

    Both amounts have to be in the same unit first. There are 10 millimetres in 1 centimetre, so

    \(4 \times 10 = 40\) millimetres

    The ratio is \(9 : 40\). Dividing both parts by their highest common factor gives

    \(9 : 40 = 9 : 40\)

    • M1both amounts in the same unit, \(4\) centimetres as \(40\) millimetres, giving \(9 : 40\)
    • A1simplified to \(9 : 40\)
  8. Either paper 2 marks

    In a car park the ratio of red cars to blue cars is \(12 : 168\).

    Write this ratio in the form \(1 : n\).

    Show answer
    \(1 : 14\)

    To make the first part of the ratio 1, divide both parts by 12:

    \(12 \div 12 = 1\)

    \(168 \div 12 = 14\)

    So \(12 : 168 = 1 : 14\)

    • M1both parts divided by 12
    • A1\(1 : 14\)
  9. Either paper 4 marks

    Neither of these ratios is written with whole numbers.

    Write each of them in its simplest form.

    (a)

    \(0.8 : 2.8\)

    (b)

    \(\frac{ 2 }{ 9 } : \frac{ 6 }{ 11 }\)

    Show answer
    (a) \(2 : 7\)
    (b) \(11 : 27\)

    A ratio in its simplest form is written with whole numbers, so multiply both parts by 10 first.

    \(0.8 \times 10 = 8\) and \(2.8 \times 10 = 28\)

    The highest common factor of 8 and 28 is 4, so divide both parts by 4:

    \(8 : 28 = 2 : 7\)

    Multiply both parts by 99, the lowest common multiple of 9 and 11.

    \(\frac{ 2 }{ 9 } \times 99 = 22\) and \(\frac{ 6 }{ 11 } \times 99 = 54\)

    The highest common factor of 22 and 54 is 2, so

    \(22 : 54 = 11 : 27\)

    • (a) M1both parts multiplied by 10, \(8 : 28\)
    • (a) A1simplest form \(2 : 7\)
    • (b) M1both parts multiplied by 99, \(22 : 54\)
    • (b) A1simplest form \(11 : 27\)
  10. Either paper 3 marks

    A box contains only green, black and pink pencils.

    There are 24 green pencils in the box.

    There are three times as many black pencils as green pencils.

    There are 66 fewer pink pencils than black pencils.

    Write the ratio of green pencils to black pencils to pink pencils in its simplest form.

    Show answer
    \(4 : 12 : 1\)

    Work out how many of each colour there are.

    There are 24 green pencils.

    Three times as many black pencils: \(3 \times 24 = 72\)

    66 fewer pink pencils than that: \(72 - 66 = 6\)

    So the ratio is \(24 : 72 : 6\).

    The highest common factor of 24, 72 and 6 is 6, so divide every part by 6:

    \(24 : 72 : 6 = 4 : 12 : 1\)

    • M1black pencils \(72\)
    • M1pink pencils \(6\)
    • A1simplest form \(4 : 12 : 1\)