Higher June 2022 Paper 3 Q15
15 A town has
a population density of 278 people per km2
and
a population of 158 460
\(\text{population density} = \dfrac{\text{population}}{\text{area}}\)
The population increases to 168 720
Work out the population density after the increase. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(158\,460 \div 278\) or 570 | M1 | |
| \(168\,720 \div\) their 570 | M1dep | |
| 296 | A1 | |
| Alternative method 2 | ||
| \(158\,460 \div 168\,720\) or 0.939… or 0.94 | M1 | |
| \(278 \div\) their 0.939… | M1dep | |
| 296 | A1 | |
| Alternative method 3 | ||
| \(168\,720 \div 158\,460\) or 1.0647… or 1.065 or 1.06 | M1 | oe eg \(1 + \dfrac{168720 - 158460}{158460}\) or \(1 + \dfrac{10260}{158460}\) |
| \(278 \times\) their 1.0647… | M1dep | |
| 296 | A1 | |
Additional guidance
| \(278 \times 1.065 = 296\) | M1M1A1 |
| \(278 \times 1.065 = 296.07\) with 296 on answer line is evidence of premature rounding in their working | M1M1A0 |
| \(168720 \div 158460 = 1.06,\ 278 \times 1.06 = 294.68\) with answer 294 | M1M1A0 |
| Embedded answer eg \(168720 \div 296 = 570\) | M1M1A0 |