Compound units and direct/inverse proportion – Year 10 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. 3 h 30 min \(=3.5\) h (1). \(168\div3.5=\) 48 mph (1) Change the time into hours, then divide the distance by the time. The lorry's average speed is 48 mph.
   2. 1 h 40 min \(=1\tfrac23\) h (1). \(42\times1\tfrac23=\) 70 km (1) 40 minutes is \(\tfrac{40}{60}=\tfrac23\) of an hour. Distance \(=\) speed \(\times\) time, so the bus travels 70 km.

   A common error in (a) is to divide by 3.3 instead of 3.5, which gives about 50.9 mph.

2. Answer to Question 2 (4 marks)

   1. \(200\times15\times10=\) 30 000 cm³ (1) The beam is a cuboid, so multiply the three dimensions. Its volume is 30 000 cm³.
   2. Mass \(=0.65\times30\,000\) (1) \(=19\,500\) g (1) \(=\) 19.5 kg (1) Use mass \(=\) density \(\times\) volume, then divide by 1000 to change grams into kilograms. The beam has a mass of 19.5 kg.

   Check with sensible values: a beam this size weighing about 20 kg is realistic, since the wood is lighter than water.

3. Answer to Question 3 (4 marks)

   1. \(0.6\times0.4=\) 0.24 m² (1) Change centimetres into metres before multiplying: 60 cm \(=0.6\) m and 40 cm \(=0.4\) m.
   2. \(480\div0.24=\) 2000 N/m² (1 for dividing, 1 for the answer) Pressure \(=\) force \(\div\) area. The pressure on the floor is 2000 N/m².
   3. This side has area \(0.18\) m², which is smaller, so the pressure is bigger (same force on a smaller area) (1) The new pressure is \(480\div0.18\approx2667\) N/m².

4. Answer to Question 4 (4 marks)

   1. 1 ticket \(=94.50\div7=13.50\) (1). 11 tickets \(=\) £148.50 (1) Use the unitary method: find the cost of one ticket, then multiply by 11. So 11 tickets cost £148.50.
   2. \(5\times12=60\) machine-hours (1). \(60\div8=\) 7.5 hours (1) This is inverse proportion: more machines take less time. Divide the total of 60 machine-hours by 8. Eight machines take 7.5 hours.

   In (b), check: 8 machines for 7.5 hours is again 60 machine-hours.

5. Answer to Question 5 (4 marks)

   1. \(y=kx^2\), so \(45=9k\) and \(k=5\) (1). \(y=5x^2\) (1) Write \(y=kx^2\) and substitute \(x=3\), \(y=45\). Since \(3^2=9\), \(k=45\div9=5\).
   2. \(5\times36=\) 180 (1) Substitute \(x=6\): \(6^2=36\) and \(5\times36=180\). So \(y=180\).
   3. \(125=5x^2\), \(x^2=25\), \(x=\) 5 (1) Divide by 5, then take the positive square root. So \(x=5\).

   Check (c) with the formula: \(5\times5^2=125\). In (b), doubling \(x\) multiplies \(y\) by \(2^2=4\), and \(45\times4=180\), which confirms the answer without the formula.

Total : 20 marks