[discia] GCSE Maths › Ratio and Proportion › Speed and Density › Practise
  1. Either paper 4 marks

    A lorry travels along a motorway at a steady speed.

    12345501001502002500time (hours)distance (km)
    (a)

    Calculate the speed of the lorry. Give your answer in kilometres per hour.

    (b)

    The lorry keeps going at the same speed.

    Work out how long it takes the lorry to travel \(350\) km.

    Show answer
    (a) \(50\ \text{km/h}\)
    (b) \(7\ \text{hours}\)

    The line passes through \((0, 0)\) and \((5, 250)\), the top right corner of the grid.

    \(\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{ 250 }{ 5 } = 50\) km/h

    \(\text{time} = \frac{\text{distance}}{\text{speed}} = \frac{ 350 }{ 50 } = 7\) hours

    • (a) M1distance \(250\) km divided by time \(5\) hours
    • (a) A1\(50\) km/h
    • (b) M1\(350\) km divided by speed \(50\) km/h
    • (b) A1\(7\) hours
  2. Either paper 3 marks
    Two solids, J and K, have the same density.
    JK
    Mass108
    Volume122
    Density
    Complete the table. Include units in your answers.
    Show answer
    Density \(9\) g/cm³ (both cells); Block K's missing mass is \(18\) g

    Both solids have the same density, so work it out from Block J, where the mass and volume are both known.

    \(\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{ 108 }{ 12 } = 9\) g/cm³

    Multiply the known volume of Block K, \(2\) cm³, by the density to find its missing mass — the same density applies to both solids, so the same working that found it from J can be used on K.

    Block K's missing mass is \(18\) g, and both density cells are \(9\) g/cm³.

    • M1density from Block J, \(\frac{ 108 }{ 12 } = 9\) g/cm³
    • M1Multiply the known volume of Block K, \(2\) cm³, by the density to give \(18\) g
    • B1units given for the density and the missing quantity
  3. Either paper 2 marks

    No real values are given in this question.

    (a)

    The distance of journey \(A\) is 9 times the distance of journey \(B\), and its time is 12 times the time of journey \(B\).

    Complete the sentence. The average speed of journey \(A\) is ______ times the average speed of journey \(B\).

    (b)

    For a journey, the distance is multiplied by 5 and the time is divided by 4.

    What happens to the average speed? Circle your answer.

    multiplied by 5/4

    multiplied by 20

    multiplied by 4/5

    divided by 20

    no change

    Show answer
    (a) \(3/4\)
    (b) multiplied by 20

    The average speed is the distance divided by the time.

    Let \(B\) have distance \(M\) and time \(V\), so its average speed is \(\frac{ M }{ V }\).

    Then \(A\) has distance \(9M\) and time \(12V\), so its average speed is

    \(\frac{ 9M }{ 12V } = \frac{ 9 }{ 12 } \times \frac{ M }{ V }\)

    \(9 \div 12\), written in its simplest form, is \(3/4\).

    So the average speed of journey \(A\) is \(3/4\) times the average speed of journey \(B\).

    The average speed is the distance divided by the time.

    Take the original distance to be \(x\) and the original time to be \(y\), so the average speed starts as \(\frac{ x }{ y }\).

    The distance is multiplied by 5, so it becomes \(x \times 5\). The time is divided by 4, so it becomes \(y \times 1/4\).

    The new average speed is \(\frac{ x \times 5 }{ y \times 1/4 }\), so the average speed is multiplied by \(5\) and divided by \(1/4\), a net factor of \(20\).

    So the average speed is multiplied by 20.

  4. Either paper 3 marks

    A cuboid of copper has length \(14\) cm, width \(4\) cm and height \(8\) cm. The copper has a density of \(9\) g/cm\(^3\).

    Find the mass of the cuboid in grams.

    Show answer
    \(4032\ \text{g}\)

    Find the volume of the cuboid first, by multiplying its three dimensions.

    \(14 \times 4 \times 8 = 448\) cm\(^3\)

    Density is mass per cm\(^3\), so multiply the volume by the density to get the mass.

    \(\text{mass} = \text{density} \times \text{volume} = 9 \times 448 = 4032\) g

    • M1volume \(448\) cm\(^3\)
    • M1volume multiplied by density \(9\)
    • A1\(4032\) g
  5. Either paper 3 marks

    A lorry travels \(54\) km at an average speed of \(30\) km/h.

    Calculate how long the lorry takes to travel this distance.

    Give your answer in hours and minutes.

    Show answer
    \(1\) hour \(48\) minutes

    Time is distance divided by speed.

    \(\text{time} = \frac{ 54 }{ 30 } = \frac{9}{5}\) hours

    That is a number of hours, so whatever is left over the whole hours is a fraction of an hour and not a number of minutes. Multiply by \(60\) to turn the whole thing into minutes.

    \(\frac{9}{5} \times 60 = 108\) minutes

    \(108\) minutes is \(1\) hour and \(48\) minutes, because \(1 \times 60 = 60\) and \(108 - 60 = 48\).

    • M1time = distance ÷ speed: \(54\) ÷ \(30\)
    • M1changes the time to minutes, \(108\) minutes
    • A1\(1\) hour \(48\) minutes
  6. Either paper 3 marks

    Sam and Priya are two runners in a 216-metre race.

    Sam completes the race in 29 seconds.

    Priya's average speed for the race is 28.8 kilometres per hour.

    Who wins the race? You must show your working.

