Data: averages and charts – Year 7 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. The mode is the value that occurs most often. The number 5 appears twice while every other value appears once, so the mode is 5. (1 mark)
   2. Put the data in order first: 5, 5, 7, 9, 12, 14, 20. With 7 values, the median is the \(\frac{7+1}{2}=4\)th value, which is 9. (1 mark)
   3. Add all the values: \(12+5+9+20+5+14+7=72\). Divide by the number of pupils: \(72\div7=10.28\ldots\), so the mean is 10.3 messages to 1 d.p. (1 mark)
   4. The range is largest minus smallest: \(20-5=\)15. (1 mark)

   The mean is higher than the median because the value 20 pulls it upwards.

2. Answer to Question 2 (4 marks)

   1. The modal number of pets has the highest frequency. The biggest frequency is 9, which belongs to 1 pet, so the mode is 1 (not 9). (1 mark)
   2. Multiply each number of pets by its frequency: \(0\times4=0\), \(1\times9=9\), \(2\times7=14\), \(3\times3=9\), \(4\times2=8\). The total is \(0+9+14+9+8=\)40 pets. (1 mark)
   3. Mean = total number of pets ÷ number of pupils \(=40\div25=\)1.6 pets. (1 mark)
   4. There are 25 pupils, so the median is the 13th value. Running totals: 4 pupils have 0 pets, then \(4+9=13\) pupils have 0 or 1 pet. The 13th pupil has 1 pet, so the median is 1. (1 mark)

3. Answer to Question 3 (4 marks)

   1. The whole circle is 360° and represents 60 pupils, so each pupil is shown by \(360\div60=6°\). Walkers have 150°, so \(150\div6=\)25 pupils walk. (1 mark)
   2. The car sector is 72°, so \(72\div6=\)12 pupils travel by car. (1 mark)
   3. The cycling sector is 48°, which is \(48\div6=8\) pupils (1 mark). As a percentage of the class: \(\dfrac{8}{60}\times100=13.33\ldots\), which is 13.3% to 1 decimal place (1 mark).

   Check: the bus sector gives \(90\div6=15\) pupils, and \(25+15+12+8=60\). The percentage can also be found directly from the angle: \(\dfrac{48}{360}\times100=13.3\%\).

4. Answer to Question 4 (4 marks)

   1. Mia's times total \(38+40+37+39+38+44=236\), so her mean is \(236\div6=\)39.3 s and her range is \(44-37=\)7 s. Noah's times total \(36+41+35+42+34+46=234\), so his mean is \(234\div6=\)39 s and his range is \(46-34=\)12 s. (1 mark for each swimmer)
   2. In swimming a lower time is better. Noah's mean (39 s) is lower than Mia's (39.3 s), so Noah is slightly faster on average. (1 mark)
   3. A smaller range means the times are closer together. Mia's range of 7 s is smaller than Noah's 12 s, so Mia is more consistent. (1 mark)

5. Answer to Question 5 (4 marks)

   1. Take the other three sports away from the 50 pupils: \(50-18-12-10=\)10 choose athletics. (1 mark)
   2. The 360° of the circle are shared among 50 pupils, so each pupil gets \(360\div50=7.2°\) (1 mark). Football \(18\times7.2=129.6°\), netball \(12\times7.2=86.4°\), swimming \(10\times7.2=72°\) and athletics \(10\times7.2=72°\) (1 mark for all four angles).
   3. Netball has 12 out of 50 pupils: \(\dfrac{12}{50}\times100=\)24%. (1 mark)

   Check the pie chart before drawing it: \(129.6+86.4+72+72=360°\), so the sectors fill the whole circle exactly. Use a protractor and measure each sector from the end of the previous one.

Total : 20 marks