Foundation November 2019 Paper 1 Q6
6.
(a)
The table shows the number of road closures in Hayshire during 6 months of 2018.
| March | April | May | June | July | August |
|---|---|---|---|---|---|
| 14 | 15 | 22 | 21 | 12 | 18 |
For these six months, calculate each of the following.
(i) The range of the number of road closures. [1]
(ii) The mean number of road closures per month. [3]
(b)
The table shows the populations of some places in Hayshire at the end of 2018.
| Place | Population |
|---|---|
| Tanham | 12 212 |
| Copley | 4658 |
| Pinestow | 619 |
| Elmvale | 3600 |
(i) Write the populations in order of size.
Start with the smallest. [1]
Start with the smallest. [1]
(ii)
The population of Elmvale is predicted to be 4700 by the end of 2019.
To work out the number of houses to build for the extra people, the builders use the rule:
Build one house for every 4 extra people.
How many houses should they build? [3]
| Scheme | Marks | Comments |
|---|---|---|
| (i) 10 | B1 | |
| (ii) 14 + 15 + 22 + 21 + 12 + 18 (= 102) | M1 | or e.g. 12 + 14 + 15 + 18 + 21 + 22 |
| 102 ÷ 6 | m1 | FT ‘their 102’ |
| 17 | A1 | CAO |
| Scheme | Marks | Comments |
|---|---|---|
| (i) 619, 3600, 4658, 12212 | B1 | Allow place names/abbreviations in the correct order (Pinestow, Elmvale, Copley, Tanham); If figs and place names stated, answer line takes precedence |
| (ii) (4700 − 3600) ÷ 4 | M2 | M1 for sight of 4700 − 3600 |
| 275 | A1 | If no marks, award SC2 for sight of (4700 − 619) ÷ 4 and answer 1021 or (4700 − 4658) ÷ 4 and answer 11 or SC1 for sight of (4700 − 619) ÷ 4 or (4700 − 4658) ÷ 4 |
| (8) |