    Show answer
    Priya wins the race - Sam takes 29 s and Priya takes 27 s.

    Sam's time is given directly as 29 s, but Priya's performance is given as a speed in km/h, so it must first be converted to a time over the same 216 m.

    Divide Priya's speed by 3.6 to change it to metres per second:

    \(28.8 \div 3.6 = 8\) m/s

    Use this speed to find how long Priya takes to cover 216 m:

    \(216 \div 8 = 27\) s

    Sam takes 29 s and Priya takes 27 s, so Priya wins the race.

    • M1convert Priya's speed to m/s, \(28.8 \div 3.6 = 8\) m/s, or find Sam's speed, \(216 \div 29 = 7.45\) m/s
    • M1find the time for Priya over 216 m, \(216 \div 8 = 27\) s, or convert Sam's speed to km/h, \(7.45 \times 3.6 = 26.82\) km/h, or compare 7.45 m/s with 8 m/s
    • A1Priya wins, comparing 29 s with 27 s, or 26.82 km/h with 28.8 km/h, or 7.45 m/s with 8 m/s
  7. Either paper 6 marks

    Ben cycles 144 m in 48 seconds at a constant speed.

    (a)

    On another day, he cycles \(4.5\) km, assuming the same speed as before.

    Calculate how long this takes. Give your answer in minutes.

    (b)

    In fact, on the second day Ben cycled at a slower speed than on the first day.

    Is the time he actually took:

    less than the answer to part (a)

    the same as the answer to part (a)

    greater than the answer to part (a)

    not possible to say

    Tick the correct answer.

    Show answer
    (a) \(25\ \text{minutes}\)
    (b) greater than the answer to part (a)

    Speed on the first day:

    \(\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{ 144 }{ 48 } = 3\) m/s

    Convert the second distance to metres so it matches the speed's units:

    \(4.5\) km \(= 4500\) m

    \(\text{time} = \frac{\text{distance}}{\text{speed}} = \frac{ 4500 }{ 3 } = 1500\) seconds

    Convert to minutes:

    \(1500 \div 60 = 25\) minutes

    The distance on the second day is fixed. A slower speed over the same distance means more time is needed than the same-speed answer in part (a), without recalculating it.

    The time is greater than the answer to part (a).

    • (a) M1speed on the first day \(3\) m/s
    • (a) M1second distance in metres \(4500\) m
    • (a) M1time in seconds \(1500\) s
    • (a) A1\(25\) minutes
    • (b) M1a slower speed over the same distance means more time is needed
    • (b) A1greater than the answer to part (a)
  8. Either paper 3 marks

    A car makes a journey of \(220\) miles in \(4\) hours 10 minutes.

    For the first \(100\) miles of the journey the average speed of the car is \(60\) mph.

    Find the average speed of the car for the rest of the journey.

    Show answer
    \(48\ \text{mph}\)

    Time for the first leg: \(\frac{ 100 }{ 60 } \times 60 = 100\) minutes

    Time for the rest of the journey: \(250 - 100 = 150\) minutes \(= 2.5\) hours

    Distance for the rest of the journey: \(220 - 100 = 120\) miles

    \(\text{average speed} = \frac{ 120 }{ 2.5 } = 48\) mph

    • M1time for the first leg \(100\) divided by \(60\), \(100\) minutes
    • M1distance for the rest \(120\) miles and time for the rest \(2.5\) hours
    • A1rest speed \(120\) divided by \(2.5\), \(48\) mph
  9. Either paper 4 marks

    A, B, C and D are junctions on a motorway, in that order.

    ABCDBC = 20 miles CD = 2 × AB (not drawn to scale)

    Callum drives from A to C. He drives for \(30\) minutes at an average speed of \(60\) miles per hour.

    Work out the distance AD.

    Show answer
    \(50\ \text{miles}\)

    A to C is driven for \(30\) minutes, which is \(\frac{ 30 }{60}\) of an hour.

    \(AC = 60 \times \frac{ 30 }{60} = 30\) miles

    AC is made of AB and BC, so take BC off:

    \(AB = 30 - 20 = 10\) miles

    CD is \(2\) times AB:

    \(CD = 2 \times 10 = 20\) miles

    AD is AC and then CD:

    \(AD = 30 + 20 = 50\) miles

    • M1AC = speed x time = \(\frac{ 30 }{60} \times 60 = 30\) miles
    • M1AB = AC - BC = \(30 - 20 = 10\) miles
    • M1CD = \(2 \times 10 = 20\) miles
    • A1AD = AC + CD = \(30 + 20 = 50\) miles
  10. Either paper 3 marks

    The first stage of a journey takes \(2\) hours at an average speed of \(75\) km/h. The second stage takes \(3\) hours at an average speed of \(45\) km/h.

    Work out the average speed for the whole journey.

    Show answer
    \(57\ \text{km/h}\)

    Average speed is the total distance divided by the total time. It is not the average of the two speeds, because the two stages do not last the same length of time.

    Distance for the first stage: \(75 \times 2 = 150\) km

    Distance for the second stage: \(45 \times 3 = 135\) km

    Total distance: \(150 + 135 = 285\) km

    Total time: \(2 + 3 = 5\) hours

    \(\text{average speed} = \frac{ 285 }{ 5 } = 57\) km/h

    Averaging the two speeds would have given \(60\) km/h, which is wrong.

    • M1total distance \(285\) km
    • M1total time \(5\) hours
    • A1\(57\) km/h